Fluid Mechanics in Channel, Pipe and Aerodynamic Design Geometries 2 9781786301406, 1786301407

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Fluid Mechanics in Channel, Pipe and Aerodynamic Design Geometries 2

...To our family Depy Andreas Nikos Giannis Lilian …as without their support none of this would have ever been possible for us

Engineering, Energy and Architecture Set coordinated by Lazaros E. Mavromatidis

Volume 3

Fluid Mechanics in Channel, Pipe and Aerodynamic Design Geometries 2

Christina G. Georgantopoulou George A. Georgantopoulos

First published 2018 in Great Britain and the United States by ISTE Ltd and John Wiley & Sons, Inc.

Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA. Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address: ISTE Ltd 27-37 St George’s Road London SW19 4EU UK

John Wiley & Sons, Inc. 111 River Street Hoboken, NJ 07030 USA

www.iste.co.uk

www.wiley.com

© ISTE Ltd 2018 The rights of Christina G. Georgantopoulou and George A. Georgantopoulos to be identified as the authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. Library of Congress Control Number: 2018932756 British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library ISBN 978-1-78630-140-6

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

Chapter 1. Pipe Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.1. Introduction . . . . . . . . . . . . . . . . . . . . . . 1.2. Calculation of pipe networks . . . . . . . . . . . . . 1.3. Problem-solving methodology for pipe networks . . 1.4. Overall approach for the network calculation . . . . 1.5. The Hazen–Williams equation for network analysis 1.6. Hazen–Williams and Darcy–Weisbach identity . . . 1.7. Hardy–Cross method . . . . . . . . . . . . . . . . . 1.8. Formulae . . . . . . . . . . . . . . . . . . . . . . . . 1.9. Questions . . . . . . . . . . . . . . . . . . . . . . . . 1.10. Problems with solutions . . . . . . . . . . . . . . . 1.11. Problems to be solved . . . . . . . . . . . . . . . .

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1 3 10 12 13 15 18 21 23 23 53

Chapter 2. Open Channel Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

2.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . 2.2. Non-dimensional parameters in open channels . . . . . 2.3. Open channel types of flow . . . . . . . . . . . . . . . . 2.4. Open channels’ geometrical shapes . . . . . . . . . . . 2.4.1. Channels of rectangular cross-sectional area . . . . 2.4.2. Channels of trapezoidal cross-sectional area . . . . 2.4.3. Channels of circular cross-sectional area . . . . . . 2.5. The hydraulic jump . . . . . . . . . . . . . . . . . . . . 2.6. Calculation of the depth flow after the hydraulic jump . 2.7. Velocity distribution . . . . . . . . . . . . . . . . . . . . 2.8. Velocity distribution at the vertical level . . . . . . . . 2.9. Uniform flow in open channel equations – Chezy type

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57 58 59 61 61 62 63 65 65 68 68 70

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Fluid Mechanics 2

2.10. Best hydraulic cross-sectional area . . . . . . . . . . . . . . . . 2.11. Specific flow energy . . . . . . . . . . . . . . . . . . . . . . . . 2.12. Channels of rectangular cross-sectional area. . . . . . . . . . . 2.13. Open channels’ more adequate cross-sectional areas . . . . . . 2.13.1. Rectangular cross-sectional area . . . . . . . . . . . . . . . 2.13.2. Trapezoidal cross-sectional area . . . . . . . . . . . . . . . 2.14. Non-uniform flow in open channels . . . . . . . . . . . . . . . 2.15. Channels of non-rectangular cross-section area . . . . . . . . . 2.16. Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.16.1. Channels of rectangular cross-sectional area formulae . . . 2.16.2. Channels of trapezoidal cross-section formulae . . . . . . 2.16.3. Channels of circular cross-sectional area formulae . . . . . 2.16.4. Channels of rectangular cross-sectional area formulae . . . 2.16.5. Channels of non-rectangular cross-sectional area formulae 2.17. Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.18. Problems with solutions . . . . . . . . . . . . . . . . . . . . . . 2.19. Problems to be solved . . . . . . . . . . . . . . . . . . . . . . .

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75 79 80 83 83 85 88 90 92 93 93 94 99 100 101 103 116

Chapter 3. Boundary Layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

119

3.1. Introduction . . . . . . . . . . . . . . . . . . 3.2. Laminar incompressible boundary layer . . . 3.3. Characteristic values of the boundary layer . 3.3.1. Thickness δ of the boundary layer . . . . 3.3.2. The thickness displacement δ1′ . . . . . 3.4. Flow types in the boundary layer . . . . . . . 3.5. Formulae . . . . . . . . . . . . . . . . . . . . 3.6. Questions . . . . . . . . . . . . . . . . . . . . 3.7. Problems with solutions . . . . . . . . . . . . 3.8. Problems to be solved . . . . . . . . . . . . .

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147

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Chapter 4. Flow Around Solid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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119 120 125 125 126 127 131 133 133 145

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4.1. Introduction . . . . . . . . . . . . . . . . 4.2. Geometrical characteristics of an airfoil . 4.3. Kutta–Joukowski equation . . . . . . . . 4.4. Aerodynamic paradox . . . . . . . . . . . 4.5. Pressure distribution in an airfoil . . . . . 4.6. Lift curve . . . . . . . . . . . . . . . . . . 4.7. Drag force and drag coefficient curve . . 4.7.1. Drag of skin friction . . . . . . . . . 4.7.2. Form drag . . . . . . . . . . . . . . . 4.7.3. Induced drag . . . . . . . . . . . . . .

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147 148 150 152 153 156 158 158 159 161

Contents

4.8. Parameters that influence the drag coefficient CD. . . . 4.8.1. Dependence of CD on the body’s shape . . . . . . . 4.8.2. Dependence of CD on relative roughness . . . . . . 4.8.3. Dependence of CD on the Reynolds number . . . . 4.9. External flow around industrial solid bodies . . . . . . 4.9.1. Car’s motion . . . . . . . . . . . . . . . . . . . . . . 4.9.2. Surface vessel’s motion . . . . . . . . . . . . . . . 4.9.3. Wind flow in ground constructions . . . . . . . . . 4.9.4. Airplane’s motion . . . . . . . . . . . . . . . . . . . 4.10. Drag in fluid drops and gas bubbles in creeping flow . 4.11. Formulae . . . . . . . . . . . . . . . . . . . . . . . . . 4.12. Questions . . . . . . . . . . . . . . . . . . . . . . . . . 4.13. Problems with solutions . . . . . . . . . . . . . . . . . 4.14. Problems to be solved . . . . . . . . . . . . . . . . . .

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vii

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163 163 165 170 177 177 178 179 181 190 191 193 196 209

Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

211

Appendix 1. Symbols and Units . . . . . . . . . . . . . . . . . . . . . . . . . . . .

213

Appendix 2. Tables and Diagrams of Natural Values . . . . . . . . . . . . . .

219

Appendix 3. Symbols and Basic Conversion Factors . . . . . . . . . . . . . .

237

Appendix 4. International Standard Atmosphere, SI Units . . . . . . . . . .

239

Appendix 5. International Standard Atmosphere, in BS . . . . . . . . . . . .

251

Appendix 6. NACA Airfoils’ Diagrams . . . . . . . . . . . . . . . . . . . . . . . .

259

Bibliography. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

287

Index. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

289

Preface

This book presents an extended and detailed analysis of both the flow phenomena in closed and open channels and the flows around solid bodies. It comprises two volumes. This book is a specialized resource for those students, engineers and researchers who want to focus on the industrial applications of flows and study the fascinating world of internal and external flow phenomena. We have both had extensive experience in teaching, studying and researching fluids since the completion of our respective PhD theses. We felt that it was time to write about the practical and analytical aspects of flow applications, all of which can be applied in industrial flows, to support researchers, engineering students and industrial engineers in the field of fluids in order to optimize their work in “flows”. For the first author, the “fluids direction” began in the early stages of her PhD thesis study in Computational Fluid Dynamics in 1998 at the National Technical University of Athens. The second author’s knowledge of the fluids’ path is very extensive, obtained from more than 45 years of studies and work involved in his PhD thesis and further research work at the University of Patras, as well as through his position as Professor of Aerodynamics at the Hellenic Air Force Academy, spanning more than 35 years. We have both gained substantial experience in Fluid Mechanics research through numerous publications, presentations at international conferences, academic textbook authoring, teaching through international experiences and collaborations. However, we felt that more should be offered to the Fluid Mechanics community, and hence this book. Although we both have experience in writing for academic textbooks, this is our first publication that caters to international students, researchers and engineers, considering the industrial phenomena that are met in international industries and we

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Fluid Mechanics 2

have tried to present most of the applications in flows inside or around bodies. This book is based on books written previously by us on Fluid Mechanics and on Aerodynamics, but for the first time our work focuses on the practical aspects of industrial internal and external flows. Christina, the first author, offers this textbook to the Bahrain polytechnic engineering students and all the industrial delegates who have worked with her in “flows” for many years. She also wishes to express her appreciation for her colleagues, namely Payal Modi, for the thousands of hours of constructive discussions and collaborations in fluids aspects, to Lazaros E. Mavromatidis for his support during the publishing procedure, to her father George who has been her mentor for all these years and to Stephanie Sutton and Amerissa Kapela for their continuing support with the quality of the academic English language. Additionally, George, the second author, wishes to share his more than 40 years of experience in fluids with the fluids community around the world and support them in their “flows” work as best he can. We both have a special sentimental feeling for this book in that we are extremely proud that we have been able to write, publish and offer it to you, hoping that it will really support you in your fluids journey. We have both worked on fluids with a passion not only for our students, but also to honor our colleagues around the world. We are equally happy to say that the Fluid Mechanics community has been served by the same family for more than 40 years. We hope that we will be physically and mentally healthy to continue to serve our students and support our colleagues in the fluids aspects in the future. We hope that you will enjoy this book and be engaged with the fascinating world of flows. Christina G. GEORGANTOPOULOU George A. GEORGANTOPOULOS February 2018

1 Pipe Networks

1.1. Introduction Pipe networks, which are interconnected and often have a netting structure, are used for transportation or even distribution of fluids from their storage or production areas to various other areas for certain purposes. These networks are used in everyday life, such as plumbing networks of water transportation for productive applications such as fire extinguishing, networks of natural gas transportation, networks of fluid and gas transportation, and networks of water pipes, wet waste and compressed air. Some of these networks are simple piping systems equipped with flow adjustment devices, whereas others are complicated, such as fluid distribution networks. Some of the most complicated networks are fire extinguishing networks (because of the fluid used for fire extinguishing), water distribution networks and plumbing networks. Depending on the complication of mixed networks and the form they take for the feasibility of the distribution service, we distinguish them into the following three categories: 1) The tree-system-type pipe networks: these networks are characterized by the presence of a central pipe from which other pipes are branched (pipes of distribution) with a gradual decrease in their cross-section, taking the form of a tree (tree system), as shown in Figure 1.1. 2) The grid pattern: during development the pipes are formed into this pattern, resembling a chessboard, which covers the whole area of the distribution, with a decrease in their cross-sections with respect to the distance (grid system), as shown in Figure 1.2.

Fluid Mechanics in Channel, Pipe and Aerodynamic Design Geometries 2, First Edition. Christina G. Georgantopoulou and George A. Georgantopoulos. © ISTE Ltd 2018. Published by ISTE Ltd and John Wiley & Sons, Inc.

2

Fluid Mechanics 2

Figure 1.1. Tree-system-type pipe network

Figure 1.2. Grid pattern networks

3) The loop pattern: this pattern contains a central pipe, forming loops with smaller pipes at low flow rates (loop system), as shown in Figure 1.3.

Figure 1.3. Loop pattern network

Pipe Networks

3

Of the three forms of networks, the loop and grid systems show high reliability due to their flexibility, expandability and ability to offer multiple paths to the fluid. In general, in all the aforementioned network systems, we distinguish three groups of pipelines: 1) Transportation lines: these pipes transfer the fluid from the storage or production area to the distribution area. 2) Central pipes: these pipes transfer the fluid to the target area, e.g. transfer of water to a town or village or transfer of natural gas to an industrial installation. 3) Supply lines: these pipes of small diameters transfer the fluid from central pipes to the users. Thus, the entire distribution system consists of pipes, valves and pumps. The fundamental aim of a network system is to supply sufficient amounts of fluid to target areas with desired pressures and flow rates. Therefore, the choice of materials, the diameter of pipes and the formation of pipelines in networks are mostly influenced by the necessity of ensuring sufficient pressures and flow rates, despite installation costs and operations. 1.2. Calculation of pipe networks Simple pipe networks have procedures for connecting the pipes in a row or in parallel, as described in Chapter 5 of Volume 1 [GEO 18]. However, the same is not true for complicated pipe networks. The schematic representation of a typical plumbing network is shown in Figure 1.4.

Figure 1.4. Schematic representation of a pipe network

The geometric convergence of three or more pipes is called a network node. In technical applications, nodes with more than four branches do not exist. Nodes with

4

Fluid Mechanics 2

three branches (which are most common in practice) are classified into branch nodes (Figure 1.5a) and convergence nodes (Figure 1.5b). In branch nodes, the incoming branch with a flow rate Q is divided into two branches with flow rates Q1 and Q2, while in convergence nodes, two branches with flow rates Q1 and Q2 converge into one branch with a flow rate Q. When the fluid passes through the node, there is some kind of energy loss, which can be attributed to a decrease in the cross-sectional areas of branches 1 and 2, so that the average velocity of the fluid in all the three branches of the node is approximately the same (V≈V1≈V2), and the walls of the branches are rounded instead of being sharp, to avoid higher energy losses.

Figure 1.5. Pipe network nodes

Minor energy losses at a node can be calculated either by the method of losses coefficient or by the method of equivalent lengths. In Table 1.1, typical values of the losses coefficients for nodes T−90° with constant diameter are given. Furthermore, in Table 1.2, representative values of the equivalent lengths in diameters for tees are given. For a flow in parallel connection, the following two basic rules are applicable: 1) For an incompressible flow in a node, the algebraic sum of flow rates in its branches is zero. This means: i =k

Q = Qi

[1.1]

i =1

for the flow rates of the k branches of the node. During the addition of flow rates, streams entering the node are considered positive, while those leaving the node are considered negative. By applying this rule at node α, shown in Figure 1.6, we get:

Q = Q1 + Q2 + Q3

[1.2]

Pipe Networks

Q2/Q

BRANCH NODE

CONVERGENCE NODE

( Q 0 = Q1 = Q 2 )

( Q1 = Q 2 = Q 0 )

0

0.2

0.4

0.6

0.8

1,0

Q2/Q

0

0.2

0.4

0.6 0.8

5

1.0

Km,1

0.04 0.08

0.05 0.07 0.21 0.35

km,1

0.04 0.17 0.30 0.41 0.51 0.60

Km,2

0.95 0.88

0.89 0.95

km,2

-1.20 -0.40 0.08 0.47 0.72 0.91

h

m ,0 → 1

= k

V

2 0

h

m ,1 2 g

m ,0 → 2

1.10 1.28

= k

V

2 0

m ,2 2 g

h

m ,0 → 1

= k

V

2 0

m ,1 2 g

h

m ,0 → 2

= k

V

2 0

m ,2 2 g

Table 1.1. Coefficient of losses for T-90 nodes

e / d

KIND OF APPLIANCE

e / d

Sudden dilatation d1 / d2 = 1/ 2

20

Tee, entrance from main line

20

Sudden systole d2 / d1 = 1/ 2 Borda mouth Mouth with sharp lips

(

12

Tee, entrance from branch

60

28 18

Valves (totally open): back flow with swing

135

45° curve

16

Angular

145

Standard 90° curve

30

Hydrant

18

90° curve of large radius

20

Butterfly ( d ≥ 6in)

20

90° angle

60

Sliding

13

180° curve

65

Spherical

340

KIND OF APPLIANCE

(

)

)

Table 1.2. Equivalent length values

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Fluid Mechanics 2

Relationship [1.1] is a mathematical expression of the node theorem, also known as the first law of Kirchhoff in the theory of electrical networks 2) For a steady flow in a hydraulic network, for example, in the network shown in Figure 1.6, the total head loss h between the nodes α and b is the same as the respective head losses, h,i , in each branch i of the network, which means:

h = h,i = h,1 = h,2 = h,3

[1.3]

where h,1 , h,2 and h ,3 are the heads of the energy losses in branches 1, 2 and 3, respectively, for the corresponding flow rates Q1, Q2 and Q3. Equation [1.3] constitutes the mathematical statement of the energy principle of the hydraulic network

Figure 1.6. Nodes and branches of a pipe network

The procedure followed for obtaining the solutions of these problems depends on the type of information asked. Therefore, if the total head loss of the flow is known, then it is easy to calculate individual flow rates Qi and finally their sum. The reversed problem is solved with successive approximations because the distribution of the flow rate Q in the individual branches of the network is not known. In fact, in the first approximation, we consider zero energy loss at the nodes and, if they are considered important, we add them in the second approximation. In more complex hydraulic networks, the balance sheet method, head losses in loops and flow rates in nodes are applied. It is difficult to obtain the solution of such network problems; it can be obtained with a suitable computer program.

Pipe Networks

7

However, in addition to this form of network, there are more complicated distribution networks, particularly the grid and loop patterns. A loop with one inlet and two exits, as shown in Figure 1.7, is impossible to obtain by the methods that we developed in Chapter 5 of Volume 1 [GEO 18].

Figure 1.7. Loop pattern network

Before we develop a method for solving these types of networks, we should state the basic relation of these calculations. Therefore, considering that a network consists of pipes (branches) with nodes and forms closed circuits or loops, we will individually examine the relations of the nodes and the loops. Viewing a network macroscopically and applying the continuity equation to the network, we have:

Σm ent = Σm ex

[1.4]

and for incompressible fluids, the equality of flow rates is also applicable:

ΣQent = ΣQex

[1.5]

A respective relationship is applicable for every node of the network. If we conventionally pre-label the flow rates (+ for entrance to the node and − for exit from the node), we have:

ΣQnode = 0 For the node shown in Figure 1.8:

[1.6]

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Fluid Mechanics 2

Q1 − Q2 + Q3 − Q4 − Q5 = 0 or Q1 + Q3 = Q2 + Q4 + Q5

[1.7]

In addition to the points of the branch or the nodes, a network is characterized by closed circuits or loops. A loop is a closed path formed by the sum of successive pipes, which will lead us back to the starting point if we follow them. Thus, in the pipe shown in Figure 1.9, starting from Α and following a clockwise route, we will return to A again: Α–Β–C–D–Α. The pipes that constitute the loop are called branches. For the formation of a loop, at least two branches are required.

Figure 1.8. Loop pattern of network

The loop shown in Figure 1.9 constitutes four branches (ΑΒ, ΒC, CD and DΑ) and four nodes (Α, Β, C and D). If we consider the flow path in the pipes shown in Figure 1.9, by applying the Bernoulli equation between points Α and C of the fluid paths ΑΒC and ADC, it is given that: A

B

Δ

Γ

Figure 1.9. Closed-loop pattern of a network

Pipe Networks

9

hABC = hADC  hAB + hBC = hAC + hDC   hAB + hBC − hAD − hDC = 0

[1.8]

If we conventionally pre-label the losses, putting the sign + when the fluid flow is clockwise in the loop and the sign − when it is opposite, equation [1.8] becomes:

hAB + hBC + hAD + hDC = 0   Σhloop = 0

[1.9]

Equation [1.9] is general and applicable for any closed system of pipes (loop). In combination with equation [1.6] of the branch points (nodes), it is the key for the solution of the network problems. We emphasize that both these relationships have resulted from conventional pre-labeling of the flow rates and the losses. In their specific application, we must take into account the rules of pre-labeling and specify the respective relationships. For example, for the nodes shown in Figure 1.8, the equation of the nodes will take the form [1.5], while for the node shown in Figure 1.9, the loop equation will take the form [1.8]. Finally, we remind ourselves that for each branch, we have the losses equation (Darcy–Weisbach):

hi =

8 π ⋅ g ⋅ di2 2

   ⋅  fi ⋅ i + ΣKi  ⋅ Qi2  di 

which can take the form:

hi = ai ⋅ Qi2 where ai =

[1.10]

   8 ⋅ f ⋅ i + ΣKi  2  i π ⋅ g ⋅ di  di  2

[1.11]

The analysis and calculation of the distribution networks is a complicated and time-consuming procedure, which is based on equations [1.6], [1.9] and [1.10] (nodes, loops and branches, respectively).

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Fluid Mechanics 2

1.3. Problem-solving methodology for pipe networks Let us consider a pipe network in which the pipes have a common point, that is, a node or a bend, and the other side is connected to a tank, whose level is at a height h, as shown in Figure 1.10. In this category of problems, the kinematic viscosity ν, the lengths of the pipes

1 ,  2 ,  3 , their diameters d1, d2 , d3 , their heads h1 , h2 , h3 and their roughnesses ε1 , ε 2 , ε3 are usually given while their flow rates Q1, Q2 , Q3 are asked. The following methodology is used to solve this category of problems: 1) We assume a value for the head hΑ , where:

hΑ =

pΔ

γ

+ hΔ = pressure head

[1.12]

Figure 1.10. Pipe network with a common point for the pipes

2) From the Darcy–Weisbach relationship, the given data of the problem and the Moody diagram, we find the flow rates Q1 , Q2 , Q3 , ..., Qn . For example:

hn = hn − hA = f n

 n Vn2 ⋅ dn 2 g

where fn is found by various tests, and Vn can be found by the relationship:

[1.13]

Pipe Networks

Vn =

2 ghn dn fn n

Therefore, Qn =

11

[1.14]

π dn2 2 ghn dn 4

fn  n

[1.15]

3) After calculating all the flow rates, we may use the continuity equation. This means that the values that we find for Q1 , Q2 , Q3 , ..., Qn have to satisfy the continuity equation. If the flow rates Q1, Q2 and Q3 are higher than the flow rates of branches, then we obtain a higher value for hA, and we repeat steps 1, 2 and 3. If the flow rate to the branch is lower than the flow rate from the branch, then we use a smaller value for hA. It is clear from the above procedure that for these problems, the following conditions are applicable: 1) For every pipe, the Darcy–Weisbach relationship must be satisfied. 2) Q → branch = Q → branch. 3) There has to be a flow from the higher tank to a lower one. This means that one of these relationships must be applicable:

Q2 = Q1 + Q3

[1.16]

or Q3 = Q1 + Q2

[1.17]

As hA < h1 , we have a flow toward a tank, so it is given that:

Q2 = Q1 + Q3

[1.18]

4) Regardless of the losses in the flow in any pipe, equivalent lengths will be able to express them, which are added to the real length, as mentioned in section 1.2.

12

Fluid Mechanics 2

1.4. Overall approach for the network calculation In the beginning of this chapter, we mentioned that the basic purpose of a distribution network is to ensure the necessary flow rate and pressure for the exit of the fluid from the network. Consequently, a water supply network must ensure a flow rate higher than 1  / s for every user and a pressure of 2 bar. The flow rate is related to not only the selection of the pipes with suitable diameters but also the differences of the energy heads between the nodes of exit and inlet, as well as the presence or absence of pumps in the network. The collateralization of the required pressure is related to the energy heads: the pressure head at the entrance nodes, the heights and the height that the pumps attribute. More specifically, if we assume a distribution network that is supplied by a tank, as shown in Figure 1.11, the Bernoulli equation, if applied between the surface of tank Α and the node Ε, will give:

Figure 1.11. Pipe network connected to a tank

y A − yE



VA2 − VE2 p A − pE + = ΣhAE − hp  Y 2g

pE V 2 − VE2 p A = y A − yE + A + − ΣhAE + hp Y Y 2g

[1.19]

As the variation of the kinetic energy is negligible because the flow velocities are smaller than 3 m/sec, as well as zero at the surface of the tank, the above relationship becomes:

pE p = y A − yE + A − ΣhAE + hp Y Y

[1.20]

Pipe Networks

where

h

AE

13

= hAB + hBΓ + hΓE , with the losses of every branch pre-labeled

according to the assumption described in section 1.2. 1.5. The Hazen–Williams equation for network analysis In the previous sections for the solution of the network problems, we used the Darcy–Weisbach losses equation as follows:

h1 = f

 V 2 8 f ⋅ Q2  = 2 5 = 8 f 2 d −5Q2 d 2g π d g π g

[1.21]

Considering that the flow is turbulent, we calculate the friction coefficients f from this relationship and then the flow rates.

Figure 1.12. Pipe network for water distribution

Thus, when the flow is turbulent, the losses along a pipe with respect to the flow rate, the friction coefficient and the pipe’s dimensions can be expressed by the following relationship:

h1 = CQn

[1.22]

where the coefficient C is a parameter that depends solely on the dimensions and the roughness of the pipe. Relationship [1.22] can be derived from [1.21] after we express the friction coefficient f with respect to the flow rate, which means that:

f = KQa

[1.23]

14

Fluid Mechanics 2

Substituting [1.23] into [1.21] gives [1.22]. After a series of experiments, Hazen and Williams, who worked with pipes of various diameters and roughnesses, found that the value of the average water velocity in pipes is proportional to the hydraulic radius if it is raised to a power and to the square root of the inclination value of the pressure gauge line, which means that: V ∞RhX S

[1.24]

where Rh is the hydraulic radius and S is the inclination of the pressure gauge line = h1 /  , with  being the length of the pipe and h1 being the losses during the length



of the pipe.

Therefore, depending on the unit system (SI or British), the Hazen–Williams equation in its final form is given by the relationship: 1) In the metric system (SI): V = 0,849Wm Rh0,63 ⋅ S 0,54 in m / sec

[1.25]

If Rh is in m, the coefficient Wm , depending on the type of the pipe, has the following values: No. Kind and situation of pipes

Value of coefficient

1

Small, not of a good structure

up to 40

2

Old, in bad situation

60–80

3

Old, made of cast iron, nailed

95

4

Made of cast iron or mud brick

100

5

New, steel, nailed

110

6

Glazed, in good situation

110

7

Wooden, smooth

120

8

Very smooth

130

9

Made of concrete, of large dimensions 130

10

Asbestos tubes

140

11

Very smooth and rectilinear

140

Table 1.3. Values of the coefficient Wm

Wm

Pipe Networks

15

2) In the British system:

V = 1,318WE Rh0,63S 0,54 in ft/sec

[1.26]

WE

Table of values of the coefficient No. Kind and situation of pipes

Value of coefficient WE

1

Made of cast iron, corroded

80

2

Made of cast iron, after some years of function 100

3

Glazed pipes of drainages

110

4

Average situation of cast iron

110

5

New, steel, nailed

110

6

New, smooth, made of cast iron

130

7

Very smooth and rectilinear

140

Table 1.4. Values of the coefficient WE

If Rh is in ft, the coefficient WE , depending on the type of the pipe, has the above values. 1.6. Hazen–Williams and Darcy–Weisbach identity These two expressions that offer the possibility of solving complicated problems of pipe networks have some common points. Therefore: 1) If we assume relationship [1.25]:

V = 0,849Wm Rh0,63 S 0,54

[1.27]

where S = h1 /  , and substituting S and rearranging, we have:

 h1    

0,54

=

V Rh−0,63 0,849Wm 1/0,54

 V  h1 =   Rh−0,63   0,849Wm 

1,852

 V  Rh−0,63  =   0,849Wm 

16

Fluid Mechanics 2

For a cylindrical pipe, Rh =

d . Then, we have: 4

 ⋅V 1,852 d  h1 = 1,852   (Wm 0,849) 4

−0,63×1,852

 ⋅V − 0.148 V 2 2 g  d  = ⋅ ⋅   0,7385 2g Wm1,852  4 

−1,167

2 Now, if we set g = 9,81m / sec , we have:

h1 =

41,167 × 2 g  V 2 V −0,148 134d −0,019    V 2 1 ⋅ ⋅ 0,167 = =   1,852 1,852 0,148 0,148 0,7385Wm d 2 g d Wm d  d  2g V

h1 =

 ⋅V 1,852 d  1,852   (Wm 0,849) 4

=

−0,63×1,852

=

 ⋅V − 0.148 V 2 2 g  d  ⋅ ⋅   0,7385 2g Wm1,852  4 

−1,167

134d −0,019  ν 0,148      V 2      = ν 0,148Wm1,852  V 0,148 d 0,198   d   2 g  2 134d −0,019 L U2   V = f 0,148   D 2g Re ν  d  2g

Wm1,852 0,148

However, because f =

134d −0,019 , we have: ν Re0,148

Wm1,852 0,148

2

  V h1 = f    d  2g

[1.28]

2) If we consider relationship [1.27] and substituting S and rearranging, we have d for a cylindrical pipe, where Rh = : 4 d  V = 1,318WE   4

So, solving for

0,63

 h1    

h1 , we find: 

0,54

=

4Q πd2

Pipe Networks

17

1

  0,54 4Q   2 h1  πd   = 0,63   d   1,318WE      4   1,852

 41,63    1, 318   h1 =   WE     

⋅

1,852

 2,3136  h1 =    WE 

⋅

 Q1,852 d 4,87

 d

4,87

Q1,852

[1.29]

where all the lengths are expressed in ft and the flow rate in ft/sec. Comparing [1.29] and [1.22], we see that: 1,852

 2,3136  C =   WE 

 d

4,87

[1.30]

and n =1,852. Combining relationship [1.29] with the nomograph of Hazen–Williams, we obtain solutions in various cases of pipe connection, which are much easier and faster than those obtained using the Moody diagram. Moreover, the Hazen–Williams relationship is simpler than the Darcy–Weisbach relationship because the calculation of the coefficient C is easier. This can be solved by using tables or graphs, and therefore this method has been commonly used in the calculation of water networks. However, it also has some disadvantages of providing less accurate results than the Darcy–Weisbach equations, and it is applicable only if the fluid is water. Moreover, both relationships are empirical, but the Darcy–Weisbach equation is theoretically more close to an analytical method and compatible with the conclusions of the two-dimensional analysis.

18

Fluid Mechanics 2

1.7. Hardy–Cross method Another method by which we can solve problems of pipe networks is the Hardy–Cross method, which is a relatively simple procedure. According to this methodology, we initially assume flow rate distribution under only one condition, which is to satisfy the mass conservation in each node. Continuing, we make corrections to these flow rates at each loop of the network, with corrections every time a new value of the flow rate comes up at every circle of calculations so that finally there is a better balance of flow rates in the network than the value we assumed before. If we get in the beginning a good value for the flow rate distribution, we can have convergence in the final value after two or three attempts. For simple networks, as shown in Figure 1.12, the solution can be obtained easily using a calculator, while for more complicated ones, the assistance of a computer is necessary. The procedure for finding a solution according to the Hardy–Cross method is as follows: 1) We initially assume the flow rate distribution Q01, Q02 , Q03 ,... with the only limitation at each node of the network being the mass conservation applicable, meaning that the amount of the mass of water that goes in is the same that comes out. 2) We calculate the losses at each pipe of the network based on the relationship:

h10 = C1Q0n

[1.31]

Therefore, if there are seven pipes in the circuit, we will find seven losses using the Hazen–Williams nomograph. n h101 = C1Q01 n h102 = C2Q02

h107

⋅ ⋅ ⋅ ⋅ n = C7 Q07

3) Choosing any direction and paying attention not to make a mistake in the sign, we find the following sum in each loop of the network:

Pipe Networks

m

m

i =1

i =1

 h10i =  Ci Q0ni

19

[1.32]

where m is the number of pipes in the loop, n = 1,852 and i is the sum indicator, which takes the values 1, 2, 3,…,m. 4) We also calculate the sum of the absolute values of the terms for each loop: m

 nCi Q0ni−1 i =1

m

= n  Ci Q0ni−1

[1.33]

i =1

5) In order to correct the values of the flow rates in each loop, so that at each m

loop we finally have

 h1 i =1

i

= 0 , we calculate the value of the corrective flow rate

ΔQ of each loop, where: m

ΔQ = −

 Ci ⋅ Q0ni i =1 m

[1.34]

n ⋅  Ci Q0ni−1 i =1

This value of the corrective flow rate is obtained with the following analysis. Assume that Q = Q0 + ΔQ , where Q is the real flow rate of each pipe, Q0 is the initial flow rate that we assumed and ΔQ is the corrected flow rate value. However, because we have that:

(

)

h1 = CQ n = C ( Q0 + ΔQ ) = C Q0n + nQ0n −1ΔQ + ... , n

if ΔQ is relatively small with Q0 , we will not take into account the terms that are powers of the ΔQ , and thus it is given that:

CQn = CQ0n + CnQ0n−1ΔQ whereas if the loop has m pipes, the following relationship is applicable:

20

Fluid Mechanics 2

m

m

m

i =1

i =1

i =1

 Ci Qin =  Ci Q0ni + Ci nQ0ni−1ΔQ

[1.35]

m

 Ci Qin = 0

[1.36]

i =1

Moreover, because ΔQ is the same for all the pipes of each loop, we can take it out of the sum, so we have: m

ΔQ = −

 Ci Q0ni i =1 m

n⋅ i =1

m

but as

 h1 i =1

0i

[1.37]

Ci Q0ni−1

m

=  Ci Q0ni is valid, [1.37] becomes: i =1

m

ΔQ = −

 h1

0i

i =1

m

h10 i

i =1

Q0i

1,852 m

h10i

i =1

Q0i

 1,852



[1.38]

m

= 1,852 Ci Q0ni−1 since n = 1,852

[1.39]

i =1

6) We calculate the new flow rates for the pipes of each loop from the relationship:

Qi = Q0i + ΔQ and we repeat the procedure (ii) until we achieve the desired accuracy, for example:

Qi = Q0i < 10−2 flow rate units

[1.40]

Pipe Networks

21

1.8. Formulae 1) Continuity equation for the node networks: ΣQnode = 0

where

Q

node

is the sum of the flow rates of all the pipe networks.

2) Losses of a network: Σhloop = 0

where

h

loop

is the sum of the losses networks

3) Losses equation for the network (Darcy–Weisbach)

hi =

and if ai =

8 π g ⋅ d i4 2

   ⋅  f i ⋅ i + ΣK i  ⋅ Qi2 di  

  8  f ⋅ i + ΣK i  4  i di π gd i   2

then hi = ai ⋅Qi2

where hi is the losses height, fi is the friction operator,  i is the pipe’s length, di is the pipe’s diameter, g is the acceleration due to gravity, Ki is the losses operator and Qi are flow rates. 4) Average flow velocity in a network:

Vn =

2ghn dn fn ⋅ n

22

Fluid Mechanics 2

where g is the acceleration due to gravity, hn is the losses, d is the diameter, f is the friction coefficient and  is the pipe’s length. 5) Network flow rate:

Q=

π ⋅ dn2 2ghn dn 4

fn  n

where all the variables have known values. 6) Bernoulli equation for networks: PE

γ

= y A − yE +

PA

γ

− ΣhAE + h p

where PE is the pressure in the position E, γ is the specific weight of the fluid, yA is the height of point A, yE is the height of point E, PA is the pressure of point A, hAE is the head losses and hp is the pressure head. 7) Losses equation of Hazen–Williams:

VaRnx S hi is the inclination of the pressure gauge line,   is the pipe’s length and hi is the losses in the length  of the pipe.

where Rn is the hydraulic radius, S =

a) In the SI:

V = 0,849Wm Rh0,63 ⋅ S 0,54 in m / sec and R h is in m. b) In the British system:

V = 1,318WE Rh0,63 ⋅ S 0,54 in ft / sec and R h in ft .

Pipe Networks

23

1.9. Questions

1) What are pipe networks, where are they used and what are the categories of networks? Describe each category. 2) How many and which pipes’ groups may we consider for hydraulic systems? 3) What are nodes of the networks, in which categories do we distinguish them and what are their effects on the fluid flow? How many types of node connection are there and which rules are applicable for a parallel connection? 4) Define node theorem. What is the energy condition of a hydraulic network? 5) In a closed circuit of loops and branches, which laws are applicable? 6) What is the solution methodology of a network that consists of a node and a branch, while the other end of the pipe connects to three tanks that lie at a height, different from the node for each one? 7) What is the Hazen–Williams equation for the network analysis, how is it generally expressed and which form does it take in the SI and the British system? 8) Is there any identity (connection) between the Hazen–Williams equation and the Darcy–Weisbach equation, and how is it proved in the SI and the British system? 9) What is the Hardy–Cross method for the solution of network problems and what is the solution procedure according to this method? 1.10. Problems with solutions

1) In a network of tubular pipes made up of galvanized iron of diameter 6 inches, we expect the velocities to vary from 0.5 to 2 m/sec. Calculate the coefficients r and m m in the exponential expression of the loss of the load, hf = r Q per 100 m of the

pipe’s length. The coefficient of the kinematic viscosity is given as n = 1,31×106 m2 / sec for water at 10°C. Solution: The relative roughness is:

ε d

=

0,01525cm = 0,001 15,25cm

The expected Reynolds numbers for the two values of the velocities are:

24

Fluid Mechanics 2

Re1 =

Re2 =

V1d

ν

=

0,5 × 0,1525 = 5,82 ×104 1,31×10−6

=

2 × 0,1525 = 2,328 ×105 1,31×10−6

V2d

ν

From the Moody diagram, we find: f1 = 0024 and f 2 = 0215

Therefore, from the equation:

hf = f

hf =

0,083(0,024)Q12  ⋅V 2  hf = f = rQ1m 5 d 2g d

0,083(0,0215)Q22

d

5

= rQ2m

Dividing the above equations by terms, we get: m

2

 Q2  f 2  Q2  0,0215 2 (4) = 14,33   hf =   = f1  Q1  0,024  Q1  4m = 14,33 From which we take:

m=

because

λογ 14,33 = 1,92 λογ 4

Q2 V2 2 = = =4 Q1 V1 0,5

The value of r is taken from one of the above exponential equations:

(0,083)(0,0215)(100)Q2 2 (0,1525)5

= rQ21,92

Pipe Networks

r=

(0, 083)(0,0215)(100)Q20,08 (0,1525)5

because Q2 =

πd2 4

V2 =

25

= 1660

(3,14)(0,1525) 2 2 = 0, 0365 4

1,92 Finally, the exponential equation hf = 1660Q gives the load loss per 100 m

length of the pipe. 2) Two open water tanks Α and Β are connected to a horizontal pipe (see Figure (2)) with an internal diameter of 15 cm, a length of 500 m and a friction coefficient of 0.02. The altitude range of the water’s free surfaces in the two tanks is 8 m. The secondary losses of the flow are considered to be negligible. a) Calculate the volumetric flow rate of the water. b) In order to increase the flow rate of the water to 25%, a second pipe of length 200 m is placed, as shown in Figure (2). The new pipe has the same friction coefficient as the initial one. Calculate the required diameter of this additional pipe. Solution: a) The flow rate Q of the water is calculated from the production of the average velocity V1 of the transverse cross-section of the pipe: Q = V1 S1 = V1

π d12 4

(1)

For the calculation of velocity V1 , we apply the equation of mechanical energy, in heads form, for an incompressible, steady flow between the positions α and b: Ζα +

pα

γ

+

pβ Vβ 2 Vα 2 = Ζβ + + + h 2g γ 2g

(2)

or if we take into account that the pressure pα = pb = pb and we consider that the velocity Vα ≈ Vb ≈ 0 :

h = Za − Zb = 8m

(3)

26

Fluid Mechanics 2

Figure (2)

Since the secondary losses of the energy are considered to be negligible, the head h is equal to the friction head h f of the flow in the pipe. The head h f is calculated from the Darcy–Weisbach equation:

hf = f

(1 +  2 ) V12 d1 2g

(4)

Solving the last equation for V1 , we have:

V1 =

2 gd1h f f (1 +  2 )

=

2(9,81m / s 2 )(0,15m)(8m) = 1,53m / sec (0,02)(500m)

(5)

Substituting the values of V1 and d1 from equation (1), we have:

Q = (π / 4)(1,53m / s)(0,15m)2 = 0,027m3 / s

(6)

So, the volumetric flow rate of the water is 27  / S . b) After the addition of pipe 3, the volumetric flow rate of the water to pipe 1 is 25% higher than the initial value, which means that it is:

Q1 = 1,25(0,027m3 / s) = 0,03375m3 / sec The new flow velocity in pipe 1 is:

(7)

Pipe Networks

V1 =

Q1 4Q1 4(0,03375 m3 / s) = = = 1,91m / sec S1 π d12 π (0,15m)2

27

(8)

From the application of the node theorem in the node γ, arises the relationship: Q1 = Q2 + Q3

(9)

where Q 2 and Q 1 are the flow rates of the water in pipes 2 and 3, respectively. For the fluid that flows from tank Α to tank Β through pipes 1 and 2, equation (2) takes the following form:

Zα − Z β = h = h,1 + h,2 = f1

1 V12  V2 + f2 2 2 d1 2 g d2 2 g

(10)

In this equation, the only unknown quantity is velocity V2 of the water in pipe 2, which can be calculated as: (8m) =

2 2 (0, 02)(300m)(1,91m / s )2 (0, 01)(200m)(1,91m / s ) V2 + 2(0,15m)(9,81m / s 2 ) 2(0,15m)(9,81m / s 2 )

which means: V2 = 0, 64m / sec Q2

(11)

So, the flow rates of the water in pipes (2) and (3) are: Q2 = V2 S2 = υ2 (π d 22 / 4) = (π / 4)(0, 64m / s ) − (0,15m)2 = 0, 01131m3 / s

(12)

Q3 = Q1 − Q2 = (0, 03375m3 / s ) − (0, 01131m3 / s ) = 0, 02244m3 / s

(13)

For the fluid that flows from tank Α to tank Β through pipes 1 and 3, equation (2) takes the following form:  V 2  V2 Ζα − Ζb = h = h,1 = h,3 = f 1 1 + f 3 3 d1 2 g d3 2 g

where V3 is the velocity of the water in pipe 3:

(14)

28

Fluid Mechanics 2

Q 4Q3 4(0, 02244m3 / s ) 0, 02857 3 V3 = 3 = m / sec = = 2 2 S3 π d 2 π d d 3 3 3

(15)

By combining the last two equations, we have:  3 (0, 02857 d3−2 ) 2 1 V12 Ζα − Ζb = f +f d1 2 g d3 2g

(16)

Replacing in this equation the values of the known quantities, it arises that: (8m) =

−2 2 (0, 02)(300m)(1,91m / s )2 (0, 02)(200m)(0, 02857 d3 ) + 2(0,15m)(9,81m / s 2 ) 2d3 (9,81m / s 2 )

Moreover, after the relating calculations we receive: d3 = 0,197m

(17)

So, the internal diameter of the additional pipe has to be 19.7 cm. 3) The three pipes, shown in Figure (3a), have lengths  AB = 10m,  AC = 12m and  BC = 8m and diameters d AB = 5cm, d AC = 5m, d BC = 7cm, respectively. These pipes are made up of galvanized iron, and the flow rates of the water to and 3 from the system are QA = 0, 010m2 / sec (inflow), QB = 0,006m / sec (outflow) and QC = 0, 004m3 / sec (outflow). If the secondary losses of the system are

considered to be negligible, calculate the flow rates of these three pipes ( V = 10−6 m 2/sec ε ΑΒ = 0, 00015m , ε ΑC = 0, 00015m , ε Β C = 0, 00015m ).

Figure (3a)

Pipe Networks

29

Solution: Here, we have a network with a loop and three branches. We prepare the sketch as shown in Figure (3a) according to steps 1, 2 and 3. We will calculate the relationships of losses – flow rates of the three pipes. Pipe ΑΒ: (ε ΑC / dΑC ) = 0,003AIT → f ΑC = 0,0262 (Moody) and replacing in the losses equation the f ΑΒ , we have: 2 aΑΒ = 69,18 (SI )  hΑΒ = 69.180 ⋅ QΑΒ (SI )

(1)

Pipe ΑC: (ε AC / d AC ) = 0,0025 AIT → f AC = 0,025 (Moody) 2

a AC = 31, 720 ( SI )  hAC = 31.720 ⋅ QAC ( SI )

(2)

Pipe ΒC: (ε BC / dBC ) = 0,0021AIT → f BC = 0,02394 (Moody) 2 aBC = 9,385 (SI )  hBC = 9.385 ⋅ QBC (SI )

(3)

First testing: we assume a random but logical flow rate distribution. In general, larger diameters lead to higher flow rates. If the diameters are the same, of smaller length, it leads to larger flow rates. A better criterion is the comparison of the coefficients α of the above relationships. The larger the α, the smaller the flow rate. i) Starting from node A and with the coefficients as a criterion, we have:

QAB = 3 / sec = 0,003m3 / sec , so it occurs that: QAC = 7 / sec = 0,007m3 / sec (10 = 3 + 7) For node Β, we have:

QBC = Q − QAB = 3 / sec = 0,003m3 / sec (in flow to Β).

30

Fluid Mechanics 2

The random distribution of the flow rates is shown in Figure (3b) Until now, we have made only one assumption:

QAB = 3 / sec This agrees with what we have mentioned in the theory. ii) Formulation of checking relationships: As there is only one loop, we have one checking relationship. According to the signs convention and Figure (3c), we have: Σ h = hAB − hBC − hAC = 0

(4)

iii) Assumption check: [1]  hAB = 0, 62m [2]  hAC = 1, 55m [3]  hBC = 0, 08m [4]  Σh = −1, 02 m ≠ 0

iv) Flow rate correction: ΔQ =

−Σh −( −1, 02)  ΔQ =  2 ⋅ (69.180 ⋅ 0, 003 + 31.720 ⋅ 0, 007 + 9385 ⋅ 0, 003) Σ 2 ⋅ α i ⋅ Qi

 ΔQ = 0, 0011m3 / s = 1,1 / sec

We correct the flow rates (paying attention to whether the direction is the same or opposite to the clockwise route) as: AB :

Q AB = (3 + 1,1) / s = 0, 0041m3 / sec

ΑC :

Q AC = (7 − 1,1) / s = 0, 0059 m 3 / sec

ΒC :

QBC = (3 − 1,1) / s = 0, 0019 m3 / sec

Second testing: we repeat the calculation of the losses h, Σh and the required correction Δ Q using the corrected flow rates.

Pipe Networks

Figure (3b)

31

Figure (3c)

From (1)  hAB = 1.17m (2)  hAC = 1.10m (3)  hBC = 0.03m (4)  Σh1 = +0, 03m ≠ 0 and  ΔQ = −0, 00004m / sec = −0, 04 / sec So, the correction that came up is very small. We correct the flow rates, and the results are very close to the flow rates that we are looking for:

QAB = 0,00406m3 / sec = 4,06 / sec QAC = 0,00594m3 / sec = 5,94 / sec QΒC = 0,00194m3 / sec = 1,94 / sec 4) For Figure (4), calculate the flow rates Q1, Q2 and Q of the three pipes. The satisfactory solution is the one in which the flow rate to branch D and the flow rate from D agree to the second decimal numeral.

32

Fluid Mechanics 2

Figure (4)

The data given are as follows: ο −6 2 Water of temperature 12 C ν = 1,244 ×10 m / sec

d1 = 12′′ , 1 = 2000m, ε1 = 0, 00048 Ft , y1 = 50m d 2 = 8′′ ,  2 = 1000m, ε 2 = 0, 00048 Ft , y2 = 80m d3 = 8′′ ,  3 = 3000m, ε 3 = 0, 00048 Ft , y3 = 100m

Solution: We consider that y A is equal to the height of the intersection point of the pressure gauge lines of the three pipes. So: If yΑ > (of the height of the middle tank) y2 then  we have : Q3 = Q1 + Q2

Pipe Networks

33

But if yΑ < y2 then  we have: Q3 + Q2 = Q1

In general, the solution is found after a number of iterations of approximate solutions by applying the following theory: if the flow rate to branch D is such that it is larger than the flow rate from the Δ at the end of an iteration, then we conduct a new iteration in which the new assumed value of yA is larger than that of the previous iteration. If the flow rate to D is found to be smaller than the flow rate from D in the end of an iteration, we conduct a new iteration of calculations by choosing a smaller yA. We stop the iterations of the approximate solutions when finally we find that the flow rate to and from D agree to the second decimal numeral. First iteration: assume that y A = 75m Then: a) for pipe (1): 1/2

h1 = yΑ − y1 = f1

 2h gd  1 V12  V1 =  1 1  d1 2 g  f11 

and because h1 = 75m − 50m = 25m 1/2

 2 × 25 × 9,81×12 × 0, 0254  then: V1 =   2000  

ε d1

=

× f1−1/2 = 0, 2734 f1−1/2 in m/sec

0,00048 ×12′′ = 0,00048 12′′

Assuming that f1 = 0, 017 , −1/2 then V1 = 0,2734(0,017) m / sec = 2,0969m / sec

so Re1 =

V1d1

ν

=

2,0969 ×12 × 0,0254 = 5,14 ×105 −6 1, 244 ×10

34

Fluid Mechanics 2

So, with: ε 5  = 0, 00048 και R e1 = 5,14 × 10   f1 = 0, 0175 (Moody)  d1  −1/2 So, V1 = 0,2734(0,0175) m / sec = 2,0667m / sec

and Re1 =

V1d1

ν

=

2,0667 ×12 × 0,0254 = 5,06 ×105 1, 244 ×10−6

So, with: ε 5  = 0, 00048 και Re1 = 5, 06 × 10   f1 = 0, 0175 (Moody)  d1 

So, V1 = 2, 0667 m / sec and Q1 = V1

or Q1 =

π d12 4

2, 0667 × πΧ(12 × 0, 0254) 2 3 m / sec = 0,1508m3 / sec 4

b) For pipe (2): 1/2

 2h gf   V2 h 2 = y A − y2 = f 2 2 2  V2 =  2  d2 2 g  f2 2  So, with h2 = 75 − 80 m = 5m 1/2

 2 × 5 × 9,81× 8 × 0, 0254  V2 =   1000   and

ε d2

=

× f 2 −1/2 = 0,1412 f1−1/2 in m/sec

0,00048 ×12′′ = 0,00072 8′′

Pipe Networks

Assuming that f 2 = 0, 019 , −1/2 m / sec = 1,0244m / sec then V2 = 0,1412(0,019)

so Re2 =

V2 d2

ν

=

1,0244 × 8 × 0,0254 = 1,67 ×105 1, 244 ×10−6

So, with: ε 5  = 0, 00072 and Re 2 = 1, 67 × 10   f 2 = 0, 02 (Moody) d  2  −1/2 = 0,9984m / sec then V2 = 0,1412(0,02)

so, Re2 =

V2 d2

ν

=

0,9984 × 8 × 0,0254 = 1,63 ×105 −6 1, 244 ×10

So, with: ε 5  = 0, 00072 and Re 2 = 1, 63 × 10   f 2 = 0, 023 (Moody)  d2  −1/2 = 0,9310m / sec then: V2 = 0,1412(0,023)

so, Re2 =

V2 d2

ν

=

0,9310 × 8 × 0,0254 = 1,52 ×105 −6 1, 244 ×10

So, with: ε 5  = 0, 00072 and Re 2 = 1,52 × 10   f 2 = 0, 023 (Moody)  d2 

then V2 = 0,931m / sec and Q2 = V2

π d2 4

=

0,931× π × (8 × 0, 0254) 2 3 m / sec = 0, 0302m3 / sec 4

35

36

Fluid Mechanics 2

c) For pipe (3): 1/2

h3 = y A − y3 = f3

 2h gd   3 V32  V3 =  3 3  d3 2 g  f3 3 

So, with h3 = 75 −100 m = 25m 1/2

 2 × 25 × 9,81× 8 × 0,0254  V3 =   3000  

ε d3

=

f3−1/2 = 0,1823F3 −1/2 in m/sec

0,00048 ×12′′ = 0,00072 8′′

Assuming that f 3 = 0, 019 , −1/2 = 0,1823(0,019)−1/2 = 1,3225m / sec then V3 = 0,1823 f3

Re3 =

V3d3

ν

=

1,3225 × 8 × 0,0254 = 2,16 ×105 1, 244 ×10−6

with ε 5  = 0, 00072 and Re3 = 2,16 × 10   f3 = 0, 0197 (Moody)  d3  −1/2 = 1,2988m / sec then V3 = 0,1823(0,0197)

So, Re3 =

V3d3

ν

=

1, 2988 × 8 × 0,0254 = 2,12 ×105 1, 244 ×10−6

So, with: ε 6  = 0, 00072 and Re3 = 2,12 × 10   f3 = 0, 0197 (Moody)  d3 

so V3 = 1, 2988m / sec

Pipe Networks

and Q3 = V3

π d 32 1, 2988 × πΧ (8 × 0, 0254) 2 4

4

37

m 3 / sec = 0, 0421m 3 / sec

So, from the results of the first circle, assuming that y A = 75m < y2 = 80m , we have:

Q1 = 0,1508m3 / sec Q2 = 0,0302m3 / sec Q3 = 0,0421m3 / sec thus, it must always be Q3 + Q2 = Q1 . However, we see that:

Q2 + Q3 = 0,0723m3 / sec < Q1 = 0,1508m3 / sec So, we conduct another iteration by choosing a value of y Α smaller than the first one. Second iteration: assume that y A = 65m a) For pipe (1): h1 = 65 − 50 = 15m 1/2

 2h gd  V1 =  1 1   f11 

ε d1

=

1/2

 2 ×15 × 9,81×12 × 0,0254  =  2000  

0,00048 ×12′′ = 0,00048 12′′

Assuming that f1 = 0, 017 , −1/2 = 1,6243m / sec then V1 = 0,2118(0,017)

f1−1/2 = 0, 2118 f1−1/2

38

Fluid Mechanics 2

Re1 =

V1d1

ν

=

1,6243 ×12 × 0,0254 = 3,98 ×105 1, 244 ×10−6

so, with: ε 5  = 0, 00048and Re1 = 3,98 × 10   f1 = 0, 018 (Moody)  d1  −1/2 = 1,5787m / sec then V1 = 0,2118(0,018)

Re1 =

V1d1

ν

=

1,5787 ×12 × 0,0254 = 3,87 ×105 1, 244 ×10−6

so, with: ε 5  = 0, 00048and Re1 = 3,87 × 10   f1 = 0, 018 (Moody) d  1 

b) For pipe (2): h 2 = 80 − 65 = 15m 1/2

 2h gd  V2 =   2 2   f2 2 

−1/2 or V2 = 0,2445 f2 in m/sec

ε d2

=

0,00048 ×12′′ = 0,00072 8′′

Assuming that f 2 = 0, 02 , −1/2 = 1,7289m / sec then V2 = 0,2445(0,02)

Re2 =

V2 d2

ν

=

1/2

 2 ×15 × 9,81× 8 × 0,0254  =  1000  

1,7289 × 8 × 0,0254 = 2,82 ×105 1, 244 ×10−6

f 2−1/2

Pipe Networks

39

so, with: ε 5  = 0, 00072 and Re 2 = 2,82 × 10   f 2 = 0, 0195 (Moody)  d2  −1/2 = 1,7509m / sec so V2 = 0,2445(0,0195)

and Re2 =

V2 d2

ν

=

1,7509 × 8 × 0,0254 = 2,86 ×105 1, 244 ×10−6

So, with: ε 5  = 0, 00072 and Re 2 = 2,86 × 10   f 2 = 0, 0195 (Moody)  d2 

c) For pipe (3):

h 3 = 100 − 65 = 35m, 1/2

 2h gd  V3 =  3 3   f3 3 

ε d3

=

0,00048 ×12′′ = 0,00072 8′′ 1/2

 2 × 35 × 9,81× 8 × 0,0254  =  3000  

f3−1/2 = 0, 2157 f3−1/2 m / sec

Assuming that f 3 = 0, 019 , −1/2 then V3 = 0,2157(0,019) m / sec = 1,5646m / sec

Re3 =

V3d3

ν

=

1,5646 × 8 × 0,0254 = 2,56 ×105 −6 1, 244 ×10

So, with: ε 5  = 0, 00072 and Re3 = 2,56 × 10   f3 = 0, 0195 (Moody)  d3  −1/2 then V3 = 0,2157(0,0195) m / sec = 1,5447m / sec

40

Fluid Mechanics 2

so Re3 =

V3d3

=

ν

1,5447 × 8 × 0,0254 = 2,52 ×105 1, 244 ×10−6

So, with: ε 5  = 0, 00072 and Re3 = 2,52 × 10   f3 = 0, 0195 (Moody)  d3 

So, the second iteration gives us the following results:

Q1 = V1

Q2 = V2

Q3 = V3

π d12 4

=

π d22 4

π d32 4

1,5787 × π × (12 × 0,0254)2 m3 m3 = 0,115 4 sec sec

=

1,7509 × π × (8 × 0,0254)2 m3 m3 = 0,57 4 sec sec

=

1,5447 × π × (8 × 0,0254)2 m3 m3 = 0,050 4 sec sec

The following relationship must always be satisfied: Q2 + Q3 = Q1

The results of the second iteration give:

Q2 + Q3 = 0,107m3 / sec < Q1 = 0,115m3 / sec So, we also conduct a third iteration by choosing a value of than the one we assumed at the beginning of the second iteration. Third iteration: assume that yA = 63,8 m. Then: a) For pipe (1):

h1 = yΑ − y1 = 13,8m,

ε d1

=

0,00048 ×12′′ = 0,00048 12′′

yΑ

slightly smaller

Pipe Networks

1/2

 2h gd  V1 =  1 1   f11 

1/2

 2 ×13,8 × 9,81×12 × 0,0254  =  2000  

f1−1/2 = 0, 2031 f1−1/2

Assuming that f1 = 0, 018 , −1/2 then V1 = 0,2031(0,018) m / sec = 1,5141m / sec and

Re1 =

V1d1

ν

=

1,5141×12 × 0,0254 = 3,71×105 1, 244 ×10−6

So, with: ε 5  = 0, 00048and Re1 = 3, 71× 10   f1 = 0, 018 (Moody) d  1 

b) For pipe (2):

h1 = yΑ − y2 = 16, 2m, 1/2

 2h gd  V2 =  2 2   f 2 2 

ε d2

=

0,00048 ×12′′ = 0,00072 8′′ 1/2

 2 ×16,2 × 9,81×8× 0,0254  =  1000  

f21/2 = 0,2541 f2−1/2

Assuming that f 2 = 0, 0195 , −1/2 then V2 = 0,2541(0,0195) m / sec = 1,8199m / sec and

Re2 =

V2 d2

ν

=

1,8199 × 8 × 0,0254 = 2,97 ×105 −6 1, 244 ×10

with ε 5  = 0, 00072 and Re 2 = 2,97 × 10   f 2 = 0, 0195 (Moody)  d2 

41

42

Fluid Mechanics 2

c) For pipe (3):

h3 = yΑ − y3 = 36, 2m, 1/2

 2h gd  V3 =  3 3   f3 3 

ε d3

=

0,00048 ×12′′ = 0,00072 8′′ 1/2

 2 × 36, 2 × 9,81× 8 × 0,0254  =  3000  

f3−1/2 = 0, 2193 f31/2

Assuming that f3 = 0,0195, −1/2 = 1,5707m / sec then V3 = 0,2193(0,0195)

V3d3

and Re3 =

ν

=

1,5707 × 8 × 0,0254 = 2,57 ×105 1, 244 ×10−6

So, with: ε 5  = 0, 00072 and Re3 = 2,57 × 10   f3 = 0, 0195 (Moody)  d3 

So, the results of the third iteration are the following:

Q1 = V1

π d12

Q2 = V2

Q3 = V3

4

π d22 4

π d32 4

=

1,5141× π × (12 × 0,0254)2 m3 m3 = 0,110 4 sec sec

=

1,8199 × π × (8 × 0,0254)2 m3 m3 = 0,59 4 sec sec

=

1,5707 × π × (8 × 0,0254)2 m3 m3 = 0,051 4 sec sec

So, we have:

Q2 + Q3 = 0,059 + 0,051 = 0,110m3 / sec = Q1 = 0,110m3 / sec

Pipe Networks

43

This means that we have an agreement to the third decimal, so finally the flow rates will be:

Q1 = 0,110m3 / sec to D Q2 = 0,059m3 / sec to D Q3 = 0,051m3 / sec to D 5) Calculate the flow rates in the various pipes of the distribution network of water shown in Figure (5) with the Hardy–Cross method using the Darcy–Weisbach equation and the Moody diagram. The characteristic values of the pipes are given (absolute roughness ε, diameter d and length  ), with the average temperature of the water (10°C), the kinematic viscosity (v = 1,31×10−6 m2/sec),the entrance flow rate in the node A(QA = 30 l/sec), the exit flow rates in the nodes D and Z(Q∆ = 10 l/sec, QZ = 20 l/sec) and the coefficient of the absolute roughness (ε=0.01524 cm). Solution: i) First, we make the diagram (Figure (4)) of the network and place in it the facts of the problem. The loops are numbered in Roman numerals (Ι and ΙΙ) and the pipes in Arabic numerals (1, 2, 3, 4, 5, 6 and 7). ii) We proceed with a logical distribution of the flow rates Qα to value and direction in all the pipes in such a way that in each node, the continuity equation is satisfied. Therefore, in this problem, the first “logical” distribution of flow rates is as follows. The flow rate of 30 l/sec that enters node Α is distributed into 16 and 14 l/sec to pipes 1 and 4, respectively. The flow rate of 16 l/sec of pipe 1 is distributed into 9 and 7 l/sec in pipes 5 and 2, respectively. From the flow rate of 14 l/sec of pipe 4, the 10 l/sec exits the node Δ and the rest 4 l/sec constitutes the flow rate of pipe 3. The flow rates 7 and 4 l/sec of pipes 2 and 3, respectively, enter as a flow rate 11 l/sec of pipe 7. The flow rate of 9 l/sec of pipe 5 goes on to pipe 6 and with the flow rate of 11 l/sec of pipe 7 exits as a flow rate of 20 l/sec from the node Ζ. iii) The Darcy–Weisbach equation given by the relationship h f = f r = 0,083 f

 gives: d2

 V2 and d 29

44

Fluid Mechanics 2

Figure (5)

h f = 0,083 f  Q2 / d 5 = r Q2

(1)

where r = 0,083 f  / d 5

(2)

For the above equations, the values of length  and diameter d are given. For each pipe, the friction coefficient f is taken from the Moody diagram as a function of the relative roughness ε / d and the Reynolds number:

Re =

Vd 4Q 4 Q Q = = = 9,724 ⋅105 −6 v π vd 3,14 ⋅1,31⋅10 d d

(3)

Therefore, we calculate the relative roughness ε / d of each pipe once from the given values of ε and d, while we calculate the Reynolds number Re of each pipe from equation (3) in every calculation iteration for the respective values of the flow

Pipe Networks

45

rates Qα. Based on the calculated values of ε / d and Re, we take the values of the respective friction coefficients f from the Moody diagram. If two successive values of the friction coefficient f do not differ significantly between them, the last values of f are used as constants for the next iterations of calculations. iv) From equation (2), the value of the coefficient r of each pipe is calculated. v) From equation (1), we calculate the value of the height losses hfa of each pipe in each loop separately. The value of hfα is positive if the flow inside the pipe is to the direction of the indicators of the clock and negative if the flow inside the pipe has the opposite direction. For every loop, we get the algebraic sum Σhfα vi) For every pipe, we obtain the quotient hfa/Qa. For every loop, we obtain the sum ∑hfa/Qa.  Σf a vii) The correction ΔQ is given by the equation  ΔQ = −  mΣ h fa / Qa  which, in this problem for m = 2 , has the specific form:

(

(

ΔQ = − Σh fa 2Σ h fa Qa

)

Table (5a). Calculating tables of the water distribution example with the Hardy–Cross method, first iteration of calculations

)

   

(4)

46

Fluid Mechanics 2

Table (5b). Calculating tables of the water distribution example with the Hardy–Cross method, second iteration of calculations

Table (5c). Calculating tables of the water distribution example with the Hardy–Cross method, third iteration of calculations

Pipe Networks

47

viii) We correct the flow rates we calculated in step (ii) of each pipe of the same loop, adding algebraically the value of ΔQ found from equation (4). Since pipe (2) is common for both loops I and II in its flow rate, we make a second correction of the ΔQ΄, which is opposite to the value of (4) from the other loop. ix) The values of the flow rates that were corrected this way are used for the second iteration of calculations, with repetition of steps (iii) to (viii). The correcting values of the flow rates of the second iteration are used for the third iteration of calculations, by which is possible to bypass steps (iii) and (iv) and to use the values of r of the second iteration, repeating in this way steps (v) to (viii) in the fourth and, if necessary, in the fifth as well. Usually, four to five calculations give satisfactory results. 6) Figure (6) shows a part of a hydraulic network, the facts of which are given in 3 Table 6. If the flow is QΑ = QΕ = 0,5m / sec and the kinematic viscosity of the flow is ν = 1, 2 − 10 − 6m2 / sec , calculate the flow rates of the pipes.

Pipe

L (m)

D (cm)

κs (mm)

1

3.000

30

smooth

2

1.000

30

0,1

3

940

20

1

4

1.000

35

0,1

5

2.000

40

smooth

6

450

20

0,3

Table (6). Network data

Figure (6)

Solution: The directions of the flows are known (Figure (6)), and the calculation is probably simple. Since, generally, it is  / d > 1.000 , the local losses are not

3 calculated, but tests are performed. Assuming that Q1 = 0,2m / sec , then, we have: 3 Pipe 1: Q1 = 0,2m / sec, V1 ≅ 2,83m / sec

Re1 ≅ 7 ⋅105 ,

f1 ≅ 0,0125

48

Fluid Mechanics 2

h f 1 ≅ 51m Pipe 2: Q2 = 0, 2m3 / sec, V2 ≅ 2,83m / sec

Re2 ≅ 7 ⋅105 , f 2 ≅ 0, 016,

ε / d2 = 0,1/ 300 ≅ 0,0003

h f 2 ≅ 21, 8 m

So, for pipes 1 and 2, it is h f 1,2 = 51 + 21,8 = 72,8 m Pipe 3: hf 3 = hf1,2 = 72,8 1

R e 3 ⋅ f 32 ≅ 9 , 2 ⋅ 1 0 4 ,

f 3 ≅ 0, 032,

ε / d 3 = 1 / 200 = 0, 005 ,

V3 ≅ 3,1 m / sec ,

Q3 ≅ 0,10 m 3 / sec

Pipe 4: Q 4 = Q 2 + Q3 = 0, 2 + 0,1 = 0, 3 m 3 / sec V 4 ≅ 3,11m 3 / sec, R e 4 ≅ 9 ⋅ 10 5

ε / d 4 ≅ 0, 0003,

f 4 ≅ 0, 016

h f 4 = 22, 5 m

Pipe 5: Q 5 = Q 4 = 0, 3 m 3 / sec, V 5 ≅ 2, 38 m / sec , R e 5 ≅ 7, 9 ⋅ 10 5 ,

f 5 ≅ 0, 012

hf5 ≅ 17,3 m

So, for pipes 4 and 5, it is: h f 4,5 = 22, 5 + 17, 3 = 39, 8 m

and

Pipe Networks

49

hf 1,2,4,5 = 51+ 21,8 + 39,8 = 112,6 m Pipe 6: h f 6 = 112, 6 m ,

1

R e 6 ⋅ f 62 ≅ 1, 6 ⋅ 10 5 , ε / d 6 ≅ 0, 0015

f 6 ≅ 0, 022 , V6 ≅ 6, 6m / sec

and Q 6 ≅ 0, 2 m 3 / sec

For the testing, it is: Q 5 + Q 6 = 0, 3 + 0, 2 = 0, 5 m 3 / sec = Q A

which means that the first choice of QA was correct. 7) In Figure (7), three tanks are shown, of which Α has a seasonally variable level. In pipe ΒC, there is a sliding valve, while QBC = 50 lit/sec. a) With the maximum flow rates and QBD = 100 lit/sec, define ZA and dBD. b) If Ζ Α = 116,75 m, calculate the value at which the valve will close so that QBD = 100 lit / sec. . In Figure (7),

AB

 = 1.200 d = 300 ΒC

 = 209 d = 150 ΒD

 =1.650 ε = 1 mm (all the pipes) ν = 1.1·10–6 m2/sec

 is in m and d in mm.

50

Fluid Mechanics 2

Figure (7)

Solution: The pipes have Qmax when the valve is fully open. a) In pipe CΒ, it is:

 / d = 209 / 0,15 ≅ 1.393 > 1.000 and the valve is fully open. So, the total losses are not calculated and therefore we have: V = 4 ⋅ 0, 05 / π ⋅ 0,15 2 ≅ 2,83m / sec Re = 2, 83 ⋅ 0,15 / 1,1 ⋅ 10 −6 ≅ 3, 8 ⋅ 10 5 ,

ε / d = 1/150 ≅ 0, 006 , so: f ≅ 0, 032 ,

hf = 0,032⋅ (209/ 0,15) ⋅ 2,832 / 2⋅ 9,81 ≅ 18,2m

Pipe Networks

So, it is: Η B = Η C + h f = 76 + 18, 2 = 94, 2 m

In pipe ΒD, it is: h f = H B − H D = 94, 2 − 55 = 39, 2 m

It is n = 0,012 , and with an absolutely rough pipe, J E = 39, 2 /1.650 ≅ 0, 0237 1

3

while d = 1, 548 ⋅ (0,1 ⋅ 0, 012 / 0, 0237 2 ) 8 ≅ 0, 250 m which means that (testing of n):

ε / d = 1/ 250 ≅ 0,004 Also,  / d = 1.650 / 0, 25 = 6.600 > 1.000 . In pipe ΑΒ, it is:

 / d = 1.200 / 0,3 = 4.000 > 1.000 , and

Q = 0,05 + 0,1 = 0,15m3 / sec So, V = 4 ⋅ 0,15 / π ⋅ 0,32 ≅ 2,12m / sec

Re ≅ 5,8 ⋅105 , ε / d = 1/ 300 ≅ 0,003 f = 0, 027

and h f = 0, 027 ⋅ (1.200 / 0,3) ⋅ 2,122 / 2 ⋅ 9,81 ≅ 24, 74m

51

52

Fluid Mechanics 2

So, it is: Η A = Z A = H B + h f = 94, 2 + 24, 74 = 118, 94 m ,

which means that: Ζ A ≅ 119 m

b) Since the valve is not fully open, we will take into account the energy losses from it, independently of the ratios  / d . With ΖΑ = 116,75m for pipe ΑΒ, as QΒΔ = constant, h f = 116, 75 − 94, 2 = 22, 55 m 1 2

and R e ⋅ f or, R e ⋅ f

1 2

3

1

= (0, 3 2 / 1,1 ⋅ 10 − 6 ) ⋅ (2 ⋅ 9, 81 ⋅ 22, 55 / 1.200) 2 , ≅ 9 ⋅ 10 4 ,

ε / d = 0, 003 ,

f ≅ 0, 027 V ≅ 2 m / sec, Q ≅ 0,14 m 3 / sec

For pipe ΒC, it will be: Q = 0,14 − 0,1 = 0, 04 m 3 / sec

V = 4 ⋅ 0, 04 / π ⋅ 0,15 2 ≅ 2, 26 m / sec Re ≅ 3,1 ⋅ 10 5 , ε / d = 1 / 150 ≅ 0, 006

f ≅ 0, 032,

h f = 11, 6 m

Between node Β and tank C, it is: Η = Η B − ΗC = 94, 2 − 76 = 18, 2m

and, 18, 2 = 11, 6 + Κ τ ⋅ V therefore, Kτ ≅ 25, 4

2

/ 2 ⋅ g = 11, 6 + 0, 26 ⋅ K τ ,

Pipe Networks

53

However, it is: (relative opening) ≅ 0,15 = (opening)/0,15 , which means that the opening is 0.0225 m = 2,25cm . The valve will be closed at: (1 − 0,15) = 0,85, or 85%

The decrease in linear energy in the position of the valve will be:

hf = 25,4⋅ 2,262 / 29,81 = 6,6m which is 6.6/18.2 = 0.36 or 36% of the total losses in the pipe. 1.11. Problems to be solved

1) Calculate the flow rates in the loop of the figure if QA= 0.32 m3/s (inflow), QΒ= 0.28 m3/s (outflow) and QΔ= 0.10 m3/s (inflow). All the pipes are made up of cast iron, of diameter 25 cm with ΑΒ = ΓD =200 m and ΑD=ΒΓ=100 m.

2) The five pipes of the horizontal network, shown in the figure, have a friction coefficient f = 0.025, lengths ΑΒ = CD = 600 ft, ΑC = ΒD = 450 ft, BC = 750 ft and diameters. If the water inflow in the network is QA= 3 ft3/s, calculate the flow rates and the flow directions in all the pipes. If the pressure at point Α is 120 psi, calculate the pressure at points Β, C and D.

54

Fluid Mechanics 2

3) Calculate the pressures of the network shown in the following figure. The pipes are made up of cast iron with lengths: ΑΒ = ΒΕ = DΕ =ΑD = 100 m CD = 50 m; AC= 112 m.

Secondary losses are negligible: ν = 10−6 m2/s. 4) Calculate the water flow rates and the losses in the pipes made up of cast iron for the network shown in the figure. The given data are as follows:

5) In the water network shown in the figure, the pipes are made up of cast iron. The lengths and diameters of the pipes are given, as well as the flow rates to and from the network. Calculate a) the flow rates in all the pipes of the network and b) the pressure difference between Α and Ζ if yA– yΖ = 4 m. The given data of the pipes are as follows:

Pipe Networks

55

Figure (5)

6) The network shown in the figure is supplied with water from tanks 1 (y1 = 128 m) and 2 (y2 = 122 m). All the pipes are made up of cast iron. The outflows from the network are: QB= QΔ = 60  / s , QΓ= QΕ = 40  / s . Calculate the flow rates in each node of the network. The given facts are as follows: yΒ= yE= 40 m yC= 40 m yD=28 m  AB = 500 m EF = 300 m CF = 150 m, BC = BD = CE = DE = 120 m d AB = d EF = 20cm d BC = d CE = 10cm

56

Fluid Mechanics 2

7) In the network shown in the figure, the pipes are made up of asphalted cast iron with lengths and diameters given in the following table. Calculate the flow rates if the altitudes are: y1 = 150 m, yΒ= 50 m, yC= 53 m, yD= 45 m yΕ= 65 m, yF= 60 m, yG= 30 m, yH= 35 m

2 Open Channel Flow

2.1. Introduction A channel transferring a fluid is called the open channel flow or channel of free flow or canal when the fluid that flows inside its free surface is below the atmospheric pressure. Forces that create the flow are influenced by the gravity and the inclination of the channel. The air resistance and the forces of the surface stresses on the fluid’s free surface are considered to be negligible. Forces that slow down the flow are caused by the viscous shear stress of the fluid, and frictions are created by its contact with channel walls. The open channels are distinguished into two categories, namely natural channels and artificial channels. The first category includes naturally existing channels, such as rivers, streams and various natural currents, while the second category includes artificially created canals and ditches for irrigation, drainage and other purposes, which vary in size, shape and surface roughness. Because of their construction process, artificial channels have known geometry and different roughness values. They are also called prismatic, when the cross-sectional area and the inclination of their bottom are constant. According to the geometrical shape of their cross-sectional area, the prismatic channels are: i) rectangular; ii) trapezoidal; iii) triangular;

Fluid Mechanics in Channel, Pipe and Aerodynamic Design Geometries 2, First Edition. Christina G. Georgantopoulou and George A. Georgantopoulos. © ISTE Ltd 2018. Published by ISTE Ltd and John Wiley & Sons, Inc.

58

Fluid Mechanics 2

iv) semi-circular; v) parabolic. 2.2. Non-dimensional parameters in open channels The flow of a fluid in open channels is influenced by gravity, cohesion and friction forces on the channel walls. The forces of compression and pressure are negligible. The non-dimensional parameters that express these forces are: a) the Froude number Fr and b) the Reynolds number. i) The Froude number expresses inertia forces and gravity forces and is given by the relation: Fr =

inertia forces V , = gravity forces gh

[2.1]

where V is the average flow velocity, g is the acceleration due to gravity and h is the depth of the flow. As the Froude number is non-dimensional, the denominator gh must have dimensions of velocity. Indeed, this term shows the velocity of transmission of infinitesimal surface waves. ii) The Reynolds number expresses inertia forces and viscosity forces and is given by the relation: Re h =

inertia forces Vd h = v viscosity forces

[2.2]

where V is the average flow velocity, ν is the kinematic viscosity and d h is the hydraulic diameter of the channel, which is equal to four times (not two times) the hydraulic radius, Rh , which is the quotient of the surface area of the channel’s cross-sectional area and the length of the perimeter that surrounds it, which is:

Rh =

hw h⋅w = , h + w + h 2h + w

[2.3]

where h is the depth of the channel and w is the length of the channel, as shown in Figure 2.1.

Open Channel Flow

59

Figure 2.1. Dimensions of an open channel

2.3. Open channel types of flow

The flow of fluids in open channels can be categorized as follows: i) steady or unsteady; ii) uniform or non-uniform; iii) laminar or turbulent; iv) sub-critical or super-critical. i) Steady flow: this occurs when none of its variables vary (velocity, depth, etc.) with time, while unsteady or unstable flow occurs when time is considered to be an independent variable. In natural channels such as rivers and streams, the flow is never steady, while in artificial channels, during their normal function, the flow is steady very rarely and for a very small period of time. Variations that occur due to the demands of hydro-electric factories, irrigation needs and so on, result in the instability of the flow that becomes apparent, mostly because of the variation of its depth, the increase of which causes an imbalance in the hydrostatic pressures on the installations. Consequently, this leads to their destruction. Moreover, the variation of the flow depth influences the flow rate, in both the main channel and the secondary one, and as a result, the automation of the distribution system becomes difficult. As the analysis of the unsteady flow is difficult, hydraulic studies are conducted according to the principles and the admissions of the steady flow. ii) Uniform flow: this occurs when the depth of the fluid remains constant along the channel, while non-uniform flow occurs when the depth varies along the direction of the flow (dh/dx ≠ 0). The non-uniform flow is divided into two categories, namely gradually non-uniform flow, in which the rate of change of depth

60

Fluid Mechanics 2

rate h with distance x along the length of the channel is low ( ℎ/ < 1), and rapid non-uniform flow, in which the rate of change of depth rate with distance x along the length of the channel is not low ( > 1). Figure 2.2 shows the various flow types in an open channel. ΤΑΡ

ΒΑΡ

Ομοιόμορφη Ροή

ΤΑΡ

ΒΑΡ

h > hn ΒΑΡ = Βαθμιαία Ανομοιόμορφη Ροή ΤΑΡ = Ταχεία Ανομοιόμορφη Ροή

h = hn h ® hn

Figure 2.2. Types of flow in open channels

In general, we have a uniform flow in every linear, or straight, channel of long length, which has a steady transverse cross-sectional area, or otherwise in a channel of certain roughness, cross-sectional area and inclination, there is only one depth of the fluid hn for which the flow will be uniform, which is called the regular or uniform depth. In a uniform flow, the fluid moves with a constant velocity; in order for that to occur, the dynamic energy, which reduces, due to the decrease in height along the canal, has to balance exactly the loss of the energy that is consumed in order to overcome the friction forces. iii) Laminar or turbulent flow: as in the closed pipes, this flow is determined according to the values that the Reynolds number takes. Laminar flow occurs when Re 2000. The latter is more frequent and occurs in more than 99% of the flow problems in open channels. iv) Sub-critical or super-critical flow: in open channels, this flow is related to the Froude number, Fr. Thus, sub-critical flow occurs when Fr < 1, while critical flow occurs when Fr = 1 and super-critical flow occurs when Fr > 1. In the sub-critical area, the fluid velocity is low, while in the super-critical area, the fluid moves with high velocity. Therefore, if at some point of the flow a

Open Channel Flow

61

disturbance is caused, the waves created will be able to move both upstream and downstream if the flow is sub-critical, while if it is super-critical, they will not be able to move upstream because the velocity of the wave is lower than the velocity of the flow. 2.4. Open channels’ geometrical shapes

Natural open channels such as rivers and streams do not have a definite and specific geometrical shape, in contrast with artificial open channels, like canals, which have a certain geometrical shape, depending on the needs of the hydraulic systems they are used in. In addition, natural channels have an irregular crosssection, while the artificial ones have a regular cross-section. Regular cross-section refers to the cross-sectional area whose shape does not vary along the channel, while irregular cross-section refers to the one whose geometry varies. The three most common geometries of open channels are: i) rectangular cross-section; ii) trapezoidal cross-section; iii) circular cross-section. 2.4.1. Channels of rectangular cross-sectional area

This is the simplest form of an artificial open channel. There are three geometrical characteristics that are interesting in this category: a) Wetted transverse cross-sectional area S, which is given by the relation:

S = b⋅ h ,

[2.4]

where b and h are the width and the depth of the channel, respectively. b) Wetted perimeter Π, which is given by the relation:

Π = b + 2h .

[2.5]

c) Hydraulic diameter dh, which is given by the relation:

4bh S dh = 4   = .  Π  b + 2h

[2.6]

62

Fluid Mechanics 2

Figure 2.3 shows a rectangular cross-sectional channel.

Figure 2.3. Rectangular cross-sectional channel

2.4.2. Channels of trapezoidal cross-sectional area

This cross-sectional area has one more variable: the inclination of the lateral walls b0 of the channel. Therefore, we have: i) Wetted transverse cross-sectional area S, which is given by the relation: 1 S = b ⋅ h + (m1 + m2 )h 2 . 2

[2.7]

ii) Wetted perimeter Π, which is given by the relation:

Π =b+

(

)

1 + m12 ⋅ 1 + m22 h .

[2.8]

iii) Width of free surface b0, which is given by the relation: b0 = b + (m1 + m2 )h

[2.9]

Open Channel Flow

63

where m1 is the ratio of the horizontal to vertical variation of the wall of one side and m2 is the perspective quantity of the other side of the channel. Moreover, the above equations can describe the geometry of triangular channels if we set: b=0

[2.10]

and if we set: m1 = m2= 0,

[2.11]

we have the perspective equations of a channel of rectangular cross-section. Figure 2.4 shows a trapezoidal cross-sectional channel.

Figure 2.4. Trapezoidal cross-sectional channel

2.4.3. Channels of circular cross-sectional area

Figure 2.5 shows a circular cross-sectional channel, which has several technical applications.

64

Fluid Mechanics 2

Figure 2.5. Circular cross-sectional channel

If d is the internal diameter of the circular channel shown in Figure 2.5, then we have the following geometrical characteristics: i) Wetted transverse cross-sectional area S:

1 S = (2θ − sin 2θ )d 2 8

[2.12]

ii) Wetted perimeter Π:

Π =θ ⋅d

[2.13]

iii) Width of the free surface of a circular cross-sectional area b0: b0 = d ⋅ sin θ ,

[2.14]

where θ is the angle associated with the depth h of the flow by the relation:

θ = cosθ −1 [1 − 2( h / d ) ]

[2.15]

Finally, the geometry of natural channels can be described in a functional form as: S (h), Π (h), bh (h), b0 (h).

Open Channel Flow

65

2.5. The hydraulic jump

In an open channel of depth h1, under suitable circumstances, the flow can be released and create a new depth h2. This phenomenon is called the hydraulic jump. The hydraulic jump is created when the flow varies from super-critical to sub-critical. Generally, the creation of a hydraulic jump is accompanied with energy loss, significant increase of the depth h of the flow and intense turbulence of the fluid. Therefore, it is often used as an energy diffuser downstream of a barrier or a spillway, for the restriction of the destructive action of a fluid at very high velocity, as well as for the effective mixing of fluids, for example in wastewater treatment and drainage systems. When a hydraulic jump occurs, the bottom of the channel must be very resistant to mechanical exhaustion so that it is not destroyed. The basic parameter for the hydraulic jump to be created is the Froude number of the upstream flow, while the Reynolds number barely influences it. Based on the Fr, five types of hydraulic jump have been ascertained that exist in a horizontal rectangular channel. Therefore, 1 < Fr < 1.7 represents undular jumps, 1.7 < Fr < 2.5 represents weak jumps, 2.5 < Fr < 4.5 represents oscillating jumps, 4.5 < Fr < 9.0 represents constant jumps and Fr > 9.0 represents strong jumps. The most effective type of hydraulic jump is the one with 4.5 < Fr < 9.0. Finally, it has been experimentally proved that right after the increase in the flow head, the distribution of velocity in a cross-sectional area is uniform. 2.6. Calculation of the depth flow after the hydraulic jump

We consider a horizontal canal with rectangular cross-sectional area, as shown in Figure 2.6.

Figure 2.6. Horizontal channel with a rectangular cross-sectional area

66

Fluid Mechanics 2

Let V1 and h1 be the velocity and depth of the flow, respectively, before the hydraulic jump in the channel, as shown in Figure 2.6. Now calculate the height h2 after the jump. Because at the cross-sectional areas 1 and 2 the flow lines are practically parallel, and as the fluid is considered to be ideal (friction losses on the wall at the length 1 → 2 are close to zero), the pressure at 1 and 2 is hydrostatic. If the width of the jump is W, then from the sum of the Χ forces, we have: p1C S1 − p2C S 2 = ρ Q (V2 − V1 ) , where: p1C =

h1 h ρ g , p2 C = 2 ρ g , S1 = Wh1 , S2 = Wh2 , 2 2

Q is the flow rate = V1 S1 = V2 S 2 and ρ is the fluid density, so: V2 =

V1 S1 V1h1 = . S2 h2

Solving for h2, we have:

ρ 2 or

gh12W −

ρ 2

h  gh22W = ρV1h1W (V2 − V1 ) = ρV12 h1W  1 − 1 h  2 

1 g ( h12 − h22 ) h2 = V12 h1 ( h1 − h2 ) . 2

[2.16]

Relation [2.16] is a cubic equation of h2, but it has one solution for h1=h2 when the flow is uniform and with no jump. If h1 ≠ h2 , then relation [2.16] gives: 1 g ( h1 + h2 ) h2 = h1V12  2

h22 + h1h2 −

2h1 2 V1 = 0 g

This quadratic equation gives:

h2 =

 h1  V12 − 1 .  1+ 8 2  gh1 

Open Channel Flow

V1

Substituting Fr1 =

h2 =

h1 2

gh1

67

, we have:

{ 1 + 8F − 1} 2 r1

[2.17]

where Fr1 = ο (non-dimensional Froude number calculated at position 1). During a jump, the flow, where h = h1 , becomes turbulent and then the flow energy is lost as heat energy. So, the amounts of energy at 1 and 2 transferred by the fluid are different, with the total amount of energy at 2 lower than that at 1. Thus, from the equation:

gh1 +

V12 V2 = gh2 + 2 + Eθ , 2 2

where Εθ is the energy loss per mass unit, with V2 =

  1 Εθ = gh1 1 −  2 

(

)

F2 1 + 8 F − 1 + r1 2 2 r1

  1 −  

V1h1 and using [2.17], we have: h2

  . 2   2   1 + 8Fr1 − 1  4

(

)

[2.18]

Equation [2.18] that gives the value of Εθ is always non-negative, so: i) When Fr1 = 1 h1 = h2 and Εθ = 0 when there is no jump and therefore no losses. ii) When Fr1 < 1

h2 < h1 and Εθ < 0, which means increase in flow energy instead of decrease in flow energy, which is inappropriate. iii) When Fr1 > 1 h1 < h2 and Εθ > 0, so we have a jump.

68

Fluid Mechanics 2

2.7. Velocity distribution

In general, in the open channels, the flow is three-dimensional, but in some cases, like the mouth of a river, it can be considered to be two-dimensional or even one-dimensional, like in rivers and artificial channels. Here, we will limit our study to the one-dimensional flow in channels, which have constant transverse cross-sectional area and straight axis, and which are the prismatic channels. Generally, the distribution starts with the zero value at the solid boundary under the non-slipping condition, and it increases as we move away from it. Because of the secondary flows that are created, its maximum value is not at the free surface of the fluid, but below this distance, from 0.05 to 0.25, of the canal’s depth. A typical distribution of the fluid velocity at the center of the open channel and the equal-velocity curves to a transverse cross-sectional area of two open prismatic channels with a rectangular (a) and a random (b) cross-sectional area, is shown in Figure 2.7.

Figure 2.7. Rectangular and random cross-sectional areas of channels

2.8. Velocity distribution at the vertical level

The distribution of velocity in an open channel depends on the type of the flow as follows: i) Laminar flow: when the flow passes through a channel of large width and medium depth h, the distribution of the velocity is given by:

V ( y) =

g y  h −  y tan θ , ν  2

[2.19]

where θ is the inclination of the channel. In the relation [2.19], the velocity becomes maximum for y = h at the free surface, where θ is the inclination angle, which is generally very low; g is the

Open Channel Flow

69

acceleration due to gravity; ν is the kinematic viscosity; h is the depth of the channel and y is the distance of a random point from the solid boundary.

Figure 2.8. Laminar flow in an open channel

From this relation, the average velocity is given by:

Vm =

gh2 tan θ . 3ν

[2.20]

This value results from the integration of relation [2.19] with respect to y from 0 to h. ii) Turbulent flow: when the flow in an open and wide channel is turbulent, the distribution of velocity is expressed by the relation:

V ( y ) = 2,5

τ0  y  ln   ρ  y0 

[2.21]

From this relation, the velocity takes its maximum value for y = h = flow depth, where τ0 is the shear stress at the bottom of the channel:

y0 =

ν , τ0 9 ρ

with ρ being the density and ν being the kinematic viscosity. In practice, as mentioned in section 2.7, the maximum value of the velocity is not at the free surface (y = h), but it is usually at a depth of 0.05–0.25 of the total depth

70

Fluid Mechanics 2

under the surface, as shown in Figure 2.9. A practical method used for the calculation of the average velocity is the relation: Vm = average velocity =

V (0.2 h ) + V (0.8h ) 2

[2.22]

Figure 2.9. Velocity distribution in an open channel

2.9. Uniform flow in open channel equations – Chezy type When a straight channel has a constant transverse cross-sectional area and is of a long length, then there will be a uniform flow. In this flow, the fluid moves with constant velocity and the flow depth is constant along the whole channel. In this case, the hydraulic line of the flow, the free surface of the fluid and the bed of the channel are parallel. For this flow, the fluid’s velocity was suggested by Chezy and named after him (1768). In order to calculate this type, we consider the channel shown in Figure 2.10.

Figure 2.10. Non-uniform flow in open channels

Open Channel Flow

71

Let ABCD be the volume of a fluid of constant cross-sectional area S and length  . Because the flow is steady, this volume does not accelerate and therefore the sum of the forces to x will have to be zero. So, the hydrostatic force at AB = FAB = ρ g The hydrostatic force at CD = FCD = ρ g

h (inflow). 2

h (outflow). 2

[2.23]

[2.24]

Moreover, the component of the fluid’s volume weight in the direction is × = W sin θ = ρ g S  sin θ (inflow).

[2.25]

The shearing force at the channel walls = τ 0 S  (outflow)

[2.26]

and Ρ = length of the wetted surface of the cross-section, where P = h + W + h = 2h + W .

[2.27]

Figure 2.11. Dimensions of the cross-sectional area of the open channel

So from [2.22]–[2.27], we have: ΣF x = ρ g

or τ 0 =

h h − ρ g + ρ gS  sin θ − τ 0 pl = 0 2 2

ρ gS P

sin θ

As the hydraulic radius is defined as Rh = small, then sinθ ≅ tanθ , and thus:

τ 0 = ρ gRh tan θ .

[2.28] S and if the channel inclination is P

[2.29]

72

Fluid Mechanics 2

Moreover, we find that:

τ 0 = ρ C1V 2 , so ρ g Rh tan θ = ρ C1V 2 or V =

g Rh tan θ = C Rh tan θ . C1

[2.30]

Thus, the Chezy type is:

V = C Rh tan θ ,

[2.31]

where V is the average velocity of the fluid and C is the coefficient of Chezy. The coefficient C is given by various relations named after the researchers who developed them. From the Chezy type, the flow rate Q can be found: Q = (average velocity of the flow) × (surface area of the flow’s cross-sectional area):  Q=VS=CS Rh tan θ 

[2.32]

Q = CS Rh tan θ . The dimensions of the coefficient C of Chezy are:

[C ] = L1/ 2 T −1 . For a laminar flow, the coefficient C is:

C=

8g 8g = = 64 f Re

g Re = 0.3536 g Re , 8

so: V = 0.3536 gRh Re tan θ = average velocity

[2.33]

and Q = 0.3536S gRh Re tan θ = flow rate.

[2.34]

Open Channel Flow

73

The coefficient C has the following types, which are named after the researchers who stated them: i) The Manning type

CM =

1.486 1/6 Rh . n

[2.35]

ii) The Bazin type

CB =

157.6 . m 1+ Rh

[2.36]

iii) The Kutter type

0.00281 1.811 + tan θ n Ck = . 0.00281  n  1+  41.65 +  tan θ  Rh  41.65 +

[2.37]

The parameters n and m in the above types for the C take values depending on both the roughness of the surfaces of the open channels and their size and shape, and have been calculated experimentally for flows in which the fluid is water. The values of n and m are given in Table 2.1. NOTE.– i) When Rh > 10 ft , the values of n and m must increase by 10–15% of the values given in the table. ii) The types and the tables require that the average velocity be expressed in ft/s and the hydraulic radius in ft. iii) When both the hydraulic radius Rh and the cross-sectional area are given in SI units, the Manning type is expressed as: CM = kRh1/ 6 .

[2.38]

The values of the Manning roughness coefficient are given in Table 2.2a, with the hydraulic radius Rh expressed in m and the average velocity V in m/s.

74

Fluid Mechanics 2

Among the types, the most commonly used is the Manning type. From the Manning type and the relation of Chezy, we have:

V=

1.486 2/3 Rh tan θ or V = kR h2 / 3 tan θ n

and Q =

[2.39]

1.486 2/3 SRh tan θ or Q = kSRh2 / 3 tan θ . n

[2.40]

To reiterate, these relations are applicable for steady and uniform flow in open channels. Construction material and finishing of the channel’s surface

n

m

0.010

0.11

0.012

0.20

0.013

0.29

Concrete of medium roughness

0.015

0.40

Brickworks

0.016

0.65

Rivet (nailed) surfaces of steel

0.018

0.92

Metallic surfaces with stripes

0.022

1.43

Smooth coating of cement Well-planed wood Planed tree logs Coated cast iron Bottled channel of smooth surface Brickwork of good construction Channels with cement coating of medium roughness Smooth metallic channels Clay channels of medium roughness Cast iron channels of medium roughness

Rectilinear earthen canals with smooth surfaces

0.023

1.54

Earthen canals opened with an excavator without further smoothing of the surfaces

0.027

2.36

Rivers’ canals in good condition

0.030

3.00

Earthen canals with stones and plants (vegetation)

0.035

3.24

Canals opened in rocky grounds

0.040

3.50

Table 2.1. Values of n for the Manning and Kutter types and m for the Bazin type

Open Channel Flow

Form of walls

75

k

1. Natural open channels – Rivers with steady bed with no irregularities

40–42

– River with some debris material

35–38

– River with vegetation

30–35

– River with cobblestones and irregular bed

28–30

– Coastal area of river bed with a little vegetation

20–25

– Stream with large cobblestones and some debris material

25–28

– Stream with large cobblestones and intense debris material movement

19–22

2. Earthy open channels – Bottom and slopes coated with cast asphalt

70–75

– Coherent material

60

– Coherent sand with a little clay or debris

50

– Bottom of sand and gravel, sidewalls and stone-coats

45–60

– Thin gravel

40

– Large gravel

35

– Large stones

26–30

– Sand, clay, gravel in large content

26–30

3. Built open channels – Well-jointed brickwork

80

– Chipped rectangular stonework

70–80

– Pruned built stonework

70

– Common stonework

60

– Dry stonework

50

– Slope from dry stonework, bottoms of sand and gravel

45–60

Table 2.2a. Values of the roughness coefficient k according to Manning (Rh in m, V in m/s)

2.10. Best hydraulic cross-sectional area

We proved in the previous section that the flow rate of an open channel in a uniform steady flow is given by the Manning type and the Chezy type from the equation:

Q=

1.486 2/3 Rh ⋅ S tan θ n

[2.41]

76

Fluid Mechanics 2

where h is the parameter that depends on the roughness and geometrical characteristics of the channel surface, Rh is the hydraulic radius, S is the surface area and θ is the channel inclination. 4. Canals of concrete – Coat of mortar

100

– Concreting with iron-types

90–100

– Very smooth concreting

90–95

– Smoothed concrete

20–90

– Good wood-type with no coating

65–70

– Old concrete, clean surfaces

60–65

– Rough concrete

55

5. Steel channels – No internal coating d = 2.0… 1.2 m

96–98

d = 1.2…0.8 m

100–102

d = 0.8…0.4 m – Medium deterioration – Quite deteriorated – With totally smooth internal coating 6. Wooden pipes – New smooth canals – Planed totally applied planks – Un-planed planks – Old wooden canals

104–107 90–98 67–80 120 95 90 80 65–70

7. Closed Pipes and tunnels made of concrete – Concrete with iron-types – Concrete with wood-types and coating

90–100 80–90

– Concrete with wood-types and no coating

65–70

– Old rough concrete with no joints

60–65

– Connected channels, pruned joints

85–99

– Connected channels, not pruned joints during construction d 0.50 m d 0.50 m

70–80 60–70

Table 2.2b. Values of roughness coefficient k according to Manning (Rh in m, V in m/s)

Open Channel Flow

77

Equation [2.41] shows that for a certain surface area of wetted cross-section of the bottom inclination and surface roughness, the flow rate in the channel will be S maximum if the wetted perimeter P is minimum, because: Rh = . P The wetted cross-sectional area that has this minimum perimeter is called the best hydraulic cross-sectional area.

Figure 2.12. Open channel with trapezoidal cross-sectional area

In order to find the best hydraulic cross-sectional area of an open channel of regular cross-section, we consider a channel in the shape of a trapezium. For the trapezoidal channel shown in Figure 2.12, the wetted cross-sectional area is: S = bh + mh 2

m = cot θ

[2.42]

and the wetted perimeter is: P = b + 2a = b + 2 h(1 + m 2 )1/ 2 .

[2.43]

Deleting b between equations [2.42] and [2.43], we have:

P=

S − mh + 2h(1 + m2 )1/ 2 . h

[2.44]

For the definition of the minimum perimeter P, we calculate the derivative

dP / dh (with the values S and m being constant) and equate it to zero. The final result is: S = [2(1 + m 2 )1/ 2 − m ]h 2 , P = 4h(1 + m 2 )1/ 2 − mh , Rh =

1 h. 2

[2.45]

78

Fluid Mechanics 2

The last expression is very interesting because for every angle θ, the best hydraulic cross-sectional area occurs when the hydraulic radius of the channel is equal to half of the flow depth. Because the rectangle becomes a trapezoid with m = 0, the best hydraulic crosssectional area of the channels is the rectangular cross-sectional area, such that: S = 2h 2

P = 4h

Rh =

1 h b = 2h . 2

[2.46]

We note that, in the open channels of rectangular cross-section, the best hydraulic cross-sectional area occurs when the depth of the flow is equal to the half of the width of the bottom, regardless of the size of the channel. Equations [2.45] are applicable for every value of m. Now, we will have to calculate the best value of m = cotθ as well; the depth of the flow h and wetted cross-sectional area S are given. For its calculation, we get the derivative dP/dm from equation [2.44], keeping the values S and h constant, and we equate it to zero. The final result is: 2 m = (1 + m 2 )1/ 2

[2.47]

or, if we perform the sums ( m = 3 / 3) and we consider that m = cot θ :

θ = 60ο .

[2.48]

This means that, in the open channels of trapezoidal cross-sectional area, the very best hydraulic cross-sectional area is half-hexagonal. In the same way, it is proved that, for open channels of circular cross-sectional area (Figure 2.5), the very best hydraulic cross-sectional area is the semi-circular ( h = d / 2) one. Indeed, of all the cross-sections of open channels, the semi-circle has the smallest perimeter for a given wetted cross-sectional area. Therefore, an open water channel of semi-circular cross-sectional area will carry more water than any other channel of different shape as long as the surface area of the wetted cross-sectional area, the bottom inclination and the surface roughness are the same. To conclude, we must say that the best hydraulic cross-sectional area is not the only criterion for the successful design of open channels. Other parameters that must be taken into account are the kind of excavation and, concerning an inclining

Open Channel Flow

79

channel, the stability of the inclination of the side walls and the possibility of bottom corrosion. Moreover, in channels where the flow rate may produce high variances, like in sewers, it is desirable that the velocity of the fluid, when the flow rate is small, is kept high enough that the deposition of sediments is avoided, and that the velocity is not so high, when the channel is full, as to cause too much dilapidation on the channel walls. 2.11. Specific flow energy

Specific flow energy is called the total energy per unit weight. However, as it has length units, it is considered to be the head measured from the bottom of the channel, which is given by the following relation, where Ε is the special energy = flow depth + pressure gauge head of the velocity: Ε= y+

V2 2g

[2.49]

where V = average velocity =



=

and S = Wy .

For a uniform flow, the line of the specific energy remains at the same head along the channel. In the case where the flow is non-uniform, the head of the line from the bottom of the channel can be increased or decreased. Because: V =

Q Q = , S Wy

then the pressure gauge head of the velocity is given by: V 2 Q2 / W 2 y 2 Q2 = = , 2g 2g 2W 2 gy 2 so: Ε = y +

and

Q2 1 W 2 2 gy 2

or

Q = 2 gy 2 ( E − y) . W

Q2 = 2 gy 2 ( E − y ) W2 [2.50]

80

Fluid Mechanics 2

Figure 2.13. Dimensions and relationships in a cross-sectional area of an open channel

This relation expresses the flow rate in relation to the depth and specific energy. 2.12. Channels of rectangular cross-sectional area

For a flow in channels of rectangular cross-sectional area, the specific energy can be written as a function of only the depth h of the flow and specific flow rate q, which is defined as the ratio of the total flow rate Q of the fluid to the width b of the channel:

q=

Q = Vh b

[2.51]

where V is the average flow velocity in the channel. Therefore, for a flow in a rectangular channel, the specific energy in a cross-sectional area is given by the equation: Ε = h+

q2 . 2 gh 2

[2.52]

Equation [2.52] is shown graphically in Figure 2.14 for a certain value of q. We note that the curve Ε(h) comes through a minimum (at point c) and, for small values of h, it tends to infinity along the axis E, while for large values of h, it asymptomatically approaches the line Ε = h, which forms an angle of 45° with the axis Ε.

Open Channel Flow

81

Figure 2.14. Specific energy variation

The specific energy of the flow that corresponds to point c is the minimum possible for the given q, and is called the critical energy Εc. The respective depth hc, is called the critical depth. The cross-sectional area of the current with h = hc is called the critical cross-sectional area. Therefore, the critical depth of a rectangular open channel of constant cross-sectional area is called the depth y, which gives rise to a maximum value of Q and a minimum value of Ε. In order to calculate it, we get the derivative of Ε to y. Setting it equal to zero and solving for y, we have: Ε= y+

Q2 1 W 2 2 gy 2

dE Q 21 = 1− = 0 → y = yk ρ = dy gW 2 y 3 yk ρ =

3

Q2 gW 2

3

Q2 gW 2 [2.53]

82

Fluid Mechanics 2

Substituting the value of y k into the relation of Ε, we have: 2

Ε = Ε k ρ = yk ρ +

or Εk ρ

1 Q ⋅ = W 2 2 gyk2ρ

3

Q + gW 2

Q2 gW 2

1 3

 Q2  = + = 2  2 2  gW  3  Q  2 2   gW 

so Ε k ρ =

Q2 W2

2

2

 Q2 2g  3  gW 2 

Q2 Q2 + gW 2 gW 2

   

2

1

3  Q2 3 =   2  gW 2 

2

 Q2  3 2 2   gW 

3 yk ρ . 2

[2.54]

Similarly, the average velocity of the flow at a critical depth yk ρ is given by: Ε k ρ = yk ρ +

or

Vk2ρ 2g

=

Vk2ρ 2g

yk ρ 2

,

Vk2ρ 2g

= Ek ρ − yk ρ =

so Vk ρ =

gy k ρ or

3 1 yk ρ − y k ρ = yk ρ 2 2

Vk2ρ gyk ρ

= 1.

Moreover, because the Froude number is Fr =

when Fr =

when Fr =

Vk2ρ gyk ρ

[2.55] V gy

,

= Fk ρ = 1 , the flow is called critical;

V2 > 1 , the flow is steep and is called super-critical; gy

and when Fr < 1 , the flow is calm and is called sub-critical.

Open Channel Flow

83

2.13. Open channels’ more adequate cross-sectional areas

A cross-sectional area is more adequate for a flow in open channels when it gives the lowest possible wet perimeter, when the flow rate is known. We will examine when this happens for the following two channels: i) rectangular cross-section; ii) trapezoidal cross-section. 2.13.1. Rectangular cross-sectional area

In a rectangular cross-sectional area, the most adequate Sπ = x, y is the one for which the wet perimeter P = 2 y + x becomes minimum. We can assume a cross-sectional area, shown in Figure 2.15, with dimensions x and y, whose values of Q, n and tan θ are known.

Figure 2.15. Rectangular cross-sectional area of an open channel

From the Manning type, we have that the flow rate is given by the relation:

Q=

1.486 2/3 SRh tan θ . n

[2.56]

Because Q, n and tanθ are known: – the surface area is S = xy , – the perimeter is Ρ = x+2y and – the hydraulic radius is: Rh =

S xy = . P x + 2y

84

Fluid Mechanics 2

So, solving [2.56] to SRh2/3 , we have:

SRh2/3 =

Qn 1.486 tan θ

= C1 = constant

S So: S = C1 Rh−2/3 = C1   P

−2/3

= C1

5/3 2/3 2/3 or S = C1 P ή S = C1 P 

S −2/3 P −2/3

3/5

finally, S = C2 P 2/ 5

[2.57]

This last relation shows that the values of x and y that make the surface area S minimum also make the wet perimeter P, and vice versa. From the relations of Ρ and S, solving for x, we have: x = P − 2y .

so: S = xy = ( P − 2 y ) y , but also S = C2 R 2/ 5 , so: ( P − 2 y ) y = C2 P 2/5 .

[2.58]

The last relation will give the value of y that makes Ρ minimum. This value will dP be calculated by setting the derivative equal to zero: dy So, from [2.58], we have: Py − 2 y 2 = C2 P 2/ 5 

Pdy + ydP − 4 ydy = C2

2 −3/5 P dP  5

 2  ( P − 4 y)dy =  C2 P−3/5 − y  dP ,  5 

Open Channel Flow

85

 2  dP so: P − 4 y = C2 P −3/5 − y   5  dy With

dP = 0 , we have P = 4 y = 0 , dy

so P = 4 y Moreover, because P = x + 2 y , x = P − 2y = 4y − 2y = 2y ,

so: x = 2 y

[2.59]

This means that, when the shape of the cross-sectional area is a rectangle, the width of the channel must be twice the depth of the flow. 2.13.2. Trapezoidal cross-sectional area

Assume that the cross-sectional area of the trapezoidal flow, as shown in Figure 2.16, has an inclination equal to the non-parallel sides Α. The surface area of the wet cross-sectional area is: S = β y + Ay 2

Figure 2.16. Trapezoidal cross-sectional area of an open channel

[2.60]

86

Fluid Mechanics 2

The wet perimeter is: P = 2 y 1 + A2 + β .

[2.61]

Solving [2.61] for β and substituting it into [2.60], we have:

)

(

S = P − 2 y 1 + A2 y + Ay 2 .

[2.62]

However, from the Manning relation, we find: S = CP 2/5

[2.63]

From [2.62] and [2.63], we have that: Py − 2 y 2 1 + A 2 + Ay 2 = CP 2 / 5 .

[2.64]

Taking the differential of both sides of [2.64] and keeping the inclination of Α constant, we have:

Pdy + ydP − 4 ydy 1 + A2 + 2 Aydy =

or P + y

And setting

2C 3/5 P dP 5

dP 2C −3/5 dP P − 4 y 1 + A2 + 2 Ay = . dy 5 dy dP = 0 , we have: dy

P = 4 y 1 + A2 − 2 Ay .

[2.65]

Afterward, differentiating [2.64] and keeping y constant, we have: ydP − 2 y 2

y

1 2 AdA 2C −3/5 P dP + y 2 dA = 2 2 1+ A 5

dP 2 y2 A 2C −3/5 dP P − + y2 = . 2 dA dA 5 1+ A

[2.66]

Open Channel Flow

Moreover, setting

87

dP = 0 , we have: dA

2Α = 1 + Α 2

or 4Α2 = 1 + Α2 or 3Α2 = 1 so A =

3 . 3

Substituting [2.66] into [2.65], we find: P = 4y 1+

3 3 4 3 8 2 3 −2 y = y 12 − 2 y= y 3= y 9 3 3 3 3 3

so P = 2 y 3 .

[2.67]

Substituting [2.66] and [2.67] into [2.61], we have:

β = P − 2 y 1 + A2 = 2 y 3 − 2 y 1 +

 β = 2y 3 −

 β = 2y

3 2 12  = 2y 3 − 9 3

4y 3 3 3 3 = 6y − 4y = 2y  3 3 3 3

3 . 3

[2.68]

Substituting [2.68] and [2.67] into [2.60], we find: S = β y + Ay 2 = 2 y 2 or S = y 2 3

3 3 + y2 = y2 3 3 3 [2.69]

88

Fluid Mechanics 2

3 , having a vertical inclination of 3 non-parallel sides equal to 30°, we take the best trapezium, which must have non-parallel sides with an angle of 120°. From [2.66], it can be noted that with Α =

From [2.67] and [2.68], we note that P = 3β , which means the lengths of the non-parallel sides and the width of the bed are equal. Therefore, the trapezium that gives the smallest wet perimeter is one half of the regular hexagon (Figure 2.17).

Figure 2.17. Trapezium open channel that provides the smallest wet perimeter

2.14. Non-uniform flow in open channels

With gradual change of the depth of a flow, roughness of the walls, surface area, inclination of the bottom, etc. along a natural canal, the flow varies gradually. In order to calculate and find the solution of the flows of this type, we take as granted that the inclination of the energy line at a certain point is given by the Manning relation, which is, for low angles: n2Q 2 = sin θ ≅ tan θ . 2.21S 2 Rh4/3 From the energy equation for positions 1 and 2 of Figure 2.18, we have: (Total energy at position 1) − (total energy at position 2) = losses.

[2.70]

Open Channel Flow

89

From Figure 2.18, we have:

 V12   V22   Ζ1 + y1 +  −  Z 2 + y2 +  = h . 2g   2g  

[2.71]

2

1 θ

V 12 2g

ΑΔl

Γραμμή Ενέργειας

V 22 2g

y1 Δl θ0

Α0Δl

z1

y2 z2 z=0

Figure 2.18. Non-uniform flow in an open channel

Therefore, if θ is the angle between the energy line and the horizontal level and θ0 is the angle between the bottom and the horizontal level, then we have, for low values of θ and θ0: ( Ζ1 − Ζ 2 ) = (Δ ) sin θ 0 ≅ (Δ ) tan θ 0 and h ≅ ( Δ ) tan θ .

[2.72]

Thus, if we substitute these values into relation [2.71], we have:

V2 V2  (Δ ) tan θ 0 + ( y1 − y2 ) +  1 − 2  = (Δ) tan θ  2g 2g  or Δ =

(V

2 1

− V22 ) / 2 g + y1 − y2 tan θ − tan θ 0

, where tan θ =

n 2Q 2 . 2.21S 2 Rh4/3

[2.73]

90

Fluid Mechanics 2

For the solution of problems where the flow is non-uniform, we try various values of the unknown and find the values that satisfy relation [2.72]. In the case where the flow conditions in position 1 are known and we seek the depth y2, we follow the following procedure: i) We choose a value for y2 and calculate the surface area and velocity at position 2. ii) We calculate the value of tanθ from the Manning relation, where for S and Rh , we have taken the average of S1 and S2, and for Rh the average of the wet perimeters Ρ1 and Ρ2. iii) We replace in [2.72] and solve for Δ . iv) If the value of Δ is not the correct one, we choose a new value for y2 and repeat the procedure i → iv. 2.15. Channels of non-rectangular cross-section area

When a channel has a non-rectangular cross-sectional area, for example a triangular cross-section, the specific energy Ε of the flow is usually written as a function of the total flow rate Q and the wetted transverse cross-sectional area S. Thus, we have the expression: Ε = h+

Q2 . 2 gS

[2.74]

Because the surface S is a function of depth h for a given cross-section, we can draw a curve of specific energy, almost the same as the one shown in Figure 2.14a for a value of the total flow rate Q. Then, the point of the minimum energy of the curve Ε(h) will lie where the derivative of the special energy to the depth is equal to zero, which is (dE / dh) = 0 for a steady total flow rate Q. If we use this condition in equation [2.73], after we take into account that the surface S is a function of the head (S = S(h)), then we will have: 3  dS  gS . =   2  dh  Q

[2.75]

Open Channel Flow

91

If we have infinite variations of depth h, the respective variations of the surface S will be dS = b0 dh , where b0 is the width of the free surface. Therefore, equation [2.74] becomes:

Sc =

b0 Q 2 , g

[2.76]

so the velocity of the fluid V at the critical cross-sectional area will be:

Q = Sc

Vc =

gSc b0

[2.77]

and the Froude number d  for a flow in non-rectangular cross-sectional area channels is: Fr =

V = Vc

V gS c / b0

[2.78]

Hydraulic depth hh of a non-rectangular cross-sectional area channel is defined as the ratio of the surface area S of the wetted cross-sectional area to the width b0 of the free surface of the fluid. This means: hh =

S . b0

[2.79]

Critical inclination θc in non-rectangular cross-sectional area channels is the value of the inclination θ 0 of the channel’s bottom, beyond which there will be a critical flow. We have a critical flow for a uniform flow when Fr = 1. Therefore, when θ 0 < θ c , the uniform flow is sub-critical ( Fr < 1) ; when θ 0 > θ c , the uniform flow is super-critical ( Fr > 1) . The critical inclination of the channel is calculated from the equation that arises from the combination of equation [2.76] and the Manning equation, which is given by the relation:

V=

κ 2/3 Rh θ0 , h

[2.80]

92

Fluid Mechanics 2

where κ is the Manning coefficient and h is the parameter that depends on the geometrical characteristics of the channel. Thus, we have:

θc =

gh2 Pc 4/3 ⋅ . b0κ 2 Sc1/3

[2.81]

Finally, if the inclination of the channel is lower than the critical inclination (θ 0 < θ c ), then it is called light; if it is higher than the critical inclination (θ 0 > θ c ), then it is called steep. 2.16. Formulae 1) Froude number, Fr:

Fr =

V gh

where V is the average flow velocity, g is the acceleration due to gravity and h is the flow depth. 2) Reynolds number, Re:

Re =

Vdh v

where V is the average velocity, ν is the kinematic viscosity and dh is the hydraulic diameter of the channel. 3) Hydraulic radius, Rh: Rh =

h⋅w 2h + w

where h is the depth of the channel and w is the basis length of the channel.

Open Channel Flow

93

2.16.1. Channels of rectangular cross-sectional area formulae 4) Wetted transverse cross-section, S:

S = b⋅ h where b is the width of the channel and h is the depth of the channel. 5) Wetted perimeter, Π:

Π = b + 2h where b and h are the width and the head of the channel, respectively. 6) Hydraulic diameter, dh: 4bh 5 dh = 4   =  Π  b + 2h

where S is the wetted transverse cross-sectional area, Π is the wetted perimeter and b and h are the width and the depth of the channel, respectively. 2.16.2. Channels of trapezoidal cross-section formulae 7) Wetted transverse cross-section, S: 1 S = b ⋅ h + ( m1 + m2 ) h 2 2

8) Wetted perimeter, Π:

Π =b+

(

)

1 + m12 + 1 + m22 h

94

Fluid Mechanics 2

9) Width of the free surface, b0:

b0 = b + ( m1 + m2 ) ⋅ h where m1 is the ratio of horizontal to vertical variation of the wall on one side of the channel, m2 is the respective quantity on the other side of the channel and b and h are the width and the depth of the channel, respectively. 2.16.3. Channels of circular cross-sectional area formulae 10) Wetted transverse cross-section, S:

1 S = (2θ − sin 2θ )d 2 8 11) Wetted perimeter, Π:

Π =θ ⋅d 12) Width of the free surface, b0:

b0 = d ⋅ sin θ 13) Angle, θ: θ = cos θ −1 [1 − 2( h / d ) ]

where θ is the angle associated with the depth h and the width b of the channel. 14) Depth of the flow after the hydraulic jump, h2: h2 =

h1 1 + 8 Fr1 − 1 2

where h1 is the depth of the flow before the hydraulic jump and Fr1 is the Froude number calculated at depth h before the jump.

Open Channel Flow

95

15) Energy loss, Εθ, because of the hydraulic jump:

 1 Εθ = gh1 1 − 2 

(

)

1 + Fr12 − 1 +

Fr12 2

  4 1 −  2  ( 1 + 8Fr1 ) 

  

where g is the acceleration due to gravity, h1 is the flow depth before the jump and Fr1 is the Froude number calculated at position 1. 16) Velocity distribution of the laminar flow, V(y):

V ( y) =

g y  h −  y ⋅ tan θ ν  2

where g is the acceleration due to gravity, ν is the kinematic viscosity, h is the flow depth, y is the distance of a random point from the solid boundary and θ is the channel inclination. 17) Average velocity in the laminar flow, Vm:

Vm =

gh2 tan θ 3v

where g is the acceleration due to gravity, h is the flow depth, θ is the channel inclination and ν is the kinematic viscosity. 18) Velocity distribution of the turbulent flow, V(y):

V ( y ) = 2,5

τ0  y  n   ρ  y0 

where τ0 is the surface stress, ρ is the fluid density, y is the distance of a random point from the solid boundary, y0 =

ν and ν is the kinematic viscosity. τ0 g ρ

96

Fluid Mechanics 2

19) Average velocity in the turbulent flow, Vm:

Vm =

V(0.2 h ) + V(0.8h ) 2

where V(0.2 h ) is the velocity of the flow at depth (0.2h), V(0.8 h ) is the velocity of the flow at depth (0.8h) and h is the flow depth. 20) Chezy type:

V = C Rh tan θ where V is the average velocity of uniform flow, Rh is the hydraulic radius and C is the Chezy coefficient. 21) Channel’s flow rate for a uniform flow from the Chezy type, Q:

Q = CS Rh tan θ where S is the surface area of the wetted surface, Rh is the hydraulic radius, θ is the channel inclination and C is the Chezy coefficient. 22) Chezy coefficient for the laminar flow:

C=

g Re = 0.3536 g Re 8

where g is the acceleration due to gravity and Re is the Reynolds number. i) Average velocity for the laminar flow

V = 0.3536 gRh R e tan θ and

Open Channel Flow

97

ii) Flow rate for the laminar flow

Q = 0.3536S gRh R e tan θ 23) Manning type:

CM =

1.486 1/6 Rh h

24) Bazin type:

CB =

157.6 m 1+ Rh

25) Kutter type:

0.00281 1.811 + tan θ n Cκ = 0.00281  n  1+  41.65 +  tan θ  Rh  41.65 +

where n and m are parameters that take values depending on the roughness, size and shape of the channel, and are given in Tables 2.1 and 2.2b; θ is the channel inclination and Rh is the hydraulic radius. 26) Manning type in SI units: CM = KRh1/ 6

where Κ is the Manning roughness coefficient and Rh is the hydraulic radius. 27) Average velocity and flow rate combining the Manning and Chezy types: i) Average velocity: V = Κ Rh2 / 3 tan θ

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Fluid Mechanics 2

ii) Flow rate: Q = KSR h2 / 3 tan θ

where Κ is the Manning roughness coefficient (Table 2.2a), Rh is the hydraulic radius, S is the wetted surface and θ is the channel inclination. 28) Best hydraulic cross-section: i) Trapezoidal channel Rh =

1 h 2



best hydraulic cross-section

ii) Rectangular channel h=

b 2



best hydraulic cross-section

29) Specific flow energy, Ε:

Ε = y+

V2 2g

where y is the head of the line of specific energy from the bottom of the channel, V is the average flow velocity and g is the acceleration due to gravity. 30) Channel’s flow rate in relation to the depth and specific energy of the channel, Q:

Q = 2 gy 2 ( E − y) W where W is the length of the channel, g is the acceleration due to gravity, Ε is the specific energy and y is the head of line of specific energy from the bottom of the channel.

Open Channel Flow

99

2.16.4. Channels of rectangular cross-sectional area formulae 31) Specific flow rate, q:

q=

Q = Vh b

where Q is the total flow rate of the channels, b is the width of the channel, V is the average flow velocity and h is the depth of the channel’s bottom. 32) Specific energy, Ε:

Ε = h+

q2 2 gh 2

where h is the depth from the bottom, q is the specific flow rate and g is the acceleration due to gravity. 33) Critical depth, ycr:

ycr =

Q2 gW 2

where Q is the total flow rate, y is the acceleration due to gravity and W is the length of the channel. 34) Critical specific energy, Εcr:

Ε cr =

3 ycr 2

where ycr is the critical depth. 35) Average velocity, Vcr:

Vcr = gycr where g is the acceleration due to gravity and ycr is the critical depth.

100

Fluid Mechanics 2

2.16.5. Channels of non-rectangular cross-sectional area formulae 36) Specific energy, Ε:

Ε = h+

Q2 2 gS

where h is the flow depth, Q is the flow rate and S is the wetted transverse cross-section. 37) Wetted transverse cross-sectional area S at the critical point:

Sc =

3

b0 O22 g

where b0 is the width of the fluid’s free surface, Q is the total flow rate and g is the acceleration due to gravity. 38) Average velocity at the critical point, Vc:

Vc =

3

gSc b0

where g is the acceleration due to gravity, Sc is the wetted transverse cross-sectional area at the critical point and b0 is the width of the fluid’s free surface. 39) Froude number at the critical point: Fr =

V gSc / b0

where V is the flow velocity, g is the acceleration due to gravity, Sc is the wetted surface at the critical point and b0 is the width of the flow’s free surface.

Open Channel Flow

101

40) Hydraulic depth, hh:

hh =

S b0

where S is the surface area of the wetted surface and b0 is the width of the fluid’s free surface. 41) Critical inclination, θc:

θc =

gh 2 Π c4/3 ⋅ b0κ 2 S c1/ 3

where g is the acceleration due to gravity, h is the bottom depth, Πc is the critical perimeter, κ is the Manning coefficient and S is the critical wetted surface. 2.17. Questions

1) What are the open channels and how many categories do we have? 2) Which are the prismatic open channels and how many categories of them do we have? 3) How is the Froude number defined and how is the Reynolds number for open channels defined? 4) Which flow forms are created in open channels? How is each one of them defined and in which open channels are they created? 5) What do you know about the laminar and turbulent flows in open channels and when does each one appear in relation to the velocity of the flow? 6) When is a flow sub-critical, critical and super-critical in open channels? 7) What shapes do the open channels have? Describe each one of them. 8) What are the geometrical characteristics of the rectangular open channels, and how is each one of them defined? 9) What are the geometrical characteristics of the open channels of a trapezoidal cross-section? 10) What are the geometrical characteristics of the open channels of a circular cross-sectional area and how are they defined?

102

Fluid Mechanics 2

11) What do we call hydraulic jump, when is it created and what phenomena accompany it? Which basic parameters create them and what forms of jumps do we have? 12) How is the depth of the flow after the hydraulic jump calculated? Give an example of the calculation. 13) How is the energy loss because of the jump calculated? Which equation gives it and how is it associated with the Froude number? 14) Which is the velocity distribution inside an open channel, how does it develop from the bottom to the free surface of the fluid and when do we have its maximum value and why? Give schematically a distribution of this kind for a prismatic cross-sectional area of an open channel and for one of random cross. 15) Calculate the velocity distribution at the vertical level of an open channel a) for a laminar flow and b) for a turbulent flow. 16) To which flow does the Chezy type give the velocity distribution and how is it calculated? 17) Calculate the flow rate of a uniform flow in an open channel using the Chezy type. How is the coefficient C of Chezy defined, what are its dimensions and what is its value for a laminar flow? 18) How are the Manning, Bazin and Kutter types defined, what are they and which one of them is used the most, what is its value when the hydraulic radius and the cross-sectional area are given in the SI system and what are the expressions that give the distribution of the velocity and of the flow rate in combination of the Manning and Chezy types? 19) How is the best hydraulic cross-sectional area defined and calculated for the best trapezoidal open channel? 20) What do we call specific flow energy in open channels, what does it express, what does it give and how do we calculate the flow rate of the channel in relation to the depth of the flow and of the specific energy? 21) How is the specific flow rate for open channels of rectangular cross-sectional area defined and what relation gives it? 22) Show the specific energy of a rectangular channel graphically and from this, define the critical energy, the critical depth and the critical cross-section. 23) How is the critical depth of an open rectangular channel defined and how is it calculated? Moreover, calculate the critical special energy and the average velocity of the flow as a function of the critical depth.

Open Channel Flow

103

24) Which is the relation that expresses the Froude number with the critical velocity of the flow and which flow does this number characterize to the unit (1)? 25) When is a cross-sectional area more adequate for the flow of a fluid in open channels, and when for a rectangular cross-section? Prove if Q, n and tanθ are known. 26) The same for a trapezoidal cross-section. 27) In which open channels is the flow variable and why? What condition do we take for the solution of these flows and how does it take place? 28) Which is the expression of the special energy in open channels that do not have a rectangular cross-sectional area and which condition should be considered applicable for the solution of these problems, with the surface S being a function of the flow depth, which means S = S(h)? 2.18. Problems with solutions

1) A canal of rectangular cross-sectional area is made up of concrete of medium roughness, with inclination θ = 0.0009; its dimensions are 2 m and 3 m, as shown in Figure (1). Calculate its flow rate.

Figure (1)

Solution According to the given facts of the problem, we have: S = 2 × 3 = 6 m 2 = 64.58 ft 2

Rh =

S 6 = m = 2.81ft 4+3 7

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Fluid Mechanics 2

From Table 2.1, we see that for concrete of medium roughness n = 0.015, the ft 3 is: flow rate Q in s

Q=

1.486 ft 3 × 64.58 × (2.81)2/3 (0.0009)1/ 2 0.015 s

or Q = 382.17

ft 3 m3 = 10.82 . s s

2) An open channel of trapezoidal cross-sectional area with slope inclination of 1:2 coated with concrete has a bottom width of 0.60 m and a bottom depth of 0.90 m. If the flow rate of the water is 0.90 m3/s, calculate: i) the inclination along the canal and ii) the average velocity of the water (if the Manning coefficient n = 0.012). Solution

Figure (2)

i) We use the Manning equation, which is given by the relation:

V=

1 2/3 1/ 2 Rh θ n

For a channel of trapezoidal cross-sectional area, it is applicable that: For the hydraulic radius Rh =

S (b + my ) y = Π b + 2 y 1 + m2

Open Channel Flow

And for the velocity V =

105

Q Q = S (b + my ) y

Solving the Manning equation to θ, we get:

θ=

n2Q 2 n 2 Q 2 [b + 2 y 1 + m2 ]4/3  = 2 4/3 S R (by + my 2 )10/3

0.012 2 ⋅ 0.9 2 (0.6 + 2 ⋅ 0.9 1 + 2 2 θ= (0.6 ⋅ 0.9 + 2 ⋅ 0.9 2 )10/3

θ = 0.0000688 ii) V =

V=

4/3



θ = 0.0688%.

or

 0.6 ⋅ 0.9 + 2 ⋅ 0.9 2  1 ⋅  0.012  0.6 + 2 ⋅ 0.9 1 + 2 2 

2/ 3

⋅ 0.00006881/ 2 

0.472/3 ⋅ 0.00826 = 0.415m/s  0.012

V = 0.415 m/s. 3) Calculate the flow rate of a uniform flow of a perimeter moat of a complex trapezoidal cross-section (Figure (3)), when the Manning coefficient n = 0.025 for the smallest bed and n = 0.032 for the flood bed. The bottom inclination is 0.0001 (the dimensions shown in the figure are expressed in meters).

Figure (3)

106

Fluid Mechanics 2

Solution

The flow rate is given by the continuity equation: Q = VS = V1 ( S1 + S3 ) + V2 S2 , where S1 = S3 = ( ABCL ) = ( MFGH ) = (5 + 8 + 8)

3 = 31.5 m 2 2

S2 = ( LCFM ) + (CDEF ) = = 3(4 + 9 + 4) + (4 + 9 + 4 + 9) S1 + S3 = 2 ⋅ 31.5 = 63 m 2

V1 =

S 2 = 103 m 2 ,

1 1 Rh12/3 θ 01/ 2 , V2 = Rh 2 2/3 θ01/ 2 , n1 n2

where Rh1 =

Rh 2 =

S1 31.5 31.5 = = = 2.28 m , Π1 8 + 52 + 32 13.83

S2 103 103 = = = 5.05 m . 2 2 20.3 Π2 9 + 2 4 + 4

So V1 =

V2 =

4 = 3.17 + 2.26 = 51 + 52 = 103 m 3 2

1 1.73 2.282/3 ⋅ 0.00011/ 2 = = 0.54 m/s , 0.032 3.2

1 2.94 5.052/3 ⋅ 0.00011/ 2 = = 1.178m/s 0.025 2.5

and Q = 0.54 ⋅ 63 + 1.178 ⋅103 = 34 + 121.2 = 155.2 m 3 /s 

Open Channel Flow

107

Q = 155.2 m 3 /s .

4) An open channel of rectangular cross-section, of width 4 m, has regular depth 0.50 m. The channel is made up of cast iron and its bottom has an inclination of 0.0049. Define: a) the average velocity of the fluid in the channel; b) the specific energy of the flow; c) the type of the flow (sub-critical or super-critical); d) the critical depth of the flow and the critical inclination of the bottom. Solution i) The average velocity V of the fluid in the channel is calculated by the Manning equation:

V=

κ 2/3 Rh θ n

(1)

where the constant κ = 1 m1/3/s and the inclination of the bottom θ = 0.0049. The Manning coefficient for cast iron is n = 0.013 (from Table 2.1). For a rectangular cross-section, the wetted transverse cross-section S and the wetted perimeter Π are given by equations [2.4] and [2.5], respectively, so the hydraulic radius Rh of the channel, for b = 4 m and h = 0.50 m, is: Rh =

S bh (4m)(0.50m) = = = 0.40m . Π b + 2h (4m) + 2(0.50m)

(2)

Substituting into equation (1) the values of κ, n and Rh, we have: V=

(1m1/3 / s) (0.40m) 2/3 (0.0049)1/ 2 = 2.92 m/s . (0.013)

(3)

ii) The specific flow rate q is given by equation [2.51], which is: q = Vh = (2.92 m/s) (0.50 m) = 1.46 m2/s

(4)

and the specific energy Ε of the flow from the equation: Ε = h+

V2 (2.92m / s) 2 = (0.50m) + = 0.935 m. 2g 2(9.81m / s 2 )

(5)

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Fluid Mechanics 2

iii) In order to answer for the kind of the flow, we use the Froude number, Fr, as a criterion, whose value is calculated from the equation:

Fr =

V

(2.92m / s )

=

gh

(9.81m / s 2 )(0.50m)

= 1.32.

(6)

Therefore, the flow is super-critical, because the Froude number is greater than 1. iv) The critical depth, hc, of the flow is calculated from equation [2.53]: 1/ 3

 q2  hc =    g 

1/ 3

 (1.46m 2 / s ) 2  =  2  (9.81m / s ) 

= 0.60 m.

(7)

The critical inclination θcr of the bottom is calculated from the equation:

q = Vc hc =

κ hc Rc2/3 θcr n

(8)

where Rc is the critical hydraulic radius of the channel: Rc =

bhc (4m)(0.60m) = = 0.46 m. b + 2hc (4m) + 2(0.60m)

(9)

Substituting the values of q, κ, n, hc and Rc in equation (8), we have:

(1.46m 2 / s) = [(1m1/3 / s) / (0.013)](0.60m) 2/3 θ cr .

(10)

Then, after we perform the sums:

θ c = 0.0028.

(11)

We note that hc > h(= 0.50m) and θ c < 0(= 0.0049). These results were expected, because the given flow is super-critical (Fr > 1). v) The conjugate depths of the flow arise from the solution of the equation: Ε = h+

q2 , 2 gh 2

(12)

Open Channel Flow

109

which, for q = 1.46 m2/s, E = 0.935 m and g = 9.81 m/s2, finally takes the form: h3 − 0.935h 2 + 0.109 = 0.

(13)

This equation has two positive solutions: h1 = 0.50m and h2 = 0.73m. The depth h1 is the real depth of the flow and h2 is its conjugate. Finally, the velocity V2 of the fluid for the depth h2 is: V2 = g / h2 = (1.46 m 2 /s) / (0.73 m) = 2 m/s .

5) In an open channel of triangular cross-sectional area, water flows with a constant flow rate of 3 m 3 /s . The inclination of the channel sides is 45° and the channel has a depth of 2.5 m, as shown in Figure (5). Calculate: i) The critical depth of the flow and the critical cross-sectional area of the current. ii) The length  of the sides over the free surface of the water so that the channel can transfer 50% more water.

Figure (5)

Solution The wetted transverse cross-sectional area S, the wetted perimeter Π, and the width of the water’s free surface b0 of this channel, which is of triangular cross-sectional area, are given by the respective equations of a channel of trapezoidal cross-sectional area if we set in them: b0 = 0, m1 = m2 = 0

110

Fluid Mechanics 2

Moreover, because cot 45o = 1 , we have the equations S=

1 ( m1 + m2 ) h 2 = h 2 , 2

(1)

)

(2)

and b0 = (m1 + m2 )h = 2h .

(3)

Π=

(

1 + m12 + 1 + m2 2 ⋅ h = 2 2h ,

Therefore, the hydraulic radius Rh of this channel with relations (1) and (2) is:

Rh =

S 2 = h. Π 4

(4)

i) At the point of the minimum energy, where the depth of the fluid is equal to the critical depth of the flow (h = hc ) , the following equation is applicable: gS c3 = b0 Q 2 .

(5)

Substituting S c = hc2 and b0 = 2hc in this equation and solving it for the critical depth hc, we find: 1/5

 2Q 2  hc =    g 

1/5

 2(3m3 / s ) 2  = 2   (9.81m / s ) 

= 1.13m.

(6)

The critical cross-sectional area, Sc, of the current is: S c = hc2 = (1.3m ) 2 = 1.275 m 2 .

(7)

ii) For a uniform flow in open channel, the volumetric flow rate Q is given by the Chezy–Manning equation, which for the triangular canal of the figure is written as: Q=

1.486 1.480 2  2  ( h )  S ⋅ Rh2/ 3 tan θ =  n n  4 

2/ 3

⋅ tan θ = α h8/ 3 ΄

(8)

where α is constant. The same relation also gives the new flow rate Q0 of the channel: Q0 = α h08/ 3 .

(9)

Open Channel Flow

111

Dividing by terms equations (8) and (9) and taking into account that the flow rate Q0 = 1.5Q , we have:

 h0  h  

8/3

= 1.5.

(10)

From this equation, for h = 2.50 m , arises the new depth of the flow:

h0 = 2.91m.

(11)

From the geometry of Figure (5) arises the relation that gives the length  0 :

0 =

h0 − h (2.91m) − (2.50m) = = 0.58m. sin θ sin 45o

(12)

6) A canal with a rectangular cross-sectional area of bottom width 3.0 m and water depth 1.50 m carries water at a flow rate of 7.75 m3/s: i) Calculate the special energy. ii) Is the flow sub-critical or super-critical? iii) Calculate the inclination of the channel when the velocity becomes critical (n = 0.014). Solution i) We know that the specific energy as a function of the flow rate is given by the type: E = y+

Q2 2 gS 2

or E = y +

(1)

Q2 7.752 = 1.5 + = 1.5 + 0.15 = 1.65 m. 2 2 2 gx y 19.62 ⋅ 32 ⋅1.52

So, E = 1.65 m. ii) The most important parameter that characterizes the flow is the Froude number. The number Fr in relation to the flow rate is given by the equation: Fr 1 2 =

Q2 x2 , gS 3

(2)

112

Fluid Mechanics 2

so: Fr 1 2 =

Q2 7.752 = = 0.2 < 1 gx 2 y13 9.81 ⋅ 32 ⋅1.53

Thus, the flow is super-critical. iii) For V = Vc and y = yc , the inclination is θ 0 = θ c . We know that the relation

θ c = g 10/ 9 ⋅ n 2 ⋅ q −2 /9

(3)

is applicable for channels of large width for which R = y. Thus, R=

2y S y ≠ 0. where = x 2 y Π   1 +  x  

Equation (3) becomes:

θc =

g 10/9 ⋅ n 2 q 2/9

or θ c =

 2y  1 +  x  

g 10/ 9 ⋅ n 2 q 2/ 9

4/3

 q 2/ 3   1 + 2 1/ 3  g x 

4/3

9.8110/ 9 ⋅ 0.014 2  2.5852/ 3  1 2 + θc =   (7.75 / 3) 2/ 9  9.811/ 3 ⋅ 3 

=

12.643 ⋅1.96 ⋅10−4  1.883  1 + 2  1.235 6.422  

θc =

4/ 3

=

4/3



20.065 ⋅1.84 = 0.00369 . 104

So, θ c = 0.00369 . 7) In a triangular canal, the inclination of the bottom is 0.6% and the Manning roughness coefficient n is 0.012. For a uniform flow of 500 lt/s, check whether the flow is sub-critical, critical or super-critical (slope inclination 1:1).

Open Channel Flow

113

Figure (7)

Solution The most important parameter that characterizes the flow is the Froude number. As we know that for: Fr 2 = 1 , the flow is critical Fr 2 > 1 , the flow is super-critical Fr 2 < 1 , the flow is sub-critical

where Fr 2 =

Q2 x gS 3

(1)

For a triangular channel and m = 1, we have: Fr 2 =

Q 2 (2my ) 2Q 2 y 2Q 2 = = 5 . g ⋅ (my 2 )3 gy 6 gy

In order to find the Froude number, the flow depth y must be calculated. Using for the velocity the Manning equation, the flow rate takes the form:

Q = VS = S

1 2/3 1/ 2 R θ  n

1 S  Q = my ⋅   nΠ

2/ 3

2

=

1 y 8/ 3 S 1/ 2 , n (2 2) 2/3

θ

1/ 2

 y2  1 = y 2 ⋅   n  2y 2 

2/3

θ 1/ 2 =

114

Fluid Mechanics 2

because S = my 2 , x = 2 my and Π = 2 y 1 + m 2 . Solving for y, we have:  (2 2) 2/ 3 nQ  y=  S 1/ 2  

3/8

 2 ⋅ 0.012 ⋅ 0.5  y=  1/ 2  0.006 

 2nQ  =  1/ 2  S 

3/8

3/8

 0.012  =   0.077 

 3/8

= 0.1563/8 = 0.498 

y = 0.498m . 2 So: Fr =

2 ⋅ 0.52 = 1.664 > 1 ; the flow is super-critical. 9.81⋅ 0.4985

8) Water flows in an open rectangular channel with constant flow rate 12 m3/s and depth 30 cm. The channel is horizontal and has a width of 5 m. At some position of the channel, the flow creates a hydraulic jump. Define: i) the type of the hydraulic jump; ii) the depth, the velocity and the Froude number of the flow in the hydraulic jump.

Figure (8)

Open Channel Flow

115

Solution i) In order to define the type of the hydraulic jump, we must know the value of the Froude number, Fr , of the flow before the hydraulic jump. 1

Fr 1 =

V1 gh1

(1)

,

where g is the acceleration due to gravity, V1 is the average velocity and h1 is the depth of the water before the hydraulic jump (Figure (8)). The velocity V1 is calculated from the equation: V1 =

Q Q (12m3 / s) = = = 8 m/s . S bh1 (5m)(0.30m)

(2)

Substituting the values of the quantities g, V1 and h1, into equation (1), we obtain:

(8m / s)

Fr 1 =

(9.81m / s 2 )(0.30m)

= 4.66.

(3)

Therefore, the hydraulic jump is constant because the Froude number of the flow before the hydraulic jump in the area is 4.5< Fr υ

[3.4]

2) From the boundary surface to the end of the boundary layer we have a variation of the velocity from 0 to V , we therefore conclude that: ∂u ∂u >> ∂y ∂x

[3.5]

3) The thickness of the boundary layer is very small, we therefore conclude that: x >> y

[3.6]

4) = ( ), we therefore conclude that the pressure variation along the y-direction: ∂p , ∂y

is very small.

[3.7]

122

Fluid Mechanics 2

Therefore, based on the previous conditions, we have: x0 =

x  (1) L

[3.8]

u0 =

u  (1) V

[3.9]

y0 =

y δ =  (δ 0 ) 0   ∂x  ∂y  y = 0

Figure 3.4. The three flow areas inside the boundary layer

[3.30]

128

Fluid Mechanics 2

The distance of the point of separation Α from the projection lip of the body, which is the length of the laminar boundary layer, depends on: 1) the form of the body; 2) the smoothness of the body surface, which increases with it; 3) the initial turbulence of the flow stream, and it is higher the smaller it is; 4) the vibrations of the body; 5) the flow velocity V and the density ρ of the fluid with which it varies in inverse proportion; 6) the viscosity coefficient μ , with which it varies proportionally. We expected the phenomenon of the flow separation because the energy and momentum losses of the fluid that osculates with the body, combined with the application of a shear stress on it, justify a progressive deceleration of the fluid. From this constant deceleration of the fluid, the fluid finally appears motionless, while the boundary layer makes its separation from the body surface, as shown in Figure 3.5. Γραμμές ροής

Α

Διανομές ταχυτήτων

Περιοχή διαχωρισμού

Κατανομή ταχυτήτων σε στρωτή ροή

Κατανομή ταχυτήτων σε τυρβώδη ροή

Figure 3.5. Velocity distribution and flow separation

Generally, we can say that the creation of the turbulence comes from instability of the flow. In the case of the boundary layer, an instability appears because its external flow velocity is higher than the velocity of the fluid inside the boundary layer. This increase in the external velocity results in a higher increase in the inertial forces from the friction forces and therefore the latter are not able to defuse the disorders, which increase and finally lead to turbulent flow. Therefore, in the transition of the flow from laminar to turbulent, the ratio of the inertial forces

Boundary Layer

129

to the friction forces plays a significant role. These values are thus combined in the characteristic Reynolds number: Re =

Vl v

[3.31]

Thus, as already mentioned, a great flow criterion is the Reynolds number. The laminar flow is characterized by low Reynolds numbers, while the turbulent flow by high ones. The critical value of the Reynolds number for the transition from the laminar to turbulent flow, when it occurs around circular pipes, has been found from Schiller’s experiments (1922) as:

Recr = 2.320,

[3.32]

while for the flow around a body: Re cr = 5 ⋅105

[3.33]

Completing the study of the separation of the boundary layer, we can say that the separation of the flow never appears in the area of the pressure decrease (accelerating flow), but only in the area of the pressure increase (decelerating flow). Consequently, in a friction layer that flows in an area with pressure increase, there is always the danger of flow separation. In this case, the value of the pressure and the flow condition inside the boundary layer play the most important role. Therefore, a sudden pressure increase, such as at the back side of a blunt body, leads to a faster separation of the flow than a smooth increase in the pressure at the back side of a thin body. The turbulent boundary layer is combined with higher pressure increase than in the laminar boundary layer. From this, we conclude that, in order to avoid the separation of the flow and to fulfill favorable resistance, the back side of the bodies flowing must end up to thin cross-sections. The separation of the flow also plays a significant role in the creation of the lift force on a wing. At small projection angles (up to approximately 10° ), the part of the flow on the top and bottom sides does not show any separation, and with a good approach, it comes near to the flow without friction (Figure 3.6).

130

Fluid Mechanics 2

Figure 3.6. Flow over an airfoil without turbulence

At higher attack angles, there is a danger of separation at the side of absorption of the airfoil, where the pressure increase is very sudden. At a certain attack angle, approximately 15° , separation of the flow may appear (Figure 3.7). The position of the separation point lies slightly away from the front side of the wing.

Figure 3.7. Flow over an airfoil with separation areas, where x, L, Rex and ReL have been previously defined

Boundary Layer

131

3.5. Formulae 1) Non-dimensional Navier–Stokes and continuity two-dimensional unsteady incompressible flow:

equations

for

a

where t is the time, u, υ are velocity components, p is the pressure and Re =

VL V

2 2 ∂u0 ∂u ∂u ∂p 1  ∂ u 0 ∂ u0  + u0 0 + υ 0 0 = − 0 +  2 + 2  ∂t0 ∂x0 ∂y0 ∂x0 R e  ∂x0 ∂y 0 

2 2 ∂υ0 ∂υ ∂υ ∂p 1  ∂ υ0 ∂ υ0  + u0 0 + υ 0 0 = − 0 +  2 + 2  ∂t0 ∂x0 ∂y0 ∂y0 R e  ∂x0 ∂y 0  ∂u0 ∂υ0 + =0 ∂x0 ∂y0

is the Reynolds number. 2) Equation of the boundary layer for an unsteady two-dimensional flow of incompressible fluids:

∂u0 ∂u ∂u ∂ 2 u0 1 ∂p0 + u0 0 + υ0 0 = + v ∂t0 ∂x0 ∂y0 ρ ∂x 2 ∂y02 −

∂p0 =0 ∂y0

∂u0 ∂υ0 + =0 ∂x0 ∂y0

with boundary conditions if the boundary surface is motionless:

y0 = 0 :

u0 = 0 , υ 0 = 0

y0 → ∞

u0 → V ( x , t )

132

Fluid Mechanics 2

where t is the time, u0 , υ0 are velocity components in the boundary layer, ρ is the fluid density, p0 is the fluid pressure and V ( x, t ) is the free rheumatic velocity. 3) Separation condition of the boundary layer:

∂p  ∂u  >0   = 0 or ∂x  ∂y  y = 0 where u is the velocity component inside the boundary layer and p is the fluid pressure. 4) Shear stress of a fluid, τ:

 ∂u    ∂y 

τ =μ

where μ is the fluid’s velocity and u is the fluid’s velocity in the boundary layer. 5) Displacement thickness of the boundary layer, δ1: δ

 0

δ1′ =  1 −

u  dy V

where u is the fluid’s velocity in the boundary layer and V is the free stream velocity. 6) Equation of a two-dimensional turbulent incompressible unsteady boundary layer: ∂u ∂υ + =0 ∂x ∂y ∂u ∂u ∂u 1 ∂p ∂  ∂u  +u +υ = +  (v + εt )  ∂t ∂x ∂y ρ ∂x ∂y  ∂y 

Boundary Layer

133

where t is the time, u , υ are velocity components of the fluid in the boundary layer, ρ is the fluid density, v is the kinematic fluid viscosity and ε t is the μ  kinematic turbulence viscosity, ε t =  τ  , with μτ being the turbulence viscosity.  ρ 

3.6. Questions

1) What is the non-dimensional form of the Navier–Stokes equations? 2) What is a boundary layer, who introduced it and when? 3) What are the simplified boundary layer equations, who made it first and on what facts did they base those simplifications? 4) What are the conditions applicable inside the boundary layer? 5) Based on the above conditions, find the order value of each term of the Navier–Stokes and continuity equations. 6) Write the equations of the boundary layer and the boundary conditions that are applicable for them, in the case of a motionless boundary surface. 7) What is the distribution of the velocity of the fluid inside the boundary layer? 8) How do we define the thickness of the boundary layer? 9) How is the thickness of the boundary layer varied along the surface of a flat boundary surface? 10) How many types of coherent boundary layers are there and what are their main characteristics? 11) What are the most important qualities of the boundary layer? 12) How is the separation of the boundary layer created? 13) Which form of the boundary layer do we usually pursue in practice? 3.7. Problems with solutions

1) The wing of an airplane FAIRCHILD REPUBLICA-10A (two-engine support) has a rectangular shape with an opening 17.5 m and a chord

3 m. The airplane flies above the sea surface with a velocity of 200 m/s. a) If the flow is laminar, calculate the thickness of the boundary layer at the outflow lip and the total friction resistance.

134

Fluid Mechanics 2

b) If the flow is turbulent, calculate the thickness of the boundary layer at the outflow lip and the total friction. c) If the critical Reynolds number for the transmission is 106 , calculate the friction resistance of the wing.

(p

∞

= 1, 225 kg/m3 , μ∞ = 1.789 × 10−5 kg/ms )

Solution a) The thickness of the laminar boundary layer is given by the relation:

δx =

5.2 ⋅ x

(1)

Rex

At the outflow lip of the wing, where x = 3 m , Rex gives: Rex =

ρ ∞V∞ x (1.255 ) ( 200 ) ⋅ ( 3) = = 4.1× 107 −5 μ∞ (1.7894 ×10 )

(2)

Combining (1) and (2), we have:

δx =

5.2 ( 3)

= 0.0024 m = 0.24 cm

4.1×107

(3)

Now, D of the one side of the wing is given by the relation: D f = q∞ ⋅ S ⋅ C D f

However, q∞ = and CD f =

1 1 2 ρ ∞ ⋅ V∞2 = (1, 225 )( 200 ) = 2.45 × 10 4 N/m 2 2 2

1.328 4.1× 107

= 0.00021

(4) (5)

(6)

Therefore, combining (4), (5) and (6), we have: S   D f = ( 2.45 × 10 ) ( 3) (17.5 ) ( 0.00021) = 270 N 4

(7)

Boundary Layer

135

However, since both sides of the wing are exposed to the flow, Dολ f is given by: Dtotf = 2 ⋅ 270 = 540 N

(8)

b) The thickness of the turbulent boundary layer is given by the relation:

δ=

0.37 ⋅ x Rex

=

0.37 ⋅ ( 3)

= 0.033 m = 3.3 cm

( 4.1× 10 ) 7

(9)

However, from the theory, we know that δ of the turbulent boundary layer is larger than the one of the laminar, so combining (3) and (9), we have:

δ turb 3.3 = = 13.75 δ lam 0.24

(10)

Moreover, the resistance of the turbulent boundary layer on one side is given by the relation: D ftur = q∞ ⋅ S ⋅ C Dturb

(11)

However, the coefficient of the resistance of the turbulent boundary layer is given by:

CDturb =

0.074

(R )

0.2

ex

=

0.074

( 4.1×10 )

7 0.2

= 0.0022

(12)

So, substituting (5) and (12) into (11) gives: D fturb = ( 2.45 ×104 ) ( 3)(17.5 ) ( 0.0022 ) = 2.830 N

(13)

and the total is: Dtot fturb = 2 ⋅ 2830 = 5660 N

(14)

Furthermore, since Dtot fturb is also greater than Dtot flam , by combining relations (8) and (14), we have:

Dtot fturb Dtot flam

=

5660 = 10.5 540

(15)

136

Fluid Mechanics 2

c) Because Rexcτ is 106 , we have to find the distance, xcτ , on the wing. Thus, we have: Rexcτ

6 −5 Re τ μ∞ (10 )(1.7894 × 10 ) ρ ∞ ⋅ V∞ ⋅ xcτ =  xcτ = =  μ∞ ρ∞V∞ (1.225 ) ( 200 ) xc

(16) −2

xcτ = 7.3 × 10 m

We will now find the resistance of the wing in both areas Α and Β. Thus: For area Α, we have: D f Alam = q∞ ⋅ S ⋅ CD flam = q∞ ⋅ S

1.328

( Rcτ )

=

= ( 2.45 × 104 )( 7.3 × 10−2 ) (17.5 )

1.328 106

Thus, D f Alam = 42 N

(17)

For area Β, we will find the resistance of the turbulent boundary layer for area Β and subtract it from the resistance of the turbulent boundary layer of the whole wing that we found from relation (13). One side is: D f Bturb = D fturb (13) − D f Aturb

(18)

So: D f Aturb = q∞ ⋅ S A CD fturb = ( 2.45 × 104 )( 7.3 × 10−2 ) (17.5 ) 0.074 Recτ

= ( 2.45 × 104 )( 7.3 × 10−2 ) (17.5 )

0.074 106



Boundary Layer

 D f Aturb = 146 N (on one side)

137

(19)

Combining (18) with (13) and (19), we have: D f Bturb = 2830 − 146 = 2684 N (turbulent)

(20)

So D f Bturb = 2684 N

(21)

So, the total resistance of the wing will be (on one side): D ftot = D f Alam + D f Bturb

and using (17) and (21), we have: D ftot = 42 + 2, 684 = 2, 726 N

and for both sides: D ftot = 2 ⋅ 2726 = 5, 452 N

(22)

Comparing the results of these questions with those of the previous ones, we see that the flow above the wing is more turbulent, which is usually the case for a real airplane that is flying. 2) The distribution of the velocity of a laminar boundary layer is approximately parabolic. Having this as a base, calculate the shear stress on the wall of a flat plate that lies inside the water, when the thickness of the created boundary layer is δ = 1.5 mm and the velocity of the free flow is V = 2 m/s. The temperature of the water is 20 °C and the viscosity is μ = 1.00 × 10 −3 kg/m×s . Solution

As the velocity inside the boundary layer is expressed approximately as a parabola, we assume the following secondary polynomial:

138

Fluid Mechanics 2

u  y  y = a +b +C   V δ   δ 

2

(1)

which will verify the following boundary conditions:

y = 0: u = 0

(2)

and y = δ : u = V

(3)

Substituting (2) into (1), we have: a=0

(4)

while substituting (3), we have:

1= a+b+c

(5)

and because of (4), we have: 1= b+c

For y = δ ,

(6)

∂ (u V )

∂ ( y δ y)

= 0, because the velocity outside the boundary layer is

constant. So, for y = δ , (1) becomes: 0 = b + 2c

(7)

Solving the equation systems (6) and (7), we have: b = 2 and c = −1

(8)

Substituting the values of b and c from (8) into (1), we have: u  y  y = 2 −  V δ  δ 

2

(9)

Boundary Layer

139

The shear stress is given by the relation:

 ∂u    ∂y  y = 0

τ0 = μ 

(10)

u ∂  ∂u ∂u V  V  can also be written as However, , hence because of (9), we = ⋅ ∂y ∂y δ  y ∂  δ  have:

∂u V = ∂y δ

  y   2 − 2   δ  

(11)

So, because of (11), (10) becomes:

τ0 = μ

2 Nsm/s V  V  y  2 − 2   = 2 μ = 2 ⋅1, 0 × 10−3 = δ  δ δ 0.0015 m2 m   y =0

= 2.67 N/m 2

3) A flat plate of chord 0.6 m, with a long expanse, is put in a fluid current of velocity 45 m/s, at a zero projection angle. The distribution of the velocity on the surface of the plate is given by the relation: u = 2y − y2,

where u = u / V and y = y / δ . i) Examine whether the boundary layer around the plate is laminar by comparing the obtained results with those in the given diagram. It is given that δ =

CD =

2 D0 x ⋅ 1/ 2 , I Rex

1 32 D0 I  du  , where D0 =   and I =  u (1 − u ) dy Re  dj  y = 0 0

ii) Also, calculate the transmission thickness of the boundary layer.

140

Fluid Mechanics 2

Solution i) In order to evaluate whether the boundary layer is laminar in the whole length of the plate, we have to calculate the Reynolds number of the flow and the value of the resistance coefficient. So, from the given facts of the problem, we calculate:

 ∂u  a) D0 =   = ( 2 − 2 y ) y = 0  D0 = 2  ∂y  y = 0

(1)

b) The integration: 1

1

0

0

I =  u (1 − u )dy =  ( 2 y − y 2 )(1 − 2 y + y 2 ) dy  I =

2 15

(2)

The Reynolds number of the flow at x = 0.6 m is: Re =

V ⋅x 45 ⋅ 0.6 = = 1.85 ⋅106 v 14.6 ⋅10−6

(3)

So, using the relationship that the problem gives us for the resistance coefficient, we find: CD =

32.2 ⋅ ( 2 /15 ) 32 ⋅ D0 ⋅ I = = 0.00215 Re 1.86 ⋅106

(4)

In the given diagram, which shows the variation of the CD of the plate, as a function of the Reynolds number, the areas of the laminar, transitional and turbulent flows are distinguished.

Boundary Layer

141

Comparing the obtained value of CD with its respective value, we find that the boundary layer is laminar. ii) The transmission thickness δ ∗ is given by the relation: 1

δ .∗ = δ  (1 − u )dy = δ ⋅ I1

(5)

0

1

1

0

0

However, I1 =  (1 − u )dy =  (1 − 2 y + y 2 )dy  I1 =

1 3

(6)

So, substituting (6) into (5), we have:

δ∗ =

δ

(7)

3

Substituting in the δ as given by the problem, we have:

δ∗ =

x 1 2 D0 ⋅ 1/ 2 3 I Re x

or δ ∗ =

1 2⋅2 x ⋅ 3 2 Re1x/ 2 15

(8)

Substituting in the value of Rex from relation (3) and by setting x = 0.6, the value that corresponds to the outflow lip, we have:

δ∗ =

1 0.6 2 ⋅15 ⋅ 1/ 2 3 (1.85 ⋅106 )

or δ ∗ = 0.8 ( mm ) 4) For a laminar steady flow, in which there is a thin flat plate of length L, develop an expression for the thickness δ of the boundary layer, with the assumption that the velocity distribution is:

 2 y y2  u =V  − 2 δ δ 

142

Fluid Mechanics 2

where u is the velocity of the fluid inside the boundary layer and V is the velocity of the free stream. Solution

As we know, steady flow means that the velocity of the fluid does not vary with ∂u time, that is, =0 ∂t Moreover, it is known that δ < x. The mass dm of the fluid that passes through any intersection of the boundary layer is:

dm = ρ dV or dm = ρ dSdh or dm = ρ dydzudt and per plate and time unit:

dm = ρ udy Integrating this relationship from 0 to δ, we have: δ

m =  ρ udy

(1)

0

Now, the variation in the velocity at any point is equal to the difference of the free stream velocity V and the velocity u inside the boundary layer, which is V − u.

Boundary Layer

143

This variation of the velocity inside the boundary layer results in the variation of the momentum, which is given by the relation: dJ = Vdm dt or dJ = Vdmdt or dJ = dm (V − u ) dt and because of (1), the relationship per time unit becomes: δ

δ J =  ρ u (V − u )dy

(2)

0

However, this variation in the momentum is caused only by the shear force dFD , which is given by the relation:

dFD = τ 0 dS

(3)

and which if expressed per width unit becomes:

dFD = τ 0 dx

(4)

So, we have FD = δ J , therefore because of (2), we have: δ

FD =  ρ u (V − u )dy 0

Moreover, by substituting in it the value of the velocity u, which is given by the problem, we have: δ  2 yV y 2V   2 y y 2  + 2 V  − FD =  ρ  V −  dy = δ δ   δ δ2   0 δ  2 y y2   2 y y2  = ρV 2   1 − + −   dy δ δ 2  δ δ 2  0

and by performing the sums, we have:

144

Fluid Mechanics 2

FD =

2 ρV 2 δ 15

(5)

Moreover, we know that the shear stress, τ 0 , is given by the relation:

 du    dy  y = 0

τ0 = μ 

Or τ 0 = μ

or τ 0 =

(6)

d   2 y y2 − V  dy   δ δ 2

 2V μ  y  1−   = δ  δ  y = 0   y =0

2V μ

(7)

δ

Solving (4) for τ 0 and by equating it with (7), we have: 2V μ

δ

=

dFD dx

and because of (5), we have: 2V μ

δ

=

δ

2 dδ ρV 2 15 dx

or  δ d δ = 0

or δ 2 =

or

or

δ x

δ x

=

=

x

15μ dx ρV 0

30 μ x ρV 30v xV

5, 48 Rex

Boundary Layer

145

3.8. Problems to be solved

1) Oil flows with a free stream velocity V = 3 m/s over a thin plate of width 1.25 m and length 2 m. Define the thickness of the boundary layer and the shear stress at the middle of the length, and calculate the total resistance of both sides of the plate. It is given that:

ρ = 860 kg/m 3 , v = 10 −5 m 2 /s and CD =

1.33 Re

2) A smooth flat plate of width 3 m and length 30 m is dragged inside a motionless water of temperature 20 °C and velocity 6 m/s. Calculate the resistance that is developed on the plate, as well as the resistance that is developed on the first three meters of the plate. It is given that:

ρ = 1, 000 kg/m 3 , μ = 10 −3 kg/m×s and CD = 0.455 [ log10 Re ]

−2.58

3) The wing of an airplane has almost the shape of a rectangle with wingspan of 17.5 m and a chord of 3 m (parallel to the flow). The airplane flies at the surface of the sea at a standardized atmosphere with a velocity of 200 m/s. If there is a laminar flow along the whole length of the chord, calculate the thickness δ of the boundary layer at the outflow lip and the friction resistance. 4) Water flows over a horizontal flat plate along the x-direction. The water reaches the beginning of the plate (at the position x=0) with a constant velocity of 0.05 m/s. The boundary layer that is formed on the plate is always laminar. The water has a density of 998 kg/m 3 and a kinematic viscosity of 1.0 ×10−6 m2 /s. Calculate: a) the thickness δ of the boundary layer at the position x = 5 m; b) the maximum value of the velocity vertically to the plate at the position 0.84V x = 5 m when it is given by the relation: Vmax = . Rex 5) Water (of temperature 20 °C ) flows over a horizontal rectangular plate of length 2 m and width 1m. The velocity V of the free stream is 1m/s. The distribution of the velocity inside the boundary layer is given by the equation: Vx V∞ = ( y δ )

1/ 2

146

Fluid Mechanics 2

Moreover, it has been experimentally shown that the stress τ w is given by the relationship:

τ w = 1.66μV∞ δ Calculate the thickness of the boundary layer and the respective transmission thickness. 6) During the air flow over a thin flat plate of length 3m, a laminar boundary layer is formed. The velocity V∞ of the free stream is 2 m/s. The air has a kinematic viscosity of 1.5 × 10 −5 m 2 /s. Calculate the thickness of the boundary layer along the plate. 7) Water (of temperature 20 °C ) flows over a horizontal rectangular plate of length 1m and width 0.5 m. The velocity V∞ of the free stream is 1m/s. Calculate: a) the thickness δ of the boundary layer at the position x = 0.3 m ; b) the velocity of the fluid at the point ( x = 0.3 m, y = δ ) ; c) the fluid dynamics resistance that is applied on each side of the plate.

4 Flow Around Solid Bodies

4.1. Introduction When a solid body of random shape moves with a constant velocity inside a fluid, a flow force R and a momentum Μ are often applied on it. From the technological perspective, the analysis and study of these forces is one of the significant interests of the applied fluid mechanics community. Therefore, in this chapter, we will deal with some basic theorems, which offer us expressions for the calculation of these forces, as well as with the definition and analysis of the forces and momentums applied on a body during its movement inside the fluid. The study of this flow, which is called external flow, in contrast to the fluid’s flow inside pipes, is of technological interest for many cases, such as the flow around airplanes, cars, submarines, chimney buildings, blades or airplane wings. The flow around solid particles, bubbles and water drops is also very interesting, which is important in environmental technologies, the process of mass and heat transfer, and heterogeneous chemical reactions. The geometry of the flow around the body depends on the form of a surface of the body immersed in a fluid, which is a criterion to characterize a body as two- or three-dimensional when the flow around it is two- or three-dimensional, respectively. So, if there is a cylinder of very large length, when the flow projects vertically to its axis, it will be considered as a two-dimensional body because the flow around its edges does not influence the geometry of the flow field and the distribution of the pressure around the central part of the body. In comparison, if there is a cylinder of small length, it will be considered as a three-dimensional body because the influence of the flow around its edges is significant. For the study of this flow, we consider the flow around an airfoil of an airplane’s wing.

Fluid Mechanics in Channel, Pipe and Aerodynamic Design Geometries 2, First Edition. Christina G. Georgantopoulou and George A. Georgantopoulos. © ISTE Ltd 2018. Published by ISTE Ltd and John Wiley & Sons, Inc.

148

Fluid Mechanics 2

4.2. Geometrical characteristics of an airfoil By airfoil, we mean the intersection of the wing from a parallel level to the symmetry level of the airplane (Figure 4.1). Based on the previous definition, the airfoil is parallel to the level xz and, generally, it is a function of y. The airfoil is characterized by the following geometrical parameters (Figure 4.2): a) Main camber line is the line that connects the centers of the circles inscribed in the airfoil. b) Leading edge is the front part of the airfoil frame. c) Trailing edge is the back part of the airfoil frame. d) Chord (c) is the linear or straight part that connects the leading edge to the trailing edge of the airfoil. e) Maximum thickness (t) is the diameter of the inscribed circle to the airfoil. f) Maximum camber (f) is the biggest banking of the airfoil frame from its chord. g) Upper surface of the airfoil is the curve that connects the leading edge to the trailing edge, which is surrounded by the sky, while lower surface is the curve turned toward the earth.

WRIGHT brothers

P-36 (subsonic) P-51 (subsonic)

F-104 (supersonic)

Figure 4.1. Airfoil types

Flow Around Solid Bodies

149

Figure 4.2. Airfoil’s geometrical characteristics

When the airfoil is placed at a moving fluid stream, generally, an aerodynamic force R is created, which is a force caused by the relative motion, as well as an aerostatic force because of the body’s placement in the fluid. The second force is very small compared to the first one, and hence it is not considered. The aerodynamic force R acts along a straight line, whose intersection point Α on the chord of the airfoil is called the center of pressure, as shown in Figure 4.3. The force R is usually analyzed in two components, namely a vertical one and a parallel one to the direction of the motion. These components are called lift L and drag D, respectively. Therefore, we define lift L as the force that is vertical to the direction of the motion and moves upward when the body is in straight horizontal motion, while we define drag D as the force that is parallel to the direction of the motion and perpendicular to the direction of the velocity motion (Figure 4.3). The direction and value of aerodynamics R, as well as of the lift L and drag D forces, depend on the angle of attack α. By the term angle of attack, we mean the angle formed from the chord of the airfoil and from the direction of the vector of the fluid velocity. The angle of attack α is positive when the stream of the air projects the airfoil to the lower surface.

Figure 4.3. Aerodynamic forces on an airfoil

150

Fluid Mechanics 2

For each airfoil, there is a certain angle of attack α, for which the lift L is zero. The direction of the velocity that corresponds to this angle of attack is called the direction of zero lift (Figure 4.4).

Figure 4.4. Direction of zero lift depends on the airfoil type and shape

4.3. Kutta–Joukowski equation In each airfoil that lies inside a two-dimensional flow field, a force is applied vertical to the velocity vector. This force is called lift L, which is given by the Kutta–Joukowski equation. In order to prove this suggestion, we consider a part of an infinite wing of width b an infinite lane dx parallel to the leading edge, as shown in Figure 4.5. Thus, the surface area is:

Figure 4.5. Characteristics of a part of an infinitive wing

dS = bdx

[4.1]

and because of the pressure differences between the bottom and top surfaces of the wing, a force is created: dL = ( pu − p0 ) ds .

[4.2]

Flow Around Solid Bodies

151

The dL is considered to be vertical to the direction of the flow, and integrating it, the total lift is given by: C

L=

 ( pu − p0 )dS = b  ( pu − p0 )dx .

(S )

[4.3]

B

The pressure difference between the lower and upper surfaces of the wing can be expressed by the Bernoulli equation as a function of flow velocities. Therefore, if on the top surface of the wing the velocity is V + ΔV and on the upper surface V − ΔV , the Bernoulli equation gives: p+

ρ 2

ρ

V 2 = pu +

or pu − p0 =

ρ 2

2

(V − ΔV ) 2 = p0 +

(V + ΔV ) 2 −

ρ 2

ρ 2

(V + ΔV )2

(V − ΔV ) 2

or pu − p0 = 2ρV ΔV ,

[4.4]

[4.5] [4.6]

It is proved that the quantities of the velocity variation on the lower and upper surfaces of the wing are equal. This means: ΔV

u

= ΔV

[4.7]

0

So, from equation [4.3], the lift is given by: C

L = 2 ρbV  ΔVdx .

[4.8]

B

Now, the circulation Γ along the wing’s surface is given by: Γ=

C

B

C

Β ,0

C ,u

B

 ΔVdx −  ΔVdx = 2  ΔVdt

[4.9]

where the first integration in the first equation is taken along the top surface of the wing, and the second one along the bottom surface.

152

Fluid Mechanics 2

Thus, from relationships [4.8] and [4.9], the lift is given by:

L = ρ bV Γ

[4.10]

This relationship constitutes the Kutta–Joukowski equation for the lift of a wing’s airfoil. This expression was initially proposed by Kutta in 1902, and then independently by Joukowski in 1906. From the proof of the Kutta–Joukowski theorem, we conclude that the circulation Γ is the necessary condition for the creation of the lift L. However, in order for the circulation to exist, we must have velocity difference between the upper and lower surfaces of the wing, in fact, high velocity on the upper surface and low velocity on the lower surface, so that the circulation is positive. This means, according to the Bernoulli equation, there will be a smaller pressure at the upper surface than that at the lower surface. These pressures can turn into forces, as shown in Figure 4.6. The resultant of these forces gives us the total force R, which if analyzed gives us the lift L and the drag D. The application point of L is called the center of pressure.

Figure 4.6. Forces distribution along an airfoil

4.4. Aerodynamic paradox As mentioned in the introduction of this chapter, when a body of random shape moves inside a fluid, a force R is applied on the body that comes from the effect of the fluid on the body. This force is generally analyzed in two components, namely a component parallel to the direction of the flow, drag D, and a component perpendicular to the direction of the flow, lift L (Figure 4.7).

Flow Around Solid Bodies

153

Figure 4.7. Aerodynamic force and components on a body inside air flow

Because of the important role played by the Reynolds number in the greatest applications of the motion problems of bodies inside the fluid, we could expect agreement between the theory and the experiment if we overlook the viscosity, which is for a flow without friction. This occurs only for certain types of flow (symmetric flows), especially for the study of the lift and pressure distribution. There lies a great difference between the flow theory for the flow without friction and the observations for large Reynolds numbers for all the shapes of the bodies, in the drag problem. The dynamic flow around a random body inside an infinite extended calm fluid does not give a component to the direction of the motion, which means there is a zero drag. This phenomenon constitutes the known aerodynamic paradox or D΄Alembert paradox, according to which, in this case of flow, we only have the appearance of a vertical force to the direction of the flow, which is lift L. This strongly contrasts with the observations that for each body, there is a drag. D΄Alembert’s paradox is a result of the solution of the motion equations in a flow without any friction. However, there is in fact no real flow without friction; thus, for the motion of each body inside a fluid, there is a drag. 4.5. Pressure distribution in an airfoil When a fluid’s stream surrounds an airfoil, which is placed in it at a certain angle of attack, the narrowing of the streamlines appear at the upper surface and their widening at the lower surface of the airfoil, as shown in Figure 4.8. This variation of the distances of the streamlines, according to the continuity equation, implies acceleration of the flow at the upper surface of the airfoil and deceleration at the lower surface. According to the Bernoulli theorem, these local variations of the velocity around the airfoil also imply variations of the static pressure. The distribution of this pressure defines the lift production, pitch moment,

154

Fluid Mechanics 2

drag and position of the center of pressure, and it is usually expressed using a pressure coefficient Cp, which is given by the relationship: Cp =

V  p − p∞ = 1−   1 2  V∞  ⋅ ρ ⋅ V∞ 2

2

[4.11]

where p and V are the pressure and velocity, respectively, at a certain point close to the airfoil’s bound, and and are the pressure and velocity at a point very far away from the airfoil, where the flow is undisturbed (free-stream conditions). In order to further explain this concept, the following definitions related to the pressure distribution around an airfoil are necessary. Stagnation point at a flow field is the point at which the flow velocity is zero. According to the Bernoulli theorem, at the stagnation point the pressure takes high values and the pressure coefficient Cp takes the maximum value of 1, as is clearly shown by relationship [4.11]. The appearance of stagnation points at an airfoil plays a significant role in the distribution of pressures in it, and their position on the airfoil varies with the angle of attack. For the airfoil shown in Figure 4.8, the stagnation point is S. We must note here that, in order to have a stagnation point, the presence of a solid body in the flow is not always necessary. When two or more flow streams meet at some point, the resultant velocity of the flow is zero, and this point is called the stagnation point.

Figure 4.8. Stagnation point

A positive pressure coefficient means that the pressure at this area of flow is higher than that of the free flow, while a negative pressure coefficient means that the pressure is lower than that of the free flow. Based on the previous experience and the fact that there is no constant pressure but a pressure distribution around an airfoil, because of the over-pressures and the

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sub-pressures of the lower and upper surfaces, respectively, a graph for this pressure coefficient is considered necessary. Starting from a surface of the airfoil, we draw straight lines at this point, whose length is proportional to the pressure coefficient at this point, and which are vertical to the airfoil’s surface. The edges of these lines join with curves, as shown in Figure 4.9. From this figure, we see that the decrease in pressure at the upper surface of the airfoil with the pressure of the undisturbed flow contributes more to supporting the motion of the fluid rather than the increase in the pressure at the lower surface. Interestingly, Figure 4.10 shows the experimental pressure forces that act on the airfoil’s surface, indicating the local pressure distribution on the respective surface. From this figure, we can see that the forces that push the wing on the lower surface are larger than those that push the back. The resultant force of all those sub-component forces gives us the aerodynamic force, which if analyzed gives lift L vertical to the direction of the flow and drag D parallel to the direction of the flow.

Figure 4.9. Pressure distribution around an airfoil

Figure 4.10. Pressure force distribution around an airfoil

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4.6. Lift curve Lift L is the desired component of the total aerodynamic force R because it ensures the lifting of the body. The lift is given by the relationship: L=

1 ⋅ ρ ⋅ V 2 ⋅ S ⋅ CL = q ⋅ CL ⋅ S 2

[4.12]

where q is the dynamic pressure, CL is the lift coefficient and S is the surface of the body. The lift coefficient CL is a non-dimensional parameter that takes values up to approximately 1.5 and mostly varies with the angle of attack α from −6° to 15°. For angles larger than 15° and smaller than −6°, CL is zero, so there is no lift. Thus, the body falls under the stall condition. The variation of CL with the angle of attack for a symmetrical airfoil is given by the curve shown in Figure 4.11. From this figure, we also note that for α = 0°, CL = 0. Figure 4.12 shows the CL−a curve for an asymmetrical airfoil.

Figure 4.11. Lift curve for a symmetrical airfoil

According to the above lift curves, the following conclusions can be drawn: a) The angle of attack of zero lift, a0, is negative for an asymmetrical airfoil. In general, its quantity is equal in degrees with the percent (%) of the airfoil’s curvature. This means that an airfoil with 2% curvature will have α0 = −2ο . For symmetrical airfoils, we have α0 = 0 . b) As the angle of attack increases from the value of the zero lift, the lift coefficient increases linearly with it for a long period. This occurs up to a certain

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value of the angle of attack, which is called the stalling angle of attack or the critical angle of attack acrit., where the lift coefficient takes its maximum value, CL max . . For every increase in the angle of attack, after acrit, there is a decrease in the lift coefficient. This occurs due to the detachment of the boundary layer from the upper surface of the airfoil and then the appearance of the stall phenomenon, which will be discussed later.

Figure 4.12. Lift curve for an asymmetrical airfoil

A characteristic stalling angle of attack is approximately 15° and a characteristic CL max . for a flat wing, with no mechanisms of high-lift, is approximately 1.2–1.4. c) For the linear part of the curve, it is applicable that:

CL = a(a − a0 ) ,

[4.13]

where α is a constant. Then, we have:

dCL =a da

[4.14]

and this derivative is called the slope of the lift’s curve. Its theoretical value is 2π, but experiments have proved that a characteristic value for a two-dimensional airfoil is approximately 5.7 per rad or approximately 0.1 per degree.

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4.7. Drag force and drag coefficient curve Total Drag D is defined as the force corresponding to the quantity of the momentum decrease in a direction parallel to the fluid velocity V. Therefore, the drag is clearly a resistant force to the motion and produces the whole work, which obstructs the motion. The total drag is a sum of: 1) the parasite drag, which consists of the form drag and the drag of skin friction; 2) the induced drag of the body, especially of an airplane’s wing. Therefore, the total drag will be equal to the sum of the parasite drag and the induced drag while these two components do not vary proportionally. 4.7.1. Drag of skin friction From the theory of the boundary layer, we know that the velocities of the fluid’s molecules inside it are not the same at all its points. More specifically, as the distance from the solid boundary increases, the velocity of the fluid’s molecules also increases. Because of this difference, the forces of internal friction appear between the successive layers, which react to the motion of the body and, when added, give a resistant force to the motion of the body, which is called the frictional drag Df. The frictional drag generally depends on the surface S of the body that surrounds it, the density of the fluid ρ, the velocity of the fluid V and the viscosity coefficient μ. This dependence of the frictional drag on the above quantities is not the same for the laminar boundary layer and the turbulent one but is higher than (Δu / Δy)0 of the turbulent boundary layer. Because of this, the turbulent boundary layer produces a higher frictional drag, almost five times higher, than that produced by the laminar boundary layer. In general, this leads us to the conclusion that we should prefer the laminar boundary layer and keep it as much as possible to a greater length, which means that the detachment point should move to the back side as much as possible. This observation is applicable for decreasing the frictional drag. However, there are cases where the turbulent boundary layer is preferred. This preference is based on the significant involvement of its layers and its synchronized ability to absorb energy from the external flow (free-stream flow). Therefore, it lies at a higher energy level than the laminar boundary layer, and hence it can deal more effectively with the abnormalities of the flow around the body. The significance of the turbulent boundary layer is shown when there is a fluid’s flow around certain shapes. Therefore, as shown in Figure 4.13, a flow around a sphere is considered, for case (a), with small Reynolds numbers, Re < 3,85 ⋅105 . If there is a laminar boundary layer of low energy level, the detachment occurs at an

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angle of approximately 84°, from the leading edge of the sphere. In comparison, for case (b), with larger Reynolds numbers, Re > 3,85 ⋅105 , if there is a turbulent boundary layer of high energy level, the detachment occurs at an angle of approximately 140°, and thus the area of turbulence is limited significantly. In fact, in this case, there is an increase in the frictional drag, but this is very small compared to the desired result to avoid an unwanted balance in the flow around the wing. The possibility of existence or maintenance of a turbulent boundary layer takes place with a relative buffing of the body’s surface or the creation of small local irregularities. Therefore, the whole surface of a golf ball’s sphere is constructed with uniformly distributed dents in order to achieve reduction of the Re number to 105 values, so with diameter d = 41 mm and initial velocity V > 35 m/s, we have a turbulent boundary layer, and as a result, the range of the sphere increases from 46 to 210 m. Finally, in other cases, where problems of detachment are dealt with, like in wings or wheels of airplanes, the local irregularities of the surfaces are suitably placed in crucial points of the body so they cause the transition of the boundary laminar layer into a turbulent one.

Re < 3,85 ⋅105 Re > 3,85 ⋅105 α

β

Στρωτό

Τυρβώδες

Figure 4.13. Laminar and turbulent boundary layers around a sphere

Finally, the drag of skin friction is given by the relationship: Df =

1 ρV 2 SCDf , 2

[4.15]

where CDf is the coefficient of the frictional drag. 4.7.2. Form drag From the previous analysis, we note that the frictional drag would not appear if the boundary layer did not exist. However, it is not the only force that resists the motion of a body inside a fluid because of the boundary layer. Because the thickness

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of the boundary layer is not constant, which, in fact, increases in the upper surface of the body, the real flow of the fluid behind the body is not symmetric. The result of the asymmetry of the flow is the creation of different pressures of the fluid between the windward and leeward points, and as a result, a resistant force to the motion appears, which is called the form drag. Although the form drag appears because of the creation of the boundary layer, its variation and intensity depend exclusively on the body’s shape because of the fact that it defines the flow field of the fluid behind the body. We mention indicatively that in the case of a flat surface, of very little thickness, the form drag will be zero. The form drag is given by the relationship:

Dp =

1 ρV 2 SCDp , 2

[4.16]

where CDp is the drag coefficient, which depends solely on the shape of the body. Therefore, based on the previous analysis, the total drag for a solid body is the parasite drag, which is written as: Dπ = D f + Dp

[4.17]

or generally: Dπ =

1 ρV 2 SCDπ 2

[4.18]

where CDπ is the coefficient of the parasite drag. The shape of the body plays a basic role in the variation of the parasite drag. Thus, for a certain surface S, aerodynamic shape produces the minimum parasite drag and therefore has the lowest possible coefficient of parasite drag. In general, the aerodynamic shape is characterized by the ratio of its length  to its largest diameter d, which is called the aspect ratio λ. Therefore:

λ=

 . d

[4.19]

Also, the aerodynamic shape depends on the shape of its edges. Based on relationship [4.18], the actualization of an aerodynamic shape involves minimization of the parasite drag. Therefore, for the actualization of the minimum form drag Dp, a successive decrease in the body’s cross-sections behind its maximum thickness is needed, which means a long tail, while for the decrease in the

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frictional drag, sufficiently limited surface S is demanded, which is surrounded for a given frontal surface or diameter. These two requirements are generally conflicting. And this is because the first one demands an extremely long, fish-shaped body, which at the boundary will have  / d → ∞ , while the second one demands the presence of a flat disk, vertical to its motion ( / d → 0) . Based on this observation, we can take the aerodynamic characteristics of an airfoil of the small subsonic velocities. Therefore, the aerodynamic subsonic figure should have a rounded leading edge, a sharp trailing edge and maximum diameter at 1/3 of its length from the leading edge. The variation of the parasite drag with its aspect ratio is given by the curve shown in Figure 4.14.

Figure 4.14. Variation of the parasite drag with aspect ratio

4.7.3. Induced drag

Because of the sub-pressures and the over-pressures that appear at the upper surface and lower surfaces of a wing of a solid body (airplane) at its edges, two vortices are created with the motion of the airplane, which expand backward along the whole length of the motion’s orbit. Because at each time a part of these vortices must be created from the beginning, according to the principle of conservation of energy, it is clear that for this purpose a work is constantly consumed. This work is consumed so that a resistance to the motion is surmounted, which is called the induced drag. From this, we conclude that even in a flow with no friction, further drag is created. From the fact that the total drag D is the sum of the parasite and the induced drag, we also conclude that its coefficient will be a function of the coefficients of the previous components of the drag. Therefore, according to experiments, it has been

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observed that the coefficient of the drag is a parabolic function of CL, which is given by the relation:

CD = CD0 + kCL2 ,

[4.20]

where the coefficient CD0 expresses the value of the parasite drag, while the coefficient kCL2 constitutes the induced drag, which is the drag that in the finite wing, as already mentioned, expresses the losses of the flow that are created because of the appearance of the vortices at its edges. Finally, the quantities CL, CD and α in a double-rectangular axis system is shown in the polar diagram. Therefore, these two basic aerodynamic curves of the wing, which are the curves of the lift coefficient CL and the drag coefficient CD with the angle of attack α, can be represented in one unique curve, which is called the polar curve of the wing. This curve has axes CD (horizontal) and CL (vertical) with the respective angles of attack at various points, as shown in Figure 4.15. If we draw the tangent ΟΑ of the curve, then we can immediately take the following measurements: a) from the tangent point, the optimum angle of attack aopt, which is the angle at which the ratio CL/CD = L/D is maximum;

Figure 4.15. Polar curve

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163

b) from the angle θ, the value of the ratio CL/CD is given as: tan θ =

CL . CD

[4.21]

Especially:

C  tan θ MAX =  L  .  CD  MAX

[4.22]

The optimum angle of attack aopt is of significant interest for the performance of an airplane. The airplane that flies with aopt gains the maximum possible ratio of slipping and the smallest slipping angle, which is especially interesting for the non-thrust descent phase of the flight. Regarding the two forces through which the aerodynamic force is analyzed, of particular interest is the lift L of the body that moves in the air, which has a characteristic shape, called the aerodynamic shape. This is because it is the force that effectuates the lifting of the body and the maintenance of its motion (flight) in the air, while the body is of either defined shape (e.g. sphere, cylinder) or undefined shape, which move in fluids where, interestingly, the drag force D obstructs the motion. However, we will study this force later. 4.8. Parameters that influence the drag coefficient CD

As mentioned in the previous section, the drag force and therefore its coefficient CD for a flow around solid bodies depend on the shape of the body, on the relative surface roughness of the body  / d , on the Reynolds number and for a compressible flow on the Mach number. Thus, the drag coefficient as a function of these parameters is given by the relationship:

CD = f (Shape,  / d , Re , M).

[4.23]

In the following, we will study the effect of each one of the above parameters separately. 4.8.1. Dependence of CD on the body's shape

The drag coefficient CD depends on the shape of the body and more specifically on its form at the back side of the body. It also depends on the length parameter if the body is two-dimensional, which means that it is infinitely long perpendicular to

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the direction of the flow, or three-dimensional, which means that it is of a small length. For non-spherical bodies, the size, the orientation of the body to the direction of the flow and the geometrical shape must be defined. For the definition of the form of a non-spherical body, we choose the characteristic length as the main dimension and the other important dimensions of the body as ratios to this dimension. These ratios are called aspect ratios. Thus, for small cylinders, the diameter is usually the characteristic dimension and the length-to-diameter ratio is the aspect ratio. Figure 4.16 shows the coefficient CD as a function of the aspect ratio,  / d , for an incompressible flow around a flat plate and a circular cylinder. The two bodies are placed in such a way that the direction of the flow is perpendicular to the plate and the axis of the cylinder. It is obvious that the small length of the body allows the fluid to move more easily around it and therefore it limits its resistance. Thus, the coefficient CD of the body with a ratio  / d < ∞ is smaller than that with a ratio /d =∞. Tables 4.1 and 4.2 present the typical values of the drag coefficient for two- and three-dimensional bodies, respectively.

Συντελεστής οπισθέλκουσας, C D

2,0

V¥

l/d=¥

d

1,5

l l/d=¥

1,0

V¥

0,5

0

d

5

10

V d 5 Re = ¥v @ 10

l 15

20

25

30

Συντελεστής σχήματος, l / d

Figure 4.16. Variation of drag coefficient with aspect ratio

Bodies that do not have a regular shape and those for which there are no experimental data available can be considered as “cylinders”, “spheres” or “disks”, depending on the value of the aspect ratio,  / d , which can be used for the respective bodies of a regular shape. We distinguish three types of bodies: Thin bodies with a ratio  / d > 10 : these bodies can be considered as cylinders 2 with equivalent diameter de = d and frontal surface S0 = π de / 4 . The equivalent

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165

diameter of a non-spherical body is defined as the sphere’s diameter, which has the same volume as the body. Short bodies with a ratio  / d  1 : these bodies can be considered as spheres with equivalent diameter de = (d1d2 )1/2 , where d1 is the minimum diameter and d2 is 2 the maximum diameter of the body and the frontal surface S0 = π de / 4 . ΣΧΗΜΑ

ΡΟΗ

CD

ΣΧΗΜΑ

CD

ΡΟΗ

ΟΡΘΟΓΩΝΙΚΗ ΡΑΒΔΟΣ

ΔΟΚΟΣ

1,2

2,05

1,6

1,80 1,65

ΗΜΙΑΓΩΓΟΣ

ΗΜΙΚΥΛΙΝΔΡΟΣ

1,2 2,3

1,2 1,7

ΤΡΙΓΩΝΙΚΗ ΡΑΒΔΟΣ

ΕΛΛΕΙΠΤΙΚΟΣ ΚΥΛΙΝΔΡΟΣ

θ θ

30° 60° 90° 120°

a/b 1,0 1,4 1,6 1,7

1,8 2,1 2,2 2,0

a

b

1 2 4 8

Στρωτή

Τυρβώδης

1,2 0,6 0,35 0,25

0,3 0,2 0,15 0,1

Table 4.1. Drag coefficient values

Very wide bodies with a ratio  / d  1 : these bodies can be considered as disks with equivalent diameter de = (d1d2 )1/2 , where d1 is the minimum diameter and d2 is the maximum diameter of the diameter of the body and the frontal surface S0 = π de2 / 4 . 4.8.2. Dependence of CD on relative roughness

Relative roughness of the surface of a solid body has a significant effect on the drag coefficient CD. As is known from the study of flow in pipes, the relative

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roughness of the wall plays a significant role in the modulation of the value of the resistance coefficient λ, not in the laminar flow but in the turbulent one. We have also proved that, in the fully developed turbulence, where Re > Rec