Fundamentals of Fluid Mechanics [second ed.] 9789380578859

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Fundamentals of Fluid Mechanics

Citation preview

Fundamentals of Fluid Mechanics

Fundamentals of

Fluid Mechanics Second Edition

G.S. SAWHNEY B.Tech. (IIT Madras), ME, Ptsc Professor & HoD GNIT, Greater Noida Formerly, Professor and HoD Lord Krishna College of Engineering Ghaziabad

I.K. International Publishing House Pvt. Ltd. NEW DELHI

•

BANGALORE

Published by I.K. International Publishing House Pvt. Ltd. S-25, Green Park Extension Uphaar Cinema Market New Delhi–110 016 (India) E-mail : [email protected] Website: www.ikbooks.com ISBN: 978-93-80578-85-9 © 2011 I.K. International Publishing House Pvt. Ltd. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or any means: electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission from the publisher. Published by Krishan Makhijani for I.K. International Publishing House Pvt. Ltd., S-25, Green Park Extension, Uphaar Cinema Market, New Delhi–110 016 and Printed by Rekha Printers Pvt. Ltd., Okhla Industrial Area, Phase II, New Delhi–110 020.

PREFACE TO THE SECOND EDITION

In the second edition, the book has been thoroughly revised and enlarged. Additional typical problems based on the examination papers of various technical universities have been included with solutions for easy understanding by the students. I thank all the faculty members of Mechanical Department of GNIT, Greater Noida for their assistance in completing the second edition of the book. I once again request students and teachers to send constructive suggestions and criticism by emailing at [email protected]

G.S. Sawhney

PREFACE TO THE FIRST EDITION

Fluid Mechanics is an important subject which has been given equal weightage in Mechanical, Civil, Chemical, and Aeronautical undergraduate engineering curriculum. It deals with the flow of fluids. This book is designed to explain the fundamentals of fluid mechanics in the areas of properties of fluids, pressure and its measurements, hydrostatic forces, buoyancy, dimensional analysis, hydraulic similitude and model studies, fluid kinematics, fluid dynamics, laminar and turbulent flows, boundary layer analysis, flow through pipes, ideal fluid flow and flow past submerged bodies. Efforts have been made to cover the syllabi of several universities. Based on my teaching experience, I have tried to explain principles and concepts of Fluid Mechanics in simple and clear terms. The endeavour is to present the subject matter in the most comprehensive and useable form. The derivation of fundamental relations has been kept as simple as possible. Diagrams have been used abundantly to elucidate the difficult concepts which cannot be explained effectively. The theory is further supported with illustrations and suitably worked out examples. The book has an easy-to-read style, as it is written in question-answer form that is going to benefit the readers immensely. I express my gratitude to K.K Aggarwal, Chairman, and Dr. A.M. Chandra, Director, Lord Krishna College of Engineering, Ghaziabad for their support and encouragement extended to me during the compilation of this book. I am grateful to my doctoral guide, Dr. Prasad, Galgotia Engineering College, Greater Noida for moral support extended to me. Above all, I wish to record my sincere thanks to my wife, Jasbeer Kaur for her patience shown throughout the preparation of the book. I would appreciate constructive suggestions and objective criticism from students and teachers alike with a view to enhance further usefulness of the book. They may mail me their views at channi_sawhney@hotmail com. G . S. Sawhney

CONTENTS

Preface to the Second Edition .................................................................................... v Preface to the First Edition ...................................................................................... vii 1. Fluids: Definitions & Properties ............................................................................ 1 2. Pressure and Head ............................................................................................ 64 3. Hydrostatic Forces .......................................................................................... 104 4. Buoyancy and Floatation .................................................................................. 187 5. Fluid Masses Subjected to Acceleration ........................................................... 238 6. Dimensional Analysis ........................................................................................ 268 7. Similitude and Model Analysis .......................................................................... 287 8. Fluid Kinematics .............................................................................................. 310 9. Fluid Dynamics-I ............................................................................................. 372 10. Fluid Dynamics-II ............................................................................................ 450 11. Laminar Flow .................................................................................................. 493 12. Turbulent Flow ................................................................................................ 545 13. Boundary Layer Analysis ................................................................................. 583 14. Flow Through Pipes and Compressibility Effects ............................................... 629 15. Ideal Fluid Flow .............................................................................................. 692 16. Flow Past Submerged Bodies .......................................................................... 707 Question Papers ....................................................................................................... 737 Index ....................................................................................................................... 753

Chapter

1

FLUIDS: DEFINITIONS & PROPERTIES

KEYWORDS AND TOPICS Ù Ù Ù Ù Ù Ù Ù Ù

CONCEPT OF CONTINUUM IDEAL FLUID REAL FLUID VISCOSITY KINEMATIC VISCOSITY NEWTON’S LAW OF VISCOSITY NEWTONIAN FLUIDS NON-NEWTONIAN FLUIDS

Ù Ù Ù Ù Ù Ù Ù Ù

COMPRESSIBILITY VAPOUR PRESSURE COHESION ADHESION SURFACE TENSION CAVITATION DROPLET & PRESSURE CAPILLARITY

INTRODUCTION Fluid includes liquid and gaseous state of a matter. In spite of larger mobility and spacing of molecules in a fluid, the fluid is considered to be continuum for mechanical analysis. This assumption is justifiable as the fluid has large number of molecules with small intermolecular distance. Fluid mechanics is that branch of science which deals with the behaviour of fluids at rest and in motion. The study of fluids at rest is called fluid static. The study of fluids in motion due to pressure forces is called fluid dynamics. The study of fluids in motion without pressure forces is called fluid kinematics. A solid can resist tensile, compressive and shear forces upto a certain limit while a fluid has no or negligible tensile strength. A fluid can resist compressive forces only when it is kept in a vessel. A fluid deforms continuously when subjected to a shearing force. The fluid therefore changes shape or it flows as it offers little resistance to shearing stresses. There exists, however, shearing stresses between the adjacent fluid layers which try to oppose the movement of one layer over the another. The magnitude of this shearing stress in a fluid depends on the rate of deformation of the fluid elements. As during rest, the fluid has no deformation resulting in nonpresence of any shearing force when fluid is stationary.

2

Fundamentals of Fluid Mechanics

In ordinary conditions, a gas can be compressed easily while a liquid is difficult to be compressed. Hence, liquids are regarded as incompressible. 1. What do you understand by the composition and intermolecular forces in the matters? How does intermolecular force vary? Every matter in nature is made up of molecules. Each molecule has positively charged nucleus and negatively charged electrons. The molecule is electrically neutral as negative charge in electrons is equal to the positive charges of its nucleus. When two molecules come close to each other, the distribution of charges in them becomes such that the force of attraction between opposite charges (positive charge of nucleus of one molecule and negative charge of electrons of other molecule) becomes greater than the force of repulsion between similar charges (nucleus – nucleus and electron – electron of one and other molecule). This net force of attraction between the molecules is called “intermolecular force”. As the distance between molecules increases, the net attractive force between them decreases and vice versa. However at mean distance r0, no net force acts between them. In case distance is decreased below r0, then a repulsive force is generated. 2. The diagram as given below shows a graph of distance (r) between two molecules against the intermolecular potential energy (U). State with proper argument; (i) which point of the graph indicates the equilibrium state of the molecules? (ii) which part of the intermolecular force is attractive?

Intermolecular Potential Energy

A

r0 O h

B

r E

F

D

The equilibrium state of the molecules is at point D because it is a point of minimum potential energy. In portion DBA of the curve, the potential energy increases as r decreases i.e. intermolecular force is repulsive. In portion DEF of the curve, U increases as r increases. 3. What are the different states of a matter? A matter can exist in three states, namely, (1) solid, (2) liquid and (3) gas. The liquid and gas states can be grouped together, which is called fluid state. Spacing of molecules (r0) is large in gas, small in liquid and extremely small in a solid. Hence intermolecular bonding forces are very weak in gas, sufficient in liquid and very strong in a solid. Due to the strong intermolecular forces, a solid is very compact and rigid in form. A liquid has sufficient intermolecular cohesive bonding forces to hold it as a continuous mass but without a shape. This is the reason why a liquid adjusts to the shape of the container. The intermolecular bonding forces are very weak in gases and a gas can fill up the whole of a container as it does not have any definite volume. For a given mass, a solid has definite volume and shape,

Fluids: Definitions & Properties

3

a liquid has definite volume while a gas has neither shape nor volume. Solids are incompressible, liquids are slightly compressible and gases are compressible.

Liqid

No volume, no shape

4. Explain the responses of solids and fluids to external forces. Fluids differ from solids in their responses to external applied forces. Within elastic limit, deformation occurs in solids on the application of shear force and deformation disappears when intermolecular force between molecules restores the molecules to their original position on the removal of the shear force.

B¢

For examples a solid twists when a torque (T) is applied by shear angle f, which disappears on the removal of the torque (T). However, twist behaviour of fluid is completely different. The fluid can sustain only normal forces and deforms continuously when shear force is applied. Fluids do not acquire equilibrium on the application of shear force like solids but it continues to deform as long as the shear force is acting even when it is small. For example, Fixed cylinder

Outer cylinder

4

Fundamentals of Fluid Mechanics

if a torque (7) is applied on an external cylinder which is rotatable and separated from the fixed cylinder by a thickness of fluid as shown in the figure, the outer cylinder does not have equilibrium position with the applied torque (7) and the outer cylinder starts picking up velocity from the stationary state. However the outer cylinder will attain an equilibrium velocity depending upon the magnitude of the torque and the properties of the fluid (specially viscosity) separating it from the fixed cylinder. If T is large then the equilibrium velocity will be high and equilibrium velocity will be small when T is made smaller. It is evident from this that the fluid does ( resist the applied torque as external cylinder achieves the equilibrium rate d0 of deformation — or equilibrium velocity. The deformation (0) in fluid unlike solids does dt not disappear when torque (7) is removed. 5. What do you understand from the fluids? Fluid is a substance, which cannot acquire any static equilibrium under the action of any shear force of even a small magnitude. In other words, fluids cannot acquire a static equivalent deformation (0) in order to achieve a equilibrium with the external applied torque. However the fluid resists the applied torque by attaining the constant value of the rate of change of deformation (" 1± /and then the acceleration of fluid becomes zero. Hence a fluid dt is a substance which can (1) attain the shape of the container as it has no definite shape of its own; (2) deform continuously under the action of shear force; (3) attain the equilibrium rate of deformation and in which (4) the deformation does not disappear on the removal of the torque. 6. What is viscosity of a fluid? The property, which characterizes the resistance which a fluid offers to applied shear force is called viscosity. ( The resistance does not depend upon the deformation (0) but on the rate c10 of deformation — . When there is a relative motion between different layers of fluids, dt then there is a tangential friction force in between the layers. Viscosity is less in gases but larger in liquids. 7. Differentiate between a solid and a fluid? Features

1. Spacing of molecules 2. 3. 4.

Intermolecular force External shape Shear force

Solid Closely spaced Large Does not change Can withstand elastic limit

Fluid Spacing is large Low Shape changes and it can flow Cannot withstand and fluid flows in direction of the shear force

cond.

Fluids: Definitions & Properties 5.

Extent of deformation

6.

Deformation

7.

Rate of deformation

8.

Motion on shear force

5

Deformation is constant Continuously with force and force dependent Disappears on the removal Deformation does not disappear of shear force Deformation does not The rate of deformation change with time i.e. rate becomes constant of change of deformation is zero Solid deforms without Fluid is in motion in direction of motion shear force. It comes in rest on the removal of shear force.

8. What is fluid mechanics? Fluid mechanics is a science in which we study the behaviour of the fluids, which are either in rest or in motion. The study of the fluids under static conditions (fluids at rest) is called fluid statics. For example, the study of water in a dam or other static conditions are done in fluid statics. Aerostatics is the term used for the study of incompressible gases under static conditions. The study of mechanics of fluid in motion is called fluid dynamics. The flows of fluids through pipes and channels due to shear force are studied in fluid dynamics. Force =>

Force

Force = 0

//1//////i dO

/1/71/7/1/ Static Deformation of Solid (Deformation= 0 )

= MAX

dt d0 =i dt d0 =0 dt

Continuous Deformation of Fluid dO) Deformation =

dt

9. What is fluid kinematics? Fluid kinematics deals with the motion (translation and rotation) and deformation of the fluid elements without any consideration to the force or energy which is causing such motion or deformation. Hence kinematics deals with the velocities, accelerations and flow patterns of the fluid. 10. What is fluid kinetics? In fluid kinetics, we study the relationship of the velocities/accelerations developed by the fluid when an external force is applied on the fluid or the forces exerted by the moving fluid with certain velocities / accelerations.

6

Fundamentals of Fluid Mechanics

11. What is hydraulics? Hydraulics is a branch of fluid mechanics in which we study the behaviour of water (incompressible) in motion or rest. 12. What is Pneumatics? Pneumatics is a branch of fluid mechanics in which we study the behaviour of compressible fluids, which are in motion or rest. 13. Describe same applications of fluid mechanics. The applications of fluid mechanics are: 1. The design of dams, canals and weirs. A dam is a masonry structure to store water and the main forces acting on a dam is static water pressure (F) against upstream face and the weight of the masonry (W). The resultant of these two forces must pass through the middle third of the base of the dam for its stability. Canals are meant to take water from one place to another by gravitational action. Weir is any angular obstruction in open stream over which the flow takes place.

Weir Heigh water

Middle third DAM

WEIR

2. Design of pumps, overhead reservoirs and pipelines for transporting water to domestic service lines. There is pressure head loss due to: (1) friction in pipeline and (2) lifting of water against gravity. Overhead Reservoir Service tank

Pump

/Ns T\ \

Underground water supply line Water supply installation and distributions

Fluids: Definitions & Properties

7

3. Design of fluid control devices such as hydraulic gates, valves, cocks and taps. These devices are used for controlling the flow of the fluids. A plug cock is a simple device in which the fluid passage is a hole in a rotatable plug, fitted in the body of the cock. In screw down stop valve, the stop has disc with washer to make the actual contact with seat to stop the flow of water. Angle valve also works on the same arrangement. Hydraulic gates are used in dams to allow the water out of the dam by rotating the gate. Hole in plug

Plue Cock MOM

I

Disc & washer

Meg' seat

Close

Open Stop Cock

Disc & washer

Gate Pin at centre of curvature of the gate Anele valve

Hydraulic Radial Gate

Head race Dam Penstock

DA Tailrace

DAM 4. In pressure measurement using manometers. The difference of the level of the fluid in the limbs of U-tube is an indicator of the pressure of the liquid in the vessel or pipe.

8

Fundamentals of Fluid Mechanics

Pipe with filled liquid Fluid 5. In measuring flow of the fluids in the pipes by venture or orifice meter.

Orifice Metering

Venturi Metering

6. In designing an airplane. As an airplane is a mechanically driven machine and is heavier than air, it is supported by the dynamic reaction of the air over its wings, which are in the shape of aerofoils to provide the required lift. The air stream moves at higher speed over the upper surface of aerofoil as compared to air stream moving at lower surface, thereby providing lift. Lift

vi VI>V,

Propulsive thrust

Drag

V, Weight

Aerofoil

Lift

FLYING OF AIRCRAFT

7. In using wind power for moving sail boat. It is possible to move the boat in the desired direction by using sails properly when the wind is approaching from different directions.

Fluids: Definitions & Properties

from different directions.

9

Mast

Wind

Mast

Propulsive Thrust Side force (resisted by keel)

Lift on sail

Resultant force

8. Swinging of a cricket ball is achieved by making one side smooth and other side rough. The air stream moves with lesser velocity on rough side as compared to smooth side, thereby providing side force.

> where v = velocity V,, h-,

where h = pressure head side force oc hi — h1

Swing of Cricket Ball Spinning of table tennis ball and cricket ball is possible with change of flow of air stream on rotating ball. Vi, Lift

v2,

Spinning

Vi> v2 h2> h1

v2

Without Spin

Vi=V2

I 0 Fundamentals of Fluid Mechanics

9. Using the property that the fluids exert same pressure in all directions in designing hydraulic ram or press. It is possible to lift a heavy load with the application of small effort by the method shown below Load = 1 Ton

Effort = 50 kg Lift 1"1

NLN.N.N.

10. Using stream lined body to reduce drag while designing the hull of a ship or increasing the drag to act as airbrakes in an aircraft as well as in the removal of chaff from the grain in a separator. Stream lined hull

Hull

Lower Drag

Hither Drag

Stream lining to reduce drag Part of wing is raised

Grains with chaff 000000 0 o 000 o- 0..

—01" 000 000

88arsi6o0% 0 0000 0 000 00 000

DO

•

°

°°° 0 G O

:K DD

chaff grins Airbrake of an Aircraft Separator Application of drag

Fluids: Definitions & Properties

11. To understand the circulation of blood in arteries, veins, heart and lungs. The blood has to be oxygenated and supplied to all tissues of the body so that they can perform living functions. Human heart is a two-stage four-chamber pump. Artificial heart machine is used as a cardio pulmonary bypass to enable the operation of a heart. Heart Lung

Oxygenation Artery IN•11•••1 I.•••••I I••••••I I.•.•••I

Deoxygenated

Artificial Blood Oxygenation 12. To design pumps for pumping liquid through pipes. The principle is to produce suction at one side to draw the fluid and to create excess pressure on the other side to force the liquid out. Pumps can be: (1) piston type, (2) centrifugal type and (3) jet pump, which utilize the energy of flowing fluids. The centrifugal pump produces delivery head by centrifugal acceleration of fluid in the rotating impeller.

Operating Principle of a Piston Pump

I2

Fundamentals of Fluid Mechanics

Impeller

Diffuse:

Inlet

13. To design water turbines to obtain mechanical work from the kinetic energy of falling water. The oldest type of water turbine is the water wheel in which wheel is rotated

Upstream

downstream

Water Wheel

Pelton Wheel

Vanes

Guide Vanes

.?. Francis Turbine

Propeller

Kaplan Turbine

Fluids: Definitions & Properties

I3

by the difference of water level in a stream. The pelton wheel is also based on the principle of the old water wheel. Francis & Kaplan turbines are used where flows and heads are small. The principle of these turbines is that the water is diverted to obtain change of momentum inside the turbine. The diversion takes place at right angles to the direction of entry, causing turbine rotor to rotate. Guide vanes are provided to ensure maximum rotary motion of the turbine rotor. 14. What are the differences between liquids & gases? Liquids 1. 2. 3. 4. 5. 6.

Gases

Liquids have definite volume Liquids have significant cohesive forces due to closely packed molecules. Liquids cannot expand. Liquids are almost incompressible. The change of temperature and pressure does not alter the volume of liquid. Liquids form a free surface in gravitational field.

1. Gases do not have definite volume 2. Gases have weak cohesive forces due to widely spaced molecules. 3. Gases can expand to fill any volume. 4. Gases are compressible. 5. The volume of gas changes with temperature & pressure. 6. Gases do not form free surface. •: • : • : • : : : :

Free surface

•:•

: : : • • •- • • • : : : • : • : • : • : • : • : •:• : • :

15. What are the differences between ideal fluid & real fluid? Ideal Fluids 1. 2. 3. 4. 5.

Have no viscosity. Have no surface tension. Incompressible. Shear force is zero during flow as viscosity is zero. Ideal fluid does not exist.

Real Fluids 1. 2. 3. 4.

Have viscosity Have surface tension Slightly compressible Shear force is acting during flow.

5. All fluids are real

16. What is continuum? Is air a continuum? Does it always remain so? (UPTU 2002 — 03) A substance is composed of a vast number of molecules. In the concept of continuum, the substance is considered free from any kind of discontinuity. The concept is valid as most engineering systems are concerned with the macroscopic or bulk behaviour of a substance rather than the microscopic or molecular behaviour. Hence in most cases, it is convenient to think of a substance as a continuous distribution of medium or a continuum. As the scale of analysis is large, the discontinuity of the order of intermolecular spacing or the free mean path is negligible The properties such as pressure, temperature and velocity, which are

I4

Fundamentals of Fluid Mechanics

measured at infinitely small points at different places, are considered to vary continuously from one point to another if they are not the same. However, there are certain instances where pressures are low and high accuracy is required (such as rocket propulsion), the concept of continuum cannot be applied.

Molecules

Continuous

Air is considered a continuum in most engineering applications. The atmosphere air has mean free path of about 5 to 7.5 x 10-6 cm. The molecular spacing of air increases under low pressures and this large spacing between the molecules cannot be neglected. In that case, the concept of continuum cannot be applied on the air while determining its properties. 17. What do you understand from the properties? What are the different types of properties? Every fluid has certain characteristics which describe the physical condition of the fluid. Such characteristics are called properties of the fluid. The properties of any fluid are pressure, temperature, density, mass and volume etc. The properties can be intensive or extensive. Intensive properties do not depend upon the mass of the fluid. Pressure, temperature and density are extensive properties. The extensive properties depend up the mass of the fluid. Volume and weight are extensive properties. We can convert the extensive properties into intensive properties by dividing them by the mass of the fluid. Hence volume and weight are extensive properties but they can be converted into intensive properties by dividing them by mass, which give us the properties as specific volume and specific weight. 18. What do you understand from the following fluid properties: (1) density (2) specific weight (3) specific volume and (4) relative density? 1. Density. It is defined as the mass of the fluid per unit volume. The density of water is 1000 kg/m3. kg Density (p) — mass = volume m3 2. Specific Weight. It is defined as the weight of the fluid per unit volume. weight Specific Weight — volume

Fluids: Definitions & Properties

I5

_ mass x g volume =pxg =Y The specific weight of the water = 1000 x 9.8 = 9800 Newton/m3 = 9.8 kN/m3 3. Specific Volume. It is defined as the volume of the fluid per unit volume v — volume =1— m3/kg mass p 4. Relative Density (Sp. Gravity). It is the ratio of specific weight of the fluid to the specific weight of standard fluid. Water is the standard fluid for liquid while air or hydrogen is the standard fluid for the gases: Specific gravity of liquid —

Specific weight of the liquid Specific weight of water

_ Y = Y y, 9800 19. 3 litre petrol weighs 21 N. Calculate (1) sp weight (2) density (3) sp volume and (4) sp gravity. Guidance: All dimensions are to be in MKS. Litre has to be changed into m3. Volume (V) = 3 litre = 3 x 10-3 m3 Weight = 21 N Weight = 21 Sp weight (y) — vl 3 x 10-3 = 7 x 103 N/m3 Sp. Weight = p x g p _Y = 7x103

g 9.81 713.6 kg/m3 Sp volume = 1 P 1 713.6 1.4 x 10-3 m3/kg 713.6 Sp gravity — P Petrol _ AWater 1000 0.7136s

16

Fundamentals of Fluid Mechanics

20. A liquid has sp gravity = 0.90. Find its (1) density (2) sp volume and (3) sp weight. P liquid = 0.9 P water Pliquid = 0.9 x 1000 = 900 kg/m3 1 Sp volume = 1 900 P liquid = 1.22 x 10-3 m3/kg Sp weight = ()liquid x g = 900 x 9.81 N/m3 = 8829 N/m3 Sp gravity —

21. A blower delivers 8m3 air per second at 27°C and one atmospheric pressure (1 bar). Find mass of the air delivered if molecular weight of the air is 30. Also find (1) density (2) sp volume and (3) sp weight of the air being supplied. Guidance: The mass of air can be calculated by the gas equation PV = mRT where P = Pressure, V = Volume, m = Mass, R = Gas constant & T = temperature in Kelvin, universal gas constant R = MR where M = molecular weight and R = gas constant. The value of R = 8314 Nm/kg mol K Gas constant (R) for air —

R M air

8314 30 277.13 T = 273 + 27 = 300 k P = 1 bar = 1 x 105 N/m3 V = 8 m3 Mass of air — m — PV RT iX10 5 x8 = 9.62 kg 277.13 x 300 Density (p) = mass volume 9.62 8 1.2 kg/m3 Sp volume (v) = 1 P 1.2

Fluids: Definitions & Properties

I7

= 0.833 m3/kg Sp weight (y)= pxg = 1.2 x 9.81 = 11.772 N/m3 22. What is the importance of fluid mechanics? Fluids like water and air are readily available everywhere in nature. We employ these fluids in many engineering applications. Many such applications involve the flow of these fluids. The flow through pipes; energy conversion by pump and turbine, the storage capacity and stability of dams, the movement of aircraft, submarine and rockets can be analysed by the fluids mechanics. Three basic principles which are used in the fluid mechanics for analysing fluid flow problems are: (1) the principle of the conservation of mass flow (2) the principle of the conservation of energy and (3) the principle of the conservation of momentum. 23. What do you understand from the viscosity of the fluids? When a solid mass slides over a surface, a friction force is developed to oppose the motion. Similarly when a layer of a fluid slides over another layer of the same fluid, a friction force is developed between them opposing the relative motion. This tangential force is called the viscous force. Suppose a fluid is moving in streamlined manner on a fixed horizontal surface AB as shown in the figure below. The layer of the fluid in the contact of the surface remains stationary (at rest), while the velocity of other layers increases with distance from the fixed surface. In the figure, the length of arrows represents the increasing velocity of the layers. It can be seen that there is a relative motion between the different layers. If we take three parallel layers a, b, and c from the fixed surface, then their velocities are in the increasing order. The layer a tends to retard the layer b and layer b tends to retard the layer c. Therefore each layer tries to decrease the velocity of the layer above it. It also means that each layer tries to increase the velocity of the layer below it. There is therefore an internal tangential resistive force acting between two layers which is opposing their relative motion. This force is called viscous force. In order to maintain the flow of the fluid, we have to apply an external force to overcome this tangential resistive or viscous force. In the absence of external force, the viscous force will resist the flow and bring the fluid to rest.

tic > Ub > Ua

(

4

l

Velocity a C / ///// /7/7 ////////// )

B

I8

Fundamentals of Fluid Mechanics

24. Define viscosity. Give examples to show the existence of viscosity. The property of the fluid by virtue of which it opposes the relative motion between its adjacent layers is known as viscosity. The property of viscosity can be seen in the following natural phenomenon:1. We cannot walk fast in water as compared to air as water has comparatively larger viscosity. 2. A stirred liquid comes to rest on account of viscosity. 3. If we pour honey and water on a table then the honey will stop flowing soon while the water will flow for a large distance as it has small viscosity as compared to honey. 25. Derive the Newton's equation of viscosity. (UPTU 2007 — 08) A thin layer of fluid is contained between two parallel plates as shown above. The upper plate is moved by tangential force P while lower plate is kept stationary. The movement of upper plate is resisted by the liquid through its viscosity. Therefore the magnitude of velocity of upper plate will depend upon the viscosity of the fluid. In no slip condition, the fluid layer on the lower surface remains stationary whereas the top fluid layer moves with the speed of the upper plate. Hence a velocity gradient will act, resulting a vertical fluid line OA deforms to the position OA'. A point B on line OA at a height dy from 0, will move to B1 due to its velocity du.

Now Also . or

BB1 = velocity x time = du x dt BB1 = dy x dB du x dt = dy x de dO = du dt dy

The angle of deformation de or strain increases with time but the time rate of shear strain de ) remains constant as long force P is unchanged. As shear stress (r = P/A) per time (— dt

Fluids: Definitions & Properties

I9

depends upon the applied force (P) and area of contact (A) , it remains constant and it depends upon the time rate of shear strain

, dO — dt a du

dy

= du dy This is called the Newton's equation of viscosity. The constant of proportionality p is called viscosity. A fluid in which the shear stress (z) is linearly proportional to time rate of strain is called Newtonian fluid. 26. Enunciate Newton's law of viscosity. Distinguish between Newtonian and nonNewtonian fluids. (UPTU 2001 — 02) Newton's Law of Viscosity. The tangential shear stress existing between two adjacent fluid layers is directly proportional to the velocity gradient existing in the fluid in a direction perpendicular to the fluid layers. Shear stress (r) a du

dY

r= du

or

dy Where

du = velocity gradient which is equal to time rate of shear strain ( c )

dy p = viscosity

Newtonian Fluids 1. Newtonian fluids follow Newton's law of viscosity

dt

Non-Newtonian Fluids 1. Non-Newtonian fluids do not follow Newton's law of deformity Cound.

20

Fundamentals of Fluid Mechanics

du

du

# LI ' ndy

= LI ' dy

2. The value of L is constant and it does not change with the rate of deformation.

2. The value of 1/ is not constant and it changes with the rate of deformation. A tan 62 > tan°,

P1> P2> P3

28. What are dilatant or shear thickening fluids? Fluids in which the apparent viscosity increases with increasing rate of deformation are called dilatant or shear thickening fluids. Suspension of sand, butter, starch and sugar solutions are shear thickening fluids. = g(du) and n < 1 d

Fluids: Definitions & Properties

2I

tan 03 > tan G2 > tan C., r

P3 > P2 > Pi

du dy —O. 29. What are Bingham plastics or ideal plastic fluids? The fluids, which act like solids by withstanding a definite shear stress without strain change and later rate of strain increase with shear stress after yielding. Oil paints, jellies, drilling sand and sewage sludge are ideal plastic fluids.

i

du —10dy 30. What are thixotrophic or plastic fluids? The fluids, which possess a definite yield shear stress and later their shear stress varies nonlinear with the rate of strain are called thixotropic or plastic fluids Lipstick and printer ink are plastic fluids.

z I

du _ii... dy z=

Ty

+

µl

dy

and n> 1

22

Fundamentals of Fluid Mechanics

31. Show the curve of solid and inviscid or ideal fluid on shear stress versus shear strain rate diagram. The solids deform or undergo strain on application of shear strain but their strains do not d(/) = du = vary with time i.e. 0. Hence solids are shown along stress axis. dt dy

Solid

du — —1,dy Behaviour of Solids The ideal or inviscid fluid has zero viscosity. Hence it is shown along zero shear stress axis. 32. Flow of a fluid is given by: (du 1 _— Pr d—y The velocity distribution is given by:

)1.5

u= um" [2 3' — Y 3 1 3 h h3 where umax = max velocity and h = film thickness. If viscosity = 0.5 Ns/m2 then find (1) shear stress at solid surface for umax = 0.3 m/s and h = 10 mm (2) what Newtonian fluid can induce same shear stress for same velocity gradient and maximum velocity? du can be found out by differentiating given — ( dy relationship of velocity distribution. Shear stress can be found out by using shear stress and velocity gradient relationship. Guidance:

The velocity gradient

u — um' [2 31 3 h

Y3

h3

du _ tima. [2 3y2 dy 3 h h3

Fluids: Definitions & Properties

23

Ideal or non-viscous fluid

du dy Behaviour of Ideal Fluids At solid surface y = 0, umax x 2 (du dy) .0 3 h y 0.3

du ( dy) y=0

3

2

x

0.01

20s1 ( du TO —

)1.5

T y

= 0.5 x (20)1.5 = 24.56 N/m2 For Newtonian fluid du dy 24.56 = x 20 24.56 Ns A — 20 — 1.23 2 m =

Newtonian fluid of viscosity 1.23 N';: can induce same shear stress. m 33. What do you understand from (1) poise (2) centipoise (3) specific viscosity and (4) kinematic viscosity? or Differentiate between dynamic and kinematic viscosity (UPTU 2004-05) Poise is the unit of viscosity in CGS system. One poise = 0.1 N2 . Poise is a relatively large unit. Hence centipoise is used which is equal to 0.01 poise. Water has viscosity at 20°C = 1 centipoise.

24 Fundamentals of Fluid Mechanics Kinematic viscosity is the ratio of the viscosity of fluid to its density. Kinematic viscosity =

viscosity µ (in2 ) density p s

The kinematic viscosity in CGS system is stoke (=10-4 m2/s). Stoke is a relative large unit and centistoke (cst) which is equal to 0.01 stoke is used. The kinematic viscosity of water at 20°C and air at STP are 1 cst and 15 cst respectively. Air has higher kinematic viscosity than water. Dynamic Viscosity

Kinematic Viscosity

1. It is the ratio of shear stress produced by unit rate of shear strain T =

1

.

(du) dy 2. It has unit of poise in CGS system and

Ns

m

in S1 units

104 poise =

Ns2

Poise = 100 centipoises 3. It is an indicator of only viscous force of the liquid.

4. It is used where viscosity forces are predominant.

1. It is the ratio of dynamic viscosity by unit density v = ii P 2. It has unit of stoke in CGS system or m2/s in S1 units 104 stokes = 1 m2/s stoke = 100 centistokes. 3. It is the ratio of viscous force to the inertia force of the fluid. If viscous force is large, kinematic viscosity is large and vice versa. 4. It is used where both viscous and inertia forces are predominant.

34. What are the effects of temperature and pressure on viscosity? Increase of temperature causes a decrease in the viscosity of liquid while it increases the viscosity of gases. The viscous forces in a fluid are produced by the intermolecular cohesion and molecular moment transfer. The molecules of a liquid are comparatively more closely packed and molecular activity is rather small. Hence viscosity of liquid is primarily due to molecular cohesion. The molecular cohesion decreases with elevated temperature which results into drop in viscosity of liquid. The molecular cohesive forces are very small in a gas and the viscosity is primarily due to the molecular momentum transfer. The rise of temperatures increases the molecular activity thereby increasing the viscosity of gas. High pressures also affect the viscosity of a liquid. The viscosity is found to be increasing with rising pressure. Viscosity increases, as more energy is required for the relative movement of liquid molecules at elevated pressure. 35. A plate 0.05 mm distance from with a fixed plate fluid in between moves of 1 m/s. It requires a force of 2 N/m2 to maintain the speed. Determine the viscosity.

Fluids: Definitions & Properties

25

du= lmls 1 0.05

•▪.----------------------------------------------------------------------

I =2

----------------------------------------------. - - - - - - - - - - - - - - - - - - - -- - -.

//////////////////// Guidance: We have to apply the Newton's law of viscosity, which relates velocity gradient to the applied shear stress. Shear force and velocity gradient are given. du T=g dy 2=px

1 0.05 x 10 3 = 2 x 0.05 x 10-6 Ns/m2 = 1 x le poise

36. A plate having size 100 x 100 mm is pulled with velocity of 0.05 m/s over a fixed plate at distance of 0.25 mm. Find (1) force and (2) power to maintain velocity if fluid has fl = 1 poise. Guidance: The shear stress is force divided by area of the plate. Newton's law of viscosity is used to find the shear stress from the velocity gradient (du and dy are given) du dy = 0.1 x

0.5 — 200 N/m2 0.25x10-3

F =

A F=AxT = 100 x 100 x 10-6 x 200 = 20 kN Power = Force x Velocity du = 0.05 m/s I

0.25

/ / / / / / / / / / / // // / / / / /

26

Fundamentals of Fluid Mechanics

= 20 x 103 x 0.5 = 10 kW 37. A skater weighing 900 N is skating with speed of 20 m/s. The skates have area of 9 cm2. Find the thickness of water layer between skates and ice if it = 1 x 10-4 poise and dynamic coefficient of friction of the ice = 0.02. Guidance: The shear force enabling skater to move is friction force. The thickness of water layer can be found out from Newton's law of viscosity. ICP

Friction force

_____ ------------------------

t

/777 7 /7 /I 7-7 7717 7 7 7 7

Friction force = x mg = 0.02 x 900 = 18 N „ du 7 Rwater 4 x20 18 = 1.1 x 10-5 m = 1.1 x 10-2 mm 1x10-5

38. A plate of size 25 cm x 25 cm and weight 1000 N slides down on inclined surface inclining 30° to the horizontal which has certain thickness of lubrication with it = 0.1 poise. This attains velocity of 0.5 m/s over the lubricated surface. Find thickness of lubrication. Guidance: The component of the weight acting along the inclined surface will act as shear force. Shear force = w sin 30 r = 1000 x - = 500 N 2 500 _ r- F , - 0.8 x 104 N/m2 A 25x25 x 10 -

Fluids: Definitions & Properties

27

u

"C = 12 cL

dy 8 x 103 —0.01x 0.5 dy _ 5 x10-3 dy 8x103 = 0.625 x 10-3 mm 39. A piston having 100 mm diameter and 150 mm length moves inside a cylinder of 100.4 mm diameter. The weight of the piston is 50 N and annular space between piston and cylinder has lubrication oil of g = 0.5 NS/m2. Find the velocity with which the piston will slide inside the cylinder.

ll_

100.4

—11°'

Piston

A'

Cylinder A1 ---------,A'

/11 1

Guidance: It is the weight of the piston, which is providing shear force. The surface of shear force is the surface area of the piston. Shear force = Weight of piston = 50 N

28

Fundamentals of Fluid Mechanics

Surface area = 71" DL = 71" X

100

X

150

X

10-6 1112

= 4.71 x 10-2 m2 Oil film thickness dy

=

100.4 — 100 x 10_3 2

= 0.2 x 10-3 "I- —

50 , — 10.6 x 102 N 4.71 x 10-2

-I- — ti du dy 10.6 = 0.5 x

du 0.2 x10-3

du — 10.6 x 102 x 0.2 x 10-3 0.5 = 0.424 m/s 40. Determine the viscosity of a fluid having kinematic viscosity 10 stokes and specific gravity of 2. Guidance:

Kinematic viscosity = Li , ti can be found out from known kinematic viscosity P

and density. Sp gravity =

P Pwater

P = Pwater x Sp gravity = 100 x 2 = 2000 kg/m3 Y=

Or

1-1

P p=pxy = 2000 x 10 x 10-4 = 2 Ns/m2

(1 stoke = 10-4 m2/s)

41. Through a very narrow gap of height h, a thin plate of large extent is pulled at a velocity v. On one side of the plate is oil of viscosity att1 and on other side oil of viscosity /Iv Calculate the position of the plate so that the pull required to drag the plate is minimum. (UPTU 2003 — 04)

Fluids: Definitions & Properties

29

y V

Guidance: When plate is pulled, drag forces F1 & F2 will be exerted at upper and lower surface of the plate by fluids. The total drag (F1 + F2) will be minimum, when d (F1 + F2) _ 0, where y dy

distance of the plate from the upper surface of the gap. F1 = drag force by exerted upper fluid V = pi xA 2

F2

V

x

A

F = vA(h — ± y 1-1 2

)

b— y

y dF =0 dy

For minimum F,

dF _ TiA ( dy

P1

P2

y

b—y

)—0

(h — y)2 _ P2 2

h2 +y2 —thy _,u2 2 Pi ) + (1 P2

2(

Y) (Y

0

Y 2_ 4 — 4 (1— 41-12 ) h _ 2 h = 1 ± 1-12, kt

(I; has to be >1)

h 1 + VP2 Pi 42. Two large plane surfaces are 24 mm apart. The space in between filled with glycerin. What force is required to move a very thin plate having surface area 0.2 m2 between the two plane surface with velocity of 1 m/s when (1) thin plate is in the middle of two (2) thin plate is at 8 mm from one surface. Take g= 0.5 NS/m2.

30

Fundamentals of Fluid Mechanics

Guidance: The shear force is the sum of the shear force of upper and lower fluid. //11/ 11//////

///////////// :•:•:•:•:•:•:

dy1 = 12

dyi = 16 du - 1 mls

..

dy2= 12

du = 1 m/s

dy2= 8 ••:•:::•:

•.•

\\\\\V\\\\\ Case 1

\\\ \\ \\\\ Case 2

Case 1 dyi = 12, dy2 = 12

ZI

=Z2 or Fi = F2

= 0.5 x

1

12 x10-3 = 41.367 N/m2 F1 =Ax1-1 =0.2x41.67N = 8.334 N F = + F2 = 2 x 8.334 =16.668 N Case 2 dyi = 8, dy2 = 16 ti =0.5x

1

8x10-3 = 62.5 N/m2 F1 = 62.5 x 0.2 =12.5 N F = + F2 =12.5 + 6.25 =18.75 N

T2 = 0.5x

F2

1 16 x10-3

= 31.25 N/M2 = 31.25 x 0.2 = 6.25 N

43. A square plate 2 m side and 1 mm thick weighing 60 N is to be lifted through a vertical gap of 21 mm of infinite extent. The oil filling the gas has EL = 3 Ns/m2 and sp gravity = 0.9. The plate is moved with 0.5 m/s, find force and power required for moving the plate.

\\\\\\\\\\\\\

Fluids: Definitions & Properties

31

u =0.5

0 bA

0 Plate

Vertical gap

li

Fs= wi

411-

Guidance: The plate has to be moved against drag force

/V du acting both surfaces dy and weight (w) but the buoyancy force w1 (weight of liquid displaced) is acting upwards. Weight of liquid displaced (w1) = volume of the plate x sp gravity of liquid X density of water =2 x 2 x 1 x 10-3 x 0.9 x 1000 = 3.6 N Buoyancy force = 3.6 N Drag force = T x 2A = 2 x 4 x du dy (dy _ 21-1) 0.5 =2x4x3x 2 10 x10-3 = 1200 N Force required = Drag force + weight - Fp = 1200 + 60 - 3.6 = 1256.4 N Power required = force required x velocity = 1256.4 x 0.5 = 628.2 watts 44. The velocity distribution in a viscous flow over a plate is given as u = uy - y2 for y 2 m where u = velocity (m/s) at distance y from the plate. If au = 1.5 Ns/m2. Find shear stress at y = 0 and y = 2. Guidance: By differentiating the velocity equation, we can find velocity gradient and shear stress. u = 4y - y2 du — = 4 - 2y dy

32

Fundamentals of Fluid Mechanics

u = 4y-y2

Y

Plate (du _ 4 & (du) dy) y=0 dy y=2

=0

p (du ) ro = p (du dy ) y=0 dy y=2 = 1.5 x 4 = 6 N/m2 "T2 = 0 45. A fluid kinematic viscosity 2.5 stokes flows over a horizontal plate with area 1 m2. The velocity is given by u = 2y — 2y3 where y is distance from the plate. If shear force measured on plate is 0.4 N, find sp weight and sp gravity of the fluid. Guidance: The viscosity gradient is to be found out by differentiating velocity equation. Y t

— 2y-y3 //// f)( 7-7---Th = 0.4 N (due to flow) Plate F0 = To X A=

To x 1

u = 2y — 2y3 du = 2 — 6y2 dy (du) =2 dy y=0 „ = p (du co dy)

y=0

and F0 = Ico x A z0 x 1

F0 = p x 2 Ns p = 0.4 = 0.2 2 2 m Y= — P 2.5 x 10-4 = 0.2

Fluids: Definitions & Properties

P

33

0.2 kg/m3 2.5 x 10—A = 800 kg/m3

Sp gravity =

P P water

800 — 0.8 1000

46. The velocity of a fluid over a horizontal plate is varying as parabola with vertex at 10 cm from the plate where the velocity is 0.1 m/s. If tt = 0.9 NS/m2, find velocity gradient and shear stress at y = 0, and y = 0.05 m from the plate. Guidance: The velocity profile can be assumed as u = aye + by + c as it is varying parabolic. There are three unknowns and we require three conditions. At plate when y = 0, then u = 0, at vertex y = 0.1, velocity u = 0.1 m/s (given). Also velocity gradient ( du

dy y=o1 at the vertex. U=0.1 m/s

Velocity profit of the fluid

u = ay2 + ay + c

(1)

0=c u = ay2 + by

(2)

0.1 =0.01 x a + 0.1b a + 10b = 10

(3)

putting y = 0, u = 0

putting y = 0.1, u = 0.1 or puttting

Or

duY

=0 y=0I

(du = 0 =2 xax 0.1 + b dy) _ 01 y a = — 5b From equation (3) and (4) — 5b + 10b = 10

(4)

34

Fundamentals of Fluid Mechanics

b=2 a = — 10 u = — 10y2 + 2y

and

(du) = dy

20y + 2

(du) = 2 (du =— 1 + 2 = 1 dy y=0.05 dy y=o _ = (du „ (du co d dy) y=0 To.o5 — —Y)y=0.05 = 0.9 x 2 = 0.9 x 1 = 1.8 N/m2 = 0.9 N/m2 N Z is moving over a fixed surface with height = 0.1 m. m A float moves 0.01 m in 2 seconds. Find the velocity gradient and shear stress acting on the float.

47. A fluid of viscosity m = 0.2

Float

1

h=0.1h 1 6

1 11 1

I /

X417 X4I.(/ /// 0 2

Fixed surface

Guidance: The velocity gradient is equal to time rate of strain. x2 — X0 d e tane = h — Rate of strain — dt dt 2

0.01 — 0.05 per sec. 0.1 x 2 de —du = 0.05 per sec. dt dy But

z= ,u du — 0.2 x 0.05 = 0.01 N dy

48. Find the torque and power required to turn 10 cm long, 5 cm diameter shaft at 600 rpm in a 7 cm diameter concentric bearing flooded with lubricating oil of viscosity = Ns 1 m2 •

Fluids: Definitions & Properties

35

Oil

Shaft Bearing

dy

Guidance: The problem is similar to a thin plate moving between two fixed plates. The velocity can be calculated from rpm and when it is divided by the thickness of fluid, the velocity gradient is found out. dy =

dbearing dshaft

2 _ 0.07 —0.05 2 = 0.01 m irdN rcx 0.05 x 600 du = = 60 60 =1.57m 1 s du T =F.i dy =lx L57 =157 N/m2 1x10-2 Area = lr dL =71- x5x10x10-4 =15.7x10-3m2 F= irx A =157 x15.7 x10-3 = 2.465 N d Torque = F x— 2 5 x10-2 = 2.465x= 6.16 x10-2 Nm 27r NT Power = 60 2rx 600 x 6.16 x10-2 60 = 3.868 watts

36

Fundamentals of Fluid Mechanics

49. A vertical cylinder of diameter 800 mm rotates concentrically inside a bigger cylinder of diameter 200.4 mm, both having height = 300 mm. The gap between the cylinders contains fluid with viscosity unknown. Find viscosity if torque of 30 Nm is applied to rotate the inner cylinder with rpm = 200. T= 30 Nm Fluid

1— 200 200.4 —10; Guidance: Velocity gradient can be found from the rpm and fluid gap. By Newton equation of viscosity, viscosity can be calculated. r dN du — 60 rc x 0.2 x 200 60 = 2.093 m/s do — di _ 200.4 — 200 X 0-3 dY 2 2 = 0.2 x 10-3 m Force = T x A Torque —

TxAxd 2

A = irdL = x 200 x 300 x 10-6 = 18.84 x 10-2 m2 30 = r x 18.84 x x 200 x 10-3 2 30 x 2 18.84 x 200 x10-5 = 1.592 x 103 N/m2 = p du dy

2.093 0.2 x10 3 1.592x103 x 0.2 x10-3 2.093 = 0.152 Ns m2

1.592 x 103 = p x

Fluids: Definitions & Properties

37

50. A circular disc having radius R is kept at small height above a fixed surface by a layer of oil having viscosityµ Determine the expression for the viscous torque on the disc.

Disc -4—

Oil

R

// // //

h t

In order to find viscous torque, we take an elementary ring at radius r and thickness dr. If w is angular velocity of the disc, the tangential velocity at ring is u = r.w and area of ring = 2 iv .dr du T=P dy

_ p x rw h dF,. = TT x area of ring _ x rw x 27r rdr —

241w

2 x r dr

Torque = Force x radius 3 dTT — 27Tuw x r dr Total Torque =

Jo

dTr — 22rhliw Jr R r3 dr o T—

Drkiw R4 x but R= D h 2 4 2.7r,uw x D4 4x16 h

ripwD4 = 32h

3$

Fundamentals of Fluid Mechanics

51. A thrust bearing has 10 cm diameter is separated by an oil film of height 2 mm from fixed base. Find power dissipated in the bearing in case it rotates at 400 rpm. Given µ = 0.9 Ns2 M

N=400 rpm

h=2x 10-3 M 1410.1 M 1

Guidance: Viscous torque T —

-i

repwD4 can be calculated. Power is equal T x w 32h w = 27-cN/ 60 27r x 0.1x 400 — 60 = 41.86 rad / sec. 7rµwD4 32h = 7r x 0.9 x 41.86 x (0.1)4 32x 2 x10-3 = 0.1851x 10-3 Nm

T=

Power =T xw = 0.1851x 10-3 x41.86 = 7.74 x 10-3 watts 52. A solid cone of radius R and vertex angle 2f) is mode to rotate at angular velocity w in the conical cavity containing oil with viscosity p. If h is the gap between the cone and cavity, find torque T to rotate the cone.

Fluids: Definitions & Properties

R

Take a slant height ds between a radius r and r + dr dr sin 0 = — ds or

ds —

dr

sine

Velocity at radius r = u = wr Velocity difference du at r from stationary outer cone surface = wr du Shear stress at strip = Tr = — A T y =

wr

Force at strip = dF = Tr X dA dr dr = ,u wr x 271-r x as ds = sin° sine 2 — 27rIlw r dr h sin Torque at strip = dF x r 27cAlw 3 r dr dTr. — hsin9 2,7r,uw dT,. = r3 dr Torque = Jo hsin0 .10 27ryw R4 7rA1wR4 x 4 2h sin 0 h sin 0

39

40

Fundamentals of Fluid Mechanics

53. What is the compressibility and coefficient of compressibility of the fluid? Compressibility. Fluids can be compressible or incompressible. Incompressible fluids have constant density while incompressible fluids have variable density. In compressible fluids, when pressure is applied, fluids contract and when pressure is reduced, they expand. Compressibility is the property of a fluid, which characterizes its ability to change its volume under pressure.

dp

V Coefficient of compressibility. It is the ratio of the change of volume per unit volume i.e. volumetric strain to change of pressure. If there is change of volume — dv from the original volume V and the change of pressure is dp, then dv Coefficient of compressibility (P) —

dp

54. What is bulk modulus of elasticity? Find the value of bulk modules for isothermal and adiabable processes. The bulk modulus of elasticity (k) is the ratio of compressive stress to the volumetric strain k= dp _ 1 dV R V The bulk modulus can be also expressed in relative change in density m = pV dm = dp x V+px dV dV _ dp V p d K =P dp

Fluids: Definitions & Properties

41

Isothermal process can be expressed as: PV = constant pdV + Vdp = 0 dp —k P= V Bulk modulus of isothermal process is equal to the pressure of the fluid. Adiabatic process can be expressed as PVT constant PyV1 1 dV + VY 'p = 0 _ dp =k yp

Bulk modulus in adiabatic process is equal y time the pressure. 55. An increase in pressure of a liquid from 7 MPa to 14 MPa results into 0.2% decrease in its volume. Find (1) bulk modulus and (2) coefficient of compressibility P = 14 — 7 = 7 x 106 N/m2 dV = 0.2 % = 0.002 V dp =7x106 k — .002 V = 3.5 x 109 N/m2 _1 = 1 k 3.5x109 = 0.286 x 10-9 m2/N 56. A cylinder contains 0.4 m3 air at 100 KN/m2 which is compressed to 0.08 m3. Find bulk modulus if compression is done (1) isothermally and (2) adiabatic. Given y= 1.4. Guidance: Bulk modulus in isothermal process is equal to P2 which it is yP2 for adiabatic process Pi VI = P2 V2 for isothermal process D

/2

P1V1 100 X 103

V2

=

0.08

= 500 x 103 N/m2 Bulk modulus k = P2 = 500 x 103 N/m2 for adiabatic process VI P1VT = P2 P2 — P1 (W)

x 0.4

42

Fundamentals of Fluid Mechanics

0'4 )1.4 x103 0.08 = 9.52 x 105 N/m2 K = yP2 = 1.4 x 9.52 x 105 N/m2 = 100 x

57. Explain: (1) Evaporation (2) Vapour pressure and (3) boiling. Evaporation: Evaporation means a change of phase from liquid to gaseous. The rate of evaporation depends upon the pressure and temperature. Evaporation increases with rise of temperature or lowering of pressure. Vapour Pressure: Consider a liquid, which is enclosed in a closed space. The molecules of the liquid which attain high energy leave the liquid phase to vapour phase. However some of the molecules in vapour phase have a tendency to rebound back and get absorbed in the liquid. Hence there is continuous interchange of molecules between the liquid and the vapour above it. The vapour pressure will have a constant value when the molecules leaving and entering into vapour state is same. The constant vapour pressure exerted by the vapour on liquid is called the saturated vapour pressure. Higher the vapour pressure, more volatile is the liquid. Petrol has vapour pressure of 30.4 kN/m2 while water has vapour pressure of 2.35 KN/m2 at 20°C. Hence petrol vaporizes faster than water. Mercury has very low vapour pressure and high density. Due to this, it is very suitable as fluid for manometer. Boiling: Evaporation of liquid ceases when saturated vapour pressure has been achieved on the liquid. In case vapour pressure decreases below the saturated vapuor pressure, evaporation of liquid starts again and it continues until new equilibrium condition is attained. If the vapour pressure falls considerably, then the molecules leave the liquid surface very rapidly to vapour state and this is called boiling. The boiling point of a liquid is the temperature at which its vapour pressure equals the external pressure. For water, the vapour pressure becomes 1.01 bar at 100°C, which is equal to atmospheric pressure, resulting boiling of the water. At lower atmospheric pressure like in hill stations, water attains vapour pressure equal to lower atmospheric pressure at lower temperature and water boils at lower temperature. 58. What is cavitation? What is its effect? How is it avoided? The fluid starts boiling even at low temperature if pressure on the fluid falls to its vapour pressure at that temperature. Such boiling often occurs in flowing fluids specially on suction side of the pump. When such boiling does occur in flowing fluids, vapour bubbles begin to grow in local regions of very low pressure. The formation of bubbles of vapour in flowing fluid is called cavitation. The bubbles of vapuor are formed due to cavitation in the region of low pressure and bubbles are carried by the fluid flow to other regions such as delivery side of the pump where the pressure is high. In the region of high pressure, bubbles suddenly collapse and vapour condenses, resulting drop in volume occupied. The space occupied by the bubbles is now available for the fluid to rush in to fill it.

Fluids: Definitions & Properties

43

at Pam 0 a C

Vapour Pvapour

00 0

0 6

0 0

a

00 0

0 as 0

a

a a ooa 0

0

0

-R-Z-R

Patm>Pv

'vapour >Pa

a

a 00

,„ 0

0 0

- 0 -0- 0 -0- 0 -0- 0.

Vapour

Bubbles

The rushing of fluid leads to generation of noise and vibrations. If bubbles collapse on the impeller vanes, the rushing fluid corrodes the vanes and other surfaces of the pump. Cavitation can also occur if a liquid contains dissolved air or other gases n liquid. The solubility of gests in a liquid decreases as the pressure is reduced. Gas or air bubbles are released in the same way as vapour bubbles with the same damaging effects. The cavitation occur in hydraulic machines having flow of high speed fluids such as turbines, pumps, propellers of ship and pipeline. The cavitation has to be avoided by ensuring that pressure on the fluid should not fall below its vapour pressure in any region of the flow by proper designing the fluid flow.

iaaazzaajz _ : Pfluid < Pvapour _-_-_-_-_-_-

iff -7-1-11/17-7-1-1-?./ Bubble

Vapour

Fluid

Cavitation

Collapse of Bubble

59. What are differences between cohesion and adhesion?

1. Cohesion is the intermolecular attraction between the molecules of same fluid (like molecules). 2. The fluid tends to resist the shear stress due to cohesion. 3. If cohesive force is more than adhesive force liquids form droplets without

1. Adhesion is the force of attraction between alike molecules i.e. the molecules of fluid and the molecules of solid boundary surface in contact with the fluid. 2. The property of adhesion enables a liquid to stick to another surface. 3. If adhesive force is more than cohesive force, liquids spread out on the contact surface

44 Fundamentals of Fluid Mechanics wetting the contact surface. Example — mercury. 4. For higher cohesion, contact angle is more than 90° for most surface.

and wet the surface. Example — water 4. For higher adhesion contact angle is less than 90°.

60. What are wetting and non-wetting liquids? Contact surface

Adhesion> P f Pgauge = Plghg

Pressure and Head

73

(b) Vacuum Pressure: Tabs

± hf P f g ± hvPlg = Patm

Patm — Pabs Pvacuum

= hf P f g ± hvPlg

= hf P f g ± hvP lg

If Pi >> Pf Pvacuum

= hvPlg

16. What are the micrometers? Why are they called sensitive manometers? Micrometers are modified form of double column manometer in which the cross-sectional area of the limb connected to vessel is so large with respect to other limb opened to the atmosphere that any change in the level in the large limb can be neglected. Hence the gauge pressure inside the vessel is directly proportional to the level in the open limb. In micrometers, either one limb is made much larger or a reservo it of large cross-section area is introduced in one of the limbs. In case the cross-sectional area of larger diameter limb (aiarge) is 100 times than the cross-sectional area of smaller diameter limb, then volume moving from one limb to other has to be same. ... Change of volume of large diameter limb = Change of volume in small diameter limb

MICROMETER dh xa=hx asmall dh _ asmall = 1 -30 h 100 alarge Pgai„, = pgh — pgdh _ por h _ dh) ° h) = pg(h — 0) = pgh

74

Fundamentals of Fluid Mechanics

Hence we can directly read the pressure in the vessel by the level in the smaller diameter limb. The change of the level in the limbs is also sufficient for all pressures. Hence such instruments are also called sensitive manometers. As the pressure can be read from the smaller diameter limb, this limb is required to be made of glass and larger diameter limb can be made of any metal. 17. What is the difference between vertical and inclined single column micrometer? Vertical Single Columns Micrometer

Inclined Single Columns Micrometer

1. Narrow limb is vertical as shown in the figure. 2. The distance moved by the liquid in the narrow limb is small 3. The distance moved by the liquid

1. Narrow limb is kept inclined (angle 0) to the horizontal. 2. The distance moved by the liquid in the narrow limb is more. 3. The distance moved by the liquid is 1 where

is h 4. Less sensitive 5. Fixed sensitiveness

1= h . As 0 < 1, hence 1 > h. sine 4. More sensitive. 5. Sensitiveness can be improved by inclining more the narrow limb

- --- ----- --------------------_-_---_--_--_--_--_--_-:-

Vertical Single Column Micrometer 18.

What are differential manometers?

Inclined Single Column Micrometer (UPTU 2003 — 04)

or Derive an expression for the absolute pressure to be measured in pipe carrying highly pressurised water with the help of a double U-tube manometer having mercury in each tube with different levels in the four limbs. (UPTU 2009 — 10) Differential manometer is an instrument, which is used to measure the difference of pressures between any two points in a pipeline or the difference of pressures between two pipes. Differential manometers can be: (1) Two piezometer tubes method (2) U tube differential manometer and (3) Inverted U tube differential manometer. 1. Two Piezometer Tubes Method: Two separate piezometer tubes are inserted at two points 1 and 2, between which the pressure difference is to be found out. The pressure

Pressure and Head 75 difference between the points is equal to the difference of the levels of the liquid in the two tubes i.e. dp = h1 — h2.

Flow

Flow

Two Piezometer Tubes Method 2. Inverted U Tube Manometer: In case two piezometer tubes are joined at top, they form an inverted U tube manometer. An opening at the top of the inverted U tube is provided to drive away air present in the U tube. The upper part of the manometer can also be filled with a lighter measuring fluid, which is lighter than the fluid flowing through the pipeline. This helps in restricting the height of the limbs. AP = — h2 = h or AP = pgh pg For manometer having measuring liquid (ml) AP = hl pg (I— Pm/ = hl pg(1 — smi) if water is flowing, then Smi = Sp gravity of measuring liquid. Measuring liquid (m, Smi) —r h ' — --X

1—> Flow Inverted U Tube Differential Manometer

1=> Flow Differential Manometer with Lighter Fluid

76

Fundamentals of Fluid Mechanics

3. U Tube Differential Manometer: In case pressure difference between two points is high, then length of piezometer tubes and inverted U tube differential manometer is long and unsuitable. In such cases. U tube differential manometer is used to measure the pressure difference. The measuring liquid (generally mercury) used in such manometer is heavier than fluid flowing through the pipe or pipes. The pressure difference can be given as:

Fluid density = sp gr=S

X Hg (density = pzig sp gr=SH8.) U Tube Differential Manometer: One Pipe

U Tube Differential Manometer: Two Pipes

(a) One Pipe Line dP = hp Hgg — hpg = hpgr —1)= hpg(SHg —1) (b) Two Pipe Lines AP = hpg( P i I g p

11

krlg

= P g[h(SH g —1) — k] 19. What are differences between simple and differential manometer? Simple Manometer 1. It is used to measure pressure at a point. 2. One limb is connected to the point and other is open to air. 3. The pressure at a point is obtained in terms of difference of level of fluid flowing through pipe.

Differential Manometer 1. It is used to measure pressure between two points 2. Limbs are connected to the two points 3. The difference of pressure is obtained in term of measuring the difference of the levels of manometer liquid.

Pressure and Head

4. It can be manometer like piezometer or simple U-tube type.

77

4. It can be manometer like differential U-tube, inverted U-tube or two piezometer tubes.

20. Why mercury is used as manometeric liquid? The mercury is used as manometric liquid for following reasons: 1. 2. 3. 4. 5.

It has high density so that mercury rise is reasonable for a pressure. It is immiscible with the liquid flowing in the pipes. It does not stick to the surface of the tubes of the manometer. The vapour pressure is negligible at room temperature. It does not react chemically with most of the liquids flowing through pipes.

21. What is a micromanometer or two liquids double columns enlarged ends manometer? Micromanometer is used for vary high precision measurement of pressure difference. It is a U-tube manometer with two ends widened or two reservoirs provided at the ends. In order to magnify the readings, two liquids of different specific gravity are used for the reservoirs and U-tube. These two liquids and fluid in pipes are required to be immiscible. Initial levels in reservoirs and U-tube are x — y and y — y. On application of pressures, the rise of liquids in U-tube is h, which is directly proportional to the difference of pressures.

h,

dh • •

•

• 1 . •

•

• • • . • • . • .

t

Manometer liquid 2 (Sp gravity = S9) y

y

z

z

Micro Manometer

Manometer liquid 1 (Sp gravity = S1)

Volume of liquid fall in reservoir = Volume of liquid rise in tube h x a A = Cross-sectional area of reservoir and dh x A = — 2 a = Cross-sectional area of tube

78

Fundamentals of Fluid Mechanics

axh dh = — — —> 0 A 2 The pressure above the line z — z is equal in both limbs.

it — dh)

P1 + p1 x g(hi + dh)+ p , x g(h2 + l

= P2 ± pi x gh+ p2 x g h2 —11 + dh + p f x g(hi —dh) ( ) P1 — P2 = Plgh — P2gh P1 — P2 = h (Pi P2 where p density of water Pg P P = h(S1 — S2) 22. If atmospheric pressure is 100 kpa and pressure gauge A reads 140 kpa, pressure gauge B reads —60 kpa, then find the absolute pressure of H2 and 02. Also find the reading of pressure gauge C.

H2

02

B

C The pressure gauge A is reading gauge pressure of 112 as its one end is open to atmosphere. The absolute pressure of H2 is — Pabs = Pgauge + Patm

= 140 + 100 = 240 kpa Hence absolute pressure of 112 is 240 kpa. The pressure gauge B is reading the vacuum pressure of 02 as its one end is open to atmosphere. The absolute pressure of 02 is — Pabs = Patm + Pvacuum = 100 — 60

= 40 kpa Hence absolute pressure of 02 is 40 kpa. The pressure language C is reading gauge pressure of H2 w.r.t. 02. If it is reading Pgauge, then — PH2 — Pgause + PO2 Pgause = PH2

P02

= 240 — 40 = 200 kPa The pressure gauge C will show 200 kpa.

Pressure and Head

79

23. If the readings of a barometer at top and bottom of a hill are 660 and 760 mm of Hg, then find the height of the hill. Take specific weight of the air (y) = 12 N/m3. P1

Pressure of top of hill = h1 x p Hg x g = 1000 76° x13.6 x103 x 9.81 = 101.4 KN/m2

P2 = Pressure at bottom of hill = h2 x pHg x g = 1000 66° x13.6 x103 x9.81 Now

= 88.06 KN/m3 P1 P2 = yh

or

h—

(101.4 — 88.06) x 103 12

13.34 x 103 12 =1112 m 24. A tank is filled with water and mercury with air at top as shown in the figure. If gauge 1 reads 400 kpa (absolute), find (1) height of the water and (2) reading of gauge 2.

Pr = 230 kpa

Water Hg 1 120 cm

= 140 kpa

The pressure read by gauge 1 is P1 = 140 kpa. Now pressure P1 will be equal to pressure exerted by 1.20 cm of Hg, height h of water and air. Therefore is equilibrium — = 1.20 x plig X g +hXpX g + Pair 120x13.6 x103 x9.81 hx103 x9.81 + 400 = • +230 103 103 =160+9.81xh+230

10

h = — = 1.02 m 9.8

80

Fundamentals of Fluid Mechanics

Now pressure at pressure gauge 2 is sum of the pressure exerted by water and air (1.02 +1.2) x103 x 9.81 + 230 P2 = 103 = 21.78+230 = 251.78 kpa 25. A tank contains air, oil and water as shown in the figure. The manometer attached to the tank has mercury and mercury rise is 50 cm. Find the air pressure Pa if Pa = 700. Air 150 cm

_

h2 t hl I

.1.

50, cm

5: Hg h2 = 50 + 150 = 200 cm hi = h2 — 90 = 200 — 90 = 110 cm As fluid is in balance in both limbs from datum x — x Pressure in left limb = Pressure in the right limb

Pa -Fh1xPwxg+(h2 — hi)Poaxg=

50 xplig xg

100

Pa + 110 X 103 x 9.81 + 90x 700x 9.81 100 103 100 x 103 Pa +10.8+6.18=66.7 Pa = 49.73 kpa

05x13.6x103 x 9.81 103

26. A U tube is used to measure the water pressure in the pipeline. The mercury level is 100 mm below the centerline in left limb connected to it and 200 cm above the centerline in the right limb open to the atmosphere. Find the gauge pressure of the water in the pipeline.

Pressure and Head

8I

Gauge Pressure Let P1 = absolute pressure of the water in the pipeline during equilibrium — Pressure in the left limb from datum = Pressure in the right limb from datum p ±

100 0 xpiv xg = (10 0 + 1 0 0 ) x p

i

1

1 00

x lig

± p g

aint

000

0.2 x 13.6 x 103 x9.81 0.1x103 x g x 9.81 1000 1000 Pgauge = 26.68— 0.981

P1

_ Pa`"`= —

Pgauge = 25.7

kpa

27. The figure shows a conical vessel having a U tube manometer attached to its outlet at A. When the vessel is empty the reading of the manometer is given in the figure. Find the reading of manometer when the vessel has been completely filled with water.

When the vessel is empty, the pressure in both limbs is same from datum line x — x Pressure left limb = Pressure in right limb

82

Fundamentals of Fluid Mechanics

Pat, + 0.2 x pit x g=Hx p, x g + Pam,

H _ 0" 2 x13 6x103 x 9.81 , 10' x 9.81 = 2.72 m

When the vessel is filled, assume the mercury moves down by h meter in the right limb as shown in the figure

h+H+3 x

t +211

- 4p- x h y

y

Jr

Now equating pressure in both limbs w.r.t. datum line y — y. Pressure left limb = Pressure in right limb Palm + (0.2 ± 2h)f) fig X

g = Pa,. + (h + H + 3)pw x g

(0.2 + 2h) x 13.6 x 103 x g = (h +5.72) x g x pi, 113.6(0.2 + 2h) = h +5.72 2.72 + 27h = h + 5.72 26h = 3 h=

m 16 =115.4 mm

The manometer reading will be — = 200 + 2 x 115.4 = 430.8 mm 28. Determine the differential reading h of an inverted U tube manometer having measuring liquid of density 800 N/m3 when connected to point 1 and 2 of the pipelines containing fluids with densities 1.2 x 103 and 1 x 103 N/m3. All fluids are immiscible and pressure at point 1 and 2 are equal and located at same datum level. Take the datum line at x — x. The pressure on the datum line is same from left and right limb. P1 +0.24xp1 xg+hxp,a/ xg= P2 +(0.24+h)p2 xg Pi = P2 , pi =1.2 x 1 03 N3 , P2 =1 x103 N3, p m m

,ii = 800 N/m3

Pressure and Head

83

0.24x1.2x103 xg+hx0.8x103 xg = (0.24+h)1x103 0.288 + 0.8h = 0.24 + h 0.2h = 0.078 h = 0.39 m = 390 mm Measuring liquid, pml

Pressure =P1

24 cm

p2 Pressure =P2

(1)

(2)

29. Liquids of sp gravity 1.0 flows through pipes A and B at 0.5 and 0.2 bar respectively. Pipe A is 2 m higher than pipe B. What would be the difference of levels of U-tube manometer connected to A and B in case sp gravity of manometer liquid is 13.6 and when liquid is at higher level in the limb connected to B.

Equating pressure on left and right limb from datum line x — x Pax( 2±11 ±k)P1X g =

Pb

+hp2g +hipig

0.5 x105 + (2 + h) x103 x9.8 = 0.2 x105 +h x13.6 x103 x9.8 0.3x102 +(2+h )9.81 = 13.6x9.81xh

84

Fundamentals of Fluid Mechanics

30 + 19.62 + 9.81h = 133.4h 123.6h = 49.62 h — 49'62 = 0.401 m 126.6 30. Water flow downward in a pipe inclining at 30° with the horizontal as shown in the figure. Find the pressure difference between point A & B if 1 and h = 0.14 m. The manometer fluid is mercury (S = 13.6).

Water

During equilibrium, the pressure in left and right limb is same from datum line x — x PA +1x103 x 9.81(tsin30+hi + 11 )=13.6x103 xhx 9.81+1x103 x(h4x 9.81+PB PA Pg =10 3 x 9.81(13.6x.14 —1— 0.14) = 9.83 x103(0.764) = 7.5 kpa 31. Determine the pressure difference (PA — PB) when the reading of manometer is as shown in the figure. RD = 0.8

10 cm tw

15 cm

RD = 13.6

Pressure and Head

85

Guidance: To simplify such multi tubes problem, it is best to divide the setup in two or more parts as shown below

B

f. 10c t

15 cm

10 cm

Taking equilibrium from datum x — x for left part and datum y — y for right part PA ± 0.24 x 0.8 x 103 x g

+ 0.10x.8 x 103 x g

= 0.12 x 13.6 x 103 x g +

= 0.15 x 13.6 x 103 x g

or P1 = PA + 1.88 x 103 —16 x 103

+0.10 x 0.8 x103 x g + PB

= PA —14.12 x 103

Now

or Pi = Pg + 20.01 x 103

P1 = PA — 14.12 x 103 = PB ± 20.01 x 103 PA PB = 34.13 kpa

32. A multi tube manometer is employed to determine the pressure in a pipeline. The levels inside the tubes are as shown in the figure. What would be the length of single mercury U-tube to record this pressure?

Guidance: Divide the tubes in three parts to simplify the problem as shown below:

86

Fundamentals of Fluid Mechanics

P,

P2

P2 +

P1+.5X1X103 xg PA = 0.5 x13.6 x103 X g + Pi PA = 66.7 x103 + P1

=.5 x13.6 x 103 x g + P2 P1 = 61.8 x103 + P2

= 66.7 x103 +pain, ±

0.5 xlx103 xg

= Patm+. 5 x13.6 x103 xg P2 = Patm + 61.8 x103

= 61.8 x103 ± Patni ±

123.6 x103

61.8 x103

= 200.3 x103 + Patm

A = Patm +123.6 x103

PA — Patin = "gauge = 200.3 kPa

In case we have ample U tube manometer, then, Pa, 4

Hg g 200.3 x 103 = h x 13.6 x 103 x 9.81 PA — P atm — hi") Hg X

h —

200.3 — 1.5 m 13.6 x 9.81

33. Find the pressure difference between pipe A and B for the reading of differential manometer as shown below

Pressure and Head

87

‘.... 20 cm \ i Water

30 cm Y

Water

t

--f - - 45 cm

67.5 cm

Hg x

x

30 cm

1. Apply equilibrium from datum line x — x

la +0.2x1x103 xgsin45+0.3x13.6 x103 xgsin45 +0.675 x13.6 X 103 X g = 0.45 x 1 x103 Xg+ PA PA — Pa = 1.39 x103 + 28.3 x103 + 90.06 x103 — 4.14 x103 = 115.61 kpa 34. Obtain PA — PB in terms of the quantities shown in the figure. Find also Pc — PA.

Guidance: Divide the tube system in two parts as shown in the figure

88

Fundamentals of Fluid Mechanics

PA =hiP2X g- F Pi from eqn (1)

PI3 + P2hg = p ihg + Pi PI= 4 + het) 2 — pl) — eqn (1)

PA = hiP 2 x g + Pg ± hg(p2 — pi ) PA — PB = (P2 — Pi )gh +p2hig For finding PA — Pc consider the part of the tubing as shown below

PA ± h1P2g ± h2P1 X g = Pc PA — Pc = —(14P2g + h2P1g) 35. Two pressure points in a water pipe are connected to manometer which has the form of an inverted U-tube the space above the water in two limbs of manometer is filled with toluene (sp gr 0.875). If the difference of level of water columns in two limbs reads 12 cm, what is the corresponding difference of pressure? (AMIE Mech. Engg.) Guidance: Consider two parts for simplifying PA +(h1 + h)xlx103 x g = P1 PB + h1 x 1 xio3 xg+hx.875 x103 x g=P1 P1 = Pi PA - F (111 +11)X1X103 X g = PB +hx.875x103 x g +hi x1x103 X g PA — PB = h x 103 x g(1 — 0.875) = 0.12 x103 x 9.81x 0.125 = 0.147 kPa

Pressure and Head

36. Compute the pressure difference between A & B in the figure. Sp gr = 0.92

1 30 cm

Guidance: The tubing system can be divided in two parts as shown below P1 P1 4 4 " - -f

30 cm

PA +hx1.15x103 xg =P1-F.20 x 0.92 x103 xg

P1 +(.50 — h) x1.15 x103 xg=PB

Now P1 = P1 PA +hx1.15x103 xg-0.2x0.92x103 xg =PB —(.5—h)x1.15x103 xg PA -PB =0.2x0.92x103 xg-0.5x1.15x103 xg =1.8 x103 — 5.64 x103 = —3.84 x103 kPa 3.84 x103 1x103 x9.81 = —391 mm of water

89

90

Fundamentals of Fluid Mechanics

37. The pressure at the point m in the figure is given below is to be measured by the open manometer as shown. Fluid A is oil (sp gr 0.80) and fluid B is mercury. Height y = 75 cm, z = 25 cm. (Punjab University)

Guidance: Divide the tubing in to two system as shown above: Pm =Pi -FYx0.8x103 Xg

li = Patm + Z x13.6 x103 x g P1 = Patni ± 0.25 x13.6 x103 x 9.81

=P1 + 0.75 x 0.8 x 103 x9.81 Now

P1 = P1 Pm — 0.75 x 0.8 x 103 X9.81 = PM

—

Patm

Parm + 0.25 x 13.6 x 103 X9.81

= "gauss

Pgause =

0.25 x 13.6 x 103 X9.81+ 0.75 x 0.85 x103 X9.81

= (33.35 + 6.25) x103 = 39.6 x103 Pa Pgause —

39.6 kPa

Panspite = Pgause + Patm

= 39.6 + 101 = 140.6 kPa 38. Show that PA — PB = 1+ aA (r2 — ri )gh for the manometer as shown below

(

Pressure and Head

9I

In case we neglect the move of fluid in the left limb, then PA — PB = (P2 — Pl)gh a (a = tube Reservoir type manometers are used for accurate measurement and the ratio of — A area, A = reservoir area) is assumed as zero and a correction is made to the scale which reads h on the vertical limb. Hence dh= a h A The correct expression for the pressure difference is PA — PB = (P2 — P1)(h+ dh)g = (P2 — P1)(11 + ih) i =(1± l(p2

A

- Pi )gh

... Correction factor = (1+ i) A

39. A single tube manometer has a well of cross-sectional area 200 times larger than the tube. The manometer liquid has sp gr = 1.6. On application of pressure on the well, the rise of manometeric liquid in the tube is 10 cm. Find the pressure of the gas. What is the error involved as the change of liquid level in the well is neglected? When dh is neglected, we get P = hpg = 0.1 x 1.6 x 10 3 x9.81 = 1.57 kPa When dh is considered, then c/ )hpg Pc = (1+ A

= (1+ 200) x 137 = (1+ 005) x1.57 = 1.578 Percentage error = PAP— P x100 = 1.578571.57 x100 1.

92

Fundamentals of Fluid Mechanics

0.008 x 100 1.57 = 0.51 % +P

p =1.6 40. A closed tank A is filled with a liquid (sp gravity = 2.25) and has a small air space at its top as shown. A manometer is connected between liquid and water tank (B). The pressure at water tank = 100 101/m2. Find the pressure of the air in tank A. Air space

60 cm M11•1111•111

t6

S = 2.25

can

90 Gm

4 cm x

S=1 eNilepfze.re.ez4 1,,,,

S = 13.6

Hg

Pressure at left side = Pressure at right side Pair

+1.5 x 2.25 x 103 x g +0.45 x13.6 x 1033 x g 103 10 _L

Xg 10 0 Pair = 100 — 33.1 —60.04+4.41 = 11.27 kPa iN 1 Pwater .45x

Pressure and Head

93

41. A hydraulic jack consists of a handle cum lever of 40 cm long and various dimensions of assembly as shown in the figure. A vehicle of 15 KN is supported by jack by exerting force P. Find P if distance between plunger and fulcrum of the lever is 3 cm.

4130 —IP

15 KN 60 roati

----- ----A = area of cross-section of larger piston x (0.06)2 = 0.0028 4 Pressure developed in the fluid by load = 15 kN is — P —

15 x 1000 = 5.36 x 106 N/m2 0.0028

a= area of plunger = = .000314 Force on the plunger = a x P =.000314 x 5.36 x 106 F =1684 N

x (0.02)2

94

Fundamentals of Fluid Mechanics

Taking moment w.r.t. point 'A' F x 30 = P x 400 F x 30 _ 1684 x 30 400 400 = 126.3 N

P—

42. Calculate the gauge pressure at point A, B, C and D in the water and oil storage arrangement as shown in the figure.

/ ," V" -..." ," ...,,,,-...,...." ..," ...., .....,.......W ..".." •."w...... *.r..."....,...",,,,,,,,,,, ..., ..."..","•.".."...",".." 4.".."..., •-••• se

5 ..... • •-,

Z::: Oil 'V. ZZV ..,,, ..."............","...."," se

...-....,...--,-...,-..., .--

Sp gr = 0.9

▪ ..e...",,,d, J..", • V' • ......."..........."7 : . ...,,,, ....",,,,,"• V' Ad•••-••• ,,l's ..."," ./ ....• .." . ....Vs," V' . 7 /

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

de

/

Guidance: The atmospheric pressure is acting on the level point 0. Any level above this level will have vacuum gauge pressure and below this level will have gauge pressure. At point A, pressure is — PA = — pgh At point A, h = — 80 cm = — 0.8 PA= — X 103 x 9.81 x 0.8 = —7.85 kN2 m At point B, h = + 70 cm PB = 0.7 x 1 x 103 x 9.81 = 6.867 kN/m2 At point C, Pc = PB as air column will already change the pressure. Now at point D, we have pgh + PD = 0.9X 10P3B X9.81X2.0 6.867 103 = 17.658+ 6.87 = 24328 m2 ki\T

Pressure and Head 95 43. What is the pressure of air in tank as shown in the figure if /1 = 40 cm, 12 = 120 cm and /3 = 90 cm?

Oil (sp gr=0

NNNNN MNN N N NN N NN

ve,

NNN

V".V,v.,

yr, yrk

v.. V,

NNN

•a" •••.••.',"

NNNNN

Guidance: The system can be divided in two parts as shown in the figure. Pair = i1XPxg -1-

Patin + t3PHgg =

0.4 x 0.8x103 x9.81+Pi

1o3

= Patm

=3.14+P1

0.9 x13.6 x103 x 9.81 103

= Patm + 120.07 From eqn (1) put the value of P1 Pair = 3.14 + 120.07 + P aim Pair — Patm = 12321 kPa Pgause = 123.23 kPa

44. In inclined manometer was used to find the pressure of water in a pipeline as shown in the figure. Find the absolute water pressure in the pipe in case the inclination of limb to the horizontal is 30°. The ratio of area of reservoir to the tube is 20. The sp gravity of measuring liquid is 1.6. Flow of volume from reservoir to tube remains constant dhxA=lxa

A= area of reservoir a= area of tube Ah = ±1 ( = 20 x 0.8 = 0.04 m A h= 1 sin° = 0.8x

2 Applying the condition of equilibrium on datum line x — x

96

Fundamentals of Fluid Mechanics

PA+ Pw (0.3+dh)x g = Pi x(h+ dh)x g PA +

lx103(0.3+ 0.04) x 9.81 _ 1.6 x 103(0.4 + 0.04) x 9.81 103 103 PA ± 0.34 x 9.81 = 1.60.44 x 9.81 PA ± 3.34 = 6.91 "A = 357 kPa(gause) "absolute — "atm + "A = 101.33 + 3.57 = 104.90 kPa

45. In a chemical reactor of height 80 m, find the pressure difference between top and the bottom if density varies with height (y) as p = 1000[1+ y + y 500 1000 )2] 1 Y dy

4_

+

P

---

4 1

80 At'

P+dP

2

I

////// /

1

Considering the equilibrium of an elementary vertical prism of height dy and cross-sectional area dA. During equilibrium PdA — (P + dp)dA = Weight of prism =dyxdAxpxg

Pressure and Head

97

)2

— dpdA = 1000 [1+ 2

or

—5 dp

dyxdAxg

500 ± (103100)

)2

= 1000 gf [1+

1

100 ± (103100)

clY 1 0

Pi — P2 = 1000 x 9.81 [y + Y2 + Y3 200 3 x 106 0 = 9810 x [80 + 6400 + 512 x103 1 200 3 x106 = 9810[80 + 32 + 0.171] = 1100.4 kPa 46. If the density of sea water at the surface is 1030 kg/m3 and the average bulk modulus is 2.4x109 N/m2, find the density of sea water at a depth of 1500 m.

--ps =1030 1500 mm-_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ---------------

P=?

Or

Bulk modulus k = dP dp P dp = dp or di) = p dP K K P

Now change of pressure dP = pgh (1030) x 9.81x 1500 2.4 x 109 _ 1.06x106 x 9.81x 1.5 x103 2.4 x 109 = 65 kg/m3 ... Density at 1500 = Ps + dP = 1030 + 6.5 = 136.5 kg/m3

98

Fundamentals of Fluid Mechanics

47. The barometric pressure at sea level is 76 mm of Hg, while that on a mountain top is 73.5 cm. If the density of air is assumed to be constant as 1.2 kg/m3, prove that the elevation of the mountain top is 284.4 m. (UPTU — 2004-05) P2 = 73.5

cm of Hg

= 76 cm of Hg

Given: P1 = 73.5 cm of Hg, P2 = 76 cm of Hg, p = 1.2 kg/m3 PI — P2 = Pairgh h= Pi — P2 Pg 0.76 x pHg x g — 0.735 x pHg x g Pairg (0.76 — 0.735) x13.6 x103 1.2 _ 0.025 x 13.6 x103 1.2 = 284.4 m 48. A U-tube manometer with small quantity of mercury is used to determine the static pressure difference between two locations A and B in a conical section through which an incompressible fluid flows. At a particular flow rate, the mercury column appears as shown in the figure. The density of mercury is 13,600 kg/m3 and g = 9.81 m/s2. Which of the following is correct?

Pressure and Head

99

(a) Flow direction is A to B and PA — PB = 20 kPa (b) Flow direction is B to A and PA — PB = 1.4 kPa (c) Flow direction is A to B and PB — PA = 20 kPa (d) Flow direction is B to A and PB — PA = 1.4 kPa (GATE — 2005) As per continuity equation— AA X VA = AB X VB where A = area and V = velocity AA > AB VB > VA As velocity head at '13' is more than at 'A', hence PA > PB and the flow takes place from `A' to '11'. Now PA — PB = P•g• AB = 13.6 x 103 x 9.81 x 150 x 10-3 = 20 kPa Option (a) is correct. As

49. The manometer shown in the given figure connects two pipes, carrying oil and water respectively. From the figure one. Oil

Water Horizontal plane

Horizontal plane Mercury

(a) can conclude that the pressure in the pipes are equal (b) can conclude that the pressure in the oil pipe is higher (c) can conclude that the pressure in the water pipe is higher (d) cannot conclude the pressure in the two pipes for want of sufficient data. (IES 1996) Now balancing the liquid columns and pressure in both tubes with respect to horizontal plane, we have— Pail po$•H = Pwater + Pwg1-1 or Poet — Pwater = (Pw po)•g•H As pa Pwater Option (b) is correct.

r

50. A differential manometer is used to measure the difference in pressure at point A & B in terms of sp weight of water (W). The sp gravities of liquids X, Y and Z are

I 00 Fundamentals of Fluid Mechanics respectively S1, S2 and S3. The correct difference (PA — Ps) is given by— WW

liquid y'

I

h3 liquid x' I

11

11jj

D

liquid 'z' (a) h3S2 — h1S1 + h2S3 (b) h1S1 + h2S3 —h3S2 (c) h3S1 — h1S2 + h2S3 (d) h1S1 + h2S2 — h3S3 From datum line CD, we have for both limbs— PA —

P AP B W W

PB =— =

h3S2

(IES 1997) h3S2 h2S3

—

h1S1

h3s3

Option (a) is correct. 51. In the figure shown, air is contained in the pipe and water in the manometer liquid. The press at 'A' is approximately

Patm 340 mm 200 mm

(a) 10.14 m of water absolute (b) 0.2 m of water (c) 1.2 m of water vacuum (d) 4901 Pa. (IES 1998)

Pressure and Head I 0 i

Take Patm = 10 m of water From datum CD, we have— PA = (340 — 200) x 10-3 m of water + Pan, = 0.14 m + 10 m = 10.14 m Option (a) is correct. 52. A U-tube manometer is connected to a pipeline conveying water as shown in the figure. The pressure head of water in the pipeline is—

.; Atmosphere

Benzene (Sp Gr = 0.88)

Water (Sp Gr = 1)

Mercury (Sp Gr = 13.6)

(a) 712 m (b) 6.56 m

(c) 6.0 m

(d) 5.12 m (IES 2000)

Taking datum CD, we have— P + 0.56 x 1 = 0.45 x 13.6 + 0.50 x 0.88 P = 6 m of water or Option (c) is correct. 53. Three immiscible liquids of specific densities of p, 2p and 3p are kept in a jar. The height of the liquids in the jar and at the piezometer fitted to the bottom of the jar are as shown in the figure. The ratio H/h is—

.!. 3h

p

#

1.5h i_

2p H

#

h

i

(a) 4

(b) 3.5

3p

(c) 3

(d) 2.5

(IES 2001)

I 02

Fundamentals of Fluid Mechanics

Taking base as datum, we have 3h x p + 1.5h x 2p + h x 3p = H x 3p or 3h + 3h + 3h = 3H or 3h = H H=3 h

or Option 'c' is correct.

54. To measure the pressure head of the fluid of specific gravity 'S' flowing through the pipeline, a simple micro-manometer containing a fluid of specific gravity `Si' is connected to it. The readings are as indicated in the diagram. The pressure head in the pipeline is—

T i S

ih

d

f

(a) h1S1 — hS — Ah(Si — S) (b) h1S1 — hS + Ah(Si — S) (c) hS — h1S1 — Ah(Si — S) (d) hS — h1S1 + 6,18(S1 — S) (IES 2003) Balancing both limbs, we have— Ps + (h + 6,11)S = (h 1 + All) S1 or Ps = hiSi — hS + Ah(Si — S) Option (b) is correct. 55. The balancing column shown in the diagram contains 3 liquids of different densities pi, P2 and p3. The liquid level of one limb is `121' below the top level and there is a difference of 'h' relative to that in the other limb. What will be the expression of 'h'?

Pressure and Head

103

P3

(a)

— 132 hi P1 — P3

(b) P2 — P3 hi P1 — P3

(c) P1 —P 3 h1 P2 — P3

(d) Pi — P2 hl P2 — 1-'3 (IES 2004)

Balancing the limbs— or

P hiPi = h x P2 ± h(192. — P3) = hi(Pi — P3)

or

h — hi(Pi — P3) (P2 — P3)

h)P3 P

Option 'c' is correct. 56. The reading of gauge 'A' shown in the figure

Air 0.25 m Oil Relative density of memory 13.6 Gauge A Relative density of oil 0.8

(b) — 1.962 kPa

(a) — 31.392 kPa As per manometer, PA

Option (b) is correct.

(d) + 19.62 kPa (IES 1999) 0.25 x 13.6 = — 3.4 m of water Pair = — P air + 4 x 0.8 = — 3.4 + 3.2 = — 0.2 m of water = — 0.2 x 9.81 x 103 = — 1.962 kPa. (c) 31.392 kPa

Chapter

3

HYDROSTATIC FORCES

KEYWORDS AND TOPICS A A A A A A A A

HYDROSTATIC FORCE TOTAL PRESSURE PRESSURE INTENSITY CENTRE OF PRESSURE FREE SURFACE IMAGINARY FREE SURFACE PIEZOMETRIC SURFACE HORIZONTAL SURFACE

A A A A A A A A

VERTICAL SURFACE INCLINED SURFACE CURVED SURFACE PRESSURE DIAGRAM STABILITY OF DAM SLUICE GATE LOCK GATE GRAVTIY & ARCHED DAM

INTRODUCTION A force is exerted by the fluid on a surface, either plane or curved, when a static mass of the fluid comes in contact with the surface. This fluid force always acts normal to the surface and it is known as total pressure. The point of application of total pressure on the surface is known as the centre of pressure. The intensity of total pressure increases with the depth of a point in the immersed surface. As the pressure intensity is non-uniform, the total pressure force will not act at the centre of gravity of the vertical surface but at other point below the centre of gravity. This point is called the centre of pressure. In case of horizontal surface, the intensity of pressure is uniform throughout the surface, resulting the centre of pressure moving up and merging with the centre of gravity. The total pressure and centre of pressure have to be determined while designing hydraulic structures such as dams, sluice gates and lock gates. 1. Define the terms 'Total Pressure' and 'Centre of Pressure'. (UPTU 2001-02) or

Hydrostatic Forces

I 05

Define the term of centre of pressure of plane area immersed in a fluid. What relation has it got with the centre of gravity of the area? Do the centre of pressure and centre of gravity ever coincide and if so under what conditions? (UPTU 2009-10) Total Pressure: When a fluid is at rest and there is no relative velocity or velocity gradient between the layers, then the fluid has no tangential shear force. It follows that the force exerted by stationary fluid has no component along the surface. The force exerted by the static fluid on the surface in contact with fluid is called 'total pressure'. It is always normal to the contact surface. The surface can be horizontal, vertical, inclined or curved. However water must be in contact with the surface.

Total Pressure Forces: Normal to surface Centre of Pressure: The point at which total pressure is supposed to act is called centre of pressure. The intensity of pressure increases with the depth of a point on the immersed surface. Had the pressure intensity been the same at every elementary surface area, the resultant pressure force would have passed through the centroid of the immersed area. It happens so for horizontal contact surface. In all other surfaces, the resultant of pressure force shall lie near the deeper end of the surface. The centre of pressure of the contact surface from free surface is given by:

where

— _ /Gsin2 19 h =x + Ai x = centre of gravity (CG) from free surface 1G= moment of inertia about C.G. A = Area 0 = Angle of inclination of the contact surface

P„,, Free surface

I 06

Fundamentals of Fluid Mechanics

2. What is total pressure acting on a plane horizontal surface?

-- EEEUEEVIVEEEEE: A plane horizontal surface, which is lying in the fluid of density, (p) has same depth of water over its entire surface area. Hence pressure acting on the entire surface is the same. Intensity of pressure = h x p x g Force acting on elementary area dA is dF = h pg x dA F=

dF = h pgdA

F= h pg dA F = hpgA 3. Find the total pressure acting on a rectangular plate of size 50 cm x 200 cm placed horizontally at the depth of 5 m below the free surface of water. For horizontal plate F = hpgA Area = A = 0.5 x 2 = 1 m2 p = 1 x 103

kg

3

F = 5 x 1x 103 x9.81 x1 =40.05 lcl•T 4. What is the total force or total pressure on a plane vertical surface submerged in liquid having area A and depth of CG = x from the free liquid surface?

. - - - -

••••••1,1•141,••

- -

/7/7/77"17/ Front View

Side View V ertical Surface

Hydrostatic Forces

I 07

Consider a vertical surface submerged in water as shown above. Take an elementary horizontal skip of thickness dr and width b at a depth of x from the surface. The intensity of pressure on this strip is: P = xpg where p = density Force dF = P x dA 5 dF= fxpgxbxdx 5 dF = pg 5 bxdx F = pgi x A 5. Find the total pressure on a rectangular surface of size 60 cm x 200 cm immersed vertically in water. The longer edge is vertical and upper edge is 5 m below the free surface. F = pgA x x

---------------------------------------------------------------------------------

2 i=5+— =6m Area (A) = 2 x .6 = 1.2 m2 F= 1 x 103 x 9.81 x 1.2 x 6 = 70.632 kN 6. What is the total pressure on a inclined surface immersed in liquid at angle 0 from the free surface? Show that total pressure is independent of angle of inclination of the contact surface.

Plate surface \ ,' _ 7•Z.Za-mZ...Z.N.7•7.••.7.. --...---... %_%..._.=_%..

dA = bx dx :-.4e

)1.'4 Total Pressure on the Inclined Surface

I 08

Fundamentals of Fluid Mechanics

Let an inclined surface making an angle 0 to the free surface be immersed in liquid as shown in the figure. Let an elementary strip of thickness dr and breadth b at depth of x is taken. The intensity of pressure acting normal at this strip is: P = p.x.g = py sin fag dF = P x dA = py sin Og x bdy as dA = bdy F = f dF = 5 py sin 0 gbdy F = p x sin Og i bydy but f bydy = yA F = pg sin 0 yA F = pg YA as Y/ sin 0 = X Hence total pressure is independent of angle of inclination of the contact surface. 7. What is the total pressure on a rectangular plate of size 60 cm x 200 cm which is immersed in the water at angle of 30° to the free surface? The shortest edge of the plate is horizontal and its top edge is at a depth of 5 m from the free surface.

F = pgiA i=5+

sin 30 = 5 + 0.5 = 5.5 m 2 A = 0.6 x 2 = 1.2 m2 F= 1.0 x 103 x 9.81 x 5.05 x 1.2 = 64.75 kN

8. What do you understand from the imaginary free surface or piezometric surface? The vertical depth of centroid of the immersed surface is taken from the free surface of the liquid. If the liquid surface is not a free surface and it has a pressure intensity (P1) acting on it, then imaginary free surface of the liquid can be assumed to be located at a height of

Hydrostatic Forces

Pl Pi x g surface.

I 09

from the actual liquid surface. This imaginary free surface is also called piezometric

Imaginary free surface

Air with pressure Pi

x —P

Actual free surface

Plate Plate

9. What do you understand from the centre of pressure? Find the expression for the depth of the centre of pressure for a vertical surface immersed in a liquid from its free surface. How is it related to the depth of CG of the vertical surface from the free surface? What happens if the contact surface is horizontal instead of vertical? The intensity of pressure increases with depth of a point on the immersed surface. Hence on a vertical contact surface, the pressure intensity continuously increases as we move down from the top edge to the bottom edge. As the pressure intensity is non-uniform, the total pressure force will not act at the centre of gravity of the vertical surface but at other point below the centre of gravity. This point is called the centre of pressure of the vertical surface.

Point A = CG Point B = Centre of pressure h>.X Increasing Pressure Intensity on Vertical Surface

/////////// Expression for Centre of

Pressure

0 Fundamentals of Fluid Mechanics

A vertical surface is immersed in liquid as shown above. Take an elementary strip of depth dx and width b at the depth of x from the free surface. Pressure intensity P = xpg dF = force on the strip = PdA = xpg b x dx = pgbxdx F = f dF = .1 pgbxdx = pgiA Moment of force dF acting on the strip with respect to the free surface dM = dF x x = p.g.b. x2dx Now sum of moment of all such pressure forces on the vertical contact surface with respect to the free surface can be calculated by integration f dM = $ p.g.b.x2 dx Area

Or

M= pg 5 x2.c/A

But 5 x2 . dA = moment of inertia with respect to the free surface = Ifs M = P•g• IFS

(i)

In case C is the centre of pressure of the vertical surface and its depth is h from the free surface, then the moment of the pressure force acting on the vertical surface with respect to free surface is: M=Fx h = pgiA x h from equation (i) and (ii) M = pgIfs = pg YA x h If, = iA h from the parallel axis theorem /f, = IG + A (x)2 where IG = moment of inertia from axis through CG iAh = IG ± A(X)2 IT, = I G + A(702 Ax Ax h=

/

It is obvious that h > i and h—= G Ai

I G +Y Ai

Hydrostatic Forces

The centre of pressure always lies below the centre of gravity by a distance of -± . In case Ax i is large when the vertical surface is immersed for a greater depth, then - tends to be Ax zero, resulting in the centre of pressure moving up and merging with the centre of gravity. In case the plate is made horizontal then the intensity of pressure is uniform throughout the surface of the plate. In this case, the centre of gravity and the centre of pressure is same i.e. h =X . 10. Find the relation between the centre of gravity and centre of pressure for an inclined surface immersed in a liquid.

Inclined Surface At any depth, the relation between x and y is x = y sine Consider an elementary strip of thickness dy and width b. The intensity of pressure acting on this strip due to depth x is P = xpg but x = y sin 0 =y sin 0 pg dF = PdA = y sine pgbdy as dA = dyb F= f dF = pg sin Of ybda = pg sin 0 YA = pg .iA

(i) (ii)

Moment of the pressure force (dF) acting on the contact surface with respect to axis 0 - 0 is: dM = ydF = pg sin 0 y2bdy M = .1 dM = pg sin Of y2bdy = pg sin 0 /00 In case depth of centre of pressure is by from the axis 0-0, then

II2

Fundamentals of Fluid Mechanics

(iv)

M=Fxhy = pg sine yithy

from eqn (iii) and (vi) M= pg sine /00 = pg sine yAhy Ioo = yAhy

using parallel axis theorem 40

= IG ± A(y)2

IG ± A(Y) 2 = yAhy

Now IG

or

by —

A

(x)2 —_ sin e

h

sine i

and y = .x sine xAx

Sin 0

_

h sine

/ sine e Ai

h = x+ G

11. Tabulate CG, area and moment of inertia of important plane surfaces. S. N. Plane Surface

1

Square

Area

a x= —

a2

a x = ,_ V2

a2

2

Moment of inertia about CG a

4

12

G .- — — —

x

14—

2.

CG from base

a

—*I

Square with diagonal

horizontal /

a

4

12

Hydrostatic Forces 1 13

3.

d 2

Rectangle

X= —

bd

3 bd 12

IG

d

x b

4.

Triangle with base

2

1 bd 2

x = — 3

1 bh 2

x =

h.b

3

48

vertical

5.

Triangle with base

3 bh 36

horizontal

d x =— 2

6.

Circle

7.

Semicircle

x

=

2d 37F

x d

= 0.21d

TLC!

2

ird

4

4

64

2 ird 8

4 ird 128

114

Fundamentals of Fluid Mechanics

8.

Trapezoidal

x_(

2a+b)h a+ b 3

a+b xh 2

(a2 + 4ab + b2) 2 h 36(a+ b)

I41-- al --►

f

fh c

14--

't

b ---19

12. Find the total pressure on a triangular surface immersed in water at an inclination of 30° to the free surface of water as shown below. The top edge is at a depth of 2 m from the free surface.

AG = 3x 2 =2m i = 2+2 sin30 =2+1=3 Area of surface in a triangular form = x 3x 2 2 _ 3 m2 Total pressure force = F = pgAi = 1 x 103 x 9.81 x 3 x 3 = 88.3 lcisl 13. A trapezoidal plate is immersed in a liquid of sp gravity = 0.9 at an inclination of 30° to the free surface of the liquid. The trapezoid has height = 2 m and sides of 2 and 3 m. The depth of top edge of the plate is at 2 m from the free surface. Find the hydrostatic force on the plate.

Hydrostatic Forces 1

15

The CG of the plate — h(2a + b. ) 3 a +b ) 2(2x 2+3) Y = 3 2 +3 2(4+3) Y= 3 5 ) 2 7 14 Y = X = 3 5 15 Area = a +b xh= 2 3 x 2 = 5 m2 2 2 Now

•i = 2 + 2(2 — 14 ) sin 30 15 6 = 2 +L x 15 2 = 2 + 8 = 233 15 Total pressure force = F = pgitt = 0.9 x 103 x 9.81 x 2.53 x5 = 111.69 IN

14. Does the centre of pressure varies with different liquids? The centre of pressure is given by: h=Y+

IG sin2 0 Ai

As h does not depend upon the sp gravity of the liquid, hence h remains the same for a particular surface for all liquids.

116

Fundamentals of Fluid Mechanics

15. A rectangular sluice gate is situated on the vertical wall of lock. The vertical side of sluice is d & breadth is b. The centroid of the sluice gate is at the depth of P meter from the free surface. Prove that the centre of pressure on the sluice gate is at the depth of: h = P + d2 12P

rd L

x

=P

/////////// The center of pressure h is h = . k + IG AY Hence x = P , area A = b x d and IG =

h=P+

= P+

bd3 12

3 bd 12 xbxd x P b2 12P

16. A rectangular gate is 2 m wide and 4 m high is located in a vertical plane in water and other side is air. Find total pressure and depth of centre of pressure from the free surface if (1) top edge of the gate concoid with free surface and (2) top edge is at depth of 4 m from the free surface.

Case-1

Case-2

Hydrostatic Forces 1

17

Case-1 F= X xAxpxg F=ixAxpxg Here

4 =2m,A=4x2=8m2 i=— 2 F= 2x8x1x103 x9.81 = 157 kN

Centre of pressure from free surface is: h=Y+ IG Ai I bd 3 2 x 43 -G 12 — 12 =10.7 h= 2+ 103 8x2 = 2 + 0.67 = 2.67 m Case-2 F= XxAxpxg i=4+2=6 F= 6x8x1x103 x9.81 = 471 kN (Note: Total pressure has increased with depth) h = .Tc+ IG Ax 6+ 103 = 8x6 = 6+0.22=6.22 m 17. Prove the distance between the centre of pressure (h) and centroid of a rectangular plate located vertically having water at one side and air at other side is — where d 6 is height of the plate and b is width of the plate.

-•G -•C 14- b—oi

ci—

1I8

Fundamentals of Fluid Mechanics

i —_ d I, = bd 3 2' 12

andA=bxd

h = i +1G AX =d+ bd3 2 bdx x12 2 d d = — +— 2 6 d 2d = x (3+1)=6 3 Distance = h_x=qd-1 d=d 3 2 6 18. Flow in a pipeline of 4 m in diameter is controlled by a circular gate. The sp gravity of the fluid is 0.9. The pressure at the centre of gate is 180 kN/m2. Find (1) total pressure force on the gate and (2) the depth of centre of pressure from the free surface.

-_-_-_-_-_-_-_-_-_-_-_-_-

f I

------------------------_ _-_-_-_-_-_-_-_-_-_-_-_-_ ------------------------

_v

_-_-_-_-_-_-_-_-_-_----------------

Gate

Pipeline

Side view

Pressure at Y = 180 kN/m2 P= xxpxg Or

180x103 = ix 0.9 x103 x 9.81 Y — 180 0.9 x 9.81 = 20.39 m Pressure force = ixpxgx A = 20.39x.9 x103 x 9.81x 7(4)2 4 = 2261 kN

Hydrostatic Forces

119

The depth of centre of pressure is: - _

h = x+ G

Ai

71-(d ) 4 64

7X

(4)4

4i =1256 m4 64 = 12.56 h = 20.39 + 1256 x 20.39 = 20.39 + 0.05 = 20.44 m G

19. A triangular plate is used as gate and it is located vertically to stop the flow of water as shown in the figure. Find (1) total pressure and (2) depth of centre of pressure.

///////7/// = 2+x3= 2+2 =4 m 3 1 A = — x6x3=9 m2

,

iG

bh3 6 x 33

36

h = +

—

36

= 4.5 m

A

4 + 4.5 9x4 4 + 0.125 4.125 F= xApg 4x9x1x103 x9.81 353.2 1c1•1 20. A square plate of 2 m side is used as gate to stop the flow of water as shown in the figure. Find (1) hydrostatic force and (2) depth of centre of pressure from the free surface.

I 20

or Or

Fundamentals of Fluid Mechanics

Side a = 2 m Diagonal = d (2±(d2)2 4 d) 2d2 = 16 d = -,./8 = 2.83 m i = 3+ 2.83 2 = 3 + 1.42 = 4.42 A = 2 x 2 = 4 m2,

,4 24 IG = '' = —= 1.33 12 12

h = . i + IG AY = 4+ 1.33 4x4.42 = 4.42 + 0.075 = 4.495 m F = 5Apg = 4.42x4x1x103 x9.81 = 173.44 kN 21. A square plate with side 4 m has a hole of 2 m diameter is located in water vertically as shown in the figure. Find (1) total pressure and (2) depth of centre of pressure.

a = 3, d = 2a sin45= 2 x3 =4.24m If Al = area of square and A2 = air of circle, then net area A is —

Hydrostatic Forces

A= Ai —A2 = 16

(2)2

4

12 I

— 16 — 3.14 = 12.86 m2

Net moment of inertia (I) I = Isquare — Icircle _ a4 12

64

81 it '12 64 = 6.75 — 0.79 = 5.96 = ?-' 4 =3+2.12=5.12m Total pressure F = = 5.12 x12.86 xlx103 x 9.81 = 646 KN The depth of centre of pressure is — h=

_

+ G A

= 5.12 +

12.86 x 5.12 = 5.12 + 0.09 = 5.21 m 22. A vertical square having area lmxlm submerged in water with layer edge 50 cm below the free surface. Find distance of horizontal line from upper edge of the square so that the forces on the upper and lower portions are equal on the square.

I 22

Fundamentals of Fluid Mechanics

k0.5m —+— , Y (1) •• 4,

,

portion Horizontal line Lower portion

Guidance: We can solve the problem considering two rectangular surfaces at depth of 0.5 m and (0.5 + y) which have same total pressure force depending upon their areas. F1 on area 1 = xiPAig = (0.5+1x1x103 x(yx1)x9.81 2 F2

on area 2 = x2pA2g = (05-Fy+ 1-2 1x1x103 K(1—y)xlix9.81

Now

F1 = F2

(0.W)X1X103 X(yX1)X9.81 = (0.5+ Y

2

lx1x103 x(1—y)9.81 2

‘,2

•Y 22 x (1— y)

0.5y + 2

y 2 +y = —y 2 —y+2 2y2 +2y-2 =0 y 2 + y-1 =0 Y

—1± V1+ 4 = 0.62 m 2

23. A circular opening of 4 m diameter in a vertical wall of a water tank is controlled by a circular disc of 4 m diameter and which can rotate about its horizontal diameter. Find (1) total pressure on disc (2) torque required to maintain the disc in vertical position when the free surface is 6 m from its centre. Thydrostatic B

Liplied

Hydrostatic Forces

I 23

Guidance: Total pressure force (F) acts below the CG at centre of pressure. The torque to stop the disc from rotating due to F is equal hydrostatic torque = F x (h F=.

13 >

=

4 = 739.3 KN

x lx 103 X 9.81

TC

,4

= 64

= 6+

7C

.4

64

= 47r

jr

= 6 + 0.167 = 6.167 m h —x = 6.167 — 6= 0.167 )xF Hydrostatic torque = (h = 0.167 x 739.3 = 123.5 Nm (anticlockwise) However torque is to be applied in clockwise direction = 123.5 Nm 24. A square door with side 1 m is provided in the side wall of a tank which is filled with water. What force must be applied at the lower end of the gate so as to keep the hinged door closed? The upper edge of the door is at 4 m from the free surface.

hinge 4 TT A

_ P= applied IF a=1J force = 4+ i =4.5 A=1x1=1 F=ixAxpxg =43x1x1x103 x9.81 = 44.12 kN 1 h = i + IG =43+ 12 lx 43 = 4.5 + 0.019 = 4.519

hinge "...,,, A .

h —4 —h P

I 24

Fundamentals of Fluid Mechanics

Now taking moment of forces acting the gate from hinged point (A) Fx(h-4) =Px 5 44.12(4519 —4) = 5P 44.12 x 0519 5 = 4.58 IcN _

P

25. A plate of size 4 m x 2 m is hinged at bottom and supported at midpoint by a prop so that it remains inclined in water at 60° to horizontal as shown in the figure as fully immersed. Find the compressive force in the prop if it is inclined at 45° horizontally. (Jadavpur University) 4-

I

_ 2_ sin 6D

1

'G ;C

2 \/4 2 2 F= i x Ax px g

x = — sin 60 = — = 0.87

= 0.87x(4x 2)xlx103 x9.81 = 68.3 lth 4 x 23 x (0.87)2 1G sin2 0 12 h = x+ — 0.87 + Ax 4x2x0.87 0.87 + 8 x 0.87 x 0.87 _ 0.87 + 0.29 3x8x0.87 = 1.15 Yi = distance from hinge to centre of pressure along the gate surface = 2 h sin 60 2 1.15x 2 = 13 = 2 — 1.32 = 0.68 2 Y2= i = 1

Hydrostatic Forces

I 25

Taking moment about hinge (B) F x yi = sinx y2 where F, = compressive force in the prop. 68.3 x 0.68 = Fc. x 0.71 x 1 F _ 68.3 x 0.68 0.71 = 65.4 kN

26. A flash board is rectangular of size 3 m x 1 m immersed in water and hinged as shown in the figure. Find the depth of water x and compressive force in the strut when water is about to trip the flash board. (IIT Kharagpur)

hingg

3 sint0

Guidance: The flash board will trip when total pressure force starts acting above the hinge. sin 60 is the condition for tripping. Hence h Suppose the height of water is x h = Centre of pressure = x + =

xx1 A= = 1.16x 2 sin 60 ,„

x

)3 = 0.13x3

x sine 60 1.16x x 05x = 0.5x +0.17 x = 0.67x Now for tripping of flash board. h = 05x+

h = x—lxsin60 0.67x = x — 0.87 0.33x = 0.87 x— 0.87 =2.6m F= ixAxpxg

0 Ax

I 26

Fundamentals of Fluid Mechanics

= 2x1.16xx1x103 xg = 2 x1.16x2.6x1x103 x9.81 2 = 38.46 IcN 27. A rectangular gate of size 6 x 2 is hinged at the base at inclination angle of 60° with the horizontal. The gate is kept in position by a weight of 6 ton by an arrangement as shown in the figure. Find the water level (x) when gate about to open. Neglect the weight of the gate. fil

—

h= x+

2

IG sine 0 Ai )3

x 2

2x

(sinx60

12 x 2 x

x (sin60)2 x xx sin60 2

2x 12 = 0.5x + 0.167 = 0.667x Force (P) acting on top edge to hold the gate = 05x+

P = 6 x 1000 x g= 58.86 KN h = x — 0.667x sin60 sin 60 = x— 0.767x = 0.233x

Y —x

F= 'x2x x xlx103 xg=0.233x 2 sin 60 During equilibrium, the moment from hinge is zero i.e. EM= 0 Fxy = Px6 11.28x2 x 0.767x = 58.86 x 6

Hydrostatic Forces

I 27

58.86 x 6 — 40.8 11.28 x 0.767 x = 3.44 m

x3 =

The gate will trip when height of water x 3.44 m . 28. A vertical rectangular gate 6 m high and 4 m wide has water to a depth of 5 m at upstream and 2 m at downstream. Calculate (1) total pressure exerted at upstream and downstream (2) the resultant total pressure and its location with respect to the bottom edge: 4

6

F1 = ii xAi xpxg = 5 x5.4x1x103x9.81 = 490.5 1(1•1 h=x+ —

5 2

I G Ax 4 x (5)3 5x 4 x

2

x12

= 25 + 5 6 = 3.33 oryl = 5 — 3.33 = 1.67 m F2 — X2 X A2 xpxg = 1x2x4x1x103 x9.81 = 78.48 IcN —

_

IG

h2 = X2 ± A _ ti2X2

= 1+

4x1 12 x 4 xlx1

I 28 Fundamentals of Fluid Mechanics = 1.083 or y2 = 2 — 1083 = 0.917 Resultant (R) = F1 — F2 R = 490.5 — 78.48 = 412.02 lcl•T Let resultant R acts at distance y from the base R x y = Fiyi — F2Y2 412.02 x y = 490.55 x 1.67 — 78.48x.917 Y—

819.1— 77.42 412.02

741.68 412.02 = 1.8 m Hence resultant force is 412.02 kN, which is acting towards downstream at 1.8 m from the bottom of the gate. 29. The pressure of air over a tank is 0.6 kg/cm2 (58.86 KN). There is an opening in the tank with flap of 1 m square with centre of square at 2 m from the free surface. The flap is hinged at the centre. Find the force P is to be applied vertically at bottom edge of the flap at point B to hold the water.

Flap -_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_ -_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_ - - - - - - - - - - - - - - - - ----------------------------------

Hinged at centre

Since air has pressure of 0.6 kgf/cm2 = 6 m of water, hence imaginary free surface is 6 m above the actual free surface. Force

Y=6+2=8m F= ii xAxpxg = 8x(1x1)x lx103 x g = 78.48 kN

For the square plate from free surface (actual)

Hydrostatic Forces

—

I 29

sin2 19 Al =2+ lx1xsin260 12x1x1x2 0 = 2 + '75 .24 = 2.03 _

h2 = x +

1G

Now Force (F) is balancing against the applied force (P) about the hinge. ... EM = 0 from the hinge. (h — i) — P x sin30 x 0.5 sin 60 (2.03 — 2) x 2 1 — P x — x 05 78.48 x 2 114 Fx

P — 78.48x2x0.03x2 0.5 x •%/i = 10.82 IN 30. A vertical dock gate of 10 m height is reinforced by three horizontal beams in such a way that each carries equal total pressure due to depth of water. Find the location of beams from free surface.

Guidance: The gate dock is divided in three portions (h1, h2 & h3) so that total pressure in each portion is equal (F1 = F2 = F3) and beams are located at centre of pressure given by hl, h2 & h3 F= iApg=5x1Oxbxpxg 2

h1xbxpx g where b= width F1 = — 2

1 30

Fundamentals of Fluid Mechanics

F3 —

F1 = F2 = F3 =

Now h12

2 Or Now put

(h32 — h221 2 bpg

bpg —

(h32 — h22 ) (h22 — h12 ) bpg = bpg =50bpg 2 2 3

h? = (h2 2 hi 2)_ (h3 2 h21_ 100 3 10, we get h1 h3 = = 5.77 m, h2 = 8.16 m h1 = h1 2

'G

A x i' 2

= 1" 1 2

1. x hi3

hi 12xhixlx2 = h1 +h1 6 2 2 = — 14 3 2 x5.77 = 3.85 m =— 3

The position of resultant force F2 can be found out by taking moment from the free surface (F1 +F2 )xxh2 = F1 x3hI +F2 xh2 F1 = F2 =

F

2 x 2 xh2 = F x2 hi + F xh 3 3 3 3

Hydrostatic Forces

I3I

4x8.16-2x5.77 3 _ 32.64-11.54 3 = 7.03

10 = h3

Now again taking moment from free surface 2 2 (Fi +F2 +F3)x 3x10 = (F1 +F2 )x 3xh2 +F3 xh3 F xhi = 3Fx2x10 2F x 2 x816 3 - 3 3 3 3 . Or

k _ 60-32.64 3 3 = 9.13 m

31. A circular plate of 4 m diameter is immersed in water in such a way that its greatest and least depth below the free surface are 4 m and 2 m respectively. Determine the total pressure on one face of the plate and position of the centre of pressure. (UPTU — 2002-03)

1 32

Fundamentals of Fluid Mechanics

The angle of inclination is 4— 1 2= El = 30° 4 2 x = 2 + 2 sin 30 = 3

sin 0=

A = ird 2 = g x42 4 4 ird 4

IG =

64

h=Y+

= 3+

I G sine 0 A+ 704 x sin2 30 ,r,./ 2 64x' x3 4

16x1 48 x 4 = 3.883 F = xApg = 3+

= 3x g 442 x 1 x103 x 9.81 = 369.64 kN 32. An angular gate ABC is designed to tip automatically when water rises above certain level in the tank. The gate is hinged at point B and AB is vertical while BC (= 1 m) is horizontal. Find the height h from the hinge (B) when water rises to tip the gate to allow the water to flow out.

I

1111111111111111111

Guidance: When level h increases, the force F2 on vertical plate increases. When F2 y1 F1x1, the gate will tip. Take width of gate = 1 m. Fl =

h X Ahorizontal x p < g

Hydrostatic Forces

I 33

= hx1x1x1x103 x9.81 = 9.81 h F2 = h x Avenicai xpxg 2 = xhx1x1x103 x9.81 2 = 4.9 h2 x1 = distance of CB from hinge = 1 = 0.5 2 = h—h=h— 3 h= 3 h X Xi = F2 X yi 9.81h x 0.5 = 4.905h2 x h 3 h2 = 3

Or

h= = 1.732 m The gate will automatically tip when water rises above 1.732 m from the hinge B. 33. A tank is filled under pressure (17 kPa) with an oil having sp gr = 0.87. It has been a cover of size 1.2 m x 1.2 m hinged at top edge & 30° to horizontal and held in position by a force P as shown in the figure. Find (1) force P and (2) reaction at the hinged point. Imaginary free surface 7 kPa A

1.2

Free surface

1.2

/ • -7t7-77•7-. Gauge pressure of the oil = 17 kPa 17 = hxSp grx g Or

h=

17 = 2 m of oil 0.87 x 9.81

X = 2+ 12 x sin 30

I34

Fundamentals of Fluid Mechanics

= 2 + 0.3 = 2.3 m F= ixAxpxg=2.3x1.2x1.2x0.87x9.811cN =28.27kN IG sine 0 Axx

h=

1.2 x (1.2) 3 x

4 12 x 1.2 x 1.2 x 2.3 = 2.3 + 0.013 = 2.313 m = 2.3+

Taking moment from the hinge (point A) PxAB=FxAC AC— h-2 = 0.313 =0.626 sin 30 2 P x 1.2 = 28.27 x 0.626 28.27 x 0.626 1.2 = 14.75 kN The reaction at point A is R which is — P

R=F—P = 28.27 —14.75 =13.52 kN 34. A tank with square vertical side of 2 m and depth of 1.5 m contain water upto a height of 0.6 m and oil of sp gr 0.8 upto 0.9 m as shown in the figure find (1) total pressure on the vertical side and (2) centre of pressure from the free surface. A Pressure distribution

Oil ,,--,--".,...W ...W . W and submarine will float. The submarine will submerge if weight of submarine is increased to its buoyancy force (FB). Hence if V m3 of water is to be pumped to submerge the submarine, then we have W + pumped water = FB 9x 103+

VXpXg — 11.772 x 103 103

Buoyancy and Floatation

V xlx 103 x 9.81 103 or

203

— 2.772 x 103

V=

2.772 x 103 — 282.57 m3 9.81

34. A rectangular pontoon is 8 m long and 4 m wide. The weight of the pontoon is 400 kN. Find the position of centre of gravity above the base of the pontoon such that it does not overturn in still water. Guidance:

For stability, metacentre M should be above G or M should concide with G.

Weight of pontoon = Weight of water displaced 400 = Volume of water displaced x p x g Vxlx 103 x 9.81 103 or

V=

400 =40.77 m3 9.81

V = length x width & height of pontoon inside the water 40.77 = 8 x 4 xh or

h _ 40.77 _ 1.27 m 2 4

(t)

G

6B Pontoon in Water

Distance of the centre of buoyancy B from the base is OB — h _ L27 _ 0.635 m 2 For stability of the pontoon BM =

where I = moment of inertia about longitudinal axis of the pontoon & Vs is Vs submerged volume.

1-

length x (width)3 12

204

Fundamentals of Fluid Mechanics

8 x 43 — 42.67 m3 12 42.67 BM= — 1.05 m 8x4 x1.27 Now OM= OB + BM = 0.635 + 1.05 = 1.685 m For stability, CG must lie below the metacentre or at least at same level. Therefore the maximum height of the centre of gravity from the base is OG = OM= 1.685 m 35. Find out the sp. gr. of a log of 3 m diameter and 5 m length if it has depth of 2 m inside the water.

D ////////////////////////////////// LOG IN WATER

OC = 2

—

1.5 = 0.5

cos 0 — OC — 05 — 0.333 1.5 OB 0 = 70.5° Wetted area —

2 360 - 20 x 7rr + 2 AAOC 360

(Note: hatched portion

is wetted area of log) 360 — 20 x 360

R-

x 1.52 + 2 x 6 x 0.5 x 1.5 sin 0

— 20 = 7.065 x 3606 + 0.75 sin 0 3 0 = 7.065 x 0.608 + 0.75 x 0.629 = 4.3 + 0.47 = 4.77 m2 Sp. gravity — Volume of displaced water Volume of log wetted area x length irr 2 x length

Buoyancy and Floatation

205

= 4.77 7r x1.52 = 0.675 36. A ship of weight 15 MN has horizontal cross section of water line as shown in the figure. If it has its centre of gravity is at a depth 70 cm from the free surface, find its metacentre height for rolling (about x-x axis) for pitching (about y-y axis). The ship has centre of buoyancy at a depth of 320 cm below the free surface. Depth of centre of gravity = 0.7 m Depth of centre of buoyancy = 3.2 m BG = 3.2 — 0.7 = 2.5 m Volume displaced — Weight pxg 15 x 106 103 x 9.81 = 1529 m3 Y 0.7

0G 0B

1

x-

------- —+— : 3.2

x 20 - - - - - - - - - - - - 1- - - - - - - - - - - -

1 40 Y

Water line of the ship

Centre of gravity & buoyancy

Metacentric height is given by GM=

— v

—BG

/„ — 40 x 123 + 4 x 1 x 8 x 63 12 12 = 15760 + 576 = 6336 /"'

— 1 x 12 x 403 + 2 x 1 x 12 x 83 + 96 x (22.67)2 12 36

206

Fundamentals of Fluid Mechanics

= 64000 + 341.33 + 49337.2 = 113678.53 m4 For rolling: GM —

Ixx BG V

6336 — 2.5 1529 = 4.14 — 2.5 = 1.64 m For pitching GM = IYY V

- BG

113678.52 — 2.5 1529 = 74.35 — 2.5 = 71.84 m 37. A rectangular pontoon is 5 m long, 3 m wide and 1.2 m high. The depth of immersion of the pontoon is 0.8 m in sea water. If the centre of gravity is 0.6 m above the bottom of the pontoon, determine the metacentric height. The density of sea water is 1023 (Delhi University 1992) kg/m3. Guidance:

The centre of buoyancy is at the half height of immersion. The center of gravity

I is given. Metacentric height can be found out by the formula GM = — — BG. V From the figure, the height of CG and center of buoyancy is OG = 0.6 m (given) OB — 08 _ 0.4m Y

I x

0.8

5

x

oG oB 0 3

-0.-

3

y

Buoyancy and Floatation

207

Immersion & Water line

BG = 0.6 — 0.4 = 0.2 m I = 12 x 3 x 53 YY

= 11.25 m2 V = Volume of water displaced = length x breadth x height of body immersed = 5 x 3 x 0.8 = 12 m3 GM= VYY —BG 12.25 0.2 12 = 0.74 m 38. A cylindrical buoy has 2 m diameter and 2.5 m length. It weighs 2.2 x 103 kg. If density of sea water is 1025 kg/m3, show that the buoy cannot float with its axis vertical. Guidance: Depth of immersion can be found out from the displaced water. The centre of buoyancy is half of the depth of immersion. The height of centre of gravity will be the half the height of the buoy. BG can be found from these two. GM can be calculated. If GM is negative the buoy is instable in floatation with the axis vertical. Weight of buoy = mg = 2.2 x 103 x 9.81 = 22.56 lcI•I If h is the height immersed when the buoy is floating with its axis vertical Volume displaced = ' x h 4 x 22 xh 4 y

x- -

///////////1177777///-/////// Plan of water surface at water line.

- -x

208

Fundamentals of Fluid Mechanics

= gh Weight of the buoy = Weight of water displaced 22.56 =7-chxpxg Irx1025x 9.81 xh 1000 Or

h=

22.56 x 103 — 0.714 m g x1025x 9.81

OB = height of centre of buoyancy _ h _ 0.714 2 2 = 0.357 m OG = height of centre of gravity

.

= 25 = 1.25 m 2 BG = 1.25 — 0.357 = 0.893 m 1=

a4 _ Ex24

64 64 0.785 m4 Volume of displaced water V = rch = TC x 0.714 = 2.24 m3 I — BG GM = V — = 035 0.893 2.24 = 0.335 — 0.893 = —0.51 m Since GM is negative, the buoy is unstable in flotation in vertical axis. 39. A hollow cylinder with ends open has internal and external diameter of 30 cm and 50 cm. The length of the cylinder is 100 cm and it weighs 700 N. Find whether the cylinder is stable in flotation in vertical axis.

Plan of water surface at water line.

Buoyancy and Floatation

209

Area of cross-section of cylinder = 7-1-(r(2, — r?) = Tc(0.252 — 0.152) = 0.126 m2 Weight of displaced water = A x h x p x g A = area & h = height immersed Weight of cylinder = Weight of water displaced 700 = 0.126 x h x 1000 x 9.81 h—

700 — 0.566 m 0.126 x 1000 x 9.81

I—

7rcg 64

7rcri 64

= L (0.54 - 034) = L (0.0625 — 0.0081) = 2.67 x 10-3 m4 V = 0.126 x 0.566 = 7.13 x 10-2 m2 OB — h — 0566 2 2 = 0.283 m 1 OG = — = 0.5 2 BG = 0.5 — 0.283 = 0.217 m I GM= — — BG V 2.67 x10-3 , 7.13 x 10-2

0.217

= 0.037 — 0.217 = —0.18 m Since GM is negative, hence the cylinder is unstable in flotation in vertical axis. 40. A rectangular pontoon is 10 m long, 8 m wide and 4 m high. It weighs 800 kN. It carries on its upper deck an empty boiler 4 m in diameter which weighs 400 kN. If the centre of gravity of the pontoon and boiler are at their geometric centre on the normal axis, find the stability of the pontoon.

2I0

Fundamentals of Fluid Mechanics

Guidance: The CG of the system can be found out by Wx OG = W1 CG1 + W2 0G2 where CG, 0G1 and 0G2 are distance from the reference point. The centre of buoyancy can be found out from the depth of immersion.

8

1-4— 8 ////////////////////// Plan of water surface at water line.

W x OG = W1 OGi + W2

0G2

(800 + 400) x OG = 800 x (24 ) + 400(4 + — 42 ) 1200 x OG = 1600 + 2000 OG — 3600 — 3 m 1200 Weight of (pontoon + boiler) = Weight of water displaced 1200 = 8 x 10 x h x 9.81 x 1000 1000 h—

1200 — 1.53 m 8 x10 x 9.81

OB — h _ 1.53 _ 0.765 m 2 OG = 3 BG = OG — OB = 3 — 0.765 = 2.235 m 10 x 83 — 466.67 m4 12 V = 10 x 8 x 1.53 = 122.4 m2 =

GM= — — BG V 466.67 2.235 122.4 = 3.813 — 2.235 = 1.577 m Since metacentric height is positive, the pontoon is stable in its floatation in vertical axis.

Buoyancy and Floatation 2 I 1 41. A ship weighs 2 x 107 kg and its water line cross-section is as shown in the figure. The centre of buoyancy is 2 m below the surface. Find the maximum permissible height of the CG for static stability. Guidance: The stability has to be checked for rolling and pitching i.e. about x-x and y-y axis. However, instability is more critical about x-x axis as I, < In,.

Water line cross-section of ship.

I, = I, of ABCD + IL, CBF + IL, ADE 60 x 203 + 2 x 10 x 103 + 1 HX104 12 12 2 64 456 x 102 m4 Weight of the ship = Weight of the water 2x 107 xg= Vxpxg .

Now

V—

BM — —

2x107 103 2 x 104 m4

Ixx V 456 x 102 ,, — 2.28 m 2 x 10'

Since buoyancy is located at depth of 2 m from the free surface, it means that the metacentre is located at 0.28 m (2.28 — 2 = 0.28 m) above the free surface. For stability, metacentre has to be either above the CG or at least at same height. Hence maximum possible height of CG is 0.28 m above the free surface. 42. A cone having base radius R and height H floats in water with its vertex downwards. 1 [ R 2 s '-'3 14 Prove for stable equilibrium (1) sect 60> L I and (2) H < 1/ 3 where h is depth h 1— S

2I2

Fundamentals of Fluid Mechanics

of immersion, 0 is semi vertex cone angle and S is sp. gr. of the cone material. (UPTU 2006-7, 2007-8) 3 The CG of a cone is located on normal axis at a height of — th. 4 OG = H 4

x

--

0

Cross-sectional area at water line. Similarly if h is the height immersed, then the centre of buoyancy is at the CG of the water displaced OB = h 4 Now

BM =

v 2

=

if

4

V = —1 xr2 h 3

Now

BM= irr4 1 trr 2h 4x— 3 r = h tan 61 BM —

xh

3 4

r2 h

2 tan2 0

3 xh =— tan2 0 4 OM = OB + BM = —3 h + —3 h tan2 El 4

Buoyancy and Floatation

4

2I3

h (1 + tan2 0)

=

h sec2 4 For stability, metacentre must be above CG OM > OG 4

h sec2 0 > sect 0 >

or

4

H (1)

h

Weight of cone = Weight of displaced water

= 1- n-r2h pi,

2

1n-R H x S x

Also

R = H tan & r = h tan 0 h = H 5113

From eqns. (1) & (2) H sec2 B > h 1 S1" 1 + tan2 0 > S1 1/3 Or

tan2 B >

1—s1/3 SU3

But

Or

R tan 0 = H R2

1- S1"

H2

Sin

H2

S1/3

2 < R1—S13 [R2 s1/3 1/2

Or

H
0143 • H2 — 0.0875 • 1.634 or Or

D H

H D

• 1.278 • 0.782

57. A wooden cylinder sp. gr. = S of circular cross-section having length = 1 and diameter = d floats in water. Find the maximum possible lid ratio so that the cylinder may float in stable equilibrium with its axis vertical.

Buoyancy and Floatation

229

Weight of the cylinder = Weight of water displaced rd 2

4

2

x/xSx 103 xg — — xhx 1 x103 xg 4 h = SI

64 ,r,./ 2

V

4

xh= 71-d 2 xSxl 4

OG = — 2 OB = h — = S1 2 2 1 (1 — S) BG = 2 irtex 4 BM—I;— V 64 x ncl 2 x SI

d2

1651

OM = OB + BM = S1 + d 2 2 16S/ For stable equilibrium, metacentre must be at higher distance from 0 as compared to OG OM > OG SI + d 2 > 1 2 16 S1 2 or


11112S(1— S) where S is sp. gr. of cylinder material. Further also show that if R>

V2

the cylinder is stable for all values of S.

230

Fundamentals of Fluid Mechanics

H

G B

t h

I It has been already shown that h = SH H OG = — 2 S OB — h H 2 2 I (2R)2 BM — ' — 16SH V OM > OG for stability H SH + (2R)2 > 2 16SH 2 (2R)2

16SH

> 11(1 — S) 2

R2 > 2 SH2 (1 — S) R > HV2S(1— S) h For any values of S, put S = — H R2 > H2(2h 1 H) H H R2 > 2hH — 2h2 Now differentiating w.r.t. h to get maximum value of h. d (2hH — 2h2) = 0 dh 2 H— 4 h =0 or Now Put

H 2 R2 > 2 hH — 2 h2 h = HI2 h=

Buoyancy and Floatation

23 I

2 R2 > H2 >H or

R>

59. A cylindrical buoy is 2 m in diameter and 2.5 m long and it weighs 23 x 103 N. Show that the buoy will not float vertically. Given sp. wt. of sea water = 10 kN/m2. Find the weight to be attached with chain to its base to make it float vertically.

Without Weight

With Weight

W= 23 x 103 N Displaced volume of sea water=

w

23 x 103

Pxg

10x103

= 2.3 m3 V h = height of immersion —ird2

2.3 — 0.732 m

4 OB — = 0 732 — 0.366 m 2 "4

BM— 1' = — 0.341 m 2.3 V OM = OB + BM = 0.366 + 0.341 = 0.707 m OG = 25 2

=1.25

As OG > OM, hence the buoy is not stable. To make the buoy stable, we attach a additional weight W, at bottom of the cylinder

232

Fundamentals of Fluid Mechanics

Now

WT= W + Wa +W Displaced water = W a — 0.1 WT kN 10 h=

0.1WT — 0.0318 WT 2 — it x 2 4

0.0318 x WT OB' — h _ 2 2 = 0.0159 WT 23x1.25+Wa x0 WT

28.75 WT

71-(2)4 B'M' = I' = 64 x 0.1WT V OM' = OB' + B'M'

7.85 WT

OG' =

= 0.0159 WT + 7.85 WT OM' OG' for stability 0.0159 WT + 7.85 > 28.75 WT — WT WT > 20.9 T 0.0159 1314.46 WT 36.25 lth Wa = 36.25 — 23 = 13.25 kN 60. A rectangular floating body 20 m long is 5 m wide. The water line is 1.5 m above the bottom. If the centre gravity is 1.8 m from the bottom, then its metacentric height will be approximately: (a) 3.3 m

(b) 165 m

(c) 0.34 m

(d) 0.30 m (Civil Services 94)

Buoyancy and Floatation

233

1 x20x54 1 BM= 11 '— 2 — 1.4 m 20x5x1.5 V

l5 ' = 0.75 2 OB = 0.75 BG = BG — OB = 1.8 — 0.75 = 1.05 m Metacentric height MG = BM — BG = 1.4 — 1.05 = 0.35 m h=

Answer (a) is correct. 61. A wooden plank (sp. gr. = 0.5) 1 m x 1 m x 0.5 m floats in water with 1.5 kN load on it with 1 m x 1 m surface horizontal. The depth of plank lying below water surface will be — (a) 0.178 m

(b) 0.250 m

(c) 0.403 m

(d) 0.500 m (Civil Services 95)

Weight of wooden plank and load = Weight of water displaced lx 1 x 0.5 x0.5 x 103 xg+ 1.5 x 103 =1 x 1 xh103 xg h = 0.5 x 0.5 + 13 9.81 = 0.25 + 0.153 = 0.403 m Answer (c) is correct 62. A symmetrical right circular base of wood floats in fresh water with axis vertical and apex down most. The axial height of the centre is L. The submerged portion has a height h, measured upwards from the apex. What would be the height of the centre of buoyancy from the apex? (b) ; h

211

(c) — 3

(0

I (IES 98)

234

Fundamentals of Fluid Mechanics

0

The centre of buoyancy is centroid of the displaced water. The displaced water is cone OAC and the centriod of the displaced water is 3 OB = — x h

4

Hence answer (d) is correct. 63. A body weighs 30 N and 15 N when weighed under submerged conditions in liquids of relative densities 0.8 and 1.2 respectively. What is the volume of the body? (a) 12.50 litres

(b) 3.82 litres

(c) 18.70 litres

(d) 75.50 litres (IES 2008) As per Archimedes' principle, the loss of weight of body is equal to the weight of liquid. Hence, W - V • pi • g = 30 (i) Also W - V • p2 • g = 15 (ii)

W - Vpig _ 2 or

W - Vp2g W - V x 0.8 x 9.81 x 103 = 2(W - V x 1.2 x 9.81 x103)

or W = 15680 V Putting the value of W in equation (i) 15680 V - V x 800 x 9.81 = 30 or V = 3.82 x 10-3 m3 = 3.82 litres Option 'b' is correct. 64. A wooden rectangular block of length of '1' is made to float in water with its axis vertical. The centre of gravity of the floating body is 0.15 x 1 above the centre of buoyancy. What is the sp gravity of the wooden block? (a) 0.6

(b) 0.65

(c) 0.7

(d) 0.75 (IES 2007)

Buoyancy and Floatation

235

Centre of buoyance (B) from bottom of the block = 0.5/ — 0.15/ = 0.35/ Hence, submerged block length is— = 2 x 0.35/ = 0.7 x I Hence, weight of water displaced is— = 0.7 x/xAxpw xg The weight of the block is— =IxAxpb xg From the law of floatation, we have— /xAxpb xg= 0.7 x/xAxpw xg or Pb/Pw = 0.7 Option 'c' is correct. 65. A cylindrical body of cross-sectional area A, height 'H' and density 'Ps' is immersed to a depth 'h' in a liquid of density `po and tied to the bottom with a string. The tension in the string is— (a) p . g . h . A

(b) (Ps — MghA

(c) (p — ps)ghA

(d) (ph — Ps H)gA (GATE 2003)

T h

String

Forces acting down = Force acting up T+ m • g = Buoyancy force where m = mass T+(ps xHxA)xg= Volume immersed xpxg =(Axh)xpxg T = gA(p • h — ps • H) Option (d) is correct.

236

Fundamentals of Fluid Mechanics

66. A hydrometer weighs 0.03 N and has a stem at the upper end which is cylindrical and 3 mm in diameter. It will float deeper in oil of sp gravity 0.75, than in alcohol of sp gravity 0.8 by how much amount? (a) 10.7 mm

(b) 43.3 mm

(c) 33 mm

(d) 36 mm (IES 2007)

We have when hydrometer floats in oil0.75 xpxgxAxxi = 0.03 where x1 = depth of float 0.75 x 103 x 9.81 x x (3 x 10-3)2 x x1 = 0.03 4 or xi = 0.577 m When hydrometer floats in alcohol, then we have0.8 x 103 x 9.81 x x (3 x 10-3)2 x x2 = 0.03 where x2 = depth of float 4 or x2 = 0.541 m Now x1 — x2 = 0.577 — 0.541 = 0.036 = 36 mm Option `d' is correct. 67. A barge 30 m long and 10 m wide has a draft of 3 m when floating with its sides in vertical position. If its centre of gravity is 2.5 m above the bottom, the nearest value of metacentric height is— (a) 3.28 m

BM =

(b) 2.78 m

I

displacement of water

(c) 1.78 m

/ x b3 12 30x10x3

30 x 103 12 30x10x3

(d) Zero (IES 2001)

where I = moment of inertia

= 2.778 m 3 AB = — = 1.5 m

2

AM = AB + BM = 1.5 + 2.778 = 4.278 m GM = AM — AG = 4.278 — 2.5

Buoyancy and Floatation

237

1.78 m Option 'c' is correct. 68. A solid cylinder of 2 m diameter of height 2 m is floating in water with its axis vertical. If the sp gravity of the cylinder is 0.65, find its metacentric height and state whether the equilibrium is stable or unstable. (UPTU 2008-9)

B = centre of buoyancy G = centre of gravity

Depth of cylinder in water is— = Sp gravity x height = 0.65 x 2 = 1.3 m The height of centre of buoyancy from bottom point 'A' is1.3 AB = = 0.65 m 2 Distance of centre of gravity from 'A' is— height 2 — —1 m AG — 2 2 BG = AG — AB = 1 — 0.65 = 0.35 m I Metacentric height GM = — — BG where I = Moment of inertia and V = immersed V volume n-c14 64 0.35 7rd 2 x 1.3 4 4 0.35 = 0.192 — 0.35 16 x 1.3 = — 0.158 Negative sign indicates that metacentre is below the centre of gravity. Hence, the cylinder has unstable equilibrium.

Chapter

5

FLUID MASSES SUBJECTED TO ACCELERATION

KEYWORDS AND TOPICS A A A A A A A

STATIC BODY RELATIVE EQUILIBRIUM TRNASLATION ACCELERATION HORIZONTAL ACCELERATION VERTICAL ACCELERATION INCLINED FREE SURFACE CONSTANT PRESSURE LINE

A A A A A A A

D'ALEMBERT'S PRINCIPLE ROTATIONAL ACCELERATION INCLINED ACCELERATION PARABOLOID FREE SURFACE FORCED VORTEX VERTEX AND RISE OF LIQUID IN VORTEX ROTATION OF CLOSED VESSEL

INTRODUCTION When a container containing a fluid is made to move with a constant acceleration, then the liquid particles initially will move relative to each other and after some time there will not be any relative motion between the liquid particles and the walls of the container. The liquid will take up a new position under the effect of acceleration imparted to the container. A container containing a fluid may be subjected to (i) translatory acceleration in horizontal or vertical direction and (ii) rotational motion at constant acceleration. d'Alembert's principle helps in changing the dynamic equilibrium of fluid mass subjected to acceleration into static equilibrium. The conditions for the relative equilibrium of a liquid mass subjected to acceleration are (i) no shear stress in liquid (ii) no motion between fluid particles and (iii) no motion between the fluid and its container. 1. What do you understand by a static body? A static body is a body which does not move. In other words, the body is at rest. 2.

Is anybody at rest in universe?

Fluid Masses Subjected to Acceleration

239

In universe, none of the body is at rest or in static state as universe is moving and everybody in it is moving. 3. How do we say that a body is at rest or in static state? A body on earth is said to be at rest or in static state when it does not change its position with respect to other bodies existing on the earth. It can be appreciated that the body at rest or in static condition is in fact is moving with the same velocity as that of the earth. Therefore we are considering the body to be at rest or in static state relative to the movement of the earth. 4. What do you understand by relative or dynamic equilibrium in moving fluid mass? If the fluid particles in the moving fluid mass do not move relative to each other, then they are said to be in static position and in this situation dynamic or relative equilibrium exists between the fluid particles under the action of accelerating force. If a container containing a liquid is made to move with a constant acceleration, the liquid particles initially will move relative to each other and after some time there will not be any relative motion between the liquid particles and the walls of the container. The liquid will take up a new position under the effect of acceleration imparted to the container. The entire liquid mass moves as single unit. The liquid attains static equilibrium in new position relative to the container and law of hydrostatic can be applied to find out liquid pressure. 5. How can fluid be subjected to horizontal translation acceleration? The fluid can be subjected to translation acceleration without relative motion between fluid particles by moving the vessel containing the fluid with horizontal acceleration. A fluid in vessel transported by a vehicle is subjected to horizontal translation acceleration as shown in the figure. Fluid is sloped

Acceleration =a

h2

//////// /////////////// //////// Free surface on horizontal acceleration from right to left

Free body diagram of liquid mass

The free body diagram of entire mass of liquid is as shown in the figure. If we consider the equilibrium of entire mass of liquid, the vertical component of hydrostatical pressure force at rear end is F1 and at front end is F2. Since the level of liquid at rear end is more than front end, hence F1 > F2. For equilibrium, EPx = 0. and

F1 — F2 = 171 • a, where m = mass a = acceleration

242

Fundamentals of Fluid Mechanics

Also

F1 = ii Ai pg and F2 = X2A2Pg

Here Al and A2 wetted areas at rear and front walls of the tank. ii and x2 are centriods at wetted areas at rear and front walls of the tank. Slope of free surface of the liquid Consider a liquid particle A on the free surface. As there is no relative motion between liquid particles, no shear force is acting on any layer. Hence pressure force F exerted by the surrounding liquid on point A is normal to the free surface as shown in figure. If we resolve this force (F) along vertical and horizontal direction and apply the conditions of equilibrium i.e. EPx = 0 and EPy = 0. Free surface ..

m.a Free surface

EPx = 0, F sin 0 = ma where 0 is the angle of slope of free surface from horizontal. EPy = 0, F cos 0= mg F sin 0 = ma F cos 0 mg tan 0 =

g

13. What are the lines of constant pressure in liquid mass when subjected to constant horizontal acceleration?

Lines 1 — 1, 2 — 2, 3 — 3, & 4 — 4 are pressure lines P1 < P2 < P3 < P4 ---

Constant pressure lines

When a liquid mass is given a constant horizontal acceleration, the slope of all points on the free surface of the liquid remains constant. Hence the free surface of the liquid is inclined at a constant angle 0 to the horizontal. The lines parallel to the free surface are constant

Fluid Masses Subjected to Acceleration

243

pressure lines as all points on each line have same pressure. However the intensity of pressure will increase on lines as they distant away from the free surface of the liquid. 14. What happens when a liquid is subjected to constant vertical acceleration? When a liquid is subjected to constant vertical acceleration, the free surface of the liquid remains horizontal as it was earlier without subjecting the liquid to the acceleration. However the pressure intensity now at any point in the liquid is greater than the static pressure (pgh) by an amount given by pgh (i c where a is acceleration in vertical direction. g Height

h

h mxa

Ir pgh Pgh(1 + Static liquid & static pressure

—.I

Liquid subjected to vertical acceleration

15. Find the pressure variation in a liquid when it is accelerated upwards and downwards with acceleration of a. m.a 4

h h mg

to

1F=13, dA Liquid Subjected to Vertical Acceleration.

Upward Acceleration Consider a small column of liquid having height h as shown in figure. As there is no relative motion of fluid particle, the column of the liquid is in stable equilibrium. If we apply the conditions of equilibrium, then EPx = 0, and EPy = 0 EPy, = 0, F — mg = ma PdA — mg = ma Now m = pdAh P•dA — p•dAh•g = pdAha

244

Fundamentals of Fluid Mechanics

P — pgh = pah P = pgh (1+1 Downward Acceleration

4 h

ma

T h mg 1

,

1

Forces acting on downward acceleration on the column of the liquid are as shown in the figure. mg — F = ma m=pxhxdA F = P x dA phdAg — PdA = phdAa P= phg (1-1 Hence during downward acceleration, the pressure at any point in liquid is less than static pressure (phg) by an amount equal to a 16. What will happen if a liquid is subjected downward acceleration equal to the coefficient of acceleration (g)? (UPTU 2009-10) The pressure at any point with downward acceleration is P = pgh(1-1 Now a = g, then P = 0 Hence the pressure intensity in the liquid does not vary and it is same at every point of the liquid. The pressure is same as atmospheric pressure. 17. What happens when a liquid is subjected to constant acceleration while moving up along an inclined plane? Find the angle of slope of the free surface of the liquid. (UPTU 2009-10) When a liquid container on a transport is standing on a inclined plane, the free surface of the liquid remains horizontal in spite of the normal axis of the liquid sloping with angle (inclination angle) with the vertical.

Fluid Masses Subjected to Acceleration

245

Vertical Normal axis Liquid surface horizontal

Inclined plane or road

Liquid at rest on inclined plane.

When transport moves up with uniform acceleration a along the inclined plane, we can resolve the acceleration in two axes viz along the horizontal as ax = a cos q) along the vertical as a, = a sin 0. The liquid can be considered to be simultaneously subjected to constant horizontal and vertical accelerations. Horizontal acceleration will make the free surface of the liquid to slope upwards at angle 0 at rear end of the container while the vertical acceleration will increase the pressure at any point in liquid above its static pressure Horizontal level at rest Sloping free surface due to acceleration

Free surface on acceleration

Forms on point A

Consider a liquid particle A on the free surface of the liquid. Since there is no relative movement of the liquid particles, the conditions of equilibrium can be applied i.e. EPx = 0, and EPy = 0 Now

EPx = 0, F sin H= max EPy = 0, mg + ma, = F cos 6 F sin 0 ma, F cose mg + ma, tan 0

ax g + aZ

246

Fundamentals of Fluid Mechanics

18. What happens when a liquid is subjected to constant acceleration while moving down along an inclined plane?

When transport moves down with uniform acceleration a along the inclined plane, the free surface slopes up toward the rear wall of the container as shown in the diagram. During equilibrium, we have EPx = 0, F sin 0 = max EPy = 0, F cos 0 = mg — ma, tan 0 —

ax g — a,

19. The transport tanks for carrying liquid in mountainous region are specially designed to avoid any spillover of the liquid. Elucidate. The transport tanks for transporting liquid are specially designed in such a way that liquid depth at both ends remains the same when the transport is at rest on an inclined road as shown in the figure. The tank has rear end raised up to ensure that base of the tank is parallel to the horizontal. Free surface horizontal --Base is almost horizontal Tanks for inclined plane

20. Explain why the ends of railway wagons used for transporting petroleum or other liquids are made hemispherical.

Fluid Masses Subjected to Acceleration

247

Railway wagons for petroleum The slope of the free surface of the liquid changes with the magnitude of the acceleration. The slope of liquid will smoothly change in case end walls are hemispherical. The hydrostatic pressure forces on the end walls will also be less in case the ends are hemispherical (projected area on vertical plane is small). 21. What happens when a liquid is subjected to radial acceleration? When a liquid is made to rotate by a forced vortex, the whole liquid mass rotates with constant angular velocity, thereby attaining the situation of relative equilibrium i.e. no relative motion of liquid particles.

Free surface

„ = ex412g Datum

W z0 = 0

Forced vortex in open vessel. The form of the free surface of a liquid in a rotating open vessel is similar to that of a paraboloid of revolution. Any vertical plane passing through the axis of rotation which cuts 2 2

the liquid will produce a parabola. The equation of the parabola is z = w x where w = 2g angular velocity and x is radial distance from the vertex and z is height from the vertex. 22. What are constant pressure surfaces when a liquid is subjected to a forced vortex? When a liquid is rotated in a open vessel, its free surface attains the form of paraboloid. Any surface in the liquid having all its points at equal distance from the free surface will have

248

Fundamentals of Fluid Mechanics

constant pressure. Hence all such surfaces are called constant pressure surfaces as shown in the figure.

Free surface Constant pressure surfaces

23. An open vessel partly filled with a liquid rotates about a vertical axis at constant angular velocity. Determine the equation of the free surface of the liquid after liquid has acquired the angular velocity of the vessel.

Tangent 'I

Free surface

C=mw2x

When a liquid is subjected to a forced vortex, the free surface attains the shape of paraboloid. As there is no relative motion between the liquid particles, the relative equilibrium prevails in the liquid. Consider the equilibrium of a small mass m of the liquid at a point A on free surface. It is subjected to following forces: (1) centrifugal force c = mw2x acting along x axis i.e. horizontally (2) weight of the fluid particle acting downward i.e. mg and (3) fluid pressure force F acting normal to the free surface. Now applying the conditions of static equilibrium i.e. EPx = 0 and EPz = 0 F cos 0 = mg EPx = 0, F sin 0= c = mw2x Fsin 8 _ w2x Fcos0 g .2,

tan 0 —

g

Fluid Masses Subjected to Acceleration

But

249

tan 0 — dz dx dz _ w 2x dx g

or

„,2, dz — '" '' x dx g rx w2 x rz j dz = dx o Jo g

z—

w2 x2

2g

The above is the equation of free surface of the liquid subjected to a radial acceleration. 24. What are axial depth and rise of water? How is the rise of water is related to original level? When is the axial depth of water zero?

Original level

Rise of water

Axial depth of water

4

When a liquid is subjected to radial acceleration, the free surface attains the shape of paraboloid as shown in the figure. The distance OA is called axial depth of water and OB is called the rise of water. Also OD = BD where D is on original level of liquid. The axial depth of water is zero when vertex of the paraboloid is at the bottom of the container. 25. What is the volume of empty space in the paraboloid?

250

Fundamentals of Fluid Mechanics

The empty space in the paraboloid depends upon the radius. The empty space at radius r1 (volume OA" AA') is equal to 1 2 irri 2 Zi (Z1 = OA). Similarly the empty space at radius r2 1 2

2 7 7 r2 L.,2 kz,2 =

mil \

26. What is spill out? When does it happen?

H = height of vessel from vertex

When a liquid is subjected to radial acceleration, the rise of water is given by Z as 2 2

W r

2g If Z> H, then liquid as shown in shaded will spill out. After spilling, the paraboloid surface will satisfy the equation w2R2 H— 2g

27. What happens when liquid is subjected to radial acceleration and sufficient height for liquid to rise is unavailable in the closed vessel?

Fluid Masses Subjected to Acceleration

25 I

Lid

z'

Modified free surface paraboloid

r r

If sufficient height in the vessel is unavailable, the water cannot rise above this height and spill out is impossible as the vessel is closed at the top. The liquid attains another shape of paraboloid for its free surface as shown by dotted line in the figure. The paraboloid has radius r' instead of r and height z' instead of z. Since the volume of air in the vessel remains the same as the well as volume of the water (no spill out), hence we have Volume of air in the vessel = volume of space in modified paraboloid 7112

2

=

1

2

,2 ,

irr z

,2 2

_ rw 2g

but

r'2 = 2gz'lw2

or 71

r2 =

Z

,2

2

x x

2gz' w

2

xz

,

r 2Z w2

2g =

M

2

W2 where m =

1/-2Z 1

2g

z' = mw2 28. A rectangular open tank 8 m long; 2 m deep contains 1 m of water. If the horizontal acceleration of 3 m/s2 is applied, then find (1) forces acting on the ends of the tank and (2) show that the difference between these forces equal to the magnitude of force to accelerate the water mass.

252

Fundamentals of Fluid Mechanics

B •

1m

2 Acceleration = 3 m/s

I

IC

A F.,

When horizontal acceleration is applied, the free surface of the water will slope by angle 0 as shown in the figure tan 0 — a 3 — 0.3058 g = 9 .81 0= 17° Rise and fall of water = 6 tan 17 2 = 0.917 m AB = 1 + 0.917 = 1.917 m CD = 1 — 0.917 = .083 m AB 1 x Area = — x 1 x 103 x 9.81 x 1.917 x (1.917 x 2) 2 2 = 36.06 kW

FAB — pg

FCB = pg x

CD x Area = 1 x 1 x 103 x 9.81 x 0.083 x 0.083

= 0.00 kW F = FAB — FCD = 36.06 — 0.06 = 36 kW M= Mass of Water = Volume x p =6x1x2x1x103 = 12 x 103 kg Fa „ = M X a = 12 x 103 x 3 = 36 kW F = Fa „ 29. An open tank 10 m long and 2 m deep is filled with 1.5 m of oil having sp. gr. 0.9. The tank is subjected to horizontal acceleration to the velocity of 20 m/s. What is the smallest time to attain this velocity without spilling out? The oil surface will slope at angle 0 on the application of acceleration tan 0=

Fluid Masses Subjected to Acceleration

253

a 9.81 AB = 2 — 15 OB 5 = 0.1

tan B=

Also

0.5 5

a 9.81 a = 0.981 m/s2

tan = 0.1 —

Acceleration = a

Now apply velocity equation V = U + at U = 0 & V = 20 m/s 20 = 9.81 x t 20 t — 20.38 sec. 0.981 30. A rectangular open tank of size 3 m length, 2 m wide and 2 m height is completely filled with water and is subjected to horizontal acceleration of 0.8 m/s2. Find the volume of water spilled out.

/0 t

Acceleration 2 a = 0.8 m/s

3

The free surface slopes by angle O as shown and water rises by AB resulting water in volume AOB spilling out. tan

0_a _ _ 0.0815 g 9.81

254

Fundamentals of Fluid Mechanics

Also

AB tan 9 — AO = 0.0815

AB = —3 x 0.0815 2 = 0.122 Volume of water = Area of triangle AOB x width 3 x— x 0.122 x 2 2 2 = 0.183 m3 =

31. An open tank has 2.0 m height and it contains 1.8 m of water. How high must its side be if no water to be spilled out when subjected to horizontal acceleration of 4.5 m/s?

2 Acceleration = a

1.8

I

k The free surface will slope when subjected to acceleration 4.51 _ 0.5 tan 9 _ a g _ 9.8

Also

tan 0 — AB=AB = 0.5 AO 1 water rise = AB = 0.5 m Height of tank = 2 Height needed = CB = 1.8 + 0.5 = 2.3

Hence side of tank must be 2.3 m to avoid any spill out 32. A rectangular open tank is 6 m long, 2 m deep and 2 m wide is containing water upto 1 m height. If the tank is subjected to horizontal acceleration of 3 m/s2. Find (1) slope of surface (2) maximum and minimum pressure intensity at the base and (3) pressure forces acting on each end walls.

Fluid Masses Subjected to Acceleration

255

Acceleration 2 3 m/s

tan 0 — a — 0.3 g —9.81 Also

or .

E tan 0 = B = 0.3 BO 0 = 16.7° BE =

x 0.3 = 0.9 2 AE = 1 + 0.9 = 1.9 m CF = 1 — 0.9 = 0.1 m

Maximum pressure intensity at point A of the base PA = 1.9 xpxg = 1.9 x 1 x 103 x 9.81 = 18.64 kN/m2 Minimum pressure intensity at point C of the base P, = 0.1 x 1 x 103 x 9.81 = 0.981 kN/m2 Maximum pressure force at rear end wall Fmax = x Apg = 2 x (1.9 x 2) x 1 x 103 x 9.81 = 35.41 kN Minimum pressure force at front end wall Fmin = X APg = al x (0.1 x 2) x 1 x 103 x 9.81 = 0.0981 lcIsT 33. A rectangular container 10 m long, 5 m wide and 4 m deep is containing oil upto top. Oil has sp. gr. = 0.9. Find the pressure difference at the lid existing at rear and front end when container is subjected to horizontal acceleration of 4 m/s2.

256

Fundamentals of Fluid Mechanics E D

! 0'

I

C F

Acceleration 2 4 m/s

B 10

If the container was not having lid, then oil would have risen at lower end and angle of slope is given by tan 0 — a _ 4 _ 0.407 g 9.81 ED ED tan 6' = OD = = 0.407 5 ED = 2.035 m AE = AD + DE= 5 + 2.035 = 7.0.35 m BF = BC — CF = 5 — 2.035 = 2.965 m Height difference between point E & F is h = AE — BF = 7.035 — 2.965 = 4.070 m PD — PC= hPoiig = 4.07 X 0.9 x 103 x9.81 = 35.93 kW/m2

Also

34. A open rectangular tank 6 m long 3 m high and 3 m wide contains oil (sp. gr. = 0.9) upto height of 2 m. Find the total pressure force on the base of the tank when the tank is moving with acceleration of

g 2 — M/S

2

(i) vertically upwards and (ii) vertically

downwards. Acceleration

2 g I

Case 1. The oil is subjected to the acceleration of Total pressure at base of tank P = pgh (1+ ci)

2

or

upwards.

2

Fluid Masses Subjected to Acceleration

257

g = 0.9 x 9.81 x 2[1+

g

3 = 0.9 x 9.81 x 2 x — x103 2 = 26.49 kW/m2 F = P x base area = 26.4 x 6 x 3 = 476.77 kW Case 2. Acceleration is applied downwards P = pgh (1— :,)

P = 0.9 x 103 x 9.81 x2 (1

g12) g

1 3 = 0.9 x 0.981 x 2 x — 10 2 = 8.830 kW/m2 Pressure force on the base F=PxA = 8.83 x 6 x 3 = 158.94 kW 35. An open cylindrical tank is 2 m high, 1 m in diameter contains 1.5 m of water. If the cylindrical tank is rotated about its normal axis, find (1) maximum angular velocity without spill out of water and (2) The pressure at centre and edge of the cylinder at base. The water level was at AD initially before the rotation of the cylinder. On rotation the free surface attains the shape of paraboloid with vertex at 0. The rise of the water at AE and DF from original level AD is AD = 2 — 1.5 = 0.5 m If we take origin at 0, then y = OK = EG = 2 x AD = 1 m x at top = EK or FK = 0.5 Now

2 2 W X

Y — 2g

258

Fundamentals of Fluid Mechanics

1—

W2 X (05)2 2Xg

_ 2xg — 7.85 25 w = 2.8 rad/s

w2

E

F

K

D

A

G

H

O

1.5

B

H

1

Maximum pressure is at point c at base Pc = hpg = 2 x 1 x 103 x 9.81 = 18.62 kN/m2 Minimum pressure at point 1 at base P1 = hpg

= 1 x 1 x 103 x 9.81 = 9.81 1c1•1/m2 36. Find out total pressure force on lid and base of fully filled liquid when subjected to radial acceleration. Where liquid is subjected to radial acceleration, the free surface cannot attain the paraboloid shape as vessel is closed with lid. Hence there is a pressure head (z) given by: z

2 2 W X

2.g

Since pressure (P) = pgz 2

P— W2X

2

Fluid Masses Subjected to Acceleration

T

259

Total pressure line z ------

,

--------------

- -

Lid

-

- - - - - - - - - - - - - - - - -

Bottom

Now take a small circular strip of thickness dx at distance x from centre.

Pressure force on this strip is dF = p x area = p x 27rxdx 2 2 W X

2p

x 27rxdx

dF — Kw2p j*x x3 dx

-

irw 2 p x 4 4 Hence F is total pressure force acting on the lid of the cylinder due to radial acceleration. Now total pressure force at base of the vessel is sum of force at the top (lid) and weight of liquid in the vessel Ftop

Fbase = Ftop + pgH

KW 2 p x 4 + pgH 4

260

Fundamentals of Fluid Mechanics

37. A cylindrical vessel of 1 m height and 15 cm diameter is filled upto 0.667 m with water. It is subjected to radial acceleration, find (1) speed of rotation when water rises to its top (2) the speed when vertex touches the base.

—0.15 —•1

Case 1 When the liquid rises to the top, We have and

z = 2 x (1 — 0.67) = 2 x 0.33 m = 0.66 m x

_ 0.15 _ 0.075 m 2

z—

0.66 — w2

2 2 W X

2g w2 x (0.075)2 2 x 9.81 0.66x 2 x9.81

(0.075)2 = 2302.08 w = 47.98 rad/sec 27rN — 47.98 60 60 x 47.98 N= 2 x Tr = 458 rpm Case 2 When vertex touches the base Volume of paraboloid = Volume of empty space before rotation 1 rcri2 x H = 1.7-cr2(H — 0.67) — 2

Fluid Masses Subjected to Acceleration

26 I

r12 = 2 x (0.075)2 x 0.33 = 3.71 x 10-3 r1 = 0.061 22

Z H

w1 2g

1 — (.061)2 x w? 2 x 9.81 2 2 x 9.81 w— 371 x 10-3 w2 = 52.88 x 102 w = 72.72 rad/sec 27r N — 72.72 60 60 x 72.72 N= 2x ic = 694.8 rpm 38. A closed cylindrical vessel of 2 m diameter and 3 m height is completely filled with water. If vessel is rotated at 120 rpm, find the total pressure on the top, bottom and sides of the cylinder.

2

w—

27rN

60

2ir x 120 = 4rc = 12.56 rad/sec 60

Pressure force at top is Ftop

irpw 2 X 4 4 zxlx 103 x (1256)2 x 14 4 = 123.8 kW

Pressure force at bottom is

262

Fundamentals of Fluid Mechanics

Fbottom =

Ftop + weight of water

ir x 22 x 3 x 1 x 9.81 4 = 123.8 + 92.4 = 216.2 kW = 123.8 +

Pressure force at sides 22

q

= pAg ( x w +1 2g 2 2 x 12562 14) =1 x 103 x(3 x2 xH)xgx 1 2x 9.81 ±— = 6rc x 9.81(8.04 + 1.5) IcN = 1763 kN 39. A glass tube 5 cm in diameter and sufficiently high, open at the top containing a liquid rotates about its vertical axis at 700 rpm. Find the depression of the lowest point of free liquid surface below the original surface of liquid when at rest. (London University)

z

—

w—

2 2 W X

2g 27rN 60

2Tcx 700 — 73.26 rad/sec 60

(73.26)2 x (13.5)2 z

—

2 x 9.81

5.37 x 103 x 6.25 x 10-4 2 x 9.81 = 1.7 x 10-1 m

Fluid Masses Subjected to Acceleration

263

= 0.17 m Depression of free surface is OA — z = 017 ' — 0.085 m 2 = 8.5 cm 40. A closed cylinder 30 cm diameter and 2.3 cm deep is completely filled with water. It is rotated about its axis which is vertical at 240 rpm. Calculate the total pressure of water at each end. (Roorkee University) )4 240 rpm

2.5 cm

[4— 30 Icm

irpw2 X 4 Ptop

4

2x/rx240 27rN 60 60 = 25.12 rad/sec

w=

n x lx 103 x (25.12)2 4( 1)4 Ptop —

4

x 6.31x 103 x 5.06 x 4 = 2.51 x 102 N = 251 N Pbottom = Ptop + Weight of water in cylinder = 251 + x (.15)2 x 2.5 x 10-2 x 1 x 103 x 9.81 = 251 + 17.32 = 268.32 N 41. A rectangular open container is partially filled with water and is shown below at rest. If the container is moved horizontally to the right at constant uniform acceleration, the pressure at A will (a) decrease (b) increase (c) stay the same (d) first decrease (IES 90, BHEL) than stay the same.

264

Fundamentals of Fluid Mechanics

---------- Water ----------

Acceleration

• A

When water is subjected to acceleration, the free surface slopes, thereby increasing the level of water at rear end and decreasing the level of water at front end as shown below.

T

h

hinitial _L.

• A

As the height of water at A rises due to the acceleration of water the hydrostatic pressure increases P = pghA h initial < hA P is more when acceleration is applied as compared to static pressure. Answer (b) is correct. 42. A one meter U-tube acceleration meter having its vertical ends open to atmospheric. The cross-sectional area of tube is uniform throughout. The meter is placed in a vehicle moving horizontally towards right at a constant acceleration of g/2 m/sec2. Find reading on the left side scale of the acceleration meter.

Acceleration 2 9 = m/s

=>

spgr = 1.5

hinitial

1."

1

Since U-tube liquid is continuous in both limbs, it will behave similarly to a container containing liquid which is when subjected to horizontal acceleration. The liquid at front end will fall and rear end will rise i.e. liquid level rising in right limb as compared to left limb. tan g=

=

R12

= 0.5

Fluid Masses Subjected to Acceleration

265

AB =0.5 OB Or AB = 0.5 x 0.5 = 0.25 Rise in right limb = 0.25 m Fall in left limb = 0.25 m ...

43. Find the slope of free surface of liquid in a tank (a) moving horizontally, towards right (b) moving downwards, along at 20° inclined plane and (c) moving upwards along 20° inclined plane if the constant acceleration of liquid is 4 m/s2. Case 1 Moving horizontally tan 0 — a _ 4 _ 0 41 g 9.81 ' 0 = 22.18° Case 2 Moving downwards at 20° inclined plane.

The horizontal component of acceleration is a cos 20 and a sin 20 is acting against gravity tan 0 =

a cos 20 g — a sin 20 4 x 0.939 9.81 — 4 x 0.34

= 3.756 8.45 = 0.444 0 = 24° Case:

Moving upwards at 20° inclined plane. tan 0 —

a cos 20 g + a sin 20 4 x 0.939 9.81 + 1.36 4 x 0.939 11.17

266

Fundamentals of Fluid Mechanics

= 0.336 = 18.58° Note: The sloping of free surface of liquid is always more when tank car is moving down on an inclined plane. 44. An open circular cylinder 1.2 m height is filled with a liquid to its top. The liquid is given a rigid body rotation about the axis of the cylinder and the pressure at centre line at the bottom surface is found to be 0.6 m of liquid. What is the ratio of volume of liquid spilled out of the cylinder to original volume?

1 (a) 4

(b)

(d) 3/4

(c)

(IES 2007)

0.6 m

E N

2r

Original volume of liquid (V1) = xr2h = rr2 x 1.2 m3 1 ( rr2 x 0.6) Volume of liquid spilled out (V2) = — 2 = irr2 x 0.3 m 172

Now

V1

gr2 x 0.3 nr2 x L2 1 4

Option 'a' is correct. 45. A cylindrical vessel having its height equal to its diameter is filled with liquid and moved horizontally at an acceleration equal to acceleration due to gravity. The ratio of the liquid left in the vessel to the liquid at static equilibrium is— (a) 0.2

(b) 0.4

(c) 0.5

(d) 0.75 (IES 2001)

Fluid Masses Subjected to Acceleration

I

267

> a=g

The liquid at front end will fall and liquid at rear end will rise. tane = But

g

=

g

= 1 = 45°

tan t9 — ED _ ED _ 1 dI2 OD

ED = d —= h 2 2 Hence, tank will be empty by half. Option 'c' is correct. ... or

46. An open rectangular box of base 2 m x 2 m contains a liquid of specific gravity 0.80 upto a height of 2.5 m. If the box is imparted a vertically upward acceleration of 4.9 m/s2, what will be the pressure on the base of the tank? (a) 9.81 kPa

(b) 19.62 kPa

p = p• g •

(c) 36.8 kPa

h • 11 + ) S/

0.8 x 103 x 9.81 x 2.5 x 1+ 49:9 ( 8) = 29.4 kpa Option 'd' is correct.

(d) 29.40 kPa (IES 2004)

Chapter

6

DIMENSIONAL ANALYSIS

KEYWORDS AND TOPICS A A A A A A A A

FUNDAMENTAL DIMENSIONS SECONDARY DIMENSIONS HOMOGENEOUS EQUATION DIMENSIONAL FORM NON-DIMENSIONAL FORM RAYLEIGH'S METHOD it THEOREM VARIABLES

A A A A A A A A

DEPENDENT VARIABLES REPEATING VARIABLES DIMENSIONLESS NUMBERS REYNOLDS' NUMBER FORDE' S NUMBER WEBER'S NUMBER MACH'S NUMBER EULER' S NUMBER

INTROIDUCTION Dimensional analysis is a mathematical technique which uses the dimensions of quantities in solving the engineering problems. Dimensional analysis helps in determining a possible arrangement of variables in a physical relationship. Physical relationship is rational in case the dimensions of left hand of the relation is equal to the dimensions of right hand. All physical relationships are dimensionally homogeneous. In fluid mechanics, a fluid particle is subjected to a number of forces such as (i) inertia force (ii) viscous force (iii) gravitational force (iv) pressure force (v) surface tension force and (vi) elastic force. Dimensionless parameters are useful in fluid mechanics when any two forces are predominant. The dimensionless parameters are (i) Reynolds number (ii) Euler's number (iii) and Weber's number. 1. What do you understand from dimensional analysis? Dimensional analysis is a branch of mathematics which deals with the dimensions of quantities. Dimensional analysis helps in determining a possible arrangement of variables in

Dimensional Analysis

269

a physical relationship. This is achieved by forming a number of non-dimensional group free from units from a given number of dimensional quantities in such a way that variables can be reduced and the physical relationship can be expressed in a possible arrangement of remaining variables. 2. What are the uses of dimensional analysis? The uses of dimensional analysis are: (1) To ascertain whether any equation of physical phenomenon is rational or not. If the equation is dimensionally homogeneous (dimensional units at both sides of the equation area same), the physical phenomenon is rational. Otherwise it is impossible. (2) To ascertain whether the relationship between various quantities in an equation signifying a physical phenomenon. (3) To reduce the number of variables involved in a physical phenomenon. The performance of experiment becomes easy due to this reduction of variables. (4) To convert the theoretical equation into a simple dimensionless form. (5) To solve any complex physical phenomenon. (6) To convert the units of quantities from one system to the another system. (7) To facilitate model testing by reducing the number of variables into four primary quantities. 3. What are the fundamental dimensions or primary quantities? Mass, length, time and temperature are fundamental or primary dimensions. These four physical quantities are called fundamental dimensions or primary quantities as the unit of these quantities does not depend on the unit of any other physical quantities. 4. What are the derived or secondary dimensions? The unit of physical quantity which depends on the unit of other physical quantities is called derived or secondary dimensions. 5. Find the dimensions of (1) force (2) angular velocity (3) discharge (4) torque (5) power (6) density and (7) viscosity Force = mass x acceleration L F = M x — = MLT-2 T2 Angular Velocity — Angles in radian time w= =T-1 T

Discharge = Velocity x Area L x L2 Q=— T

=

L3 7-1

270

Fundamentals of Fluid Mechanics

Torque = Force x distance = (mass x acceleration) x distance T=Mx

L x L = ML2 T-2 T 2

Power — Work done Time _ Force x distance Time _ Mass x acceleration x distance Time Mx L xL T-2 _ AlL 2 7-3 P= T Density — P= Viscosity —

Mass Volume M _ AIL-3 L3 Shear stress Velocity gradient Force/Area Velocity/distance

_ Mass x acceleration x distance Area x Velocity —

M x LT-2 x L L2 x LT-1 -1

ki =

7.-1

ML

6. What is dimensional homogeneity? What are variables and dependent variables? or What do you understand by a dimensionally homogenous equation? Write three dimensionless parameters. A physical phenomenon is given in variables in the form of an equation. The physical phenomenon is rational in case the dimensions of left hand of the relation is equal to the dimensions of right hand. If it is so, we say that equation is dimensionally homogenous. For example, take the expression of static pressure: P = pgh

Dimensional Analysis

27 I

Mass x acceleration Area = M x LT-2/L2 = ML-1 r2

Pressure = Force/Area -

p = density - Mass

Volume

= M = ML-3 L3 g = acceleration = L7-2 h = height = L P = pgh AIL 1 - 2 = ML-3 X L T2 X L = AE-1 T-2 dimensions of left side = dimensions of right side In the above equation, P is dependent variable and p, g and h are variables. 7. What are the applications of dimensional homogeneity? The applications of dimensional homogeneity are: (1) To ascertain whether the equation is homogeneous, thereby finding out whether phenomenon is rational. (2) To ascertain the dimensions of a variable. (3) To facilitate easy conversion of units from one system to another. (4) To help in dimensional analysis and model testing. In fluid mechanics, a fluid particle is subjected to a number of forces such as inertia force, viscous force, gravitational force, pressure force, surface tension force and elastic force. Dimensionless parameters are useful in fluid mechanics when any of two forces are predominant. Hence dimensionless parameters are the ratio of two forces which are predominant. The dimensionless parameters are: (a) Reynolds number. It is the ratio of inertia force to viscous force in fluid. ni2v2 R

e

lin

_ pVL (b) Euler 's number.

It is the ratio of inertia force to pressure force in the fluid. n i2v2 EAPL2

- pV 2 AP

272

Fundamentals of Fluid Mechanics

(c) Weber's number.

It the ratio of inertia force to surface tension force. ni2v2, pv2L W= a/

8. What are dimensional form and non-dimensional form of homogenous equation? In dimensional form of homogenous equations, the dimensions of left side are equal to dimension of right side. For example: Q = 8 CdV2g tan 0 H5I2 5 7-1 = L3 7-1 On reduction L3 In non-dimensional form of homogenous equations, the left side and right side have no dimension. For example, the above homogeneous equation can be written in nondimensional form as: — 8 2 Cd tan 6 15 •H"2

Q

9. What are different methods of dimensional analysis? There are two methods of dimensional analysis viz. (1) Rayleigh's method and (2) Buckingham's method or Buckingham's 7rtheorem. 10. Explain dimensional analysis by Rayleigh's method. The Rayleigh's method is used for determining the expression for a dependent variable which depends upon maximum three or four variables or otherwise it becomes difficult to find the expression for the dependent variable. IfX is a variable which is dependent on three variables X1, X2 and X3, then the expression can be written as X = X2, X3). This expression can be simplified by Rayleigh's method by writing X= m•Xla X2b X3' where m = constant and a, b and c are powers. The values of a, b and c can be found out by comparing fundamental dimensions of both sides. Hence the expression for dependent variable can be obtained. 11. Prove by the method of dimensional analysis that R, the resistance to be motion of a sphere of radius r falling with a velocity V through a fluid of viscosity q is given by R = rV where k is a dimensional constant. (London University) Guidance: The dependent variable depends upon three variables which makes it suitable for Rayleigh's method of dimensional analysis Rocifixrb xVx1f 1 Putting fundamental dimensions

Dimensional Analysis

MLT-2 oc (IVE-3)a x (L)b x (LT1 )c MLT-2 oc ma+d x L-3a+b+c—d x T

x

273

(ML-1 rl)d

Now equate powers of both sides, we get three equations as under 1=a+d 1=-3a+b+c—d —2 = —c — d

(1) (2) (3)

From equation (1), (2) and (3), we can find the values of a, b and c in term of d as under— a=1—d b =2— d c =2— d Putting the values to the equation R « p" r" V" lid yi = k pre V2

(pnri 7

Putting d = 1 R = kr? rV 12. State Buckingham's theorem. What are repeating variables? How are these selected in dimensional analysis? (UPTU-2004-5) or Describe Buckingham's theorem? Why this theorem is considered superior to Rayleigh's method for dimensional analysis. (UPTU-2005-6) BUCKINGHAM'S THEOREM Buckingham's theorem states that if there are n number of variables and m number of fundamental dimensions in a dimensionally homogeneous equation, then the variables can be arranged in (n — m) number of dimensionless terms. These dimensionless terms are called 71- terms. Let Or

Xi = fix2, x3 . — x„) fixi, x2 ... xn) = o.

If there are n variables with 3 fundamental dimensions, then equation can be written as — ,(g1, 762 • • • 76n-3) = 0 Each 71- terms is formed with 4 variables in which three variables are repeating variables (say X2, X3 and X4) Hence Tr terms can be written as — xi = X201 X3b1 X4cl Xi 71-2 = X202 X3b2 X4c2 X5 and so on but upto nn-3 = X20n-3 X3bn-3 X4cn-3 Xn_3

274

Fundamentals of Fluid Mechanics

The above equations are solved by the principle of dimensional homogeneity. When the values of irk , 7r2 ... ir„-3 area known, the required expression can be written as: f (xi, 7-c2 - 7l 3) = 0 REPEATING VARIABLES These are variables which are selected to appear in each ir term. The numbers of repeating variables are equal to the number of fundamental dimensions appearing in the variables of the equation. The choice of repeating variables is made on basis of following: (1) The dependent variables should not be selected. (2) Variables having different categories of property instead of one should be selected. For example, variables should be one each from geometric property (length, height and diameter), flow property (velocity, acceleration) and fluid property (viscosity, density). (3) Should not be dimensionless. (4) Should have different dimensions. (5) Should have same number of fundamental dimensions. Superiority of x theorem over Rayleigh's method: The Rayleigh's method of dimensional analysis becomes more laborious as number of variables becomes more than fundamental dimensions. Such problem is not faced when 71- theorem is used. 13. The drag force F on a partially submerged body depends on the relative velocity V between the body and the fluid, characteristics linear dimension 1, height of surface roughness k, fluid density p, the viscosity g and the acceleration due to gravity (g). Obtain an expression for the drag force using the method of dimensional analysis. (UPTU-2001-2) Guidance: There are six variables which suggest to use Buckingham's n- theorem for solving the expression. Let the expression be f(F, I, k, V, µ, p, g) = 0 Total number of variables (m) = 7 Total number of fundamental dimension (n) = 3(M, L & 7) Number of ir terms = 7 — 3 = 4 Each ir term consists of M + 1 variables i.e. 3 + 1 = 4 We select repeating variables as L, V and p. 7ri = 1,611 ribl p° 1 F u 7r2 = L°2 Vb2 i:f2 j 7/-3 = La3 Vb3 113 g 7r4 = La4 Vb4 p' 4 k Now xi = Lai(LT-1)11. (ML-3)°i (AILT-2) mOLO TO = La1+b1-3c1+1 112`1+1 741-2 —b1 — 2 = 0 or bi = —2 + ci + 1 = 0 or c1 = —1 ai + bi — 3c1 + 1 = 0 or ai = —2

Dimensional Analysis

= L-2 r2-1

275

pF

F pL2V2 pc2p, = La2(LT-1)b2 (ML-3)C 2 (ML-1 T-1) .moLoTo = La2+b2-3c2-1 Mc 2+1 T-b2-1

Now

11-2 = La2

c2 + 1 = 0 or c2 = —1

—b2 — 1 — 0 or b2 — —1 a2 + b2 — 3c2 — 1 =0 or a2 — —1 7r2. = L-1

Now

ri Pri = LVp

7r3 = La3 Vb3 Pc3 g M°L°P) = La3(LT-1)1'3 (ML-3)c3 (LT-2) La3+b3-3c3+1 mc3 ra3_2

c3 = 0 —b3 — 2 = 0 or b3 = —2 a3 + b3 — 3c3 + 1 — 0 or a3 — 1 L

71- Ll v-2 a 3

V2 6 7r4 = La4 Vb4 pc4 k 7c4 = La4 (LT-1)b4 (A4L-3)c4 L = La444-3c4+1 Apt T-b4.

Now

c4 = 0, b4 = 0, a4 — —1 11-4

= L-1 k=

fOrt, 7r2, 11-3, 7r4) = 0

F 1-1 L k pL2 V2 LVp' V2 g' L.)

pL2 V2 or

Lg k )

_ th (

F

LVp V2

L

F = pL2V2 x 0( P

LE? k ) LVp' V 2 L

14. Show by 76 theorem that a general equation for discharge Q over a weir of any shape is given by — 5 1 Q = H 2 g2

a // 2/3g1/2

H2

gP

276

Fundamentals of Fluid Mechanics

where H = head over weir, v = kinematic viscosity, p = density of the fluid g = acceleration due to gravity and a = surface tension. (UPTU-2006-7) F(Q, H, p, v, g, cr) = 0 Select H, p and g as repeating variables. Hence we will have M + 1 = 3 + 1 = 4 variables in each it term and there will n — m = 6 — 3 = 3 x terms Tri = Ha1gb1Pc1Q = Lai. (LT-2)bl (14L-3)cl L3 T-1 = La+bi-3c1+3 Mc T-2b-1

c1 = 0, bi =

d al = —3/2 1 and 2

Q _ Q xi _ H-5/21/2 S H5/2 1/2 g

Now

7r2 = /122 gb2 if2 v

= La2 (LT-2)b2 (ML-3)c2 L2r1 = La2+b2-3c2+2 Mc T-2b2-1 c2 = 0, b2 = —1/2 and a2 = —1/2 = H-3/2 g-1/2 V 7C2 = 7c3 Ha3 gb3 pc3 a = La3(LT-2)b3 (4-L-3)c3 (MT 2) = La3+b3-2c3 T-2b3-2 Af3+1

.

C3 = —1, b3 = —1 and 71- _ H-2 g-1 p-1 6 3

v cr F [ Hg 5/F 1/2 ' 3/2 1/2 ' 2 Hg H gp 3

—

a3 = —2

0

Q H5/2

g

1/2

— 0(

'2 13 ) H 3/2vg1/2 H gp

Q = H512 g1/2 ih[

V

Y H3/2 gin '

Cr

H2 gP

15. What do you mean by dimensionless numbers? Derive an expression for any two dimensionless numbers. (UPTU-2002-3, 2007-08) or What is the importance of dimensionless numbers? Derive their expressions and write their applications. Fluid is generally subjected to various forces such as (1) Inertia force (2) Viscous force (3)

Dimensional Analysis

277

Gravitational force (4) Pressure force (5) Surface tension force and (6) Elastic force. The ratio of any two forces is a dimensionless parameter. These parameters have great physical significance as fluids having same parameter have dynamical similarity. The most important dimensionless numbers are: (1) Reynolds number (2) Froude's number (3) Euler 's number (4) Mach's number and (5) Weber's number. (1) Reynolds Number: If inertia force and viscous force are predominant in any flow phenomenon, then Reynolds number is used to compare model and prototype. Reynolds number is the ratio of inertia force to the viscous force. Inertia force Viscous force Inertia force = mass x acceleration = volume x density x acceleration = L3 x p x LT-2 R e

x L2 x p but

= V= Velocity

= V2L2p Viscous force = Tension x area

-

dV x area

-

LT-1 x L2

dy

= 42 7-1 = AVL R

e

L2V 2 p pVL LVp

Reynolds number is used for study of (1) motion of completely submerged bodies like submarines, aeroplanes and automotives and (2) incompressible flow through pipes, bends and turbines in which inertia and viscous forces are predominant. (2) Froude's Number: Froude's number is used where inertia and gravitational forces are predominant. It is the square root of the ratio of inertia force to the gravitational force. Inertia force = mass x acceleration = V2L2p Gravitational force = mass x g = volume xpxg = L3 pg F=

Inertia force Gravitational force

278

Fundamentals of Fluid Mechanics

v

= if

2L2p

L3 pg

= 1172 1 Lg V Lg Froude's number is used for the study of fluid flow from open hydraulic structures such as spillways, weirs, sluices and open channels where gravitational forces are predominant. (3) Weber's number: It is the square root of the ratio of the inertial force to the surface tension force in the fluid. W= i

Inertia force

l Surface tension force

Inertia force = V2L2p Surface tension force = cr x L 1172L2p

TV — (FL

1

V lipaL The Weber's number is used for the study of (1) capillary movement of water in soil and (2) flow of blood in arteries and veins when the surface tension forces are predominant in the flow. (4) Euler's number: It is the square root of the ratio of inertia force to the pressure force in the fluid. E

Inertia force Pressure force Inertia force = V2L2p Pressure force = PA = PL2 E_

iv 2.1,2 p \ PL2 V

117 ; The Euler's number is used for the study of fluid where pressure force is predominant

Dimensional Analysis

279

such as (1) water hammer effects in penstocks of hydro power plants and (2) discharge coefficients of orifices, sluice gates and mouthpieces. ( 5 ) Mach's Number: Elastic force becomes predominant in compressible fluids. Mach's number is significant in finding dynamical similarity of the fluids where elastic forces are predominant. Mach's number is the square root of the ratio of inertia force to the elastic force in the fluid. M

Inertia force - 11Elastic force V2 L2 p 11 E x L2

_

where E = Volumetric modulus of elasticity

lpV 2 I E

A1= V iE l0

Mach's number is used for the study of flow at high speed such as (1) flow in pipes (2) water hammer effect (3) high speed motion objects like aeroplanes, projective and missiles in atmosphere where the elastic forces are predominant. 16. Develop an expression for the distance travelled by a freely falling body in time (7). Take it to be dependent on mass of the body, the acceleration of gravity and time. Guidance: There are 4 variables viz. distance (S), mass (M), acceleration (g) and time (7). Hence we can use Rayleigh's method of dimensional analysis. S = f(M, g, 7) M°L1T° = k x ./VP x (L7-2)b x (7)c where k = constant moLi To = mae r2b+c .

a = 0, b = 1, c = 2 S = k g T2

17. Derive an expression for an ideal fluid flow (Q) while flowing through an orifice which is dependent on density of fluid, the diameter of the orifice and the pressure difference. Q = AP, I), P) = k rig Db PC M°L37-1 = k(ML-3)a (L)b (ML-17-2)c = k A/17+c L-3a+b-c 7-2c 1 c= — a + b = 0 or a = -1/2, -3a + b - c = -1 2'

280

Fundamentals of Fluid Mechanics

or

b=2 Q = k p-1/2 D2 P1/2 = kiD2 II P

p

18. Derive an expression for the power delivered to a pump which is dependent on the sp. wt. of the fluid, the flow and the head delivered. Power delivered = P = f(v, Q, H) mL2T-3 = k(vE-2 7-2.)a (L3 T-1)b mc = kM2 L-2a±3b±c r20-b a = 1, b = 1, c = 1 P = k VQH 19. Using dimensional analysis, find the expression of dynamic pressure exerted by a flowing incompressible fluid on an immersed object if it is dependent on density and velocity. Pressure = P = f(pV) P = k pa Vb

or

Airi T-2 = k (M I (LI e T = k M2 1,-3a±" b=2& a=1 P = kpV2

rb

20. Using n theorem show that the shear stress of the pipe wall is given by — „I_ = pv2 it

j

where

pVd) II )

p = fluid density, /1 = viscosity, V = average velocity, d = diameter of pipe

The relationship can be expressed as

AT, 19, 14 V, d) = 0 Here m = 5 and n = 3. Hence there will be two n terms with each n term having four variables. Select d, V & p as repeated variables

.

.

xi = da Vb Pc XT

lin.°7° = LV7 -1)b (ML-3)c AIL7-2

Dimensional Analysis

28 I

= mc+1 La+b-3c+1 74-2 C = -1, b = -2 and a = 0 = do v-2p-1 t V 2p

Now

7r2 = da Vb Pc 1,1

= La(LT-1)b (E-3)c

ML-1 La+b-3c-1 rb-1 M°L°T° _ b = -1, c = -1, and a = -1 71-2 = crl J-1 p 1 µ dVp =0

V 2p ' dVp Or

dVp

V2 p

or

z = V2p tfi ) vp

21. Assuming that thrust T of a screw propeller is dependent upon the diameter d, speed of advance V, fluid density p, revolution per second n and coefficient of viscosity p, show using the principle of dimensional homogeneity that it can be represented by

T= pd2V2 (pe = Vd1. , c1V.11) (Allahabad University) Let the expression be given by F(T, p, d, V, it, n) = 0 Here m = 6 and n = 3. Hence there will be three 7t terms and each ic term will have four variables. Select p, V & d as repeated variables = pa vb dcT = (ML-3)a (LT-1)b (L)c mu-2 moLoTo = ma-pi L-3a+b-pc+i a = -1, b = -2 and c = -2 = p-1 v-2 ci-2 T

_

T

pv 2 d2

282

Fundamentals of Fluid Mechanics

Now

= Pa Vb p = (ML-3)a (LT-1)" (L)c ML-1 T-1 M°L°T° = Ma+1 L-3a±b±c-1 T4-1 a = —1, b = —1 and c = —1 7r2

Tc2. = P-1 V-1 d-1 Now

F

( T

7c3 = pa Vb do r = (ML-3)a (LT-1)" (L)c T-1 = M a L-3a+b-Eb 7.-b-1 a = 0, b = —1, c = 1 2. = n c3 d.n V c1.1

PV 2 d 2 PVCI V

or

P pVd

T 2 pV 2d

=0 p do ) , pVd V

T

= p v2 d2 0

dn) pVd V

(u

22. Show by the method of dimensional analysis that the volume rate of flow of a gas through a sharp edge orifice is given by 2 = d2 1P ,t,(17 x 1 P p 1 p d where d is the diameter of the orifice, p is the pressure difference between the two sides of the orifice, p and V are the density and kinematic viscosity of the gas respectively. (London University) Let the expression be given by F(Q, d, P, p, V) = 0 Here m = 5 and n = 3. Hence there will be two it terms and each term is having four variables. Select d, P and p as repeated variables. Pb Pc Q == da (L)",2)b (mL_3)c L27, moLoTo = mb+c La-b-3c+2 7-2b-1

Dimensional Analysis

b= Rl

Now

=

283

—1 1 ,c= and a = —2 2 2 d-2 p-1 /2 pl/2 Q

Ph pc v = (L) (ML-1 ry (ML-3)c L2r1 ° moLo To = mb+c La-b-3c+2 T-2b-1 7r2 = da

b = —1/2, c = Tc2

2

and a = —1

= (I A p-1 /2 pl /2

= V JP d VP Q

F

v

=0

d 2 1) d

VP or

Q =d2

611I pr Ap)

23. Prove by method of dimensions that in the rotation of similar discs in fluid with turbulent motion, the frictional torque T of a disc diameter D rotating at speed N, in a fluid of viscosity µ and density p. T = D2N2p (p(

11 ) D2 N17

Let the expression be F(T, D, N, p, µ)= 0 Here m = 5 and n = 3. Hence there are two g terms and each 7-4- term has four variables Trl = D° Nb Pc T = (.0a (7.1)b (ML-3)C mOLO TO = mc+1 La-3c+2 T-b-2 C = —1, b = —2 and a = —2

Tcl

Now

T D 2 N 2p

ir2 = Da Nb Pa = (L)a (T-1)b (ML-3)c M°L°T° = ivf±i La-3c-1 T-b-1 C = —1, b = —1, a = 2

2 T-2

284

Fundamentals of Fluid Mechanics

/-1

jr2

D2

Np

F(, T „P ) — 0 D' Ar`p' D'Arp T = D2N2p 0(

Or

D'Arp

24. In turbomachinery, the relevant parameters are volume flow rate, density, viscosity, bulk modulus, pressure difference, power consumption, rotation speed and a characteristic dimension. According to Buckingham's pi(n) theorem, the number of independent non-dimensional groups for this case is — (a) 3

(b) 4

(d) 6

(c) 5

(IES 93) Here total numbers of variables (m) = 8 Number of primary dimension (n) = 3 Hence number of independent non-dimensional groups = m — n = 8 — 3 = 5 Answer (c) is correct. 25. Using Buckingham's 7r-theorem, show that the velocity through a circular orifice is given by—

where

D ti )1 V = ‘12g H fn(H , i pVH H = Head causing flow D = Diameter of the orifice p. = Coefficient of viscosity p = Mass density g = Acceleration due to gravity

(UPTU 2006-7) We have— V = f(H, D, p„ p, g) or F(V, H, D, Ix, p, g) = 0 Select H, g and p as repeating variables Ix + 1 = 3 + 1 = 4 while . Variables in each a- term will be— n —m=6—3=3 xi = Hal • gbi • pc1 • V 71 .. = Ha2 . gb2 . pc2 . D 2

n=6

Dimensional Analysis

7r3 =Ha1 gb

pc 3

Now putting dimensions in 7r1, we have— moLoTo = (LT-2)b

(moci

(L7-1)

Equating the power of M, L and p on both sides, we have— c1 = 0 al = — 1/2 bi = — 1/2

1r1

Or

_ H

2

—

V

•

g-1/2

po

V

Putting now dimensions in /r2, we have— 2 (LT-2)b 2 (M M°L°7° _ -3)`2 (L) Equating the powers of M, L and p on both sides, we have— c2 =0 a2 = — 1 b2 = 0 ir2 = H-1 g° p° D =

H Putting now dimensions in it3, we have— = gb3 •pc3 M° Lo La3 •

Equating the power of M, L and z on both sides, we have— C3 = —1

a3 = — —3 2 b3 = —1/2 7r3 = H-3/2 g-112

p-1

1-1

p• H3/ 2 • Vi

v• H • p =

or

X

x

gH

V•H•p

F

7C2, 1C3) = o

F

D (V „Ki x VHp gH H

—0

285

286

Fundamentals of Fluid Mechanics

Or

or

or

F ,V

D p.

gH ' H ' VH p — 0(D , p

V ..\IT-1 V=

0

)

H VH p

II T-1

0( 1) , ki H pVH

= ,j2gH ,,,i(

D p ) H ' pVH

Chapter

7

SIMILITUDE AND MODEL ANALYSIS

KEYWORDS AND TOPICS A A A A A A A A

PROTOTYPE MODEL MODEL TESTING MODELANALYSIS HYDRAULIC SIMILITUDE GEOMETRIC SIMILARITY KINEMATIC SIMILARITY DYNAMIC SIMILARITY

A A A A A A A A

SIMILARITY LAW MODEL LAW REYNOLDS MODEL LAW FROUDE'S MODEL LAW EULER' S MODEL LAW WEBER'S MODEL LAW MACH' S MODEL LAW DISTORTED MODEL

INTRODUCTION The behaviour of the hydraulic structures and machines can be predicted by performing tests on their models. Since prototypes are costly as compared to the models, any failure of model does not therefore involve much loss of material and human labour. Geometric similarity implies similarity of shape between model and prototype. Kinematic similarity means the similarity of motion. Dynamic similarity implies that the forces acting on the matching points of the prototype and its model are equal in magnitude and direction. Model testing is used in design of (i) aeroplanes, rockets and missiles (ii) harbours (iii) ships and submarines (iv) skyscrapers (v) turbines, pumps and compressors (vi) dams, weirs, spillways and canals and (vii) flood control measures. 1. What is a model? A model is small scale replica of the actual machine or structure. 2. What is a prototype?

288

Fundamentals of Fluid Mechanics

The actual structure or machine is called prototype. 3. Can a model be bigger than its prototype? It is not necessary that the models should be smaller than the prototypes. However in most of cases, models are smaller than prototypes. 4. What is the model analysis? Or What is difference between a model and prototype? (UPTU-2006-7) It can be appreciated that when hydraulic structure or machine has been built, it becomes very difficult to modify it to meet the desired requirements. In order to predict the performance of hydraulic structures or machines before they are built, it is advantageous to make their models and perform experiments on them to see that the performance is as required. In case any changes are necessary to modify the performance, it is very convenient to incorporate in the model. Hence model making and model analysis is a very useful technique in the fluid machines. 5. What are the advantages of model testing? The advantages of model testing are: (1) The behaviour and performance of the structures or machines can be predicted by performing tests on their models. Since prototypes are costly as compared to the models, any failure of model does not involve much loss of material and human labour. (2) It is possible to make models of a prototype based on alternative designs which facilitates selecting most economical, safe and sound design of the prototype. (3) Safety and reliability of a structure or machine can be ascertained by model testing before actually constructing or putting into the use of the prototype. (4) Model testing also helps in identifying the defects in existing structure or machines. 6. Which are the fields where model testing can be used? Model testing is used in following fields (1) Design of aeroplane, rockets and missiles. The models are tested in wind tunnels by subjecting them to similar air flow. (2) Design of harbour (3) Design of ships and submarines by testing their models (4) Design of skyscrapers by subjecting their models to wind loads (5) Design of irrigation channels (6) Design of turbines, pumps and compressors (7) Design of civil engineering structures such as dams, weirs, spillways and canals

Similitude and Model Analysis

289

(8) Design of flood control measures. 7. What is hydraulic similitude? For ascertaining the soundness and performance of the hydraulic structure or machine, it is most essential that the model should represent its prototype completely in all aspects. This similarity between the prototype and its model is known as hydraulic similitude. 8. What is the outcome of the hydraulic similitude? The outcome of hydraulic similitude is that the results of model tests can be successfully applied to the prototype of the hydraulic structure or machine. Similitude help us to assume that the flows of the fluid are mechanically similar for both prototype and its model. 9. What are the different similarities which should exist between models and prototypes? Or Discuss geometric, kinematic and dynamic similarities. Are these similarities truly attainable? If not why? (UPTU-2005-6) Models must reproduce the behaviour of the prototype. Hence it must possess following similarities: (1) Geometric similarity (2) Kinematic similarity (3) Dynamic similarity A model which satisfies all above similarities with its prototype is known as completely similar and true model. In practice, it is not possible to achieve complete similarity in models. It is, therefore, common to consider only those forces which are predominant in a phenomenon and design the model such that the same forces influence the flow phenomenon in the model also. The effects of all other forces which are insignificant are either neglected or considered by a correction factor based on experimentation. 10. What is geometric similarity? Geometric similarity implies similarity of shape between the model and prototype. The model is an exact replica of the prototype having identical shape but smaller in size. The model and prototype have same ratio for all corresponding linear dimensions but all included angles are same. For geometric similarity — hp = W P hm Wm

And angle

£Xp =

UM

= 60° in this case

LP

4„

Ls = length scale

290

Fundamentals of Fluid Mechanics

WP Proto Type

Model

Similarity leads to — (1) Area scale ratio: A,5 =

A

P— Am

L xW :— 4 xw

(2) Volume scale ratio: V — V

Vm

L xW xh s x wP x — 43

Lm

hm

11. What is kinematic similarity? Kinematic similarity means the similarity of motion. For obtaining similarity of motion, both the model and its prototype must produce identical time rates of change of motion. In other words, the ratio of velocities and accelerations of a fluid particle at certain points in the model and at the matching point of prototype must be same in magnitude and direction. For kinematic similarity, we must have: (1) Same velocity ratio, V, = VvP (2) Same acceleration ratio, a, —

a

p

am

(3) Same direction of velocity VP

vn, Aro.

Prototype

Model

Similitude and Model Analysis

29 I

12. What is dynamic similarity? The dynamic similarity implies that the forces acting on the matching points of the prototype and its model are equal in magnitude and direction. Dynamic similarity can be said to exist between the model and its prototype if the ratios of all forces acting on corresponding fluid particles or corresponding boundary surfaces of the model and prototype are identical. Both geometric and kinematic similarities are prerequisites for dynamic similarity For dynamic similarity: (a) Same force ratio, F, =

Fp F.

(b) Same direction of forces.

Prototype

Model

F 3

F W R 13- 13P F. W. P.

—

13. What are similarity laws or model laws? The similarity is difficult to be achieved in all respects as : (1) For the dynamic similarity, the ratio of corresponding forces acting at the matching points of the model and the prototype must be equal. (2) The dimensionless numbers should be having same magnitude for the model and prototype. But it is very difficult to have all dimensionless numbers such as Reynolds number, Froude's number, Euler's number, Weber's number and Mach's number same for both model and prototype. (3) The models are therefore designed on the basis of the few forces out of all forces such as inertia force, viscous force, elastic force, pressure force and gravitational force which are predominant in the flow situation. The laws on which models are designed for dynamic similarity are called model laws or similarity laws. The model laws are: (1) Reynolds model law (2) Froude's model law (3) Euler's model law (4) Weber's model law and (5) Mach's model law. 14. What is Reynolds model law? If the model is designed on the basis of Reynolds number then the model is said to be based on Reynolds model law. Reynolds model law is used whenever inertia and viscous forces

292

Fundamentals of Fluid Mechanics

are predominant in the fluid as compared to other forces. (Re)prototype = (ROmodel pp VP Lp

— pmV„, L„,

Pp P p Vp Lp

Pm

P. V. L. Pp

1

Pm

Pr Vr

4

1

Pr

r = scale ratio

15. Find the values of following by the Reynolds model law (1) acceleration scale ratio (2) force scale ratio and (3) discharge such ratio. (1) Acceleration scale ratio ar = but

a

aP a.

= Velocity _ V

Time

T

VP

.

Tp

a,.

_

Vm

VP Vm

T. = Vr Tr (2) Force scale ratio :

Fp F.

F,. _

mP aP mm am L3 a

— PP P P P

p m em am = P,. 4.3 ar3 (3) Discharge scale ratio,

Q, -

Qp

Qm

- pp

4 VP

Pm L,3,Vm = Pr 4.3 Vr3

-Tm Vr

Tp 1 ;

Similitude and Model Analysis

293

16. Where can we use Reynolds model law? The Reynolds model law can be used for the following fluid flow situations : (1) Flow of fluid in pipes, flow meters and fans (2) Flow around the submerged bodies such as aeroplanes and submarines. 17. Which is Froude's model law? Whenever inertia and gravitational forces are predominant in the fluid, the Froude's number for the model and the prototype must be equal for dynamic stability. The Froude's model law is based on the Froude's number for establishing similarity between prototype and its model. (nprotype = (flmodel

V.

Vp

1

1 gLp

1Wn

Vp V. _ 1 /Lp .\ Lm

or

V,. =1

Or

V4

or

V, = X

18. Find the following by the Froude's model law: (1) acceleration scale ratio (2) discharge scale ratio (3) force scale ratio and (4) pressure scale ratio. 1. Acceleration scale ratio:

or

ar

a am

—P

ar —

(V ) T p

Vp Vm

1

TP T.

M. =

,/4. 1

- 1 VL,.

Lp As

Vm

—

TP

L„,

T.

— Lr X

Tr '

294

Fundamentals of Fluid Mechanics

VP

but

Vm

Hence 2. Discharge scale ratio

P Qr-Q Qm

= V 1.,,

and

7; = V 4

[4) x ( T.) _ iLpy . Tp

ern

1,m )

1

(t)

=L 3 x 1 — L"2 r VT ,. r 3. Force scale ratio F_ r

L2 V 2 FP_ p P PP Fm Pm `Mt .2

Generally pp = pm = density of the fluid 2 L ( vp 2 F —( 1 x . r Lni v.) = 1,2 • Lr = Lr3

4. Pressure scale ratio Pp —

Pr — Pm

pp VP2 p,Vm2

Generally pp = pm = density of fluid ...

V

2

pr= (V) m P = Lr

19. What is Euler's model law? Euler's model law is applicable to the models which are designed on the basis of the Euler's number. Wherever pressure and inertial forces are predominant in a fluid, for dynamic stability the Euler's number of the model and its prototype must be equal. (E)prototype = (E)model

VP

Pp Pp Generally pp = pm = density of the fluid

Vm Ai Pm Pm

Similitude and Model Analysis

V

Vm

x

295

1

— 1 1p p Al Pm

20. Where can Euler's model law be applied? Euler's model law is applicable in the following fluid mechanics situations: (1) (2) (3) (4) (5)

To avoid water hammer phenomenon Ascertaining pressure distribution on ship To avoid cavitation phenomenon To ascertain pressure forces on aircraft wings and fan blades To ascertain fully turbulent flow in case of closed pipes.

21. What is Weber's model law? The models which are based on the Weber's number, they obey Weber's model law. Whenever inertia and surface tension forces are predominant in a fluid, for dynamic similarity Weber's number for the model and its prototype must be equal. (W)protoype = ( W)model Vp — Vm I 6P I

6mm ii Pm 1-m i

p p Lp

22. Where can Weber's model law be applied? Weber's model law can be applied for: (1) (2) (3) (4)

Study of capillary movement of water in soil Study of flow over weirs for small heads Study of droplets and very small jet Study of capillary waves in channels

23. What is Mach's model law? Mach's model law is applicable for the models which are based on Mach's number. Whenever inertia and elastic forces are predominant in the fluid, Mach's number for model and its prototype must be equal for dynamic stability. Wprototype = Mmodel Vp - Vm

I

IE p

Em

Pp

Al Pm

where E = Elastic modulus

296

Fundamentals of Fluid Mechanics

24. Where can Mach's model law be applied? The Mach's model law can be applied for: (1) (2) (3) (4) (5)

Study of flow of air on the aeroplanes with supersonic speed Study of movement of torpedoes underwater Overcoming the problem of water hammer phenomenon Study of movement of rockets and missiles aerodynamic testing.

25. Oil of kinematic viscosity 5 x 10-5 m2/s is to be used in a prototype in which both viscous and gravity forces dominant. A model scale of 1:4 is also desired. What viscosity of model liquid is necessary to make the both the Froude's number and Reynolds number the same in model and prototype? If Reynolds number is same, then Pr lir 4

Pr Also Froude's number is same.

—1

or

Vr Lr _ 1

Vr

V, = 11 4. L, = 4

Given

V,. = X. = Vi =2 Now

Vr 4

Vr

—1

2x4 —1 Vr V = 1 r

or

8

vp = 1 8 v„„ V„,= 8vp = 8 x 5 x 10-5 m2/s = 40 x 10-5 m2/s

26. A 1:20 model of submarine is to be tested in towing tank containing salt water. If the submarine moves at 30 km/h, at what velocity should the model be towed for dynamic stability. Since the submarine is fully submerged, hence viscous force is predominant and Reynolds model rule is applicable. 1,,. = 20 Given Pr = 1

Similitude and Model Analysis

297

Air = 1 Now

4 =1

Pr Vr Pr

1 x Vr X20

1

—1

or

1 Vr 20

or

Vp = 1 Vn., 20 V„, = 20 x Vp = 20 x 30 = 600 km/h

or

27. A model of reservoir is drained in 5 min by opening a sluice gate. The model scale is 1:256. How much time would it take to empty the prototype? Here gravitational force is predominant and we have to apply Froude's model law. Vr

— 1 and given 4 = 256

114 yr = V256 = 16 But,

4 - Vr Tr

L,. = 256 _ 16 Vr 16

T = r

T

—P — Tr T.

T P

5

— 16

Tp = 5 x 16 = 80 min 28. A rectangular pier in river is 2 m wide and 4 m long. The average depth of water is 3 m. A model is built to a scale of 1:16. The velocity of flow in the model is 1 m/s and the force acting on the model is 5 N. Find (a) the values of velocity and force on prototype (b) the height of the standing wave at the pier if it is 0.09 m in model and (c) coefficient of drag resistance: As gravitational force is predominant, Froude's model law is applicable

V,. = L. 4 = 16 (given) .

V,.= ,5 =4

298

Fundamentals of Fluid Mechanics

.

V =4 V„, Vp = 4 x V„, = 4 x 1 = 4 m/s

Force acting is gravitational force which is

. Now

F = mg = pL3g F,.= 4.3 = 163 = 4096 Fp = 4096 x 5 N = 20480 N V,. = .,/,,. = XX =4 X, = 4 x

or

= 4 x A/0.09 = 4 x 0.3 = 1.2 m hp = 1.44 m = standing wave height Drag force = CD pA v 2 2 Drag force = 5 = CD X 1 x 103 x ( 2 x 3 ) x 12 16 16) 2 ,

5x2x16x16 103 x 6

k--1)

= 42.67 x 10-2 = 0.427 29. A model of reservoir emptied in 10 min. If the model scale is 1:25, the time taken by the prototype to empty itself could be: (a) 250 min

(b) 50 min

(c) 6250 min

(d) 2 min (IES-94)

Tr= 71'' = 114

L,.= 25 (given) 7',. = A25 =5

Similitude and Model Analysis

299

T Tr =

=5

Tm

Tp = 5 X Tm = 5 x 10 = 50 min

Answer (b) is correct 30. A ocean liner 250 m long has maximum speed of 15 m/s. The towing speed of a model 10 m long to simulate the wave resistance should be: (a) 15 m/s

(b) 3 m/s

(c) 5 m/s

(d) 0.5 m/s (NTPC 90) Guidance: The wave resistance is a function of Froude's number. In order to determine the model speed, Froude model law is used V, _ 1 AT ,r L,. =

250 — 25 10

V,. = 25 =5 Vp lin,

=5

Vm = 5 = 3 m/s Answer (b) is correct. 31. A hydraulic model of a capillary is constructed with a scale 1:10. If the prototype discharge is 2048 m3/s, then the corresponding discharge for which the model should be tested is (a) 1 m3/s

(c) 4 m3/s

(b) 2 m3/s

Here Froude model law is applicable

V, _ 1 lilt V. = .11 =4 Qr=

L3 Tr

L,. = 16, T,. —

4 16 —

V,.

4

=4

(d) 8 m3/s (Civil Services 96)

300

Fundamentals of Fluid Mechanics

Q,. — 16 3 = 1024 4 Qp

Q.

= 1024

Q _ 2048 _ 2 m3/s Qm 1024

... Answer (b) is correct

32. What flow rate in m2/s is needed using a 20:1 scale model of a dam over which 4 m3/s of water flows? (a) 0.010

(b) 0.0068

(c) 0.0047

(d) 0.0022 (IES 90)

Froude model law is to be used. Vr 11L,

_1

V, = X. = VY) = 4.472

4 Tr - Vr Qr =

20 - 4.472 4.472

(4)3

203

T,.

4.472

= 1.789 x 103 Qp

= 1.789 x 103

Qm

Qm —

4 — 2.2 x 10-3 1.789 x 10'

= .0022 m3/s Answer (d) is correct 33. In 1:20 model of a stilling basin, the height of the hydraulic jump in the model is observed to be 0.2 m. The height of the hydraulic jump in the prototype will be: (a) 1 m

(c) 4 m

(b) 2 m

(d) 8 m (IES 90)

Given Now

L,. = 20 h hm

= L = 20 r

Similitude and Model Analysis

.

30 I

hp = 20 x 0.2 = 4 m Answer (c) is correct

34. The model of a propeller, 3 m in diameter cruising 10 m/s in air, is tested in a wind tunnel on a 1:10 scale model. If the thrust of 50 N is measured on the model at 5 m/s wind speed, then the thrust on the prototype will be: (a) 20,000 N

(b) 2000 N

(c) 500 N

(d) none (IES 95)

Fr = prV 2Lr2 0\2 ( 0\2 =1x( 5) 5) 104 5 Fr — Fp = 104 F, 52 FP

Fm X 104 25 = 20,000 N

—

50 x 10000 25

Answer (a) is correct 35. What do you understand by distorted model? Distorted model is one which has its one or more characteristics not similar to the corresponding characteristics of prototype. A model having different horizontal and vertical scale ratio is a distorted model. In order to predict the performance of prototype, the law of distortion have to be applied to the results obtained from the model test. The distortion can be (1) geometric distortion i.e. horizontal and vertical scales are different (2) hydraulic distortion i.e. fluids in model and prototypes are different (3) material distortion i.e. materials in model and prototype are different and (4) configuration distortion i.e. slope in model and prototype is different. 36. How are scale ratios for distorted models utilized for finding (1) velocity (2) vertical cross-section (3) discharge and (4) time of flow. In case the horizontal scale for non-tidal model be 1: m, then LP _ Bp = m (where L = length & B = breath) L, An In case the vertical scale is 1: n, then h

P hm

=

n (where h = height)

302

Fundamentals of Fluid Mechanics

Generally the horizontal scale is greater than vertical scale i.e. m > n (1) Velocity ratio: As per Froude's law

V.

= V, =

h ( i' ) = VT2 hin

A is vertical cross-section, then (A _ Bp x hp — mxn APm vertical B. x kn. If Q is the discharge, then Qp _ Vp x AP — .1 , 2 XmXn Q. V. x A. = m n3/2 If T is the time of flow, then LP Tp = VP LP v — x m T. L. L. VP V.

(2) If

(3)

(4)

T —

Tr

m AITI

37. A horizontal model has horizontal ratio of 1:2500 and a vertical scale ratio of 1:225. If the flood peak requires 10 hours to travel a distance y 1 km in the model, find time the flood peak will take in the actual river (Patna University)

m = 2500 & n = 225

Here

T — 2500 r V225 TP

T. TP

2500 15

_ 500 3 500 x 10 — 1666.67 hours 3 = 69 days 11 hours 40 min

38. A model of spillway is built to a scale of 1:36. If the model velocity and discharge are 1.25 m/s and 2.5 m3/s respectively, what are the corresponding values for the prototype? (Punjab University) Applying Froude's model law

Similitude and Model Analysis

Vr =

303

Lr

=6 = Qr = Vr(4)2 = 6 x 362 = 7776 V =6 „, Vp = 6 x 1.25 = 7.50 m/s

V,. =

Yr =

Qp

Qm

= 7776

Qp = 7776 x 2.5 m3/s = 19440 m3/s 39. A model of a reservoir is drained in 6 min by operating the sluice gate. How long should it take to empty the prototype if the scale ratio is 1:256? (Poona University) Applying Froude's model law Vr = I 4. = .1256 = 16 L T Tr — r — 256 — 16 16 VP = 16

Tr = m

or

Tp = 16 x 6 = 96 min

40. A geometrical model of a surface vessel is tested in a laboratory. The linear scale of the model is 1/49. It is observed that with a speed of 10 m/s, the resistance of the model is 3 N. The liquid used for the test is the same as that one which the surface vessel is to sail. Calculate the corresponding speed and the resistance of motion of the surface vessel. Consider the effect of gravity only. (Punjab University) As effect of gravity is to be considered, the Froude's model law is applicable. V,. = AiLr = 149 = 7 V P = 7, or Vp = 7 x 10 = 70 m/s Vm Fr = PT V,2 1,2 FT = resistance force scale

304

Fundamentals of Fluid Mechanics

= 72 x 492 Fp = 49 x 492 x 3 as Fm = 3 N (given) = 353 lth

Now

41. Estimate (a) the speed of rotation (b) the thrust produced (c) the torque developed and (d) the efficiency of propulsion by a 3 m diameter propeller to cruise at 10 m/s if a 1/10 scale model produced the following results V„, = 5 m/s, N„, = 750 rpm Fm = 50 N and T„, = 10 Nm For dynamic similitude N, dr _ , Vr

1

dr = 10 (given) Vp 10 V — — =2 r V. 5

N, x10 _ 1 2

Np = 5 x 750 = 150 rpm Now

Fr. = Pr Vr2 Nr2 where Fr = thrust ratio scale. = 1 x 22 x 102 = 400 F p = 400 x 50 = 20 kN where Fp = thrust produced Tr = Torque ratio = pr N,?4,5 2

= 1 x (7 15 0)

X

(10)5

100 X 103 25 = 40 lcN Prototype efficiency

Fp xVp T x2n-N160

20 x 10 x 60 40x2rcx150

= 31.8% Model efficiency

F. x V. T x 2irN 60 = 31.8%

50 x 5 x 60 10 x 27 x 750

Similitude and Model Analysis

305

42. The drag of a small submerged hull is desired when it is moving far below the surface of water. A 1/10 scale model is to be tested. What dimensionless group should be duplicated between model and prototype and why? If the drag of the prototype at 1 knot is desired, at what speed should the model be moved to give the drag to be expected by the prototype? Would this result still be true if this prototype were to be closed to the surface? Explain. When the body is moving fully submerged, the viscous forces are predominant and Reynolds model law is applicable. If the prototype is moving close the surface, then gravitational forces are also acting due to wave resistance, thereby drag force also contributes to the resistance Using Reynolds model law PrlIr

Pr Now Lr = 10 and p,.= 1 and pr = 1

4 =1

V,L, = 1 Or

Now

1 V = r 10 Vp _ 1 V,, 10 Vp = 1 Knot V„, = 10 Vp Knot = 10 Knot Fr = Drag ratio = p,. V,.2 4.2 =1x

100

x 100

=1 Hence Drag force experienced by the prototype at 1 Knot and model at 10 Knot is same. 43. Calculate the speed of rotation, the torque produced and the power of a windmill of 5 m diameter in a wind speed of 30 km/hr from the performance of a 1/10 scale geometrical model in a wind tunnel with 10 m/s free stream. The model rotated at 1200 rpm produced a torque of 2.5 Nm. Apply speed parameter for the model & the prototype. Nr Dr _ 1 Vr Dr =

D P =10 Dm

Vp = 30 kMh

306

Fundamentals of Fluid Mechanics

30 x 103 — 8.33 m/s 3600 V. = 10 m/s V,= — VP — 8'33— 0' 833 . 10 V Now

N, Dr _ 1 V, N = Vr r Dr _ 0.833 — 0.0833 10 Np = 0.0833 A Iry, Np = 1200 x 0.0833 = 100 rpm Tr = Torque ratio = Pr Nr2 D,5 5 = 1 x ( 100 2 (10) 1200) L=T X X 105 m 144 ` = 2'5 x 100 kNm 144 = 1.736 kNm

27r Power of wind mill =TP xw=T x — Np P 60 g = 1736 x 100 x 2 60 = 18.2 kW 44. An offshore oil drilling platform is expected to encounter waves of 4 m height at 0.1 Hz frequency and a steady current of 1 m/s. Determine the parameters for the model wave channel where a one-sixteenth model of the platform can be tested. Froude model scale is to be applied as gravitational force is predominant V,.

Xr

— 1 or V,. = .X.

Similitude and Model Analysis

307

V,. = 16 =4 Vp = 1 m/s (given)

Now

V P — 0.25 m/s '1' 4 hr = L, = 16 V

—

hp 4 h=— = = 0.25 m '1' 16 16 Frequency ratio

=

V, f= I Tr —4

, 4 1 r 16 4

J

f

1 4 f„, = 4 x fp = 4 x 0.1 = 0.4 Hz Jr

45. An automobile moving at a velocity of 40 km/hr is experiencing a wind resistance of 2 kN. If the automobile is moving at a velocity of 50 km/hr, the power required to overcome the wind resistance is— (a) 43.4 kW

(b) 3.125 kW

(c) 2.5 kW

F=

or

p•

(d) 27.776 kW (IES — 2000)

V2 • d2

F2

p V22 d 2 = (172 )2

F1

P 1712 d 2

F2 = 2

(50)2 40

171

= 3.125 kN () 3 Power = 3.125 x 50 x10 kW 3600 = 43.4 kW Option 'a' is correct. 46. A 1 : 256 scale model of a reservoir is drained in 4 min by opening the sluice gate. The time required to empty the prototype will be

308

Fundamentals of Fluid Mechanics

(a) 128 min

(b) 64 min

(c) 32 min

(d) 25.4 min (IES — 1999)

t tp tn,

p hm \ 256 / 1

tp = 256 x tm tP = 16 x 4 = 64 min Option (b) is correct. 47. A ship model 1/60 scale with negligible friction is tested in a towing tank at a speed of 0.6 min/s. If a force of 0.5 kg is required to tow the model, the propulsive force required to tow prototype ship will be— (a) 5 MN

(b) 3 MN

(c) 1 MN

As per Froude's law for dynamic similarity, we have— = Vgi or Now or

Vm VP

/m i 1 /P = \ 60

F = p v2 L2 )2

[Fmj_ (Vmj 2 FP VP

Lm

L. )2

Fc = Frn x

Vm

= 0.5 x 9.81 x (VW3)2 x (60)2 = 0.5 x 9.81 x 63 x 103 = 1.06 x 106 N

(d) 0.5 MN (IES 1999)

Similitude and Model Analysis

309

= 1.06 MN 1 MN Option 'c' is correct. 48. A model is to be conducted in a water tunnel using a 1 : 20 model of a submarine, which is to travel at a speed of 12 km/h deep under sea surface. The water temperature in the tunnel is maintained so that its kinematic viscosity is the half that of seawater. At what speed is the model test to be conducted to produce useful data for the prototype? (a) 12 km/h

(b) 240 km/h

(c) 24 km/h

(d) 120 km/h (IES 2002)

(Re)model = (Re)prototype

(pVd) = [13.174

/-1

1).

dp

VP, = VP

[gm)

pp

pp

pm

[ dm

= 12 ( 2°) (1 ) 1 2 = 120 lan/h Option 'd' is correct.

Chapter

8

FLUID KINEMATICS

KEYWORDS AND TOPICS A A A A A A A A A A A A

FLUID KINEMATICS STREAM LINE STREAM TUBE STREAK LINE PATH LINE EQUIPOTENTIAL LINE ROTATION DISTORTION VORTICITY CIRCULATION ROTATION & IRROTATIONAL FLOW STEADY & UNSTEADY FLOW

A A A A A A A A A A A A

UNIFORM & NON-UNIFORM FLOW LAMINAR & TURBULENT FLOW CONTINUITY EQUATION ONE DIMENSIONAL FLOW TWO-DIMENSIONAL FLOW THREE-DIMENSIONAL FLOW STREAM FUNCTION POTENTIAL FUNCTION LAPLACE EQUATIONS FLOW NETS FREE VORTEX FORCED VORTEX

INTRODUCTION Fluid kinematics is a branch of fluid mechanics which deals with the study of velocity and acceleration of the particles of fluid in motion and their distribution in space without considering any force or energy involved. Fluid kinematics provides an idea about (i) the rate of flow and (ii) type of flow. The fluid motion can be described using (i) Lagrangian method and (ii) Eulerian method. The Eulerian method is commonly adopted. The motion of fluid is also described by various lines of flow such as (i) path line (ii) stream line (iii) stream tube (iv) streak line and (v) potential line. The flow can be (i) uniform or non-uniform, (ii) laminar or turbulent, (iii) compressible or incompressible, (iv) rotational or irrotational (v) steady or unsteady and (vi) onedimensional, two-dimensional and three-dimensional.

Fluid Kinematics

311

1. What is fluid kinematics? Fluid kinematics is a branch of fluid mechanics which deals with the study of velocity and acceleration of the particles of fluids in motion and their distribution in space without considering any force or energy involved. Kinematics provides — (1) an idea about the rate of flow which is also called discharge (2) an idea about different types of velocities of flow. 2. What is the first and fundamental equation of flow? The equation of continuity is the first and fundamental equation of flow. 3. What is fluid motion? What are the methods used to describe fluid motion? In fluid motion, each fluid particle has its own velocity and acceleration at any point of time. Velocity and acceleration of a particle may change with time or in change the position of the particle during the flow. Two methods are used in describing fluid motion viz. (1) the Langrangian method and (2) the Eulerian method. 4. What is the Langrangian method of describing fluid motion? In Langrangian method, the motion of each particle is defined with respect to its location in space and time from a fixed position before the start of the motion. The movement of single particle is observed which gives the path followed by the particle. The position of the fluid particle in space (x, y, z) at any time from its fixed position (a, b, c) at t = 0, shall be x = fi (a, b, c, t) y = f2 (a, b, c, t) z = f3 (a, b, c, t) From the position, velocities, u, v and w and acceleration ax, ay and az can be given as —

ay & w = — ax (1) Velocities u = ax — v= atat , at a2 a2x , ay — at2' & az — ate at2 The resultant of velocities and acceleration can be calculated. The resultant of pressure and densities can also be worked out. a2x

(2) Accelerations a, —

5. What is the Eulerian method of describing fluid motion? In this method, instead of concentrating on single fluid particle and its movement, concentration is made on all fluid particles passing through a fixed point. The changes of velocity, acceleration, pressure and density of the fluid particles as they pass through the fixed point are observed. Hence the Eulerian method describes the overall flow characteristics at various points as fluid particles pass through them. The velocity at any point (x, y, z) can be given as u = f1(x, y, z, t), v = f2 (x, y, z, t), w = f3 (x, y, z, t)

3I2

Fundamentals of Fluid Mechanics

6. What is the different between Langrangian and Eulerian method? Or Differentiate between Eulerian & Langrangian approach (UPTU-2004-5) Langrangian

Eulerian

1. Observer concentrates on the movement of single particle 2. Observer has to move with the fluid particle to observe its movement 3. The path and changes in velocity, acceleration, pressure and density of a single particle are described 4. Not commonly use 2...,/,/z/z.zz/z/z,/,//z7_,,

1. Observer concentrates on various fixed point particles 2. Observer remains stationary and observes changes in the fluid parameters at the fixed point only 3. The method describes the overall flow characteristics at various points as fluid particles pass. 4. Commonly used o

Po' p

Po

/A.,/

p

P ///////////////////////

7. What are various lines of flow to describe the motion of the fluid? The various lines of flow are: (1) path line (2) stream line (3) stream tube (4) streak line or filament line and (5) potential lines. 8. What is a path line in fluid motion? What happens to path line during steady and unsteady flow? The path traced or taken by a single fluid particle in motion over a period of time is called its path line. Hence path line is a locus of a fluid particle as it moves along. The path line shows the direction of velocity of the particle as it moves ahead. During steady flow, path line concides with the stream line as there is no fluctuation in velocity. However the path line fluctuates between different stream lines during an unsteady flow. 9. What is a stream line? A stream line is an imaginary line drawn through a flowing fluid such that the tangent at each point on the line indicates the direction of the velocity of the fluid particle at that point. For example, if ABC is a stream line, then the direction of velocities of A, B and is C as shown

C A A Stream line

Velocities of A, B & C

Fluid Kinematics

3I3

10. Explain the following (1) the value of velocity at right angle to the stream line (2) can there be any flow across the stream line and (3) can two stream lines cross each other? Since velocity of the fluid particle at any point on the stream line is tangential to the stream line, there cannot by any component of velocity normal or right angle to the stream line i.e. the component of velocity at right angle to the stream line is zero. There cannot be any flow across the stream line as the flow is always tangential to the stream line Two stream lines cannot cross each other as otherwise there would be two velocities at that point, one each tangential to the stream lines. This is inconsistent with the definition of a stream line. 11. How does a stream line behave in the vicinity of a solid surface? The stream line in the vicinity of a solid surface conforms to the outline of the boundary surface. For example, the stream lines around a solid cylinder and within a closed conduit are as shown in the figure ////////////////

,,,iymocow ,_,------,------

Stream line around a solid cylinder

Stream lines within a closed conduct

12. What happens to stream lines during steady and unsteady flow conditions? In steady flow, the pattern of stream lines remains unchanging with time. However the pattern of stream lines may or may not remain same with time for unsteady flow. If unsteady flow is due to the change of magnitude of velocity, the stream line pattern remains invariant (unchanging) with time. However if the unsteadiness due to the change in the direction of the velocity, the stream line pattern cannot remain same. 13. What is a stream tube? The stream tube consists of stream lines forming its boundary surface. The stream tube is defined as a circular space formed by the collection of stream lines passing through the perimeter of a closed curve in a steady flow. As stream tube is bounded on all sides by stream lines therefore no fluid can enter or leave the stream tube from the sides except from the ends. Hence stream tube behaves as a solid surface tube. The general equation of continuity can be applied on stream tube though it has no solid boundaries. Stream tube may be of regular or irregular shape. The velocity is uniform throughout the cross-section of the stream tube which necessitates small cross-section for the stream tube. In steady flow with

3I4

Fundamentals of Fluid Mechanics

uniform velocity, all stream lines are straight and parallel. The contents of stream tube are called current-filaments.

Stream tube

14. What do you understand from streak line or filament line? It is an instantaneous picture of the positions of all fluid particles in the flow which have passed or emerged from a given point. The line of smoke from a cigarette or from a chimney is nothing but a streak line 15. What is a potential line? The lines of equal velocity potential are called potential lines. These potential lines cut stream lines orthogonally i.e. a stream line and a potential line are at right angle to each other. 16. Distinguish between stream lines, streak lines and path lines (UPTU 2001-2) Stream lines 1. Imaginary lines showing positions of various fluid particles 2. Particle may change stream line depending on type of flow 3. Stream lines cannot intersect each other, they are always parallel 4. No flow across stream line

Path lines

Streak lines 1. Real line showing instantaneous positions of various particles 2. May change from instant to instant

1. Real line showing successive position of one particle. 2. Particle may cross its path line

3. Streak line changes with time. Two streak lines may intersect each other 4. Flow across streak line is possible

3. Two path lines for two particles may intersect each other 4. Flow across a path line is possible by other particles.

17. What are different types of displacements of fluid particles? Any fluid element can be subjected to translation, rotation or distortion during its course of motion. In translation the fluid element moves to another position in the same direction. For example, translation takes place when liquid flows through pipes or in open channel. Pure translation does not cause any stress in the element.

Fluid Kinematics Element at original position

Original position D

Element at new position

D

3I5

New position

D'

A' Translation

Translation

In rotational displacement, the fluid element is subjected to the rotation as shown in the figure. No stress is caused as there is no relative movement of the element with respect to the rotating system. AB and AC are rotating in the same direction to AB' and AC'. Angular deformation is zero

de — de au) — 0 and shear stress is zero as strain rate = 2 — —0 2 1 ax ax ay D' C

C'

\\ I

\ de

D

B'

- de) A

B

Distortion can be (1) angular or (2) volume distortion as shown in the figure. Stresses are produced when fluid element is distorted. In angular distortion, stress induced is shear stress due to shear strain as given by the angle of distortion. Angular distortion is de

l + c/92 and 2

shear strain rate = 1( v — 2 ax ax). C

D

C

C'

del

A

D'

A Angular distortion

Volume distortion

3I6

Fundamentals of Fluid Mechanics

18. What are different types of fluid flow? The fluid flows can be : (1) steady or unsteady (2) uniform or non-uniform (3) laminar or turbulent (4) compressible or incompressible (5) rotational or irrotational (6) one-, two- or three-dimensional 19. Define steady and unsteady flow with one practical example (UPTU-2002-3) The flow in which fluid characteristics like velocity, acceleration, pressure or density do not change with time at any point in the fluid is called steady flow. A flow of water with constant discharge through a pipeline is an example of steady flow. For steady flow, we have the following: a2 v dP ap aV —0 =0—=0 =0 at ' at' at ' at The flow in which fluid characteristics like velocity, acceleration, pressure and density change with time at any point is called unsteady flow. The water with varying discharge through a pipe is an unsteady flow. For unsteady flow, we have: aV #0 a2 v #0 dP #0 & ap #0 — ' at at ' ate ' at In steady flow, the path line and stream line will concide. In unsteady fluid flow, the path line of successive particle will be different and stream line pattern of the flow will be changing at every instant.

20. What do you understand by uniform and non-uniform flow? Uniform flow is the flow in which velocity of the flow does not change along its direction of flow at any point of time. A flow through a constant diameter pipe line is an uniform flow. For uniform flow, we have av — = 0 where v = velocity and s = distance in direction of flow. as Non-uniform flow is the flow in which velocity of the flow changes in the direction of the flow at any instant of time. Flow through a pipe having variying cross-section is a nonuniform flow. For non-uniform flow, we have: dv #0 ds

-----------4///////////////////////////// Uniform flow

Non-uniform flow

Fluid Kinematics

3I7

21. Define laminar and turbulent flow with one practical example. (UPTU-2002-3) A laminar flow is one in which the fluid particles move in layers with each layer sliding over the other. There is no movement of fluid particles from one layer to another. Flow of blood in small veins and oil flow in bearings are examples of laminar flow. It is a smooth regular flow where velocity of flow is very small

Laminar flow

The flow in which adjacent layers cross each other and the layers do not move along the well defined path is called turbulent flow. The flow through rivs or canals, smoke from chimney or cigarette are turbulenter flow. ///////////////////////////

///////////////////////////

Turbulent flow 22. How can flows be classified laminar or turbulent or transit flow? The flow can be classified by Reynolds number. The Reynolds number is the ratio of inertia force to viscous force. The viscous force tends to make the motion of fluid in parallel layers while inertia force (mass x acceleration) tends to diffuse the fluid particles. Higher viscous flows are laminar flow which have lower value of Reynolds number. Low viscous force (higher increase in velocity or inertial force) are turbulent flow which have higher values of Reynolds number. pVL Reynolds number = Inertia force = Viscous force P For flow in pipes, the flow can be: (a) Laminar flow if Reynolds number < 2100 (b) Transit flow if Reynolds number is in between 2100 to 4000 (c) Turbulent flow if Reynolds number > 4000 23. What do you understand by compressible and incompressible flows? The flow in which the volume and thereby the density of fluid does not remain constant

3I8

Fundamentals of Fluid Mechanics

during the flow is called compressible flow. Gases are most compressible. The compressibility related problems are observed when gases flow through orifice, nozzle, turbine, compressor and in flight of planes. The flow in which the changes in volume and thereby in the density of the fluid are insignificant, the flow is said is to be incompressible. Flow of liquids are incompressible flow. For gases, in subsonic aerodynamic conditions air flow can be considered incompressible. 24. What are rotational and irrotational flows? The rotational flow is a flow in which the fluid particles also rotate about their own axis while flowing along stream lines. Motion of a liquid in a rotating cylinder is a rotational flow. The irrotational flow is a flow in which fluid particles do not rotate about their own axis while flowing along stream lines. Flow of water in emptying wash basin is a irrotational flow. The other example of irrotational motion is that of carriages on a giant wheel used for amusement rides. Each carriage describes a circular path as the wheel revolves but it does not rotate with respect to its own axis.

,B) Element AB circling bot not rotating Irrotational motion

I

Element AB circling and rotating

Rotational motion

25. Differentiate one-, two- and three-dimensional flow. 1. One-dimensional flow: One-dimensional flow is a flow in which the velocity of the flow is a function of time and one space coordinate (x, y or z). The flow through a pipe is one-dimensional flow. For one-dimensional flow, we have for (a) Steady flow: V = f(x) (b) Unsteady flow: V = f(x, t)

One dimensional flow

Fluid Kinematics

3I9

2. 7Wo-dimensional flow: Two-dimensional flow is a flow in which the velocity of the flow is a function of two space coordinates (xy, xz or yz) and time. The flows between two parallel plates of large extent, flow over a long spillway and flow at the middle part of the wing of an aircraft are two-dimensional flows. For two-dimensional flow, we have for: (a) Steady flow: v = f(x, y) (b) Unsteady flow: v = f(x, y, t)

X

Two dimensional flow

3. Three-dimensional flow: Three-dimensional flow is a flow in which the velocity of the flow is a function of three space coordinates (x, y, z) and time. Flow in converging or diverging pipe section is a three-dimensional flow. V= f (x, y)

X

Three dimensional flow

For three-dimensional flow, we have for — (a) Steady flow: v = f(x, y, z) (b) Unsteady flow: v = f(x, y, z, t) 26. What is discharge or rate of flow (Q)? The discharge is the amount of fluid flow per unit time. Hence discharge is Q _ Volume Time Area x Length Time = Area x velocity =AxV The discharge can also be defined as cross-sectional area of the flow multiplied by the velocity of flow.

320

Fundamentals of Fluid Mechanics

27. What is the principle of conservation of mass? The principle of conservation of mass states that matter cannot be created or destroyed in non-nuclear processes. 28. What is the equation of continuity of flow? The equation of continuity of flow is based on the principle of conservation of mass. It states that the mass of fluid entering the stream tube from one end must be equal to the mass of fluid leaving the stream tube at the other end per unit time and there is no accumulation of fluid in the stream tube. 29. Derive the equation of continuity for one-dimensional steady flow based on stream tube concept. The continuity equation can be derived based on stream tube with the following assumptions: (1) steady flow (2) uniform flow (3) incompressible flow and (4) onedimensional flow.

D 1-1 section

Consider a stream tube as shown above. Fluid enters tube from section 1-1 and exits from section 2-2. Let ds be distance between section 1-1 and section 2-2. Rate of increase of mass within the stream tube = mass of fluid entering — mass of fluid leaving Mass of fluid entering = pAV where A = cross-sectional area and V = velocity Mass of fluid leaving = pAV +

a

(pAV) ds

Rate of increase of mass within stream tube = pAV — (pAV + dd (pAV)ds s

a (pA•V)ds — as But rate of increase of mass between section 1-1 and section 2-2 =

a at

(density x volume)

Fluid Kinematics

=

at

321

(p x Ads)

By law of conservation of mass, we have: Rate of mass increasing in stream tube = Difference of fluid entering and leaving the stream tube

a

pAds = —

a as

(pAV)ds

—a 0A) + aas (pAV) = 0 at This is continuity equation. For different conditions, we have: (a) Steady flow:

a (pA) = 0, hence a a (pAV) = 0 a t s

(b) For steady and incompressible flow, p is constant As

a (pAV) = 0 — as a

or or or

a s

(AV) = 0

(p = constant)

AV = constant (on integration) A1 V1 = A2V2

30. Derive the continuity equation in three-dimensional in cartesian coordinates (UPTU 2003-4) or Derive the expression for the continuity equation for the steady-state 3D flow of a compressible fluid. (UPTU 2009-10) Derivation is based on the principle of conservation of mass which states that the quantity of fluid per second remains constant when fluid flows through any section of the pipe.

x

Fluid Element

Consider a fluid element as shown in the figure above, having lengths dx, dy and dz in x, y and z direction. Assume u, v and w are inlet velocities along x, y and z direction respectively. p is the density of fluid element at a particular instant.

322

Fundamentals of Fluid Mechanics

Mass of fluid entering from face ABEF =pxux(dyxdz) Mass of fluid leaving from face CDGH = (pudydz) +

x

(pu • dy dz) dx

a u • dy dz)dx] — (pudy • dz) Rate of mass increase in x-direction = [(pu • dy dz) + —(p ax — ax (pu)dxdydz Similarly, we can find for y and z direction as under Rate in mass increase in y-direction = Rate in mass increase in z-direction =

a ay

a

az

(pv)• dxdydz (pw)dxdy dz

[a

a

a

Total rate in mass increase = dx • dy • dz —(pu) + — (pv) + ..(pw) a ay dZ As there is no accumulation of fluid mass, there is no mass increase as per the law of conservation of mass.

Hence

or

(Pu) + a(Py) + (Pw) 1 dx • dy • dz = 0 LPa ay az a(pu)

a(pv)

ax

ay

a(pw) — 0 az

If p = 0 for incompressible fluids au ± Dv aw _ 0

ax

ay az

This is continuity equation for three-dimensional flow. For two-dimensional flow, velocity variation along z-direction is zero i.e.

az

=0

au + av 0 is continuity equation in two-dimensional flow. ax ay = 31. Derive the continuity equation in two dimensions in polar coordinates for incompressible fluids.

Fluid Kinematics

Vo

av + age

dO Vr +

323

avr dr ar

r

H

r + dr

Consider a fluid element ABCD between radius of r and r + dr which is subtending an angle dB at origin 0. Left Vr is radial velocity and Vo is tangential velocity. Assume the thickness of the fluid element as unity. Considering the flow in radial direction: Fluid entering from AB face = p x Velocity V. x Area x time = p (x V,.(rd0 x 1)dt Fluid leaving from CD face = p Vr + sa dr (r + dr) dO dt Dr Mass of fluid accumulated = pV,.r dO • dt —[p(Vr +7; dr)(r + dr)d0. dt]

= — [pvr dr • de. dt+ p

av r xrxdrxdOxdt ar i [neglect (dr)2 term]

a vr = —p[Vr + r ar idr • dO • dt vr a vr = —p[ + ] x dr x (rd0)•dt r ar Now considering the flow in tangential direction in the selected fluid element ABCD, we have Mass entering from face BC = pVe • dr • dt Mass leaving from face AD = p (V6, +

ave de)• dr •dt ae

324

Fundamentals of Fluid Mechanics

Mass accumulated = pVQ. dr -dt — p (vo + aa1708 do) x dr • dt = —p x = —p x

a y, ae

x dO x dr x dt

alle (rd0 x dr x dt) r ae

_ p aVe (rde x dr x dt) r ae

a (mass) x dt Change of fluid mass in time dt in the element = — at =

at

(p x volume) x dt

= at (p x rde x dr) x dt

By the law of conservation of mass

(p x rde x dr)• dt — p[ rr ±

aV r ] dr(rde)dt t p r ar

aye (rde x dr x dt) ae

a If dt = 1, — = 0 for steady flow and p = constant

at

vr + or

aa r (vo x r + a ae (ve) = o a t.

(r x Vr ) +

( V6 ) = 0

32. What do you understand from rotation and vorticity of fluid particles? A flow is said to be rotational if fluid particles are rotating about their own mass centres. The rotation of a fluid particle at a point is specified by funding the average angular velocity of two perpendicular linear elements of fluid particle. Rotation of the fluid particle takes place about an axis which is perpendicular to the plane formed by these two elements of the fluid. The sense of rotation is given by the right hand rule for the motion of a right-hand screw. Vorticity of a rotation flow has twice the value of the rotation of the flow. Both rotation and vorticity are vectors. 33. Derive the expressions for rotation and vorticity in terms of velocities in x, y and z directions.

Fluid Kinematics

\ ,

,......- ......--11 D", \ de

325

\ , .

,----

,-I B A Fg- dx Rotation about z axis

Consider a rectangular elemental sheet of fluid in x-y plane as shown in the figure above. During the time dt, the perpendicular fluid element AB and AD have rotated in x-y plane by small angle de. The velocity A in x and y direction is u and v respectively. Hence we have (1) The velocity components in x-direction at A and D = u and u + au • dy . aY C ) (2) The velocity components in y-direction at A and B = v and (v + a v • dx) ax Since the velocities at D and B are different from point A, element AB and AD will move relative to point A. The fluid sheet will rotate by de in time dt and take new position at AB' C'D' Distance BB' = [(v + Dv • dx — v dt ax )

]

av ax

= — dx • dt

Distance DD' = [u —(u + au -s • dy)1 x dt 0Y =—

au • dy • dt ay

dO — BB' —DD' dx dy

a v •dx • dt ax

dx av • dt — ax

— au • dy • dt ay dy au • dt ay

326

Fundamentals of Fluid Mechanics

de Now angular velocity w = — dt

WAB

a v • dt _ av _ ax ax dt —

au • dt

ay

au

WAD — dt

ay

The average of wilB and WAD will give the rotation of fluid flow about z axis 1, wz = — 2 lwAB + WAD) = 1 aV 2 ( ax

au) ay

We can also find out the component of rotation about x and y axis. 1 r aw Wx = 2 ay

avi az

1 aU 2 [ az

awl

wY— Also

ax

W = Vwx2 + Wy2 + Wz2

Rotational vector: w = wxi + wyj + wzk Vorticity = 2 x rotation SZ = 2 w Three components of vorticity are — S2x = 2wx — aw av az ay 0.5, = 2wy —

au aw ax az

S2z = 2wz — av

ax

Note: Vorticity in polar coordinates is SI =

au ay

a Vo + Vo ar

r

1 a Vr r as

34. What are the conditions for irrotational flow? The irrotational flow will have both rotation and vorticity as zero.

Fluid Kinematics

or

327

w = 0 and SZ = 0 wx = wy = wz = 0, and nx = ny = 14 = 0

The above gives the conditions of irrotation of the flow as: aw _ av ay az au _ aw ax az aw — au

ax

ay

Note: The conditions in polar coordinates is SZ = 0 which gives

a Vo 0 1 DV + 17 —+ — r=0 Dr r r ao

35. What is circulation? How is it related to vorticity? Circulation is defined as the line integral of the tangential velocity of the fluid around a closed curve ..

CirculationF== Vx ds c Where Vo = tangential velocity & ds = small length of curve.

u+ D

au • dy ay

C

T

dy , v A k

av

iv.

B"

v + — • dx ax

dx —Pol

Circulation

Consider a number of stream lines in a closed curve. Let tangential and normal components of velocity V are Vo = V cos 0 and V. = V sin O. If we calculate circulation around a rectangular rotating element of size dx x dy, then we get — .

I" =

Vcos 0 ds

av , au = udx + (v + ay ax dy _ u + dy dx — v dy aY (ay au) dx • dy ax ay

328

Fundamentals of Fluid Mechanics

= Slz x dx x dy = SZz x area enclosed by closed curve in plane at right angle to z-axis Hence vorticity is the circulation per unit surface area. .. •

F 0=— A

or

0—

where A =are enclosed by closed curved

F

(ay au)

dxdy

ax ay

36. Explain the physical significance and use of the term 'stream function' (UPTU 2001-2, 2007-8, 2008-9) The flow rate (q) between two stream lines per unit thickness and time is known as stream function (v). Physical significance of stream function. Consider a flow between two stream lines AA' and BB' which is two-dimensional steady and incompressible. The flow (q) per unit time and thickness between these stream lines AA' and BB' is stream function IV. The flow across points A and B will remain same = iv- = q irrespective of their joining line which may be AB or ACB or ADB. If we assume ty = 0 at point A, then ill at point B will be flow per unit time and thickness = q = ilf Y

Streamline

x Physical concept of V'

Stream function is also defined as the scalar function of space and time whose partial derivative with respect to any direction gives the velocity component at right angles to that direction. Hence in steady two-dimensional flow, iv = f(x, y) which gives

aIII ax

= v and

av — u ay

37. Derive the equation of stream line. (UPTU 2009-10)

Fluid Kinematics

329

Stream line

0

Equation of stream line Consider a stream line having velocity at point A (x, y, z) as V. The components of this velocity is u, v and w in x, y and z direction. After time dt, the distance moved by the particle on three major axes are: dx = udt dy = vdt dz = wdt dx:dy:dz=u:v:w u _ v _ z dx dy dz This is known as equation of stream line or

38. Explain the mathematical concept of stream function. Or alp ay, = v and (2) yi along a stream is constant. Prove (1) that = —u and ay ax Or Prove that a stream function v represents the equation for a stream line. Stream line

y A

T dy

u

v

I 0 Mathematical concept

Consider two-dimensional steady flow in which tp = f(x, y). The point 'A' and 'B' on the stream line separated by a small distance dx. This distance ds can be represented by dx and dy on x and y axis respectively. Let u and v are velocities at point A. The equation of streamline is:

330

Fundamentals of Fluid Mechanics

udy — vdx = 0

(1)

The discharge across dy will be dyr = udy Or

U

=

ay

Similarly

v = ay, (Hence proved) ax Putting the values of u and v in the eqn. of streamline (eqn.1) ay

dy +

dx = 0

ax

or or

chi/ = 0 yr = constant (Hence proved)

The stream function along a stream line is constant. The discharge between two stream lines is the difference of the stream functions. discharge (q) = Now to prove that stream function represents stream line, we know that stream line equation can be given as: udy — vdx = 0 Also

u——

ay

and v =

ay. ax

Therefore, we get:

ay ay

dy +

atif ax

or

dx = 0 dyr = 0

Hence yi represents the stream line equation. 39. Show that stream function satisfies the equation of continuity. The equation of continuity for a two-dimensional flow is

au + — av = 0

ax As per stream function, we have

(1)

ay

u— v—

ayl ay ax

Put the values of u and v in the continuity equation (eqn. (1))

Fluid Kinematics

a ax.

33I

—avi + a (+ al =0 ay ay Dx a2,,,, ' + a2 ' — 0 ax ay ay ax

or

The above is satisfying as LHS = RHS. Hence stream function satisfies the equation of continuity. 40. How yr and 0 in polar coordinates are used to find velocity components in radial and tangential directions. V = f(r, 0) 1 aYr Radial velocity u,. = r as Tangential velocity uo — aV ar 0 = f(r, 0)

ao

Radial velocity u,. = — ar Tangential velocity Ito = 1 ao r as Note: Following may be used as simple way to remember — (1)umue,vau,. (2) dx m rde, dy m dr 41. Derive Laplace equation of stream function for irrotation flow. For two-dimensional flow in (x, y) plane, the vorticity S2 is given by ax

(1)

ay

For stream function Iii, we have

v = + ax Put the values of u and v in eqn. (1)

a= '(') ax ax

a i atif) ay j

ay

332

Fundamentals of Fluid Mechanics

n2,„.

n2 „,

axe

ay2

— For irrotation flow, SL = 0 412

111

a2 axe

ay2

0

This is Laplace equation 42. Enumerate the properties of stream function yr. The properties of stream function are: (1) lir is constant at all places at a stream line. (2) The flow around any path in the fluid is zero (3) The velocity vector at any point on the stream line can be found out by differentiating its stream function i.e. u =

and v = ay ax (4) The discharge between the stream lines is equal to difference of their stream functions i.e. q= — (5) The rate of change of stream function with distance is proportion to the component of velocity normal to that direction (6) If two stream lines superimposed, then

as 2 _ am-Eli/2) as

as

as

43. What do you understand from the velocity potential? (UPTU 2007-8, 2008-9) The flow takes place as in pipeline when there is a difference of pressure. The flow always takes place from higher to lower pressure side. This potential difference resulting into the flow of fluid is known as velocity potential and it is denoted by (1). The velocity potential is defined as a scalar function of space and time such that its negative derivative with respect to any direction gives the fluid velocity in that direction. Hence for steady flow, if velocity potential is given by q) = f (x, y, z), then

u=

ax

,v=-

and w _

ao az

(Note: negative sign indicates that flow is taking place in direction inwhich velocity potential decreases.) 44. Derive the Laplace equation for the velocity potential. For steady flow, the continuity equation is:

au av ax

aw — 0

ay az

Fluid Kinematics

Now u =

333

aao v = — ao and w = — — ao . Put these values in continuity equation. x ay az ax

(_ 1))

1- 11

a20 axe

a20 + a 20 = 0 aye

+ =13 ax ay ay az az ) (-')

This is Laplace equation

45. What are the properties of potential function? The properties of potential function are: (1) Flow having potential function q) are irrotational (2) Flows are steady, incompressible and irrotation if velocity potential satisfies the Laplace equation. 46. What is the relation between stream function (v) and velocity potential (0? As per definition, the derivatives of stream and velocity potential give components of velocities as under: u— Also

ax

_ —DO u _— — D ,V—

ay

ax

u

and

v=

ay

=— av =— a° ay ax

v =

ax

ao = -ay

Hence, we can write —

ax and

=

ay

ao _ ay, ay ax

47. Prove equipotential line and stream line are orthogonal to each other. (UPTU-2004-5,2007-8) Equipotential lines have potential function as constant. = f(x, y) = constant

334

Fundamentals of Fluid Mechanics

do =

a DO • dx + —° dy = 0 ax ay

DCb = — u and ° = —v, we get

ax

ay

—udx — vdy = 0 dy ___u dx v

or

It is slope m1 of equipotentail line Now stream function along a stream line is constant. ty = f(x, y) = constant or

As

alit ax

— v and

ail/

ay

chir —

atif ax

dx +

ali f

ay

dy = 0

— u, we get vdx — udy = 0 dy _ v dx u

Hence the above is slope m2 of stream line v u x — Now mi x m2 = — — = —1 v u Hence equipotentail line and stream line are orthogonal. 48. What is a flow net? What is the condition of flow net to be a set of square? A flow net is a graphical representation of stream lines and equipotential lines. The stream lines in flow net show the direction of flow which the equipotential lines join the equal velocity potential (0) points in the flow. Flow net provides a simple graphical technique for studying flows which are two dimensional and irrotational as compared to mathematical calculation techniques which are difficult and tedious. In flow net, the stream lines are spaced such that the rate of flow q is same between each successive pair of streamline. The flow net can be drawn for irrotational flows. tp is constant along stream line and (p. is constant along equipotential line as shown in the figure. If q is the flow and V1 & V2 are velocities along streamline tyi and tv2 q = V1 An1 = V20n2 = constant Where An1 and An2 are the distances between two streamlines V1 _ One An1 V2

Fluid Kinematics

But

335

6.0. = ViAsi = V2As2 = constant Vi _ As2 V2

Asi

One = As2 Ani Asi or or

Ant An]. — As2 asi An = constant As

In case An = 1, then An = As. We get a set of square in the flow net As 49. What are the methods used for drawing the flow nets?

(UPTU — 2004-5)

Three methods are used for drawing the flow nets which are: (1) analytical method (2) graphical method and (3) electrical analogy method. In analytical method, flow net can be drawn be finding the expression for 0 and iv in terms of x and y which can be plotted. In graphical method, the flow passage is equally devided to draw stream lines and then equally potential lines are drawn orthogonally to the stream line. In electric analogy method, equipotential lines are drawn using null method. Stream lines are later drawn orthogonally. 50. Explain graphical method of drawing flow net. The graphical method is carried out as under: (1) The flow passage between the boundaries is devided into suitable number of equal parts and a set of stream lines are drawn enclosing these equal parts. (2) Equipotential lines are drawn orthogonally to those stream lines such that each block of flow net is almost a square.

336

Fundamentals of Fluid Mechanics

Converging pipe

Pipes or parallel plates

Flow

Bend pipe

Outlet from tank

Flow nets: Graphical method

51. There is similarity between flow of fluids and flow of electricity. What are these similarities? Flow of fluids

Flow of electricity

1. Velocity potential co is essential for fluid flow. Hence 6 is analogous

1. Electricity potential (v) is essential for flow of current

to voltage (v) 2. Velocity of flow results with the magnitude of velocity potential.

2. Electric current flow due to potential difference

Velocity is analogous to current (/) 3. Phenomenon takes place in homogeneous fluids 4. Darcy-Weisbach equation is

3. Flow of current needs homogeneous conductor 4. Ohm's law is applicable

applicable 5. It is possible to divide flow by pipe network

5. It is possible to divide current by electrical network

Q Q1 ± Q2

=

Fluid Kinematics

337

52. Describe the electrical analog method of drawing flow net. Conducting boundary

Vessel

Water with electrolyte

Conducting boundary

III 10 volt Electrical analogy method

The apparatus consists of a vessel with non-conducting walls containing electrolyte which can conduct electricity. Two copper plates with voltage supply are used to obtain current flow. The applied voltage (10 volt.) gradually falls from positive plate to negative plate in the electrolyte. The equal potential lines can be ascertained with the help of galvanometer by finding null points on the potential meter with the help of probe. After drawing equal potential lines, stream lines which are orthogonal to the equal potential can be ascertained. By drawing stream lines and potential lines, flow net can be made. 53. Describe the uses and limitations of the flow nets? (UPTU 2003-4) The uses of flow nets are: 1. The flow problems in which boundary configuration and flow conditions are such that the mathematical analysis is difficult can be solved by flow nets. Flow through a Francis water turbine rotor is analysed easily by flow nets. 2. Once the flow net is ascertained, it can be used for different discharges. 3. It is possible to ascertain the velocity of different places in flow net, if velocity at some points in the flow net are given 4. Flow nets can be used to ascertain the seepage and upwards hydraulic pressure force acting below the hydraulic structure like in a dam. 5. Efficient boundary shapes of blades of turbine can be designed by the use of flow nets. 6. Flow nets help in the identification of likely points of stagnation and separation which may occur in the flow so that the design of boundary shape can be improved to avoid these points of stagnation & separation in the flow of fluid.

338

Fundamentals of Fluid Mechanics

The limitation of flow nets are: 1. Flow net analysis cannot be applied where viscous forces are predominant such as close to the boundary surface. 2. Flow net analysis cannot be applied to the flow in sharply diverging flow passages. 3. Flow nets cannot show accurately the formation of wakes, resulting from the separation and eddy formation. 54. What is the difference between velocity and acceleration vector of fluid particle? The velocity vector is always in tangential direction to the path of the fluid particle. However the acceleration vector may also be in the normal direction in addition the direction of velocity vector. For example, when fluid is flowing in a curved path, the velocity at any point is tangential but the acceleration is both tangential and normal. Velocity vector is given by V (x, y, z, t) and acceleration vector by a = dV dt or

d —

ai7 dx ax dt

+ av ay # av az + ai7 ay at Dz ax at

av ai7 + v +

w DV+DV ax ay at az Acceleration of fluid particle occurs due to (1) change of position and (2) change in flow with time. Acceleration due to change in position is known as transport or convective acceleration. The acceleration due to change in flow with time is called temporal or local —u

acceleration. In steady flow at = 0, hence convective or local acceleration is zero. In Dt uniform flow as = 0, hence transport acceleration. However for both steady and uniform Ds flows, acceleration is zero. 55. What is normal acceleration when a fluid particle moves along a curvilinear path? Find acceleration vector also. When a fluid particle moves along a curvilinear path, its velocity (Vs) continuously changes direction along the path. Hence the moving particle has to have an acceleration component normal to the direction of flow while moving on the curvilinear path.

s

x 6,1

Radius = r of curvature

.19

— (V, + Mid (V, + dVs) cos 0

vs = ye

sin 0

Fluid Kinematics

339

Consider fluid particle A moving to point B in curvilinear path. Change of tangential velocity V0 = (V, + dVs) cos 0 — V, Cos 0 = 1 if c/0 is small Hence c/Ve = Vs + dVs — V, = dVs Similarly change of velocity normal to Vo Now

dV, = (Vs + dVs) sin dO sin dO = dO for small de dV, = V,d0

Now

(neglecting dVs x dO)

dVs _ aV0 avn dt at + at =

avs at

+ V a0 S at

s a = w = angular velocity = 7 But — dt a

_ ( dV, + 17 s i f) dt)0 ( n v2 = (Vs Ns ± Ns j + r as at

56. What are the equations of kinematics which are applicable for the analysis of a twodimensional steady ideal flow in horizontal plane? 1. Continuity

(a) Cartesian coordinates

au + av — o ax ay u=

alif v = — ay. ay '

ax

(b) Polar coordinates

a(rur ) + au, = 0 arae 1 acvay'

ur —

r ae ' ue

ar

2. Irrotation

(a) Cartesian coordinates

au

av — o ay ax

340

Fundamentals of Fluid Mechanics

. — ax ao ', — ao ay (b) Polar coordinates

ar

(ruo) —

aur =0

ae

ur— ao ue — r1 ao ar ' ae 3. Fluid flow in two dimensions (a) Possible if stream function iv exists (b) Irrotational if velocity potential 0 exists a2vir

a2w. 4. Laplace equation for irrotation in iv function is

ax2

+

ay2

0

,./., 5. Laplace equation of continuity in 0 function is a2,..), ' + a2.i. — 0. In polar coordinates, ay2 ax2

1 ao

it is —

r

ar

20 ± 1 20 a =0 +a a r2 r ar2

57. Distinguish between forced vortex and free vortex flow.

(UPTU-2005-6)

Forced vortex flow: Fixed vortex flow is defined as a flow in which some external torque is used to rotate the fluid mass. The fluid mass in forced vortex rotates at constant angular velocity w. Then the tangential velocity of a fluid particle is Vo = w • r Where r is the radius of fluid particle from the axis of rotation. Examples of forced vortex are: (1) flow of fluid inside the impeller of centrifugal pump (2) flow of water through the runner of a turbine and (3) a liquid mass in a container rotated about its central axis with constant angular velocity.

Rise of fluid

I

1

IV?

Initial level Z =

Fall of fluid —

Forced vortex

Fluid Kinematics

34I

Free vortex flow: When the fluid mass rotates without application of external torque, then it is called free vortex flow. The fluid in free vortex rotates due to the rotation imparted to the fluid previously. Example of free vortex are: (1) a whirlpool in a river (2) flow of fluid through a hole provided at the bottom of a vessel (3) flow around a circular bend and (4) flow in casing of centrifugal pump. 58. At a point on a stream line, the velocity is 3 m/s and the radius of curvature is 9 m. If the rate of increase of velocity along the stream line at this point is 1 m/s2, then the total acceleration at this point would be: (a) 1 m/s2 (b) 3 m/s2 (c)1 m/s2 (d)

m/s2

(IES 94)

a = V2 = = 1 111/S2 9 n r cro = 1 m/s2 a = I ao + a, = 111+1 =' m/s2 Answer (d) is correct 59. If u and v are the components of velocity in the x and y directions of flow given by u = ax + by, v = cx + dy. Then the condition to be satisfied is: (a) a + c = 0 (b) b + d = 0

(c) a+b+c+d=0 (d) a + d = 0 u = ax + by

v = cx + dy

au = a ax

av = d ay

Flow has to satisfy continuity equation

au + — av = 0 ax ay a+d=0

Answer (d) is correct 60. x component of velocity in a two-dimensional incompressible flow is given by u = y2 + 4xy. If Y-component of velocity v equals zero at y = 0, the expressions for v is given by — (a) uy (b) 2y2 (c) —2y2 (d) 2xy (GATE 96) u = y2 + 4xy au = 4y ax

342

Fundamentals of Fluid Mechanics

From continuity equation au ± av _ 0

ax

ay

av _

au = 4y

ay

ax

V

4y 2 + 2

—

C

= —2y2 + c

Condition y = 0, v = 0 c=0 v = —2y2 Answer (c) is correct 61. Given the x-component of velocity u = 6xy — ax2, the y-component of flow v is given by (a) 6y2 — 5xy (b) —6xy + 2x2 (c) —6x2 — 2xy (d) 4xy — 3y2 (Civil Services 1995) The continuity equation is

au + — av = 0 ax ay u = 6xy —2x2 au = 6y — 4x ax av — au — —6y + 4x ax ay If we carryout integration, we get — V = —6 Y

2 + 4xy + c

= — 3y2 + 4xy + c Answer (d) is correct. 62. The velocity component representing the irrotational flow is (a) u = x + y, v = 2x — y (b) u = 2x + 3y, v = —2y2 + x (c) u = x2, v = —2xy (d) u = —2x, v = 2y (Civil Services 1995) For irrotation flow,

au ay

av — 0 ax

Fluid Kinematics

343

For choice (a): a u— 1, and v = 2 ay ax au — av = 1 ax ay

2 = —1 # 0

For choice (b) a u— + 3 and av — 1 ax ay au ay

—

av = 3 — 1 = 2 # 0 ax

For choice (c) au — 0, av — = —2y y ax

au ay

av = + 2y # 0 ax

av 0 For choice (d) a u — 0 —= ay ax au _ av _ 0 ay ax Also

av + ax

u = —2 + 2 = 0 ay

Hence flow is irrotational Answer (d) is correct 63. If the stream function is yi = 2xy, then velocity at a point (1, 2) is equal to (a) 2 (b) 4 (c) 20 (d) 16 (Civil Services 1996) tg = 2xy

u—

av _ —2x ay

v—+

av ax

... (u, v) at point (1, 2) = (-2, 4) Resultant = -Ni u2 + v2 = j4+16

= Vkl Answer (c) is correct.

2y

344

Fundamentals of Fluid Mechanics

64. The given figure shows the flow net for two-dimensional constructions. The size of mesh square at '0' is 7 mm. If velocity at '0' is 5 m/s, than the velocity at 'A' is (a) 15 m/s (b) 10 m/s (c) 5 m/s (d) 20 m/s

ommmmwsitimommi. iMM==Mlikagoill EMMM•MillammmENN 11/ofluiii

••••••

Ano = 7 x 10-3 m AnA = 3.5 x 10-3 m Vo x Ano = VAX AnA VA — An°

AnA

x Yo

7x10-3

X Yo x 10-3 = 2 V0 = 2 x 5 = 10 m/s The answer (b) is correct 65. Which of the following is rotational flow? (a) u = y, v = 3/2 x (b) u = xy2, v = x2y (c) Both (a) and (b) (d) none of the other (IES 90) For choice (a), we have au = 1, and Y

ax

— 3/2

au av = 1 — 3/2 0 ay ax It is rotation flow For choice (b), we have au = 2xy and

av — 2xy

y

au — av — = 2xy — 2xy = 0

ay ax

Fluid Kinematics

345

It is an irrotational flow Answer (a) is correct. 66. A two-dimensional flow is described by the velocity components u = 5x2 and v = — 15 x2y. The stream function at (1, 2) will be: (a) 2 m2/s (b) 5 m2/s (c) 10 m2/s (d) 20 m2/s (IES 92) u = 5x2 V = —15 x2y

a v = aaaxx

dx +

aiff

ay

dy

= + vdx + u dy = + 15x2y dx + 5x2dy 1 5x3

y + 5x2y 3 = 10x3y (')1,2 = 10 x 1 x 2 = 20 m2/s V1 — +

Answer (c) is correct 67. Of the possible irrotation flow functions given below, the incorrect relation is: (a) iv = xy (b) v= A (x2 — y2) (c) 0 = urcos 0 — `3 sin 0

(d) 0 = (r —2 ) sin 0 r (IES- 95) a2 i, . a2,,,, 2 + ay2). — 0 and Laplace equation Guidance: The Laplace equation of irrotation in vi ls ax a 20 + a20 for continuity in 0 is

x2

y2

0

or

1 ao + a20 r ar a r2

For choice (a) =x

aty _ ax y

a2V — 0

U 4— 0

aX2

ax2

all/

ay

2_

+ 1 a20 — o r2 ae

346

Fundamentals of Fluid Mechanics

D 2 x a2,1, ax2 aye

0

Hence iv = xy represents a possible irrotation flow For choice (b) cif = A (x2 — y2) ax

ax2

=2Ax

ay

=2A ay

— —2Ay

— 2A

a 2m.

a2 — 2A — 2A = 0 + aye axe Hence lif = A(x2 — y2) represents a possible irrotation flow Choice (c) u = urcos 0 — cos 0 r

ao ar

a 20 r2

= ucos 0 — 11 cos 0 r

— + 23 cos 0 r

ao = —ursin e— — DO

a zo a0 2

sin 0

—urcos B — = 11 cos 0

1 a2 0 = 0 r2 ao2

a20

Laplace equation = ;a0 + a r2 ar

LHS = —1 u cos 0 — 2 cos 0 I + 3 cos 0+ 1 (—urcos 0 — = 11 cos 0 r r = v cos 0 (—1

1

+ 3— r —

= 0 = RHS Hence 0 represents a possible irrotational Choice (d)

r

1) r-

Fluid Kinematics

347

0= (r — ' ) sin 9 t.

a° = (1+ 22 ) cos 0 Dr r a20 art

4 sin 0

r3

: = (r —1 cos 0 r D20 — ae2

(r 2r )sin 0

a20 ± 1 a20 = 0 y + Laplace equation 1— DO r r r2 aci2 art LHS = (1+ 22 ) sin 0 r r =sin BI r

4 sin 0 3

r

1 ( 2 ) sin 0 2 r

r

r

± 2 4 1± 2 r 3 r 3 r r3

=0 Hence 0 represents a possible irrotational flow. 68. The velocity components in a flow fluid in x and y direction are u = xy and v = 2yz respectively. Examine whether these velocity components represent two or three dimensional incompressive flow with these dimensions. Determine the z-component. (UPTU-2001-2)

u = xy du _ ax Y

v = 2yz av = 2z ay

Continuity equation for two dimensions Du Dv = 0 +— ay ax y + 2z * 0 The flow is not a two-dimensional flow. It is therefore a three-dimensional flow and it must satisfy continuity equation Du ± Dv ± aw =0 ax ay az

348

Fundamentals of Fluid Mechanics

aw y + 2z + — = 0 aZ dw — = —2z — y

or

az

or

w= —z2 — yz + ci c1 = constant

69. A velocity function in a two-dimensional flow is yr = 2xy. Determine the corresponding velocity potential ep (UPTU-2003-4) tv = 2xy

Now

atV = —2x ay

u—

v — ax = 2y ax do=

ax yx

• dx +

ao

dy

= —udx — vdy = 2xdx — 2y dy

th _ 2x 2 2y 2 T 2 — 2 =c1 = x2 —y2 + c1 70. Check whether the function ii, = A (x2 — y2) represent the possible irrotational flow phenomenon (UPTU-2002-3) v = A (x2 — y2)

av ax

= 2Ax = v

a/If ay = —2Ay = —u u = 2Ay

Now

au =2A ay

Also

v = 2Ax av =2A ax

For irrotation flow, 0 must be zero.

Fluid Kinematics

0 — av ax

349

au ay

= 2A — 2A =0 Hence flow is irrotational 71. Velocity for a two-dimensional flow is given by — V = (3 + 2xy + 4t2) i + (xy2 + 3 t)j

Find the velocity and acceleration at point (1, 2) after 2 see. Guidance: The velocity is expressed in x, y and z direction by vector as V = ui + vj + wk. Hence u, v and w can be found out from the vector equation u= 3 + 2xy + 4t2 u at(1, 2)andt= 2 =3 + 2 x 1 x 2 + 4 x 22 = 3 + 4 + 16 = 23 m/s v = xy2 + 3 t ✓at (1, 2) and t = 2 = 1 x 22 + 3 x 2 = 4 + 6 = 10 V = V14

2

+ V 2 = 1/232 +102

= 25.1 m/s Now

a—

av ax

av av — u av • + av v + dt

ax

ay

at

— 2yi + y2j

aV = 2xi + 2xy j ay

aV — 8t • i + 3 • j at a = (3 + 2xy + 4t2) (2yi + y2 .j) + (xy2 + 3 t) (2x • i + 2xy • j) + (8t. i + 3 .j) Now put x = 1, y = 2 and t = 2 a = 128i + 135j a = 11282 + 252 = 186.04

350

Fundamentals of Fluid Mechanics

72. A flow is given by

V = 2x3 i — 2x2yj Clarify: (1) flow is steady or unsteady and (2) flow is two or three dimensional. Find velocity, local acceleration and convective acceleration at point (2, 1, 3).

Guidance: The velocity of flow has no term of time which indicates that flow is nondependent of time i.e. it is steady flow. The velocity is also non-dependent on the flow in z direction as w = 0. Hence it is a two-dimensional flow. Local acceleration is time av n this case is zero independent as at i Put

V = 2x3. i — 2x2y - j x = 2 and y= 1 V = 2 x 23 .i — 2.22.1.j = 16i — 8 • j V = V162 + 82 = V256+ 64 = 17.89 m/s a_ — u

av ax

+v

u = 2x3 and

at

ay

at

au = 6x2.i - 4xy] ax

v = —2x2y and

au

av+av

ay

— 2x2.j

= 0, hence local acceleration = 0

Put

a = 2x3(6x2i — 4xy j) — 2x2y(— 2x2 . j) = 12x5i — (8x4y — 4x4y)j = 12x5i -4x4yj x=2&y=1 a = 384i — 64j a = V3842 + 642 = I14.75 x 104 + 0.41x 104 = 389.4 m/s2

73. A one-dimensional steady flow through a converging nozzle is having linear velocity distribution u = 4(x). At entrance u = Vo and at exit u = 3V0. Find the expression for acceleration. If nozzle has length 1000 mm and velocity Vo = 10 m/s, calculate acceleration at entrance and exit

Fluid Kinematics

35I

—0-3V0

The velocity (u) will vary linearly as under u = Vo + 3

VO — VO

/

Xx

= Vo (1+ 2x ) i d — u du + du

ax

at

2x ) x Vo x 2 /

= V0(1+

—

or_x=o = ax=i =

2V02 (., _L 2) v+ 1 ) 2 x 102

1

(1) = 250 m/s2

2x102 x 3 = 600 m/s2 1

74. Water is flowing through a pipe of 2 m diameter with average velocity of 1 m/s. What is its discharge? If flow takes place in the section having diameter 4 m, what will be the velocity? Q = Airli —

7r cl? x Vi 4

x 22 x1 4 = 7C = 3.14 m3/s Al V1 = A2V2 =

7Cd?

4

XV1-

7C

7Cdi

4

X V2

352

Fundamentals of Fluid Mechanics

22 X 1 = 42 X V2

4 — 0.25 m/s V2 = 16 75. A pipe AB branches into two pipes C and D as shown in the figure. The pipe has a diameter of 120 cm at A, 90 cm at B, 50 cm at C & 2.0 cm at D. Determine the discharge at A if the velocity at A is 2 m/s. Also determine the velocities at B and D if velocity at C is 4 m/s. C = 0.5 m V, = 4 m/s

A d,= 1.2 m Va = 2 m/s

db =0.9m

cy 0.2 m

QA —

7rd 2 a XV 4

a

ix L22 4 Now

X

2 = 2.26 m3/s

Aa Va — AbVb

Vb -

2.26 =45 m/s x 0.9' 4

QB =

(QB = QA)

Qd

2.26 = Ac VV + AdVd 7t X

0.52

X 4+

Tr

x4.22

4 = 0.785 + 0.0314Vd Vd = 46.97 m/s

x Vd

76. A stream function is give by: tif = 4 x2y + (2 + t)y2 Find the velocity vector and its value at position vector r = 1.1 + 2 j — 3k when t = 3 sec. 111 = 4x2y + (2 + t)y2

a4i — —v = 8xy ax

Fluid Kinematics

aay

353

= + u = 4x2 + 2 (2 + t)y

V = ui + vj = [4x2 + 2 (2 + t)y] i — 8xy.j Now put x = 1, y = 2 z = —3 and t = 3 [4 + 20] i — 16j = 24i — 16j

V(1,2,-3,3) —

V = V242 +162

=.J576 + 256 = 28.89 m/s 77. If for a two-dimensional potential flow, the velocity is given by 0 = x(2y — 1) Determine the velocity and stream function at point P(4, 5) (AMIE 1977) = x (2y —

1)

ao

u = — yx. = —(2y — 1) = 1 — 2y v — — a° ay

Put

= —2x

V = ui + vj = (1 — 2y) i — 2x.j x=4&y=5 V4, 5 — —9 i — 8j V = 1192 +82 = 12.04 m/s

= f(x, —

ax

dx +

ay

dy

= vdx — udy = —2xdx — (1 — 2y)dy —2x —

Put

2

2

2y Y +

= —x2 + y2 — y x = 4, y = 5

2

2

354

Fundamentals of Fluid Mechanics

1114,5 = —16 + 25 — 5

=4 78. Sketch the stream line represented by yr = x2 + y2. Also find out the velocity and direction of point (1, 2) (AMIE 1980) iv = x2 + y2 u—

a — 2y ay

v = av = 2x ax V = ui + vj — 2y.i + 2xj V1, 2 = —2 x 2.i + 2.1j = —4i + 2j V = )142 +22 = J20 = 4.47 m/s tan 0 = — v = 2 — 0.5 u 4 0 = 26.5° with x — axis tif = x2 ± y2 = r2 If r = 2, 3, 5 ... so on then yr= r2 = 4, 9, 25 ... and so on

Stream lines

79. Determine the circulation I' around a rectangle defined by x = 1, y = 1, x = 5 and y = 4 for the velocity of fluid u = 2x + 3y and v = —2y

Fluid Kinematics C(1,4)

y

355

y=4 B (5, 4) x=5

x= 1

A (5, 1)

y= 1

D (1, 1)

0

x

Specified Rectangle Circulation is a line integral of velocity around a closed curve = =

uds =

Scurve

udx + vdy

(2x + 3y)dx —2y.dy

ABCD

= J AB

SABCD

— 2ydy +

JBc

— 2ydy +

(2x + 3y)dx +

1

= 4 —

2ydy + — (2x + 12)dx + 5

.11

DA

(2x + 3y)dx

4 — 2ydy + J.5 (2x — 3)dx 1

5

1

12x) + ( 2x2 + 3x1 2 2 1 5 = 1 + 12 — 25 — 60 + 25 + 15 — 1 — 3 = 53 — 84 = —36 =

(2./C2

Another method = vorticity = 2w =

av

au ax ay

= 0 — 3 = —3 Area = 4 X 3 = 12 Circulation I' = S2 x A = — 3 X 12 = —36 80. A two-dimensional incompressible flow in polar coordinates is given by: V, = 2r sin 0' cos 0' and V0 = —2r sin2 Find whether these components represent a possible flow. Guidance: The velocity components must satisfy the continuity equation. In polar coordinates, the continuity equation is: 1 ay, r ae

avr ar

vr

=0 r V,. = 2r sin 0 cos 0

356

Fundamentals of Fluid Mechanics

DV'. = 2 sin 0 cos 0 ar Vo = —2r sin2 6 DV°

— —2r x 2 sin 0 cos 0

De

LHS of continuity equation is:

1(-2r x 2 sin cos 0) + 2 sin 0 cos 0 r

— 2 rsin 0 cos 8 r = 2sin 0 cos 0(-2 + 1 + 1) = 0 = RHS

Hence the flow is a possible flow. 81. In a two-dimensional flow, the tangential component of the velocity is: Asin 0 where A = constant r2 Find (a) radial velocity Vr (b) magnitude and direction of the resultant. V9 = —

Guidance: Use continuity equation to find radial velocity.

Or

1 avo ± Dv,. ± vr = 0 Dr r r ae V0 a (r V.) = 0 ao + Now

Ve — D Vo _

DO

A sin 0 r2 A cos 0 r2

Dr17, _ Acos 0 ar r2 r V,. —

A cos 0 r

A cos 0 Vr — r 2 Now

V = Vo ± Vr V = VVe2 ± Vr2

Fluid Kinematics

357

A2 sin2 0 A2 cos2 9

r4

r4

A 2

82. A fluid flow is given by V = (.1

a2 ) sin 0, V, — (/

a2 cos a

Show that it represents a possible flow and find whether it is irrotation flow.

Guidance: Use continuity equation in polar coordinates to show the flow is possible. For rotational flow, vorticity will be zero. V,. = (1— )cos 0 r V,. = (r —

cos 0

1

a (rVr) — (1+ a2 ) COS

ar

T7 = — (1+ a2 ) sin

0

av8 = —(1+ 2 cos o a0 r Continuity equation is —

avo ae

arVr ar LHS = — (1+ a2 ) cos 0 + (1+ a2 cos2 0

)

= 0 = RHS Hence flow is possible. Now

S2 =

ar

(r Vy) — aVr

v,, = —( 1+

ae

a,) sin 0 r`

rrio = —(r +1 sin 0

358

Fundamentals of Fluid Mechanics

ar

(rite) = —(1—

r

sin e

V,. = (1— .2 )cos 0 aaVer — (1— Now

0=

a ar

ra)

(r Vo)

— (1

sin 0 aVr ao

a ) sin 0 + (1 a2 sin 0 r2 r

=0

)

Hence flow is irrotational 83. Do the following velocity potentials represent possible flows. If so, determine the stream functions (a) 0 = y + x2 — y2 (b) (fr= u r cos 9 + (C) 0 = u (r

u cos 8 r

„2

+ -`` ) cos 8+ e r 27r

Case 1. 0=y+x2 -y2

aax° —2x a2C6 = 2 axe

aay' = 1 — 2y a2(P — —2 ay

a 20 axe

a202 =2-2=0 2—0 ay

Fluid Kinematics

359

Hence Laplace equation is satisfied and flow is possible D iv DO — +2x u=+ -+ ay ax v=+=— ' ay ax dip

—

— 2y)

ay. dx + ayr dy ay

ax = —(1 — 2y)dx + 2xdy = —x + 2xy + 2xy = —x + 4xy = 4xy — x

Case 2. 0 = 4cos 9+ 4 , cos 0

Continuity equation —0 —+ r ar

a2 r2

1 a20 r2 a r2 — 0

ao = 4cos 6 — = 11 cos 9 Dr r2 a 2th ± 2u3 cos 9 r art

DO a 20

ae2

= —ursin 0—

r

sin 0

u cos 0 — —urcos 0— — r

LHS of continuity equation = —1 uu cos 0 r C

u, cos 0 + 2,u cos 0+ 1, ( rcos 0 — 14 cos 0) r ` r' r`

= u cos 0 [1 r = 0 = RHS

1 +Y 1 r3 r

11 r3

Hence flow is possible and irrotational u = r

ao ar

ao

1 DV u cos 0— = 11 cos 0 r De = r2

— aty _ 1 ( ur sin 0 — 1= 1 sin 0) r ra 0 Dr r2 dig = ur• r • do + uo dr u0 —

360

Fundamentals of Fluid Mechanics

= (ur cos 0 — L4 cos

o)

d0 — u sin 0 (1+

r3

• dr

On integration, we get — (if = u sin 0 (r — 1) —u sin 0 r

1 2r2

1 —r+— 1) = u sin 0 (r -r2 = 1 — — sin 0 — — 1 ur (21r ) Case 3. =u

+ a) —2 cos O+

27r

= u (1 a2 ) cos 0 r2

ar

a zd, =

are

2a2u cos 0 r3 ,2

DO

a0

=— u r +

a2° — ao2

sin 0+ 1 27

u Cr + a2 ) cos 0

ao +--a— Continuity equation: 2+ " =0 r r id r2 ae2 LHS =

2 [u 1 1 a2 cos

2a2

u cos 0— 2 r3 r

1 a2 ] = u cos 0 [1 a23 2a2 3 r r3 r =0 uo —

Ur

=

ao rai9

r

ar

_+ -u a2 ) r

ao ar

rae

U (1

sin

a22 ) cos 0 r

0+

1 27rr

„2 + `) cos 0

Fluid Kinematics

dip"— =

Dr

dr +

V

ao

u r+a2 2

36I

d0 2

sin

x dr +

2

dr

= — u sin 0 (1+ a ,, `

dr + ur (1 `" 2 ) cos 19 x de 21r r r 2

1

r dr + u cos 0 r a2 + 27r

d0

On integration, we get — = — U sin 0 (r

= — u sin

= 2 u sin

(r

a2 + r

log r

a 2 + r a2

) + u sin 0 r a2

log r 27c

a 2 r + log r — 27r r

84. For given 0 = 4(x2 — y2), find yf

(UPTU — 2006-7)

u= v= dip —

°— 8 — x

ax

ay

—

ax

8 — atv Y ax dx + Dig dy

ay

= 8y dx + 8x dy = 8xy + 8xy = 16xy 85. A stream function is given by = 3 x2y + (3 + t) y2 Find the flow rates across the faces of the triangular prism at t = 5 sec if prism thickness is 3 m in z direction

Prism: 3 m thick

362

Fundamentals of Fluid Mechanics

Prism: 3 m thick lif = 3x2y + (3+ t) y2

At point A, x = 0, y = 2, t = 5 VA = 3 x 0 x 2 + (3 + 5) x 4 At point B, x = 3, y = 0, t = 5 tifB = 3 x 9 x 0 + (3 + 5) x 0 = 0 At point 0, x = 0, y = 0, t = 5 tvo = 0 Flow rate = (pi — th) x thickness Flow from AO = (VA — 110 x 3 = (32 — 0) x 3 = 196 m3/s Flow from BO = (VB — 'o) =0 Flow from AB = (VA — ViB) x 3 = (32 — 0) x 3 = 96 86. If the velocity field is given by u = 16y — 8x, v = 8y — 7x, find the circulation around the closed curve defined by x = 4, y = 2, x = 8, y = 8 SI =

av au — ax ay

u = 16y — 8x au =16 ay v = 8y — 7x av ax

7

0 = — 7 — 16 = — 23 Area = (Y2 — Yi)(x2 — x1) = (8 — 2) (8 — 4) = 24 T=S2xA =— 7 x 24 =— 148

Fluid Kinematics

363

Y

.

8 r

,

4

8

2

87. A pipeline 60 cm diameter bifurcates at a y-junction into two branches 40 cm and 30 cm diameter. If the rate of flow in the main pipe is 1.5 cu m/s and mass velocity of flow in 30 cm diameter pipe is 7.5 m/s, determine the rate of flow in the 40 cm diameter pipe. (UPTU-2004-5)

6,B = ec ± op oc = 9B — eD

g x (0.3)2 x 7.5 4 = 1.5 — 0.53 = 0.97 m3/s = 1.5

88. The stream line spacing is 5 cm at a section having a velocity of 6 m/s in two dimensional flow. If at another section, the spacing is 3 cm, what is velocity? (a) 10 m/s (b) 3.6 m/s (c) 100 m/s (d) 36 m/s (e) 50 m/s V1An1 = V2An2 6 x 0.05 = V2 X 0.03 V2

6 x 0.05 — 10 m/s .03

Answer (a) is correct 89. If velocity along a stream line is given by — V = 25 + t + 3 at s = 2 Then the acceleration after 1 sec is given by (a) as = 16, at = 1 (b) as = 8, at = 0 (c) as = 7, at = 1 (d) non of a, b and c

364

Fundamentals of Fluid Mechanics

V= 2s + t + 3

av as

ay at

=2 =1

a, = V x

as as

— (25 + t + 3) x 2

= (2 x 2 + 1 + 3) 2 = 16 a, _

av _ 1 at

Hence answer (a) is correct 90. A fluid particle moves along a curved path of radius 1 m with a velocity of 2 m/s. Its normal acceleration in m/s2 is (a) zero (b) y (c) 2 (d) — 1 (c) none of above 2 V

a,.

2

r = 4 4

Answer (b) is correct 91. Which of following stream functions satisfies Laplace equation? (a) x2 + y2 (b) 2xy (c) x3 + y3 (d) x2 y2 a2V =2 V = x2 + Y2 .% ax a =2x, ax2

aaytff = 2y, a2 2 — 2 Y

a2 Y

ax2

±

Now

a2ur .r = 2 + 2 = 4 # 0. Does not satisfy Laplace equation ay2 yi = 2xy,

ay/ ax

=2y and

a m, a2 r, ''''' = 0 ‘t. = 2x, ay2 ay

a2V — 0 ax2

Fluid Kinematics

365

a2thr a2,,, '''' + aye 'r = 0, satisfies Laplace equation a2 x2 a 2 v — 6x v= x3 ±Y3, av ax — 3 ' axe aw

zi, _ 3.„2 - /- = 6, , -Y ay '' ay2

a2 H1 a ,,, '''' + aye ‘r = 6 (x + y) * 0. Does not satisfy Laplace equation axe tit = x2y2

Now

,

a ip

ay

a lif a

a2v — 2y2

aX2

= 2x2y, aa2y2 l if = 2x2

a2mr a2ii, Y'r + 1- = 2(x2 + y2) * 0. Does not satisfy Laplace equation ay2 a2 Answer (b) is correct. 92. A fluid flow is represented by the velocity field V =a•x• i +a•y• j where 'a' is a constant. The equation of stream line passing through a point (1, 2) is(a) x - 2y = 0, (c) 2x - y = 0,

(b) 2x + y = 0 (d) x + 2y = 0 (GATE 2003)

Hence, and Equation of a stream line is-

V =a•x i +a•y• j ux = a•x uy = a• y dx _dy uy ux

or or

dx _ dy — a •x a•y dx _ dy x y

On integration, we havelog x= logy + logc

366

Fundamentals of Fluid Mechanics

where c = constant or x= c•y Since the stream line passed through point (1, 2), hence1=2•C or C= 1/2 The equation of stream line is— 1 x= 2 y 2x — y = 0

or Option (c) is correct.

93. The velocity components in 'x' and 'y' directions of a two-dimensional potential flow are `u' and 'v' respectively. Then

av

(a) — ax

au is equal to— ax

av ax

(b) — —

(c)

av

ay

av ay

(d) -(GATE 2005)

Continuity equation is satisfied in 2-D flow. Hence, au av =0 + ax ay or

au ax

av ay

. Option (d) is correct. 94. Show that the velocity vector 'V' is tangential to the stream lines everywhere in the x-y plane. The stream function in x-y plane can be represented as— V (x, y) = constant The stream function can be expressed as— dip = Now

tif(x, y) =

a iv ax

dx +

ax

ay

dy

constant. Hence, a tif = 0

dx +

a yr ay

dy = 0

Fluid Kinematics

But

Vx

—

ay/ ay

367

—ay!

and Vy — ax

Vx dy — T/5, ak = r0 dy _ Vy dx vx

Or

The above indicates that velocity vector V is tangent to lines of v(x, y) = constant. 95. A velocity field can be given as u = VcosO, v = VsinO and w = 0. Determine the expression of the stream lines of this flow. The streamlines can be expressed as— dx _ dy _ dz u v w dx = dy = dz Vcos0 Vsine 0

Or

0 dy = V sin e _ tan dx VcosO

Or

or

dy = tan() • dx On integration, we get— y = tane • x + c Hence, streamlines are group of lines which inclined at angle `0' to the x-axis.

96. A two-dimensional flow field is specified by V= 3yi + 3xj. Find whether the flow is steady, irrotational as well as feasible. Determine the stream function and volume flow rate passing between streamlines through the points (1, 3) and (3, 3). Given

vx = 3y and T5, = 3x and

Now

SL —

a vy

—

av dt

=0

avx

ax ay =3—3=0 Therefore, the field is irrotational. The field is feasible if it obeys continuity equation which is given by—

avx + OVy _ Ox But

ay

0

OVx = a (3y) _ 0 ax

ay

(i)

368

Fundamentals of Fluid Mechanics

avY= ay D (3x) = 0

and

ay Putting these values in equation (i), we get LHS = RHS. Hence, the field satisfies the continuity equation and it is feasible. Now we have—

at!,

thy — ax dx +

a yr

ay = — Vy dr + Vx dy = — 3xdx + 3ydy

dy

On integration, we get 3 " 2312- 2x2±c The discharge between streamlines (1, 3) and (3, 3) is- iff(3, 3) — V(1, 3) 3 (32 _ 32) _ 3 2 _ 32) =2 2 (1 = 12 units 97. Determine whether the two-dimensional flow field as given below is rotational. V = 2x3yi +

-v'

2

a Vy

a Vx

ax

ay

we know

V=

Here

Vx = 2x3y

j

aVx 2x3

and

ay 4 v=x Y 2

2x3 ax yr = 2x3 — 2x3 = 0

DV)?

Hence, the given flow is irrotational. 98. Of the possible irrotational flow functions given below, the incorrect relation is (where 111 = stream function and q) = velocity potential) (a) lif = xY (b) lif = 24(X2 — y2)

Fluid Kinematics

(c)

= u • r cos° + u cos° r

(d)

= Cr — — sin() )

369

(IES 1995) If stream function Op) satisfies the Laplace equation, then flow is irrotational, i.e.

a 2,„, axe

a2V ayz 0

(a) Now Iv = xy ay/ ax

a2 vi 0

and ax2 2 ay/ — X and ay2 — 0 ay

Similarly,

Hence, tif = xy satisfies the Laplace equation and the flow is irrotational. (b) Now iiif = A(x2 — y2) a 2 u,

a Iv

— 2Ax and

ax

— 2A

a 2 m,.

ati f _ ay

Also

ax2

2Ay and 1.2 =

2A

Hence, yi = A(x2 — y2) satisfies the Laplace equation and the flow is irrotational. (c)

Now

0 = ur cos() + ± 1 cose r

vr = - ao— = - u coo + u cosO ar

and Now

r

ao

u

Tio — 1 — u sin() + 2 sine r ae r 1 a a2 2Wz = — — [v [V9] — 1 [Vr] r ar r ae r ar

u [usine +— cosi)] r2

2u sin 0

r4 Hence, flow is not irrotational. 2 (d) Now = Cr — ) sin()

u sin 0 r2

1

r

a ao [—ucoso+ u-2 COS 01

u . sm0,--- 0 r4

r

+

370

Fundamentals of Fluid Mechanics

Vr —

v,= —

ao — [1+ 2I sin° 2 ar

r

1[ 1 ao —=—— r r ao = — [1-

Now

1a r

r

r

2 2

cos)

COSO

1 a (rV,) + —

d

r

ao (vv) is—

=1 a r

r

ar

= 1 ,(_..i _ r

r

2 r

sin 01 +

1

a

r

ar

[ — (-1— 2 cos 0' L

j . 1 ( 2) sm0 + — 1— smu r

r

= 0 and flow is rotational.

The relation in (c) is incorrect. 99. For steady incompressible flow, if the u-component of velocity is u = Aex, then what is the v-component of velocity? (a) Ae

(b)Aexy

(c) — Aexy

(d) — Aex (IES 2008)

Continuity equation is— du + dv = 0 dx dy

or

d(Aex) dv 0 dx dy

or

Aex +

Or

dv = — Aexdy

dv— dy

0

On integration, we get v = — Ae'y

Option 'c' is correct. 100. For a steady two-dimensional flow, the scalar components of the velocity field are Vx = — 2x, Vy = 2y, Vz = 0. What are the components of acceleration? (a) ax = 0, ay = 0 (b) ax = 4x, ay = 0 (c) ax = 0, ay = 4y (d) ax = 4x, ay = 4y

(IES 2006)

Fluid Kinematics

371

du du du du ax — u + v + w + dx dy dy dt ax = (— 2x) (— 2)+0+0+0 = 4x Also

dv dv dv dv aY = u — + v + w — + dx dy dz dt = 0 + 2y • 2 + 0 + 0 = 4y

Option 'cl' is correct. 101. The stream function tp = x3 — y3 is observed for a two-dimensional flow field. What is the magnitude of the velocity at point (1, —1)? (a) 4.24

(b) 2.83

(c) 0

a 2 = vx ax — ax (x3— Y3) = 3x

41

=

ay a ay

(x3

—

y3) = — 3y2 = — Vy

V = VVx2 + Vy2 = V32 +32 = 4.24 Option 'a' is correct.

(d) 2.33 (IES — 2004)

Chapter

9

FLUID DYNAMICS-I

A A A A A A A A A A

CONTROL VOLUME REYNOLD TRANSPORT THEORY NAVIER-STOKE'S EQUATION BERNOULLI'S EQUATION HYDRAULIC COEFFICIENTS FREE FLOW ORIFICE SUBMERGED ORIFICE PARTIALLY-SUBMERGED ORIFICE VENA CONTRACTA PITOT TUBE

A A A A A A A A A A

EXTERNAL CYLINDRICAL MOUTHPIECE CONVERGENT-DIVERGENT MOUTHPIECE BORDA'S MOUTHPIECE FREE JET LIQUID WEIRS RECTANGULAR NOTCH TRIANGULAR NOTCH TRAPEZOIDAL NOTCH END CONTRACTIONS VELOCITY OF APPROACH

INTRODUCTION A fluid in motion is subjected to several forces due to which the flow phenomenon of the fluid has variation in its acceleration and its energy content. Hence, fluid dynamics is concerned with the study of the fluid motion along with the forces and energies involved in the flow. The forces acting on a fluid mass can be (i) body or volume force (ii) surface force and (iii) line force. The body forces are proportional to fluid volume which include forces such as (i) weight and (ii) centrifugal force. The surface forces depend upon surface area which include forces such as (i) pressure force (ii) shear force and (iii) force due to turbulence. The linear forces depend upon the length. Surface tension is an example of linear force. The resultant force on any fluid element is equal to the product of the mass and the acceleration of the fluid element as per Newton's second law of motion. The acceleration and resultant force acting on the fluid element must be along the same line of action. In Euler 's equation of motion, forces acting on the fluid mass due to gravity and pressure are taken into consideration. In an ideal, incompressible and steady flow, the total energy

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at any point of the fluid remains constant and is called Bernoulli's equation. The rate of flow can be measured using devices whose workings are based on Bernoulli's equation. The devices are (i) venturi meter (ii) orifice meter and (iii) Pitot tube. 1. What is the difference between kinematics and fluid dynamics? Kinematics is the study of the velocity and acceleration of fluid particles in a fluid flow without taking into the consideration the forces causing the flow. The fluid dynamics is the study of fluid motion with the forces causing the flow. 2. What law is used for studying dynamic behaviour of the fluid flow? What are the assumptions made for the fluid while applying this law? The dynamic behaviour of the fluid flow is analyzed by the Newton's second law of motion which relates acceleration of the fluid flow with the applied forces on the fluid. The fluid is assumed to be (1) incompressible and (2) non-viscous while applying this law. According to Newton's second law of motion, the net force 'F' acting on a fluid element is equal to mass `in' of the fluid element multiplied by the acceleration in the direction of the applied force. F=mXa

3. What are the forces acting on the fluid flow? What these equations of motion are called when (1) force of compressibility is negligible, (2) force of turbulence is negligible, and (3) force of viscous force is negligible? In a fluid flow, the forces acting on a fluid element are: (1) Gravity force (Fg) (2) Viscous force (Fr) (3) Pressure force (Fp) (4) Turbulence force (Fe) (5) Compressibility force (Fe) Ftotal= Fg Fy ± Fp ± Ft ± Fc

1. In case compressibility force is negligible, then total force acting is Ftotai = Fg Fy ± Fp ± Ft This resulting equation of motion is known as Reynolds's equation of motion. 2. In case turbulence force `F1' is negligible, the resulting equation of motion is known as Navier Stoke's equation. The forces due to gravity, pressure and viscosity are taken into consideration. 3. In case viscous force is negligible when the fluid is ideal, the resulting equation of motion is known as Euler's equation of motion. The force due to gravity and pressure are taken into consideration. 4. What are the different forms of energies in a flowing liquid? (Osmania University)

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There are three forms of energy in flowing liquid: (1) Potential Energy. The potential energy is due to the position of the fluid element from a datum line. PE = mgh where h = height from datum line. (2) Kinetic Energy. The kinetic energy is due to the velocity of flowing liquid. If ' is the velocity, then kinetic energy (KE) — m v2 2g (3) Pressure Energy.

Pressure energy is due to the pressure of liquid.

The pressure energy = Pressure x Volume = P x m 2

Total Energy = mgh + my + mp 2 Energy/unit fluid weight = P + V 2 + h 2g pg Total head — P + 1 V 2 + h pg 2 g Fluid element

h

/ Datum 5. What is the concept of control volume and control surface? The laws of physics are basically stated for a particle or a control mass system. The statements of the conservation of mass, momentum and energy are based on control mass or closed system. However fluid flows cannot have control mass. Therefore, the concept of control volume becomes necessary so that the principles of conservation of mass, momentum and energy can be applied on fluid flow. Reynolds transport theorem is used to relate the control volume concept with the concept of a control mass system in terms of general properties of the system.

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Control volume ----------

Control surface

Control Volume A control volume is defined as an identified volume fixed in space. The control volume is bounded by a closed surface as shown in the figure which is called control surface. The shape of the controlled volume may be taken as parallelopiped or hollow cylinder or any other convenient shape. The fluid can enter and leave the control volume across the controlled surface. The fluid can also transfer mass, momentum and energy through the controlled surface. The net rate of change of fluid mass, momentum and energy across the control volume is called mass flux, momentum flux and energy flux. 6. What is Reynolds transport theorem ? The laws of motion are stated for a body having a lump of mass. In fluid mechanics, it is not possible to identify a lump of mass since it undergoes continuous deformation. It is easier to consider a control volume. Reynolds transport theorem is meant to relate the control volume concept with that of a control mass system in terms of general properties of the system. Reynolds transport theorem states that the time rate of increase of property N within a control mass system is equal to the time rate of increase of property N within the control volume plus the net rate of afflux of the property N across the control surface. If N = extensive property depending upon mass = Then

np dV where n = intensive property independent of mass

dN) k.at

)Closed Mass

at

11.1

Control Volume

np x dV +

ff

np V, dA

Control Volume

V = Velocity of fluid, D A = element area vector on control surface V, = V — Vc, V, = fluid velocity relative to control volume. 7. Show the application of Reynolds transport theorem to the conservation of Mass. Conservation of Mass

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The Reynolds transport theorem is:

m

C at )ells = If Mass is conserved, then

a np dV + ff np Vr dA at J .v c cv

dN r)

as

=0 at )cms N = (Mass) and n = 1

p av+ ff eVr dA = 0 a ~~~ cv cv At steady state

iff cv

p dv = 0

p Vr dA = 0

Hence cv This is continuity equation

8. What do you understand by Navier-Stoke's equations? What are the applications of Naiver-Stoke's equations? Navier-Stoke's equations are the fundamental equations for the analysis of viscous flows. In incompressible flow without turbulence, there are four unknown characteristics of the flow viz u, v, w and P. Hence three Navier-Stoke's equations with continuity equation

( aau + av+ a-aw .0 xay z

are the sufficient conditions for the determination of the flow

characteristics. The Navier-Stoke's equations are:

Bx—

p

ap _ du dt x

B Y

B Z

p

a

—

p

ap = dv

ay

dt P

ap _ dw dt p az

a 2u

a2u a2u 2 ay 2 az

a2v + a2v +a2v

ay 2

az

2

[a2 w a2 w a2 w1 ay 2 az2

B, By and B, are components of body force per unit mass. The solutions of Navier Stoke's equations are possible for such flow conditions where the fluid characteristics such as viscosity and density are constant and boundary conditions are simple. The applications of Navier-Stoke's equations are generally used for the analysis of: (1) (2)

Laminar flow in circular pipes Laminar unit directional flow between stationary parallel plates

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(3) Laminar flow between concentric rotating cylinders (4) Laminar unit directional flow between parallel plates having relative motion. 9. Derive Euler's equation of motion along a stream line for an ideal fluid stating clearly (AMIE 1984) the assumptions. (UPTU 2007-8) Or State the assumptions made in Bernoulli's equation

(UPTU 2004-5)

In Euler's equation of motion, forces due to gravity and pressure are taken into consideration. The assumptions are (1) fluid is ideal i.e. non-viscous (2) flow is incompressible (3) flow is non-turbulence (4) flow is steady (5) flow is irrotational (6) flow is one dimensional i.e. along stream line and (7) velocity is uniform over the cross-section. Eurler's equation is derived considering the motion of the fluid element along a stream line. Consider a stream line in which flow is taking place in S-direction as shown in the figure. Take a cylindrical element of cross-sectional area dA and length dS on the stream line. The forces acting on the chosen fluid element are: (1) pressure force. (PdA) acting in the direction of flow (2) pressure force ( P +

dp ds x dA acting opposite to the direction as )

of the flow and (3) the weight of the fluid element pg dA • ds. The resultant force on the fluid element is S-direction is equal to the mass of fluid multiplied by acceleration in S-direction (as) P + dP •

ds). dA

Stream line

pg . dA . S

c). aP

Forces on Fluid Element

ap as

P • dA — (P + — • ds)dA — p•g•dA ds cos 0 = p • dA • ds • as

—ap as

But

pg cos o = p x as a — dV as V = velocity = f(s, t) dt _

av ds

av

as dt

at

(i)

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Fundamentals of Fluid Mechanics

=V• If flow is steady,

aV at

av + as

as as =V Dt

as Dt

—0

v av

a, — at

Substituting the value of a, in equation (i)

-ap DS

pg cos 0 — p •V•

av as

1 + g cos 0 + V av — p asp as pa

or

0

dz as shown in figure. But cos 9= — ds 1 OP + dz + vaV =0 p as g ds as ap + gdz + VdV = 0 P The above is Euler's equation of motion (Note: The assumptions for Euler's and Bernoulli's equation are same.) 10. Write Euler's equation of motion along stream line and integrate it to obtain (AMIE 1990, UPTU-2002-3) Bernoulli's equation The Euler's equation along a stream line is: dP + gdz + VdV= 0 P Now integrating this equation, we get dP J

or

p

+ 5 gdz + 5 vdv = Constant 1 •P+ g.z + V 2 = Constant P

P +V 2 + z = Constant pg 2g This is Bernoulli's equation of motion. or

11. State Bernoulli's theorem for steady flow of an incompressible fluid.

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Bernoulli's theorem. Bernoulli's theorem states that in an ideal, incompressible and steady flow, the total energy at any point of the fluid remains constant. The total energy of the fluid consists of : (1) pressure energy (2) Kinetic energy and (3) potential energy, Hence: P V 2 + z = Constant pg 2g 12. What are modified Bernoullie's equations when: (1) real fluids are used, (2) Kinetic energy correction factor is applied, (3) energy is taken from flow and (4) some energy is added to the flow? Real fluids. Real fluids are viscous and they offer some resistance to the fluid flow. Hence there is some loss of energy head between two section of the fluid flow. If hf is the loss of energy head due to viscosity, then modified Bernoulli's equation is 2

Vi

PI pg

2g

+ Zi

P2

pg

2 V2 _,_ z2 + hf

2g

Kinetic Energy correction factor 'a'. The velocity of flow is assumed to be uniform in Bernoulli's equation. In actual flow the velocity is not uniform across the cross-section and it is larger than what is calculated by the average velocity. In case kinetic energy correction factor a is applied to the velocity head, then the modified Bernoullie's equation is: 2 2 VI + z1 PI P2 + v 2 + z2 + a1 a2 2g 2g Pg Pg When energy is taken out. The modified Bernoulli's theorem when some energy has been out from the flow between two sections is: PI + V12 + pg 2g

P2 + V22

pg 2g

z2 + energy taken out

When energy is added. The modified Bernoulli's theorem when some energy has been added to the flow between two sections is: 2

P1

V1

pg

2g

2

+ z1 + energy added = P2 + V2 + z2

pg 2g

Energy taken out is through turbine while energy added is through pump. From Bernoulli's theorem, we get HT = head turbine and Hp = head pump. Energy produced by turbine = rjp g Q HT, while energy consumed by pump =pgQHplq where n = efficiency. 13. Describe measuring devices whose workings are based on Bernoulli's equation Following are devices whose working are based on the Bernoulli's equation: 1. Venturi meter. A venturimeter is a device used for measuring the rate of flow of any fluid through a pipe. 2. Orifice meter. An orifice meter is also a device used for measuring the rate of flow of any fluid through a pipe.

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3. Pitot tube. It is a device used for measuring the velocity of a flow at any point in a pipe or channel. 14. What is a pitot tube? How is it used to measure velocity of flow at any point in a pipe (UPTU - 2001-2, UPTU - 2005-6) or channel? A pitot tube is a simple device used for measuring velocity of flow at any point in a pipe or channel. It works on the principle that if the velocity of flow at any point becomes zero, the pressure at that point increases due to the conversion of kinetic energy into pressure energy. This is the reason why pitot tube is also called stagnation or impact tube.

Static head = H Dynamic head = h Stagnation head = H + h Flow

PITOT-TUBE

In the simplest form, a pitot tube consists of a transparent glass tube bent through 90° with both ends open as shown in the figure above. The diameter of tube is kept large enough to avoid any capillary effect. One end of the pitot tube (called body) is inserted into the flow and it is aligned with the direction of the flow. The other end (called stem) is vertical and it is open to the atmosphere. The liquid impacting the body of the pitot tube stops flowing and loses its kinetic energy which results into the increase of the static pressure at the end of the body. The liquid in the pitot tube starts rising due to this increased static pressure. The velocity can be determined by measuring the liquid rise in the pitot tube. Consider point `1' in free flow and point '2' at inlet to the pitot tube. If we apply Bernoulli's equation: T72

T7

P1 + v 12 +z1 — P2 + v 2 +Z2 pg 2g pg 2g Now z1 = z2 as both points are at the same level, V2 = 0, h= rise in pitot tube and H = statical head (pressure head at point '1') H+

Vi2

2g

—h+H 2

Or

or

h= iii 2g V1 = ,1 2gh = flow velocity

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V1 is theoretical flow velocity which should be determined by the pitot tube. However actual velocity measured is found to be less than V1. A correction is therefore applied which is called coefficient of velocity (Cy). Hence flow velocity: V = Cv V 2gh Measurement of velocity in pipes: For measuring velocity of flow, pitot tube is used with piezometer tube in following arrangements: (1) Pitot tube with piezometer tube at upstream is shown below. The piezometer tube measures static heart of the fluid by rise of the level in it while difference of level in pitot tube and piezometer gives dynamic head.

4 V2 =h 2

P V pg

Dynamic head = h=

2 gi

2g

2

2g

Static head = H= P

Pg

' P

H.

Stagnation head = H + h

Pg

'1

/////// ///////////////////// Pitot Tube with Piezometer

(2) Pitot-tube connected with piezometer tube to read dynamic head as shown below.

-4— h

Dynamic head = h

Flow -11P-

Connected Pitot tube & Piezometer

(3) Connected pitot tube, piezometer and differential manometer to measure dynamic head as shown below:

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Fundamentals of Fluid Mechanics

Measuring liquid, Sm!

S gr = S P Dynamic head = x [ S fli—1

Sn,

Flow ----- ---

Connected pitot tube, piezometer and differential manometer

(4) Pitot tube consisting of two concentric tubes measuring dynamic head with the help of differential manometer as shown in the figure below.

Dynamic head = x [ S .11I—1 Sm

Flow

r

Opening

Pitot tube

15. What is a pitot static tube?

The pitot tube is suitable for measuring velocity of the flow in open channel or used for measuring speed of flying aircraft. Special type of pitot tube known as pitot static tube is used with aircraft. This consists of two concentric tubes. Inside tube faces the flow of air and it acts similar to pitot tube i.e. converting the kinetic energy of air into high pressure energy. The outer tube has openings perpendicular to the flow of air and it measures the static pressure of air (low pressure side). These two tubes are connected individually to each limb of the manometer so that the manometer level difference is the dynamic head (h) Speed of aircraft = V =

VVI

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Opening in outer tube

Outer tube (low pressure)

Air_,, flow

Inner tube (high pressure)

Pitot Static Tube 16. Define the term Cd, C,, C„ and derive the equation Cd = Cc x C„ where Cd = coefficient of discharge, Cc = coefficient of contraction and Ci, = coefficient of velocity. (UPTU - 2002-3, UPTU-2006-7)

Cd, C, and C„ are hydraulic coefficients which are used to find the actual discharge, contraction and velocity of the liquid passing out of a orifice in the form a jet. The jet goes on contracting from the mouth of orifice upto a distance of about one half of the orifice diameter. Later it begins to expand. The place where jet has least cross-sectional area is known as vena contracta. The velocity at vena contracta is maximum. Coefficient of Velocity (C,,). It is the ratio of the actual velocity of the liquid jet at venacontracta to the theoretical value of the liquid jet. Since theoretical velocity is determined without any loss, its value is always larger than the actual velocity of the liquid jet. Therefore the coefficient of velocity is always less than one.

C„

— Actual velocity of jet at vena contracta Theoretical velocity of the jet V 271/

The value of C„ varies from 0.95 to 0.99. Coefficient of contraction (Cc). It is the ratio of the area of the jet at vena-contracta to the area of the orifice. Since the jet keeps on converging after passing orifice upto venacontracta, the cross-section area of the vena contracta is minimum. The coefficient of contraction is always less than one.

C,

— Area of jet at vena contracta Area of the orifice

= a

A The value of coefficient of contraction varies from 0.61 to 0.69.

384

Fundamentals of Fluid Mechanics

Coefficient of discharge (Cd). It is the ratio of the actual discharge to the theoretical discharge. Since theoretical velocity and area of orifice is more than the actual velocity of the jet and area at vena contracta, the theoretical discharge is always larger than the actual discharge. Hence the value of coefficient of discharge is always less than one. Cd — (IQ —

( 1 )( a A) tli Vth "

Cd = CvC, The value of Cd varies from 0.61 to 0.65. 17. Define vena contracta. A pitot tube is used to measure the velocity in a pipe with water as the running fluid. The stagnation pressure head is 12 m and static pressure head is 9 m. Calculate the velocity of flow assuming the calefacient of pitot tube equal to 0.985. (UPTU — 2006-7) Vena contracta. The liquid flowing past through an orifice in the form of a jet of liquid is converging type. The jet has a minimum area of cross-section at a distance of about half the diameter of the orifice and this section is called vena contracta. At this section, the stream lines are straight and parallel to each other besides being perpendicular to the plane of the orifice. Beyond this section of vena contracta, the liquid jet diverges and it bends towards downward direction due to the effect of gravity. Static Head (H) = 9 m Stagnation Head = (H + h) = 12 m h = dynamic head h = 12 — 9 = 3 m

where

V = 0 112,gh (I)= Coefficient of pitot tube V= 0.985 ,I2 x 9.81x 3 = 0.985 V58.86

h Vena centrata

Jet of liquid

Vena Contracta

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= 0.985 x 7.672 = 7.556 m/s 18. Describe how coefficient of velocity (Cr) is determined experimentally? (IIT, Kharagpur) Or Explain the method for the determination of coefficient of velocity, coefficient of contraction and coefficient of discharge respectively. (UPTU - 2005-6)

Velocity = V Measuring tank Tank with water at constant head

1. Coefficient of discharge The experiment setup for determining coefficient of velocity is as shown in the figure. It consists of water tank in which water is maintained at constant head. The water is allowed to flow out from the water tank through an orifice and the flowed out is collected in a measuring tank for a definite time 't'. The height of collected water in the measuring tank in time duration of 't' is measured. The actual discharge in time 't' is: Volume collected in measuring tank time Axh t where A = area of cross-section of measuring tank and h = height of water collected in it.

Q-

Theoretical discharge Hence

QM = area of orifice x J2gH

Cd =

Q a x 112gH

2. Coefficient of Velocity (Cr) In case C-C section represents the vena contracta of the jet from the orifice under static head 'H' in the water tank, the horizontal distance travelled by the jet in time 't' is— x = Vt

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Fundamentals of Fluid Mechanics

where

V = horizontal velocity at vena contracta t=

V The gravity is trying to move the jet in downwards direction. The initial velocity of the jet at vena contracta (CC section) towards downwards direction is zero (Vy = 0 t = 0). Hence vertical distance travelled by the jet is: 1 ,2 ,gi

y = Vyt +

Vy = 0 at t = 0

1 y= —2 gt2 But

t_ — v 1 2

Y or

x2 g

v2

V = li 'i = actual velocity 2y

The above is actual velocity while the theoretical velocity is Vth = V 2gh .

Coefficient of velocity —

Vactual

Val

—

x

21b-1 ,

3. Coefficient of contraction Cd = G

x Cv

C — Cd c C, As Cd and C„ is now known, c can be found out. or

19. What is an orifice? Under what heads are the orifices classified? Orifice. Orifice is an opening in a wall of a vessel or tank through which fluid discharges from high pressure side to a low pressure side. The orifice is provided to measure the discharge of the fluid from a tank or reservoir. The orifice may be provided for this purpose at the side or at bottom of the vessel or tank. For a long pipeline, the orifice is placed between the flanges to measure the flow.

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Classification. The orifice can be classified according to their size, shape, shape of upstream edge and discharge conditions as described below. (1) Size. Orifices can be small or large. (2) Shape. Orifices can be circular, rectangular or triangular. (3) Shape of upstream edge. Orifices can be sharp edged, round cornered or bell mouthed. (4) Discharge conditions. Orifice can be (1) discharging free when it has liquid at one side only. Such orifice is discharging in free atmosphere as shown in the figure (2) drowned orifice when it has liquid at both sides while discharging. The orifice may be partially drowned or fully drowned (submerged or partially submerged).

20. Find the expression for velocity when orifice is discharging free.

1 Datum ORIFICE DISCHARGING FREE

When an orifice is discharging in open, the outlet of the orifice is exposed to atmosphere i.e. atmospheric pressure is acting on the liquid jet as shown in the figure when as orifice is discharging to atmosphere. Applying Benoulli's equation to point 1 at the surface and point 2 downstream of vena contracta, we get: z 2 V2 Pi + Vi + Z1 _ P2 + — +z

pg

2g

pg

But

PI = P2 = Patin, h=

v2 2 2g

2g 2 zi = h + z2, V1 = 0

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Fundamentals of Fluid Mechanics

or

V2 =

/2gH

The above equation is known as Bernoulli's equation 21. Derive an expression for a discharge through a large rectangular orifice.

I H2

T

C

h

J_

iI —*1

b

d

LARGE RECTANGULAR ORIFICE

A large rectangular orifice is located at the side of the tank discharging fluid freely into atmosphere under a constant head of the water in the tank as shown in the figure above. The rectangular orifice has width = b and depth = d. The height of water from lower and upper edges of the orifice is H2 and H1 respectively. The jet of fluid will have similar rectangular shape of the orifice. Take an elementary horizontal strip of fluid of depth `dh' at a depth of 'h' from the fluid free surface. Area of the strip = b x dh Theoretical velocity of the fluid in the strip = Fluid discharge through the fluid strip = dQ = Cd x area x velocity dQ = Cd

Or

X

b x dh x ‘12gh

Discharge through the orifice is: Q= =

H2 S

dQ

112

H,

Cd X b x V2gh dh

= Cd b 12g

= Cd b 12g

3

Cd b

112

h1/2 dh

JHl

[h 3/2 3/2 ]

2

[H'2 — HP]

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22. Find an expression for discharge through a fully submerged orifice.

-4— H

I

I

4

d

1-4— b --IT

1 Z2

I

Datum line FULLY SUBMERGED ORIFICE

An orifice fully submerged is discharging from the reservoir having higher fluid level to the reservoir having lower fluid level as shown in the figure above. Applying the Bernoulli's equation at point 1 and point 2, we have: 2 P1 + Viz + Zi _ P2 + V2 + Z2 pg 2g pg 2g V1 = 0, Z1 = Z2 ± H + h, Px — P2 = H Pg z2 + H + h — P2 — P1 + V2 +z2 2g Pg Vi2 = H + = + z2 2g or

V22 _ h 2g

or

V22 = 2gh

Or

V2 = jg71 Qty, = Area of orifice x Velocity =bxdx V2gh Q = Cd b x d 112gh Note: d can also be expressed as H2 — H1 where the depth of liquid is at lower and upper edge of the orifice is H2 and H1 respectively.

23. Find an expression for discharge through a partially submerged orifice.

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Fundamentals of Fluid Mechanics

I-I jci 1-.— b —.IT

PARTIALLY SUBMERGED ORIFICE

Consider an orifice which is partially submerged in the fluid upto depth of H from the free surface of the fluid in the tank having higher level of the fluid. H1 and H2 are the depth of the fluid from the free surface to the upper and lower edge of the orifice. The orifice can be considered to be consisted of two small portions or orifices i.e. one is discharging freely and other is discharging as drowned orifice. Discharge through free discharging portion is2 2 3/2) Cd• b .i.g (H3/2 — HP =— 3 Discharge through submerged portions is— Q2 = Cd• b • (H2 — H) 11 2 gH Total discharge can be gives as— Q = Qi + Q2 2 = — Cd• b 1127g (H3/2 — HP) + Cd• b(H2 — H)..1 2 gh 3 24. Find an expression for time to empty a tank through an orifice fitted at bottom of the tank.

I

1 — dh 1— h

I

I , ±

After time T

Time to empty a tank

A tank has liquid to a height `H' which is being emptying by a discharge through an orifice fitted at bottom of the tank. Let it take time `T' for the level to fall from height H1 to H2 due to the discharge taking place from the orifice. Let A = area of the tank and a = area of the orifice. Volume of liquid leaving the tank in time dT when level falls by dh is given by

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39 I

QdT = A x dh But theoretical velocity at depth of h is V= j2gh Hence discharge through orifice per unit time is— dQ = Cd x area x velocity = Cd X a x j2gh Discharge in time dT is— dQ• dr = Cd Xax 2gh dr In time dT, the discharge from the tank is Q•dT = A x dh or or

—A dh Cd X a x j2gh • dr dT =

—A

. h-indh

a•Cd ,I2g

Total time for level to fall from H1 to H2 dT —

T-

—A

f H2

a• Cd 2g Jill 2A Ca •a•V2g

VH2 ]

To empty completely the tank, time required can be found out by putting level H2 = 0 Hence

T—

2A V Hl Cd •aXX

25. What are different mouthpiece and their uses? (IIT Kharagpur) Or What is a mouthpiece? Why mouthpieces are preferred over orifices? How can coefficient of contraction be increased? Mouthpiece is an extension of an orifice, whose length is two to three times the jet diameter. Mouthpieces are preferred as the discharge can be increased by making them run full of liquid, thereby increasing the coefficient of contraction G. The different types of mouthpieces are1. Short tube or cylindrical external mouthpiece. The vena contracta takes place at a point about one half diameter from the orifice. The mouthpiece in a from of short tube is fixed to the orifice having length of about 1.5 to 2 times the diameter of orifice so that vena contracta of the jet occurs inside the short tube and also jet is fully developed inside the tube before discharging out. Hence the coefficient of contraction becomes unity.

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Fundamentals of Fluid Mechanics

L=

1.5 to 2 times 'cf

___ Liquid

_L Vena contracta

Short Tube Mouthpiece

The mouthpiece is made convergent as shown in the figure. As now the boundary is converging, loss due to sudden enlargement is avoided, thereby increasing the coefficient of discharge to about unity. The nozzles of fire fighting hoses are generally made of convergent mouthpieces as more discharge of water is required.

2. Convergent mouthpiece.

L=0.625 d -.1

Liquid

C

CONVERGENT MOUTHPIECE 3. Convergent divergent Mouthpiece. The

mouthpiece is made converging upto the vena contracta and diverging afterwards. Due to divergence after vena contracta, the velocity at vena contracta increases. Hence partial vacuum is created at vena contracta due to divergence. Both coefficient of contraction and coefficient of velocity increase in this type of mouthpiece.

Liquid ----d

A Convergent Divergent Mouthpiece.

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4. Reentrant or Borda Mouthpiece. A short tube is attached to the orifice projecting inside instead of outside of the liquid tank as shown in the figure. Reentrant or Borda mouthpieces are two types viz (1) running free (2) running full. In running free Borda mouthpiece the length of the tube is equal to the diameter. The jet coming out from the mouthpiece in this case does not touch the sides of the mouthpieces. This is the reason why it is called running free mouthpiece. In running full Borda mouthpiece, the length of the tube is kept three times to the diameter of the tube. The diameter of the liquid jet coming out from the tube has same diameter. This is the reason why it is called running full Borda mouthpiece. If the mouthpiece is running full, vacuum pressure is created at vena contracta which increases the velocity as well as discharge.

d

roil i

A Running free Borda mouth Diece

Running full Borda mouth Diece

Fig. 9.21

26. Distinguish between orifices and mouthpieces

1. Orifice is a small opening of a cross-section on the side or at the bottom of a tank through which liquid is discharging 2. It is used for measuring discharge of liquid. 3. It can be an opening of any type of cross-section i.e. circular, rectangular or triangular 4. It has lesser value of Cd as losses are more 5. It can be (1) discharging free (2) partially submerged and (3) fully submerged

(UPTU- 2004-5)

1. A mouthpieces has length of about two to three times its diameter which is fitted to tank containing liquid. 2. It is also used for measuring flow of liquid 3. It is a tube which is circular

4. It has high discharge as losses are less 5. It can be (1) short tube (2) convergent mouthpiece (3) convergent divergent mouthpiece and (4) Borda mouthpiece.

27. Derive an expression for (1) coefficient of velocity (2) coefficient of discharge and (3) pressure head at vena contracta for flow through a short tube or external cylindrical mouthpiece. Take C, = 0.62

394

Fundamentals of Fluid Mechanics Atmospheric 7---------pressure head = Ho

The figure above shows an external cylindrical mouthpiece of cross-sectional area 'a' fitted to the side of a liquid tank. The jet of liquid entering the mouthpiece converges to form a vena conctracta at section C-C in the mouthpieces and again expands to fill the diameter of the tube at section 1-1. Applying the Continuity equation at C-C and 1-1, we get

a,V, = where a, = area of cross-section at vena contracta, a1 = area of cross-section of mouthpiece, V, = velocity at vena contracta and V1 = velocity at outlet of the mouthpiece. or

V, — al a,

= Cc. V1 where Cc = coefficient of contraction

As vena contracta has minimum area, hence velocity at vena cotracta is maximum. As > V1, there is loss of velocity head due to sudden enlargement of the jet. The head loss is 171 (Vc — Vi) 2

hL

v 1

Cc

2g

2

2g

=V1_ [ _11 2: C', Now C, = 0.62 (given) _ v12 1 IT 2g [0.62 J 0.375 V12 2g Now apply Bernoulli's equation between point A on liquid surface and 1-1 PA 171

pg 2g

+ z, -

+

vi2

pg 2g

+ zi +

Fluid Dynamics-I

395

zA - z1 = H, P1 - 0, VA - 0, PA - 0 Pg Pg

Now

0 375 V2 V2 H= + 1 2g 2g 1.375 V12 2g V1 = 0.855112gH Theoretical velocity at the outlet of the mouthpiece isVth = V 201 Hence coefficient of velocity for the mouthpiece isCv

0.855V2gH V 1 Vth

= 0.855 Since the jet completely fills the mouthpieces at the outlet of the mouthpiece, hence 1. Now we known C, and C„, we can find out Cd using them. Cd = Cv X C, = 0.855 x 1 = 0.855 Pressure head of vena contracta Apply the Bernoulli's equation between point A on liquid surface and C-C, we have PA + VI? ± z. - Pc + Vc2 + zc

pg 2g - pg 2g PA = Ho, Er

n

vA

Pg

ZA Zc

2

Ho + 0 + H = Hc + V2 C 2g Now

Ho

2g H - Hc

V, - 171 0.62 Vc3' _ 17? 2g (0.62)2 x 2g

TT PC

Pg

TT

=

396

Fundamentals of Fluid Mechanics

Ho ± H — 1--4 —

172 1

(0.62)2 x 2g

We have already found out— 2

H = 1.375 V1 2g 2

Hence

Ho + H — ii, — or

V1

(0.62)2 x 2g

H, = Ho + H — 0.894

The above is the expression of pressure head of vena contracta in a mouthpiece. 28. An external cylindrical mouthpiece of diameter 120 mm is discharging with a constant head of 6 m. The coefficient of contraction = 0.62 and coefficient of discharge = 0.86 and atmospheric pressure head = 10.0 m of water. Find (1) discharge and (2) absolute pressure head of vena contracta. Area of mouthpiece (a) or

— 7C d 2

4

a = L x 0 122 = 1.13 x 10-2 m2 4 .

Discharge Q = Cd X a x J2gH = 0.86 x 1.13 x 10-2 x 1/2 x 9.81x 6 = 0.105 m2/s Pressure head at vena contracta is H, = Ho + H

Now Therefore

17c2

2g

Vi — V1 and H — c 0.62 1--4 = Ho + H — 1.89 H = Ho — 0.89 H = 10 — 0.89 x 6 = 10 — 5.39 = 4.66 m

V, —

1.375 V12 2g

29. Find the expression for the maximum outer diameter of the converging diverging mouthpiece. Find also the discharge from the mouthpiece.

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397

A convergent and divergent mouthpiece has converging shape upto vena contracta and after vena contracta it has diverging shape. The shape of the mouthpiece is converging and diverging similar to the stream lined flow in the liquid jet so that losses due to sudden enlargement of the jet are avoided. Due the divergence after vena contracta, the velocity at vena contracta increases. Hence partial vacuum is created at vena contracta due to divergence. However divergence cannot be increased much as then separation of flow may take place. There is a maximum outer diameter of the mouthpiece to avoid the separation of flow. Air pressure head = Ho

di

Applying Bernoulli's equation at point A of free surface and vena contracta C-C v2 v2 . rPa + a + z — 1-n1 + c +z pg 2g a pg 2g c Va = 0 Ho + H = Hc +

vc2

(1)

2g

Vc = \I2g (Ho + H — Hc)

Now apply Bernoullii's equation between C-C and 1-1

V Pc +

2

2

= Pi Vi + — ± zi pg 2g pg 2g ± zc Pi zc = zi, PS = Ao

He +

Putting the value of

Vc2

2g

— Ho + V12

2g

v2 c from equation (1) 2g v2

IL — (Ho + H — 1-1,) = Ho + 1 g

398

Fundamentals of Fluid Mechanics

or As per continuity equation—

V1 = J2gH acvc = aivi

or

ai = Vc ac V1 112g (Ho + H — He)

V 2gh lk+ H — H, H

=l+Ho— H il or

H,

d? —11+ Ho - He

H dc2 The discharge from the mouthpiece is given by— Q = Cd (area x velocity) at the outlet Since jet is full at the outlet, the value of Cd = 1 Q = ai x 12,11 30. The diameter of the throat and exit of a convergent-divergent mouthpiece are 50 and 110 mm respectively is fitted to the side of a tank full of water. If the maximum vacuum pressure is 6 m of water, find the maximum head of water for steady flow. The atmospheric pressure head is 10.3 m of water. Find also the discharge from the mouthpiece. The maximum vacuum pressure will be at throat only. Hence absolute pressure at throat = 6 m of water Now He = atmospheric pressure — absolute pressure at throat = 10.3 — 6 = 4.3 m Now

al _ 1 1/0 — Hc 1+ H ac Ai cl? = 1 10.3 — 4.3 1+ H do 2 ( 0.1 ) = Ii+ 6 H 0.05)

Fluid Dynamics-I

399

6 4= 1/1 + —

or

H=

6 = 0.4 m of water 5

Q = ac IC

x (0.05)2 4

x V 2 x 9.81 x 0.4

3.14 x 25 x 10-4 x 2.8 4 = 54.97 x 10-4 m3/s 31. Find the expression for hydraulic coefficient and discharge for Borda's mouthpiece running free.

Running free broda's mouthpiece

In this a short cylindrical tube is attached to the orifice in such a way that the tube is projected inwardly in the liquid tank. Hence it is also known as internal mouthpiece. In running free Borda's mouthpiece the length of tube is kept equal to the diameter of the mouthpiece. In this condition, the jet of liquid comes out from the mouthpiece without touching sides of the mouthpiece. The cross-section area of the jet (a,) is lesser than the cross-sectional area of the mouthpiece (a) i.e. a > a,. The flow through the mouthpiece is taking place due to pressure intensity at the mouthpiece which is equal to pgH. Force = area x pressure intensity = a x pgH If V, is the velocity of the jet, then mass of fluid flowing per unit time is— M = p x a, x

(1)

400

Fundamentals of Fluid Mechanics

Force = Mass x acceleration = p x a, x V, (V, — 0) = p a, v,2 a.p. 01 = p•ac.v

or

from equation (1)

Applying Bernoulli's equation between point 'A' on free surface and section 1-1, we get— PA

+ 171 +

pg

2g

z ZA = —

+ vj _, Z

pg 2g 1

zA — zi = H,

Now

H=

PA

P2

— c = 0, Pg Pg

VA = 0

V2 C 2g

V, = J2gH

Now we have found out— a•p•g•H = p•ac• VI'

Put the value of Vc from eqn. (ii), we get— a•p•g•H = p•ac•2gH a = 2 ac

or or

a

= C, = 0.5, C, = coefficient of contraction

As there is no loss of head due to running free of the liquid jet, therefore coefficient of velocity C, = 1 . Coefficient of discharge Cd = C, x Cy = 0.5 x 1 = 0.5 Discharge Q = Cd• a• 1.12gH = 0.5 a V2gH 32. Find the expression for the hydraulic coefficients and discharge for Borda's mouthpiece running full.

Fluid Dynamics-1

40 I

Running full Borda's mouthpiece.

The jet of liquid converges upto section c-c, and later it expands suddenly, thereby suffering head loss due to sudden enlargement. If V, is velocity at C-C and V1 is velocity at 1-1, then we have— Head loss — (V

—V1)2 2g

From continuity equation, we have— a,V, = al V1 but C, at vena contracta = ac = 0.5 al

Vl = ac V, a1

Or

V1 = 0.5 V, V, = 2 VI

or

-171)2 Head loss (HL) — (V 2g (2V1 —V1)2

2g Or

HL =

V2 1 2g

Applying Bernoulli's equations between point A on free surface & section 1-1 PA Pg Now

VA

2g

ZA

Pl + v12 +zL + hL

Pg 2g

ZA — = H, 2

PA — P1 = u Va = 0, Pg Pg 2

H = Vi + Vi 2g 2g

= 1712 2g

402

Fundamentals of Fluid Mechanics

2

H= 171 2g or

V1 =

V gH

V1 is the actual velocity of jet at exit for the mouthpiece while the theoretical velocity at exit is Vth = 11 zgH Coefficient of Velocity —

171 _ v ,-/ vth 2gH

= 1 — 0.707

,2 As the mouthpiece is running full, the area of jet is same as that of the mouthpiece. Hence coefficient of contraction at exit is one C, = 1 Now

Cd = Cc X Cv = 1 x 0.707 = 0.707

Q = Cd X a x 112gH = 0.707.a• V 2gH 33. An internal mouthpiece of 1.0 m diameter discharging water at constant head of 9 m. Find the discharge when (1) running free and (2) running full a=

7rd

2

— 1—r x 12 4

Running free Q= 0.5 x a x J2gH = 0.5 x

IT X 12

4

x 112 x 9.81x 9

0.5 x 3.14 x 13.29 — 5.22 m3/s 4 Running full Q= 0.707 x a x 2.-1 ir x 12 x V2 x 9.81x 9 4 = 7.37 m3/s = 0.707

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403

34. What do you understand from free liquid jet? Find (1) maximum height (2) time of flight and (3) maximum horizontal span of the jet or range. What is the condition for maximum range? 41 Vy =

0

CI Ar

ur

.--

-

U(x, y)

---, hma,

0 Ux

,i• R JET TRAJECTORY

When a jet of liquid is ejected from a nozzle, it is exposed to atmosphere. Under the action of gravity, the jet instead of going straight along angle 0 starts turning downwards, thereby jet traverse in a parabolic path or trajectory. As shown in the figure, the jet starts from point A(x, y) on the centre line of the jet with velocity U and angle '0' from the horizontal. component of velocity is towards vertical y-direction and Ux component of velocity is in horizontal direction (x-direction). Displacement in x-directions Sx is— Uy

Sx = Uxt + 21 ax t2 But acceleration

ax = 0 Sx = U, x t = U cos 9.t or

t—

Displacement in y-direction

x U cos0

where Sx = x

(Sy) is-

SY = U t + —1 a t 2 Y Y sin (9 and ay = —g

Uy = U

Sy = U sin Bt- 21 gt2 = y t

Now

—

x U cos°

y = U sin 0.

k U cos 0

1 x2 g U 2 2 cost 0

x2 8 sec2 9 2U2 This is the equation of the path of the jet and it is parabolic. Or

y = x tan 0

404

Fundamentals of Fluid Mechanics

Maximum Height At maximum height (h), the vertical velocity component

V, = 0

at point C on the trajectory;

V; = U + 2. ay• h

0 = (U sin 0)2 — 2.g. h or

h—

U 2 sine

2g

Time of flight y = Uyt — gt2 where y = vertical distance As jet goes up and comes back, hence net vertical displacement travelled is zero i.e. y = 0 UyT = 1 gr2 or

T

2.

y

g T —

T = Time of flight — 2•U sine g

2•U sine g

From this, time to reach highest point can be found out which is half of the total time of flight. The total distance travelled by a fluid particle in the jet before it falls to the ground is— 1 a t2 x= Ut +— x 2 x Ux = U cos 0, ax = 0, t = T — x = Ux x

2.U sine

2•U sine g

U cos0 x 2•U sine g

2. U 2 sin d cos 0 g U2

sin 20 as sin 20 = 2 sin 0 cos 0. g

Condition for maximum range Range = x —

U 2 sin 20

Fluid Dynamics-I

ax

For maximum range,

DO

405

=0

ax _ U 2 cos 20 x 2 = 0

ae or or or

g

cos 20 = 0 20 = 90° 61= 45°

35. A nozzle is situated at a distance of 1.0 m above the ground level and 45° to the horizontal. The diameter of nozzle is 100 mm and the jet strikes the ground at 4 m from the nozzle. Find the discharge from the nozzle.

x=4m Ux x t = 4 where

h = U t — —1 gt 2 Uy = vertical velocity Y 2 h = —1 m when it touches ground at B w.rt point A

Now

4 —1 = U • Y Ux Now

U., = horizontal velocity

1 g x 16 2 Ux2

Ux = U cos 45 =

Uy = U

,s12

cos 45 — 112

—1 = 4 — — x 9.81 x 2

16 x 2 U2

— 9.81x16 5 U= 5.6 m/s

U2 Or

Discharge Q = 4 x area of nozzle Q = 5.6 x

Ir X (0.1)2

4 = 4.396 x 10-2 m3/s

406

Fundamentals of Fluid Mechanics

36. What is a weir? What is the difference between a weir & notch? or Distinguish between weir and notch? (UPTU - 2005-6) Flow through an open channel takes place under atmospheric pressure. The motive force to create the flow is derived from the slope of the bed/ground of the channel and the kinetic energy possessed by the flowing liquid. A weir is an obstruction placed in an open channel over which the flow occurs. The weir is generally in the form of a vertical wall with a sharp edge at the top, running all the way across the open channel. When the liquid flows over the weir, the height of the liquid above the sharp edge of the weir bears a relationship with the amount of discharge across it. This is the reason why weirs are widely used for discharge measuring purpose. 1

A notch is a sharp edged device which permits the liquid to go through it while liquid downstream exposed to the atmospheric pressure. The water may be flowing out from a tank or reservoir by providing a notch in the way of a depression in the side of a tank or a reservoir. The only difference between a weir and a rectangular notch is that a weir runs all the way across the channel but a notch needs not run full way across the channel. Theoretically a weir or notch may be regarded as special form of wide orifice in which the free water surface does not touch the upper edge of the orifice. 37. What is a spillway? When water is made to flow over a dam like a weir, it is known as spillway. The flowing water is in fact excess water which cannot be stored in the reservoir. The spillway can be (1) open spillway and (2) syphon spillway. In syphon spillway, water level at upstream is connected to the downstream by the help of a syphon made with pipes. When water level rises above the crest of the spillway, water starts flowing down through the syphon. Water entering the syphon strikes at point 'X' and fills up the syphon fully upto downstream level. However there is space in the kink at point ' where air is entrapped. The flowing water in the syphon starts sucking the trapped air, thereby creating a vacuum at point Y. The vacuum increases the rate of flow over the crest of the spillway. The increased flow sucks completely the air in the kink, thereby making the syphon to operate fully.

Fluid Dynamics-I

407

SYPHON SPILL WAY

OPEN SPILL WAY

38. Explain : (1) Crest or sill of the weir (2) nappe or vein and (3) drowned or submerged weir and (4) free discharge. Crest or sill ---------

Nappe or vein

WEIR

The upper surface or edge of the weir over which the water flows is known as crest or sill of the weir: The overflowing sheet of water over the weir is called nappe or vein. If the nappe discharges into air, the weir is said to have a free discharge. If the discharge takes place above the sill or crest, the weir is said to be drowned or submerged weir. It can be totally drowned or partially drowned depending upon nappe is above or below the crest at downstream. 39. Find the expression for discharge over a rectangular weir. (UPTU 2008-9)

dh

L Weir

Free flow over a rectangular weir

The figure above shows the flow over a rectangular weir of length L, and height of liquid is `H'. Take an elemental strip of height `dh'. The area of the liquid strip is VA' = L • dh. The discharge of liquid through this strip is— dQ --= Cd x dA x V

408

Fundamentals of Fluid Mechanics

V = j2gh dQ = Cd X L•dh x 1,27 Q=

H s o

cd.L•x

•

h1/2dh

=— 3 • Cd• V 2g •L • H3I2 Note. The formula for discharge for weir is same as that of a large orifice if depth of upper edge is zero. 40. Daily record of the rainfall over a catchment area is 250 million litres. If 7% of the rain reaches the storage reservoir and then passes over a weir, find the length of the weir in order that the water shall never rise more than 60 cm above the crest. Take Cd = 0.6. (Jadhavpur University) Catchment water = 250 x 10+6 litres/day 250 x 106 103 x (24 x 3600)

m3/s

= 2.89 m3/s Q = water reaching reservoir = 100 x 2.89 = 2.02 m3/s

Discharge over the weir is given by . Cd. x .1, .H 3/2

Q=

Maximum rise over the weir = 0.6 m 2 Q= — x 0.6 x V 2 x 9.81 •L x (0.6)3/2

3

2.02 = — 2 x 0.6 ,119.62 x L x (0.6)3/2 3 Or

L = 2.46 m

41. Find the expression of discharge passing through a triangular notch/V notch. In case 0 = 90° and Cd = 0.6, what will be the value of the discharge?

Fluid Dynamics-I

409

H

A

Triangular Notch

The figure shows a flow through a notch having angle i and height of liquid is 'H'. Take an elemental strip of dh thickness at the depth of h from the free surface. Velocity (V) of the liquid in the strip is V= V2gh BD = (H — h) tan e/2

Now

Hence length of strip L = 2 x BD = 2(H — h) tan 0 Area of strip = L • dh = 2(H — h) tan 0/2. dh Discharge through strip = dQ = Cd x 2(H — h) tan 1 ( x dh x V Or

.

dQ = Cd x 2 (H — h) tan x V 2gh x dh Q = 2. Cd tan 11 • liTg IH(H — h) hil2dh o 2 i— H.h3/2 h"2

2. Cd tan 11 • I/ 2g 2 = 2. Cd tan

2

3/2 5/2

H

o

• l2g • H512 x 1+5

e • Cd tan — • .W •H5/2 2 $ The above is the equation of discharge from a triangular notch. Now Q = 90, Cd = 0.6 =

.

Q = li • 0.6. 1/2 x 9.81 .H5/2 = 1.417 H512

42.

A reservoir having a surface area of 500 m2 is emptied by a 0.5 m wide rectangular weir. How long should it take to empty the reservoir from a height of 4 m to 1.0 m above the sill? Take Cd = 0.6.

4I 0 Fundamentals of Fluid Mechanics

dh h

Rectangular weir

Discharge from weir is given by2

Q = Cd• • V 2g • L h3/2 Suppose for VT' time, the level falls by dh. If A = Surface area of the reservoir, then we have A dh = Q•dT or

.

dT — A dh Q A dh ,g • Lh312 Cd ••11T

3A •h-3/2dh 2. Cd V 2g • L T=

f dT = f H2

3A

H1 ( 2. Cd V 2g • L

) 17-3/2dh

H2

3A [h-12 4 T— 2•Cd \ I2g • L 3 x 500

[1

0.6 x 19.62 x 0.5 L.,/i

1

Ji

3 x 500 0.3 x 4.43 [1 — 0.5] = 564.33 sec = 9.4 mm 43. Find the discharge of water over a V-notch of angle 60° when the head of water over its crest is 1.0 m. Take Cd = 0.6 and neglect velocity of approach. Discharge for a V-notch is given by— Q= =

15

19 H5/2 • Cd x .X tan — 2

8 .0.6 x 1/2 x 9.81 x tan 30 x (1)94 15

Fluid Dynamics-1

4I1

8 — x 0.6 x 4.43 x tan 30 15 = 0.818 m3/s 44. What are the advantages of triangular notch over rectangular notch? Or Why is triangular weir more suitable than a rectangular weir? The advantage of triangular notch are — 1. The discharge can be measured by reading the height of the water 2. The discharge can easily be found out for notch having Q = 90° using formula Q = 1.417 H512 3. Triangular notch can give accurate results even for small discharge as compared to rectangular notch because the height of water in a V-notch is more as compared to a rectangular notch. 4. The velocity approach in case of triangular notch can be neglected 5. Triangular notch does not require ventilation at downstream side. 45. What do you understand from the ventilation of weirs? What are different types of nappe? Weir

Nappe Ventilation holes

Space where air is trapped

Ventilation of Weir When the water is flowing over a rectangular weir having no end contractions, the neppe touches the side walls of the channel. Since neppe falls away from the weir wall, a space is created by the nappe with the weir where air trapped as shown in the figure. When this trapped air is carried away by the flowing water, a negative pressure is created beneath the nappe. The negative pressure draws the lower side of the nappe towards the surface of the weir wall. Hence more water is drawn from the channel through weir due to the negative pressure below the nappe. Hence discharge is actually more than that is measured. The types of nappe are: (a) Free nappe. Holes are made through channel walls at the place just below the nappe which are connected through a pipe to the open to atmosphere. (b) Depressed nappe. The air is left below the nappe and pressure below nappe is negative. Due to this, discharge is 6 to 7% more than what is obtained from the free nappe. (c) Clinging nappe. The water stream clings to the downstream face of the weir and nappe has no air beneath it. Discharge in this case is about 25 to 30% more than what that can be obtained from the free nappe.

4I2

Fundamentals of Fluid Mechanics

46. What is end contraction? When a weir is straight and it extends over the full width of the channel, then this weir is Channel Weir with end contraction (width of nappe < width of the weir)

Rectangular Weir with end contraction

called weir without end contraction or suppressed weir. As the sides of the wall of the channel will suppress any end effect. The measurements made by the use of formula are accurate. If weir width is less than the width of the channel, it is known as weir with end contractions. The correction factors are to be applied to the calculated discharges for fmding the accurate results. The end contraction results into the decrease of the width of the neppe. It can be one end contraction, two end contractions or number of contraction as shown in the figure. The approximate value of end contraction is 0.1 x H where H is the height of the water over the crest. The effective width of the nappe = n x 0.1 H where n = number of contractions Weir as seen from top

xrwaesma maw vavek

1 0.1 H Fully suppressed weir: n = 0

One side contraction n=1

Two sides contractions n=2

Number of end contractions: n = 8

47. Derive an expression for discharge from a trapezoidal weir? How can you avoid the (11T KHARAGPUR, UPTU 2008-9) effect of end contractions in this?

Fluid Dynamics-I

4I3

(UPTU 2008-9) L

///hill/

//////////////

Fig. 9.37

Trapezoidal Weir A trapezoidal weir is a combination of a rectangular and a triangular weir. Hence the discharge through a trapezoidal weir is the sum of the discharges from rectangular and triangular weirs. Q = Qrectangular Qtriangular 2 2 8 =— 3 Cd • 1127g L•1131 + 5 Cd 2 tan 0 X H5/3 X 12. g Cippoletti weir is a trapezoidal weir which has side slope of 4 to 1(1 = horizontal & 4 = vertical). Such slope increases discharge through the triangular portion of the weir which would have otherwise decreased the discharge in case the side walls were vertical as in the rectangular weir. Hence such trapezoidal weir does not have end contractions. 48. What do you understand from the broad crested weir? Or What is the difference between a broad crested and narrow crested weir. Explain with a neat sketch. (UPTU - 2006-7) 4

Broad crested weir

Weirs which are not sharp crested are called broad crested weirs. Such weirs generally form a part of dam and these weirs are then called spillways. The thickness of weir in a broad crested weirs is more than 0.47 H (H = height of water over the crest)

4I 4

Fundamentals of Fluid Mechanics

49. What are drowned or submerged weirs? Drowned weir

Drowned or submerged weir When the downstream water level of a weir nappe is higher than the crest of the weir, the weir is said to be drowned or submerged weir. For example, a spillway of a dam when drowned during the floods becomes a submerged weir. The sluice gates installed in concrete structure in a river control act as submerged weirs when the gates are raised and water flows through the opening. Such weirs have great discharging capacity, thereby, preventing any flooding of the area at upstream. The discharge over a submerged weir is calculated by dividing the nappe in two parts. The upper part acts as a free weir with head as H1 — H2 and the lower part acts as a submerged weir under a head of 1/2. 50. What is velocity of approach? The water approaches to the notch or weir with slight velocity which is called the velocity of approach. Generally for slight approach velocity, this velocity is neglected. However when this velocity is appreciable, an additional head (ha) of

V0 2 (where V0 = approach g

velocity) is added to the static head as shown in the figure so that the actual increased discharge matches with the calculated discharge by the formula. V2 h= a 2g

Fluid Dynamics-I

4I5

Additional head due to velocity of Approach Discharge taking into consideration addition head for rectangular and triangular weirs are— QRectangula =

3 Cd

2S

.L [(H + 10312 — (ha)3/2]

9 QTri angular = — Cd s,y2g tan 0/2 [(H + 10 512 — (ha)5/21 15 51. Derive an expression for time required to empty a tank with a triangular notch.

Time for Emptying Consider a flow through a triangular notch (angle 0). Suppose the level of the liquid is at height 'h' from the vertex when the surface falls by `dh' due to discharge 'Q' of the liquid in time VT'. If surface area of the reservoir is `A', then —A • dh = Q • dr But

8 r— tan 2 .h5/2 Q = 5 Cd• v2g —Adh = ± 1 •Cd• X Stan • h5124T 2 15 H2

or

J dr =

T=

—Adh

e 8 H, — •Cd • tan — • v 2g h3/2 2 15 h"2

—15A

1112

8 Cd V2g tan I( [ —3/2 HI

—5A 4•Cd • V2g

(

)3/2

1) • tan 0 [( H2

( 1 )3/21

I\Hi

52. Water is flowing through a tapered pipe from section '1' and section '2'. The discharge through the pipe is 0.4 x 102 m3/s. Find the intensity of pressure at section `2' if di = 0.3 m, d2 = 0.15 m, p1 = 400 leST/m2, = 10 m, z2 = 6 m.

4I 6

Fundamentals of Fluid Mechanics

= 0.3 m P1 = 400 kN/m2 d1 = 0.15

Z2 = 10 m Zi= 6 m Datum

Taperd pipe a1=

4 di =

a2 =

4 d2= 4(0.15)2 = 0.0177 m2

,Irt

(0.3)2 = 0.0707 m2

Apply continuity equation Q= a1 V1 =a2 V2 Q Qi

4 x 10-2 — 0.567 m/s .0707

4 x 10-2 — 2.264 m/s .0177 az Apply Bernoulli's equation between section 1 and 2. V2

pg

Vi2 2g

+

=

Q

=

2 P2 + pg 2g 2

Z1

+

z2

4 x 103 + (0.567)2 + (2.264) 2 P2 + 10 — +6 2 x 9.81 2 x 9.81 103 x 9.81 103 x 9.81 0.408 + 0.0163 + 10 —

2 +0.261+6 103 x 9.81

, 2 — 4.2233 10' x 9.81 41.4 x 103 N = 41.4 lcN

"2 =

53. Water flows through a uniform diameter pipe of 200 mm diameter, point A and B are at elevation of 8 and 10 m respectively along the inclined pipe. Pressures are 50 kN/ m2 and 20 leST/m2 at A and B respectively. Find (1) direction of flow and (2) head loss between the points if the rate of flow is 0.06 m3/s.

Fluid Dynamics-I

4I7

10 M 8M

1 Inclined Pipe Given:

d= 0.2 m, zA = 8 m, zB = 10 m Q = 0.06 m3/s, PA = 50 1cPa, PB = 20 1cPa

Applying the continuity equation we get— Q = AVA= AVB or

VA = VB —

Total head at A =

0.06 — 1.909 m/S Jr — x (0.2)2 4

V2 + A + Za pg 2g

P A

50 x 103 + (1.909)2 +8 2 x 9.81 103 x9.81 = 13.28 m Total head at B

-

pB

q + _

pg 2g

+ZB

20 x 103 + (1.909)2 + 10 2x 9.81 103 x9.81 = 12.22 m Since head at A > B, the direction of flow is form A to B Loss of head = head at A —head at B = 13.28 — 12.22 = 1.06 m of water 54. A pump of 20 HP with 90% efficiency is discharging crude oil (sp. gr. = 0.8) to the overhead tank (EL = 30 m) from the storage tank (EL = 9 m) as shown in the figure. If the losses in the whole system is 2m of flowing fluid, find the discharge. Pressure in storage tank is 37.67 liN/m2.

4I 8

Fundamentals of Fluid Mechanics

EL - 30 Pressure = 37.67 kN/m

2

EL = 9 m

Pumping Storage to Overhead Tank

Applying Bernoulli's equation between point A &B PA

pg

2 + VA 2g

± zA± H_ P

Now

_

PB pg

+ 11 — _.k 2g

_.i_ aw ZB a Loss

zA = 9 , zB = 30 , PA = 37.67 lth VA :z, 0, VB :z, 0, PB = 0 (atmospheric), Hp = pump He

37.67 x 104 +9 +HP =30+2 0.8 x 103 x 9.81 48 + 9 + Hp = 32 or Hp = 32 — 13.8 = 182 m Pump horse power = 20 HP ri = 90% Energy supplied to the system by the pump = 0.9 x 20 = 18 HP HP — or

Q-

pg•Q•H p 0.746 1.8 x 0.746 x 103 0.8 x 103 x 9.81 x 18.2

= 0.094 m3/s 55. Brine of sp. gr. 1.15 is draining from the bottom of a large open tank through a 80 mm pipe. The drain pipe ends at a point 10 m below the surface of the brine in the tank. Considering a stream line starting at the orifice of the brine in the tank and passing through the centre of drain line to the point of discharge and assuming the friction is negligible, calculate the velocity of flow along the stream line at the point of discharge from the pipe. (AMIE 2000)

Fluid Dynamics-I

4I9

DRAINING FROM PIPE

Applying Bernoulli's equation between point 1 and 2, we getP1 ± V1 2 ± z = P2 ± V2 2± z2 pg 1 pg 2g 2g Now

z1 — z2 = 10, P1 = P2 = Palm Pg Pg Pg V2 we get 10= g Or

V1 = 0,

V2 = 112g x10 = 14 m/s

56. The figure shows a turbine with inlet pipe and a draft tube. If the efficiency of turbine is 80% and discharge is 1000 litres/s, find (1) the power developed by the turbine (2) the reading of gauge G.

CD ,

2 350 kN/m

72m

Dia = 0.4 m of inlet pipe

llrn

I

Datum Draft tube

Q = 1000 litres/s = 1 m3/s A1 = area of inlet pipe =

Ird2 7r x 0.42 — 4 4

420

Fundamentals of Fluid Mechanics

= 0.1257 m2 1 Velocity at inlet = V1 — Q — Al 0.1257 = 7.96 m/s Let HT = turbine head developed per unit weight of the water and 11L, = head losses = 0 Applying Bernoulli's theorem between point (1) at inlet and point (2) at free surface and taking free surface as datum. T,2 _, Ti P1 ± V12 _,, Z1 - P2 + v 2 _, _L Er -I- Z2 J./ T ' "L pg 2g pg 2g .

350 x 103

+ (7.96)2 + 5 = HT 2 x 9.81 103 x9.81 .. Power developed

HT = 35.68 + 3.23 + 5 43.91 m P = ii rit g HT rixQxpxgx HT 0.8 x 1 x 1 x 103 x 9.81 x 43.91 344.6 kW

57. Figure shows a pump employed for lifting water from a sump. If it is required to pump 60 litres/s of water though a 0.1 m diameter pipe from the sump to a point 10 m above, determine the power required. Also determine pressure intensities at L and M.

Dia = 0.1 m

Lifting From a Sump

Q= 60 litres/s = 0.06 m3/s A= area of pipe = L x d2 = L x (0.1)2 4 4

Fluid Dynamics-I

42 I

= 0.00785 m2 Q = AV2 where V2 = velocity at outlet Q 06 — .— 7.46 m/s A 0.00785 Applying Bernoulli's equation between point 1 at the surface of water and point 2 at the outlet, we get— or

velocity —

P1 ± 1112 P, V, +z+H — + + z2 + HL P pg 2g pg 2g

Hp = Head developed by the pump HL = Head loss in the system = 0 P2 = atmospheres, V1 = 0, V2 = 7.46 m/s Now, P1 Pg Pg = 0, z2 = 10 0 + 0 + 0 + Hp = 0 + (7.64)2+ 10+0 2x9.81 or Power to run pump

Hp = 12.97 m of water P = mg Hplq _ pxQxgx12.97 0.7 1x 103 x 0.060 x 9.81 x 12.97 0.7 = 10.9 kw

Applying Benoulli's equation between point 1 and point 'V we get z + +V i z _ + + zL pg 2g 1 pg 2g L

0+0+0 — P

pg

PL

Pg or

(7.64)2 ±4 2g

= —6.97

x 103 x 9.81 x6.97 PL = = —68.4 kN/m2

Applying Bernoulli's equation between points 1 and M, we get v2 Pm + zi + Hp +z pg 2g pg 2g

422

Fundamentals of Fluid Mechanics

P1 = 0, V1 = 0, Z1 = 0, Pg Zni = 8 m

Now

0 ± 0 ± 0 ± 12.97 =

Hp = 12.97 m

Pm (7.64)2 + +8 2g Pg

Pm — 12.97 pg

(7.64)2 8 2x 9.81 = 12.97 — 2.97 — 8 = 2 m Pm = 1x 103 x 9.81 x 2 = 19.62 kN/m2

58. A submarine is cruising at a depth of 20 m in ocean (sea water sp. gr. = 1.02). If the speed of the submarine is 10 m/s, find the reading of pitot tube and static pressure probe which is located at upstream. Static pressure Pstatic = p g H = 1.02 x 103 x 9.81 x 20 = 200.12 kPa Dynamic head

TT

d

H a

V2

2g (10)2 2x9.81

= 5.1 m Dynamic Pressure Pd = p g Hd = 1.02 x 103 x 9.81 x 5.1 = 51 kPa Pitot tube reading = dynamic head = 51 kPa Static pressure probe reading = 200.12 kPa Stagnation pressure = 200.12 + 51 = 251.12 kPa 59. A pitot tube is used to measure velocity of an aeroplane. An u-tube differential manometer gives a reading of 6 m of water. If sp. wt. of air is 11.75 N/m2 and coefficient of pitot tube is 0.98, find speed of aeroplane. Assume compressibility as zero. 11.75— 1.198 x 10-3 lx103 x9.81 SW = sp. gr. of water = 1 S,,„ x hii, = Sa x ha Sa = sp. gr. of air =

New

Fluid Dynamics-I

423

= 1, Sa = 1.198 x 10-3 h,„ = 6 cm ha -

lx6x10-2 1.198 x 10-3

Dynamic head of the pitot tube = hd = ha = 50.08 Now

V = C„\ f2ghd

= 0.98 V2 x 9.81 x 50.08 = 0.98 x 31.34 = 30.7 m/s The speed of the aeroplane is 30.7 m/s. 60. A pitot tube is placed at the centre of pipe having diameter of 0.25 m in which water is flowing. The pitot tube records 7.85 kN/m2 as stagnation pressure. The static pressure in the pipe is 0.04 m of Hg (gauge vacuum). Find discharge if mean velocity is 0.8 times the maximum velocity. Take

Cy = 0.98 Pstagnation = 7.85 kN/m2 Static pressure = - Pstatic = 0.040 m of Hg (vacuum) - 0.04 x pHg x g = -0.04 x 13.6 x 9.81 - 5.34 kN/m2 Pstagnation = Pstatic Pdynamic Pdynamic

= 7.85 - (-5.34) = 13.19 kN/m2 13.19 x 103 1x103 x 9.8

Pdynamic Hdynamic

pxg

1.34 m of water Maximum Velocity

Cd 2 -g H dynamic

0.98 112 x 9.81x 1.34 5.03 m/s Mean velocity 0.8 x Maximum velocity 0.8 x 5.03 = 4.024 m/s Discharge Q = Vmean x area of pipe 4.024 x

7t X

0.196 m3/s

(.25)2 4

424

Fundamentals of Fluid Mechanics

61. A Pitot tube having a coefficient of 0.97 is used to measure the velocity of water at the centre of a pipe. The stagnation pressure head is 8 m and the static pressure head is 4 m. What is the velocity? Hstagnation

=8

Hstatic = 4

Hdynamic = 8 — 4 = 4 V = Cd 2g Hdynamic

=0.97 1/2 x 9.81x 4 = 8.593 m/s 62. The head of water over an orifice of diameter 50 mm is 15 m. Find the actual discharge and velocity of the jet at vena contracta if Cd = 0.6 and Cy = 0.98.

T 15 m

ORIFICE : DISCHARGE & VELOCITY

Vth

=

V2gH

= j2x9.81x15 = 17.16 m/s x (0.05) 2 4 4 = 1.96 x 10-3 m2 Qth = Vth X A = 17.16 x 1.96 x 10-3 = 3.36 x 10-2 m3/s A

TCd 2

it

Qactual = Cd X Qth

= 0.6 x 3.36 x 10-2 = 2.02 x 10-2 m3/s Actual velocity = Cy x Vth or Vactuai = 0.98 x 17.16 = 16.82 m/s The above is the velocity at vena contracta.

Now

63. An orifice has a diameter of 10 cm which is discharging at the head of 6 m. The

Fluid Dynamics-I

425

diameter of the jet at vena contracta is 9 cm. If the discharge is 60 litre/s, find the hydraulic coefficients of the orifice. 7E X

Area of orifice

(0.1) 2 — 7.85 x 10-3 m2 4

Jr X (0.09)2 — 6.36 x 10-3 m3 4

Area of vena contracta

Coefficient of contraction = Cc

— Area of vena contracta Area of orifice

6.36 x 10-3 7.85 x 10-3 = 0.81 Theoretical discharge Qth = A x Tim Vth = V2gxh

However

= 112 x 9.81x 6 = 10.85 m/s Qth — 10.85 x 7.85 x 10-3 = 0.0852 m3/s Qactuai = 60 x 10-3 m3/s .*.

Cd —

Qactuai Qth 60X 10-3 85.2 x10-3

= 0.704 Cd = Cc X Cy

Now or

Cy — Cd Cc _ 0.704 0.81 = 0.86

64. A jet of water emerging from a sharp edge vertical orifice under a constant head of 20 cm at certain point has the horizontal and vertical coordinates measured from the vena contracta as 38 cm and 21 cm respectively. Find hydraulic coefficients of contraction and velocity.

426

Fundamentals of Fluid Mechanics

i H = 20 cm

T [4— x=38 cm-1.1

Jet of Water Vth = V2gH

= V2x9.81x0.2 = 1.98 m/s x = Vactual X t

(1)

y = 0 + 1gt2 From eqn. (1) y

Or

Vactual

_ 1., x g( x 2 Vactual =

x 112y NI

Or

Vactual

g

0.38 2 x .21 11 9.81

0.38 0.207

—

1.83 m/s

= 17th _ L83 _ 0.927 1.98 Vactuo Cd = Cy Cc= 0.927 x 0.6 = 0.556

cv Now

65. A tank containing water is provided with a sharp edged circular orifice of diameter = 8 mm. The height of water in the tank is 1.5 m above the orifice. The jet strikes a wall 1.5 m away and 0.5 m vertically below the centre of the orifice. If discharge is 4 litres/s, find (1) hydraulic coefficients (2) power loss in orifice. Vth = .J2gH

= /2xgx13 = 5.42 m/s

Fluid Dynamics-I

X = Vactual X

427

t

1 12 y= — 2g

or

y= 1 ( x 2 g Kauai )2 Vactuai

2y/g 15 — 4.698 m/s 112x OS 4.81

Cy

Qth

V t — 4.698 acual 5.42 th = 0.867 Ti

7rd2 X Vth 4 x (8 x 10-3)2 x 5.42 4 = 2.72 x 10-4 m3/s

Cd

actual _ Qth

2.72 x 10-4 4.0 x 10-4

= 0.68 Cd = Cy

or

X

Cc

Cc = 0.68 = 0.78 0.867

66. A tank has two identical orifices at one side in its vertical wall. The upper orifice is 4 m below the water surface while the lower one is 6 m below the water surface. In case the value of coefficient of velocity (Cy) is same and equal to 0.98, find the point of intersection of the two jets from the orifices.

T

4

6

-F Y2

x

T y1

428

Fundamentals of Fluid Mechanics

Two Orifices

Cv2 — Also

2,/ y21/2

Cyl = Cv2, therefore or

But

y2H2

21./y1 X4 21/y2X6 4y1 = 6Y2

or Or

21/

Y1 = 1.5 Y2 Yi — Y2 = 2 l.5 y2 — y2 = 2

or

Y2 = 4

=6

Now

C X

Or

2.6x4

—

2.Jy1 x4

- 0.98

0.98

or

x = 2 x 0.98 x /27y = 9.6 m The point of intersection of the two jets is at 9.6 m from the orifice. 67. Two orifices are located in a water tank in such away that lower & upper orifices are at 0.5 m and 10 m height from the ground. The distances travelled by the upper jet and lower jet are 3 to 1. Find the height of water above the upper orifice if coefficient of velocity for both orifices is same.

10 m

Y2

3x

Two Orifices

Fluid Dynamics-I

429

xl 2 ytii

Bottom orifice : Cv1

X

2V0.5x (9.5+ H) x2

Upper orifice : Cv2

11y2H2

3x 2,1110x4 Now

Cv2 or

3x 2/10x H

X

21/0.5x (9.5+H)

Or

10 H = 9 [0.5 x (9.5 + H)] = 4.5 (9.5 + H) = 42.75 + 4.5 H 5.5 H = 42.75 H = 7.77 m

68. A 120 mm diameter orifices discharge 48 litres/s of water under a constant head of 2.6 m. A flat plate held normal to the jet emerging from the orifice requires 320 N to resist the impact of the jet. Determine the hydraulic coefficients Jet: Q = 48 litre/s

T 2.6 m

Plate

Jet of Water

Force = change of momentum = mass x velocity of vena contracta = P•g• Qactual X Vactual 320 = 1 X 103 X 9.81 x 48 X 10-4X Vactual 320 9.81x4.8 = 6.8 m/s

Vactuai —

Vth

2gh

430

Fundamentals of Fluid Mechanics

= V2x 9.81x 2.6 7.14 m/s Cy — Kauai = 6.8 Vth 7.14 = 0.95 Qth = a X Vth Ird2 11- X (0.12)2 4 4 0.011 m2 Qth = 0.011 x 7.14 0.08 a=

Qactuai _ 4.8 x 10-4 0.08 Qth 0.6 Cd — Cy x Cc Cd _ 0.6 Cc — Cv 0.95 0.63 Cd —

Now

0.048 0.08

69. A tank as shown in the figure has a nozzle of exit diameter d1 at a depth H below the free surface. At the side opposite to that of nozzle 1, another nozzle is proposed at a depth H . What should be the diameter d2 in terms of d1 so that the net horizontal 2 (AMIE 2000) force on the tank is zero.

Two Nozzels Discharge velocity at nozzle 1 is— T71 = \12gH1

Fluid Dynamics-I

43 I

Qi = Ai vi & mi = pAi vi = p x L 4r •6/21 x V2gHi F1 = change of momentum = Mi V1

= p x I'

41

I/ 2gH1 x ,i2gIii

7r d? = P' 4 2gli1

Discharge velocity at nozzle 2 is— V2 =

V2gH2 = Al2g.± 1 = 2

F2 — M2 V2 —

it d?

Px

M2 — PA2 V2 —

px

4 TC d?

1

-Ni

1! gHi

glii

x gill

4 F1 = F2 i.e. not force is zero

Now

P

7C di 2g1/1 — p 2 x gill 4 4 or 2di = c6,

7C d?

Or

d2 = Ari di = 1.414 di

70. What are the practical applications of weirs? The practical application of weirs are: (1) Measurement of discharges. (2) Anicut, raised weir or barrage are constructed and regulated on this principle. (3) Discharge through bridge openings specially afflux and back water curve can be determined. (4) Orgee weir is possible as per flow. 71. What is an anicut, barrage or raised weir? An anicut is a masonry wall or dam built across a river for raising the water level on the upstream side so that raised water can flow into the canals emanating from the river. Hence anicut is nothing but a weir over which the water can flow when water level is higher than the crest of the anicut. Examples are Farraka barrage and Okhla barrage. It is possibly to raise the water level at upstream of the river with the help of wooden shutters or flash boards which can be fixed on the crest of the weir. Such weirs are called raised weirs. 72. Explain (1) afflux and (2) backwater curve

432

Fundamentals of Fluid Mechanics

A river bridge is constructed over abutments and piers. These structures are obstruction to the flow of water. As water has lesser width to flow due to the presence of piers and abutments, the level of water rises at upstream of the river. Such rise of water level is called afflux. The rise exists upto a certain distance on the upstream side and it gradually decreases upstream. Hence the water surface at upstream does not remain in horizontal plane but it curves longitudinally with concavity upwards. Such a curved surface is called a backwater curve. 73. What is an aquaduct? It can be (1) a bridge for conveying water of a canal above the ground so that a road can pass under the bridge and (2) a coarse from conveying water under the ground which may have road or a structure so that water has to flow under it. 74. Find the discharge through a fully submerged orifice having width 3 m if the difference of water level on both sides of the orifice is 60 cm. The height of the water from top and bottom edge of the orifice are 3.6 m and 3.9 m respectively. Assume Cd = 0.6. Discharge from a fully submerged orifice is: Q = Cd X b x (H2 — Hi) x V2gH Given H = 0.8, H1 = 3.6 m, H2 = 3.9 m b = 3 m and Cd = 0.6 Q = 0.6 x 3 x (3.9 — 3.6) 1/ 2 x 9.81 x 0.8 = 2.14 m3/s 75. Find discharge through a orifice 1.2 m wide and 0.8 deep in a vertical wall of a tank having water level of 2 m above the upper edge of the orifice. Water at downstream side is 0.4 m above the bottom edge of the orifice. Assume Cd = 0.6

0.4 Ft— 12m Orifice

The orifice has two portions, namely, the upper portion acts as free flow and lower portion acts as submerged flow. Qi = free flow Q2 = submerged flow

Fluid Dynamics-I

433

Q= Total flow Q = Qi + Q2 Qi

= 3 • Cd• b • /2g [H312 — H312] = 2 x 0.6 x 1.2 x V2 x 9.81 [(2.4)3/2 — (2.4)3/2] 3 = 2 x 0.6 x 1.2 x 4.43 [3.72 — 2.83] 3 = 1.892 m3/s

Q2 = Cd x

b (H2 — H1) )12gH

= 0.6 x 1.2 (2.8 — 2.4) V2 x g x 2.4 = 0.6 x 1.2 x 0.4 x 6.86 = 1.976 m3/s Q = 1.892 + 1.976 = 3.768 m3/s 76. The discharge velocity at the pipe exit in the guage head as shown in figure is 1h

t H

(a) Vi gH

(b) \ I 2gh

(c) ,‘ I 2g(H + h)

(d) 0

(GATE 98) The head between free surface and the outlet is 'h'. As per Torricelli's law, the discharge velocity is V 2gh Answer (b) is correct 77. Water is flowing with a flow rate of 0.002 m2/s. What is the average velocity at the outlet whose area is 4 cm2? (a) 50 m/s

(b) 20 m/s

(c) 10 m/s

(d) 5 m/s (NTPC)

from the continuity equation— Q = A• V where A = area & V = average velocity V—

Or

Answer (b) is correct

2 x 10-3 4 x 10-4

— 5 m/s

434

Fundamentals of Fluid Mechanics

78. If the pressure in the air is doubled, the flow rate of water from the nozzle will increase by what percentage? (a) 20%

(c) 60%

(b) 100%

(d) 41% (SAIL)

Pressure Po

Piston Water

Flow rate is velocity multiplied by area. As the area does not change, hence the rate depends upon the velocity only. Using Bernoulli's equation at section 1 and 2. v v2 p Po 2 + (Z2 — z1) u 2 + Pairs 2g 2g pwg

Taking z2 —

P2

= 0,

Pwg

— atmospheric = 0

V0 = 0 Po _ V22 2g pair x g or

PO — Pair21722

or

, v2—

2P0 Pair

If P0 is doubled, the velocity V2 will increase by Ad = 1.41. Hence percentage increase is 41% Answer (d) is correct 79. For the nozzle shown, approximate the pressure P1 (neglect friction and assume a uniform profile at section 1)

(a)

3V 2 p 2

7v 2 (b) 3P V2

(c)

2

p

(d)

15V 2 P 2 (HCL)

Fluid Dynamics-I

435

From continuity equation, we have Al V1 = A2V2 Or

Al V1

v7

/412 4 A2

V1 = 4T71 A2 Now apply Bernoulli's equation to inlet and outlet. Ti 2 P1 + 1712 v2 + Z1 — P2 + Z2 pg 2g pg 2g Now

= z2,

P2 = 0 (atmospheric)

Pg

Ti 2 Ti 2 v = P. g v 2 - • - (2g 2g = 2 016 — 1) ,2 =2 2 P Vi Answer (d) is correct 80. The flow rate from the nozzle attached to the tank can be doubled by (hold all other variables constants) —F H

J_ ••.— 100

(a) doubling D

(b) doubling T

Diameter D area A

(c) doubling A

(c)

doubling W (HCL)

436 Fundamentals of Fluid Mechanics Q=AxV V = V2gH Q= A x V2gH Hence Q oc A as H is constant. The flow can be doubled by doubling area. Answer (c) is correct 81. In an isothermal atmosphere, the pressure— (a) (b) (c) (d)

is constant with elevation decreases linearly with elevation cannot be related to elevation decreases exponentially with elevation (IES 90)

dP = —pgdz z = height Perfect gas equation as air is an ideal gas P = pRT or p — dP —

Pgdz RT

or

dP _ —g dz RT

or

dP _ RT g dz

or

log P =

P RT

RT

z put c =

RT

p = e CZ or Hence pressure depends upon elevation exponentially Answer (a) is correct

82. An L-shaped tube (Pitot tube) is used to measure the horizontal velocity in a stream as shown. The dynamic or velocity head is (neglect losses) 4

en

H V =_ /////////////////////////////////////

Pitot tube

Fluid Dynamics-I

(a) H

(b) 6,h + H

(c) Ah

437

(d) 0 (IES 900)

The pitot tube has— stagnation head = static head + dynamic head = H + Ah Hence dynamic head is Ah Answer (c) is correct. 83. The given figure represents flow of water past a body with velocity V in a lake. The curve AB is— B

((-

A

Body v

(a) An equipotential (b) a path line (c) a streak line (d) a stream line (IES 94) The curve AB represents a streak line as streak line is a curve which gives an instantaneous picture of the location of the fluid particles which have passed through a given point. Answer (c) is correct 84. If a hole is made in the Torricellis vacuum portion of barometer, then mercury— (a) level will fall in the stem and the mercury (b) level will oscillate between reservoir level and the original level of the mercury in the stem (c) will spill through the hole made (d) level in the stem will remain at the same level indicating atmospheric pressure (IES 95) When a hole is made, the mercury level will fall down in the barometer. The rushing out mercury will raise the level in the reservoir till it is stopped by the gravitational force. Now mercury starts falling in reservoir after reaching the same level and rises in the tube. This keeps on happening and mercury will oscillate. Answer (b) is correct. 85. Three reservoirs A, B and C are interconnected by pipes as shown in the figure. Water surface, elevation in the reservoir and the piezometric head at the junction J are indicated in the figure.

438

Fundamentals of Fluid Mechanics Piezometric head = 160 m EL = 200 m

EL = 180 m

EL = 140 m

Discharge Ql, Q2 and Q3 are related as— (a)G±Q2=00)Q1=Q2 -FQ3(c)Q2=G+001)G -FG -FQ3= 0 (GATE - 98) Since EL at A > head at J, hence flow Qi is from A to J Since EL at B > head at J, hence flow is from B to J. Since EL at C > head at J, hence flow is from J to B Qi ± Q2 = Q3 Answer (a) is correct 86. The outlet of a tank of water is changed as shown below, from a distance `d' to a distance `4d' from the surface of the water. Which of the following effects on the velocity of the water leaving the tank?

„

1.

_-_--Hd ___

—0-

(a) (b) (c) (d)

V0

increase the velocity by factor 2 increase the velocity by factor 4 increase the velocity only if the diameter of the outlet is increased decrease the velocity due to lower pressure

As per Torrieclli's law, velocity of discharge is V = V2gh h = head of water First case h = d, hence V1 = J 2gd Second case h = 4d, hence V2 = J 2g x 4d = 2 112gd = 2 V1

(IES 90)

Fluid Dynamics-I

439

Hence velocity will increase by factor 2 Answer (a) is correct 87. A 2 cm dia pipe transports water at 20 m/s. It exits out of two small 2 mm dia holes. The exit velocity in m/s will be— (a) 120

(b) 80

(c) 90

(d) 20 (IES 90)

Applying continuity equation Ai Vi = 100 x A2V2 7r (2 x 10 2)2

4

x 20 = 100 x

or

V2 —

It (2 x 10 3)2

4

X

V2

20 x 100 — 20 100

Answer (d) is correct. 88. In force vertex flow, velocity V and radial distance r are related as— (a) V oc r

(b) V cc

1 r

(c) V oc

1 r2

(d) V oc r2 (IES 96)

In force vertex flow, we have — V = wr Where w is angular velocity. If angular velocity is constant, then V oc r Answer (a) is correct 89. A jet of water issues from a nozzle with a velocity 10 m/s and it impinges normally on a flat plate moving away from. It at 10 m/s. If the cross-sectional area of the jet is 0.02 m2 and the density of water is taken as 1000 kg/m3, then force developed on the plate will be— (a) 10 N

(b) 100 N

(c) 1000 N

(d) 2000 N (Civil Services 94)

Force on the plate = change of momentum = Mass of water x change of velocity =(pxAxV)xV =1x103 x 0.02 x 102 = 2000 N Answer (d) is correct.

440

Fundamentals of Fluid Mechanics

90. A nozzle has velocity head at outlet of 10 m. If it is kept vertical, the height reached by the stream is— (a) 100 m

(b) 10 m

(c) V10 m

(d)

10

m (IES 92)

The stream will rise upto a height equal to its head = 10 m Answer (b) is correct 91. An external cylindrical mouthpiece has 80 mm diameter, fitted at side of a large vessel. Find discharge if the head over the mouthpiece is 8 m. (UPTU — 2001-2) Area = a =

4

d2 = ir 4

X

(.8)2 = 5.026 x 10-4 m2

Assume Cd = 0.85 Q = Cd x area x velocity = 0.85 x 5.026 x 10x 1/2 x 9.81x 6 = 46.36 x 10-4 m3/s 92. A convergent divergent mouthpiece is fitted to the side of a tank. If the diameter of the outlet and vena contracta be 52 and 37 mm respectively. Find the maximum head of water for steady flow. Given atmospheric pressure as 10.3 m of water and separation pressure as 2.5 m of water. The lowest pressure possible at vena contracta (head lic) is separation pressure which is given equal to 2.5 m of water. The atmospheric head (H0) = 10.3 m If dc = diameter at vena contracta and d = diameter at the outlet of the mouthpiece, then we have: 42 = 1+ HO — Hc H 11 d2 where H = maximum head of water at the mouthpiece.

22

[52 x 10— ) = i i + 10.3 — 25 i 37 x 10-3 H 4.21 = 1 + 7'8 H or

H-

78 ' 3.21

= 2.43 m of water

Fluid Dynamics-1

441

93. Water flowing through a convergent divergent mouthpiece is having a diameter of convergence = 40 mm and head at the mouthpiece is 4.5 m. Find the maximum diameter to the divergence of jet if the maximum vacuum pressure is 9 m of water. d2 _

d2 Y

1 + HO - He. H

d2 - li1-F 4S (40 x 10-3)2 or d2 = 2.77 x 10-3 Or d = 0.053 m = 53 mm 94. An internal mouthpiece of 80 mm diameter is discharging water under a constant head at 4.5 m. Determine the discharge when (1) it is running free (2) running full.

Area = E d2 = x 4 4 Running free condition, we have-

(80 x 10-3)2 =

5.03 x 10-3 m2

Discharge = Q = 0.5 x a x V2gH = 0.5 x 5.03 x 10-3 x „12 x 9.81x 4.5 = 23.63 x 10-3 m3/s Running full condition, we haveDischarge = Q = 0.707 x a x .J2gH = 0.707 x 5.03 x 10-3 x V2 x 9.81x 4.5 = 33.4 x 10-3 m3/s 95. The maximum flow through a rectangular channel of size 2 m wide and 1.5 m deep is 1.7 m3/s. It is proposed to install a sharp crested suppressed notch across the channel to measure the discharge. Find the maximum height of the water over the crest of the notch which is possible without water overflowing from the sides of the channel. Take Cd = 0.6.

Discharge through the suppressed weir/notch neglecting the velocity of approach isQ = Cd•L • X (H3/2)

where

3 H = height of water over the crest L = width of notch 1.7= 2- x 0.6 x 2 x V2 x 9.81 3

x43/2

442

Fundamentals of Fluid Mechanics

or or or

H"2 =

1.7 x 3 2 x 0.6 x 2 x 4.43

= 0.4796 H = (0.4796)2/3 H = 0.620 m

96. In a channel, a rectangular notch of width 0.8 m is used to measure the discharge. If water height is 0.6 m over the crest and discharge coefficient is 0.6 m for the notch, find discharge. If the rectangular notch is replaced by a triangular notch which has now height of water as 0.85 m, determine Cd for the notch. Neglect velocity of approach and end contraction. Rectangular notch

Q = 2- Cd.L. V2g H"2 3

= 2 x 0.6 x 0.8 x 112 x 9.81 x (0.6)3/2 3 = 2 x 0.6 x 0.8 x 4.43 x 0.46 3 = 0.652 m3/s Triangular notch For same flow Q, we have— Q= 0.652 =

8 d•tan 0 x li 2g 1/5/2 5C 2 15

• Cd x tan 30 x 112 x 9.81 x (0.85)5/2

15 x 0.652 8x0377x4.43x0.66 = 0.72

Cd —

97. Determine discharge of water over a triangular notch of angle 60° when the head of water over the crest is 1 m. Take Cd = 0.6 & neglect velocity of approach. Discharge triangular notch is— Q=

15

Cd 2g tan 2 x

2

H512

= 5 x 0.6 x /2 x 9.81 x tan 30 x (1)5/2 8 x 0.6 x 4.43 x 0.577 x 1 — 15 = 0.818 m3/s

Fluid Dynamics-I

443

98. A triangular notch with apex angle of 60° is located across a channel of 30 cm wide. The apex is at height of 20 cm from the bed and height of the water in the notch is 60 cm from the bed. Assume Cd = 0.6. The height of the water in the notch from the apex = 60 — 20 = 40 cm H = 0.4 m

.

Q—

8 • Cd•tan 9 • vr 2g H5/2 15 2

8 x 0.6 x tan 30 x V2 x 9.81 x (0.4)5/2 =— 5 8 x 0.6 x 0.577 x 4.43 x 0.1 — 15 = 0.0817 m3/s 99. A reservoir 120 m long and 120 m wide has a rectangular notch 2 m wide. Find the time to lower the water level in the reservoir from 2 m to 1 m if Cd = 0.6. Surface area of the reservoir = 120 x 120 = 14400 m2 Time required to lower the level of water in the reservoir with the notch is— T—

1 11 3A x 1/ Hi H2 LI Col. L • V 2g 3 x 144 x102 1 1 0.6x2xV2x9.81 [F1 11 2

3 x 144 x102 0.6 x 2 x 4.43 [1 — 0.707] = 23.8 x 102 secs = 39 min 40 secs 100.A triangular notch with apex angle 60° is used to discharge water from a reservoir of size 100 m x 100 m from the level 9 m to 4 m. Assume Cd = 0.6. Find time. A = Surface area = 100 x 100 = 104 m2 Time for lowering level by the triangular notch is3/2 3/2 1 1 A T= 5 4 Co/ tan 2 V 2g [(Hi) (H2 ) 1 ) = 5A X 4

(10)4 [( 1)3/2 ( 03/21 0) i 0.6x tan30 x 1/2 x 9.81 1_ 41.)

444

Fundamentals of Fluid Mechanics

104 [1 11 - 5 x 4 0.6 x 0.577 x 4.43 L8 27 104 19 1 —5 x 4 0.6 x 0.577 x 4.42 [8 x 27 = 0.0718 x 104 = 718 sec 101.Determine the time required to lower the level of water from 3 to 2 m in a reservoir having surface area of 104 m2 by (1) rectangular notch of width 1.5 m wide and (2) a right angle triangular notch. Assume Cd = 0.6. Case 1. Rectangular notch Time for lowering level is— T—

3A ( 1 1 Cd • L• V 2g Al H2 1[1-11 3x104 F 1 1 i 0.6x1.5xV3x 9.81 kii, -si _I 3 x 104 [ 0.6 x 15 x 4.430.707 — 0.577]

= 0.978 x 104 seconds = 9780 s. Case 2. Triangular notch Time for lowering level isT=

5 x

A

1 ,312 Cd ... x tan 2 x liTg [il2

104 F 1 = 5 x 4 0.6 x 1 x 4.43 L(2)3/2

iii Hilt 1 i

(3)3/2 i

= 0.47 x 104 [0.353 — 0.192] = 0.07567 x 104 seconds = 756.7 s Note:- Time required to lower the level is lesser in triangular notch 102.A trapezoidal notch has 30 cm base and sides inclined 30° to the vertical. If the head of water is 16 cm, find the discharge. Assume Cd = 0.6

Fluid Dynamics-I

445

H 0.30m

Trapezoidal Notch Guidance. The trapezoidal notch is equivalent to a rectangular notch and a triangular notch. The sum of discharges from these two notches is the discharge of trapezoidal notch Q

Qrect Qtri 2 =— 3

Cd•L • V4 H3/2 + 8 C d tan — Zg "2 2 H

= 3 x 0.6 x 0.3 x 1/2 x 9.81 x (0.16)3/2 +

15

x 0.6 x tan 30 x 2x9.81 x(0.16)5/2 = 0.034 + 0.00838 = 0.04238 m3/s 103. A fire brigade man is holding a free stream nozzle of 50 mm diameter as shown in the figure. The jet emerges with velocity of 20 m/s and jet is made to strike a burning window. Find the angle of elevation of the jet from the nozzle from horizontal and amount of water falling on the window.

Jet of Water

Here

x=

6, y = 8.5 — 1.5 = 7

U= 15 m/s,

Ux = cos

y= Uyt — 1 2

0,

Uy

=

4 sin 0 Ux x t = 4 or t —

1 U cos (9

446

Fundamentals of Fluid Mechanics

= U sin 9 X

1 cos 0

1 1 g 2 cos2 0

1 g sec2 0 = U tan 0 — — 2 = U tan 0—

or or Or

or

2

g(1 + tang 0)

1 x 9.81 (1 + tan2 0) 7 = 20 tan 0 — — 2 1 + tan2 0= tan 0 — 1.43 tan2 0= 4.08 tan 0 + 2.43 = 0 tan 0.—

+4.08 ± V16.6 — 9.72 2

4.08 ± 2.62 2 = 3.35 or 0.73 0= 36.3° or 73.4° Water falling on the window is equal to the discharge from the nozzle 0 = Area x Velocity =

4

d2 x u

= 4 x (0.05)2 x 20 = 0.03925 m3/s 104. Water flows over a rectangular weir 1 m wide at depth of 150 mm and afterwards passes through a triangular right angled weir. Taking Cd for the rectangular and triangular weir as 0.62 and 0.59 respectively, find the depth over the triangular weir. (AMIE, 75, Osmania University 1990) Rectangular weir Q = — x Cd X L x V2g x H3I2 3 =

x 0.62 x 1 x 3 = 0.106 m3/s

x 9.81 x (0.15)3/2

Triangular weir Q=

0 x V ,i, x H5/2 8 x Cd x tan — 2 5 2

8 x 0.59 x tan 45 x V2 x 9.81 x H5/2 0.106 = 15

Fluid Dynamics-I

or

447

0.106 x 15 8x0.59x1x4.429

H"2 —

= 0.076 H = 0.357 m

Or

105.Water flows through a triangular right handed weir first and then over a rectangular weir of 1 m width. The discharge coefficient of the triangular and rectangular weirs are 0.6 and 0.7 respectively. If the depth of water the triangular notch is 300 mm, find the depth of water over the rectangular weir. (AMIE 1990) Triangular weir Q=

15

Cd x tan lE x V2g x H5/2 2

8 x 0.6 x tan 45 x V2 x 9.81 x (0.36)5/2 — = 15 = 0.11 m3/s Rectangular weir Q= — x Cd X L x ..W x113/2 3

or or

2 0.11 = — x 0.7 x 1 x V2 x 9.81 x H3/2 3 H3/2 = 0.053 H = 0.142 m = 142 mm

106.Figure shows a stepped notch. Find the discharge through this notch if Cd remains constant and equal to 0.62.

Hi— 40 —I-I

Guidance. The given stepped notch can be considered as three rectangular notch and discharge will be sum of the discharges of these rectangular notches. Qi =

3

Cd X L x 1,g [HP — HP]

448

Fundamentals of Fluid Mechanics

= 2 x 0.62 x 0.4 x V2 x 9.81 (0.953/2 - 0.83/2) 3 = 0.0154 m3/s Q2 =

x 0.62 x 0.8 x V2 x 9.81 [0.83/2 - 0.53/2]

= 0.0530 m3/s Q3 = 2- x 0.62 x 1.2 x V2 x 9.81 [0.53/2] 3 = 0.0776 m3/s Q = Qi ± Q2 ± Q3 = 0.0154 + 0.0530 + 0.0776 = 0.1460 m3/s 107.A weir 36 m long is divided into 12 equal bays by vertical posts each 60 cm wide. Determine the discharge over the weir. If the head over the crest is 1.2 m and velocity of approach is 2 m/s (AMIE 1978) 2

Guidance. The additional head due to velocity of approach 170 = - where Tio = velocity of 2g approach. Each bay has two end contractions. Francis formula for the discharge is to be used. Length of weir = 36 m Number of bays = 12 & 11 posts Effective length = 36 - 11 x 0.6 = 29.4 m Discharge by Francis formulaQ = 1.84 [L - 0.1 x 2n (H + h0)] x [(H + 12,3,12) - hr] v2 °

2

- 0.2038 m 2g 2 x 9.81 Q = 1.84 [29.4 - 0.1 x 2 x 12(1.2 + 0.2038)] x [(1.2 + 0.2038)3/2 - (0.2038)3/2] = 75.25 m3/s

hp -

-

108.The heights of water on the upstream and downstream side of a submerged weir of 3 m length are 0.2 & 0.1 m respectively. If Cd for free and drowned portions are 0.6 and 0.8 respectively, find the discharge over the weir. Guidance Total discharge is sum of the discharges from upper free flow and lower submerged flow. If H and h are levels at upstream & downstream Q= 2 Cd x I, x V2g (H - h)312 + Cd XL Xh V2g(H -h) = 2 x 0.6 x 3 x V2 x 9.81(0.2 - 0.1)3/2 x 0.8 x 3 x 0.1 x V2 x 9.81x 0.1 3

Fluid Dynamics-I

449

= 0.168 + 0.336 = 0.504 m3/s 109. In an ornamental fountain, there are 12 nozzles each 30 mm diameter, all inclined at angle of 45° with the horizontal. The jet issuing from the nozzle falls into a basin at a point distance 1.4m vertically beneath the nozzle and 4.6m horizontally from it. The velocity coefficient of nozzle is 0.97. Find the pressure head of the nozzle and total discharge from it.

Nozzle

x = U cos 45 x t or 4.6 = 4.6 cos 45 x t or Now

t—

4.6 U cos 45

h = U sin 45 x t — 1.4 = U sin 45 x

= 4.6

2

gt2

4.6 U cos 45

(4.6)2 1 g X 2 2 U cost 45

9.81 x (4.6)2 U2

9.81 x 4.62 — 64.86 3.2 or U = 8.054 m/s 2 (8.054)2 Pressure head at nozzle = H = U (0.97)2 x 2 x 9.81 C'i x 2 g or

U2 =

= 3.51 m Discharge for 12 nozzles = Q = (ix j 0.032 x 8.054) x 12 = 0.068 m2/s

Chapter 10

FLUID DYNAMICS-II

KEYWORDS AND TOPICS A A A A A A A A A

PRINCIPLE OF DYNAMICS MOMENTUM PRINCIPLE MOMENT OF MOMENTUM LINEAR MOMENTUM ANGULAR MOMENTUM VANES FLOW EXERTED BY JET FORCES ON BENDS RECTILINEAR FLOW

A A A A A A A A A

RADIAL FLOW ROTARY FLOW FORCED VORTEX FREE VORTEX ROTATION OF SPRINKLER VENTURI METER ORIFICE METER FLOW NOZZLE ROTO METER

INTRODUCTION The fundamental principle of dynamics is Newton's second law of motion. The second law of motion states that time rate change of momentum contained in a volume of fluid is proportional to the applied force and this change of momentum takes place in the direction of force. This is called impulse momentum principle which is very useful principle in addition to the continuity and the energy principles. The impulse momentum pirnciple finds vast application in the solution of several fluid flow problems. The impulse momentum equation is used for the flow analysis of (i) pipe bends and reducers (ii) jet propulsions (iii) fixed and moving vanes (iv) propellers of ships, aircraft and helicopters (v) head loss due to sudden enlargement or contraction in pipe system and (vi) hydraulic jump during flow in open channel. 1. What is the fundamental principle of dynamics? The fundamental principle of dynamics is Newton's second law of motion. The second law states that the time rate of change of momentum contained in a volume is proportional to the applied force and this change of momentum takes place in the direction of force.

Fluid Dynamics-II

45 I

According to this law, the resultant external force acting on a mass particle along any arbitrarily chosen direction is equal to time rate change of its linear momentum in the same direction. Dynamic force = Rate of change of momentum in the same direction or

F = dT (my) where m = mass, v = velocity and T = time d

2. What is impulse momentum principle? Or Describe the momentum equation. (UPTU-2002-3, 2007-8) The impulse momentum principle states that the force F acting on a fluid mass m in short (d (mv) in the direction of force. dT ) The impulse momentum principle is based on the law of conservation of momentum which states that the net force acting on a fluid mass is equal to the change in momentum of flow per unit time in the direction of net force. interval of time dT is equal to the change of momentum

Force = change of momentum F—

ddT

(my)

or F • dr = d (my) The above is impulse momentum equation. 3. State the practical applications of the momentum equation. (UPTU 2007-8) The momentum equation is used to determine the resultant force exerted by a flowing fluid on any surface when the fluid changes its magnitude or direction or both magnitude and direction while flowing on the surface. The decrease of the momentum of the fluid means that fluid is giving energy to the surface as in turbines while the increase of the momentum of the fluid means that fluid is taking the energy from the surface as in pumps. The momentum equation is used for the flow analysis of — (1) pipe bends and reducers (2) jet propulsions (3) fixed and moving vanes (4) propellers of ships, aircraft and helicopters (5) headloss due to sudden enlargement or contraction for a flow in pipe system. (6) hydraulic jump during flow in open channel.

452

Fundamentals of Fluid Mechanics

4. What is a vane? A vane is a flat or curved plate. A number of vanes are fixed on the rim of a wheel to form a water wheel or turbine. Curved vanes Guide vanes

Water wheel

Turbine

5. Derive an expression for the force exerted on a stationary vane by a jet of water which is striking normally to the vane.

Nozzel

Vane

Stationary Vane: Jet Normal. The figure above shows a jet of water striking normally to a flat stationary vane. As per the momentum equation, the force exerted by the jet of water is equal to the rate of change of momentum of the water. Mass of water striking the vane where

= m = p•a•v p = density a= area of jet and v = velocity of jet

The velocity of water is reduced to zero from velocity v after striking the vane. Force on the vane = rate of change of momentum of water on vane surface = m x dv = (pAv)(v — 0) F = pav2

Fluid Dynamics-II

453

6. Derive an expression for force exerted by a jet of water striking inclined stationary vane. Vane V F, = F sin 0

Nozzle

Jet

F = F cos r

e

Jet inclined to vane

When a jet of water strikes inclined 0 to the stationary vane, then force exerted normal to the vane is: F= m x vsin 0= mass x normal velocity = pav2sin 9 a = area of jet and v = velocity of jet

where

The above force can be resolved into two components viz. (1) force (Fr) in direction of flow and (2) force (Fy) acting normal to the direction of flow. Hence we have:

and

F., = F sin 0 = pav2sin2 9 Fy = Fcos 9 =pav2sin 0 cos 0

7. A jet of water strikes a plate hinged at top end. The plate has weight = W and length = L. Find (a) the angle 0 to which the plate will swing on action of the jet (b) force p to exerted at the bottom of the plate to stop the plate from swinging when (a) jet is acting at L (b) jet is acting at L. 2 4 Hinged Jet

\

W Swinging of plate

Force P to hold the plate

During equilibrium, the plate has swung by an angle 0 from vertical due to force F exerted by the jet of liquid. Taking moment with respect to hinged point, we have:

454

Fundamentals of Fluid Mechanics

X Mhinged — 0, —F X t where

—WX

t Sill e = 0

F= force exerted by jet and W = weight of the plate

F =W Now we apply force p to stop the plate from swinging. Taking moment with respect to hinged point, we have or

sin 0

X Mbinged — 0, —Fxt, +PxL=0 F P_ 2

Or

In case the jet is acting at a point 1L from the hinged point, then on taking moment with respect to hinged point, we have: EMbinged-0,—FX 41 L -FpXL=0 or

3

p= F 4

8. Derive an expression for force exerted by a jet of water while striking on a moving vertical plate. Find also (1) work done (2) efficiency and (3) condition for maximum efficiency.

Nozzle

Normal jet on moving plate

Consider a jet of water striking a plate with a velocity of V. The plate is moving with a velocity of V1 in the same direction of the jet. Hence relative velocity of the jet with respect to the moving plate is V — V1. The velocity of the water of the jet is zero after it strikes. Force exerted by the jet = rate of change of momentum of the water or F= mass x change of velocity =pxax(V—V1)•(V—V1)

Fluid Dynamics-II

where or

455

p = density & a = area of jet F = p • a • (V — Vi)2

Now the force acting on the plate is given by the above equation. The work done by the force is equal to the force multiplied by distance. Work done = W = force x distance = p • a , (V — V1)2 x V1 The jet has been supplied kinetic energy by the nozzle. 1 KE = — mV2 1 = — • p , a • V V2

2

1 3 — pa V 2 Efficiency —

Work done by jet Energy supplied to jet p • a • (V — V1) 2 • Vi .

or

7/ —

3

p . a .vi

2 (V — V1) 2

' 171

V3

In order to find maximum efficiency,

DR DV

has to be equated to zero.

an _ 21/1 .2(V — Vi )V3 — 3V2 .2(V —V1)2 —0 DV V6 Or 2 V — 3. (V — VI ) = 0 or V = 3 • VI The above is the condition for maximum efficiency. The maximum efficiency is — rim

2(3V1 — V1)2 • Vi. (3V1)3 2 • 4 • 1712 • VI 27•V13 8 27

456

Fundamentals of Fluid Mechanics

9. Derive an expression for work done by the jet of water on moving inclined plate or flat vane.

F,,= F cos 01 Fx = F sin 0

Jet on moving inclined plate

The jet is striking inclined to the plate as shown in figure. The relative velocity is (V— V1) Force = F = m(V — Vi) sin 0 = p • a • (V — V1)2sin 0 Force along x-direction is F., = Fsin F, = p • a • (V — V1)2sin2 Similarly, we have F = p.a. (V — V1)2sin °cos 0 Work done by the jet is — W = F- displacement = p.a. (V — V1)2sin2 0•V1 = p•a•Vi (V — V1)2sin2 10. A 40 mm diameter water jet strikes a hinged vertical plate of 800 N weight normally at centre with 15 m/s velocity. Determine the angle of deflection 0 of the hinged plate from vertical. Also determine the magnitude of force F that must be applied at its lower edge to keep the plate vertical. (UPTU- 2003-4) Jet with dia = 40 cm

Nozzle

F V = 1.5 m/s

800 N

Area of jet

ird 2 4

x(0.4)2 4

Fluid Dynamics-II

457

= 1.256 x 10-3m2 Force exerted by jet F = pa V2 = 1 x 103 x 1.256 x 10-3 x 152 = 282.7 N Force P to hold the plate in vertical is

-

F _ 282.7 2 2 = 141.3 N

p _

Now

_ 282.7 — 0.353 F 800 W 6) = 19.46°

sin 0 —

or

11. A jet of water having diameter = 50 mm is discharging under a constant head of 60 m. Find the force being exerted by the jet on the vertical plate. Take c„ = 0.92

Jet, dia = 50 mm

I 60

I Area of the jet = a =

Trd2 4

Tr

x (0.05)2 — 1.96 x 10-3m2 4

Velocity ofjet = v = c1 ,V2gh = 0.92 x V2 x 9.81x 60 or

v= 31.56 m/s

Force exerted by the jet is: F = p• a ,v2 = 1 x 103 x 1.96 x 10-3 x (31.56)2 = 1.952 kN 12. A jet of water is emerging from a nozzle under a constant head of 30 m. Calculate the diameter of the jet, if the force exerted by the jet striking a plate normally is 2.4 kN. Assume ci, = 0.9 Velocity of the jet = cv = V2gh v = 0.9 x 1/2 x 9.81 x 30 = 21.83 m/s

458

Fundamentals of Fluid Mechanics

Force exerted by the jet is: F= p• a • v2 a—

Or

F p • V2 2.4 x 103 lx 103 x (21.83)2

= 5.034 x 10-3 m2 or

xd2 — 5.03 x 10-3 4 4 d= — x 05.03 x 10-3 Vic = 8 x 10-2 m = .08 m = 80 mm

or

13. A jet of water of 100 mm diameter emerges with a velocity of 24 m/s and strikes a plate. Find the force exerted on the plate if (1) the jet strikes normally to the plate (2) the jet is inclined at 30° to the plate. 7r x (0.1)2 — 7.854 x 10-2 m2 — 4 4 Case I: When the jet is striking normally to the plate, the force is:

Area of jet — a —

r Cl2

F= p • a •v2 = 1 x 103 x 7.85 x 10-3 x (24)2 = 4.522 kN Case II. The jet is inclined 30° to the plate, the force exerted now is: F= p• a • v2sin230 1 = 4.522 x — 4 = 1.155 kN 14. A jet of water with 30 mm diameter exerts a force of 1.2 kN in the direction of the flow on a flat plate. The jet is inclined 30° to the plate. Find the rate of discharge. The area of jet = a =

7C

x (0.03)2 — 7.065 x 10-4 m2 4

Fluid Dynamics-II

459

The force exerted on the plate is: F = p• a • v2sin230 1.2 x 103 = 1 x 10+3 x 7.065 x 10-4 x v2 x 4 2 1.2x103 x4 v — — 6.794 x 103 0.7065 v = 82.42 m/s discharge Q = a •v = 7.065 x 10-4 x 82.42 = 0.0582 m3/s or

Now

15. A jet of water with diameter of 30 mm strikes normally a plate with velocity of 25 m/s. The plate has length of 400 m and the jet strikes at the centre of the plate. What force is to be exerted at 300 mm below the hinged top to stop plate to move?

Nozzle

d2 4 F = p • a • v2

Area of the jet =

IT

x (0.03)2 — 7.065 x 10-4 x m2 4

= 1 x 103 x 7.065 x 10-4 x (25)2 = 441.56 N

Taking moment from hinged point

Or

M hinged = 0, —F x 200 + P x 300 = 0 x 200 P—F 300 441.56 X 200 300 = 294.37 N

460

Fundamentals of Fluid Mechanics

16. A jet of water having 30 mm diameter is moving with velocity of 30 m/s and impinges normally on a plate. Find the force on the plate when (1) plate is fixed (2) plate is moving with velocity of 10 m/s. The area of the jet = a =

Jr x (0.03)2 — 7.065 x 10-4 m2 4

Case 1: Plate is fixed. Or

Force = F = p • a • v2 F= 1 x 10+3.7.065 x 10-4 x (30)2 = 635.85 N

Case 2: Plate is moving with 10 m/s F = p • a • (V — V1)2 = 1 x 103 x 7.065 x 10-4 x (20)2 = 282.6 N 17. Describe differences between linear momentum and moment of momentum. When a fluid particle is moving in a straight direction, it has linear momentum. However when a fluid particle moves along a curved path such that its distance from the axis of rotation changes with times, then the distance from the axis of rotation of the particle will be different at different positions of the particle. Hence, the moment of momentum of the particle will change from the axis of rotation. The moment of momentum is also known as angular momentum. As per the equation of moment of momentum or angular momentum, when a fluid moves in curved path it exerts a torque on the curved surface which is equal to the rate of change of angular momentum of the fluid. 18. Derive an expression for the moment of momentum or angular momentum. Axis of rotation

V,.2 2

Consider a particle A is moving on a curved path with respect to the axis of rotation passing through point 0 as shown in the figure. The velocity V of the particle can be resolved into

Fluid Dynamics-II

46 I

two components viz. (1) Vr is normal velocity acting along the radius r and towards the centre of rotation 0 and (2) V0 is tangential velocity acting normal to the radius r. The angular momentums of the particle at point 1 and 2 are: Angular momentum at point 1 = moment of the momentum = r1 (m Ved Angular momentum at point 2 = r2 (m • V92) Change of angular momentum = m (r2 V02 — r1 V01) If the change of angular momentum takes place in time t Rate of change of angular momentum = 1 m (r2 V6 — r1 V0i) Now the rate of change momentum is equal to torque (T). T= —1 m(r2 V02 — ri V0i) But

m — = pgQ = mass of fluid flowing per unit time T = pgQ (r2 Ve2 — r1 V01)

The toque T is the toque exerted by some external agency on the fluid as the angular momentum is increasing. In centrifugal pump, the torque is exerted on the fluid by the centrifugal pump when it makes the water to move on the curved path. Similarly a compressor exerts torque on air and a propeller exerts torque on air. Incase, the fluid is made to move on a circular path, then r1 = r2 = r and torque is T = pgQ r (V02 — V01) In case the angular momentum of the fluid decreases, the torque is exerted by the fluid on the curved surface. For example a turbine in which torque is extracted from the fluid having higher inlet angular momentum. The equation — T = pg • Q • r (V02 — V01) is called the equation of moment of momentum or angular momentum. 19. Differentiate (1) rectilinear flow (2) radial flow and (3) rotary or vortex motion. Rectilinear flow: Rectilinear flow is one dimensional flow in which streamlines are parallel. Radial flow: Radial flow is the flow in which the fluid flow is in radial direction in such a way that pressure and velocity at any point changes with respect to distance of that point from central axis only. In this case the velocity of flow is in radial direction and the flow may take place radially inward, or radially outward from the centre. The flow is two dimensional and streamlines are not parallel.

462

Fundamentals of Fluid Mechanics

Flow

Flow inwardly (as in pump)

Flow outwardly (as in turbine)

Rotary or Vortex motion: A mass of fluid in rotation about a fixed axis is called vortex. In this flow, the velocity of rotating particles is acting in tangential direction: Also the whole fluid mass rotates about an axis. The rotating flow can be (1) free vortex and (2) forced vortex. In forced vortex, the fluid mass rotates without exerting any external torque (wr = constant). The rotation is taking due to fluid pressure or gravity or rotation already possessed by the fluid mass. The common example of free vortex are: (1) whirlpool (2) flow in centrifugal casing after emerging from propeller (3) flow in turbine casing before entering the guide vanes. In forced vortex, the fluid mass is made to rotate by means of some external agency. For example, spinning the vessel containing liquid about a vertical axis.

T z I

Original level

Forced vortex

The centrifugal head z =

W

2

r 2g

2 z is also called depth of parabola.

20. A open circular tank of diameter = 20 cm and 120 cm long contains water upto a height of 60 cm. The tank is rotated about its vertical axis at 300 rpm. Find the depth of parabola formed at the free surface of water.

Fluid Dynamics-II

w— Now

Z

271-N 60

463

2xitx300 — 31.41 rad/s 60

w2 r 2 2g (31.41)2 x (0.1)2 2 x 9.81 = 0.503 m

21. Derive an expression for force exerted by fluid flow on a bend when (1) Bend is in horizontal plane and (2) Bend is in vertical plane. Or State the momentum equation. How is it used in determining the force exerted by flowing liquid on a pipe bend. (UPTU-2005-6)

cos q

P1 A1

Forces on Bend (IN HORIZONTAL PLANE)

Case 1: Consider a flow in a bend which is in horizontal plane. Select section 1-1 and 2-2 at the inlet and outlet of the bend since flow is taking place in x-y plane, net force acting in each x and y direction is the rate of change of momentum of the fluid in x and y direction respectively by the forces acting on the bend will be in opposite direction to forces exerted by the fluid. The forces exerted by the fluid are: (1) Pressure force = (2) Force of fluid on the bend = F (3) Weight of fluid flowing = W to be considered in bend in vertical plane

464

Fundamentals of Fluid Mechanics

Applying momentum equation in x-direction, we get: P1 A1 — P2A2cos 0 + F., = pQ(v2cos 0 — vi) F., = pQ(v2cos 0 — vi) + P2A2 cos 0 — P1, Al or Similarly equating forces in y-direction, we get: Fy = pQ(v2 sin 0) + P2A2 sin 0 Resultant force F = \F.„2 + Fy The angle of resultant with horizontal 0 = tan n-

F Fx

Case 2: Forces on bend which has inlet in horizontal plane and outlet in vertical plane. Net force in x-direction will remain same but the net force in y-direction will change as the weight of fluid particle in y-direction has to be considered.

PIA.'

-I ,

Fx

= P Q (V2COS 0 - v1) + P2A2cos 0 - P1 Al

Fy = pQ(v2sin 0) +

P2A2 sin 0+ W

22. A bend in pipeline conveying water gradually reduces from 60 cm to 30 cm in diameter and the bend deflects the flow through 60°. The gauge pressure at inlet is 160 kPa. Find the magnitude and direction of the force on the bend when (1) there is no flow (2) flow is 0.8 m3/s. A1 = area at inlet =

7 C C112

4

7r X ( 0 .6) 2

4

— 0.283 m2

7C x (0 .3)2 — — 0.0707 m2 4 4 Case 1: When no flow is taking place, then the net force in x-direction is the change of momentum in x-direction.

A2 = area at outlet =

7 C di

—

Fluid Dynamics-II

P1A1 — P2A2 cos 19+

F, = pQ(v2cos 8— v1 ) 6 = 0 and P1 = P2 = P F, = P2A2cos 9 — PiAi = P(A2cos 0 — A1) = 160 x 103(0.0707 cos 60 — 0.283) = 160 x 103 (0.0353 — 0.283) = —160 x 103 x 0.2477 = —39.63 kN

Here

Similarly, we find out Fr Fy = P2A2sin 0 + p0v2cos 0 0= 0

But

Fy = 160 x 103 x 0.0707 x

2

= 9.8 kN R = VF,2 + F; = V(-39.63)2 + (9.8)2 = V15+ 05 + 96.04 = 40.82 kN 0 = tan-1 9'8 or 0 = 14° 39.63 Case 2: When flow of 0.8 m3 is taking place, we can apply continuity equation s 0= aivi — a2v2 0.8 vi — 0.283 — 2.83 m/s and

0.8 v2 = 0.0707 — 11.32 m/s

Apply Bernoulli's equation at section 1-1 and 2-2.

P1 A1 V1

465

466

Fundamentals of Fluid Mechanics

Pl

pg

+

V12

= P2 +

2g

pg

2g

160 x 103 + (2.83)2 _ P2 2 x 9.81 1 x 103 x 9.81 1 x 103 x 9.81

(11.32)2 2 x 9.81

16.31 + 0.41 -

P2 + 6.53 9.81 x 103 P2 = 10.19 x 9.81 x 103 = 99.96 kN/m2

or

Now net force in x-direction is: P1A1 - P2A2 + Fx = p0(v2cos 0 - vi) Fx = -160 x 103 x .283 + 99.96 x 103 x 0.0707 + 1 x or 103 x 0.8 (11.32 x cos 60 - 2.83) = - 45.28 x 103 + 7.067 x 103 + 2.26 + 103 Fx = -35.95 kN Now net force in y-direction is: Fy - P2A2 sin 0= pQ(v2 sin 0) FY = 99.96 x 103 x 0.0702 sin 60 + 1 x 103 x 0.8 x 11.32sin 60 = 6.12 x 103 + 7.84 x 103 = 13.96 kN R= VFx2 + =

Fy 2

35.95)2 + (13.96)2

= 1/1292.4 +194.88 = 39.84 0 = tan-1 Fy Fx = tan-1 13.96 35.95 = 21° 23. A 300 mm diameter pipe carries water under a head of 20 m with a velocity of 3.5 m/s. If the axis of the pipe turns through 45°, find the magnitude and direction of the resultant force at the bend. (UPTU 2006-7)

Fluid Dynamics-II

V2

Al = A2 = Now Since

ir x (0.3)2 4

0.07065 m2

A i Vi — A2 V2 A l = A2, hence V1 = V2 = 3.5 m/s

Applying Bernoulli's equation between 1-1 and 2-2 2 2 V2 + Z2 PI + Vi ± Zi — P2 + — 2g pg 2g pg Now Hence

zi = z2, Pi — H1, P2 = H2, V1 = V2 Pg Pg H1 = H2 = 20 PI = P2 = 20 x pg = 20 x 1 x 103 x 9.81 = 196.2 kN/m2

Now P1 A1 = P2A2cos 0 + Fx = pQ(v2cos 0 — v1 ) 196.2 x 0.07065 — 196.2 x 0.07065 x cos 45 + F., = 1 x (3.5 x 0.07065) x (3.5 cos 45 — 3.5) 1.96.2 x 0.07065 (1 — cos 45) + F., = (3.5)2 x 0.07065(cos 45 — 1) Fx = —4.06 — 0.253 F., = —4.313 kisT Similarly Fy — P2A2sin 0 = p • Q (v2 sin 0) or Fy = 196.2 x 0.07065 x sin 45 + 1 x (3.5)2 sin 45 x 0.07065 = 9.8 + 0.61 = 10.41 R= VFx2 +

F y 2

= /(4.313)2 + (10.41)2 =,j18.6+108.4

467

468

Fundamentals of Fluid Mechanics

= 11.27 kN 9= tan-1 10.41 4.313 = 68° 24. A pipe has diameter of 350 mm and right angle bend in horizontal plane. If 0.3 m3/s water is flowing, find forces acting on the bend: The pressure at inlet and outlet are 300 and 270 kN/m2 respectively. P2, A2, V2

P1 Al V1

Right hand bend (in horizontal plane)

Area A = Al = A2 = 1 x (0.35)2 = 0.0962 m2 Continuity equation or

A l V1 = A2 V2 = Q Vi = Q — Q.3 — 3.12 m/s Ai 0.0962 V2 = V1 = 3.12 m/s

Applying momentum equation in x-direction P1 A1 + F., = pQvi Fx = 1 x 103 x 0.3 x 3.12 — 300 x 0.0962 x 103 or = —27.9 kN Applying momentum equation in y-direction Fy — P2A2 = pQv2 or Fy = 1 x 103 x 0.3 x 3.12 + 270 x 103 x 0.0962 Fy = 26.91 kN R = V fx2 + fy2 = /(27.9)2 + (26 . 9)2 = 38.8 lth

Fluid Dynamics-II

469

Angle of resultant from horizontal 1 26 9 0 = tan-1 'f ' = tan fx 27.9 0= 44° 25. A pipe has diameter of 350 mm and bend of 135° in horizontal plane. If flow is 2.03 m3/s. Find the magnitude and direction of the resultant of forces. The pressure in pipe is 420 kN/m2

A bend in horizontal plane

A = Ai = A2 — Continuity equation: or Now

.7" C

x (0 .3 5) 2

4

— 0.096 m2

A1 V1 = A2 V2 = Q 0.3 Vi — V2 0.096 = 3.125 m/s P1 = P2 = 420 kN/m2

Applying momentum equation in x-direction P1 A1 + Fx + P2A2 cos El = pQ(—v2 cos 9 — vi) Or Fx = 1 x 103 x 0.3 (-3.125)(cos 135 + 1) —420 x 103 x 0.096 — 420 x 103 x 0.096cos 135 or Fx = —0.274 — 40.32 + 28.5 = —12.094 kN Applying momentum equation in y-direction — P2A2sin

or

0 + Fy = pQ(—v2sin 0) Fy = p•Q•v2 sin 135 + P2A2 sin 135 = 1 x 103 x 0.3 x sin 45 + 420 x 0.96 x sin 45 = 0.212 + 28.5 = 28.712 kN

470

Fundamentals of Fluid Mechanics

R = Al Fx2 + Fy2

= /(12.094)2 + (28.712)2 = V146.26 + 824.37 = 31.155 kN 0= tan-1 28.712 12.094 = 67° 26. Water is flowing through 0.8 m diameter pipe and reducer is fitted to the pipe which has diameter of 0.5 m at the outlet. If inlet velocity and gauge pressure at inlet of the reducer is 2.5 m/s and 400 kN/m2, find the resultant thrust in the reducer. Take frictions loss in reducer at 2 m of water. 1

"2, A2, V2 --- -

Reducer 2

Al

_ TCd2 _ lr X 0.82

4

4

—0.503 m2

7rd22

_ 7r X 032 —0.196 m2 4 4 Continuity equation = Al V1 = A2 V2 or 0.503 x 2.5 = 0.196 x v2 A2

0.503 x 2.5 0.196 = 6.4 m/s Q = Ai Vi = 0.503 x 2.5 = 1.25 m3/s

or

V2

Applying Bernoulli's equation between section 1-1 & 2-2 Pl. +

2

VI

pg 2g

±

2 _ P2 + V2 + z2 + z1 hloss

pg 2g

z1 = Z2

400 x 103 (2.5)2 + 2 x 9.81 1x103 x 9.81

(6.4)2 P2 +2 + 2 x 9.81 1x103 x9.81

Fluid Dynamics-II

Or

P2

lx 103 x9.81

—

47 I

40.77+0.318-2.087-2

= 37 p2 = 362.97 kN or Applying momentum equation in x direction PiAi P2A2 + Fx = pQ(v2 — v1) Or Fx = pQ(v2 — v1) + P2A2 — P1A1 Fx = 1 x 103 x 1.25 x (6.4 — 2.5) + 362.97 x 103 x 0.196 — 400 x 103 x 0.503 = 4.875 x 103 + 71.142 x 103 — 201.2 x 103 = —125.2 kN 27. Derive an expression for torque acting on a sprinkler using moment of momentum equation. 1i— r1

+

r2

1

,) 4 0 11--( V1

1 11 ® V2

Sprinkler

Consider a sprinkler with nozzle at 1 and 2 and hinged at point A. Momentum at point 1 = mass x velocity = pQ x v1 Moment of momentum at point 1 = p• Q•vi x ri Momentum of point 2 =pxQxv2 Moment of momentum at point 2 = p • Q•v2 x r2 Rate of change of moment of momentum = pQ(vi ri — v2r2) According to the moment of momentum equation, the rate of change of moment of momentum is equal to the torque (T) T = pQ(vi ri — v2r2) 28. A sprinkler has two nozzles, each located at 0.3 m from the centre which is hinged. The diameter of nozzles is 0.01 m and discharge is 3 x le m3/s. Find the speed of rotation if there is no friction loss. Also find torque to hold the sprinkler stationary.

1 d1 = 0.01 m

V1

0.3 m

+ Sprinkler

I

d2 = 0.01 m V2

0 0.3 m

472

Fundamentals of Fluid Mechanics

At = A2 - A —

Vi

gd 2 4

=V2 = Q 2A

it x (0.01)2 — 7.85 x 10-5 m2 4 3 x 10-3

2 x 7.85 x10-5

= 19.1 m/s v1 = v2 = wr 19.1 = w =63.66 rad/s 0.3

or

or

But

w—

27rN 60

N—

63.66 x 60 — 608.2 rpm 27r

Torque = T = pQ(viri — v2r2) v2 = - Vi T= 1x 103 x 3 x 10-3(2 x 19.1 x 0.3) = 34.38 Nm

29. An unsymmetrical sprinkler has equal flow through each nozzle with velocity of 8 ms. Find the speed of rotation in rpm Fe— 0.5

0

1m

II®

I t 8 m/s

8 m/s

Since r2 > ri Hence wr2 > wr1. Sprinkler will rotate anticlockwise. v1 = 8 + wri = 8 + 0.5w v2 = 8 — wr2 = 8 — w As per moment of momentum equation T = pQ(v2r2 — vi ri) As no external torque is applied T = 0 . v2r2 = vi ri (8 — w) x 1 = (8 + 0.5 w) x 0.5 or 8 — w = 4 + 0.25 w or 1.25 w = 4 w=

4 1.25 = 3.2 rad/s

Fluid Dynamics-II

Now

w—

2rN 60

or

N—

3.2 x 60 — 30.57 rpm 27r

473

30. A lawn sprinkler has 0.8 cm diameter nozzle at the end of a rotating arm and discharges water at the rate of 10 m/s velocity-Determine the torque required to hold the rotating arm stationary. Also determine the constant speed of rotation of the arm if it is free to rotate. 10 m/s 1 wr2

wr1

20 cm

25 cm

1 10 m/s

ird2 7r x (0.8 x 10-2 )2 — — 5.03 x 10-5 m2 4 4 Q = AV = 5.03 x 10-5 x 10 = 5.03 x 10-4 m3/s T = pQ(v2r2 - yin) = 1 x 103 x 5.03 x 10-4(10 x 0.25 + 10 x 0.2) = 5.03 x 10-1(2.5 + 2) = 5.03 x 4.5 x 10-1 = 2.26 Nm A=

If w is the angular velocity of the sprinkler. Since r2 > ri, the sprinkler will rotate clockwise as moment of momentum of section 2 is more (V1)absolute = 10 + 0.2 W (V2)absolute = 10 + 0.25 W

Now T = 0, as no external torque is applied T= PQ(r2,(V2))absolute — r 1 (V1)absolute) = 0 r2 (V2)absolute r1(V1)absolute 0.25 (10 + 0.25 w) = 0.2 (10 + 0.2 w) 2.5 + 0.0625 w = 2 + 0.09 w 0.0225 w = 0.5 w=

0.5 0.0225 — 22.22 rad/s

474

Fundamentals of Fluid Mechanics

w—

27N 60

N—

22.22 x 60 — 212.3 rpm 27r

31. A sprinkler with equal arms of 1 m but nozzles inclined 60° to sprinkler axis as shown in the figure. The nozzles have diameter of 2 cm and speed of rotation is 300 rpm. Find the flow out of the sprinkler if there is no friction loss.

it x (0.02)2 4 = 3.14 x 10-4 m2

Area of nozzle —

Now velocity component v1 sin 60 and v2 sin 60 have moment of momentum w.r.t. hinged point of sprinkler vi sin 60 x ri = v2sin 60 x r2 = wr But ri = r2 =r=lm 2irN w— 60 2 x ir x 300 60 = 314. rad/s 31.4 x 1 sin 60 = 36.25 m/s Q for each nozzle = A1 x V1 = A2V2 = 36.25 x 3.14 x 10-4 = 11.38 x 10-3 m3/s Q for both nozzles = 22.76 x 10-3 m3/s V1

v2

32. A vertically upward jet of water 10 cm in diameter issuing from a nozzle at velocity of 10 m/s strikes normal to a flat circular plate of mass 35 kg and diameter 50 cm and supports it. Find the vertical distance above the nozzle where the plate is held in equilibrium.

Fluid Dynamics-II

475

Plate

Jet

Nozzle

Force exerted by the jet = pAV2 A=

ad 2 4

a x(0.1)2 — 7.85 x 10-3 m2 4

Now plate is in equilibrium or

F = mg pave = mg

or

v2 —

35 x 9.81 1)(103 x 7.85 x10-3

= 43.74 or v = 6.6 m/s Initial velocity of jet u = 10 m/s v2 = u2 — 2gh 102 = 662 — 2 x 9.81 x h or

100 — 43.74 2 x 9.81 = 2.86 m

h—

33. Two tanks A and B on frictionless wheels are placed as shown. A jet emerging from tank A, strikes the blade on tank B. Find the forces required to hold the tanks stationary. Will the forces required will be different if either one or both of them are allowed to recede at 1 m/s? Consider control volume of tank A and jet is emerging from it with force FA. Then FA = pv2a FA = 1 x 103 x (I2gH)2 x itztd2 TC

= 103 x 4 x 9.81 x x (0.02)

4

= 12.32 N

2

476 Fundamentals of Fluid Mechanics

Jet diameter = 2 cm

FA -1.- I

1

// /// // / .7/// /// /// //

(.)

Consider control volume of tank B, then froce acting on it due to jet is (jet is deflected back) F B = - in B2 X a - pvB2 X a

= -2pvl x a vA = vB = \14 x 9.81 =6.26

Now

= —2 x 103 x 4 x 9.81 x

7E X (0.02)2 4

TB = —24.64 N The force FB is opposite to FA. If tank A is moved to left or right, the force FA remains the same i.e. reaction of the jet = 12.32 N. However if tank B is moved right, velocity vB is reduced by 1 m/s and FB becomes = 2 x 103 x (5.26)3 x

IC

x (0.02)2 — 17.37 N. If tank

is moved to the left, the velocity vB is increased by 1 m/s and FB becomes = 2 x 103 x (7.26)2 x

g x (0.02)2 4

— 33.1 N.

34. A jet engine when tested takes in air at 200 m/s & discharges the exhaust gases at 1500 m/s. The inlet and outlet cross-sections are 0.6 m2 and fuel to air ratio is 1:50. Find the force expected to act on the engine. Take density of air at entry = 1.2 kg/m3 and pi = P2

,.. V2

V1 ma -P-I

_------Jet engine

Fluid Dynamics-II

477

Applying momentum equation in x-direction FiAi — F2A2

Fx = (rha rit f)v2. — rit avi in a = P x Ai V1 = 1.2 x 0.6 x 200 = 144 kg/s . m

Since

144

- 2.88 kg/s 50 P1 = P2 and A l = A2 f

F, = (th a ± rit f) 1500 — Tit a x 200 = (144 + 2.88) 1500 — 144 x 200 = (220.3 — 28.8) kN = 191.5 kN 35. A jet engine flying at 40 m/s takes in air through 0.4 m diameter inlet in the front and discharges the exhaust gases through 0.1m diameter nozzle at 100 m/s relative to engine find (a) thrust (b) power lost (c) power output. Take pa = 1.2 kg/m3 v1 = intake air velocity = 30 m/s v2 = outlet exhaust gas velocity = 100 m/s air intake = m = pAV1 = 1.2 x

x 40.4)2

x30

= 4.52 kg/s Since fuel weight is neglected, the outlet exhaust gases = m = 4.52 kg/s F = Th (V — v2) = 4.52 (100 — 30) = 452 x 70 = 316.7 N Thrust = F = 316.7 N Power loss = kinetic energy of exhaust gases

452 x 702 2 = 11.1 kW Power output from engine = thrust x velocity of jet = 316.7 x 30 = 9.6 kW 36. What is a venturimeter? What is its principle? Venturimeter is a device used for measuring the rate of flow through a pipe.

478

Fundamentals of Fluid Mechanics

It works on the principle of Bernoulli's equation. The velocity of flow is increased by reducing the cross-section of the flow, thereby reducing the pressure head at the narrowed section. The pressure difference at normal section and narrowed section of the pipe can be measured which enables to determine the flow through the pipe. 37. Describe the construction of a venturimeter. Converging cone ..- Throat

( -I -

Diverging cone

Flow

Venturimeter

Venturimeter consists of a short length of gradual convergence and a longer length of gradual divergence. The angle of convergence is 16 to 20° and angle of divergence is 6 to 10°. A short length connecting converging and diverging section is called the throat. A pressure tapping '1' is provided at a location before the convergence commences and another pressure tapping '2' is provided at the throat section of the venturimeter. The pressure difference (pi — p2) between the two tappings is measured by means of a u-tube manometer. The manometer may contain water or mercury as the manometric fluid depending upon the pressure difference expected. 38. Describe the construction of a flow nozzle. Gauge

How

E>

FLOW NOZZLE

Gauge

Fluid Dynamics-II

479

Flow nozzle is also a device used for measuring flow through the pipes. Unlike venturimeter, the contraction of area is brought about by a nozzle. One of the pressure tappings is provided at a distance of one diameter upstream the nozzle plate and the other is provided at the nozzle exit. 39. Describe the construction of an orifice meter. Pipe AT

A

A

A

Flow

RIF ORIFICE METER Orifice meter is also a device used for measuring flow in the pipes. An orifice meter is a circular plate with an orifice at its centre. The flow in orifice meter adjusts itself such that it contracts until a point at downstream of the orifice plate and later the flow expands to fill the pipe. One of the pressure tappings is provided at a distance of one diameter upstream the orifice plate and the other at a distance of half a diameter downstream the orifice plate. The pressure difference between the pressure tappings is measured by the u-tube manometer. 40. Derive an expression for the flow for venturimeter.

Flow

VENTURIMETER Apply Bernoulli's equation between section 1 and 2 as shown in the figure 2

Pl + Vi pg 2g

2 _ P2 + V2 + z2

pg

2g

480

Fundamentals of Fluid Mechanics

But

z1 = z2 2 2 P1 — P2 _ V2 — Vi

2g

Pg

Pi — P2 = h, the pressure head difference measured by u-tube manometer Pg h

_ v22 — vi2 2g

As per continuity equation: al

vi= a2v2 where al = pipe cross-section area and a2 = throat cross section area a2 or v1 = — v2 al 2 2 2 1/2 — (a2 ) V2

h

v2

4 x2gh (al2 — a22 ) ai Igh

Or

v2 = V

Flow through pipe Or

2g

_ v3r [a? —a) l 2g a?

2 _

or

)2

al

al 2g

h—

Or

[., (a2 V2 1 —

2 (al

2 — a)

Q = v2a2 al • a2Al2gh

Q- 1 )

(ai`: — a2)

The above is theoretical discharge. The actual discharge is lesser than theoretical discharge which is: aia2 V0 Qactual = Cd 114

where

—4

Cd = coefficient of discharge

Note: If differential manometer has heavier liquid than liquid flowing through pipe, then h is to be multiplied by

sp gr of manometer liquid sp gr of flowing liquid

1 . Similarly in case differential

Fluid Dynamics-II

48 I

manometer has lighter liquid than liquid flowing through pipe, then h is to be multiplied by ( 1

sp gr of manometer liquid) sp gr of flowing liquid

41. Derive an expression for the flow through orifice meter.

ORIFICE METER Applying Bernoulli's equation between section 1 and section 0 2

Pi + vi +

pg 2g

2

_ Po + vo + zo zi pg 2g zi = Zo

Now

2 2 Po _ Vo — Vi pg 2g

P1 —

Applying continuity equation Q = aivi = aovo or

v1 =

ao Vo al 2 (ao 2 2 Vo — ) Vo

al 2g

h—

Vo2

or

[al2 — ao2 al2

—

Vo —

2gh aiNIgh Ala? — all

h

482 Fundamentals of Fluid Mechanics Discharge through orifice meter Qth = voao at ao V2gh at2 — ao2 Al

Actual discharge is lesser and given by Qactual = Cd where

at ao V2gh Ala? — 42

Cd = coefficient of discharge

42. Discuss the relative merits and demerits of venturimeter with respect to orifice meter. (UPTU-2002-3) Venturimeter

Orifice meter

1. In venturimeter the losses are less. Hence coefficient of discharge is high 2. It requires more space due to converging and diverging sections to be provided 3. It is not simple and cheaper like orifice meter

1. Losses are more and coefficient of discharge is less 2. It requires lesser space to fit on a pipe 3. Simple and cheaper

43. What is a rotometer? A rotometer is a device for measuring flow in a vertical segment of a pipe system. If consists of an accurately ground glass tube diverging upwards in the direction of flow. The tube has a float as shown in the figure. For a particular flow, the position of the float depends upon three forces: (1) drag force acting upwards (2) buoyance force acting upwards and (3) weight force acting downwards. Hence the position of the float is related to the flow. Spiral slits are cut on a part of the float to make it rotate so that it is stable against tilting. Diverging tube

Float

Ir

Drag baouaney tt

44 tt

Graduated scale

Weight

■

Flow Rofometer

Fluid Dynamics-II

483

44. A venturimeter has an area ratio of 10:1, the large diameter is 40 cm. During flow, the pressure heads are 7.75 m and 5.5 m respectively in larger and throat section. If Cd = 0.98, find flow through the venturimeter. al = 10 — 10 or al = 10a2 1 a2 h = 7.75 — 5.5 = 2.25 m

Q-

Cd ai • a2112gh V a? — 4 0.98 x 10 x aiV2x g x 2.25 V100ai —a?

0.98 x10 x a2 x6.644 9.94 = 6.55 x a2 But

K(0.4)2 1 x 4 10 = 0.01256 Q = 6.55 x 0.01256 = 0.0823 m3/s

a2 —

45. A venturimeter is used to measure the flow of oil in a pipe inclined 30° to horizontal. The area of pipe and throat area is 4 and sp gr of oil is 0.81. If the difference of mercury levels in the manometer is 50 mm, find the flow. Pipe diameter is 300 mm and Cd = 0.95

484

Fundamentals of Fluid Mechanics

s —1) Pressure head = P2 — Pl. + (z1 — z2) = x Hi pg s 13'6 1) = 0.05 (0.81 Or

h = 0.79 m of oil gd 2

al —

7r

4

a2 =

x.3)2

4

— 0.0707 m2

0.707 — 0.0178 m2 4

Q = Cd

aia2V2gh i 2 2 if a1 —a2

0.95 x 0.0707 x 0.0178 V2 x 9.81 x 0.79 A1(.0707)2 — (0.0178)2

Q = 0.068 m3/s 46. Oil of sp gr = 0.8 flows upwards with flow = 0.1 m3/s through a vertical venturimeter. Inlet diameter = 300 mm and throat diameter = 120 mm, Cd = 0.98. Vertical difference between pressure tappings is 40 cm. Find (1) if two pressure gauges are used at pressure tappings, then difference in their readings (2) if mercury manometer is used, the difference of mercury column.

t 40 cm

I Flow

Area al =

7r X (.3) 2

4

— 0.0707 m2

Fluid Dynamics-II

485

rc x (0.12)2 - 0.0113 m2 4

Area az -

al • a2 V2gh Q = cd ! \la? - 4 0.98 x 0.0707 x 0.0113V2 x 9.81 x h

0.1 -

/(0.0707)2 - (0.0113)2 0.1x V(49.98 - 1.277) x 10-4 0.98 x .0707 x 0.0113 = 8.914

V2 x 9.81xh =

h - (8.914)2 - 4.05 m

2 x 9.81 Case 1 Difference of reading of two pressure gauges is: h

Pi P2

- ( Pg Pg P1 - P2

+ (0 - 0.4) 0.8 x 103 x 9.81 pi -p2 = (4.05 + 0.4) x 0.8 x 103 x 9.81 = 4.45 x 0.8 x 9.81 x 103 = 34.92 kN/m2

-

Case 2 Difference in level of mercury manometer:

h = x( .1 -1) s

or

4.05 - x (116 1) 0.8 x - 4.05 - 0.253 m 16 = 25.3 cm of mercury

47. An orifice meter of diameter 12 cm in inserted in a pipe of 24 cm diameter. The pressure gauges fitted upstream and downstream give readings of 300 kPa and 150 ICPa respectively. Cd of the orifice meter is 0.6. Find flow of the water.

486

Fundamentals of Fluid Mechanics

Flow

'/////////////// ,///z/z/Z

al =

7rd? 4

x (0.24)2 — 0.0452 m2 4

ao

7rdg 4

7r x (0.12)2 — 0.0113 m2 4

— Po _ (300-150)x103 h= Pl 1 x 103 x 9.81 Pg 150 = 15.29 m of water 9.81 Cd X

Q

aia0V2gh

1 2 2 1. al — ao

0.6 x 0.0452 x 0.0113 x /2 x 9.81x 15.29 V(0.0452)2 — (0.0113)2 5.308 x 10-3 l(20.4 — 1.277) x 10-4

A

5.308 x 10-3 4.37 x 10-2 = 1.214 x 10-1 = 0.1214 m3/s 48. A venturimeter is to be introduced in a 25 cm diameter pipe converging flow of 0.15 m3/s. The pressure in the inlet is 6 m of water gauge find the minimum diameter of the throat for the venturimeter so that the pressure head is zero. Also calculate discharge when u-tube differential manometer reads a deflection of 20 cm. Applying Bernoulli's equation PI

pg

2

2g

—

P2

2

v2 + —

pg 2g

+ z2

Fluid Dynamics-II

Zl

— ‘,2 P2 — 0 & Pi = 6m Pg Pg

2

6+

2 = v2 2g 2g

1

x di _ it x (0.25)2 4 4

—

Q _ 0.15 — 3.06 m/s al 0.0491

vi

vz =

(3.06)2 2g 6 2 x 9.81 = 6 + 0.477 = 127.08 v2 = 11.27 a2v2 — Q — 0.15

or Now

a2

0.15 — 0.0133 11.27

ird? = 0.133 4 d2

= 0.0133 x 4

= 0.0169 d2 = .13 m Now manometer shows deflection of 20 cm h=x

—1)

=0.2 ( .6-1) 11 = 2.52 m Q

al • a2V2gh 2 V 2—a2

— 0.0491 x 0.0131/2 x 9.81 x 2.52 V(0.0491)2 — (0.013)2 = 0.097 m3/s

0.0491 m2

487

488

Fundamentals of Fluid Mechanics

49. The constant angular velocity at which a liquid rotates in a cylinder about a vertical axis such that the pressure at a point on the axis is the same as at a point 2 m higher at a radius 2 m is: (a) 2 rad/s

(c) 2g rad/s

(b) 1 rad/s

F

-

(d) Jg rad/s (Civil Services 1994)

Initial level

y

For forced vortex, we have Y

w2 r 2 2g

2=

w2 x 22 or w = 2g

The answer (d) is correct. 50. Air flows through a venturi and into atmosphere. Air density is ' p' , atmospheric pressure is `P„' throat diameter is Di, exit diameter is `D' and exit velocity is ' u' . The throat is connected to a cylinder containing a frictionless piston attached to a spring. The spring constant is 'P. The bottom surface of the piston is exposed to atmosphere. Due to the flow, the piston moves by distance 'x'. Assuming incompressible frictionless flow, 'x' is— D2 —lir D2s Dr2

(a) (p u212k)7 rD2t

(b) put l8 k(

(c) (p u212k)n- 1)2,

(d) put/8 k[ D —1 Jr DS Dr2

4

(GATE 2003) D U

Dr

•

Ds

Fluid Dynamics-II

As per continuity equation at throat and exit, we have— A1 V1 = A272 or

7t =

(A) 72 Ai

jr D2 _ 4 u g . Dr2 k. 4 2

D [A )

u

Now apply Bernoulli's equation, we have— V2 Pr r _ P+V 2 ± 2g Pg Pg 2g

P— Ps _17,2 —V 2 2g Pg

or

(P - Pt) = 2

(77 - 72) 4

_ p [(D) u2 Dr

u21

Now pressure force acting on the piston is— =

ir D s 4

X

(P — Pt)

3. i 9 x P [(14 —1 u2

4 2

Dr

Now spring force = pressure force 2 [I D Y —1 ADS Px = Pu Aj 8

x= Option (d) is correct.

Pu 8k2

3 [(D 13: )2 111°2

489

490

Fundamentals of Fluid Mechanics

51. A venturi meter of 20 mm throat diameter is used to measure the velocity of water in a horizontal pipe of 40 mm diameter. If the pressure difference between the pipe and throat section is found to be 30 liPa, then neglecting frictional losses, the flow velocity is— (a) 0.2 m/s

(b) 1.0 m/s

(c) 1.4 m/s

(d) 2.0 m/s (GATE 2005)

Applying continuity equation, we get— A1 V1 = A2V2 TC X Dig2

4 V2 = Al r — it2

TC 4

2 X Vi

x D2

(40) x Vi = 4Vi 20 2 Now on applying Bernoulli's equation for sections 1 and 2, we have— 2 2 V2 P1 + 171 — + Z1 = P2 + 2 + Z2 p• g 2g p• g g Zi = Z2,

As

— P2 - ( - V?) p• g 2

P1

= or Or Option (d) is correct.

[(u Vi)2 — Vi2]

15 V2_ 30 x 103 2 1 1x103 x9.81 V1 = 2 m/s

52. A pitot static tube is used to measure the velocity of water using a differential gauge which contains a manometric fluid of relative density 1.4. The deflection of the gauge

Fluid Dynamics-II

49 I

fluid when water flows at a velocity of 1.2 m/s will be (the coefficient of the tube may be assumed to be 1) (b) 52.4 mm

(a) 183.5 mm

(c) 5.24 mm

(d) 73.4 mm (IES 2001)

The velocity of flow is— V = 112gx( S; —1) (1.2)2 = 2 x 9.81 x x x (L' il —1)

where x = deflection S„, = sp gravity of manometric fluid

x= 183.5 mm Option (a) is correct.

S = sp gravity of flowing fluid

53. A large tank is fixed to a cart as shown in the figure. Water flows from the tank through a nozzle of diameter 30 mm at a speed of 8 m/s. The water level in the tank is maintained by adding water through a pipe. Find the tension in the wire holding the cart stationary. Tank

Wire

The force of the jet ejecting from the nozzle is— Fx = (p • A• V)• V = p • A • V2 103 it x 0.032 ) 82

4 = 45.24 N Due to this force 'Fr', the tank on the moving cart tends to move in opposite direction as reaction, thereby creating a tension T= 45.24 N in the wire. 54. A venturi meter is fitted horizontally in a 15 cm dia pipe and the pressure in the pipe corresponds to a water head of 10 m. If the maximum flow through the meter is 0.15 m3/s, find the smallest throat diameter so that the pressure does not fall below 2.45 m of water (absolute).

492

Fundamentals of Fluid Mechanics

(UPTU 2009-10) 2

Flow

Pi ± Vi_ 2 pg 2g

_ P2

pg

v22 2g

z2

= Z2

As h1 +

v12 2g

2

TT 2 v2 h = — h2 — 22 2vi g g h = 10 — 2.45 = 7.55 m

or Now

=

Now or

Q_ Q1

2 + V2 2g

h2

2 2-8,"

_ 0.15x4 0.15 = 8.5 x 0.0225 IC x 0.152 4 7.55 x 2g =

— 8.52

or

= 148.13 + 72.25 = 220.38 V2 = 14.85 m/s

Now

a2 —

Now

a2

or

d2

Q — 0.15 - 0.01 m2 V2 14.85 2 d2 4

4x 0.01 11 rc = 0.113 m = 1.13 cm

Chapter 11

LAMINAR FLOW

KEYWORDS AND TOPICS A A A A A A A A A A A

LAMINAR FLOW TURBULENT FLOW FLOW IN HORIZONTAL PIPES FLOW IN PARALLEL PLATES AVERAGE VELOCITY MAXIMUM VELOCITY HEAD LOSS FRICTION FACTOR FLOW IN INCLINED PIPES CORRECTION FACTOR FOR KINETIC ENERGY CORRECTION FACTOR FOR MOMENTUM

A A A A A A A A A A

FLOW IN POROUS MATERIAL DARCY'S EQUATION FLUIDIZATION JOURNAL BEARING FOOT STOP BEARING COLLAR BEARING DASHPOT FALLING SPHERE METHOD CAPILLARY METHOD ROTATING CYLINDER METHOD

A

FLOW IN OPEN CHANNEL

INTRODUCTION In laminar flow, the fluid particles move along straight parallel paths in layers. There is no mixing of fluid particles between two adjacent layers. The shape of lamina (layer) depends upon the shape of the boundary through which flow is taking place. In laminar flow, the fluid particles move in unmixing layers or streams in smooth continuous paths. Soldiers marching in orderly manner is an analogy to laminar flow. Laminar flow occurs at low velocity so that forces due to viscosity are predominant in comparison to inertial forces. The viscosity of fluid therefore induces relative motion within the fluid when fluid layers slide over each other. The gradient of velocity between the layers give rise to shear stresses. The shear stress in the fluid varies from point to point. It is maximum at the boundary and it gradually decreases with increase in the distance from the boundary. The shear stresses in between the

494

Fundamentals of Fluid Mechanics

layers develop a resistance to flow. The pressure of the fluid gradually drops in the direction of the flow. There is always head loss in viscous flow.

1. What is a laminar flow (viscous flow)? Laminar flow is a flow in which flow takes place in layers. There is no mixing of fluid particles between any two adjacent layers. The shape of lamina (layer) depends upon the shape of the boundary through which flow is taking place. In laminar flow, the fluid particles move in unmixing layers or streams and follow a smooth continuous path. The paths of fluid particles retain their relative positions at successive cross-sections of the flow passage. There is no transverse displacement of fluid particle. Soldiers marching in orderly manner is an analogy to laminar flow. The shape of laminae if flow takes place between two parallel flat plates are plane sheets parallel to each other as shown in figure. In case the flow takes place through a circular pipe, the laminae become concentric cylindrical sheets as shown in the figure.

Layers /////./V4/.4/./././A/A,VA,///,

Layers

Layers as plane sheets

Concentric cylindrical sheets

2. What are the conditions which help the flow to be laminar? The flow will be laminar when: (1) (2) (3) (4) (5)

Velocity of flow is low Diameter of pipe is small Viscosity of the fluid is high Density of the fluid is less Reynolds number is less than 2000 for flow in pipes.

3. Can Bernoulli's equation be applied to viscous or laminar flow? Due to the presence of viscosity, real fluids differ from non-viscous ideal fluids. Due to viscosity, each fluid layer resists the relative translation motion of adjacent fluid layers. Some energy is required to overcome this resistance which is converted to thermal energy as a loss. The Bernoulli's equation has been derived for flow of an ideal fluid (non-viscous fluid in earlier chapter). However Bernoulli's equation as such cannot be applied for viscous flow. Therefore the Bernoulli's equation in the modified form is used in analyzing the flow of real

Laminar Flow

495

fluids. The head loss is accounted in the Bernoulli's equation as given below: Pl.

pg

+

2 111

2g

2 ± Zi — P2 + — V2

where

pg

2g

+ Z2 + h/

h1 = head loss due to viscosity.

4. What are the various factors on which friction resistance in laminar flow is dependent and independent? The friction resistance depends upon: (1) Velocity of flow (2) Area of surface in contact (3) Temperature of the fluid The friction resistance does not depend upon: (1) Nature of surface in contact (2) Pressure of flow 5. What is turbulent flow? When the velocity of flow reaches a certain limit such that the fluid particles no longer move in layers or laminae. Violent mixing of fluid particles now takes place due to which they move in random manner. As a result the velocity at any point varies both in magnitude and direction from instant to instant. In turbulent flow, the motion of fluid particles is irregular. The fluid particles move along erratic and unpredictable path. The velocity of fluid particles fluctuate both along the direction of flow and also perpendicular to the flow. A crowd of commuters on a railway station rushing for boarding a train is an analogy of the turbulent flow. The flow in pipes having Reynolds number 4000 and above is always turbulent. 7--'

...---,

--____„.

---\,.."

-.__.jf --.- k \---.... -_____-0 \,..._,...,„ -----„,Turbulent flow

6. What is Reynolds number? How is it useful? Reynolds number is the ratio of inertia force to the viscous force. It is a dimensionless number and it is given by: d Re = pV where p = density, V = velocity, P d = diameter or distance and /../ = viscosity Reynolds number is very useful for:

496 Fundamentals of Fluid Mechanics (1) Predicting whether flow is laminar or turbulent (2) Finding out the coefficient of friction (1) in order to determine the loss of head due to friction. 7. Explain the experiment setup to study different types of flow. or What is Reynolds experiment to demonstrate laminar and turbulent flow? The setup consists of a tank full of water. A bell mouthed glass tube is fitted at bottom of the tank. One end of the glass tube in the water and other end have a valve for controlling flow in glass tube and flow falls in a measuring tank. A dye injection arrangement is fitted at the mouth of the glass tube. It consists of a small overhead tank containing a coloured liquid having same sp gravity as that of water. The head of water is maintained constant in the tank. The flow of water from the water tank through the glass tube is now regulated by opening the regulating valve. The velocity of flow depended upon the opening of the regulating valve. As long the velocity in the glass tube is maintained sufficiently low, the colour band remains a thin straight line flowing along the entire length of the glass tube without mixing with water. Such flow is laminar flow. As the velocity of flow is gradually increased, a stage is reached when the dye tends to develop a wavy form. This indicates that the laminar flow has become unstable. On further increasing the velocity of flow by opening the regulating valve, the dye starts mixing with surrounding water. This flow is called turbulent flow. The intermediate flow when dye develops a wavy form is called transition from laminar to turbulent flow.

Tank with water Measuring tank Reynolds experiment setup I

Laminar flow

H..—....—..—..—...1 \

/

Unstable flow ..•••*.%..............1

Laminar Flow

497

8. What is the shear stress acting on a fluid while flowing through a pipe? Draw the shear stress distribution at a cross-section.

1 R

Flow

Consider a small cylindrical fluid element of radius r and length dx as shown above. On the fluid element, pressure P is acting on face AD and pressure ( p +

ap dx on face BC. If ax )

z is resistive or shear stress acting due to flow, then shear force acting on this element is shear stress multiplied surface area i.e. T x (271-rdx). As flow had very low velocity, the inertia force (mass x acceleration) can be neglected. During equilibrium of the fluid element: P x nr2 — (p +

ap dx) x n-r2 — ax

TX

2n-rdx = 0

or

— P • dx•nr2 = z x 2rcrdx ax

or

—r ( ail) 2 ax)

T

ap Since — is constant at any cross-section of the pipe, the shear stress varies with radius. ax The shear stress is zero at r = 0 (centre of the pipe) and maximum at r = R (inside radius of the pipe). The shear stress distribution at any cross-section is as shown below. r=R

Shear stress distribution

9. Derive an expression for velocity for a flow inside a pipe. Draw the velocity distribution at a section. du As per Newton's equation of viscosity, we have shear stress z = µ — where µ = viscosity dy

498

Fundamentals of Fluid Mechanics

and

dy

is velocity gradient from stationary surface.

Therefore, the distance of fluid layer is to be measured from the boundary surface.

T R

y=R—r On differentiation, we get dy = —dr as R = Constant T

But we have found out

=

du dr

du dy =

T=— r

2 ax du = r ap dr 21.1 ax

or

f du =

or

u—

21.1a x J

rdr

1 DP r 2 2

ax 2

+ C C = constant

When r = R, u = 0 at fluid is stationary at boundary. C=—1 x R2 4p ax u=

i ( — al 0.2 _ R2 ), 4p ax

herea , AL and R are constants. Hence velocity at any section in pipe varies with the square p of radius (r). When r = 0, the velocity is maximum and when r = R, the velocity is zero. The velocity varies parabolic from zero at the boundary surface to the maximum at centre of the pipe. The velocity distribution at any section is as shown below.

Laminar Flow

499

r=

Velocity distribution

10. Derive an expression for average velocity for a flow through pipe. Find the ratio of maximum velocity to average velocity. or For a steady laminar flow through a circular pipe prove that velocity distribution across the section is parabolic and the average velocity is half of the maximum velocity. (UPTU-2006-07)

Consider the flow through a circular fluid ring element of radius r and thickness dr. The discharge through this ring element is equal to the velocity of the flow multiplied by the cross section area of the ring element. dQ = u x (2n-rdr) But

u —

1 a P (R2, _ r2.) 4/.1

1 °P (R2 - r2) x 2yrrdr dQ — 4~u ax R

Or

Q=

21r uP (R2 — r2) rdr 4,u ax

0

=-

T

2

°P [R2 4,u ax 2 7 OP

8p ax In case average velocity = u , then

R

4 -R r

4 -10

500

Fundamentals of Fluid Mechanics

—7r aaxp • R4

Q — Area of pipe gy x it x R2

a=

1 rap) R2 8,u

ax

The above is the expression for average velocity of the flow through a pipe. The velocity of flow at any cross-section is given by: u=

1 (—ap) (R2 _ r 2) 4,u ax

The maximum velocity of flow is at the centre of the pipe i.e. r = 0 max

l

1 (-a 4p ax

R2

The ratio of maximum velocity and average velocity of the flow is: 1 r_ aP) R2 u.

4p

ax ) r

8µ

p)R2 ax )

=2 Hence the maximum velocity of the flow is two times the average velocity of the flow in a pipe. 11. Derive an expression for drop of pressure for a given length of pipe when fluid is flowing through a pipe. or Derive Hagen Poiseuille formula.

0

Flow

L x2

Consider a flow through a pipe and length of pipe is L between section 1 and section 2 as shown in the figure.

Laminar Flow

50 I

The average velocity of flow through a pipe is given by: 14 —

or

ap

1 rap) R2 8µ ax

8uµ ax R2 Now integrating for section 1 to section 2

- 52 ap -

8/0 f x2

P1 — P2 —

8 up , 2 V2 - x1) R

i

or

—

R 2 Jxi

ax

814p ... , L but R`

R= D 2

32 up P1 — P2 — D2 L hf = Loss of pressure head = Pi — P2 Pg

h = 1,1 — P2 — 32uµp

...

f

pg

L pgD2

The above equation is known as Hagen Poiseuille formula. 12. What is the power required to overcome the viscous resistance to make the fluid flow through length L and discharge Q? Power required = Weight of the fluid flowing per second x head loss =mxgx hf = Q.p x g x

32WpL pgD 2

32 UpL n D2 13. What is friction factor (f)? In pipe flow, the loss of energy or friction head loss takes place between any two sections of the pipe. A rational expression for friction loss head can derived by defining a dimensionless coefficient f known as friction factor. Friction factor is:

8T° fi — pV 2

where To = shear stress at pipe wall and V is average velocity

502

Fundamentals of Fluid Mechanics

14. What is Darcy's equation? Darcy's equation is friction loss in pipe flow. The friction head is given in terms of friction factor (f) as under: hi—

Z LV2 2gD

where L = length, V = average velocity and D = diameter f = Darcy coefficient and 4 f = frictional factor = fl.

15. Derive an expression for coefficient of friction in viscous flow in terms of Reynolds number. Head loss in viscous flow by Hagen Poiseuille formula is: 32 kuL h — f p g D2 Loss of head due to friction by Darcy's equation h — f

f LV 2 2gD

In case of pipe flow, V = d i.e. velocity is always average velocity in pipe flow 32pUL _ f Lu 2 pgD 2gD or

64,u

f - pu_ D 64 pW D 11

But

Reynolds number = Re

= pvD ii

16. Derive an expression for local velocity of viscous flow between two stationary flat plates. States the assumptions made. (UPTU 2008-9) The assumptions made are: 1. Flow is two-dimensional steady laminar flow 2. Fluid is Newtonian 3. There is no slip of fluid particles at the solid boundary.

Laminar Flow aT T+ ay

///////////////////////////

•

503

dy P +r • dx

= 1 > t

dy—t— H-dx -bd —f Y

1//0//////////r//////,:i / 4_ x2

Viscous flow through two stationary parallel plates

Consider two stationary parallel plates kept at a distance t apart and a viscous flow is taking place as shown in the figure. Take a small fluid element of thickness dy, length dx and width unity at a distance y from the bottom plate. If we draw a free body diagram of this fluid element, then forces acting are (1) pressure forces p andp +

ap ax

• dx acting along x direction

at both end faces (2) the shearing forces resisting flow at lower and upper surface of the element. Shear stress is z at lower face and shear stress is (z + az — ay

at upper face. For

practical purpose, it is assumed that gravitational force is zero for horizontal flow and

ap ay

= 0. In equilibrium condition, net force acting on the fluid element is zero. p(dy x 1) —

ap • dx)(dy x 1) — dx x 1 ax

(13 +—

+(z+ d'r •dyjdx x1=0 ap = aT ax ay The above equation shows that pressure gradient in x-direction is equal to the rate of change of shear stress in y-direction. On integration dp =

=

dY dp •y + c1 where c1 = constant dx

Now as per Newton's law of viscosity, we have du dp dy dx Y ci

504

Fundamentals of Fluid Mechanics

or

du

Or

.1 Chi

or

1 dp Y dy dx

dy

p,,dy f

d x ••I 'Y

u—

dP y2

p dx

2

r P

dy

+ ci y + c2 At

As there is slip at solid boundaries, the boundary conditions are: y = 0, u = 0, hence c2 = 0 and

y = t, u = 0, hence c1 = —1/2 (d dl u=1•

dp dx (Y2 —

Since t•y > y2, we can write u= 1 (— di (t'Y — .Y2) 2p d x The above equation is a equation of parabola. Hence local velocity varies in parabolic manner when a viscous flow takes place between two parallel and stationary plates.

Local velocity distribution

17. Draw the shear stress distribution for the viscous flow through two parallel and stationary plates. The local velocity is u=

r—dp)

2/.1

dx

(tY — .Y2)

On differentiating w.r.t y du

dy But

z=

(—dp) (t —2y) 2p dx 1

du dy

Laminar Flow

1 2

T

=

T

= 0, when y =

505

(t — 2y)

dx

2 maximum when y = 0 or y = t

T=

The distribution of shear stress is as shown below.

Shear stress distribution 18. Find the expression for average velocity for the viscous flow between two parallel and

stationary plates. Find the ratio of maximum velocity of average velocity. The local velocity at distance y is given by — 1 (—d dx i (t•Y — Y2)

u=

Discharge through a small element of height dy is: dQ =uxdyxl 1 (— d) dy 2/./ d x p yt•Y — Y2) Now total discharge Q is by integrating Q=

2p

(— dp) f t (ty y2) dy 0 dx

1 (— dP

[tY 2 Y 3 1 2 3 0

2p

dx )

1

rdp)[t3

2p

1 12p

dx

t3 1 2 3

r-C1P) t.3

dx

506

Fundamentals of Fluid Mechanics

Now average velocity Ci is — d) 3 t 1 ( dx p A 12it txl

Q __ u __

1 (— dP) 12 dx

u

12p

To find maximum velocity which is at the equal distance from both the plate, put y= t2 u —

umax

1

d) dx (t .Y — .Y2)

12p

1 (_d dxp ) rt .t t2 ) ) 2 4 )

12p

1 r—dp) t2 8p( dx Now the ratio of uff,ax to

a

is 1 ( — dP) t 2 8p dx )

1 (—dp) t 2 dx

12p

3 2 19. Derive an expression for drop of pressure head of a viscous flow through two parallel and stationary plates for a length of L.

Flow

I-4— P2

Laminar Flow

507

dd p) , viscosity of fluid and distance x

The average velocity (14 ) for a pressure gradient t between the plates is given by: —

ap

or

1 rap) 12 12µ

ax

12t1p. t2

ax

12/4y. ax t2 Now integrating the equation from x1 to x2, we get or

al)

52 — 1

12up fx2 12T4

(Pi — P2) = Or

ix'

t2 2 /1 t

ax

(x2 xi)

12,u14 • L

P1 P2

2

The head loss during flow is: pl—p2

h_

12puL

pgt 2 — Note: The head loss increases with velocity and length of flow but decrease with increasing distance between the plates. pg

20. Derive an expression of local velocity for viscous flow between parallel plates when one plate is fixed and other is moving. What is the name given to this kind of flow? I

I

j _Hdx,

Moving plate

+

• dy

aP P + — dx

dy

ax •

II

/// // //// / //// / ////// /// / // /

I

This type of flow is called couette flow. In case we consider the equilibrium of the fluid element, we get

4

p•dy.1 — (19+- dx) dy.1 — ax

T. dx.1

ap _ azay ax

+ (-c + at" dy) dx.1 = 0 ay

508

Fundamentals of Fluid Mechanics

or

5

a I" = .1" ax • dY

or

t=

But

(t)

T = 1,1

Y ± CI

du dy

du= 1 ( aP) y dY ± ciaY ki ax Integrating the equation, we get: i

au — µl al f y dy + i dy ax J U = 1 (a/9 ) Y 22 ± Ci y + c2 i.i ax

(i)

Applying the boundary conditions to find c1 and c2. y = 0, y=

u = 0, hence c2 = 0

t, u = v, hence Cl

v

--—

t

1 ( al t ki ax 2

Putting the value of C1 & C2 in eqn. (i)

u= 1 (al Y2 + v_ 1 al, t) Y It ax 2 0-p ax 2)

The equation shows that local velocity depends upon

ax

and velocity (v) of the plate.

ap — ax

can be negative or positive. In case it is zero, then

v u —.y t u changes to zero from stationary plate to maximum ap velocity v linearly which is the velocity of the moving plate. In case — # 0, then the local ax The equation shows that velocity

velocity will vary parabolically.

Laminar Flow

21. Find the stress distribution in couette flow. In couette flow, the local velocity (v) is given by: u=

1 i_ap)(„„ 21.1 ax "Y

y2) + v.a, ' t l'

H v -H

Velocity distribution

r ax

Velocity distribution

r

# 0)

ax

0)

Differentiating the equation, we get — au = 1 (— a9) (t — 2y) + l' ay 2A1 ax t But

shear stress

"C — ii

au

ay

1 (— — aP) (t — 2y) + µv 2 t ax Shear stress at stationary plate is To can be found out by putting y = 0 or

T=

pl, ± rap) t

to — t

ax 2 Shear stress at moving plate can be found out by putting y = t __ = t m

yv + ( — l( ( a ax

(ap) t

yv t

ax

Shear stress at y = —t i.e. centre of two plates 2 -c = c A point where shear stress is zero is

yv t

2

t

t)

J

509

5I0

Fundamentals of Fluid Mechanics

° = P + laaxp) 2t Or

t

V

2

t

x

Y)

1 ( — ap )

The shear stress distribution is as shown below.

y

+ ,uv x

1

P1) 2 Shear stress distribution Lax y 22. A fluid of viscosity 0.72 Ns/m2 and sp. gr. 1.34 is flowing through a circular pipe of diameter 100 mm. The maximum shear stress at the pipe wall is given as 200 N/m2 Find (i) pressure gradient (ii) average velocity and (iii) Reynolds number of the flow. (UPTU-2005-6) r

Guidance: For flow through pipe, shear stress 1--

a p jr

r ax 2 . Hence for maximum shear

stress at pipe wall is when r = R. Average velocity is 14 = 1 raP) x R2 sp ax rmax —

rap )R

200 —

ax 2

ap)

0.050

ax

2

— ap _ 200 x 2 — 8 x 103 N/m ax 0.05 Average velocity (u ) is

1 (— ap) R2

81.1

ax

1 x 8 x 103 x (0.05)2 8x0.72

Laminar Flow

5I I

= 3.472 m/s Reynolds number =

u •p • D Ii

3.472 x 1.34 x 10 x 0.1 0.72 = 646.2 23. Show that for viscous flow through a circular pipe mean velocity of flow occurs at a radial distance of 0.707 R from the centre of pipe where R is the radius of pipe The local velocity is: u=

41p ( aaxP) (R2 — r2)

(i)

and average velocity is: — t4 — 1 ( a19 R2 8u ax )

(ii)

Equating equation (i) and (ii) 1 ( ap) (R2 4p ax ‘ Or

r2) = 1 ( 4) R2 8p ax

2R2 — 2r2 = R2 R2 = 2r2 or r—

or

_R 2

= 0.707 R 24. A lubricating oil of viscosity 0.1 Ns/m2 and sp. gr. 0.9 is pumped through a 40 mm diameter pipe. If the pressure drop per meter of pipe is 24 kN/m2, find the flow

rap)

Average velocity a = 1 8p ax or

a —

13

2

1 (24 x 103) x (0.02)2 8 x 0.1

a = 12 m/s Discharge Q=u x A = 12 x 7 x (0.04)2 4 = 0.015 m3/s

5I2

Fundamentals of Fluid Mechanics

25. Oil having sp. gr. 0.92 is pumped through a 20 mm diameter pipe. The discharge is 0.90 m3/min and pressure drop is 100 liPa for 800 m length of pipe. Find viscosity. Average velocity /7 = Q A 0.9 x 4 6 x x (0.20)2 = 0.4777 m/s

or

The pressure drop in a pipe for length L is — 32CI uj L P2 — P1

100 x 103 — 32

Atx 0.4777 x 800 (0.2)2

100 X 103 X 0.04 — 32 x 0.4777 x 800 = 0.327 Ns/m2

Or

26. A pressure drop of 4000 N/m2 takes place in a circular pipe of length 20 m. The Reynolds number of this flow is 120. Find drop in pressure in case flow is made double without changing liquid properties. Since Reynolds number 2000, the flow is laminar Pi — P2 —

32W,uL — 4000 D2

When flow is made double i.e.

or

Q' = 2Q we have AC(' = 2•A = 2 ,1/ „ 32 u'xpx L Pi — P2 — D2 32uxtixL D2 = 2 x 4000 = 8 kN/m2 =2x

27. A viscous flow of oil is taking place in a pipe of 12 cm diameter with discharge of 1.8 x 10-4 m3/s. What power per kilometer is required to maintain the flow if sp. gr. of fluid is 0.9 and viscosity is 0.2 Ns/m2?

Laminar Flow

5I3

1.8 x10-4 x 4 _ Q u = = A x(0.12)2 = 0.01592 m/s Pressure drop is — P2 L 32 4µL pgD pg or

hf

32 x 0.01592 x 0.2 x 1000 0.9 x 103 x 9.81 x (0.12)2

= 0.80 m of oil Power = pg• Qhf = 0.9 x 103 x 9.81 x 1.8 x 10-4 x 0.8 = 1.27 watts 28. A 20 mm diameter pipe 40 km long is used for transporting oil from a tanker in sea to shore at flow of 0.01 m3/s. Find the power required to maintain the flow. Assume = 0.1 Nm/s & sp gr = 0.9 for the oil. Flow = 4 x Area Q=4 x

ird2 4

_ 0.01 x 4 u — — 0.318 m/s x 0.2' Now to check Reynolds number R

-

pod

0.9 x 103 x 0.318 x 0.2 0.1 = 572.4 Since Re < 2000, hence flow is laminar Now

Pi — P2

32 Up L 2 D

32 x 0.318 x 0.1 x 40 x 103 (0.2)2 = 1017.6 x 103 N/m2 Power loss = (pi — p2) x Q

5I4

Fundamentals of Fluid Mechanics

= 1017.6 x 0.01 x 103 = 10176 watts = 10.176 kW 29. An oil has viscosity of 0.01 Nm/s and sp. gr. of 1.6 flows through a horizontal pipe of 10 cm diameter with a pressure drop of 6 kN/m2 per meter length of pipe. Find (1) rate of flow (2) shear stress at pipe wall (3) drag for 100 m length of pipe (4) power to maintain flow in 100 m pipe length. —ciP— 6 x 103 N/m3 dx

dL Q—

Now

IC

r aP ) R4

8p ax

7r (6 x 103) x (0.05)4 8 x 0.01 = 1.41 m3/s Now shear stress

7

__

ap r ax 2

at wall Tmax —

ap • R ax 2

1 0.65 = 6 x 10- x = 150 N/m2 The area at which the shear stress is acting is the periphery multiplied by length. Hence drag force is: shear stress x surface area =150 x (irxdx L) = 150 x 7r x 0.01 x 100 = 4.71 KN

Fd =

Power required to maintain the flow can be found out by multiplying discharge to the pressure drop.

ap •L Power = Q x — ax = 1.41 x 6 x 103 x 100 = 8.46 x 105 watts = 846 kW

Laminar Flow

5I5

30. The velocity distribution for viscous flow in a circular pipe of radius R is given by:

umax Where u is the local velocity at radius 'r' and u,„,„„ is the maximum velocity at centre. Find the values of correction factors for kinetic energy and momentum in case average velocity is used in their calculation. or Prove that for viscous flow through a circular pipe, the kinetic energy correction factor is equal to 2. (UPTU-2002-3) Guidance: We have to apply correction factors in case kinetic energy and momentum are calculated on the basis of average velocity. Local velocity (u) T R

Viscous flow through pipe

Consider an elementary area at radius r and thickness dr. Then area of the circular strip is dA = 2n-rdr A = 2Irrdr The average velocity u is

u=

A

,2 But

Q=

udA = J u•2nrdr =

,2 (1—

U

—

max

27r • r • dr R2

5

21-crdr

3 umax J

r — r2 R rR

rdr

dr

umax

1— ' R2

271-r x dr

5I6

Fundamentals of Fluid Mechanics

uniax [r2 2

r4 4 R2

-10

r2 )R 2 )0

umax 1 [2R2 — R2] •4 R2 2 — umax 2 Kinetic energy correction factor (a) is —

a-

Actual kinetic energy Kinetic energy calculated by average velocity

3 a = l s (uu ) dA A

= But

1 2 r u33 x 2nrdr IrR — u' 2 a=

1 fR u3 2 n-rdr zr R2 J0 u3 16 fR ( 1 Rr2 r • dr 2 R2 )3 3r 2 16 R 1 R 2 :10 [ R2

3ra R4

6 R6 r

r•dr

3r6 r8 R 16 [r2 3r4 R2 2 4R2 6R4 8R6 16 [0.5R2 — 0.75R2 + 0.5R2 — 0.125R2] R2 16 2 = — • R [0 125] R2

Laminar Flow

5I7

=2 The momentum correction factor 13 is: j3=

R 1 i u2 dj 1 A J -u2 — ir R2 j°

_

1 fR u2 4• 2irrdr ir R2 ° U2max

R82 foR = 1.33

2 2 r ) ( R2 1

U2

x2R-rdr -2 U

• rdr

Hence actual kinetic energy and momentum can be found out by finding these with average velocity and then multiplying the calculated values with correction factor of 2 and 1.33 respectively. 31. Oil of absolute viscosity 1.5 poise and density 848.3 kg/m3 flows through a 30 cm pipe. If the head loss in 3000 length pipe is 20 m, assuming a laminar flow determine (i) the velocity (2) Reynolds number and (iii) friction factor (Fanning's) (AMIE 2000) Pi — P2 —

32 dp L D2

32uyL hf — P1 — P2 _ D2 pL Pg —

20

20 x D2 pL 32p L

20 x (0.3)2 x 848.3 x 9.81 32 x 0.15 x 3000 = 1.04 m/s Reynolds number =

puD

848.3 x 1.04 x 0.3 0.15 = 1765 Friction factor is given as: f

16 Reynolds number 17 — 0.96 x 102 1765

5I8

Fundamentals of Fluid Mechanics

32. Crude oil of A = 1.5 poise and relative density = 0.9 flows through a 20 mm diameter vertical pipe. The pressure gauges fixed 20 m apart read 600 kN/m2 and 200 kN/m2 as shown in figure. Find direction and rate of flow through the pipe. (AMIE) 2 P2 = 200 kN/m

2 P1 = 600 kN/m

Guidance: The pressure drop includes the difference of height of the gauges. Hence add hpg at right side of the pressure drop equation. P1 —

P2 —

32 W p L D

2

± hPg

u (600 — 200) x 10'2 — 32 x x 0.15 x 20 + 20 x 900 x 9.81 (.02)2 [400 —1765] x 103 x 4 x 10-4 32 x 0.15 x 20 = 0.927 m/s

k—

Now Reynolds number is: _ Ke

pUD _ 900 x 0.927 x 0.02 p 0.15

= 111.72 Since Re < 2000, flow is laminar Now

flow rate Q = /4 x A 0.927 x K X (0.62)2 4 = 2.93 x 104 m3/s

33. A liquid of viscosity 0.07 Poise and RD of 0.86 flows through an inclined pipe of 20 m diameter. A discharge of 0.013 m3/min is to be maintained through the pipe in such a way that pressure along the length is constant. Find the required inclination of the pipe.

Laminar Flow

5I9

L sin 0

z1

.---le

_1

i

Guidance: The flow has to take from higher level to lower level so that the pressure drop head due to viscosity is equal to static head due to difference of levels for length L. In other words, hf = (z1 — z2) = L sin 0 head loss hf —

U=

32 u4L 2 D xpxg Q A 0.013

60 x Lr- x (0.02)2 4 = 0.69 m/s hf '

32 x 0.69 x 0.07 xL (0 .02) 2 x .86 x 103 x9.81

Since pressure is constant, hence h f = L sin 8 or or or

L sin 9 —

32 x 0.69 x 0.07 x L 4 x 10-4 x 0.86 x 103 x 9.81

sin 0 = 0.45 0 = 27.2

34. A tank has 4 cm diameter and 100 cm long pipe attached at its bottom. The tank has oil with sp. gr. = 0.9 and viscosity = 0.15 Ns/m2. Find flow through pipe when the height of oil in tank is 0.6 m above the pipe.

520

Fundamentals of Fluid Mechanics

0.6 m

1m Dia = 4 cm

® Applying Bernoulli's equation between section 1 and 2 2 2 Pi + vi + Zi = P2 + V2 + Z2 + hL pg 2g pg 2g

and

P — P2 = 0, Vi "=" 0, Zi — Z2 = 1.6 m Pg Pg V2 = 4 = average velocity at outlet of pipe. 2 U — = 1.6 — hi, 2g

But head loss is hL—

32 dp •

L D • pg 2

32 • u x 0.15.1 (0.04)2 x 0.9 x 103 x 9.81 = 0.34 4 Putting the value of hL in eqn. (i) 2 2g

Or

= 1.6 — 0.34 W

U— 2 + 6.67 Ft — 31.4 = 0 6.67 ± V44.49 + 125.6 2

_ u =

—6.67 ± 13.04 2 = 3.18 m/s

u —

Now Reynolds number is Re —

pD it

Laminar Flow

52 I

900 x 3.18 x 0.04 0.15 = 763.2 Since Re < 2000, the flow is laminar Q=UxA

Now

3.18 x 7t- x (0.04)2 4 = 3.99 x 103 m3/s 35. A laminar flow is taking place between two parallel and stationary plates 100 mm apart. Maximum velocity of the flow is 1.5 m/s find (1) flow per meter width (2) shear stress at the plates (3) the pressure drop for flow of 20 m (4) velocity gradient between the plates at a section (5) local velocity at 20 mm from the plate. Take viscosity 2.45 Ns/m2

T

/////////////////////////////////

t=loomm

1 Guidance:

The local velocity of the flow is given by u = 1 2µ

velocity 14 — 1

/////////////////////////////////

1 12,u (

a axP

—

(—axal (ty — y2), average

t2 and umax = 1.5 a . Pressure drop - pi P2 —

(a/ (t — 2y) ax

umax = 1.5 m/s u —

13 — 1 m/s 13

Now

Q= FA •il = 1 x (0.1 x 1) = 0.1 m3/s

Now

U= —ap ax

1 rap) 12 21.1

ax

lx12xµ (0.1)2

12u4L t2

and T

522

Fundamentals of Fluid Mechanics

12x 2.45 — 2940 N/m3 1 x 10-2 Now to find (pi — p2) between two points at 20 m apart, we know Pi — P2

12 x lx 2.45x 20 12u11L 2 t (0.1) 2 = 588 kN/m2

Shear stress is z_ 1 r—aP) (t 2y) 2 The shear stress at plates is same which can be found out by putting y = 0 or y = t wall

_ 1 r-aP) t 2 ax)

1 — x 2900 x 0.1 2 = 147 N/m2 Now to find out velocity gradient between the plates, we use Newton's equation T

or

velocity gradient =

du dy du — r dy

du _ 147 — 60 s-1 2.45 dy To find velocity at y = 20 mm U20 =

1 21

r-ap) (1Y Y2)

I

ax

2 x 12.45 (2940) [0.1 x 0.02 — (0.02)2] = 0.9 m/s 36. Two parallel plates are 4 mm apart and a steady viscous flow of oil is taking place between them. If pressure drop is 12 KN/m2 per meter length of plates and µ = 5 x 10-2 Ns/m2 for oil, find (1) flow per meter width (2) maximum shear and (3) maximum velocity of flow.

Laminar Flow

523

Drop of pressure of the flow in length L of the plates is: 12kuL P1 — P2 —

t

2

Here L = 1 m

2 — 12xWx5x10-2 xl 12 x 10' (0.004) 2 .%

U

—

12 x 103 x 16 x 10-6 12 x5x10-2

= 3.2 x 10-1 = 0.32 m/s Discharge per meter width is:

Q= a

xA= a x (t x 1) = 0.32 x 0.004 x 1 = 1.28 x 10-3 m3/s

Now maximum shear stress is rmax — ( — aP )X ti ax [ —12 x 103) x 0.004 1 2 = 24 N/m2 Now maximum velocity is: umax = 1.5

a

= 1.5 x 0.32 = 0.480 m/s 37. A masonry wall of a water tank is 1 m thick. It has developed a crack of 0.2 mm and 2 m wide and this crack extends to entire thickness of the wall. The water height is 4 m above the crack. Find leakage volume per day if g = 0.001 NS/m2.

524

Fundamentals of Fluid Mechanics

Leakage Through Tank

Guidance: The problem is nothing but the flow through two parallel stationary plates which is formed by the crack. Here t = 0.2 mm, L = 1 m and width = 2 m, hf = 4 —

pl — P2 _ 12/4p.L t 2 • pg Pg

hf = 4 —

1214,uL 2 t pg

4 x (0.2 x 10-3)2 x lx 103 x9.81 12 x (lx 10-3) x 1 4x4x108 x103 x 9.81 12 x10-3 = 13.08 x 10-2 m/s Q = flow = Ar4 = 0.1308 x (0.3 x 10-3 x 2) = 0.078 x 10-3 m3/s Leakage per day = Q x 3600 x 24 = 0.078 x 10-3 x 3600 x 24 = 6.7392 m3 38. Two horizontal plates 12 m apart contain oil with viscosity of 0.08 Ns/m2. If the upper plate is moved with 1.2 m/s and the pressure difference of two sections 100 m apart is 80 liN/m2. Find (i) velocity distribution (ii) discharge per unit width (iii) shear stress on the upper plate. Guidance: V

u=—

9/ +

The velocity distribution of the fluid in case one plate is moving is given as

(-al (ty — y2). Discharge can be found by integrating as Q = Jo udy,

shear stress by differentiating velocity and applying To = p au ay

Laminar Flow

525

Moving plate with v = 1.2 m/s

L=12mm

///////////////////////////////////11///// Stationary L- 100 m

Here

dx = 100 m dP = 80 x 103 x N/m2

ax

_ 800 x 103 — 800 N/m3 100

(1) Find velocity distribution: V u= — t

1

a/9) ax )(IY — .Y2)

1.2 2 x 10.08 (800) (0.012y — y2) 0.012 Y + = 100y + 5000(0.012y —y2) = 100y + 60y + 5000y2 = 160y — 5000y2 (2) Discharge per unit width of the plates Q = S u • dy o From equation (i) Q = fo (160y — 5000y2)dy

=

0.012

jo

(160y — 5000y2)dy 0.012

2

=[ 160 Y 5000 Y3 2 3 = 80 x (0.012)2 —

o

5000

= 0.01152 — 0.0029 = 8.62 x 10-3 m3/s

(0.012)3

(i)

526

Fundamentals of Fluid Mechanics

(3) Shear stress at upper plate T=

kl

au ay

a

= ki — (160y — 5000y2)

ay

Using equation (i) of velocity 1- = g(160 — 10,000y) Upper plate means y = 0.012 m Tupper = 0.08 (160 — 104 X 12 X 10-3) = 0.08 (160 — 120) = 0.08 x 40 = 320 N/m2 39. Laminar flow of a fluid (g = 0.8 N/m2 and sp. gr. = 1.2) is maintained in parallel plates of extensive width. The plates are inclined at 45° to the horizontal and are kept at 10 mm apart. Upper plate is moved with 1.5 m/s in direction opposite to the flow. Two pressure gauges fitted on upper plate at 1 m vertically apart reads pressure as 250 101/m2 and 80 IEN/m2 respectively. Find (1) the velocity and shear stress distribution between the plates (2) maximum flow velocity and (3) shear stress on the upper plate. Moving plate C), 250 kN/m2

Length of two section = L L —1 sin 45 or

L = -\12 m

Applying Bernoulli's equation between section 1 and 2 2

P1 ± V1 ± Z1 — P2 ± 1722 ± Z2 ± iii, pg 2g pg 2g Since pipe has constant diameter, hence V1 = V2

Laminar Flow

HL =

P1 —P2

± Zi — Z2

pg (250 — 80) x 103 +1 1.2 x 103 x 9.81 = 14.44 + 1 = 15.44 m Now

—aP ax

1-1L xpxg

—aP ax

15.44 x1.2 x103 x 9.81

as dx = L

Vi

Vi = 128.5 x 103 N/m3

The velocity distribution is:

u =V- y + t

1 2µ

(-aP)

ax ) (tY — y2)

—1.5 1 (128.5 x 103)(0.01 y — y2) 0.01 y + 2x0.8 = —150y + 80.3 x 103(0.01y —y2) The condition for maximum velocity of the flow is when au = 0, hence ay u = —150 + 80.3 x 103(0.01 — 2y) = 0 aY or

0.01 — 2y —

= 0.00186 2y = 0.01 — 0.00186 = 8.14 x 10-3 y = 4.07 x 10-3 m

or or Put y = 4.07 x

150— 1.86 x 10-3 80.3 x 103

le in velocity equation to get maximum velocity.

umax = —150 X 4.07 X 10-3 + 80.3 X 103[0.01 X 4.07 X 10-3 — (4.07 X 10-3)2] = —0.610 + 3.268 — 1.330 = 1.327 m/s Shear stress distribution is: T

= ,i V ,-- t

+ r — ap) ax

(2

,,) f

527

528

Fundamentals of Fluid Mechanics

= 0.8 x 15 + (128.5 x 103) (o.oi _ ,,) 0.01 ") = —120 + 642.5 — 128.5 x 103 x y = 522.5 — 128.5 x 103 x y Shear stress at upper plate can be found out by putting y = 0.01 Tupper = 522.5 — 128.5 X 103 X 0.01 = 522.5 — 128.5 = —762.2 N/m2 40. Derive the expression for Darcy's equation for laminar flow through porous material. Darcy proved by experiments that the velocity of a fluid through a porous material varies linearly with the loss of head (h1) and the flow is therefore laminar/viscous. Since the loss --

of head through a pipe is given by hf=

P and through parallel plates by hf=

A hence a general expression for laminar flow is:

P P

where k' = Constant P Now porous material has a number of small pores say n of diameter 'clp' hence we can write: hf =

D = n • dp Hence we say that the head loss is — hf — P u— — P

or

2 ,2

(7)

P

h, J = hydraulic gradient = i

Now

.

a = ki

k is called coefficient of permeability. The above equation is called Darcy's equation for flow of water through soil. The equation is applicable if Re < 1 41. Water at a rate of 8 x 10-7 m3/s is flowing through a soil 10 cm height and 50 cm2 crosssection area under a constant head of 10 cm. Find the coefficient of permeability. Average velocity of flow of water through the soil is: 14 — Q A

8 x 10-7 50 x (10-2)2

Laminar Flow

529

8 x 10-7 = 1.6 x 10-4 m/s 50 x 10-4 As per Darcy's equation u = ki = k( hf 4

1.6 x 10or

kx

10 X 10-2 10 X 10-2

k = 1.6 x 10-4 m/s

42. What is fluidization? Where is it used? When a liquid is forced upwards at increasing velocity through a column of granular material of size d, a state is reached when all particles are entrained or carried in the flow i.e. the bed is in fluid state. This state is called fluidization. The corresponding velocity is called fluidization velocity which is given by: V1=

(pg — p) g d 2 c2 where c = porosity of the material, pg & p are densities 18y of granules and water.

The fluidization is used in the back washing of gravity filter for water purification. By back washing impurities in the filter are washed out/removed. 43. What is a journal bearing? Find the viscous resistance and power to overcome the resistance. Bearing

Shaft

i

Thickness of oil film

L

Lubricating oil

Journal bearing.

In journal bearing, a very thin film of lubrication oil is maintained between stationary surface of the bearing and the rotating shaft. The shaft has diameter D and speed = N 27EN angular velocity = w — — rad/s 60

530

Fundamentals of Fluid Mechanics

The tangential velocity of oil at the shaft is: D

u = wR — w

2

270/ D u = — x —

TCDN

60 2 60 The velocity gradient between stationary surface of bearing and moving surface of the shaft is: du = u — 0 dy t

where

t = thickness of oil film

Now applying Newton's law of viscosity Shear stress z = p, du dy u

= 11' t

TcDN

— P 60• t ... Shear force or viscous resistance F = r x Surface area —p

- 11

;ON

x FDL

60.t

2 2 7r D N • L

60. t

Torque required to overcome viscous resistance is: D H 7r 2 D2 NI, — ' x D Torque = T= F x 2 60. t 2 _ ,u7r 2D3 N•L 120• t

Power absorbed by bearing to overcome the viscous resistance is: P = T •w —

Where

T—

2KNT

60

2 3 AUT D N • L

120• t

44. A shaft of 100 mm diameter rotates at 100 rpm in a 100 mm long bearing. The shaft and bearing are separated by a distance of 1.0 mm of oil with µ = 0.009 Ns/m2. Find the power absorbed by the bearing.

Laminar Flow

T—

53 I

,u7r2 D3 NL 120•t 0.004 x 7r2 x (0.1)3 x 100 x 0.1 120 x 1 x 10-3 4 x10-3 x7r2 x1x10-3 x10 120 x 10-3

= 0.328 x 10-2 Nm 27r x 100 x 0.328 x 10-2 Power absorbed — 27rNT 60 60 = 3.43 x 10-2 watts = 0.0343 watts 45. Derive an expression for power absorbed in foot stop bearing.

Foot stop bearing

A foot stop bearing is to provide support to a rotating vertical shaft. The space between the surface of the shaft and the bearing is filled with oil. The rpm of the shaft is N, then angular velocity w =

27r/s/ 60

Tangential velocity u = w • 27rN D 60 2

532

Fundamentals of Fluid Mechanics

irDN

60 du dy

Shear stress

T

u— 0 _ u

du =u—

dy

u

_

`-` t Shear force on the elementary ring of radius r and thickness dr dF = ,u• = l f x 27rr • dr t •

7rDN

x 27-cr • dr

60t

it • 27r rN •27rr • dr 60. t — ,u 7r 2 Nr 2 dr 15

t

dT = dF r —

15 x t

T = 51?ci 15itt t -

—

-

15 t

15 t

60 t

Nr3 dr

Nr3 dr

7r2 N

R4

4

7r2N R4

7r2 NR4

Power absorbed by bearing P = Tw —

where

T=

60• t

27cNT

60 x 7tz N• R4

46. Find power required to rotate a vertical shaft of diameter 100 mm at 750 rpm. The lower end of the shaft rests in a foot stop bearing. The end of the shaft and surface

Laminar Flow

533

of the bearing are both flat and are separated by an oil film of thickness 0.5 mm. Oil has viscosity = 1.5 poise. (Punjab University) Torque (T) =

60t 712NR4

0.15 60 x5 x10-4 = 0.23 Nm Power —

4 X 71-2

x 750

2

2.7rNT 60

27r x 750 x 0.23 60 = 18.15 W 47. Derive an expression for power consumed in collar bearing. Bearing Collar Shaft

Collar Bearing The collar bearing supports the axial thrust of the rotating shaft by separating the collar of the shaft from the surface of it the bearing with a film of oil of uniform thickness which is maintained by a forced lubricating system. If N is rpm of the shaft, then angular velocity = w —

27rN 60

tangential velocity = u = wr = du dy dF = T x dA =

2irN 60 r u t •27rr • dr

534

Fundamentals of Fluid Mechanics

= ktx

27rN x r 27rrdr 60•t

= 15.t I' le Nr2 dr dT = dF x r 15 t ir ...

2 Nr3 dr

T = 5 R2 P 2N r3dr R1 15 t 7r

R2 & R1 are radius of outer & inner surface of the collar. 4 )R2 T=

T= Power consumed —

P eN (rA

-1.

15t

Ri

P -2N (R 42 — R 41) 60 t 71 27rN •T W 60

48. A collar bearing having internal and internal diameter of 200 mm and 400 mm, oil film thickness = 0.2 mm and g = 0.8 poise, is taking the thrust of the shaft and overcoming viscous resistance when shaft rotates with 300 rpm. Find power consumed by the collar bearing. Torque required to overcome viscous resistance T = il 2N(R 42 — R 41) 60 t r 0.08 x 7r2 x 300(0.24 — 0.14) 60 x 0.2 x 10-3 8 x 10 12 x 1:3

x e x 300(16 — 1) x 10-4

= 29.58 x 10-3 x 103 = 29.58 NM Power consumed is 2958 x 2nN 60 = 928.8 W

P—

29.58 x 27r x 300 60

Laminar Flow

535

49. What is a dashpot? Explain its working and derive the equation of load.

/-

Piston

Oil

/2/////,///////// / Cylinder Dashpot Mechanism

Dashpot is a hydraulic device for damping vibrations of a machine. It consists of a cylinder filled up with viscous fluid and a piston which inside the cylinder, displacing the liquid in opposite direction to its movement. The piston is connected to the moving load w upwards and downwards. The principle of the dashpot is based on the drag force exerted by the fluid when a moveable plate is moved parallel to a stationary and parallel plates. If W is load on the piston, then difference of pressure between two ends of the piston is dp =

W 4W 2 = 7 t- i

IC

4 The flow of oil between the gap of the piston and cylinder surface is similar to the viscous flow between parallel plates and pressure drop for length L is: dp —

or

/I

t2

4W _ Jr

11 t2

u =

Wt 2

where t = thickness of gap and k = average velocity

it µ

Now suppose piston moves with velocity = vp, then flow Q is 70 2 Q = vp x 4 Also with average velocity t7 , the discharge Q through the gap is Q = 14 •ir.d.t 7rd 2 .

V X P

= U • r• cbt

536

Fundamentals of Fluid Mechanics

•cl 4t

— VP

pd

Wt2

4t 4 Wt 3

Or

50. A oil dashpot is used for damping vibrations. Piston falls 50 mm in 50 seconds in the oil of the cylinder If additional load of 1.2 N is applied, the piston falls through 50 mm in 40 seconds. The diameter of piston is 50 mm and its length is 100 mm. The gap thickness is 1 mm. Find viscosity of the oil. W _ W +1.2 Vi V V—

V' =

1x 10-3

50 x 10-3 — 1 x 10-3 m/s 50 50 x 10-3 — 1.25 x 10-3 m/s 40 W +1.2 1.25 x 10-3

1.25 W= W+ 1.2 Or Now

W _ 1.2 _ 4.8 N 0.25 p = 4Wt 3

4 x 4.8 x (1 x 10-3)3 Tt

= 0.016 Ns/m2 51. What is Stokes' law? What are the conditions and assumptions made for Stokes' law? (UPTU 2009-10) As per the Stokes' law when a small sphere moves through a viscous fluid with constants velocity (u), a drag force experienced by the sphere in a direction opposite to the motion which is given by:

Laminar Flow

537

,u The condition for applying Stokes two is that the Reynolds number of the fluid is less than one. The assumptions made in Stokes' law are: (1) (2) (3) (4)

Inertia force acting on body is small as compared to viscous force Walls of vessel containing the fluid do not affect the flow of fluid around the sphere Fluid does not slip over the sphere. The sphere is rigid

52. What are the various methods of measuring viscosity? Following are various methods for measurement of viscosity: (1) (2) (3) (4)

Falling sphere method Capillary tube method Rotating cylinder method Orifice type viscometer

The devices used for the measurement of viscosity are called viscometers. 53. Describe falling sphere method of measuring viscosity.

)arent with id

Fd = drag force BE

for coi tempe

I

Sphere ht of B

Falling sphere viscometer

- ...uoyaucy force

Free Body Diagram of sphere

Falling sphere method is based on Stokes' law. A vertical transparent tube is used which is filled with liquid, whose viscosity is to be ascertained. A spherical ball is released into the liquid. The ball attains a constant velocity at a measurable interval of time. During the state of constant velocity, the forces acting downwards must be equal to the forces acting

538

Fundamentals of Fluid Mechanics

upwards. If las = density of sphere and pi, = density of liquid, then we have: W = Fd FB where Fd = drag force, FB = bouyancy force and W' = Weight TC

6

X d3 X ps x g = 37(pd•u + •d3 •13L x g

3 TC.1 •u•d = 6d3 x g(ps — pL)

P—

gd 2 (Ps — PL) 18 u

54. Describe capillary tube method for measuring viscosity. Piezometer Liquid under test

Capillary tube of diameter V'

Measurement tank

///

Capillary Tube Method

Capillary tube method is based on the principle of head loss of the liquid while flowing through capillary tube of diameter d and length L. The head loss depends upon viscosity of the liquid which can be determined by measuring flow (Q). A capillary tube is attached to the tank as shown in the figure and discharge through the capillary is collected in the measuring tank for a given time. The flow Q is found out by measuring volume of liquid collected in measuring tube divided by time. Piezometer gives the loss of head of the liquid. Q

_

Volume of liquid in measuring tank time

—

Q — average velocity area of capillary Q _ 4Q Tc

7.cd 2,

4 The head loss is given by: ••••••••

h=

P p

Laminar Flow

Or

539

zpg hd 4 128. Q. L

µ

Viscosity can be determined from the above equation. 55. Describe rotating cylinder method for measuring viscosity. Dial to measure deflection of torsional spring

//// /////

Torsional spring Inner stationary cylinder of radius R1

Outer rotating cylinder of radius R2

Gap T of liquid

///

Rotating Cylinder Method

Rotating cylindrical method is based on Newton's law of viscosity. A liquid whose viscosity is to be determined is filled in the gap between inner stationary cylinder and outer rotatable cylinder suspended by a torsional spring. Now outer cylinder is rotated with constant speed and deflection of torsion spring is noted. If outer cylinder rotates at N rpm, then angular velocity is 27rN 60 tangential velocity u = w x R2 w

27rN R2 60 Velocity gradient in the gap is du dy

2icAT R2 60.t du — µ dy

where t = thickness of gap 2.7rN R2 µ 60•t

540

Fundamentals of Fluid Mechanics

Viscous force or drag force = r x surface area 2KR2 . N x 27-1-R1• h where h = height of liquid 60•t

Fd =

Torque (T) = Fd x radius = Fd x R1 px 2

or

T

• R2 • h • N 15•t

1 5. t n 2Ri2 R2 • h N

56. A small air bubble of diameter 1 mm rises with steady state velocity of 0.02 m/s through an oil. Find viscosity of the oil, if sp gr of the oil is 0.9. If we neglect the weight of bubble, the bubble has two forces in equilibrium when velocity is steady. Drag force = bouyant force 37rµu• d — 70 3 p x g 6 or

P—

pg • d 2 18 u 9 x 103 x 9.81x (1 x 10-3)2 18 x 0.02

or Now

p = 24.5 x 10-3 Ns/m2 R _ pud e 0.9 x 103 x 0.02 x 1 x 10-3 245 x 10-3 = 7.346 x = 0.7346

Since Re < 1 The Stokes' law is valid. 57. A small sphere is used for measuring the viscosity of an oil. If sphere (sp. gr. = 0.8) has 1 cm diameter and falls with constant velocity of 0.06 m/s through the oil (sp. gr. = 0.9), find the viscosity of the oil. gd 2 — 18 u (Ps — PL)

Laminar Flow

54 I

9.81x (1 x 103)2 (0.8 x 103 — 0.9 x 103) 18 x 0.06 Or

Now

kt = 6.45 Ns/m2 Re

pud

—

p

0.9 x 103 x 0.06 x 1 x 10-3 6.45 = 0.0084 Re < 1, hence Stokes' law is applicable. 58. In a rotating cylinder viscometer, the radius of outer and inner cylinders are 32 and 30 mm. The rpm of outer cylinder is 300. The height of liquid is 100 mm. The torsional spring dial indicates torque as 1.2 x 10-4 Nm. Find viscosity. p. — T =

15•t 2 rr x Ri x R2 x h x N 2

1.2 x 10-4 x 15 x 2 x 10-3 R-2 x (0.03)2 x (0.032) x 100 x 300

= 4.22 x 10-6 Ns/m2 59. A oil of sp. gr. = 0.8 is to be measured by capillary tube method. The tube has diameter of 40 mm and pressure drop of 0.4 m of water between two points 1 m apart. The weight of water collected in measuring tank is 494.5 N in 100 seconds. Find viscosity of the oil. Discharge = Q =

Weight of oil collected time xp x g

494.5 100 x 6.8 x 103 x 9.81 = 0.625 x 10-3 m3/s Now viscosity is given by Hp gh d 4 µ— 128.Q.L,

g x 0.8 x 103 x 9.81x 0.4 x (0.04)2 128x 0.625x10-3x1

542

Fundamentals of Fluid Mechanics

= 1.97 x Ns/m2 = 0.197 Ns/m2 60. What is an orifice type viscometer? In this method, the time taken by a certain quantity of liquid whose viscosity is to be determined, is made to flow through a short capillary tube. The coefficient of viscosity is then calculated by comparing this time with the time of other liquid whose viscosity is known

Liquid

Bath at constant temp. Capillary tube Measuring cylinder Saybolt viscometer

Saybolt viscometer is based on orifice type viscometer. The time taken by 60 cc. of liquid to flow through capillary tube is measured. From the time measurement, the kinematic viscosity is found out from the relation v = At —

B t

where

t = time and A and B are constants

61. Derive the expression for the velocity of laminar flow through an open channel. In open channel, the upper surface of the liquid is exposed to atmosphere. Hence, the pressure at every section of channel is same and no pressure gradient

DP

)can be

maintained for flow to take place. Therefore, the bed of the channel is made sloping (

ax

)

so that the head loss due to flow on account of viscous flow can be given to flow by the gravity force by sloping the bed.

Flow through channel

The pressure gradient in the direction of flow is

Laminar Flow

ap = ar ay ax r— p

and

ap

=

,,.7C

au a2u

P ay

As water is flowing due to the slope of bed i.e.

ap

ax

az

ax = P8Tx

a2 az ay2 Pg ax On integration, we get au

_

ay

Pg ( az), p ax

± -1

au = 0 at surface of water in channel where y = D = Depth of channel But ay Cl —

— 13g az) D p ax

au = Pg raz) (D — y) ay

p

ax

integrating again, we get

u=

µ —aaz x )( uY

2 y + z ) C2

Now u = 0, if y = 0 and hence c2 = 0

u= Pg 1,az )( Dy — dX )\

2)

The above is the velocity distribution equation of the flow through open channel. 62. Derive an expression for the discharge through open channel. The discharge per unit width of the channel is

dQ = velocity x area = u x (dy x 1)

543

544

Fundamentals of Fluid Mechanics

Q

rD =

udY

a

g ( — axz)(DY

2)

_ pg —az)r Dy2 y3 1D 6_10 ax )L 2 pg raziD3 ax 3 63. Derive an expression for average velocity of flow in open channel Q = t7 • D • 1 Here width is unit length 1,7

_ Q _ 1 x Ps ( az) D2 D D ax) 3

_ Pg ( az) D2 p ax) 3 64. Derive an expression for head loss in open channel Head loss in open channel = z2 — zi

—az ax where L= head drop between two sections

az L HL = — ax But and

u _ Pg ( — az) D2 p ax 3

—az = 3t7p ax

HL =

pgD2 311,u• L 2 pgD

Chapter

12

TURBULENT FLOW

A A A A A A A A A A A A A

CRITICAL VELOCITY INSTANTANEOUS VELOCITY MAGNITUDE OF TURBULENCE INTENSITY OF TURBULENCE SCALE OF TURBULENCE EDDY VISCOSITY TURBULENCE SHEAR STRESS MIXING LENGTH CONCEPT SIMILARITY CONCEPT VELOCITY DISTRIBUTION IN PIPE MAXIMUM VELOCITY OF FLOW AVERAGE VELOCITY OF FLOW SHEAR VELOCITY

A A A A A A A A A A A A A

SMOOTH BOUNDARY ROUGH BOUNDARY ROUGHNESS REYNOLDS NUMBER FRICTION FACTOR DARCY'S EQUATION HEAD LOSS RELATIVE ROUGHNESS EQUIVALENT SAND GRAIN ROUGHNESS POWER LAW LAMINAR SUBLAYER MOODY'S DIAGRAM HOTWIRE ANEMOMETER LASER DOPPLER ANEMOMETER

INTRODUCTION Laminar flow occurs at low velocity when fluid flows in a pipe. The laminar flow takes place in parallel layers or laminae which are concentric cylindrical sheets in pipe flow. If flow has Reynolds number greater than 4000, then flow in pipe is no longer in parallel layers and it is said to be a turbulent flow. The Reynolds number is the ratio of inertial force to viscous force. The viscous force tends to make the motion of fluid in parallel layers while inertial force tends to diffuse the fluid particles. The fluid particles are in the extreme state of disorder in turbulent flow and they do not move in layers. They move in haphazard manner and produce large-scale eddies, resulting complete mixing of the fluid. The pressure and velocity of the flow do not remain constant with time and they fluctuate in irregular way. Due to fluctuations in velocity, it is impossible to develop any theory for the analysis of turbulent flow. The analysis has to be

546

Fundamentals of Fluid Mechanics

therefore based on formulae derived on the basis of experimental results. Due to haphazard movement of fluid particles in turbulent flow, the velocity distribution is more uniform in turbulent flow as compared to laminar flow. However, the velocity gradient near the pipe wall is very large in turbulent flow, resulting high shear stress at the wall of the pipe. The velocity distribution is paraboloid in laminar flow while it obeys power law and logarithmic law in turbulent flow. 1. What is turbulent flow? Compare velocity distribution in laminar and turbulent flow in a pipe. A fluid flow as in pipes having Reynolds number greater than 4000 is said to be a turbulent flow. The fluid particles are in the extreme state of disorder in turbulent flow and they do not flow in layers. They move in haphazard manner and produce large scale eddies, resulting complete mixing of the fluid. The pressure and velocity of the flow do not remain constant with time but they fluctuate in irregular way. Due to fluctuations in velocity, it is impossible to develop any theory for the analysis of turbulent flow. The analysis is therefore completely based on formulae derived on the basis of experimental results. The velocity distribution in laminar and turbulent flows are as shown in the figure. The velocity distribution is more uniform in turbulent flow as compared to a laminar flow. However, the velocity gradient in turbulent flow near the pipe wall is very large, resulting high shear stress at the wall of the pipe. The flatness of velocity distribution curve (slight variation in velocity) away from the wall has resulted due to the mixing of fluid layers and exchange of momentum between the layers at macroscopic level. In laminar flow, the ratio of the average velocity to the maximum is 0.5 while in turbulent flow it is greater than 0.5 and it may increase upto 0.85. The velocity distribution is paraboloid in laminar flow while it obeys power law and logarithmic law in turbulent flow. Turbulent

Flow

Laminar

Velocity distribution in laminar and turbulent flow

2. What is dominant in laminar flow? The viscosity of the fluid is dominant in laminar flow due to which it is able to suppress any tendency to turbulent conditions. 3. What is critical velocity? The critical velocity is the velocity below which all turbulence is damped out by the viscosity of the fluid. It is found that the possible upper limit of laminar flow is represented by a Reynolds number of about 2000.

Turbulent Flow

547

4. What are factors that can change a laminar flow into a turbulent flow? A laminar flow changes to turbulent flow when the following changes are made: (1) velocity is increased (2) diameter of a pipe is increased (3) the viscosity of fluid is decreased Reynolds was first to demonstrate that the transition from laminar flow to turbulent flow depends not only on the mean velocity of the flow but also on diameter, density and viscosity pVD

. The expression

pVD

is a dimensionless quantity and this number 11 ti is called Reynolds number (Re). In case of circular pipe, the laminar flow exists upto Re < 2000 and turbulent flow exists when Re > 4000. The transition phase exists when Re is in between 2000 to 4000.

by the equation

5. Describe the mechanism of a transition from laminar to turbulent? Laminar flow becomes unstable and turbulent as velocity of flow increases. Turbulence is nothing but disorderly and irregular fluctuating motion. In laminar flow, the shear force due to viscosity is able to keep the flow in an orderly manner in layers with each layer sliding over other adjacent layer. As the velocities in laminar flows are comparatively small which account for small inertia forces and dominating viscous forces. At higher velocities, the inertia forces increase in magnitude and they start dominating over the viscous forces and a state reaches when orderly motion breaks down and turbulence starts occurring. It is also noticed that a boundary layer is laminar as a flow starts at the leading edge of a surface and it starts growing. The boundary layer tends to be unstable as it grows and passes through a region of transition before becoming turbulent. It is also noticed that a turbulent boundary layer has thin zone in the vicinity of the walls which is called laminar sublayer in which flow is laminar. Outside this sublayer, there are zones of transition and turbulence. For example, the smoke emanating from a cigarette is initially laminar and ultimately it is turbulent after passing through transition state. This shows that velocity of smoke is small initially and the velocity of smoke increases as it rises. The velocity of smoke on rise becomes critical and later velocity is so large that turbulence sets in the smoke as shown in the figure.

Turbulent flow

i Transition flow

Laminar flow

1 Rising of smoke

548

Fundamentals of Fluid Mechanics

6. Describe the mechanism of formation of vortices. If two streams with unequal velocities join each other, this phenomenon may be represented by considering a surface of discontinuity with streams of velocity V in opposite directions on either side of it. If a slight disturbance is made to the surface of discontinuity, the streamlines above and below will converge at some points and diverge at other points. The velocities at converging points are greater (peak) and diverging points are lower (valley). Peaks and valleys disturb the surface further, resulting in the formation of vortices Stream line V S

D

V V

Surface of discontinuity

Stream line Stream lines and surface of discontinuity _---4.-

S

-

-

D

-1.

Slight disturbance of surface of discontinuity

--- ---,_ __---- -----,.. S

/.-------"\,______..,'"----------

D

Higher disturbance of surface of discontinuity

Vortices formation

Transition of laminar flow to turbulent flow starts with appearance of spots of turbulence in the laminar flow. These spots encourage the formation of more spots of turbulence in the flow which change the structure of flow from laminar to turbulence. 7. What is the result of turbulence? Is it desirable to promote turbulence in the flow? Turbulence results in significant increase of mass, energy and momentum transfer between different parts of fluid flows. It is desirable to promote mixing of different parts of fluid flow so that heat transfer coefficient can be improved. In engineering applications, condensers and evaporators are designed to remove heat in an efficient way by the fluid flow with least possible sizes. Turbulence flows must take place through water tubes in boiler for efficient generation of stream. The turbulence increases mixing and reduces drag. In nature,

Turbulent Flow

549

turbulence maintains uniformity of temperature. Turbulence balances pressure differences between different locations on earth. Turbulence enables rivers to transport waste from the plains to the sea. Turbulence displaces our exhaled deoxygenated air quickly with fresh oxygenated air so that we can inhale and survive. 8. Compare the energy dissipation of laminar and turbulent flow. Turbulence is accompanied by large energy dissipation as compared to laminar flow. The turbulence motion is characterized by eddies of different sizes and wave numbers. Larger eddies have smaller wave numbers. Energy dissipation proceeds when kinetic energy of the flow is passed on from higher eddies area to area having smaller eddies and then finally from smaller eddies to area having laminar flow in the laminar sublayer near the walls. 9. Explain (1) instantaneous velocity (2) magnitude of turbulence (3) intensity of turbulence and (4) scale of turbulence.

I

Time (t)

Variation of Velocity at a Point

The turbulent flow has irregular and random nature of the movement of fluid particles. At any point of observation in a turbulent flow, the velocity and pressure fluctuate with time about a mean value. The typical variation of the velocity component in x-direction at a point in turbulent flow is shown in the figure. The instantaneous velocity (ut), mean velocity (U) and fluctuating component (u') can be related as— ut = /7 ± u' Similarly vt = 17 ± v' in y-direction wt = 1T, , ± w' in z-direction and

pt = P ± P' for pressure

As per the definition of average velocities we can write for a large interval of time (7): u' = 1I ut dt To v' = —1 iT vtdt To

550

Fundamentals of Fluid Mechanics

W=

1 rT J wtdt o

p' = T1 f T pt dt o In case we work out the averages of fluctuating components, we get: =

=

ST T

=

=0

T Jo

T

Jo

dt =0

w'dt= 0

p' = S p'dt = 0 T oT Since the mean value of fluctuating component is zero, we have to find the root mean square value of the fluctuating components, i.e. ur2 , vr2 , .1w'2 and 11 13'2 so that positive and negative fluctuating components on squaring can add up instead of cancelling and given mean value as zero. The magnitude of turbulence is the arithmetic mean of root mean square value of turbulent fluctuations in three directions Magnitude of turbulence = 1'11 (u'2 3

v'2

2 + 111' )

In case V is time average resultant velocity at any point, then we define: 11

Intensity of turbulence —

3

(u' 2 + v'2 +W2 ) V

In case the flow is one dimensional, the V = u and we can write: 1 2 2

Intensity of turbulence —

3

(u' + v' + w'2)

The average size of the eddy is also necessary in addition to the intensity of turbulence to describe the turbulence fully. The average size of the eddy can be obtained from the curve of velocity variation with time. From the curve, the average time of interval at which the curve crosses the mean value line (17) is found out and this is multiplied with the root square value of the fluctuating flow component. The size of eddy is called the scale of turbulence. Energy dissipation in flow is greater when eddy is small i.e. the turbulence is in fine scale. However, in fine scale turbulence, the intensity of turbulence is high.

Turbulent Flow

55 I

10. How is shear stress in turbulent flow expressed. Scientist J. Bousi developed expression for the turbulent shear stress. This expression is similar to Newton's law of viscosity and expressed as: du z= dy du is the velocity gradient from the dy boundary and q is turbulent coefficient of mixing and it is called eddy viscosity.

where

is the average velocity at a distance y i.e.

11. What do you understand from the eddy viscosity? What is total shear stress in a turbulent flow? Eddy viscosity is a property which is independent of the fluid properties except of density and it depends upon the characteristics of flow specially the turbulence of the flow. The magnitude of eddy viscosity is zero for laminar flow and it can increase to several thousand times of the coefficient of viscosity (µ) depending upon the turbulence in the flow. Hence eddy viscosity is not constant for any fluid as turbulence varies from point to point in the flow of the fluid. The total shear stress in a flow is the sum of viscous shear stress and turbulent shear stress. Total shear stress = viscous shear stress + turbulent shear stress

du

du

1-1 dy dy The magnitude of first term is very small to second term in a turbulent flow 12. What is kinematic eddy viscosity? The ratio of eddy viscosity (q) to mass density (p) of the fluid is called kinematic eddy viscosity. The kinematic eddy viscosity is: Eddy Viscosity _ Mass density 13. Why the concept of eddy viscosity has not been used much in engineering applications? It is not possible to express the variation of eddy viscosity (q) with other properties of turbulence. This is the reason why the concept of eddy viscosity has not found much favour in engineering applications. 14. What is the Reynolds expression for turbulence shear stress? It was shown by Reynolds that there is an exchange of transverse momentum due to velocity fluctuations between adjacent layers which develope a tangential shear force between the adjacent layers. He developed an expression for the turbulent shear stress between the adjacent layers which is expressed as:

552

Fundamentals of Fluid Mechanics = pu'v' where p = density and u' & v' are the fluctuating components of velocity in x and y directions due to turbulence.

I"

H (7H----u'. I du dy

Layer 1 (u + Le. ) -I. Layer 2

(u) Slope =

du

dy

x Momentum exchange between layers of fluid

Consider two layers i.e. layer 1 and layer 2 at distance 7' and u' and v' are the fluctuating components in x and y direction. Now mass of fluid transferred to layer 1 from layer 2 through a perpendicular area AA is pv' AA. This mass of fluid has momentum of pv'AA. 14 and on transportation to layer 1 from layer 2, it acquires as new momentum pv'AA ( 4 + u'). Hence change of momentum is pv' A/1u' which gives rise to shear stress 'I" which can be expressed as: V —

p• v'

• AA • u' — pu,v AA

The above is known as Reynolds shear stress equation for turbulent flow. The shear stress varies as the fluctuating components of velocity of turbulent flow. As usually when time average value of shear stress is considered, the equation can be expressed as: I' = p u'v' 15. Explain the concept of mixing length introduced by Prandtle and state the relationship that exists between the turbulent shear stress and the mixing length. (UPTU-2001-2) or Derive an expression for shear stress on the basis of Prandtl's mixing length theory. (UPTU-2006-7) The Prandtle mixing length theory states that the mixing length (I) is the distance between the two adjacent layers in transverse direction such that the lumps of fluid particles can move from one layer to another layer in such a way that the momentum of the particles in x-direction remains same. Prandtle assumed that velocity fluctuations in x and y directions are related to the mixing length as expressed below — diTi u' = I— dy Similarly

v' = 1

clW dy

where

I= mixing length

Turbulent Flow

= 12(

553

du )2 dy where u'v' is time average fluctuations.

As per Reynolds shear stress equation for turbulence flow = p u' v' Now use relation

we get —

u'v' = 12

(dk)2

dy )

,c = pp(d )2 dy

The mixing length (1) is analogus to the mean free path in the kinetic theory of gases. However there is a vast difference between the two as mean free path relates to the microscopic level while mixing length relates to macroscopic motion of fluid particles. 16. Find the expression of eddy viscosity in terms of mixing length. Why Prandtl's mixing length theory is useful in analysis of turbulent flow? For turbulent flow, shear stress is 2 (du )2 = p1 dy

(i)

As per Boussinesg's equation, shear stress in turbulent flow is = I/

(du) dy

Comparing equation (i) and (ii), we have pi2 (du ) ) dy The above equation indicates that mixing length (1) is a function ofy i.e. distance of the point under observation from the boundary or wall. Prandtle studied the turbulent flow in a circle and gave the expression for the mixing length as: 1 = ky where k is a universal constant. k is also known as karman constant and its value is 0.4. 17. How does the mixing length vary with distance from the wall in a turbulent flow? Show it by a sketch why Prandtl's mixing length is preferred? The mixing length varies with the distance from the wall in a turbulence flow as per expression: 1 = ky

554

Fundamentals of Fluid Mechanics

where k is some constant whose value has to be determined from the experiment data. For regions very close to the walls, the value of k = 0.4.

ro = radius of pipe

r0

Y r0 Variation of mixing length w.r.t. distance from wall

Prandtle's mixing length theory is preferred for the analysis of turbulent flow as this theory is simple to apply. 18. What is Von-karman similarity concept? What is the limitation of Prandtle's concept and karman's concept? Von karman extended the Prandtl momentum theory by establishing the dependence of mixing length in the turbulence flow on the distribution of average velocity. According to him, the mixing length is the ratio of the first derivative of mean velocity to the second derivative of mean velocity. du Mixing length — / — k dy d 2u dy e But shear stress

T=

pI 2 du dy ( uy d

= P k2

3' 2 (d 211) dy

e

The above relation relates shear stress to the mean velocity gradient in turbulent flow. Limitation: Neither Prandtl's concept nor karman's concept is valid in finding out shear stresses at the pipe centre line as at the pipe centre line, the transverse momentum transport is zero. Both the concepts are also not valid at the pipe walls as the sublayer at the walls has laminar flow. 19. A turbulent flow of water in a pipe of 0.8 m diameter has velocity profile as under

Turbulent Flow

555

1 u = 4 + — logo, 4 Shear stress at a point 0.1 m from the wall is measured as 1.2 N/m2. Find (1) turbulence viscosity q (2) mixing length and (3) turbulence constant k. 1 u = 4 + — log y 4 e du 1 dy 4y Also

d 2u dye

-1

4y2

Now apply Boussineg relation fro shear stress, we have du 1- =

71dy —

1.2 = ri x

1 4 x 0.1

q = 1.2 x 4 x 0.1 = 0.48 Ns/m2 Now apply Prandtle's mixing length concept, we get T=

p12 ( d;

)2

ci

1.2=1x103x/2 ( 1 4 x 0.1 Or

12 — 1.2 x (4 x 0.1)2 103 1.2 x 0.16 _ 0.192 103 103 12 = 0.192 x 10-3 1 = 0.0138 m

As per Prandtl's mixing length concept, we have L = ky or

k _ 1 _ 0.0138 y 0.1 = 0.138

)2

556

Fundamentals of Fluid Mechanics

Using Von karman similarity concept

p k2 T

(duy dy ) 2 ( d 2u

dy

)

e 4

- )2 1.2 = 1 x 103 x It2 (4v

[ Or

4y2

k2 = 1.2 x 16 _ 1.92 x 10-2 103 k = 0.138

20. Derive an expression for Prandtl's velocity distribution in turbulent flow in pipes. or What is velocity defect? Derive an expression for velocity defect in pipes. (UPTU-2004-5) Prandtl's assumed that mixing length near the wall depends upon the distance of the point from the wall. Hence we have: 1 = ky = 0.4

where k = karman constant

The shear stress in turbulence flow is T=

2 p12 ( d dLl )

y

2 du = P(kY) ( d y )2 or

2 (du) _ T dy A (kY)2

Or

du 1 I T dy ky 11 p

But

111 T- = shear velocity = u0 s,

Turbulent Flow

557

du uo dy ky Or

du =

or

f du =

Or

u=

uo

ky

dy

dY

k

y

logy + c where c = constant

Apply boundary condition that velocity is maximum at centre line of pipe, we have y = R, u = umax where R = radius of pipe uo

umax

= logeR + C k

C=

— logeR umax k

uo uo u = — log y + umax —— logeR k k u = umax + k (loge), — logeR) If we take

k = 0.4

Then

u = umax + 2.5 uo loge

R The above equation is called Prandtle equation for universal velocity distribution in turbulence flow. It is valid for both smooth and rough pipes. The above equation can also be written in logarithm base 10 as under u — umax = 2.5 log — uo 'R = 2.5 x 2.3 logio— R = 5.75 logio l R Since R > y — 5.75 logio — uo The term (umax — u) is called velocity defect. or

UMEIX

21. Distinguish between hydrodynamical smooth and rough boundaries.

558

Fundamentals of Fluid Mechanics Turbulent boundary layer

Turbulent boundary layer

a

Laminar sublayer

a

a—

z-

.

5'

a

Eddies

Eddies

A A / A

Laminar sublayer

Smooth boundary (5'> k)

Rough boundary (5'< k)

Hydrodynamically Smooth Boundary. The smooth boundary has the average depth of surface irregularity k lesser than the depth of laminar sublayer of the surface. The eddys which are formed in the turbulent boundary layer outside of the laminar sublayer try to penetrate in the laminar sublayer but they cannot reach to the surface irregularities due to greater depth of the laminar sublayer. Hence smooth boundary has (a) 6 < 0.25 as per Nikuradse's experiment (b)

uo kp

< 4 as per Reynolds roughness number

Hydrodynamically Rough Boundary. The rough boundary has the average depth of surface irregularity k greater than the depth of laminar sublayer. The eddies which are formed outside of laminar sublayer in the turbulent boundary layer can penetrate into the laminar sublayer of the surface. Hence rough boundary has: (a) (b) Note:

3 > 6 as per Nikuradse's experiment uo kp

> 100 as per Reynolds roughness number

The transition boundary between smooth and rough boundary has:

k (a) 0.25 < — 6,u —> U. For practical purposes, the boundary layer thickness 6 is defined as the perpendicular distance (y) from the plate where local velocity u = 0.99 U. As this is an arbitrary way to define a boundary layer thickness, alternative way is used to convey the boundary layer thickness such as displacement thickness.

Boundary thickness (b)

Boundary Layer Analysis

589

Displacement Thickness. The alternative way of conveying boundary layer thickness is the displacement thickness e. It is the distance perpendicular to the boundary surface to which the boundary surface has to be displaced outward to compensate for reduction in the discharge due to the formation of boundary layer. The displacement thickness is given by

15* =

to

(1 —

dy

14— u—.{

H 6* Displacement thickness

Boundary layer thickness

The velocity distribution in the boundary layer is as shown in the figure. Consider elementary strip at distance y from the boundary with thickness as dy. Let u be the velocity at distance y from the boundary. Now mass of the fluid flowing per second through this strips is: dmb = p • u • dA = p • u • b • dy

where

b = width of the boundary

In case the boundary layer is not existing, then mass of fluid flowing through this area dA per second is — drnf = p • U. dA = p • U. b • dy Hence reduction in mass of fluid flow per second due to existence of boundary layer from area dA is — dm,. = dmf — dmb = pU•b-dy — p•u•b•dy = p • b • (U — u) • dy Now total reduction of mass of fluid flow per second from total area of boundary layer is: 6 dm, = Al,, = j p • b (U — u)dy (i) o Now consider the plate is displaced by distance $* so that to reduce the flow of the fluid moving at constant free stream velocity U for a distance of e. Then reduction of fluid flow per second is: mr = p. u. 3*. b Equating eqn (i) and (ii), we get p • U • b • 5* = f 6 p b • (U — u)dy o

(ii)

590

Fundamentals of Fluid Mechanics

or

6* _

-u d Jo U

Y

(1- ±)dy 1 jo U Momentum thickness (0). It is the distance measured perpendicular to the boundary surface, to which the boundary has to be displaced for the reduction in the momentum of flowing fluid as it is reduced by the existence of the boundary layer. Consider a strip of area dA = b • dy at distance y from the boundary surface in the boundary layer where local velocity is u. Momentum of fluid flowing through this strip in the boundary layer is =

Momentum per second when boundary layer is existing - Massx local velocity Second = (p • u • dA) x u = pu2.b • dy where Momentum per second when boundary layer is not existing = p • u b • dy • U Loss of momentum of fluid passing through strip of area dA = p • b • u (U - u) dy 6

Total loss of momentum due to boundary layer = f p• bu(U - u)dy

b = width

(i)

-u

Loss of momentum in boundary layer

In case B is the distance to which boundary surface is to be displaced in free stream having velocity U to reduce the momentum equal to the loss due to boundary layer Total loss of momentum by displacement El = p • b • 0. U2 Equating eqn (i) and eqn (ii), we get: p•b•O•U 2 = 5,5 p•b•u(U-u)dy

0 r s u(U -u)dy = U 2

(ii)

Boundary Layer Analysis

59 I

10. What is shape factor? The shape factor is the ratio of displacement thickness to the momentum displacement Hence shape factor is: H—

displacement thickness momentum displacement (5*

53( Uu

)dy

U (1_ U)dy u

fo6 U

11. What is energy thickness (60 or Show that the energy thickness for boundary layer flow is given by 6 U 459 = J — u o

U2 2 ) dy

u

(UPTU-2005-6) It is the distance perpendicular to the boundary surface, to which the boundary has to be displaced to compensate for the reduction of kinetic energy of the flowing fluid due to existence of the boundary layer. Consider a strip of fluid in boundary layer at distance y and thickness dy with velocity u. kinetic energy of the fluid in strip 1 in presence of boundary layer = — (p• u • b • dy) u2 2 kinetic energy of the fluid strip 1 in the absence of boundary layer = — (p • u • b • dy)U2 2 Loss of kinetic energy in 1 the strip in boundary layer = — (p • u • b • (U2 — u2) dy 2 Total loss of kinetic 1 p • b u (U2 — u2)dy energy in boundary layer = — 2 o Loss of kinetic energy through 1 energy thickness Se in free stream = — p. b • 80 U3 2

(i)

592 Fundamentals of Fluid Mechanics Equating eqn (i) and (ii), we get:

1 p. b • 69. U 3

s

. . b. I 2PY

1 =—

89=

(u2 — u2)dy

(1_ 2 )th,

r U

u2

12. Find the displacement thickness, the momentum displacement and energy thickness for the velocity distribution in the boundary layer given by u = Y where u is the U 45 velocity at a distance y from the plate and u = U at y = 5 where 6 = boundary layer thickness. Also calculate the value of 6k/O. (UPTU-2001-2) Velocity distribution is:

u = U Displacement thickness (8*) (5* = 6

11— U

7)dY

=s,6)(1-)dY y

2 )6

=(_

Y

25

,2

= 6= 2 t5 Momentum thickness (9)

26

Boundary Layer Analysis

y2

y3 1 6

1

2 35 0

1 6 =

63 36 1

[522

1

[353

2.33]

6.52 1 x 53 6.52

S 6

Energy thickness (30) u (1 u2 ) d = Jo U _ U 2 Y s r Jo

3

1

— Yl dy 32

1 so ( , _ y 3 ) (, 3 o 2 dy y4 6 4.32 jo

[y2 ci

2

1 [52 :4 3 2 52 1

5.4.52

[234 — 54]

1 1 34 5 4.53

S 4

Shape factor (5*10) 6 6* =

2

3 6 =3

6 2

593

594

Fundamentals of Fluid Mechanics

13. How are the thickness of boundary layer, shear stress and drag force along the flat plate determined by Von-karman momentum equation? Consider the flow of a fluid with free stream velocity U over a flat plate as shown in the figure. Consider a small length dx of plate at a distance x from the leading edge of the plate. Now shear stress near the plate is given by Newton's law of viscosity as — To =

du dy

Momentum equation for boundary layer

The shear force or drag force acting on dx length of the plate opposite to flow direction is: dFD = Shear stress x Area = To x dx x b where b = width This drag force dFD has to be equal to the rate of change of momentum for the flow of fluid on plate for distance dx. Consider a small control volume ABCD bounded by the boundary surface and boundary layer as shown in the figure. Mass rate of flow entering the control volume through side AB is — m1 = f pxuxbxdy o Mass rate of flow leaving the control volume through side CD is —

m2 = mass through AB +

a ax

(mass through AB)dx

a p•u•b•dy + a [f0(pubdy)]dx fo ax Mass rate of flow entering the critical volume through side BC is =

6

m3 = M2 - M1

a

[r(pubdy)]dx

ax 0 Momentum of flow entering through side AB is —

(i)

Boundary Layer Analysis

595

= l (pubdy)u ob 8 =

put b. dy

(iv)

Momentum of flow leaving through side CD is — 8 = jo pu2b • dy + a x [5,05 (pu2bdy)1dx Momentum of flow entering through side BC which is entering uniform velocity u is — _

ax

[15(p • u • U • b • dy) dx ,u

= ax [r0 (p • u • U • b • dy) dx Rate of change of momentum of control volume ABCD is — = momentum of flow through CD — momentum of flow through AB —momentum of flow through BC = Jo p •u 2 • b • dy + 8 — s o

=

ax

[56

(p • u2 •

b • dyddx

pu2b • dy — L[16(p • u • U • b • dy)Jdx :,

[p • b • r(u2 _ u • u)dy]dx

= p • b ±[r(u2 _u • U)dy]dx asp and b are constant

ax

During equilibrium, the rate of change of momentum is equal to the shear force exerted by the boundary surface on the control volume. — To • dx

• b = p•ba x [56(u2 _ u • U)dyidx

= PU2 t x [cou(' U u ) dY 1 But the momentum thickness 0 =

506

(1 _ u) dy U u

= P• U Or

Z0

put

ao ax

2

DO

ax

596

Fundamentals of Fluid Mechanics

The above equation is known as Von Karman momentum equation for boundary layer flow. Thickness of laminar boundary layer. Prandtl assumed the velocity distribution in a laminar boundary layer as — 3 1y u_ 3 y 2 S 2 63 U d Any assumed velocity must satisfy the conditions: (1) y = 0, u = 0 and du = finite value, dy u and (2) y = 15, u = U and cl = 0. Substituting this into momentum equation. dv To = PU2 ci cx [506 tit (1— #) dY 1 = pU2 a [5ó (3 3' ax 0 2 b

1 y3

X (1 3

Y+1

2S 2 63

253

= 0.139 pU2 do dx But

( du) to = I-1(— dy = tl[

y=0

a [U( 3 Y 1 y3 )11

dy

2S 263

y=o

3 U — 2 [I 6 Equating both the above equations to = 0.139 pU2 do = 3 2 itt U a

dx

or

ddb = 10.78

liclx pU

Now integrating the equation, we get — 2 • Ux = 10.78 +C 2v Now

(5 = 0 at x = 0, ... C = 0 S _ 4.65 x i Ux i v

idy

Boundary Layer Analysis

But

=

597

Ux Re =Reynolds number v (5 _ 4.65 x VRe

4.65 x Re It can be seen that the thickness of the boundary layer increases with the distance x from the leading edge of the boundary layer. However thickness of the boundary layer decreases on increasing of free stream velocity. Shear stress 3 U '1-' 3= V I S U 3 — 2 'u 4.65 x Re or

5—

1.5 p ux x Ux 4.65 x V v = 0.322

0.322pU2 ,IRe

ippU 3 l x

pU 2 — Cf 2

where Cf— 0.644 and it is known as local drag coefficient. It is defined as the ratio of shear V Rex 1 pU 2 ). It is also called is coefficient of skin stress (To) to the dynamic pressure head (2 friction. Drag Force FD. Total frictional drag force FD on one side of the boundary surface of length L and unit width is — L FD = j 1-0 X 1 x dx o _ r L 0.644 = F L cf pU 2 x pU 2 dx dx 2 Jo Jo VRex _ 1.288 JReL

2

AU L 2

598

Fundamentals of Fluid Mechanics

where ReL = pUL . The drag force is generally expressed as I FD =CD

p 2

XA 2 where CD is the mean drag coefficient, A = area of the boundary surface exposed to the flow. CD = 1.288 as A = L ReL Hence mean or average drag coefficient is the ratio of total drag force to the product of the 1 pU 2 ) and exposed area (A) of the boundary surface to the flow dynamic pressure head (CD -

FD

pU 2 A 2 Note: The laminar boundary layer is stable upto Re = 3 x 105. The boundary layer changes to turbulent at Re = 5 x 105. The values of 7-0, Cf and CD change when we assume other type of velocity distribution as given below Velocity distribution

1.

coefficient of Cf

Co

5.48

0.73

1.46

4.56

0.644 1.288

2

u _ 2. _y U (5 82

-

6

3 2.

3. 4.

U U u

2 =2

6

y6

2 83

4 2y3 + y 4 5 53

5.48 0.686 1.372

Blasius Results

4.91

0.644 1.328

14. Find the expression for (1) boundary layer thickness (2) local coefficient of drag and (3) average coefficient of drag for turbulent boundary. Velocity distribution. The local velocity distribution in the turbulent boundary layer is logarithmic. Since it is difficult to solve logarithmic equation, the velocity distribution in the turbulent boundary is expressed by power law as give under =

Y )1in

5J

where n varies from 5 to 10 depending upon Reynolds number

In turbulent boundary layer in which Reynolds number varies from 5 x 105 to 2 x 107, value of n is taken as 7 and the relation is known as one-seventh power law. (y)1/7

U

=

Boundary Layer Analysis

599

Shear stress. In turbulent boundary layer, Blasius's relation for shear force for flat boundary surface is given as under — to = 0.0225 pU2[ 1-1 14

p3u

Boundary layer thickness (6) 5 = 0.37x (Rexx ) 5 Local coefficient drag (C1) Cf —

0.059 (Rex )"5

Average Coefficient of drag (CD) 0.074/ for total turbulent layer CD (Re L )"5 Or

CD

0.074 (ReL )1"

1700 for partly laminar and partly turbulent boundary layer. Re L

15. What do you mean by separation of boundary layer? What is the effect of pressure gradient on boundary layer thickness. (UPTU 2002-3) or What is separation? (UPTU 2004-5) Flow generally takes place from higher pressure to lower pressure which is called negative pressure gradient i.e. pressure is decreasing in the direction of flow or pressure gradient in ap . However, there are situations when flow takes place over a curved x boundary surface under positive pressure gradient i.e. pressure is increasing in the direction of the flow. In such cases, the flow near the boundary is continuously retarded and a point is reached when the flow starts separating from the boundary. The point at which flow separates from the boundary is called separation point. The separation takes place mainly due to positive pressure gradient which reduces the momentum of flow of the fluid within the boundary layer due to higher shear stresses in fluid having high viscosity. The change of velocity and pressure gradient are related as —

x-direction is

a2u

ap

a y2

ax

600

Fundamentals of Fluid Mechanics

ap The flow accelerates or velocity increases in case — is negative. The flow deaccelerates a

ap

or velocity decreases if — is positive.

a

Consider a flow over a curved surface causing positive pressure gradient i.e.

ap is positive. ax

The flow starts retarding as it moves forward from point 1. The flow velocity gradually diminishes to zero at point 2. The fluid particles can not move further close to boundary surface as these particles have to encounter the resistance of the boundary layer. Hence fluid leaves the boundary at point 2. The separation gap increases as the fluid moves beyond point 2 to point 3. The layer of fluid detached from point 2 has tendency to roll itself, thereby forming vortices. This region where vortices are formed due to positive pressure gradient is called wake. The kinetic energy of the flow is wasted due to reverse flow which is ultimately converted into internal energy (increase of temperature of the fluid). There is additional drag force in the wake ration. The velocity distribution helps in identification of separation as under —

(1) (du) dy) y.0

is positive, it indicates no separation

du) is zero, it indicates the flow at verge of separation (2) ( dy) y.0 du) is negative, it indicates that flow is separated. (3) ( dy) y.0 Flow Curved surface

Separation point Reverse flow

0

Separation stream line

SEPARATION OF BOUNDARY LAYER

The positive velocity gradient leads to reduction in the velocity in the boundary layer. Reynolds number decreases with velocity retardation. Since the thickness of the boundary layer is inversely proportional to the square root of Reynolds number, the thickness of boundary layer increases with positive pressure gradient.

Boundary Layer Analysis

60 I

16. What are different methods of preventing the separation of boundary layer? (UPTU 2004-5) The separation of boundary layer is undesirable and attempts are made to avoid separation by various methods. The methods for preventing the separation of boundary layer are: (1) Suction of slow moving fluid by a suction slot

Section slot to suck the retarded fluid

(2) Supplying additional energy from a blower

Air is blown in to give additional energy

(3) Streamlining of the body so that separation is avoided or delayed to take place as near to the trailing edge

----------

Small region Hof wake

Trailing edge

(4) Providing guide vanes to guide fluid to avoid separation

(5) Providing slot near the leading edge.

602

Fundamentals of Fluid Mechanics

17. Find the displacement thickness, momentum thickness and energy thickness for the velocity distribution in the boundary layer given by —

U = 2 (Y )

3

Displacement thickness (6*) 6.* = Jo (1— u )dy =

Jo

[ 1— 2 W -02 1dy

2

[3

=

3

2 y 5 2 62 + 6

5

1 y 62 3 1

0

3

3 2

Momentum thickness (0)

a= =

o # (1— t)dY l

J13

= I:

26 Y — 6Y 22 1[1- 2Y +6 Y 22 idY 6 2Y 6

4

4y2 2y3 Y2 ± 2y3 Y dy 62 63 6 (53 64

[2y 5y2 4y3 y4 1 JS

6

62

63 64

dy

Boundary Layer Analysis

2 y

2

3 y ±4

5

5 2 62 3

[

y

4

63 4

1

45

ysi6

4 5

0

1

5

=S — — .6+ 6- .6 3 5

_ 6 [15-25+15-3]

L

=

= f

U2

U

=

2 )2

Y2 1 [ (2Y y 1. 32 62 J l6

(2

6 (2y _ Y2

32

fo 5 6 [2y

= fo

dv

U2 y

0

2 .15. 15

050

Energy thickness oa

15

8y3

6

63 2

= ra ( 2 y Jo 8 2

2 4 3 i_ 4y y 4y th, + 62 64 33 f 2y5

y6 4y2 y2 4y4 — 62 + 3 4 ± 66 65 dy

3

8 y + 12 . y _ 6 . y + 1 83 4 85 6 86 a4 5

8y4 + 65 64

1 .y 82 3

4

=5 — 1.5 - 2.5+ 12 •5 - 5+ 5

—6

=

[105

—

1dy

5

6

y

7)6 7 o

4o

35— 210+252 —105+15] 105

2 105

18. The velocity distribution in the boundary layer over a plate is give by — u = 3 y 1 (I U 2 6— 2 6 Find the displacement thickness (5*)

603

604

Fundamentals of Fluid Mechanics

( 3*

=

#)dY

-

[1- (2 6 2 6Y3)] dY 2

(1 2 . 6 3 y2 +

2- (5372 )4Y 1

y4

Y 2 5.2 2.62 4 1

0

_[6- 4 6+

s 61

]

8-6+11 _ 6[ 8 =

8

.6

2 u 3 y 1 y 19. The velocity distribution in the boundary layer is given by 17. = • 6 + 2 • . 32 Find (1) Ratio of displacement thickness (6*) to thickness of boundary layer (5) and (2) Ratio of momentum thickness (0) to boundary layer thickness (S) —

6* =o (1- u )dy U y2

10 [1 (23 (5 31 1 2 62

=f

23 ; + 21 2 dY 3 /,2

= Y

2.6 2 + 2.62 3 3

= 6- 71 .6 + - 8 6 =5

1 y3 is

[12-9+2 12 _I

Boundary Layer Analysis

605

i )idy Momentum thickness 0 = [#(1_ t Jo or

(3 Y

0 = Jo 2 =

Y2 ) 252 2

5.

2

5 232

2 [1 (3 y, 2

)(I 3 . Y, 2 (

(3 y 9 y 2 ± 3y 3

E15 (2 (5 3y

Jo

2

6

262

jidy

y:,2 )dY 2 y2 + 3 y 3

y4

4 62 453 262 2 53 454) 3 2 11 y + 3 Y 4 62 2 53

Y

dy

4 )

454 ClY

2 4 3 y 11 y 3 3 y 1 y5 2.8 2 4.52 3 2.53 4 454 5 10 [ _ 6(3 11 + 3 1 .d1. 12 8 20)

=

_(90-110+45-6) 120

= 6 19

120

0 = 19 120 20. In a laminar boundary layer over a flat plate, velocity distribution is given by u = a + by + cy2. Using the boundary conditions, find the values of the constants a, b and

c. u = a + by + cy2 Boundary conditions are — (1) u = 0 at y = 0

606

Fundamentals of Fluid Mechanics

(2) u = U at y = 5 du (3) = 0 at y = 6 Now if y = 0 then u = 0, we get — 0 = a .. a = 0 u=Uaty=3 U = bo + c32

Now Now

d uy

(i)

= 0 at y = 3

du = b + 2cy dy ( du =0 (dy) y.6

Now or

b + 2co = 0

Using eqn (i) & (ii) U = —2c82 + co2 = — c62 or

c = — 3U2 U b = —2c8 = —2 x 32x8 =2 U o

Hence velocity distribution is — 2 U= 2U( Y ) —

6

u()

21. The velocity distribution in a turbulent boundary layer is as per power law — y)1/7

u _ ( U 3 Find (1) displacement thickness (2) momentum thickness (3) energy thickness (4) shape factor and (5) energy loss if U = 10 m/s, 3 = 20 mm, p = 12 kg/m3 Displacement thickness (6*)

Boundary Layer Analysis

)1/7

idy

= 16 [1-(Y

8/7

= [y

8

1.17

x

7

io

= 3[1—;] = 8

Momentum thickness 0=

u (1— )dy — Jo ° U U 1/7 [

)1/7

dy

(E 1— ,75 Jo° . /7

s

Y 2/7

Y1 8 1/7 62/7

J 1

y

Sun

dy fi

y9/7

8/7

8

32/7

9

7

_

=

I-7 L8

72

71 9 _1

•0

Energy thickness U

cSrt = — 0

U

1—

U

2

2/7

(5 y 1/7

JO S 1/7

dy

u2

1

Y

8 2/7

5 y1/7

y3/7

o

33/7

f

61/7

dy

dy

607

608 Fundamentals of Fluid Mechanics

_[ y

8/7

61/7 8

7

10/7 Y 53/7 . 10 T. 7 1

= j. I-7 _ 71 L8 10 = (5.7 [5 41 L 40 i 7 =— 40 • 0 Shape factor shape factor H = (5* 6 8 7

72 9 7 Energy loss 50= 7 • (5 = 7 x 20 x 10-3 40 40 1 Energy loss per unit width = — (poeU)U2 2 = — x 1.2 x 3.5 x 10-3 x 10 x 102 2 = 2.1 Nm/s 22. Examine the following velocity profile to find if the flow is attached or detached: u

(a) e -_

2 Y (Y)2

- (.5)

3 4 (b) u u = —2 ' + ('1 : 5) + 2 () 2 (c) ± ;J

=

2() + (Yo)

3

4 — 2 (Yo)

Boundary Layer Analysis

609

Case (a) u

y

— 2

2

(5 6 2

U

or

y

/I = U (2

2) Y —Y 6 62

du _ u ( 2 2y) dy (:5 (5

or

( du = 2U -> positive value dy) y .0 5

Or

Hence flow will remain attached

Case (b) u = UPY +(13 +2 (141

6

(5 i

du [-2 3y2 8y3 1 + dy = U 6 + 33 34 (du

Now

dy ) y .0

—

—2U —> negative value 6

The flow is separated

Case (c) u= U[ 02 03 — 2 M41 L5 5 3 i du = u [4y ± 3y2 _ 8y3 52 33 34 dy (du) =0 dy y .0 Hence flow is about to be detached.

23. Assume that the velocity distribution in the laminar boundary layer with zero pressure gradient can be give as —

6I0

Fundamentals of Fluid Mechanics

u

2•y = sin ( 7r (5

U

)

Find drag coefficient in terms of Reynolds number. The momentum integral equation with zero pressure gradient is — = pU 2 dO dx 0 = momentum thickness

To Here

= f o u (1 u ) o u u dy S

5 =5 0

,7

(',- • i) ' (1 — sin 1 •5) ' dy ( sinIC •

8 dy — 5 sin2 (Lr • 1 (i 0 26

_

r •Y cos — — 2 (5 =

Y

-5

—

7E

2.6

_

1 r 2 Jo

(1— cos

.3'

dy

-c, -5

26 — 1 =+— g 2 Y 26 it

1r sin — y 8 ir 8 -o

6 2

0=5

) 1427r

de = ( 4 — r) d8 dx l 27s ) dx

to

_ itt (du ) _ A[U cos (7 1) x 26 1 2 6 dy)y = 0 y=0 il

U • ic 26

U • rc 2 dB r° = PU Tx — ii 28

Boundary Layer Analysis

do= dx pU 25

or

or

(4 - 7r)

27r

d5 _ PA. dx 2 • pU •

or

5d5 =

,r 2

62 or

' dx pU

2r 2

"

X r-

2

4 —7r 5_

or

' (4 - 7r)

Drag force FD

pU

27r2 (4-7r)

pU

= J L To X dx o µUm dx Jo 25 µV 11(4 -7r)

FD

2. .N/r2

ipU L -1/2 7r x — x dx i it o

= µU 11(4 - ir) • p • U x EF1/2 2 µ•2 1/2 (4 - ir) • p • U 512 • µ 2 • L 2µ 11 ir) 2

pU L

Drag force both sides of the surface = 2FD -

p 2 U 4 L2

x

1.31 x

pU 2 L

.NIT ?e Drag coefficient = CD -

Drag Force pU 2 L

or

CD-

1.31

(Re) 1/2

611

6I2

Fundamentals of Fluid Mechanics

24. A uniform free stream of air moves at 10 m/s over flat plate. Find the drag coefficient and the drag for the plate of size 0.5 m long and 3 m wide. Given p = 1.2 kg/m3, = 18 x 10-6 Ns/m2 Re =

pxUxL

1.2 x 10 x 0.5 18 x 106 = 0.333 x 106 = 3.33 x 105 Hence Re is almost 3 x 105 and boundary layer is laminar, we take Blasius results CD —

L328 T ?e 1.328 V0.333 x 106

= 2.32 x 10-3 Drag for one surface = —1 pU2•il•CD 3 1 x 1.2 x 102 x (0.5 x 3) x 2.32 — 2 = 0.201 N 1,2

25. If velocity distribution u = U (2 Y, '2 , find the thickness of laminar layer, shear (5 15 stress at trailing edge and the drag force on one side of the plate 1 m long, if it is immersed in water flowing with velocity of 30 m/s. Take p = 1 x 103 and µ = 0.01 Ns/ M2

Guidance. Check that boundary layer is laminar in case Re 3 x 105. Then for the velocity distribution, the value of (5, Cf the CD can be read from table (Question : 13) ReL =

pU L

1 x 103 x 30 x 10-2 x 1 0.01

= 3 x 105 Hence the plate has laminar boundary layer throughout its length. From table, we get following b=

5.48x JRe

5.48 x 1 Al3x 105

Boundary Layer Analysis

6I 3

= 1 X 10-2 = 1 cm

TO — ki —

pU2

C — 0.73 2 f N I Re

= 0.73 x 1 x 103 x (0.3)2 2 /., = 5.997 x 10-2 N/m2 = 0.05997 N/m2 FD — ...,D CD X

PU 2

2

0.73 x 0.9 x 102 547.7 x 2

X A where CD — 1.,L6 NI Re

1.46 x 1 x 103 x(.3)2 xlx1 547.7 x 2 = 0.1199 N 26. A thin plate 1 m long and 1 m wide with velocity of 0.3 m/s is moved through water. Find the resistance offered by the plate (p = 1 x 103 and 1.1 = 0.001). Guidance. If plate has laminar boundary layer, we can use Blasius results for finding CD. pUL

lx103 x0.3x1

Y. = 3 x 105

0.01

ReL =

We can assume boundary layer to be laminar. From Blasuis results CD — 1.328 NW or

CD =

1.328

— 2.42 x 10-3

113 x 105 1 pU2 Drag force (both sides) FD = 2 x CD x — 2 = 2 x 2.42 x 10-2 x

X

A

2

x (0.3)2 x 1 x 10+3 x 1 x 1

= 2.178 N 27. Find the frictional drag on one side of the plate 1 m wide and 2 m long placed longitudinally in a stream of oil with sp. gr. = 0.9 and v = 0.9 stokes flowing with undisturbed velocity of 5 m/s. Also find the thickness of the boundary layer and shear stress at the trailing edge of the plate.

6I4

Fundamentals of Fluid Mechanics

ReL, —

UL

V

—

5x2 0.9 x 10-4

= 1.11 x 105 The flow has laminar boundary layer and we can use Blasius results CD _ 1.328 /,

1328 3.33 x 102

= 0.4 x 10-2 Drag force (one side) = FD = CD X

2

pU2 x A

0.4x10-2 x 0.9x103 x52 x(3x1) 2 = 135 N Boundary layer thickness (b) at trailing edge — S = 4.91 x

VReL

4.91 x 3 x 10 2 cm 3.33 x 102 = 4.423 cm Shear stress at trailing edge (one side) ,, p 2 To = ii x 2 r, _ 0.664 \I ReL

k-f

_ 0.664 3.33 x 102 0.9 x 103 x 52 0.664 x 2 3.33 x 10' = 22.43 N/m2

To =

28. A smooth plate 3 m wide and 5 m long moves through air at a relative velocity of 2 m/s parallel to its length. Find the drag force on one side of the plate (1) assuming laminar conditions (2) assuming turbulent conditions over the entire plate. Take for air. p = 1.20 kg/m3 and v = 1.5 x 10-5 m2/s.

Boundary Layer Analysis

ReL =

6I 5

2 2 x D, 1.5 x 10-5

pxUxL 11

= 13.33 x 105 Case 1. Considering laminar boundary layer and using Blsius results, we have — CD =

um ReL

1.328 1 13.33 x 10+5 113.33

= 0.115 x10-2 Drag force of one side — FD = CD X 1 pU2 X A 0.115x10-2 x1.2x 22 x5x3 2 = 4.14 x 10-2 = 0.0414 N Case 2. Considering turbulent boundary layer 0.074 (ReL) 1/5

CD

0.074 (13.33 x 105)1/5

0.074 — 0.0048 1.54 x 10 Drag force on one side — 1 pU2A FD = CD0.0048x1.2x 22 x5x3 2 = 0.1728 N 29. A flat plate 1 m square is held normal to flow of air at 10 m/s. It is found to experience a drag force of 300 N. Determine the drag force when plate is held parallel to the flow. Find the ratio of two drags. Take p = 1.2 kg/m3 & v = 1.5 x 10-5 m2/s. Take boundary layer to be turbulent ReL =

CD —

UxL v 0.074 vc (ReL ) ''

10x1 — 6.66 x 105 1.5 x10-5 0.074 (6.66 x 10-5)1/5

6I6

Fundamentals of Fluid Mechanics

0.074 1.373 x 10-1 = 0.00539 Drag force at both sides p 2 XA 2 = 0.00539 x 1.2 x 102 x 1 = 0.6468 N

FD = 2 x CD X

Ratio of two drag forces = 0.6468 300 = 2.156 x 10-3 30. A 3 x 2 m rectangular plate is held in water moving at 1.2 m/s parallel to its length. Assuming laminar condition in the boundary layer at the leading edge of the plate, find (1) location where the boundary layer changes from laminar to turbulent (2) thickness of boundary layer at that point (3) friction drag of the plate. NS (p=1x 103, p = x 10-2 in 2 ) ReL =

pUL

lx103 x 1.2 x 3

0.001

= 3.6 x 106 The value of Reynolds indicates that boundary layer is partly laminar and partly turbulent. The location of the end of the laminar flow condition can be found out — xo _

L xo

critical Re where critical Re = 5 x 105 full plate Re L•5 x 105 3.6 x 106

3 x 5 x 105 3.6 x 106 = 0.416 m Thickness of laminar bounder layer at xo = 0.416 m do = 4.91 x

X0

./critical Re

Boundary Layer Analysis

6I7

4.91 x 0.416 ‘15 x 105 4.91 x 0.416 7.07 x 102 = 0.00288 m Friction Drag. Friction drag of one side = Laminar friction drag for length AB + turbulent friction drag for length BC = Laminar friction drag for AB + turbulent friction drag for AC— turbulent friction drag for AB.

FNet = 2 (F1 ± F2 F3)

(Both sides)

F1 =

CD x —1 pU 2 x A (use Blasius results) 2

_ 1.328 x lxlx 103 x 1.22 x 3 x 2 VRe, 2 1.328 x 103 x 1.44 x 6 7.07x102 x2 = 8.11 N F2=

1 x — x 1 x 103 x 1.22 x 6 2 (3.6 x 106)1" 0.074

= 0.00407 x 1.44 x 3 x 103 = 17.58 N F3 —

0.074 (5 x 10' )v

X 1p

2

x 1.22 x (2 x 0.416)

= 0.00537 x 1.44 x 1 x 103 x 0.832 = 6.434 N

6I8

Fundamentals of Fluid Mechanics

FNet = 2 (F1 + F2 — F3)

= 2(8.11 + 17.58 — 6.434) = 2 x 19.256 = 38.512 N 31. A streamlined train 100 m long has carriages 3 m wide and 3 m high. If the train moves at a speed of 120 km/Hr through air having p = 1.2 kg/m3 & v = 1.5 x 10-6 m2/s. Find the power required. 3

U = 120 km/Hr = 120 x 10 x

1 — 33.33 m/s 3600

33.33 x 100 Re = pU L 1.5 x 10-6 ii = 2.22 x 107 Area can be taken as 2 x (width + height) x length = 2(3 + 3) x 100 = 12 x 100 = 1200 m2 0.074 1 pU2 x A X F D — (Re)1I5 FD =

x 1 x 1.2 x (33.32)2 x 1200 0.074 2 )1/5 (2.22 x 107

0.074 x 0.6 x 1110.9 x 1200 2.47 x 10 = 2396.3 N (Neglect the frontal resistance) Power = FD x velocity = 2.3963 x 33.33 x 103 = 79.86 kW 32. A ship model 1.5 m long with draft of 20 cm is moving at a velocity of 0.3 m/s in a basin containing water. Assuming that one side of the immersed portion of the hull may be considered a smooth plate (1.5 m x 0.2 m), find the friction drag of the hull and the thickness of the boundary layer at the stern of the model if the boundary layer is (1) laminar (2) turbulent. If the measured total drag of the model is 0.2 N, find the total drag of the prototype if the model scale is 1 : 64. ReL, —

UL 0.3x1.5 — V 0.001x10-3

= 4.5 x 105

Boundary Layer Analysis

Case 1. Laminar boundary layer 1.328 _ 1.328 AI, I V4.5 x 105

CD —

= 1.98 x 10-3 FD = 2 x CD X A x

pU 2 2

= 2 x 1.98 x 10-3 x (1.5 x 0.2) x 103 x (0.3)2 2 = 1.98 x 0.3 x 0.09 = 0.0535 N 5 = 4.91 x L AiReL 4.91 x 1.5 V4.5 x 105 = 1.09 cm Case 2. Turbulent boundary layer CD —

0.074 (ReL )115

0.074 (45 x 105)115

= 0.55 x 10-2 FD = 2xCL,xpxA LI 2 2 = 2 x 055 x 10-2 x 1 x 103 x (15 x 0.2) x (0.3)2 2 = 0.55 x 3 x 0.09 = 0.148 N _ 0.37 • L = 0.0417 m (R e)115 Now corresponding value of the prototype can be obtained by Froude law. S

UP _ U i I L pxg VLxg Up = 0.3 x II

L P L

= 0.3 x V64

6I9

620

Fundamentals of Fluid Mechanics

= 0.3 x 8 = 2.4 m/s Reynolds number of prototype is — U p x Lp

2.4 x (64 x 1.5) 0.001x 10-3 = 2.3 x 108 Flow in turbulent 0.074

CD

(ReL) 115

0.074 (2.3 x 108 )1/5 0.074 (2300 x 105)1/5 _

0.074 — 2.038 x 10-3 3.631x10

Skin resistance of prototype is — FD = 2CD X A n

2

n

,

2 = 2.038 x 10-3 x (1.5 x 0.2 x 64 x 64) x (2.4)2 x 103 = 14.425 kN

Total Drag

Total drag of model is F = Fwave Resistance + FD 0.02 x 9.81 = FwR ± 0.0535 . FWR = 0.1962 — 0.0535 = 0.1427 N Now wave resistance is — FWR oe pL2U2 (FwR)p

L2PUP2

(FwR)m

L2 U 2

x (2.4)2 (R3)2 (FwR)p = 0.1427 x 262.144 = 37.408 IcN = (64)2

Boundary Layer Analysis

62 I

Total drag of prototype = 14.425 + 37.408 = 51.833 lcN 33. Calculate frictional drag on a flat plate 45 cm long and 15 cm wide placed in longitudinal stream of fluid with sp. gr. 0.925, kinematic viscosity of 0.9 x 10-4 m2/ s and velocity of 6 m/s. Find thickness of boundary layer and shear stress at trailing edge of the plate. ReL =

UL 6x0.45 — v 0.9x10

= 3 x 104 The boundary layer is laminar Using Blasius's results, we get — S=

4.91x ReL

4.95 x 0.45 3 x 104

= 0.0127 m Shear stress at both sides of the plate — =2x

Cf X

1 pU2

Cf.= 0.664 — 0.383 x 10-2 Re To = 0.383 x 10-2 x 0.925 x 103 x (6)2 = 12.75 x 10 = 127.5 N/m2 34. A smooth flat plate is exposed to a wind velocity of 6 km per min. If the laminar boundary layer exists upto a value of Re = 2 x 105, find (1) distance from leading edge where transition takes place (2) the boundary layer thickness at transition point. Take v = 1.5 x 10-5 m2/s 3 U = 6 km/min = 6x10 60 = 100 m/s Critical Reynolds number is 2 x 105. Let x be the distance from leading edge at which transition takes place R

Ux

100 x x 1.5 x 10-5

622

Fundamentals of Fluid Mechanics

2 x 105

— 100 xx 1.5 x 10-5 2 x 1.5 x 105 x 10-5 100 = 0.03 m

x—

Boundary layer thickness using Blasius results — S_

4.91x 4.91 x 0.03 x 100 f_ — -v Re V2 x105

= 0.033 cm 35. A plate 3 m long and 1.5 m wide is kept horizontal in water which is moving at velocity = 1.25 m/s parallel to the plate. If the boundary layer is laminar at leading edge, find (1) thickness in laminar region (2) frictional drag due to both surfaces of the plate. Assume laminar boundary exists upto Reynolds number 5 x 105 and viscosity of water 0.001 NS2 m If laminar boundary layer exists upto x distance from leading edge, then Re =

5 x 105 — or

x—

pUx P ix 103 x 1.25 x x 0.001 5 x105 xlx10-3 1.25 x 103

= 0.4 m Thickness of boundary layer at x = 0.4 m is 3 _ 4.91x NIRe _ 4.91 x 0.4 115 x 105

= 0.277 x 10-2 m = 0.277 cm Frictional Drag Re =

pUL = lx103 x1.25x3 0.001 p

Boundary Layer Analysis

623

= 3.75 x 106 Flow has both laminar and turbulent boundary layer [ 0.074 (ReL )115

CD

1700 ReL

0.074 (3.75 x 106)1/5

1700 3.75 x 106

= 3.13 x 10-3 1 pU2 X A 2 = 3.13 x 10-3 x 1 x 103 x (1.25)2 x (3 x 1.5) = 22 N

Drag force = 2 x

CD X —

36. A barge with rectangular bottom surface 30 m long and 10 m wide is travelling in river with a velocity of 0.6 m/s. Laminar boundary is upto 5 x 105 Reynolds number followed up by turbulent boundary. Find (1) distance from leading edge upto which laminar boundary exists (2) thickness of the boundary layer at that point (3) total drag force on the flat bottom surface of the barge and (4) power required for the barge for drag force only. Let x be the distance upto which laminar boundary exists, then — (R

pUx Ux itic =

11

V

5x105 x1x106 0.6 = 0.833 m

or

x—

Now laminar boundary layer thickness is — 6_

4.91x

4.91 x 0.833

V(Re)critical

1/5 x 105

= 0.00578 m = 5.78 mm Overall Reynolds number is — 0.6 x 30 ReL — 1 x 10-6 = 1.8 x 107 Now coefficient of drag for both laminar and turbulent boundary layers is —

624

Fundamentals of Fluid Mechanics

CD

0.074

1700

(ReL )1/5

ReL

0.074 (1.8 x 107 )1/5

1700 1.8 x 107

0.074 0.94 x 10-4 2.38 x 10 = 3.11 x 10-3 — 0.94 x 10-4 = 3.016 x 10-3 1 FD = CD X — pU 2 x A

2

= 3.016 x 10-3 x

2

x 1 x 103 x (0.6)2 x (30 x 10)

3.016 x 0.36 x 300 2 = 162.86 N Power required is — P

FD X U = 162.86 x 0.6

= 97.71 W 1/7

37. The velocity distribution in a turbulent boundary layer is given by

U

=

What

(

is the displacement thickness 6*? (3

(c)

(b) —

8

(d) 1 • 45

8

(IES 2008)

8/7 7 y = [y — x _ 8 6ln

•— 76 =6 — 8

0

Boundary Layer Analysis

625

8 Option 'a' is correct. 38. The velocity profile in a laminar boundary layer is given by 14 = . The ratio of U momentum thickness to displacement thickness for the boundary is given by which of the following? (a) 2 : 3

(b) 1 : 2

(c) 1 : 6

Displacement thickness—

=

1(1- uidy 0

s)'

=J(1-

0

2 = (31 2 3'8 )3

= - 8= 6 2

Momentum thickness—

2

(d) 1 : 3 (IES 2008)

626

Fundamentals of Fluid Mechanics

Now

6 _ 5/6 6.

5/2 =1 3

Option 'cl' is correct. 39. A smooth flat plate with a sharp leading edge is placed along a gas stream flowing at U = 10 m/s. The thickness of the boundary layer at section r — s is 10 mm, the breadth of the plate is 1 m (into the paper) and density of gas p = 1.0 kg/m2. Assume that the boundary layer is thin, two-dimensional, and follows a linear velocity distribution, u = U x 5 Y , at the section r — s, where y is the height from plate. r

u

U —'--

The mass flow rate (in kg/s) across the section q — r is— (a) zero

(b) 0.05

(c) 0.10

Mass flow rate entering from section pq is— ms = p x volume =pxAx U = 1 x (1 x b) x 10 = 1 x 1 x 10-3 x 10 = 0.1 kg Mass flow leaving from section rs is— m2 =pxAxu =px(1 x15)xUx 3 Y = pxox 1- (UxA-FUO) 2 6 6 =pxox 11 2 = 1 x 10-3 x 10 = 0.05 kg/s 2 ... Mass leaving from section qs is— m3 = m1 — m2 = 0.1 — 0.05

(d) 0.15 (GATE 2006)

Boundary Layer Analysis

627

= 0.05 kg/s Option (b) is correct. 40. Explain the essential features of Blasius method of solving laminar boundary layer equations for a flat plate. Derive the expression for boundary layer thickness and local skin friction coefficient from this solution. (UPTU 2009-10)

y

The boundary layer along a flat plate at zero incidence The boundary layer equations for laminar flow on a flat plate are— au + av — 0 ax ay au u ax

av

" ay

_

a2v aye

The boundary layer conditions are— at y = 0, u = v = 0 and y = Dc, u = u. The solution of the above boundary equations can be easily obtained using Blasius method. The method uses similarity transformation. The independent variables x and y are transformed to single variable ri and the dependent variable iy (stream function) is transformed to function J(n) as given below— v•x and

AR)

—

,,jv • x •

This transformation helps in transforming the boundary layer equations to a single ordinary differential as given below— ! • f' + 2f" = 0 (Blasius equation) The boundary conditions which are now applicable to this equation are— n = 0, f(0) = 0, f'(0) = 0 at at —> 1.0 The differential equation can be solved numerically and the solution is given in the graph.

628

Fundamentals of Fluid Mechanics 1.0

A 0.8 0.6

Uce = f

'07) 0.4 0.2

Uoc

= Blasius's

V

•X

Equation

From the above graph, it can be seen that at ri = 5, we have— r1 =

U..

1

v•• x

= j v•x

6

—5

5 U.‘ • x v

x

5 1 (Re)x

The surface shear stress is— — P(aa;) y =o =

U„.\1

v•x

x f'(0)

= 0.332 µ UnI U— v•x The local coefficient of skin friction is— eft —

— 0.664 ^ 1 v o2 x p • U., /2 0.664 V(Re)x

Chapter

14

FLOW THROUGH PIPES AND COMPRESSIBILITY EFFECTS

A A A A A A A A A A A

MAJOR & MINOR ENERGY LOSSES SUDDEN ENLARGEMENT SUDDEN CONTRACTION FRICTION FACTOR DARCY'S EQUATION CHENZY'S EQUATION FLOW IN BEND PIPE FITTINGS TOTAL ENERGY LINE HYDRAULIC GRADIENT LINE COMPOUND PIPE

A A A A A A A A A A A

EQUIVALENT PIPE PARALLEL PIPES NETWORKING SYPHON INLET AND OUTLET LEG POWER TRANSMITTED EFFICIENCY OF TRANSMISSION WATER HAMMER SURGE TANK BRANCHED PIPE EMPTYING A TANK BY PIPE

INTRODUCTION The fluid flow in pipes is generally turbulent. Flow is turbulent if Reynolds number is greater than 4000. In turbulent flow, the velocity distribution is relatively uniform and velocity profile is much flatter as compared to laminar flow. There are irregular velocity and pressure fluctuations in turbulent flow. This is the reason why no theory has been developed for the analysis of turbulent flow. The velocity fluctuations in turbulent flow influence the mean motion in such a way that an additional shear resistance to flow is caused. The turbulent shear stress consists of viscous shear stress and shear stress produced due to turbulence. When fluid flows through a pipeline, it encounters resistance due to (i) viscosity of the fluid, (ii) the roughness existing in the interior surface of the pipeline and (iii) resistance offered by pipeline. The major energy loss in a pipeline results due to friction as the pipeline has generally long length.

630

Fundamentals of Fluid Mechanics

Total energy in pipeline decreases in the direction of flow. Hydraulic gradient line is the sum of potential energy head from a datum line and pressure head at any point in the pipeline. Syphon is an arrangement of pipe system by which fluid can be made to run up the hill utilizing the force of atmospheric pressure. Water hammer is a phenomenon which is caused in a pipeline when flowing fluid experiences a sudden reduction of velocity due to the closing of any valve abruptly in the pipeline. 1. Distinguish between pipe and pipeline. Pipe is a term used for a closed conduit of round cross-section which is used for carrying water under pressure. Incase the liquid does not flow under pressure, then the pipe is not full of liquid and the atmospheric pressure exists inside the pipe. Sewers and culverts are not full of liquids and they have atmospheric pressure inside . The flow in partly filled pipes is similar to the flow in open channel. A pipeline consists of a number of pipes which are joined together with the help of pipe fittings such as unions, couplings, flanges, elbows, bends and reducers etc. The pipes in the pipeline may have same diameter or different diameters. To regulate the flow of liquid, valves and sluices are used as pipe fittings in the pipeline. Pipelines are used generally for transporting fluids like water, oil, air and gases from the point of supply to the different places where they can be used. For example, we use pipeline for: (1) (2) (3) (4)

water supply installations to distribute water from the overhead reservoir. reservoir dams to carry water to the turbines drilling platforms to carry crude oil to refineries supply LPG to the houses.

2. Why does loss of head take place in pipelines? When fluid flows through a pipeline, it encounters resistance due to (1) viscosity of the fluid (2) the roughness existing in the interior surface of the pipeline and (3) resistance offered by pipe fittings. Turbulence is produced in fluid flow due to resistance, causing loss of energy and pressure head. 3. What are different types of loss of head in fluid flow through pipeline? The loss of head in pipeline can be: (1) constant velocity head loss and (2) variable velocity head loss due to changing cross-section. The constant velocity head loss takes place due to: (1) friction in the pipe and (2) kinetic energy loss as fluid is flowing out of the pipeline with some velocity at the exist. The head loss due to friction results due to : (1) resistance offered by the roughness of the pipe surface when fluid layer slides over the wall of the pipeline and (2) resistance offered by the fluid layers to each other. In case the length of a pipeline is more than 500 times its diameter, then pipeline is termed as long pipeline and such pipeline has head loss due to friction as a major loss. When the fluid flows out of the pipeline at the exit, it has certain velocity which is head loss due to kinetic energy of flowing out fluid. It is a minor loss. Variable velocity head loss takes place due to variable velocity resulting in a pipeline whenever there is a change of cross-section inside the pipeline. For

Flow Through Pipes and Compressibility Effects

63 1

example, there are changes in cross-section due to: (1) pipe entrance (2) sudden enlargements (3) sudden contractions and (4) many pipe fitments. 4. How is the flow in pipeline classified? The flow in pipeline can be classified: (1) laminar or (2) turbulent. It depends upon the Reynolds number of the fluid flow. If the Reynolds number of the fluid flow is less than 2000, the flow is consider to be laminar flow in the pipeline. If the Reynolds number of the fluid flow is greater than 4000, the flow is considered to be turbulent flow in the pipeline. When the fluid flow through pipe runs full, the flow is considered to be turbulent. 5. What are major and minor energy losses in a pipeline? or What are the causes of lies of energy in a pipe? (UPTU-2005-6) The major energy loss in a pipeline results due to friction as the pipeline has generally long length. It depends upon: (1) roughness inside the pipeline (2) velocity of fluid and (3) diameter of pipeline. It is determined by Darcy-Weishech formula (hf —

f V2 L where f = d • 2g

friction factor, V = velocity, L = length and d = diameter) or Chenzy's formula (hf —

V2 L 2 cd

where C = Chenzy's constant). The minor losses result due to disturbance in the flow pattern. It occurs due to: (1) sudden enlargements (2) sudden contractions (3) entrance and exit (4) valve and (5) fittings. The minor losses are taking place as shown in below: (1) Sudden enlargement

2 2 i - V2) Head loss - hL - (V 2g

(2) Sudden contraction 2 Flow

--,--_---z--1

r.

632

Fundamentals of Fluid Mechanics

head loss hi, —

1 Cv

2

2 1) V or "=" 0.5 172 2g 2g

(3) At entrance of pipeline due to vena contracta

Vena contorta

Pipe line

head loss = 0.5 v 2 2g (4) At exit of pipeline

-_-_--_ } -_--_--_--- :-_--:-

/////////////// ----- ------ ----------

7771/2/7/7777/

head loss = V2 2g (5) At pipe bends

(6) Hindrance in pipe (A — a) A = Pipe cross section a = Hindrance cross section

2 2

head loss —

(Cc (A A —a)

1) 2g

Flow Through Pipes and Compressibility Effects

633

6. Derive an expression for the loss of head due to friction in pipes (UPTU-2002-3) Consider a horizontal pipe having uniform cross-section and steady flow. Applying Bernoulli's equation between section 1-1 and 2-2, we get -

Pi + V12 8 + 1722 ± Z1 L42 2g P 2g where hf = head loss due to friction -h-

-4-

1

2

Now Z1 = Z2, VI = V2 as pipe has uniform diameter. Hence we have — 111 — Pi — P2 Friction resistance as found by Froude experimentally is given by — F = frictional resistance per unit wetted area and per unit velocity = f' x wetted area x (velocity)2 = f x (71-dL) x V2 but perimeter = P = rid =fxPxLxV2 Now net force acting on fluid in direction x of the flow between section 1-1 and section 2-2 = P1A — P2A — F where F = friction force Since velocity between section 1-1 and section 2-2 is constant, hence net force is zero. Hence

or

P1A — P2A — F = 0 F = (Pi — P2)A or (P1—P2)A=fxPxLx V2

or

Pl — P2 f'xPxLxV2

or

hf =

"

Tr 117

=

4 LV 2 P d g

634

Fundamentals of Fluid Mechanics

I

Now put

p h i= '

= f 2

where f = Darcy coefficient of friction

4 • f • L •V2 — 2g • d

fi LV 2 2 • gd

where f= friction factor

The above equation is called Darcy Weishach equation for loss of head due to friction. 7. What is the expression for loss of head due to friction in terms of discharge in pipe? discharge Q = velocity x area or

Q = ird2 — 4Q area ird2

V—Q

4 As per Darcy's equation, the head loss due to friction is — 4 f • L .V 2 h — 2g • d f 4f • L ( 4

Q )2

gd2

2• g•d 4f • L • Q2 2 2R • d 5

16 • -

.h. • L • Q

2

12.1x•d5

Friction factor

8. What are the parameters on which friction factor is dependent? What are the expressions of friction factor for laminar and turbulent conditions?

Reynolds number —,Variation of friction factor.

Flow Through Pipes and Compressibility Effects

635

Friction factor is a dimensionless term which is used for finding friction loss in flow through a pipe. It depends upon two parameter viz (1) Reynolds number and (2) relative roughness of pipe surface. The friction factor increases with the increase of relative roughness. However the friction factor decreases with the increase of Reynolds number. Beyond certain value of Reynolds number, the friction factor becomes independent of Reynolds number. Expression for friction factor 1. Laminar conditions when Re < 2000 16

f - Re 2. Turbulent conditions when Re > 4000 0.0791 Re 1/4

f

9. Derive an expression of friction factor in term of shear stress. 1

2 .1—

P1

P2

T

2

Consider the flow in pipe between two sections 1-1 and 2-2. As net force in the direction of flow is zero, we can write —

Or

PIA — P2A = to x 7rd x L To X zd x L P1 — P2 — ird 2

4 P1

—

P2 —

LITO L d

P2 _ 4f LV 2 hi, — Pi — 2gd Pg

Now

4f LV _ 4T0L 2•g•d d

or or

f-

21-0

pv2

10. What is hydraulic radius? The hydraulic radius is the ratio of the cross-sectional area of the pipe to the wetted perimeter of the pipe.

636

Fundamentals of Fluid Mechanics

Area wetted perimeter

Hydraulic ratio (m) —

ird2 4 xird

d 4 11. Derive an expression for the loss of head using Chenzy's formula. The friction head loss has already been derived as under: hf= where

pg S

P xLxV2 x— A

P= perimeter, L= length, V= velocity and A= cross-sectional area.

A hydraulic radius m = —

Now

d 4 h f— or

Or

f • 1 .L•V2

pg

m

V2 — hf • Pg.ni f'xL V=

Now we can consider following: (1)

h

f = i = loss of head per unit length

(2) 1! f, = C = Chenzy's constant

f'

Hence

V= Clii•m

The above equation is known as Chenzy's formula. 12. Derive an expression for the relation between the friction factor and Chenzy's constant C. Friction head loss as per Darcy's equation is:

Flow Through Pipes and Compressibility Effects

f • L • V2 2• g•d

h— f

637

(i)

The Chenzy's formula is: V=

m•i

m = hydraulic radius = 4

where

4

i= hf

L

V—c

or

hf d L

4

h V2 — c2. f

d

4

L

or

hf —

4•172 • L 2

C

•

d

Equating eqn (i) and (ii) f LV 2

d—

hf — 2.

g

or

4V 2 L e2 .d

8.g f

c2

13. Find the head loss due to friction in a pipe 2 km long and one meter diameter when water is flowing with a velocity of 1 m/s by using Darcy's equation. Take f = 0.020.

- 2f •• gLV• d2 0.02 x 2 x 103 x 12 2x9.81x1 = 2.038 m 14. A pipe with diameter 15 cm has been laid horizontally which shows a drop in the hydraulic gradient of 8 m per km. Find the discharge if f = 0.02. hf = 8 m for

L = 1000 m

Using Darcy's equation, we get —

hi• - f2••Lg ••v2d

638

Fundamentals of Fluid Mechanics

8_

0.02 x 1000 x V2 2 x 9.81x 0.15

8x 2 x 9.81x 0.15 0.02 x 1000 V = 1.085 m/s

or

V2

or

Discharge Q= V•A = V ird2 4 1085 x 7r X (0.15)2 4 = 0.019 m3/s 15. Water is supplied to a town of 4,00,000 inhabitants. The reservoir is 6.4 km away from the town and loss of head due to friction in pipeline is measured as 15 m. Calculate the size of the supply main if each inhabitant consumes 180 liters of water per day and half of the daily supply is pumped in 8 hours. Take friction factor for pipeline 0.03. (Jodhpur University) Water required for inhabitants in a day is: 4,00,000 x 180 x 10-3 m3 The half of the above quantity is pumped in 8 hours. Therefore discharge Q per second is: 4,00,000 x 180 x 10-3 1 m3/s 2 x 8 x 3600 = 1.25 m3/s Discharge Q= Velocity x Area Q

1.25 = V x or

V=

7rd2

1.25 x 4 _ 1.592 ird 2 d2

Now friction head loss is — h — f

f xLxV 2 2g•d

0.03 x 6.4 x 103 x ( 1.592)2 d2 15 — 2 x 9.81xd d5 _ 0.03 x 6.4 x 103 x 2.534 2x 9.81x15

Flow Through Pipes and Compressibility Effects

639

= 1.657 d = 1.1 m

or

16. Derive an expression for the loss of head due to sudden enlargement. (UPTU-2002-3)

2 Sudden enlargement

Consider a pipe line in which there is a sudden enlargement of cross-section from A l to A2 (A2 > A1). Take two sections 1-1 and 2-2. As area is suddenly increased from Al to A2, fluid does not fill up the area (A2 — A1) initially and this area is filled up by reversed flow initiated by eddies at a later stage. The pressure is determined experimentally in area (A2 — A1) which is found to be equal to initial inlet pressure P1. The resultant force acting on the liquid between section 1-1 and 2-2 in the direction of flow is: = P1A1 — P2A2 Pi (A2 — A1) = (P1 — P2)A2 The net force acting on the fluid is equal to the rate of change of momentum in the direction of flow as per Newton's second law of motion. Hence we can write: (P1 — P2)A2 = pQ(V2 — V1) Q (V2 — VT) A2 •g

P2

Or

Pg

= V2 (V2 — VT ) Q — A2 V2 g Now apply Bernoulli's equation between section 1-1 and section 2-2, we have: pg

+

V12 + 2g

But

But

P2 pg

V22

2g

+ Z2 + loss due to expansion

Z1 = Z2, he = loss due to expansion p. p_ T7 2 2 v2 he — " pg 2 " 2g — P2

Pg

—

V 2 (V2 — Vi)

640

Fundamentals of Fluid Mechanics

h,— V2 (V2 — ) h, —

or

V12 V22 2g

21722 — 2171172 + V12 1722 2g 172 -2171V2 +1712 2g (

71 — 172 )

2g

2 as V1 > V2

17. Derive an expression for the loss of energy due to sudden contraction. or Explain head loss due to sudden contraction of a pipe. (UPTU-2006-7) 3

2

3 P1

171

Eddies 2 ' 3

Consider a flow through a pipeline having sudden contraction from cross-section area Al to Az(Al > Az). Consider three sections 1-1, 2-2 and 3-3. Section 3-3 is at vena contracta and head loss occurs only after the vena contracta when the jet of liquid enlarges from section 33 to section 2-2. Applying Bernoulli's equation between section 3-3 and section 2-2. h _ (Vc —V2)2

2g

(K. 2g 072

1722 ,

1)2

As per continuity equation, we have Ac Vc = A2 V2 or where

_ A2 1 V2 Ac Cc C, = coefficient of contraction

17c

)2

r2

he

2g

1

Cc

2

2

2g

2g

Note: In case Cc = 0.62, h, = 0.375 — /72 . If Cc is not given or known, then take h, = 0.5 172 —

Flow Through Pipes and Compressibility Effects

64 I

18. Describe the flow through a pipe bend. What are the causes of loss of energy in pipe (UPTU-2001-2) bend? Vena contracta at section 2-2

1

Pressure variation at 1-1

Velocity variation at 1-1 Eddies

Flow Flow through a bend

Consider a fluid flowing through a bend. The stream lines are curved as per the boundary of the bend and a pressure gradient is created such that the pressure at the other radius where velocity is lower is more than the pressure at the inner radices where the velocity is higher. The pressure at the inner radices of the bend decreases further with the contraction of stream lines while constituting a vena contracta, resulting separation of the flow at the inner wall of the bend. From vena contracta, the expansion of fluid flow takes place accompanied with the formation of eddies and head loss similar to head loss taking place in sudden contraction in flow. The head loss for flow through a bend can be reduced by using a bend of larger radius which minimizes the effect of contraction and expansion of fluid flow. The loss of head in bend is given by — k •V 2 where k = Coefficient of bend 2g The coefficient of bend depends upon radius of curvature of bend and diameter of the pipe. hb —

19. How is the loss due to pipe fittings accounted for? It is customary to relate the losses of all pipe fittings to inlet velocity head by specifying constant k for the pipe fittings. The loss of head for a pipe fitting is — k U2 2g It is also a general practice to relate the head loss due to pipe fittings in the term of equivalent lengths of the pipeline. head loss —

head loss =

Or

Lego =

kU 2

f Leq,U 2

2g

c1.2g

d •k

642

Fundamentals of Fluid Mechanics

It is advantageous to use the concept of equivalent lengths for the pipe fittings as these equivalent lengths can be added to the length of the pipeline to get the gross length. Gross length = actual length + (Leqv)i + (4,102 20. Give the values of k for different pipe fittings. SN.

Pipe fitting

K

Velocity to be used

1. Sudden enlargement

(1 — ' .1 j A2 0.5 1 2 0.2 0.5 0 to 0.05 1

2

2. 3. 4. 5. 6. 7. 8.

Sudden contraction Elbow Tee Gate valve Entrance (sharp edge) Entrance (bell method) Exit

U1 U2

U U U U U U

21. What is total energy line or energy gradient? (UPTU-2005-6, 2006-07) The total head or energy is given by Bernoulli's equation as — + V2 + Z 2g When the fluid flows along the pipe, there is a loss of head (energy) in the pipe. Hence total energy decreases in the direction of flow. If the total energy at various points along the axis of the pipe is plotted and joined by a line, then the line so obtained is called the energy gradient (EGL) or total energy line (TEL) Total head =

P pg

22. What is hydraulic gradient line (HGL)? (UPTU-2005-6, 2006-7) The sum of potential energy head from a datum line and pressure head at any point is called the piezometric head. If a line is drawn joining the piezometric levels at various points in the flow, the line thus obtained is called the hydraulic gradient line (HGL). 23. What are the important aspects of total energy line (TEL) and hydraulic gradient line (HGL)? The important aspects of TEL and HGL are: 1. TEL always drops in the direction of flow due to loss of head occurring in the direction of flow. 2. HGL may rise or fall as it depends upon pressure changes. HGL rises during flow through a sudden expansion and falls during flow through a sudden contraction. 3. HGL lies below the TEL and vertical distance between TEL and HGL is equal to the velocity head. 4. In pipe of uniform cross-section, there is no variation of velocity. Hence the slope of TEL and HGL is same.

Flow Through Pipes and Compressibility Effects

643

5. There is no relation between TEL and the slope of pipe. 6. If a segment of pipeline lies above the hydraulic gradient line, these must be negative pressure in that segment and such segment is said to constitute a syphon. 24. Water is delivered by a 15 cm pipe at the rate 60 liters/s. Calculate the pressure difference between the two points 300 m apart on the same horizontal line. If one point is 20 m higher than the other, what will be the difference in pressure? Take f = 0.22. Draw hydraulic gradient and total energy line. P1 ± Viz ± zi — P2 ± v22 + Z1 ± hf 2g pg 2g Now Z1 = Z2, V1 = V2 as diameter is constant

pg

hf = A — P2 pg But

4PL•V2 _ fLV 2 h — 1 2g • d 2• g•d

and

V=

Q

re d 2 4 = 3.4 m/s

60 x 10-3 x 4 ir x 0.152

0.022 x 300 x (3.4) 2 2 x 9.81 x 0.15 = 25.92 m

hf—

(1) Pressure difference between two points is — P1 — P2 = pghf = 1 x 103 x 9.81 x 25.92 = 254.2 kN (2) One point (inlet) is higher by 20 m, then we have P2 — PI — hf 20 Pg = 25.92 — 20 = 5.92 The pressure difference is — P2 — P1 = 5.92 x pxg= 5.92 x 1 x 103 x 9.81 = 58.07 kN (3) Drawing hydraulic gradient and total energy line. Hydraulic grading line is the line joining the points P1 and P2 . Total energy line is Pg Pg parallel to the hydraulic gradient line and it can be obtained by adding velocity head = V2 • 2g

644

Fundamentals of Fluid Mechanics Hydraulic gradient line Total energy line

2g

Hydraulic gradient & total energy line

25. Water issues from a tank under pressure through a 5 cm diameter pipe 500 m long. The discharge end of the pipe is left at 2 m under the surface of an open channel. Calculate the discharge and sketch the TEL and HGL. Assume f = 0.02, the gauge pressure of air is 5 m of water. Minor losses such as line entrance and exit head losses are negligible as compared to friction loss in a long pipe line. h _ f LU 2

.f d • 2g Now apply Bernoulli's equation between point 1 and 2, and horizontal line through point 1 is taken as datum line. p

v2

j

pg 2g

p

Z =

pg

2g

+Z2 + h f

5xpxg 2 xpxg and , P2 pg pX g = V2 = U 2 f Lu 2 5 + 5 + 0 = 2 + u —5+ 2g d • 2g

= 0, Z2 = —5,

U

2

2g 0.05 u2 +

=5±

f Lug d • 2g

13

0.02 x 500 x u2 — 13 0.05 x 2 x 9.81 0.05 u2 + 10.19 x u2 = 13 10.24 u2 = 13 Or u2 = 1.269 or u = 1.126 m/s Q=Axu -

x (0.05)2 x 1.26 = 2.21 x 10-3 m3/s 4

Flow Through Pipes and Compressibility Effects

2 i

645

B

2g

Datum

TEL & HGL

TEL starts from point B (5 + 5 + U2 and HGL starts from A (5 + 5) from datum line. g 26. What is equivalent pipe? (UPTU-2001-2)(UPTU-2002-3) or Discuss & derive the expression for equivalent length (UPTU-2205-6) A compound pipe consists of several pipes of different lengths and diameters. These several pipes of different lengths and diameters can be replaced by a pipe of uniform diameter and one length. Hence equivalent pipe is a single pipe of uniform diameter whose discharge and loss of head are same as that of the compound pipe. The condition for an equivalent pipe of a compound pipe is: (a) Head loss is same. hfeqv = 1111 + hh + or

f LQ2 _ 12.1x c/ 5 12.1xdi

f21,2Q2 + f31-5Q2

12.1x di'

12.1x ca

(b) Discharge is same. i.e. Q is same for all lengths (c) L = Li + L2 L3... 27. What is Duput's equation for the equivalent pipe of a compound pipe? The compound pipe is generally made of several pipes of same material and smoothness. In case friction factors f = fi = f2 = f3, then we have — hfeqv = hf1

f LQ2 12.1x c/5

hf2 h13 2 + f2 Q2 f1l-4Q

12.1x di

12.1x

+ f31,3Q 12.1x

2

646

Fundamentals of Fluid Mechanics

Now apply f = A = f2 = A, then we get — 5 5 d5 Cii d2 The above equation is called Duput's equation.

5

d3

28. What are discharge and head in pipes in parallel? —°- Qi —..Pipes in parallel

If a pipeline divides into two or more pipes for some distance and these pipes again join together at some distance at downstream, then these pipes are said to be in parallel. In order to increase the discharge, pipes are connected in parallel. The parallel pipes follow the following conditions — (1) discharge in main pipe is equal to sum of discharges in each parallel pipes. Q = Qi + Q2 (2) loss of head in each parallel pipe is same. hfi = hh fiLiQi _ f2L2Q1 12.1x cg. 12.1• d2

or

29. At a sudden enlargement of a water main from 240 mm to 480 mm diameter, the hydraulic gradient rises by 10 mm. Calculate the rate of flow. (Punjab University, AMIE) The rise of hydraulic gradient is — _ ( P2 ± z ) _ ( P1 ± L ,--. j = 0.01 2 1 Pg Pg Now apply Bernoulli's equation p ± 2± i v1 z

pg

2g

where

p 2

M

2 v2

1 pg ± 2g ± Z2 + he

he = head loss due to sudden enlargement —

2 2 v1 v2he = ( P2 +Z2 ) ( Pi +z1 = 0.01 2g 2g pg pg

WI — i72 ) 2 2g

Flow Through Pipes and Compressibility Effects

or

V12

—V2 2g

(V1 — V2)2 — 0.01 2g

But

647

(i)

A1 71 — A2 72

A2

Or

1

=

( D2 )2

7_

DI

2

(00..2448)

2 72

= 4 V2 Putting the value of V1 in eqn (i) Or

161722 —V2 —9V2 — 0.01 2g 6V2 2 — 0.01 2g

or or

722 —

0.01x9.81 8

Or

V2 = 0.181 m/s

Now

Q = A2V2 = ij x

(0.48)2 x 0.181

= 3.275 x 10-2 m3/s 30. In a pipe diameter 350 mm and length 75 m water is flowing at a velocity of 2.8 m/ s. Find the head loss due to friction by using (1) Darcy's equation (2) Chenzy's formula in which C = 55. Take v = 0.012 x 10-4. Darcy's equation h

_ fLV 2 d •2g

Now

Reynolds number Re = Vd 2.8 x 0.35 0.012 x 10-4 = 8.17 x 105 > 4000

648

Fundamentals of Fluid Mechanics

0.0791 f — (Re )1/4

Now

—

2.63 x 10-2

2.63 x 10-2 x 75 x (2.8)2 0.35x 2 x 9.81 = 2.25 m

hf Chenzy's formula V = c-N/Wi c = 55

Given

m—

Area parimeter

xd 2 _ d 4 x ird 4

= 0.35 = 0. 0875 4 2.8 = 55 x.0.0875 x i or

i—

(2.8)2 1 x (55)2 0.0875

= 0.0296 =ixL hf = 0.296 x 75 = 2.22 m 31. In a 80 mm diameter pipeline an oil of sp. gr. 0.8 is flowing at the rate of 0.0125 m3/ s. A sudden expansion takes place into second pipeline of such a diameter that maximum pressure rise is obtained. Find (i) loss of energy in sudden expansion (ii) Differential gauge length indicated by an oil-mercury manometer connected between two pipes. (UPSC) Q = 0.0125 m3/s vi

_ Q

_

0.0125

i i X (0.08)2 . = 2.488 m/s A

By Bernoulli's equation P1 ± 1712 ± z _ P2 ± V2 2 ±z±h pg 2g 1 pg 2g 2 e [ 2 v_2 2 h P2 — P1 = AP = Pg vig 2 2g '

Flow Through Pipes and Compressibility Effects

As

(V1 V2)2 2g

Z1 = Z2 and he —

AP = pg [Viz V2 (V1 — V2)2 2g 2g 2g

Or

2 , 2 vi [i (v2 ) _ (1 _ (V2)2 ) 1 2g vi

T7 2

Ap _

Or

Now A1T/1 = A2 72

)2

V2

Or

D2 OP =

pV? [1 (

2

4

(1

D2 )

D01

21

D2 )

pV1z [2( Di )2 2 ( ))41 2 D2 ) D2

aAP

For max pressure rise

)

=0

D2 = 1 D2 D2 = jD1 = Vi X 0.08 = 0.113 m

or

—

— A2

0.0125

4 x (0.113)2

= 1.244 m/s and V1 = 2 x V2 = 2.488 m/s Now loss of head is — he =

(V1 — V2 )2 — [(2.488) —1.24412 2x9.81 2g

= 0.079 m of oil Now

P2 — P1

Pg

V12 V22 he

2g

649

650

Fundamentals of Fluid Mechanics

(2.488)2 - (1.244)2 2g

0.079

= 0.068 - 0.079 = 0.158 If h is reading of manometer, then P2 - P1 =h PS

[ s ll g soi i

11

0.158 - h[ 103:-1] = 16.h .

h - 0.158

Or

16 = 0.0098 m = 9.8 mm

32. Two similar pipes of the same diameter of lengths l and /2 are placed in parallel. Calculate the equivalent length of single pipe of same diameter. What should be the equivalent length if the two pipes are equal in length? 0 2

Discharge = Q= Qi + Q2 and hf = hfi = hh = head loss remains same

f 1-2Q1

_ 12.1x d1 Now

12.1x

dl = d2 _

Q2 1-2 If the equivalent length of the pipeline = Le f40

hf Now

2 12.1. d 5

LeQ2 = LiQ? = L2Qi Q =Q1+ Q2=

Q1 ± Q1111-2

Flow Through Pipes and Compressibility Effects

65

1 +1/ Ql = 1.14

but

Li

Q Le _

There

1+ A L2 1 -4

Le =

or

[1+ when

L1 )2 L2

L1 = L2 = L Le —

(2)2 = 0.25 L

33. The pipes of the same material and of equal lengths are used for connection to an overhead tank which supplies 85 x 10-3 m3/s of water. If diameter of pipes are 30 and 15 cm respectively, find the ratio of head losses if pipes are connected in series and parallel. Neglect minor losses. Series. length of each pipe line = L and Q remains same h —

f

hf —

f x Lx Q2 12 x a' f xLx (85 x 10-3)2 12 x (0.3)5 fL

— 12

f xLx(85x10-3)2

(2.97 + 95.14)

= 98.11 x fL 12 Parallel

Q = +Q2 "4

f L Q? _ fL 12 (d1)5 12 (d2)5

12 x (0.15)5

652

Fundamentals of Fluid Mechanics

A . Q? _ fL Q2 12 (0.3)5 12 (0.15)5 But

or

Qi = 5.66 Q2 . 85 x 10-3 = Qi + Q2 Q2 ± 5.66 Q2 = 8.5 X 10-3 8.5 X 10-3 = 12.7 x 10-3 m3/s 6.66

or

Q2 —

and

Qi = 5.66 x 12.7 x 10-3 = 97.2 x 10-3 m3/s hf = hfi = hf2 —

f L (97.2 x 10-3)2 x 12 (0.3)5

_fL 2.15 12 Ratio of series to parallel head loss L x98.11 — 12 L x 2.15 12 = 45.6 34. A sudden enlargement of water main is from 24 to 48 cm diameter. The hydraulic gradient rises by 1 cm. Estimate the rate of flow in the pipeline. (UPTU-2001-2) Hydraulic gradient = ( P2 + Z2 ) — ( Pi + Zi Pg Pg = 1 x 10-2 m Now using continuity equation at enlargement, we get A 1 71 = A2 72 g x (0.24)2 71 = x (0.48)2 72 4 4 or 71 = 4V2 he —

(V1 — V2 ) 2 2g

Now as per Bernoulli's equation, we have

( 4 V2 — V2 ) 2 2g

—

91722

2g

Flow Through Pipes and Compressibility Effects

P1

1712

v22 he 1 pg +2g + Z2 ±

Z — P2

Pg 2g (P2 + Z2 ) pg

653

Pi +Z1) — v 2 — v pg 2g 22:

1 x 10-2 —

161722 — v22 2g

h '

9v22 2g

6V22 2g /2g x 0.01 6 1 = 0.1808 Discharge Q = A2 V2 Or

, r2

x (0.48)2 x 0.1808 4 = 32.75 x 10-3 m3/s =

35. What water level H should be maintained in the water tank so that the discharge of 0.5 x 10-3 m3/s can be obtained from the pipe attached at the side of the tank of length 25 m and diameter 25 mm. Assume f = 0.02

L= 25 m

Q = 0.5 x 10-3 Q = AV = Area x Velocity V=

H=

0.5 x 10-3 (0.025) x 4

= 1.02 rills

= head loss —

2 +0.5 17 2g 12.1x d 5

f Le2

0.02 x 25 x (0.5 x 10-3)2 12.1x(0.025)2

0.5 x (1.02)2 2 x 9.81

654

Fundamentals of Fluid Mechanics

= 1.057 + 0.027 = 1.084 m Water level to be maintained is H = 1.084 m

36. Water level in a reservoir is 10 m above the outlet. Water is discharged through a compound pipe consisting of 30 m long pipe segment of 100 m diameter and 15 m segment of 200 mm diameter. Both segments are horizontal and f values are 0.02 and 0.022 respectively. Find (1) discharge and (2) discharge when minor losses are neglected.

T E

0

= 30 m

L2 = 15 m

As per continuity equation, we have A 1 V1 = A2 72

Or

V1 —

4 x rcd? x 4 x 1rd1

Vl

2 ( 0.2 ) 0.1) Vz

V2 = 0.25 V1 H= ± hr,2

Major loss is friction loss in both pipes. tf j

2 g • di

f2 172,2 2 g • d2

0.02 x 30 x V12 0.022 x 15 x (0.25V1)2 2x9.81x0.1 2x 9.81x0.2 = 0.306 Vie + 0.0053 V12 = 0.311 V12 Minor loss = head loss at entrance + head loss due to sudden expansion _ 0.5 V2 + (V —117)2 _ 2g 2g

Flow Through Pipes and Compressibility Effects

655

0.51712 + (V1 — 0.25V1) 2 2g 2g 0.5+ 0.563 2 V1 2g = 0.054 V12 Total head loss = head available in water tank Total head loss = major loss + minor loss = 0.311 V12 + 0.054 V12 = 0.365 V12 r,,,,.? =

or

10 — 27.397 0.365

. V1 = 5.23 m/s Discharge Q = A l V1 = 4 x (0.1)2 x 5.23 = 4.11 x 10-2 m3/s In case minor losses are neglected, then — Head available = head loss due to friction only 10 = 0.311 V12 or V1 = 5.67 m/s Now discharge Q = A l V1 = 4 x (0.1)2 x 5.67 = 4.45 x 10-2 m3/s 37. Water in a tank is maintained at constant head of 4 m. Water is discharged through a horizontal pipe 100 m long and 100 mm in diameter. There is a valve at the end of the pipe to control the discharge. When valve is half open, discharge is 12 x 10-3 m/s. Find the value of loss coefficient of the valve if f = 0.026.

T 4m

Dia = 100 mm (/)

i L = 100 m

656

Fundamentals of Fluid Mechanics

V—

Q 12 x 10-3 x 4 — = 1.52 m/s A z x (0.1)2

Total head loss = major losses + minor losses = Loss due to friction + loss at entrance + loss at valve. Total head available at tank = total head loss in pipeline

4

_ f xLxV2 + a5 2 +k V2 V 2.g•d 2g 2g

where K = loss coefficient of valve

0.026 x 100 x (1.52)2 0.5 x (1.52)2 k (152)2 + + 2x 9.81 x 0.1 I 2 x 9.81 2 x 9.81 4 = 3.06 + 0.059 + 0.117 K 0.117 K = 0.881 or K = 7.53 38. Water flows through a pipe 100 mm in diameter with discharge equal to 16 x 10-3 m3/s. The pipe line consists of straight length of 50 m, one valve (K = 10), one elbow (K = 2) and one '1' outlet (K = 1.5). What is the equivalent length of the pipeline? Find also total loss of head if f = 0.02. Q = 16 x 10-3 m3/s AV = Q where A = area, V = velocity or

V = 16x10 3 x4 — 2.038 m/s ir x (0.1)2

Major loss = friction loss = hf —

or

fLV 2 2gd

h _ 0.02 x 50 x (2.038)2 f 2 x 9.81x 0.1

Minor losses = losses due to fittings 11,7, = (K1 + K2 + K3) V2 2g (10+2+1.5) x (2.038)2 2 x 9.81 = 2.85 m Total loss of head = lif + h„, = 2.12 + 2.85 = 4.97 m Now equivalent length for minor losses

2.12 m

Flow Through Pipes and Compressibility Effects

657

le = Kd f (10 + 2 + 1.5)(0.1) 0.02 = 67.5 m Total equivalent length of the pipeline Leqv = L + le = 50 + 67.5 = 117.5 m 39. Find the diameter of uniform pipe to replace a compound pipe having three segments of 900 m, 450 m and 300 m of 45 cm, 37.5 cm and 30 cm diameters respectively. Total length of pipe is same as that of compound pipeline. 4-FL2+4 _ 4 + La + L3 d5 (115 d52 a35 900 + 450 + 300 d5

900 + 450 + 300 (0.45)5 (0.375)5 (0.3)5

1650 — 48,700 + 60,800 +12,5,00 d2 = 2,34,500 d

or

_ ( 1650 )115 2,34,500 = 0.37 m

40. For the distribution main of a town water supply, a 25 cm main is required. As pipes above 20 cm diameter are not available, it is decided to lay two parallel mains of same diameter. Find the diameter of the parallel mains. (Punjab University) fxLxQ2 12 x (cos Q = Qi ± Q2 = 2Q1 as Qt = Q2

hf— Now or

Q

Q1 = —

(Q)2 f xlxQ- — 12 xd5

f xlx

2

12 x (di)5

658

Fundamentals of Fluid Mechanics

or

(d1)5 =

x d5

1 x (0.25)5 =4 0.025 _ 0.25 1.26 (4)1/5 = 19.84 cm

or

Hence nearest standard diameter is 20 cm. Two 20 cm diameter pipes are to be laid. 41. A special pipe of diameter d carrying discharge Q at upstream is arranged to discharge q per unit length uniformly along its length. Derive an expression for head loss over a length L and head loss for entire length. qqqq

Fli

ri

Fli

r

Q

x There fall of discharge q' per unit length. If Q is the initial flow in the pipe, then remaining discharge after a distance x is — Qx=Q — q .x Suppose this discharge remains for distance dx, then head loss in this small length is — dhf

f • dx • (Q — q x)2 12d 5

By integrating, we can find head loss for complete length L. dhf =

hf —

f : 12 d 5

12 x d

5 [Q2 .x _ 2. q

23 L

2

Q ± q ;

1

0

2 L2 1 q L2 i+ q 12 •d 5 [ 3Q2 Q .i. Q2 L

=

(Q q x)2 d x

r

J

The flow keeps on discharging q per unit length till it becomes zero after length Lmin. Q = q X /min or q — 1 Q Lmin

Flow Through Pipes and Compressibility Effects

- f Lnun x Q2 (1 + 12 •d 5

3

659

—

f x Q3

36 • d5 • q 42. What is pipe networking? What are the conditions to be satisfied by pipe network?

Networking

A group of pipes which are interconnected and forming several circuits or loops is called a pipe network. Networking is a common method in water distribution system adopted by the municipalities of the cities. In networking of pipes, following conditions are always satisfied: (1) Continuity equation. At each junction of pipes, total inflow is equal to outflow as per the principle of continuity. (2) Energy equation. The loss of head due to flow in each loop in clockwise direction must be equal to the loss of head due to flow in anticlockwise direction. In the figure the head loss in the clockwise flow from A to E must be equal to the head loss in anticlockwise from AFE. In other words, the algebraic sum of head losses in any closed circuit within the network must be zero. (3) Darcy's equation. The head loss in each pipe of the network for simplicity can be expressed as hf

= r Q2 where r —

fL 12.1 x d'

The flow in each pipe must satisfy the

relation between the head loss and discharge at all times. 43. Describe the Hardy cross method for solving pipe network problems. The pipe network problems are difficult to solve analytically. Hence the method of successive approximation is used which is commonly known as Hardy cross method. Following steps are used in this method: (1) a trial distribution of flow is made for the network in such a way that the continuity equation is always satisfied at each junction or node. (2) head loss with trial distribution of flow for each pipe in network is found out. Head loss in each pipe is hf = kQ2 where k —

L 12.1. d5

660

Fundamentals of Fluid Mechanics

(3) apply the condition and see whether the sum of head losses in each loop is zero i.e. Eh.), = y, kQ2 = 0. (4) In case sum of head losses in a loop is not zero, change the value of the flow so that the sum of head loss of flow may become zero. New flow is Q = Qo + AQ and Ihf IcQ2

= y,K(Qo + A02 = 0 and AQ —

2 KQ (5) The above procedure has to be repeated till we get Ihf = 0. 44. Find the discharge in each pipe of the network as shown in the figure. The value of `k' for each pipe is also shown. The discharge from the nodes is also shown (QA = 20 x 10-3, QD = 15 x le, Q, = —40 x 10-3 and QB = 45 x le m3/s). 20

A

D

k=7

15

45 4

B

40

C

Guidance. Hardy cross method is to be used. Take loop ABC anticlockwise and CDA loop clockwise. Assume flow in each pipe of the loop. Find ly= KQ2 for each pipe. The value of hf is positive in the direction of assumed flow and negative in opposite direction. The value of Ihf of the loop is worked out. In case Ihf is zero, the assumed values of flow in each pipe is correct otherwise find change in flow AQ which is given by — y,

KQ2

I 2KQ

We assume the flow in each pipe of the loops such that the IQ = 0, at each node. The assumed flow in each pipe is as shown in the figure. 20

15

45

40

First trial Loop ABC

Loop CDA

Pipe

K

Q

hf = KQ2

2KQ

Pipe

K

Q

hf = KQ2

2KQ

AB BC

6 4

15 30

—1350 3600

180 240

CD DA

6 7

5 10

150 700

60 140 Contd.

Flow Through Pipes and Compressibility Effects CA

8

5

—200 80 1 = 2050 1 = 500 —2050

AQ

500

AC

5

8

200 80 I = 750 1 = 280

V =—

= —4

66 I

750 — —3 280

Second Trial The corrected flow i.e. Q ± AQ in each pipe of the network is as shown in the figure. 10-3=7

A ,..

20

D

15 +4 = 19

45 1

.-15

5 +3 =8

B

30-4 = 26

40

C

Loop ABC Pipe

K

AB BC CA

6 4 8

Q 19 25 6

Loop CDA KQ2

—2166 2704 —288 X = 250

Pipe

K

Q

238 208 96 I = 532

CD DA AC

6 7 8

8 7 6

250 = —0.5 532

AQ

Third Trial. figure.

2KQ

K

AB BC CA

6 4 8

2KQ

—384 96 343 98 286 96 I = —12471 = 290 —247

V

290

= —1

The corrected flow i.e. Q ± AQ in each pipe of the network is shown in the

Loop ABC Pipe

KQ2

Loop CDA KQ2

Q

19.5 —2281.5 2601 25.5 55 —242 I = 77.5

V

—72.5 526

2KQ

Pipe

K

Q

KQ2

2KQ

254 204 88 I = 526

CD DA AC

6 7 8

9 8 5.5

—486 202 242 I=8

108 84 88 X = 280

—

0.15

AQ =

—8 280

= —0.03

662

Fundamentals of Fluid Mechanics

The final flow in each pipe in the network is — 20

A

5.97

D

19.65 y

45

15

9.03

B

40

C

25.35

45. What is a syphon? Where is it used? Explain its action. Derive an expression for the length of its inlet leg. (UPTU-2002-3) Syphon is an arrangement of pipe system by which water can be made to run up the hill utilizing the force of atmospheric pressure. It is bend pipe connecting two water surfaces at different levels and separated by high ground over which the pipe is laid as shown in the figure. The water will flow from high water level surface (surface A) to the lower water level surface (surface B) even when an obstruction in way of summit C is present. The pressure at point C in pipe is less than atmospheric pressure as it lies at a level which is above the free surface of the water at point A. Since atmospheric pressure is equal to 10.3 m of water, the pressure at C can be theoretically reduced to —10.3 m of water but it is limited to —7.6 m of water to avoid air and dissolved gases separating from the water and getting collected at point C (summit) of the pipeline which are likely to obstruct line flow of the water. The inlet leg is infact the pipe line from the reservoir A to summit C and the outlet leg is the pipe from summit C to reservoir B. Summit

TEL

HGL

\\, Obstruction like bill

Syphon

Syphon can be used for: (1) To take water from one reservoir to another reservoir located at lower level but when they are separated by high obstacle like hill or ridge. (2) To draw out water from a channel without any outlet to lower ground as shown in figure.

Flow Through Pipes and Compressibility Effects

663

Syphon

Low ground Channel

Flow

Drawing out water from a channel

(3) To draw out water from a tank having no outlet as shown in figure.

Drawing out water from a tank

(4) To connect two open canals by inverted syphon as shown in figure. / HGL

i

Syphon

Obstacle like big ditch Inverted syphon connecting two channels

Principle of working Negative pressure is created at summit C of the syphon so that water can be pushed towards summit by the atmospheric pressure acting on the free surface of the water. The flow in the syphon is maintained till the negative pressure remains at summit C. The flow and velocity of the flow when started in the syphon depends upon the difference of water level at point A and B. It does not depend upon the level of C of the syphon till negative pressure is maintained at point C. On applying Bernoulli's equation between point A and B, we get PA

+

P'8 As

VA

2g

± Z _ PB ± VB2 ± Z ± losses B A p•g 2g

PA — PB — 0 and VA = VB, we have P•g P.8 ZA — ZB = losses Head loss = loss of head at entry + loss of head

664

Fundamentals of Fluid Mechanics

due to friction + loss of head at the exit = 0.5 v2 + f Lv 2 v2 2g 2g.d 2g Now apply Bernoulli's equation between point A and C, we have A ± VA2

P•g

2g

±

Vc2 + — + Zc + losses p•g 2g PA = 0, VA = 0, Vc = V z PC A

Zc — ZA

0.5172 fLACV2 2g 2• g•d

Vc 2g

=- pg = —hc T7

PC Pg

v

2

(13 + f LAC)

2g

Inlet leg and outlet leg Apply Bernoulli's equation between point A and C z

PA +

pg2g A

z h + pg2g c f

PA = Pa = atmospheric pressure Pc = 0 = absolute zero pressure VA 0 Vc = Vl = inlet velocity to syphon

hf - 2f • g • d

where ll = inlet length

2

ZA 0 +

Pa ± 0

Pg

2g

+ Zc +

f V12 2.g.d

vi2 = 2gd [pPag—(zc —ZA )1 f . ii Now apply Bernoulli's equation between point C and B or

2

(i)

2

Pc Vc B + ZB + hf — c = PB+V pg 2g + Z pg 2g Pc = 0, VB= 0, Vc = V2, PB= Pa = atmospheric pressure and 2 v2

f 1722 L2 + Zc = Pa + 0 + ZB + 2.g.d 2g Pg

T7

0+

f •V22 1-2 2g .d

Flow Through Pipes and Compressibility Effects

665

2 • g • d I -( 4. zB) Pa l (ii) Pg f • La L Now the velocity V1 must be equal or greater than V2 so that negative pressure is maintained at summit C with outlet leg flowing full always. Or

rq

V1 V2 [ Pa — (4— ZA )1 Li < pg 4 [(Zc — ZB ) — Pa l Pg

or

46. The difference in the water surface levels of two reservoirs which are connected by a syphon is 8 m. The length of the syphon is 600 m and its diameter 30 cm. Assuming f = 0.02, determine the discharge when the syphon is running full. If the vertex of the pipeline is 5 m above the surface level of the upper reservoir, determine the maximum length of the inlet leg for the pipe to run full. Neglect all losses other than that of friction. (Poona University) I

4B

H=8m

2

f I, V 2

H= 1.5 V + where H= 8 2g d 2g ' or

V—

1 2 x9.81x8 0.02 x 500 1 15 + 0.3 = 1.945 m/s Discharge Q=A•V= fi. • (0.3)2 x 1.945 = 0.1374 m3/s Inlet leg

666

Fundamentals of Fluid Mechanics

P. — (Z, — Z A) — g

4
2

—4— a i_

Karman vortex trails

During stable conditions of shedding vortices, Karman found that — a (1) = 0.281 / (2) Vortex trail moves downward with a velocity of ut which is — 0.355 F where F is the circulation around the vortex. a (3) the frequency of vortex shedding is — ut —

f = 0.198 11. (1— 19:7 ) where D = diameter of cylinder

D

Re

fD = 0 198 (1 19:7 ) Re ) U .

or f

is known as Strouhal number. As Reynolds number varies from 120 to 104, Strouhal

number varies from 0.168 to 0.198 for a cylinder. 17. What is an aerofoil? What is its purpose? An aerofoil is a streamlined body designed in such a way that a separation if occurs, it should be near the trailing edge of the aerofoil. As small wake is allowed to be produced which results into a small pressure drag acting on the aerofoil. Even when flow has high Reynolds number, pressure drag is very small but the skin friction drag is large. The resulting force on the aerofoil due to small pressure drag along the flow and large friction drag perpendicular to the direction of flow is also acting nearly perpendicular to flow. A large lift is obtained from an aerofoil. Resulting force

Force on an aerofoil

720

Fundamentals of Fluid Mechanics

18. Explain (1) chord line and chord (2) angle of attack (3) chamber (4) profile centre line (5) aspect ratio for an aerofoil and (6) stall. Mean camber line Upper camber line

Leading edge ...... Flow —1.-I.-

Angle of attack (a)

Trailing edge Lower camber line Line and chord

(1) Chord line and chord : Chord line is an imaginary line which can be draw on from leading edge to the trailing edge of a cross section of an aerofoil. The length of the line is called chord of the aerofoil. (2) Angle of Attack. The acute angle between the chord line of the aeirofoil and the direction of the flow. (3) Camber. Camber is the curvature of the aeirofoil from the leading edge to trailing edge. The camber can be (1) upper camber (2) lower camber and (3) mean camber which is the mean line that is equidistance from the upper and lower surface. In the case of a symmetrical aeirofoil, the profile centre line concides with the chord line. (4) Profile Centre Line. The imaginary line joining the mid-points of the profile starting from leading edge to exit at the trailing edge is called a profile centre line. (5) Aspect Ratio. It is the ratio of span of the aerofoil (0 to the length of mean chord (c)

Aspect ratio =

span / — mean chord c

(6) Stall. When the angle of attack of an aerofoil is greater than the angle of attack for maximum lift, the aerofoil is said to be operating in the stall condition. At stall condition, the flow separates from the aerofoil and eddies are formed, thereby the drag coefficient increases considerably.

19. Draw and explain the flow around an aerofoil when subjected to : (1) uniform flow (2) circulatory flow and (3) combination of uniform and circulatory flows. State how the stagnation points vary.

Flow Past Submerged Bodies

72 I

Aerofoil in uniform flow

Aerofoil in circulatory flow

Aerofoil in combined uniform and vortex flow

The above figure shows uniform flow around an airflow. Two stagnation points are formed under irrotational flow conditions. S1 is the stagnation point formed at the leading edge of the aeirofoil and S2 is the stagnation point formed at the upper surface of the aerofoil before the trailing edge. The flow on the upper surface of the aerofoil is unstable near the trailing edge as in this region, the flow is taking place against positive or adverse pressure gradient conditions which encourage the separation of the flow from the surface of the aerofoil. Hence stagnation S2 takes place before the trailing edge which moves to the trailing edge as the velocity of flow increases. On further increase of velocity, it gives rise to the formation of vortices. The vortices create circulation around the aerofoil and flow pattern

722

Fundamentals of Fluid Mechanics

around the aerofoil becomes similar to what is obtained when aeirofoil is subjected to combined flow of uniform and vortex flow. The rear stagnation point S2 is now at trailing edge. There is distortion of the flow pattern about the aeirofoil with stream lines are closer on the upper surface and stream lines are wide spaced at lower surface. Hence the flow pattern is asymmetrical about the central axis and aerofoil is now subjected to a lift force. 20. Define aeroplane. When will it be said to be in equilibrium? Derive expression for velocity and propulsion power. Lift

Propulsion Drag

or thrust

Weight Equilibrium of aeroplane

An aeroplane is a flying machine which is although heavier than air but it is supported by the lift provided by the flow of air on its wings having the shape of an aeirofoil. When aeiroplane is in steady level flight, it is in equilibrium as shown in the figure. In equilibrium, lift forces are equal to the weight of the aeroplane and drag force must be overcome by the thrust or propulsion provided by the engine/jet engine of the aeroplane. Lift = CI x 1pU2 X A But during equilibrium Weight = Lift W = CL x —1 pU 2 x A .

Velocity of plane = Velocity of air = u U

= 11(W A) p .2CL

Power required for propulsion P = FD X A = CD X

pU 3

XA

21. Explain the variation of the coefficient of lift and the coefficient of drag with the angle of attack during the flow over an aeirofoil.

Flow Past Submerged Bodies

723

Stall point

CDI I CL

Angle of attack (a)

As the angle of attack increases, lift forces increase and lift coefficient reaches maximum. After this, the lift coefficient decreases with higher values of the angle of attack. The aerofoil is then said to stall when CL starts decreasing. The maximum point on lift coefficient curve is known as the stalling point. The reason for decrease in the lift coefficient is due to the separation of the flow on the aerofoil. Similarly drag coefficient fristily decreases with increase of angle of attack and later is increases rapidly due to separation of flow on the aerofoil. 22. A flat plate of size 2 x 3 m is submerged in water flowing with velocity of 5 m/s. Find drag and lift if CD = 0.04 and CL = 0.2. Area of plate 2 x 3 = 6 m2 FD = CD x pU2 x A FD = 0.04 x — x 1 x 103 x (5)2 x 6

2 = 0.04 x 75 x 103 = 3 kN

1 FL = CL x — x pU2 x A 2 1 x 1 x 103 x 52 = 0.2 x — x6 = 15 IcN 23. On a flat plate of size 2 x 2 m, wind is passed with speed of 14 m/s. The coefficient of drag and lift are 0.2 and 0.9 respectively. Find (1) lift force (2) drag force (3) resultant force (4) power exerted by the air stream. Assume air density = 1.15 kg/m3. A = 2 x 2 = 4 m2 FL = CL x 1pU2 x A 1 = 0.9 x — x 1.15 x 142 x 4 2

724

Fundamentals of Fluid Mechanics

= 405.72 N FD —

CrI X 1 pU2

—2

xA

= 0.2 x — x 1.15 x 1.152 x 4 2 = 90.16 N F = V FL! + FZ, = /(405.72)2 + (90.16)2 = 1116A6 x 104 + 0.82 x 104 = 415.6 N L tan 0 — F — 405.72 FD 90.16 = 4.5 9 = 77.5°

Or Power exerted by air —

P = FD X U = 90.16 x 14 = 1.262 kW 24. A car has frontal projected area of 2 m3. It travels at the speed of 70 kmph. If CD = 0.4, find (1) power required for the car to overcome wind resistance (2) speed possible incase CD is reduced to 0.3. Assume air density = 1.2 kg/m3. Case 1. U= 70 km ph — FD = CD X —1 pU2

70 x 103 — 19.44 m/s 3600 X

A

= 0.4 x — x 1.2 x (19.44)2 x 2 2 = 181.4 N The power required to overcome drag is — Power = FD X U = 181.4 x 19.44 = 3.526 kW Case 2. CD = 0.3

Flow Past Submerged Bodies

725

Power remains same = 3.526 kW P = FD X U 3.526 x 103 = CD x 21 pU2 xAxU U3 =

Or

3.526 x 103 x 2

0.3 x — x1.2 x2 2 = 19.59 x 103 U = 26.95 m/s

or

25. A man weighing 800 N descends to the ground from an aeroplane using parachute. The parachute is hemispherical with diameter of 2 m. Find the velocity of descend of the parachute. Assume CD = 0.5 and pair = 1.2 kg/m3. Diameter = 2 m

Projected area of parachute —

7r

xd 2 4

ir x 2 — 3.14 m2 42 FD = 800 N A—

FD = CD X — x p x U2 x A 2 800 = 0.5 x 1x 1.2 x U2 x 3.14 2 Or

or

U2

800 x 2 — 05 x 1.2 x 3.14

= 849.256 U = 29.14 m/s

726

Fundamentals of Fluid Mechanics

26. A man is descending with uniform velocity = 20 m/s with the help of a parachute which has hemispherical shape with diameter of 3 m. Find the weight of the man if the weight of the parachute is 30 N. Assume CD = 0.5 and density or air = 1.2 kg/m3

Projected area A = FD —

g x d2

4 0 —

,

—

K

- PT

x 32 — 7.065 m2 4

2 , A

= 0.5 x 1 x 1.2 x 202 x 7.065 2 = 847.8 N FD = Weight of man + Weight of parachute Weight of man = 847.8 — 30 = 817.8 N 27. The total weight to be carried by a parachute is 1200 N at constant speed of descend of 20 m/s. The shape of parachute is hemispherical. If Cd = 0.5 and air density = 1.2 kg/m3, find the diameter of parachute. FD = W = 1200

Here

FD = CDX —1 p x U2 x A 2 1 x 1.2 x 202 x A 1200 = 0.5 x — 2

or

Now or

1200 x 2 0.5 x 1.2 x 400 = 10 m2

A—

x x d2 — 10 4

d2 _

4 x10 Ir

— 12.73 m2

d = 3.57 m 28. A sphere having diameter of 40 mm and relative density of 3 is suspended in a uniform stream of air with the help of a string. Air flow has velocity of 25 m/s. Find (1) inclination of string with vertical and (2) tension in the string. Assume CD = 0.5 and air density = 1.25 kg/m3

Flow Past Submerged Bodies

727

Weight W = ird3 x p —1 x x (0.04)3 x 3 x 103 6 = 0.10 N FD = CD X

2

x pU2 x A

A=

x d2 = lr x (0.04)2 4 4 = 1.256 x 10-3 m2

1 x 1.25 x 252 x 1.256 x 10-3 FD = 0.5 x — 2 = 0.245 N Applying Lami's theorem — T _ FD

sin 90 T=

or

and

W

sin (180 — 0) 0'245 sin 0

sin (90 + 6)

0.10 cos 0

tan _ 0.245 — 2.45 0.10 = 67.79° T—

0.245 sin 67.79 — 0.1 N

29. A kite weighing 0.25 kg and having an effective area of 0.8 m2 assume an angle of 15° to be the horizontal. The chord attached to it makes an angle of 45° to the horizontal. The pull on the chord during a wind of 34 km ph is 2.6 kg. Determine the corresponding coefficient of lift and drag. Density of air is 1.24 kg/m3. (UPTU-2001-2)

728

Fundamentals of Fluid Mechanics

U

T= 2.6 kg

A = 0.8 m2, U —

W= 0.25 kg

34 x 103 — 9.44 m/s 3600

T = 2.6 x 9.81 = 25.5 N W = 0.25 x 9.81 = 2.45 N FD = Tcos 45 = 25.5cos 45 =18.03 N FL = Tsin 45 + W = 25.5 sin 45 + 2.45 = 20.48 N FD

18.03 1 — x 1.24 x (9A4)2 x 0.8 2

CD 1

— pU 2 A 2

= 0.41 CL —

20.48 1 — x 1.24 x (9A4)2 x 0.8 2

FL

1 pU 2 A 2

= 0.466 30. A kite is weighing 0.5 N soars at an angle 30° to the horizontal. The string of the kite makes 30° to the horizontal and it has tension of 5 N. Find the ratio of lift to drag acting on the rate. FD = T cos 30 = 5 cos 30 = 4.33 N FL = Tsin 30 + W =5x

1 + 0.5 2

=3N FL

FD

=

3 =0.69 4.33

Flow Past Submerged Bodies

729

W=6 N

31. A kite of dimension 0.7 x 0.7 m and weighing 6 N assume an angle of 8° to the horizontal and the string attached to the kite makes an angle of 45° to the horizontal. The puli in the string is 25 N when the wind is blowing at 40 kmph. Find the lift and drag forces and corresponding lift and drag coefficients. The density of air is 1.2 kg/ 1113 . (UPTU-2004-5) A = 0.7 x 0.7 = 0.49 m2 U—

40 x 103 — 11.11 m/s 3600

W=6 N

During equilibrium of the kite, we have — FD = T cos 45 = 25 cos 45 = 17.67 N FL = Tsin 45 + W = 17.67 + 6 = 23.67 N Now

FD

C= D

2

pU 2 A

17.67 1 — x 1.2 x (1111)2 x 9.49 2 = 0.48

730

Fundamentals of Fluid Mechanics

FL 1 — pU 2 A 2

CI

23.67 1 — x1.2 x (11.11)2 x0.49 2 = 0.652 32. A passenger ship of 300 m length and 12 m draft is travelling at 45 km ph. Determine (1) total friction drag and (2) the power required to overcome the resistance. Assume the ship surface to act as a flat plate. (UPTU-2004-5) ,.,. 45 x 103 — 12.5 m/s U— 3600 1035 kg/m3 CD = 0.7 and CI = 0.2

Psea water =

For flat plate,

FD = CD X 1 pU2 • A 1 = 0.7 x — x 1035 x (12.5)2 x (300 x 12) 2 = 203.7 x 103 kN 1 pU 2 x A FL = CL x — 1 = 0.2 x — x 1035 x (12.5)2 (300 x 12) 2 = 58.2 x 103 kN F= AlFZ + FZ = J(203.7 x 103)2 + (58.2 X 103)2 = 211.8 x 103 kN Power required =FxU = 211.8 x 106 x 12.5 = 203.7 x 104 kW 33. A cylinder rotates at 300 rpm about its axis which is perpendicular to the stream of water flowing at a velocity of 20 m/s. The cylinder is 2 m diameter and 12 m long. Find (1) circulation (2) lift force per unit length and (3) position of stagnation point.

Flow Past Submerged Bodies

73 I

zDN 60

ue —

7rx2x300 — 31.4 m/s 60 Circulation. 27rRuo =2 xgx 1 x31.4 = 197.1 m/s

F=

Lift FL = L. p• U X r = 12 x 1 x 103 x 20 x 197.1 = 47.3 x 106 N At stagnation point, we have ue = 0 2 sin 0 + — U or

or

31.4 20 x 2

sin 0 —

= —0.78 = —sin 51.6° 0 = +sin (180 + 57.6) and sin (360 — 51.6) = 231.6° and 308.4°

34. An aeroplane has wing area = 30 m2 and wing span = 15 m. It weighs 150 kN. If the aeroplane is flying at speed of 360 km/h and air density is 1.2 kg/m3, find (1) the circulation around the wing (2) drag on the wing and (3) power required. A = 30 m2 360 x 103 = 100 m/s 3600 FL = W U

Now

= 150 x 103 = or

CL —

Cf

-

X

150 x 103 x 2 1.2x(102 )2 x30

= 0.83 For CL = 0.83, from clark-y wing diagram CD v--: 0.05

1 2

X PU2 X

A

732

Fundamentals of Fluid Mechanics

FD — CD X

2

pU2•A

1 = 0.05 x — x 1.2 x 104 x 30 2 = 9 kN Power = FD X U = 9 x 103 x 100 = 900 kW circulation is given by — FL = p • u• F•L 150 x 103 1.2 x 1002 x 15 = 83.33 m2/s 35. A metallic chimney on a house is 2 m high and has outer and inner diameter as 0.4 and 0.395 m respectively. If the wind velocity is 60 km/h, find the drag on the chimney. Assume density of air = 1.2 kg/m3 and CD = 0.2. 0.4k2m

A=hxD= 2 x 0.4 = 0.8 m2 U= 60 km ph — FD — CDX

2

60 x 103 — 16.66 m/s 3600

pU2 x A

1 x 1.2 x (16.66)2 = 0.2 x — x 0.8 2 = 26.64 N 36. A truck having a projected area of 6.5 m2 travelling at 70 km ph has total resistance 200 N. Of this 20% is due to rolling friction and 10% due to surface friction. The rest is due to form drag. Find coefficient of drag. Take p = 1.22 kg/m3 for air. (AMIE)

Flow Past Submerged Bodies

U = 70 km ph —

733

70 x 103 — 19.44 m/s 3600

20 Resistance due to rolling = 100 x 2000 = 400 N 10 x 2000 = 200 N Resistance due to surface friction = 100 Hence drag = 2000 — (400 + 200) = 1400 N FD — ,D _ — P T 2 „

A

1 1400 = CD X — x 1.22 x (19.44)2 x 6.5 2 CD = 0.934

or

37. As an application of Magnus effect, a ship is built having two vertical rotors 10 m high and 3 m in diameter. The rotors are spun at 250 rpm. On a day when the air temperature is 20°C and relative motion of the air to ship results in 54 km ph wind. calculate the force emitted by the spinning rotors on the ship. Take p = 1.25 kg/m3 for air. (IIT Madras) in horizontal plane

U = 54 km ph —

54 x 103 — 15 m/s 3600

7r x 3 x 250

zrDN

— 39.27 m/s 60 60 Circulation generated by rotation of each cylinder is 140

-

F

2 rci? uo 3 = 27r x — x39.27 2 = 370.1 m2/s

Horizontal lift or transverse force perpendicular to flow — FL = pLUF = 1.25 x 10 x 15 x 370.1 = 69.4 kN Total transverse force by two rotors = 69.4 x 2 = 138.8 kN

734

Fundamentals of Fluid Mechanics

38. A circular disc 3 m in diameter is held normal to a 26.4 m/s wind of density 1.2 kg/ m3. What force is required to hold it at rest? Assume coefficient of drag of disc = 1.1. (AMIE 67) FD = CD X 1 pU2 x A 2 = 1.1

X

•-, —1 x 1.2 x (26.4)- x

7r X

32

4

= 3.25 lth 39. Find the difference in drag force exerted on a flat plate of size 3 x 3 m when plate is moving at speed of 4 m/s normal to its plane in (i) water (ii) air of density 1.24 kg/ m3. Assume CD = 1.15. Case 1. Plate in water FD = CD X — x pU2 x A 2 1 = 1.15 x — x 1 x 103 x 42 x 9 = 82.8 lth Case 2. Plate in air FD = 1.15 x — x 1.24 x 42 x 9 2 = 102.67 N Difference in drag force = 82800 — 102.67 = 82797.33 N = 82.797 lcisT 40. The air is flowing over a cylinder of diameter 50 mm and infinite length with a velocity of 0.1 m/s. Find the total drag, shear drag and pressure drag on 1 m length of the cylinder if total drag coefficient is equal to 1.5 and shear drag coefficient equal to 0.2. Take density = 1.25 kg/m3 of air. Total drag force — FD = CD X —1 pU2 2

X

A

1 x 1.25 x (0.1)2 = 1.5 x — x (1 x 0.05) = 4.68 x 10-4 N

Flow Past Submerged Bodies

735

Drag due to shear FDS — CDs

s,

1

— 2

vuT 2 „ A

1.25 x (0.1)2 x 0.05 2 = 0.625 x 10-4 N = 0.2 x 1.25 x

Now we know FD = FDS FDP FDp = FD FDS

= 4.68 x 10-4 = 0.625 x 10-4 = 4.05 x 10-4 N 41. Determine the velocity of fall of rain drops of a 30 x 10-3 cm diameter, density 1.2 x 10-3 gm/cm3 and kinematic viscosity 0.15 cm2/s. (AMIE 82) p = 1.2 x 103 gm/cm3 & v = 0.15 cm2/s u =pxv= 1.2x103 x0.15 = 1.8 x 10-4 gm cm•s = 1.8 x 10-5 = kg ms Weight of rain drop = 6 x D3 x p x g = 6 x (30 x 10-5) x 1.2 x 9.81 Now

Or

= 1.66 x 10-10 N W= FD 1.66 x 10-10 =3 rcyDU U=

L66 x 10-b0 3x7rx3x10-4 x1.8x10-5

0.166 9 x x 1.8 = 3.26 x 10-3 m/s Now check Reynolds number for using Stokes' equation pUD UD 3.26 x 10-3 x 3 x 10-4 Re = v 1-1 0.15 x 10-4 = 0.0654 < 0.2

736 Fundamentals of Fluid Mechanics

42. A kite 0.8 m x 0.8 m and weighing 4 N is maintained in air at an angle of 10° to the horizontal. The string attached to the kite makes an angle 45° to the horizontal and at this position, the values of coefficient of drag and lift are 0.6 and 0.8 respectively. Find the speed of the wind, tension in the string. Take density of air = 1.25 kg/m2. (UPTU 2008-9) -4-0.8k-I

FD

T 0.8 I

Projected area

Projected area

A = 0.8 x 0.8 = 0.64 m2 In equilibrium, there are four forces which are acting viz., P, W, FL and FD. 1Fy = 0, FL = P sin45 + 4 or

CL x 1 2 p•A• u2 — ,— +4 -J2

Or

0.8 x — x 1.25 x 0.64 u2 = 0.707 P + 4 2 0.3 u2 = 0.707 P + 4 EFx = 0, FD = Pcos45

or

Or

or

Cn

(..,

1 p•A•u2 — P

X— 2

,-

0.6 x 1 x 1.25 x 0.64 x u2 — P 2 V2

or

2 U —

2P 2 (0.6 x 1.25 x 0.64)

I

= 2.95 P From equations (i) and (ii), we have0.3 (2.95 or and or

(i)

P) = 0.707 P + 4 P = 22.47 N u2 = 2.95 P u2 = 2.95 x 22.47 u = 8.14 m/s

(ii)

Question Papers B. Tech. (SEM. III) ODD SEMESTER THEORY EXAMINATION, 2009-10 FLUID MECHANICS Time : 3 Hours

Total Marks : 100

Note : Attempt all the questions. All questions carry equal marks. 1. Attempt any four of the following :

4 x 5 = 20

(a) A square plate 50 cm x 50 cm weighing 200 N slides down an inclined plane of slope 1 vertical : 2.5 horizontal with a uniform velocity of 0.40 m/s. If a thin layer of oil of thickness 0.5 cm fills the space between the plate and the inclined plane, determine the coefficient of viscosity of the oil. (b) Obtain an expressions for the height of capillary rise for a fluid of surface tension a and contact angle 0 between two parallel vertical plates at a distance B apart. (c) Define the term centre of pressure of the plane area immersed in a fluid. What relation has it got with the centre of gravity of the area ? Do the centre of pressure and centre of gravity ever coincide and if so under what conditions ? (d) Derive an expression for the absolute pressure to be measured in pipe carrying highly pressurized water with the help of a Double U-tube Manometer having mercury in each tube with different levels in the four limbs. (e) What is metacentric height ? And what are the three states of equilibrium for floating bodies ? (f) Derive an expression for the slope of the inclined surface of water in a container moving with a constant acceleration. 2. Attempt any two of the following :

2 x 10 = 20

(a) Explain and distinguish between (i) steady and unsteady flow (ii) uniform and nonuniform flow (iii) rotational and irrotational flow (iv) subcritical and supercritical flows and (v) laminar and turbulent flows. (b) Derive the expression for the continuity equation for the steady-state 3-D flow of a compressible fluid. (c) Derive the equation of a stream line for 2-D flow. Prove that the discharge between two stream lines is the difference in their stream function values. 3. Attempt any two of the following :

2 x 10 = 20

(a) What are local and convective components of total acceleration? Explain these with the help of a suitable example. Derive an expression for the equation of motion along a stream line for non-viscous and incompressible fluids.

738

Fundamentals of Fluid Mechanics

(b) A venturi meter is fitted horizontally in a 15 cm dia. pipe and the pressure in the pipe corresponds to a water head of 10 m. If the maximum flow through the meter is 0.15 m3/sec., find the smallest throat diameter so that the pressure does not fall below 2.45 m of water (absolute). (c) If both the Reynolds number and the Froude number are significant, determine the scale for kinematic viscosity in terms of the length scale if the gravitational field remains the same both for the model and the prototype. 4. Attempt any two of the following :

2 x 10 = 20

(a) Derive the equation of motion for laminar flow through pipes. Also derive the expression for velocity and shear stress distribution across the pipe. (b) A 10 cm dia. shaft runs at 900 rpm in a sleeve with a radial clearance of 1 mm. If the sleeve length is 15 cm long and the space is filled with an oil of dynamic viscosity, = 0.10 kg/m-s, determine the torque resistance. (c) The total flow in three parallel pipe system is 4 m3/sec. The pipe diameters and length of the three pipes are : d1 = 1.0 m, L1 = 1000 m; d2 = 0.5 m, L2 = 800 m; d3 = 0.75 m, L3 = 750 m. Find the discharge in the three pipes. 5. Attempt any two of the following:

2 x 10 = 20

(a) Define physically and mathematically the concept of Displacement Thickness, Momentum Thickness and Energy Thickness of a boundary layer and state their relative order these thicknesses. (b) Explain the essential features of Blasium method of solving laminar boundary layer equations for a flat plate. Derive the expression for boundary layer thickness and local skin friction coefficient from this solution. (c) State the Stokes' law for finding drag force of a sphere moving in an infmite medium. How this law can be applied to determine the viscosity of fluid ?

B. Tech. (SEM. III) EXAMINATION, 2007-08 FLUID MECHANICS Time : 3 Hours Note :

Total Marks : 100

(i) Attempt all questions. (ii) Assume missing data suitably, if any, and state the assumptions made.

1. Answer any two parts of the following :

10 x 2 = 20

(a) (i) State Newton's Law of Viscosity and derive the same. (ii) Define surface tension. Establish relationship among surface tension (0), pressure within the droplet of liquid in excess of outside pressure (p) and dia. of droplet d. (b) (i) Derive an expression for total hydrostatic pressure on curved surfaces. (ii) A cone floating in water with its apex downwards has a diameter d and vertical height h. If the sp. gravity of the cone is s, prove that for stable equilibrium

h2 < 1 d2s"3

4 1— s"3

(c) What do you mean by dimensionless numbers? Name any four dimensionless numbers and its practical applications in fluid mechanics. 2. Answer any two parts of the following :

10 x 2 = 20

(a) Explain the terms distorted and undistorted models. What is the use of distorted models? (b) (i) Define and distinguish between the following laminar and turbulent flow rotational and irrotational flow. (ii) Define stream function and velocity potential function. Prove that stream lines and equipotential lines meet each other orthogonally. (c) Derive Bernoulli's equation from Euler's equation of motion along a stream line. Write their practical applications. 3. Answer any four parts of the following :

5 x 4 = 20

(a) Describe momentum equation. Where this equation is used ? (b) A 200 mm diameter pipe carries water under a head of 15 m with a velocity of 3 m/ sec. If the axis of the pipe turns through 45°, fmd the magnitude and direction of the resultant force at the bend. (c) Prove that the loss of head for the viscous flow through a circular pipe is given by 3.2 ,u VL h — f yd2 where, p. = viscosity, V = average velocity,

740

Fundamentals of Fluid Mechanics

L = length of pipe, y = sp. weight, d = diameter of pipe. (d) Prove that the maximum velocity in a circular pipe for viscous flow is equal to two times the average velocity of flow. (e) Obtain an expression for the velocity distribution for turbulent flow in smooth pipe. (f) How would you distinguish between hydrodynamically smooth and rough pipe ? 4. Answer any four parts of the following :

4 x 5 = 20

(a) Defme displacement thickness. Derive an expression for the displacement thickness. (b) Defme the following terms with the help of sketch: Laminar boundary layer, Turbulent boundary layer, Laminar sub-layer and Boundary layer thickness. (c) What is a syphon ? Where is it used ? Explain its action. (d) Describe Reynold experiments to demonstrate the two types of flow. (e) Derive Darcy-Weisbach equation. (f) Explain the terms with sketches: (i) Pipes in parallel (ii) Equivalent size of the pipe. 5. Answer any two parts of the following :

10 x 2 = 20

(a) What is meant by water hammer ? What allowance is usually made for this in penstock design? (b) Defme the following terms in case of potential flow : (i) Uniform flow (ii) Source flow (iii) Sink flow (iv) Free vortex flow. (c) What do you understand by the following terms : (i) Total drag on body (ii) Resultant force on a body (iii) Coefficient of drag (iv) Coefficient of lift

B.TECH. CARRY OVER PAPER EXAMINATION, 2006-07 FLUID MECHANICS Time : 3 Hours Note : (i) (ii) (iii) (iv) (v)

Total Marks : 100

Attempt ALL questions. All questions carry equal marks. Use illustrations, wherever required Assume missing data suitably, if any, and state the assumptions made. Be precise in your answer

1. Attempt any four parts of the following:

(5 x 4 = 20)

(a) Explain why in a capillary tube the meniscus of water is concave upwards while the meniscus of mercury is convex upwards. (b) Differentiate between ideal and real fluids. What is the pressure within a droplet of water of 0.5 x 10-4 m diameter at 20°C, if the pressure outside the droplet is standard atmospheric pressure. (c) A thin plate of area 'A' is placed midway in a gap of height 'h' filled with a liquid of viscosity 'pi'. The plate requires a force 'f' to move with a constant uniform velocity V. The same gap is subsequently filled with another liquid of viscosity ',u2' and the same plate is positioned at a distance `h14' from one wall. Prove for same velocity 'V = (4 3) • uz (d) Differentiate between any two of the following: (i) Laminar and turbulent flow in pipes (ii) Uniform and unsteady flow (iii) Rotational and irrotational flow and force 'f',

(e) Differentiate between total, local and convective accelerations with one example each. (f) What are the limitations to draw the flow nets? For given 0 = 4 (x2 — y2), fmd 2. Attempt any four parts of the following:

(5 x 4 = 20)

(a) Derive the expression for pressure at a point in case of compressible fluid with isothermal condition. (b) Define coefficients of contraction, velocity and discharge (Cc, Cr,and Cd respectively). Hence, prove Cd = Cc x Cv. (c) Define vena contracta. A pitot static tube is used to measure the velocity in a pipe with water as the running fluid. The stagnation pressure head is 9.0 m and static pressure head is 6 m. Calculate the velocity of flow assuming the coefficient of pitot-static tube equal to 0.985. (d) A solid cylinder of 2 m in diameter and 2.0 m high is floating in water with its axis

742

Fundamentals of Fluid Mechanics

vertical. If the specific gravity of the material of cylinder is 0.65, find its meta centric height. State whether the equilibrium is stable or unstable. (e) What is the difference between a: (i) Weir and notch (ii) Free and forced vortex motion (iii) Broad crested and narrow-crested weir. Explain with neat sketches. (f) A 300 mm diameter pipe carries water under a head of 20 m with a velocity of 3.5 m/s. If the axis of the pipe turns through 45°, fmd the magnitude and direction of the resultant force at the bend. 3. Attempt any two parts of the following:

(10 x 2 = 20)

(a) What is difference between a model and prototype. Show by n--theorem that a general equation for discharge `Q' over a weir of any shape is given by: Q = (H5/2

j(

g1/2) H312

1/2

g

' H 2pg

)1

where, H = head over weir v = kinetic viscosity p = density of the fluid g = acceleration due to gravity = surface tension (b) For a steady laminar flow through a circular pipe, prove that the velocity distribution across the section is parabolic and the average velocity is half of the maximum local velocity. (c) Attempt any two pats of the following: Write a short note on hot-wire anemometer. In a pipe of 360 mm diameter having turbulent flow, the centreline velocity is 8.0 m/s and that at 60 mm from the pipe wall is 6.5 m/s. Calculate the shear friction velocity. Derive an expression for shear stress on the basis of `Prandtl's mixing length theory". 4. Attempt any two parts of the following:

(10 x 2 = 20)

(a) Explain the concept of boundary layer development over a flat plate. Derive an expression for momentum thickness. (b) In a pipe of 500 mm diameter and 2500 m length, provided with a valve at its end, water is flowing with a velocity of 1.5 m/s. Assuming velocity of pressure wave equal to 1460 m/s, find: (i) Rise in pressure, if the valve is closed in 25 seconds, 15 seconds and 5 seconds. (ii) The rise in pressure if the valve is closed in 2 seconds. Assume the pipe to be rigid one and bulk modulus of water is to be taken as 1.962 GN/m2.

Question Papers

743

(c) Write a short notes on any two of the following: (i) Simple surge tank (ii) Boundary separation and its control (iii) Water hammer 5. Attempt any two parts of the following:

(10 x 2 = 20)

(a) Explain the terms: (i) Total energy line and hydraulic gradient line (ii) Equivalent pipe (iii) Head loss due to sudden contraction of a pipe (b) (i) What do you understand by "Magnus Effect"? Why is it known as "Magnus Effect"? (ii) Draw Coefficient of Drag (CD) versus Reynolds Number (Re) diagram for a sphere and explain why CD suddenly drops at Re = 3 X 103. (c) A syphon of diameter 200 mm connects two reservoirs having a difference in elevation of 15 m. The total length of the syphon is 600 mm and the summit 4.0 m above the water level in the upper reservoir. If the separation takes place at 2.8 m of water absolute fmd the maximum length of syphon from upper reservoir to the summit. Take value of friction factor (f) = 0.004 and atmospheric pressure = 10.30 m of water.

B.Tech. THIRD SEMESTER EXAMINATION, 2004-05 FLUID MECHANICS Time : 3 Hours

Total Marks : 100

Note : (i) Attempt ALL questions. (ii) Use illustrations, wherever required (iii) Assume missing data suitably, if any, and state the assumptions made. 1. Answer any four parts of the following:

(5 x 4 = 20)

(a) Differentiate between (i) Eulerian and Lagrangian approach. (ii) Dynamic and Kinetic viscosity. (b) Assuming sap in trees rises purely due to capillary rise phenomenon, find the diameter of the tube in mm which can carry it to height of 15.0 m. Assume the sap to have the same characteristics as those of water. (c) Explain the different kinds of fluids using neat sketch of Rheogram with one example each. (d) Prove stream function OA and potential function (0) are orthogonal to each other. (e) What is flow net. What are the uses of the flow net. Enlist the different methods to draw flow nets. (f) Find the convective acceleration at the middle of the pipe, which converges uniformly from 40 cm to 20 cm diameter over a length of 2.0 m at an observed discharge of 20 //s in 30 seconds with uniform rate, fmd the total acceleration at the middle of pipe at 15 seconds. 2. Answer any two parts of the following:

(10 x 2 = 20)

(a) (i) State and prove Pascal's law. (ii) The barometric pressure at sea level is 76 cm of Hg, while that on a mountain top is 73.5 cm. If the density of air is assumed to be constant as 1.2 kg/m3, prove that the elevation of the momentum top is 284.4 m. (b) Differentiate between (any two) Stability conditions for immersed and floating bodies. Absolute, Gauge, Atmospheric and Vacuum pressure using sketch. Also given the relation between them. Reasons for the difference in Cd values of orifice meter and venturi meter. State the assumptions made in Bernoulli's equation. An orificemeter with orifice diameter 0.10 m is inserted in a pipe of 0.20 m diameter. The pressure gauges fitted upstream and downstream of the orifice meter gives readings of 19.62 x 104 N/m2 and 9.81 x 104 N/m2 respectively. Coefficient

Question Papers

745

of discharge for the orifice meter is 0.62. Find the discharge of water through the pipe. 3. Answer any four parts of the following:

(5 x 4 = 20)

(a) What do you understand by a dimensionally homogenous equations? Write three dimensionless numbers. (b) A spillway 7.2 m high and 150 m long discharges 2150 m3/s under a head of 4 m. If a 1:16 model of spillway is to be constructed, fmd. (i) Height and length of model (ii) Head over the model (iii) The model discharge (c) Give the classification of boundaries based on roughness height with neat sketch. (d) For a viscous flow through a circular pipe prove that the kinetic energy correction factor is equal to 2. (e) An oil of specific gravity 0.9 and viscosity 10 poise is flowing through a pipe of diameter 110 mm. The velocity at the centre is 2 m/s. Find the pressure gradient in the direction of flow and shear stress at the pipe wall. (f) What is velocity defect? Derive an expression for velocity defect in pipes. 4. Answer any two parts of the following:

(10 x 2 = 20)

(a) What are the main factors for pressure rise due to water hammer. Derive the relation for pressure rise with rigid pipe and sudden closure of valve. (b) Answer any two parts of the following: (i) Prove that in a simple surge tank the maximum rise in water will be sm.

= u011( As ). (LI

where all symbols have the usual meaning. What is laminar sub-layer thickness? Derive the expression for displacement thickness. What is the separation? What are the preventive measures to control it? (c) A passenger ship of 300 m length and 12 m draft is travelling at 45 km/hr. Determine the (i) total friction drag, and (ii) the power required to overcome the resistance. (Assume the ship surface at act as a flat plate) 5. Answer any two parts of the following:

(10 x 2 =20)

(a) A kite of dimensions 0.7 m x 0.7 m and weighing 6 N assumes and angle 8° to the horizontal, and the string attached to the kite makes an angle of 45° to the horizontal.

746

Fundamentals of Fluid Mechanics

The pull on the string is 25 N when the wind is blowing at a speed of 40 km/hr. Find the lift and drag forces and the corresponding lift and drag coefficients. The density of air given is 1.2 kg/m3. (b) (i) Derive an expression for the loss of head due to sudden enlargement. (ii) What is syphon? On what principle it works? (c) (i) What do you understand by the terms major and minor energy losses in pipes? (ii) A pipeline 60 cm diameter bifurcates at a rjunction into two branches 40 cm and 30 cm in diameter. If the rate of flow in the main pipe is 1.5 cumecs and mean velocity of flow in 30 cm diameter pipe is 7.5 m/s, determine the rate of flow in the 40 cm diameter pipe.

B.Tech. THIRD SEMESTER EXAMINATION, 2003-2004 FLUID MECHANICS Time : 3 Hours

Total Marks : 100

Note : Attempt ALL questions: 1. Attempt any Two parts: (10) (a) Derive the continuity equation in Cartesian coordinates. (b) Through a very narrow gap of height h, a thin plate of large extent is pulled at a velocity V. On one side of the plate is oil of viscosity gi and on the other side oil of viscosity /./.2. Calculate the position of the plate so that the pull required to drag the plate is minimum. (10) (c) A stream function in a two-dimensional flow is tif = 2xy. Determine the corresponding (10) velocity potential cp. 2. Attempt any TWO parts: (a) (i) What are Differential Manometers? (5 + 5 = 10) (ii) Define Centre of Buoyancy and Metacentre. (10) (b) State and prove Bernoulli's equation. (c) A 40 mm diameter water jet strikes a hinged vertical plate of 800 N weight, normally at its centre at 15 m/s velocity. Determine the angle of deflection O. Also determine the magnitude of force F that must be applied at its lower edge to keep the plate vertical. (10) 3. Attempt any TWO parts: (a) State Buckingham's 7-c-theorem. What are repeating variables? How are these selected (10) in dimensional analysis? (b) A lubricating oil of viscosity 1 poise and specific gravity 0.9 is pumped through a 30 mm diameter pipe. If the pressure drop per meter length of pipe is 20 kN/m2, determine (10) the discharge. (c) In a rough pipe of diameter 0.6 m and length 4500 m, water is flowing at the rate of 0.6 m3/s. If the average height of roughness is 0.48 mm, fmd the power required to (10) maintains this flow. 4. Attempt any TWO parts: (a) (i) Explain the concept of Boundary Layer. (ii) Explain the characteristics of Laminar and Turbulent boundary layers. (5 + 5 = 10) (b) The boundary layer thickness at a distance of 1 m from the leading edge of a flat plate kept over zero angle of incidence to the flow direction is 1 mm. The velocity

748

Fundamentals of Fluid Mechanics

outside the boundary layer is 25 ra/s. Calculate the boundary layer thickness at a distance of 4 m from the leading edge of the flat plate. Assume that the boundary layer is entirely laminar. (10) (10) (c) Write a short note on "simple surge tank". 5. Attempt any TWO parts: (a) Obtain the conditions for maximum transmission of power in a pipeline. For this (10) condition, what is the efficiency of transmission? (b) 300 ram main pipes are required for distribution of water in a city. Since pipes over 250 mm in diameter are not available, it is decided to provide two parallel mains of same (10) diameter. Determine the diameter of the parallel main pipes. (c) (i) Differentiate between Pressure drag and Shear drag. (ii) Explain Magnus Effect. (5 + 5 = 10)

B.Tech. THIRD SEMESTER EXAMINATION, 2002-03 FLUID MECHANICS Time : 3 Hours Note :

Total Marks : 100

(i) Attempt ALL the questions. (2) Use illustrations, wherever required. (3) Assume missing data suitably, if any, and state the assumptions made.

1. Answer any FOUR of the following:

(5 x 4 = 20)

(a) Explain the phenomenon of capillary. Obtaine an expression for capillary rise or fall of a liquid in very small diameter tube. (b) What is Continuum? Is air a continuum? Does it always remain so? (c) Define the following and give one practical example for each: Laminar flow, Turbulent flow, Unsteady flow and Uniform flow. (d) Describe the use and limitations of the flow nets. (e) Check whether the function IP = A (x2 — y2) represent the possible irrotational flow phenomenon. (f) Prove that a stream function vi represents the equation for a stream line. 2. Answer any FOUR of the following:

(5 x 4 = 20)

(a) What are the conditions of equilibrium of a floating body? Discuss with neat sketches. (b) A circular plate 4 m diameter is immersed in water in such a way that its greatest and least depth below the free surface are 4 m and 2 m respectively. Determine the total pressure on one face of the plate and position of the centre of pressure. (c) Derive Bernoulli's equation from Euler equation of motion. State the assumption also. (d) Discuss the relative merits and demerits of venturi meter with respect to orifice meter. (e) Describe the Momentum Equation. State the practical application of the momentum equation. (f) Defmce the terms Cd, Cc, Cr, and derive the expression Cd = Cc x Cr, where Cd = Coefficient of discharge Cc = Coefficient of contraction CV = Coefficient of velocity 3. Answer any TWO of the following:

(10 x 2 = 20)

(a) Derive an expression for the loss of head due to friction in pipes. (b) Prove that for viscous flow through a circular pipe, the Kinetic energy correction factor is equal to 2. (c) What do you mean by dimensionless numbers? Derive expression for any twodimensionless numbers.

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Fundamentals of Fluid Mechanics

4. Answer any TWO of the following:

(10 x 2 = 20)

(a) Define displacement thickness. Derive and expression for momentum thickness for boundary layer flow. (b) What do you mean by separation of boundary layer? What is the effect of pressure gradient on boundary layer thickness? (c) What is meant by water hammer? What allowance is usually made for this in penstock design. 5. Answer any TWO of the follwing: (a) What is a Syphon? Where is it used? Explain its action. Derive an expression for the (2 + 2 + 2 + 4 = 10) length of its inlet leg. (b) (i) Derive an expression for the loss of head due to sudden enlargement. (ii) What is equivalent pipe? (c) Diffrentiate between: (i) Stream lines body and Bluff body. (ii) Friction drag and Pressure drag.

(7) (3) (10)

B.Tech. THIRD SEMESTER EXAMINATION, 2001-02 (COMMON TO CIVIL ENGG AND MECHANICAL ENGG) FLUID MECHANICS Time — 3 Hours

Total Marks — 100

Note : Attempt ALL the questions. 1. Attmpt any Two parts of the following: (a) Enunciate Newton's law of viscosity. Distinguish between Newton and non-Newtonian fluids. 10 (b) (i) Distinguish between streamlines, streamlines and pathlines. (ii) Explain the physical significance and use of the term 'Stream Function'. 5 + 5 = 10 (c) The velocity components in a flow field in x— and y—directions are u = xy and v = 2 yz respectively. Examine whether these velocity components represent two-or threedimensional compressible flow and if, three-dimensional, determine the z-component. 10 2. Attempt any Two parts of the following: (a) (i) Define the terms 'Total Pressure' and 'Centre of Pressure'. (ii) Explain the principle of floatation. 5+5 (b) What is a pitot tube? How is used to measure velocity of flow at any point in a pipe 10 or channel? (c) Find the discharge from an 80 mm diameter external mouthpiece, fitted to a side of a 10 large vessel, if the head over the mouthpiece is 6 m. 3. Attempt any Two parts of the following: (a) The drag force F on a partially submerged body depends on the relative velocity V between the body and the fluid, characterstics linear dimension 1, height of surface roughness K, fluid density p, the viscosity u and the acceleration due to gravity g. Obtain an expression for the drag force, using the method of dimensional analysis. 10 (1)) Determine the maximum wall shear stress for laminar flow in a tube of diameter D with 10 fluid properties u and p. (c) Explain the concept of mixing length introduced by Prandtl and state the relationship 10 that exists between the turbulent shearing stress and the mixing length. 4. Attempt any Two parts of the following: (a) The velocity distribution in the boundary layer is given by :

U

=

6 ,

where u is the

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Fundamentals of Fluid Mechanics

velocity at a distance y from the plate and u = U at y = 6, 3 being boundary layer 10 thickness. Find the ratio of displacement thickness to momentum thickness. 10 (b) What is laminar sub-layer? How is the concept of laminar sub-layer useful? (c) What is meant by water hammer? Derive an expression for the rise of pressure when 10 the water flowing in a pipe is brought to rest by closing the valve gradually. 5. Attempt any Two parts of the following: (a) At a sudden enlargement of water pipe line from diameter of 24 to 48 cm, the hydraulic 10 grade line rises by 1 cm. Estimate the flow rate in the pipe line. (b) (i) What do you meant by an equivalent pipe? (ii) What are the causes of loss of energy in pipe bends?

5 + 5 = 10

(c) A kite weighing 0.25 kg and having an effective area of 0.8 m2 assumes an angle of 15° to the horizontal and the chord attached to it makes an angle or 45° to the horizontal. The pull on the chord during a wind of 34 km/hr is 2.6 kg. Determine the corresponding coefficients of lift and drag. The density of air is 1.24 kg/m3.

Index A Absolute pressure, 69 Adhesion, 43 Aeroplane, equilibrium of, 722 Aircraft, airbrake of, 10 flying of, 8 Angle of attack, 720 Angular distortion, 315 Archiedes' principle, 188 Artificial blood oxygenation, 11 Aspect ratio, 720 B Barometer, 69 Bernoulli's equation, 377, 394 Bernoulli's theorem, 378 Bingham plastics fluids, 21 Blasius results, 617 Blasius's equation, 628 Bluff body, 713 Boiling, 42, 43 Borda mouthpiece, 393 Boundary layer, analysis of, 583 growth of, 585 loss of momentum in, 590 momentum equation for, 594 separation of, 600 Boundary thickness, 588 Bourdon pressure gauge, 71 Bourdon tube, 70 Broad crested weir, 413

Bubble, collapse of, 43 Buckingham's theorem, 273 Buoyancy, 187 C Camber, 720 Capillary fall, 49 Capillary rise, 49 Capillary tube, 55 method of, 538 Cavitation, 43 Centre of pressure, 105 Centrifugal pump, 12 Chord line and chord, 720 Cohesion, 43 Cohesive force (downward on surface), 46 Collar bearing, 533 Combined uniform, aerofoil in, 721 Compressibility, 40 coefficient of, 40 Contraction, coefficient of, 383 Convergent divergent mouthpiece, 392 Convergent mouthpiece, 392 Cricket ball, swing of, 9 Curved surface, 136 Cylindrical external mouthpiece, 391 D Dashpot mechanism, 535 Density, 14

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Index

Differential manometer, 76 Dimensional analysis, 268 Discharge, coefficient of, 384 Drag, application of, 10 Drowned weir, 414 Duput's equation, 645 Dynamic similarity, 291 Dynamic viscosity, 24 Dynamics, principle of, 450 E Electrical analogy method, 337 Electricity, flow of, 336 End contraction, rectangular weir with, 412 Euler's equation, 378 Euler's model law, 294 Euler's number, 278 Eulerian method, 312 Evaporation, 42, 43 F Falling sphere viscometer, 537 Floatation, 187 principle of, 190 Floating bodies, stability for, 195 Flow nozzle, 478 Flow past a bridge pier, 712 Flow past submerged bodies, 707 Flow through capillary, 55 Fluctuating component, 549 Fluid dynamics-I, 372 Fluid dynamics-II, 450 Fluid element, 321 forces on, 377 Fluid kinematics, 310 Fluids, continuous deformation of, 5 definitions of, 1

features of, 4 flow of, 336 properties of, 1 Foot stop bearing, 531 Forced vortex flow, 340 Forces on bend (in horizontal plane), 463 Francis turbine, 12 Free vortex, flow pattern in, 699 pressure variation in, 700 velocity variation in, 700 Friction factor, variation of, 634 Froude's model law, 293 Froude's number, 277 Fully submerged orifice, 389

G Gas, 3 Gates, 169 Gauge pressure, 69, 72, 81 Geometric similarity, 289 H Head variation, 57 Hot wire anemometer, 569 Hydraulic radius, 635 Hydrodynamically, rough boundary of, 558 smooth boundary of, 558 Hydrometer, 200 working of, 191 Hydrostatic forces, 104 Hydrostatic pressure, on element of fluid, 66 with depth of, 66 Hydraulic gradient line (HGL), 642, 645 I Ideal fluid flow, 692 Ideal fluids, 13 behaviour of, 23 Ideal plastic fluids, 21 Immersed bodies,

Index stability for, 195 Inclined pipe, 417 Inclined single column micrometer, 74 Inclined surface, 111 Inlet leg, 664 Instantaneous velocity, 549 Inverted U tube manometer, 75 Irrotational motion, 318 J Jet engine, 476 Jet trajectory, 403 Journal bearing, 529 K Kaplan turbine, 12 Kinematic similarity, 290 Kinematic viscosity, 24 Kinetic energy, 374 Kutta-Joukowsky equation, 718 L Laminar boundary layer, thickness of, 596 Laminar flow, 317, 493 velocity distribution in, 546 Laminar sublayer, 587 Langrangian method, 312 Large rectangular orifice, 388 Laser Doppler anemometer, 570 Liquid surface, surface tension on, 45 Liquid, 3 Local velocity distribution, 504 M Mach's model law, 295 Mach's number, 279 Magnous effect, 716 Masonry dam, 171 Mean velocity, 549 Metallic float, 201 Micro manometer, 77 Micrometer, 73

Model analysis, 287 Molecular activity, molecule of, 46 sphere of, 46 Moody's diagram, 568 N Needle, floatation of, 47 Neutral equilibrium, 193 Newtonian fluids, 19 Non-Newtonian fluids, 19 Non-uniform flow, 316 Non-wetting liquid (forms droplet), 44 0 One-dimensional flow, 318 Open spill way, 407 Open vessel, forced vortex in, 247 Orifice discharging free, 387 Orifice meter, 479, 481 Orifice metering, 8 Orifice, 386 classification of, 387 discharge of, 424 velocity of, 424 Outlet leg, 664 Overhead tank, pumping storage to, 48 Overhead water tanks, 171 P Parallel plates, capillary rise in, 52 Partially submerged orifice, 390 Pascal law, 65 Path lines, 314 Pelton wheel, 12 Petroleum, railway wagons for, 247 Tc theorem (over Rayleigh's method), superiority of, 274 Pipe,

755

756

Index

draining from, 419 Piston pump, operating principle of, 11 Pitot static tube, 383 Pitot tube, 380 Potential energy, 374 Pressure energy, 374 Pressure intensity, 134, 161 Pressure rise, magnitude of, 670 Pressure tube, 71 Pressure variation with depth, 67 Pressures, 64 differentiation of, 70 Profile centre line, 720 R Radial flow, 461 Real fluids, 13 Rectangular notch, 406 Rectangular weir, 407, 410 Rectilinear flow, 461 Reentrant mouthpiece, 393 Relative density, 15 Repeating variables, 274 Reynolds model law, 293 Reynolds number, 277 Rheological diagram, 59 Rofometer, 483 Rotary motion, 462 Rotating cylinder method, 539 Rotational cylinder, 56 Rotational motion, 318 Running free Borda's mouthpiece, 399

Short tube mouthpiece, 391 Similitude, 287 Simple manometer, 76 Sink flow, 704 Smoke, rising of, 547 Solids, 3 behaviour of, 22 features of, 4 static deformation of, 5 Specific volume, 15 Specific weight, 14 Sphere achieving terminal velocity, 55 Sphere resistance, 55 Spherical droplet, 48 Stable condition, 194 Stable equilibrium, 193, 195 Static liquid, 243 Static pressure, 243 Streak lines, 314 Stream function, physical significance of, 328 Stream line body, 713 Stream lines, 314 equation of, 329 Stream tube, 314 Submerged body, forces acting on, 189 Submerged weir, 414 Sump, lifting from, 420 Surface tension, cause of, 45 Surge tank, 670 Syphon spill way, 407

S

Sail boat, 9 Saybolt viscometer, 542 Seawater, iceberg in, 202 Shear stress distribution, 497, 505 Ship, water line cross-section of, 211

T Taperd pipe, 416 Three-dimensional flow, 319 Total energy line (TEL), 642, 645 Total pressure forces, 105 Trapezoidal notch, 445 Trapezoidal weir, 413

Index Triangular notch, 409 Turbulent flow, 317, 545 velocity distribution in, 546 Two piezometer tubes method, 74-75 Two-dimensional flow, 319 U U tube, 53 Uniform flow, 316 Unstable condition, 195 Unstable equilibrium, 193, 195 Unsteady flow, 670 V Vacuum pressure, 70, 72 Vapour pressure, 42 Velocity distribution, 499 Velocity in pipes, measurement of, 381 Velocity, coefficient of, 383 Vena contracta, 384 pressure head of, 395 Ventilation of weir, 411 Venture metering, 8 Venturimeter, 478, 479 Vertical acceleration, liquid subjected to, 243

Vertical single column micrometer, 74 Vertical surface, 106 increasing pressure intensity on, 109 Viscometer, 56 Viscosity, Newton's law of, 19 velocity gradient due to, 17 Volume distortion, 315 Vortex flow, 704 aerofoil in, 721 Vortex motion, 462 Vortices formation, 548 W Water line, cross-sectional area of, 212, 214-215 Water supply, distributions of, 6 installation of, 6 Water wheel, 12 Water, jet of, 426, 429, 445 log in, 204 pontoon in, 203 submarine in, 202 Weber's model law, 295 Weber's number, 278 Weir, 406 Wetting liquid (spread out), 44

757