380 111 73MB
English Pages 489 [486] Year 2010
Table of contents :
1. Introduction to heat transfer
2. General heat conduction equation
3. Steady state heat conduction
4. Heat transfer from extended surfaces
5. Transient heat conduction
6. Forced convection heat transfer
7. Free convection
8. Radiation heat transfer
9. Heat exchangers
10. Condensation and boiling
11. Mass transfer
Index
Hens a NI Tr
Heat and Mass Transfer
Heat and Mass Transfer Second Edition
G. S. Sawhney Dept. of Mechanical Engineering Greater Noida Institute of Technology Greater Noida, Ghaziabad, U.P. India
I.K. International Publishing House Pvt. Ltd. NEW DELHI • BANGALORE
Published by I.K. International Publishing House Pvt. Ltd. S25, Green Park Extension, Uphaar Cinema Market New Delhi  110 016 (India) Email: infogildnternational.com ISBN 9789380578392 10 9 8 7 6 5 4 3 2 1 © 2010 I.K. International Publishing House Pvt. Ltd. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or any means: electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission from the publisher. Published by Krishan Makhijani for I.K. International Publishing House Pvt. Ltd., S25, Green Park Extension, Uphaar Cinema Market, New Delhi  110 016 and Printed by Rekha Printers Pvt. Ltd., Okhla Industrial Area, Phase II, New Delhi  110 020.
PREFACE TO THE SECOND EDITION
In the second edition, the book has been thoroughly revised and enlarged. The chapter on steady state onedimensional heat conduction has been modified to include problems on twodimensional heat conduction. Finite heat difference method of solving such problems has been covered. Modification has also been included in the text as per the suggestions obtained from various sources. Additional typical problems based on the examination papers of various technical universities have been included with solutions for easy understanding by the students. I thank all the faculty members of Mechanical Department of GNIT, Greater Noida for their assistance in the completion of this second edition of the book. I once again request students and teachers to send constructive suggestions and criticism by emailing [email protected] G. S. Sawhney
PREFACE TO THE FIRST EDITION
Heat and Mass Transfer is considered a basic and essential subject for engineering students of Mechanical and Chemical disciplines. The concepts of heat and mass transfer have a variety of applications in industry. Realizing the importance of the subject, all technical universities have introduced this subject in their syllabi. This subject is taught to undergraduate students only for one semester which is generally fifth semester. Efforts have been made to cover the syllabi of all technical universities. Based on my teaching experience, I have tried to explain principles and concepts of heat and mass transfer in simple and clear terms. The endeavour is to present the subject matter in the most comprehensive and usable form. The derivations of fundamental relations have been kept as simple as possible. Diagrams have been used abundantly to elucidate the difficult concepts which are difficult to explain otherwise. The theory part is further supported with illustrations and suitable examples. This book has easy to read style as it is written in questionanswer form and it is going to benefit the readers immensely. Above all, I wish to record my sincere thanks to my wife, Jasbeer Kaur for patience shown throughout the preparation of this book. I am grateful to Dr. S. Prasad for everlasting support always extended to me. I would appreciate constructive suggestions and objective criticism from students and teachers alike with a view to enhance further usefulness of this book. They may mail me their views at [email protected] G. S. Sawhney
CONTENTS
Preface to the Second Edition
v
Preface to the First Edition
vi
1. Introduction to Heat Transfer
1
2. General Heat Conduction Equation
30
3. Steady State Heat Conduction
50
4. Heat Transfer from Extended Surfaces
135
5. Transient Heat Conduction
172
6. Forced Convection Heat Transfer
214
7. Free Convection
266
8. Radiation Heat Transfer
301
9. Heat Exchangers
370
10. Condensation and Boiling
425
11. Mass Transfer
443
Index
477
Chapter
1
INTRODUCTION TO HEAT TRANSFER
KEYWORDS AND TOPICS A A A A A A A A A A A
HEAT ENERGY TRANSIT ENERGY STORED ENERGY INTERNAL ENERGY THERMAL EQUILIBRIUM HEAT FLUX MASS TRANSFER TEMPERATURE GRADIENT CONCENTRATION GRADIENT CONDUCTION CONVECTION
A A A A A A A A A A A
RADIATION NATURAL CONVECTION FORCED CONVECTION FOURIER'S LAW STEADY STATE THERMAL CONDUCTIVITY LOCAL COEFFICIENT OF CONVECTION AVERAGE COEFFICIENT OF CONVECTION STEFANBOLTZMAN'S LAW EMISSIVITY BLACK BODY
INTRODUCTION Heat transfer is a science which helps in determining the rate of heat energy transfer taking place between bodies having temperature differences. Heat transfer helps in determining not only how heat may be transfered but also the rate at which this heat exchange will take place under the specified conditions. The heat transfer rate forms the basis for difference between the science of heat transfer and thermodynamics. There are three distinct modes of heat transfer which are: (i) conduction, (ii) convection, and (iii) radiation. These modes have a common feature that the heat energy transfer always takes place in the direction of the decreasing temperature. However, these modes of heat transfer differ entirely in the physical mechanism and laws by which they are governed. The general mechanism of each of these modes is as shown in the figure.
2
Heat and Mass Transfer T2
q = ocT
4
Q
/////////// /////////,
To
To >To.
> T2
(b) Convection
T T = absolute temperature (c) Radiation
(a) Conduction
Modes of Heat Transfer
1. What is heat energy? When a hot body is placed in contact with a cold body, the hot body cools down while the cold body warms up. The energy transferred from the hot body to the cold body as a result of temperature difference is called heat energy. 2. What is transit energy? Energy possessed by a system which can cross its boundary is called transit energy. Heat and work are transit energy. 3. What is stored energy? Energy possessed by a system within its boundary is called stored energy. Total stored energy is the sum of potential energy, kinetic energy and internal energy. 4. Is heat energy a property of a system? Heat energy is not a property of a system. It appears as a boundary phenomenon, but it is a path function. The magnitude of heat transfer depends upon the path followed by a system during the process. Hence, heat energy is inexact differential i.e. rdQ # Q2 — Qi 1
5. How is the magnitude of heat energy measured? The area under the process or area enclosed in the cyclic process on a TS diagram is the magnitude of heat energy.
Introduction to Heat Transfer
t
t
T
T
S —.Heat transfer in process: shaded portion
3
r% S 
Heat utilised in cyclic process: shaded portion
6. What is internal energy? Internal energy is stored energy and it depends upon temperature of the system. 7. How is the internal energy related to heat energy? When heat energy enters the system, internal energy increases. Similarly, internal energy decreases when heat energy flows out of the system. 8. Does internal energy of a system change, if temperature remains constant? The internal energy depends upon temperature only. Hence, internal energy of a system does not change when temperature remains constant. 9. Why does heat transfer take place? Heat transfer takes place by virtue of temperature difference. Heat transfer takes place from a hot body to a cold body. If temperature of two bodies is same, then no heat transfer takes place. Heat transfer takes place also in a system, if there is a temperature distribution within the system. 10. What is thermal equilibrium? Two systems are said to be in thermal equilibrium with each other if no transfer of heat takes place. 11. What is the major difference between sciences of thermodynamics and heat transfer? Thermodynamics deals with the conversion of heat into work or other forms of energy and vice versa. However, heat transfer is concerned with the analysis of the rate of heat transfer. Heat flow takes place only when there is a temperature gradient in a system. 12. What is heat flux? The heat flux is the amount of heat transfer per unit area and per unit time.
4
Heat and Mass Transfer
13. What does temperature indicate? Temperature indicates the relative hotness or coldness of a body with respect to its surroundings. 14. Using an example, explain the difference between sciences of thermodynamics and heat transfer? Consider the cooling of a hot body placed in water. Thermodynamical analysis concerns with the final equilibrium temperature of the body and water. It cannot find out the temperature of the body after a certain length of time or the period of time would be lapsed before the body and water can reach the final equilibrium temperature. However, heat transfer study can predict the temperature of both the body and water as a function of time. 15. What is the reason for the lack of information obtainable from thermodynamics analysis? The main reason is the absence of time as variable. 16. What is the difference between the steady and unsteady state of heat transfer? In steady state heat transfer, the temperature at any location within the system does not vary with time. The temperature in steady state is the function of space coordinates while it is independent of time i.e. temperature is given as— T = f(x, y, z)
and
In unsteady state heat transfer, the temperature varies with time as well as location within
aT
the system i.e. temperature T = f(x, y, z, t) and — # 0. at 17. What are the purposes for studying heat transfer? The purposes for studying heat transfer are— (a) to estimate the amount of heat flow through a system under steady and transient conditions (b) to determine the temperature distribution within the system under steady and transient conditions. 18. What are different types of heat transfer based on the configuration of the system? The heat transfer can be (1) one dimensional, (2) two dimensional, or (3) three dimensional. 19. What are the applications of heat transfer study? The applications of heat transfer study are—
Introduction to Heat Transfer
5
1. Refrigeration and airconditioning system: The design of compressors, condensers and evaporators which involves the study of heat transfer. Ti> T2
T2 < Ti
Surrounding (T2)
Surrounding (Ti)
02 Refrigeration plant Cooling
Heating
2. Heat production and conversion system: The thermal design of boilers, steam turbines, condensers, heat exchangers and cooling towers are carried out on the basis of heat transfer analysis.
Vapour t
Steam (low pressure)
ir Steam Cold water: I  I
.r"
Heat (0)
r"V— Vater: Boiler
1 Fluid Condenser
u Steam (high pressure) Turbine
3. Electric Machines: When current flows through electric motors, generators or transformers, heat is generated which has to be dissipated effectively to surroundings to safegaurd these equipment. The heat transfer analysis is carried out for each of these equipment before fabricating. Stator
Pipes to cool oil
Fins to dissipate heat AC Motor
Transformer
6
Heat and Mass Transfer
4. IC Engine: The cylinder of the engine has to be kept at sufficiently lower temperature to avoid melting of piston, cylinder and other accessories. Water jacket and fins are provided to efficiently remove the heat to surroundings. This is possible by heat transfer analysis. Head Piston
Water jacket
Cylinder
AI 110
Connecting rod
Fins 14
4 IC Engine
20. What is mass transfer? Mass transfer is the movement of mass from higher concentration region to lower concentration region. 21. What is the importance of mass transfer in heat transfer study? In many applications, heat transfer processes also involve mass transfer. Mass transfer process is analogous to heat transfer process in many respects. 22. What are the conditions for the stoppage of (1) heat transfer process, and (2) mass transfer process? Heat transfer process stops when temperature gradient reduces to zero. The mass transfer process stops when concentration gradient reduces to zero. 23. Why the concentration of water molecules is higher above the water surface of a pond as compared to the main portion of air stream? Due to concentration difference, the water molecules are transported from the water surface to the air in contact with the water. Later water molecules are transported to main air stream from this region. 24. Is mass transfer by diffusion affected by molecular spacing of the substance? The mass transfer depends upon the molecular spacing. More is spacing, more is the mass
Introduction to Heat Transfer
7
transfer. Mass transfer by diffusion occurs more readily in gases as compared to liquid and least readily in solids. 25. What are different modes of mass transfer? There are basically two modes of mass transfer which are1. Diffusion mass transfer 2. Convective mass transfer 26. What are different modes of heat transfer? Heat transfer has three modes of heat transfer which are1. Conduction 2. Convection 3. Radiation Conduction usually takes place in solids, convection in liquids and gases, and no medium is required for radiation. 27. How do these three modes of heat transfer differ? What is essential for all those modes? It is essential for all these modes of heat transfer to take place only when the temperature difference must exist as heat transfer has to take place in the direction of decreasing temperature. However, the methodology of heat transfer in these modes differs. This can be explained with the help of an example. Suppose a teacher wishes to give a book to a student who is sitting at the end of the row. He can adopt any of the following methods1. He can deliver the book to the student nearest to him who delivers the same to the next and so on till the book reaches the last student. Here book corresponds to heat energy while students correspond to the molecules. Heat transfer takes place from one molecule to next molecule without the displacement of any molecule.
Book — , Teacher
44
Student
Student
Student
Student
Conduction
2. The teacher could go to the last student and deliver the book to him. Similarly, in liquid and gases, the molecule from the hot region moves to cold region and transfers heat energy and vice versa.
8
Heat and Mass Transfer Teacher Teacher Teacher Teacher
"r 4———i, Book Teacher Student Student Student Student Student ....
..
Convection
3. The teacher can throw the book so that it reaches the last student. Heat can move from hot region to cold region without any movement of molecules. No medium is required.
Book
CO
.I Teacher
Student Student Student Student Student Student Radiation
28. Explain heat transfer by conduction. Conduction is a mode of heat transfer from one part of a substance to another part of the same substance or from one substance to another in case in physical contact without any appreciable displacement of molecule forming the substance. In solids, heat is conducted by following mechanisms1. Lattice vibration: Fast moving molecules in the hottest region transfer heat to adjacent molecules by impact. 2. Transport of free electrons: Free electrons provide heat energy flux in the direction of decreasing temperature. In case of gases, molecules are in a continuous random motion exchanging energy and momentum. The kinetic energy of a molecule is a function of its temperature. When molecule from hot region collides with a molecule of cold region, it loses heat energy by collision. In liquids also, the mechanism of heat transfer is similar to that of gases. 29. Explain heat transfer by convection. Convection is a mode of heat transfer within a fluid by mixing of one portion of the fluid with another. The physical movement of fluid molecules is made possible by the existing temperature gradient.
Introduction to Heat Transfer
9
The convection heat transfer comprises two mechanisms. First is transfer of heat due to random molecular motion which is called diffusion. Second is transfer of heat due to bulk motion of the fluid which is called advection. The bulk motion is due to difference of density of fluid particles on account of difference of temperature. For example, when water is heated, the water at the bottom gets heated and its density decreases and it consequently rises. Thus, the heat is carried from the bottom to the top by the actual movement of the hot water. 30. How can convection heat transfer be classified? Convection heat transfer can be classified according to the nature of fluid flow. If fluid motion is setup by buoyancy effects, the heat transfer is said to be the free or natural convection. If fluid motion is artificially induced by a pump, fan or blower, then the heat transfer is called forced convection. Plate
Plate
Heat
t 4
i 8= Blower
Natural or free convection
Forced convection
31. Natural convection in water in lake plays important role in saving the lives of fishes when the atmospheric temperature falls below 0°C. Comment. As water at the surface of lake is cooled, it becomes denser and it goes down. The lesser cold water from the bottom of the lake rises up to the surface and gets cooled. The entire water attains the temperature of 4°C. Water has maximum density at 4°C and its density falls when temperature falls below 4°C. Surface water on cooling below 4°C remains at surface only and it keeps on getting cooled there only till it freezes. Heat is now lost to the atmosphere by the water in lake only due to conduction through the ice at the surface. As ice is a poor conductor of heat, the further freezing of water below the frozen upper surface is very slow. The temperature of water below the frozen upper surface remains at 4°C and fishes can survive. 32. The heat transfer inside a human body is carried out by a forced convection. Explain. Blood inside a human body is circulated to all parts of the body by heart acting as a pump. Heat is lost to the surroundings by the body through conduction, convection and radiation. The rate of heat loss depends upon: (1) clothing, (2) perspiration, (3) outside temperature, (4) air speed, (5) humidity and other factors. However, our body has a forced convection
I 0 Heat and Mass Transfer
system with blood flow by heart. The blood flow transports the needed amount of heat to maintain a remarkable constant temperatures within our body. Jugular vein Subclavian vein
Carotid artery Arch of the arota
Superior vena cave
Pulmonary vein
t< 0°C 0°C
Ice
 4°C water
Pulmonary artery
//////////////////////////
Inferior Survival of fishes in lake below vena cave 0°C by natural convection
Thoracic arota Abdominal arota
Forced Convection: Blood flow
33. What is radiation? Radiation is a process of the transfer of heat through space or matter other than conduction or convection. The process of radiation does not need any material medium for heat transfer. Energy is emitted by a body and thus energy travels in space just like light as electromagnetic waves. All bodies radiate heat. Transfer of heat by radiation occurs because hot body emits more heat than it receives. A cold body receives more heat than it emits. The heat from the sun reaches the earth by radiation, travelling millions of kilometres of empty space. 34. What are the properties of radiant heat? The properties of radiant heat are1. Radiation does not require any material medium for travelling. 2. Radiant heat can be reflected like light from the surface. It also obeys the laws of reflection like light. 3. Radiant heat travels with same velocity of light. 4. Radiant heat also shows interference, diffraction and polarisation like light. 5. Radiant heat also follows the law of inverse square. 6. Radiant heat is electromagnetic wave having longer wavelength. Hence, radiant heat is invisible to our eyes.
Introduction to Heat Transfer I I
35. What is the basic law of heat conduction? Or
What is Fourier's law of heat conduction? Why is a negative sign inserted in its expression? (UPTU — 2003) The basic law of heat conduction or Fourier's law of heat conduction states that the rate of heat flow by conduction in material is equal to the product of three quantities as given below1. Thermal conductivity (k) which is a property of the material. 2. Crosssectional area (A) which is normal to the direction of flow. all 3. Temperature gradient along the direction of flow of heat (
ax
dQ
Rate of heat flow = or
dt
=
q
aT —m ax
q=
The minus sign is inserted to make the heat flow rate (q) positive as temperature gradient is negative. 36. Find the expression for heat conducted (q) through a slab thickness (L), crosssectional area (A) having temperature T1 and T2 at both ends for steady state. As per heat conduction lawq = —m aT
ax
1 Temp T1 T2 x2
x1
H L H
,. x Thickness
q • dx = —kAaT In steady state, temperature distribution does not change and q is constant. In case thermal conductivity (k) is constant, then 5 q • ax = —f kAaT
I2
Heat and Mass Transfer
T2
X2
q fdx = —kA f aT
or
xi
or
T1
q(x2 — x1) = —kA(T2 — T1) (T qq _ Mai — T2) X1
=
Mai — T2) L
37. The inner surface of a plane brick wall is at 60°C and outer surface is at 35°C. Calculate the rate of heat transfer per sq metre surface area of the wall which is 220 mm thick. Thermal conductivity of the brick is 0.51 W/mk. (AMIE — 1998) Heat conduction equation is— q = kA(T — T2) L = 0.51 x 1(60 — 35) 0.22 = 57.95 W/m2 38. What are assumptions made for Fourier's law of heat conduction? The assumptions are1. 2. 3. 4.
Conduction of heat takes place under steady state conditions. Temperature gradient is constant i.e. linear temperature distribution. Heat flow is unidirectional. Material is homogeneous and isotropic i.e. thermal conductivity is constant in all directions. 5. No internal heat generation.
39. Explain thermal conductivity. The ability of a material to conduct heat is measured by thermal conductivity (k). It is a physical property of the material. It is defined as the amount of heat conducted through a body of unit crosssectional area, unit thickness and in unit time when temperature difference is unit. Materials with high thermal conductivity are good conductors of heat. Materials with low thermal conductivity are good thermal insulators of heat. Solids have higher thermal conductivity than liquids and liquids have better thermal conductivity than gases. Metals have higher thermal conductivity than nonmetals. Higher conductivity is possible as metals have a large number of free electrons which help in carrying thermal energy from one place to another.
Introduction to Heat Transfer
I3
40. Why are cooking utensils made of metals whereas their handles are made of wood or plastic? Metals have high thermal conductivity. Therefore, cooking utensils are made of metals so that heat can easily pass from burner to food being cooked in the metallic utensil. However, wood or plastic has low thermal conductivity so that heat current going to handle is smaller and it is not heated up. Utensil
Handle
Q1 >> Q2 Q1 I— —I
Utensils and Thermal Conductivity
41. Why does a rug placed in bright sun on a tiled floor help to facilitate bare foot standing? The rug and tiled floor when placed outside in sun aquire same hot temperature. But it is much easier to stay bare foot on the rug as compared to tiled floor. This is possible as the thermal conductivity of the rug is lower than the tiled floor. Hence, the heat current going into our feet is smaller in case of rug.
02 Q1 Single vs two blankets
6.
What is Wiedermann and Frantz law? The ratio of the thermal and electrical conductivities is same for all metals at the same temperature which is directly proportional to the absolute temperature of the metal.
k ... T C where k = thermal conductivity C = electrical conductivity T = absolute temperature It is apparent that good conductors of electricity are also good conductors of heat. 47. What is the basic law of convection? Or
What is the Newton's law of cooling? The rate of heat transfer by convection is—
Q = hA(T, — Toc ) or
Q = q = h(T, — Toc)
A
I8
Heat and Mass Transfer Fluid flow at temperature T„
Ts > To Q
/////////////////1//// / Hot surface at Ts Convective heat transfer
In above equation— Q = q = rate of heat transfer per unit area by convection
h = average convective heat transfer coefficient T, = temperature of solid surface T., = temperature of fluid 49. What are the parameters on which the value of convective heat transfer coefficient depends? As it can be seen from the Newton's law of cooling, the value of h depends upon— (a) (b) (c) (d)
The geometry of the solid surface The properties of the fluid in touch with the solid surface The velocity of the fluid The temperature difference between solid surface and the fluid i.e. (T, — To.).
50. What do you understand from the local and average convective heat transfer coefficients? Since the value of h depends on (i) the geometry of the solid surface, (ii) the velocity and properties of the fluid and (iii) the temperature difference (7's — T) and these quantities do not remain constant over the solid surface, hence the value of h vary from point to point on the surface. Therefore, h, denotes the value of h at a point on the surface at distance x from the leading edge of the surface and it is called local convective heat transfer coefficient. The average convective heat transfer coefficient (h) is defined as— h = f h., dA
AA
Q A(Ts —L) The average convective heat transfer h (also called film heat transfer coefficient) is defined as the amount of heat transferred for: (i) a unit temperature difference between the fluid and surface, (ii) a unit area of solid surface, and (iii) in unit time. or
h=
51. A hot plate lm x 2m is maintained at 320°C. Air at 20°C moves over the plate. If h = 40 watts/m2°C, find the rate of heat transfer.
Introduction to Heat Transfer
I9
Q = hA(T, — To j
= 40 x (1 x 2) (320 — 20) = 40 x 2 x 300 = 24 x 103 watts = 24 kW 52. An electric wire carrying current is submerged in water. The wire is 1.5 mm in diameter and 300 mm length. If h = 4500 watts/m2°C, and wire surface remains at 120°C when water boils (100°C), find electric power supplied to the wire. A = surface area of wire = MI = ir x 1.5 x 103 x 300x 103 = 14.13 x le m2 Now heat transfer rates Q = h • A • (T, — Tw)
= 4500 x 14.13 x le x (120 — 100) = 127.2 W Since 127.2 W heat is transferred from the wire, we have to supply 127.2 W power to the wire. 53. What is the basic law of radiation? Or
What is StefanBoltzman's law? StefanBoltzman's law states that the total emissive power of a black body (Eb) is proportional to the fourth power of its absolute temperature. or where
Eb oc T 4 Eb = a 74
a = StefanBoltzman constant = 5.67 x 108 W/m2k4
The rate of heat emission from a black body to the surroundings is given by Q = a A(T 14 — Ti')
where Q = Maximum rate of heat emission from a black body
A = Surface area T1 = Absolute temperature of the body in Kelvin T2 = Absolute temperature of surrounding a = StefanBoltzman constant = 5.67 x 108 W/m2K4 54. How much heat is radiated from one real surface to another? Real surface emits heat at lower rate than black body. Also, the emission of heat depends
20
Heat and Mass Transfer
upon the geometry of two surfaces. Hence, the net heat exchange between two real bodies is given by— Q12 = Cr Ai El Fi2(V — T24)
where E1= F1_2 =
emissivity of surface 1 shape factor depending upon the geometry of two surfaces.
55. What do you understand from black body and total emissive power? Out of snow and surface coated with lamp black, which is a black body? A black body is an ideal hypothetical body which absorbs all thermal radiation falling on it and it does not transmit or reflect any radiation falling on it. Total emissive power is the rate of thermal radiation emitted by the body in all directions. A surface coated with lamp black appears black but it is not a black body. Although snow appears white but it absorbs all radiation falling on it and hence it is a black body. 56. What do you understand from combined heat transfer mechanism? (UPTU — 2006 — 07) There are three mechanisms of heat transfer viz. (1) conduction (2) convection and (3) radiation. In actual practice and in many situations, heat is transferred in successive steps by different mechanisms. Heat transfer can occur— (a) By one mechanism (b) By more than one mechanism in steps. 57. Explain the combined mechanism of cooling of hot coffee kept in a thermos flask. Flask Coffee Air layer Cover
q2
q4
Room Air T.,
q6
q8 Cooling of coffee
Introduction to Heat Transfer
2I
The mechanisms for heat transfer from hot coffee to air include: (a) (b) (c) (d) (e) (f) (g) (h)
Free convection from hot coffee to the walls of the flask (q1) Conduction through wall of the flask (q2) Free convection from flask to air layer (q3) Free convection from air layer to the plastic cover (q4) Radiation from flask wall to the plastic cover (q5) Conduction of heat in the cover (q6) Free convection from cover to room air (q7) Radiation from cover to room air (q8)
58. Explain the combined mechanism of heat transfer from outside to inside of cooled room (neglect radiation). => qcond
T.,
Room temperature Tr
The pathways for heat transfer from outside to the cooled room are(a) Heat transfer by convection from surrounding (11,) to the wall at T1 qconvi = hiA(T  T1) (b) Heat transfer by conduction from outer wall (T1) to inner wall (T2) qcond = where
kA(T2  li)
L L = thickness of wall
(c) Heat transfer by convection from inner wall (T2) to room air (Tr) qconv2 = h2A(T2 — T)
The quantity of heat flow through each step is same, hence qconvl = qcond = qconv2 = q
Also we can writeT  T1 =
qconv, q hi A hi A
T1  T2 = qwnd — q kAIL kA L
(1) (2)
22
Heat and Mass Transfer qC0//1,2 _ q _ T2 — T = r h2 A h2A If we add equation (1) to (3), we get
(T — Tr) = q(
(3)
1 1 hi A + kNL + h2A
or
q = UA(L — Tr)
where
u=( 1 1+ 1j h1 +k h2
Here U is known as the overall heat transfer coefficient. 59. A black body of surface area 2 x 103 m2 is heated to 127°C and is suspended in a room having temperature 27°C. Find the initial rate of loss of heat from the black body. From StefanBoltzman law we have Q = a A(Tb4 — V)
= (5.67 x 108) (2 x 103) (4004 — 3004) = 1.98 W 60. The inside temperature of a furnace wall is 1000°C. It has k = 2.0 W/mK and thickness = 200 mm. The convective heat transfer coefficient (h) at outside is— h = 8 + 0.1 AT If surrounding temperature is 20°C, find the outer wall temperature and heat transfer per unit area.
To, = 20°C
q=
or or or or or
2.0(1000 — T2) L
10(1000 — T2) 10000 — 10 T2 T1 — 40 T2 + 400 T22 + 140 T2 — 101200
k(Ti — T2 ) = h(7, — Too
L
2
= [8 + 0.1(T2 — 20)] [T2 — 20] = 8 T2 — 160 + 0.1 (T2 — 20)2 = 8 T2 — 160 + 0.1(T22 —40040 T2) = 10(10160 — 18 T2) =0
Introduction to Heat Transfer
23
 140 ± )11402 + 4 x1012 x102 2  140 ± 236.48 = 2 = 96.48
T2 —
Now heat transfer per unit area is k q= — (1000  96.48) = 2 (903.52) 0.2 = 9035.2 W/m2 61. The walls of a house are 0.3 m thick and total surface area of the walls is 100 m2. Thermal conductivity (k) of the walls = 1 W/m2K. Temperature of outside air is 37°C and inside air is 27°C. Heat transfer coefficients inside and outside are 20 W/m2K and 10W/m2K. Calculate the inside and outside wall temperature, heat flux and heat transfer rate through the walls.
Q T.,2 = 37°C
2 = 20 w/m k
0.3 m
Overall heat transfer coefficientU=
1 L 1 ++ h1 k h2
= 1 + 0.3 + 1 10 1 20 = 0.1 + 0.3 + 0.05 = 0.45 Q= U.A. (T2  T1) = 0.45 x 100 (37  27) = 450 W Now for steady state and for interface of inside air and inside wall, we haveQ = hiA(T 1  To,1) 450 = 10 x 100(T1  27) T1  27 = 0.45
24
Heat and Mass Transfer
T1 = 27.45°C Similarly for outside air and outside wall, we have—
or
Q = h2A(T.2 — T2) 450 = 20 x 100 (37 — T2) 37 — T2 = 40 = 0.225 20 x 100 T2 = 36.775°C.
62. An electric heater emits 1000 watts of thermal radiation. The filament has surface area 0.06 m2 and may be presumed as black body. Find its temperature if a = 6 x 108 W/m2K4. 4 Q=A• a• T 1000 = 0.06 x 6 x 108 x T4 or
or
T4
_
1000 6 x 108 x 0.06
= 2.77 x 1011 T = 725.5°K
63. Wet clothes are being on a clothesline outdoor in subzero weather. After a day, the clothes are brought into the house and observed to be dry. The process of drying is best explained as— (a) vaporisation (c) melting
(b) sublimation (d) condensation (GRE)
The clothes are drying due to change of frozen water in the clothes into the vapour. Hence, it is the process of sublimation. Option (b) is correct. 64. Aluminium foil used for cooking food and storage sometimes has one shiny surface and one dull surface. Should the shiny side or the dull side be on the outside when the food is wrapped for baking and freezing respectively? (a) shiny side, shiny side (c) shiny side, dull side
(b) dull side, dull side (d) dull side and shiny side (10 L)
Shiny side can reflect the heat while dull side can absorb the heat most efficiently. We require most heat to be absorbed during baking which necessitates dull side is to kept outside. However, while freezing, we want heat to leave the food. We want heat should not be reflected back. Hence, shiny side should be outside. The option (d) is correct.
Introduction to Heat Transfer
25
65. An outer wall of a furnace is at '7'„,' Kelvin and it faces an environment at '1'e' . The wall's average convective heat transfer coefficient 'h' is 4 W/m2K. For radiation heat transfer, the wall's emissivity 'e' equals its absorbity 'a' and equals 0.8. The ratio of net heat loss by radiation to heat loss by convection from the wall is closed to— (a) 5 a
Tw4 — T4 e TH, — Te
(c) 0.5 a
(b) 2.5a
T.: — Te4 TH, — Te
(d) 0.2a T .
— Te4 TH, — Te
T4 w — T4 e TH, — Te
(IES 89) QRad = ea (T,',,I — Te4) x A = 0.8 a (Tw4 — Te4) x A Qconv = hA(T, — Te)
= 4 A(T, — Te ) QRad _ 0 .8 a (1;4,, — Te4 ) 4 (Tw — Te) Qconv —
= 0.2 a
(T 4 —7' 4)
e w (Tw — Te)
Option (d) is correct. 66. A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop, this can be replaced by a layer of diatomic earth of conductivity 0.14 W/mK and thickness:
(a) 0.5 m
(b) 0.1 m (d) 0.5 m
(c) 0.2 m
(IES — 93) Qi = —kill dT
dx1
= —0.7 x A x Q2 = —0.14 x A x Now
or
dT 0.5 dT
dx2
Qi = Q2 dT dT = 0.14 x A x 0.7 • A • 0.5 dx2 dx2 =
0.14 x 0.5 0.7
26
Heat and Mass Transfer
= 0.2 x 0.5 = 0.1 m Option (b) is correct. 67. A thin flat plate 2 m x 2 m is hanging freely in air. The temperature of the surroundings is 25°C. Solar radiation is falling on one side of the plate at the rate of 500 W/m2. The temperature of the plate will remain constant at 30°C, if the convective heat transfer coefficient (in W/m2°C) is: (a) 25 (c) 120
(b) 50 (d) 200 (IES — 93) QRad = 500 watt Qconv = hA(Tw — To0
= h x (2 x 2) (30 — 25) QRad = Qconv
500 = h x 4 x 5 or
h=
500
20
= 25 W/m2°C
Option (a) is correct. 68. A wire is plastically deformed (bent) by supplying a force of 40 N over a distance of 0.8 m (the force moves in directions in which the distance is measured). If the wire has a mass of 0.2 kg and a sp. heat of 0.5 kJ/kg°C, estimate the maximum increase in the average temperature of the wire (a) 0.03°C
(b) 0.3°C (d) 30°C
(c) 3°C
(GRE)
Now or
Work = Force x distance = 40 x 0.8 = 32 J 32 J = mCp x dT = 0.2 x 0.5 x 103 x dT dT =
32
103
= .032°C
Option (a) is correct. 69. In a long cylinder rod of radius R and a surface heat flux of 90, the uniform internal heat generation rate is
Introduction to Heat Transfer
(a) 290/R (c)
27
(b) 290
90 2R
(d) 90 R2 (GATE)
Heat flux x Area = Heat generation per volume x volume qo x 27 rR x L = qg x 4ir R2 x L or
_ qo x2nRxL qg — 47rR2 x L _ go 2R
Option (c) is correct. 70. For a given heat flow and for same thickness, the temperature drop across the material will be maximum for (a) Copper (b) Steel (c) Glass wool (d) Refractory brick (GATE 1996) dT q = kA = constant dx .
dT . Ik Hence, temperature fall is maximum when 'V is low. Since copper has lowest I', hence temperature fall for copper is maximum. Option (a) is correct. 71. A 2 kW, 40 litres water heater is switched on for 20 minutes. The heat capacity C, for water is 4.2 kJ/kg K. Assuming all the electrical energy has gone into heating the water, increase of the water temperature in degree centigrade is— (a) 2.7 (b) 4.0 (c) 14.3 (d) 25.25 (GATE 2003) Heat supplied = Heat gained Heat supplied = power x time = 2000 x 20 x 60 = 24 x 105W Heat gained = m Cp x AT
25 x 10",= 40000 x 4.2 x 1000 x AT 1000
28
Heat and Mass Transfer
or
25 x 105 = 14.3°C. 4x104 x4.2
AT =
72. A solar collector receiving solar radiation at the rate of 0.6 kW/m2 transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is used to run a heat engine which rejects heat at 315 K. If the heat engine is to deliver 2.5 kW power, the minimum area of the solar collector required would be— (a) 83.33 m2 (b) 16.66 m2 (c) 39.68 m2 (d) 79.36 m2 (GATE 2004) Solar radiation = 600 W/m2 With 50% T = 0.5 x 600 = 300 W/m2 1
11engine = or
T2
=1
315 350
W 35 •= =0.1 Q 350
or
Q=
25 x 103 = 25000 W 0.1
Now
A=
25000 Solar radiation
25000 300 = 83.33 m2 73. An electrically heated plate dissipates heat by convection at a rate of 8000 W/m2 into ambient air at 25°C. If the surface of the hot plate is at 125°C. Calculate, the heat transfer coefficient for convection between the plate and the air. (Annamalai University 20045, 20058) Q = hA(T, — To,) or
Q = h(T, — To,) — 8000 = h(125 — 25)
or
8000 100 = 80 W/m2°C
h=
Introduction to Heat Transfer
29
74. For the three dimensional object shown in the figure below, five faces are insulated. The sixth face (PQRS), which is not insulated and it interacts thermally with the ambient, with a convective heat transfer coefficient of 10 W/m2K. The ambient temperature is 30°C. Heat is uniformly generated inside the object at the rate of 100 W/m3. Assuming the face PQRS to be at uniform temperature, its steady state temperature is— (a) 10°C (b) 20°C (c) 30°C (d) 40°C (GATE 2008)
h x A x (Ts. — To ) = heat convected out heat convected out = heat generated 10 x 2 x 2 x (Ts — 30) = 100 x 2 x 2 x 1 or Ts = 30 + 10 = 40°C Option (d) is correct. Now
Chapter
2
GENERAL HEAT CONDUCTION EQUATION
A A A A A A
CARTESIAN COORDINATES CYLINDRICAL COORDINATES SPHERICAL COORDINATES THERMAL DIFFUSIVITY HEAT ACCUMULATION GENERAL HEAT CONDUCTION EQUATION
A A A A A
ONEDIMENSIONAL HEAT CONDUCTION LINEAR HEAT FLOW RADIAL HEAT FLOW LINEAR TEMPERATURE DISTRIBUTION RADIAL TEMPERATURE DISTRIBUTION
INTRODUCTION Conduction is primarily a molecular phenomenon requiring temperature gradient as driving force for heat transfer to take place. Heat conducting systems can have different shapes and geometries. Hence, heat conduction equations have to be derived using cartesian, cylindrical and spherical coordinates. The Fourier's law for unidirectional heat flow helps in derivation of threedimensional heat conduction equations. 1. Why are different coordinate systems being used? Heat conducting systems can have different shapes and geometries. Heat conducting systems can basically have rectangular, cylindrical or spherical geometries. It is therefore easier to derive the heat conduction equations by using cartesian, cylindrical or spherical coordinates system depending on the shape of the heat conducting system. 2.
How are point, surface area and volume defined in (1) cartesian coordinates (2) cylindrical coordinates, and (3) spherical coordinates?
General Heat Conduction Equation
Cartesian coordinates system
In cartesian coordinate system, we have— (a) (b) (c) (d)
Volume AV = dx • dy • dz Surface area S1 = dy • dz Surface area S2 = dx • dz Surface area S3 = dx • dy
Cylindrical coordinates system
In cylindrical coordinates, we have— (a) (b) (c) (d)
Volume AV = dr • rd0 • dz Surface area S1 = rd0 • dz Surface area S2 = dr • dz Surface area S3 = dr • rd0
Y
Spherical coordinates system
In spherical system, we have1. 2. 3. 4.
Volume dV = rd0 • dr • r sin Od0 Surface area S1 = r sin 0 • dO • rde Surface area S2 = dr • rde Surface area S3 = dr • r sin MO
3I
32
Heat and Mass Transfer
3. Derive onedimensional heat conduction equation in cartesian coordinates. (UPTU 2003) Consider a small volume of material in cartesian coordinates of sides dx, dy and dz as shown in the figure.
Qx+dx
Heat entering the control volume along the xdirection in time dt is— Qx = — kx (dy • dz) •
g •
dt
Heat leaving the control volume along the xdirection in time dt is— Qx±dx
= Qx ,ax (eodx
Net heat accumulation in the control volume for heat flowing in xdirection is— dQx = Qx — [Qx +
= —
a
ax
(Qx)dx]
x (Qx)dx
a [—k (dy • dz) aT • chi • dx
ax
ax
x
(k x
aT )dx • dy • dz • dt ax
Similarly, the heat accumulation in the control volume due to heat flow along y and z directions are— dQy =
dQz =
ay
a
1clx • dy • dz • dt
[k Y
[k
.
z az
x.
dyd
d z.
t
Now total heat accumulation in the control volume from all directions (dOx+y+z = [ax(kx ax)
a (k. ay Y
), ay
a (k aT)1d x • dy•dz•dt
az
z az
General Heat Conduction Equation
33
If qg is the heat generated per unit volume and unit time, then total heat generated in the control volume is— dQg = qg • dx • dy • dz • dt
Total heat accumulation in control volume is— dQtotal = dQx+y+z ÷ dQg Due to heat accumulation, the temperature of control volume will increase and thermal energy increase in time dt is— dQtotal = Mass x sp heat x rate of change of temperature x time
where
= (p • dx • dy • dz) x C, x f x dt t p = density and C = sp. heat
From thermal energy balance, we get—
[
( aT) + a ( al + a (kzaT)]. dx • dy • dz • dt + qg • dx • dy • dz • dt kx ax ax) ayky ay az az = p • dx • dy • dz • p • aaT • dt
t
a [ a T 1 + a [ k ay a T1 ÷ [ Ice a T1 + = aT qg ax kx ax ay az :z ay In case of homogeneous and isotropic material, k = kx = ky = kz i.e. uniform thermal conductivity. Therefore we have— or
p .
c
.
p
at
a2T
a2T a2T qg P • cp aT ax2+aye + az2 + k — k at In case of steady state condition, the temperature at any point does not change with time (aT = 0) then we have— at ' a 2T + a2T + a 2T +
qg k
=
0 az2 axe aye In case there is no heat generated in the controlled volume (qg = 0), then we have— a2T + a2 T + a2 T = 0 (Laplace equation) ax2 aye az2 If temperature varies in one direction (say along xaxis), then we have— a2 T =0 ax2
4. What is thermal diffusivity? What is its significance? Threedimensional heat conduction equation is—
a2T
a2T a2T qg ax2+aye + az2 + k —
aT k at
P • cp
34
Heat and Mass Transfer
1 aT __ — — a at where a = k
pCp
= thermal diffusivity. The significance of diffusivity is—
1. Larger is the diffusivity (a), the shorter is time required for the supplied heat to penetrate deeper into solid i.e. aT is low. at
2. Larger value of diffusivity is possible either due to high value of thermal conductivity (k) or low value of heat capacity (p • cr). 3. Lower is the heat capacity, less amount of heat is absorbed to raise the temperature and bulk heat is available for further transmission. 4. Metals and gases have high value of diffusivity and they respond quickly to any temperature change. The thermal diffusivity of nonmetals is low and they do not respond quickly to any temperature change. 5. Where can heat conduction equation in cylindrical coordinates be used? In many situations, we may have heat conducting systems having cylindrical geometries such as cylinders, rods and pipes. For these systems, we can derive the heat conduction equations by using cylindrical coordinates system. 6. Derive general heat conduction equation in cylindrical coordinates for homogeneous and isotropic materials. (UPTU 200203, 20045) Consider an infinitesimal cylindrical volume of sides dr, rdo and dz as shown in the figure. 00
"'",..
Q0+4
The element volume dV = dr • rd0 • dz Heat flow in through (z, 0) plane in time dt is— Qr = —k(dz • rdo) a T • dt
r
Heat flow out through (z, 0) plane in time dt is— Qr+dr = Qr + ar(Q0CIT
General Heat Conduction Equation
35
Heat accumulated in the element volume due to heat flowing in and out in (z, (0) plane in time dt is— (Qr)acc = Qr — Qr+dr = Qr — (Qr + a l. Q,. • dr) = — a Q • dr ar r
= a (lc • dz • rdeP • dt) dr ar ar = k • dr • d0 • dz • t (r aT ) • dt r = k(dr • d0 • dz)( aT + r a2Tjdt ar ar2 1 aT) = k(dr • rd0 • dz)( a2T r2 + a r dt r a Heat flowing in through (r, z) plane in time dt is—
aT Q0 .—k(dr • dz) rao
dt
Heat flowing out through (r, z) plane in time dt is— (Q0) • rd0 raa(1) Heat accumulated in the element volume due to heat flowing in and out in (r, z) plane in time dt is— Q0+0 = Qo +
(Q0)acc = QO — Q0+0 = Qo — Qo
= =
ra0
a (Q0) r d0 rag)
(Q0) • r • d0
rah a [—k(dr • dz) aT dti• r • d0 r •d0
= k(dr • dz • d0) •
as (1 aT •dt a)
= k(dr rd0 • dz) [ 1 a2T ) dt r 2 a2 0 Heat flowing in the element volume through (r, 0) plane is—
36
Heat and Mass Transfer
aT Q, = —k(dr • rdep) —dt az Heat flowing out from the element where through (r, 0) plane in time dt is— a
Qz+dz = Qz
az
(Qz) •
dz
Heat accumulated in the element volume due to flow in and out of the heat through (r, 0) plane in time dt is— (Qz)acc = Qz Qz+dz = Qz [Qz +
az
(Qz) •
dz]
a az (Qz) • dz = [ k(nhp • dr) • dT • dt ] dz dz a2T = k(dr • rd0 • dz) 2 dt az
a
z
Total heat accumulated in the element volume is— Qacc = (Qr)acc + (Q(p)acc + (Qz)acc If heat generation in the element is qz per unit volume, then heat generated in time dt is— Qg = (dr • rd0 • dz)qg • dt
Increase of thermal energy of the element volume in time dt is— dE = p • (dr • rd0 • dz)cp • daT • dt t p = density and cp = sp heat
where The energy balance is
Qacc + Qg = dE k[
2T aT + 1 a2T + a 2T1 dt • [dr • rd0. dz] + qg(dr • rd0. dz) • dt + ar2 r ar r2 a02 a z2 dE = p • (dr • rd0 • dz) • cp
•
dt
a aT t
or or
k[
a2T 1 aT + + 1 a2T _ L a2T ar2 r ar r2 a02 az2 1
aT qg = P cP . at
a2T
1 aT + 1 a2T
a2T
qg
ar2
r ar
r 2 atp2
az2
k
k
aT at
1 aT a at
General Heat Conduction Equation
a=
k
pcp
37
= thermal diffusivity
Note: For steady state and onedimensional heat conduction in radial direction with no heat generation (qg = 0), the conduction equation reduces to—
a2T + 1 aT = o r ar art 7. Where can heat conduction equation in spherical coordinates be used? In many situations, we may have heat conducting systems having spherical geometries such as hollow spherical vessels. For these systems, we can derive the heat conduction equations by using spherical coordinates system. 8. Derive onedimensional heat conduction equation in spherical coordinates. (UPTU 2003 — 04) Consider an infinitesimal spherical volume element of size dr, rd0 and r sin 0 thip as shown in the figure. Assume uniform conductivity as k,
density = p, sp. heat = c and uniform heat generation rate = qg per unit volume per unit time. Now
dV = dr • rd0 • r sin Od0
Now heat flow through (r, 0) plane in 0direction is—
aTedo dt r sin Heat flowing out through (r, 0) plane in 0direction isa 1 p)r sin Odo Q0+d° = Q° + r sin e ao (Q, Heat accumulated in the elemental volume due to heat flow in and out in the 0direction is— Q0 = —k(dr • rd0)
dap = Q0 — 00+0 —1 = (Q0)r sin Odo r sin 0 a0
38
Heat and Mass Transfer
=
—1
a [ — k (dr •• rde) •• 1 aT • dti• r sin 0 • chip . r sine ao rsin e ao
2
1 a = k (dr • rde • r sin 0 • d0) • r 2 sin 2 0 ao2Tdt Now heat flow through (r, 0) plane in 0direction is— Q9 = —k (dr • r sin e • d0) aaTe dt r
Heat flowing out in 0direction is— Qe+de
a
= Qe + ra 0 (Q0 • rde
Heat accumulated in the elemental volume due to heat flow in and out in 0direction is— da, = Qe — Qe+de
a (Q9) • r • dO
=
rail
=—
aT
[ k(dr • rsin 0 d0) rae dt r • de rag ]
=‘ 4 . (dr • rde • r sin 0 d0) •
1 sine
a (sin OL T )dt ae
Now heat flow in (0, 0) plane in rdirection is— Qr = —k(r • de • r sin 0 • dtP)
aT ar
• dt
Heat flow out in (0, 0) plane in rdirection— Qr+dr
= Qr
, 1
ar (Qr) • dr
Heat accumulated in the elemental volume due to heat flow in and out from the rdirection is— dQr = Qr — Qr+dr
ar (Qr)dr = =
a [ k(r de • rsin e • d0) a T dt]• dr ar r
= k • de • sin 0 • dO • dr • 1(12 aT y)dt r = k(dr • rde • r sin 0 d0) 2, 1(12 aaT )dt r 7
Now
(dQ),,,, = dQ0 + dQ 19 + dQr
General Heat Conduction Equation
39
Now heat generated in thin control volume— Qg = qg(dr • rd0 • r sin 0 • d0) x dt Rate of change of thermal energy within the elemental volume— dT dE = p (dr • rd0 • r sin 0 • d0) • cp at dt Applying now energy balance— (dQ)acc + dQg = dE a (sine ail, 1 a (r 2 al 1 a2T + 1 a02 ar ad r2 ar r2 sin2 0 r2 sin 0 ad
aT = 1 aT + qgk = p •cp k at a at
Now for steady state unidirectional heat flow in the radial direction for a sphere without internal heat generation is
ail _
1a r2 al. r al. — a 9. What is the aim in analysing a heat conduction problem in a medium? The aim in analysing a heat conduction equation is to determine the temperature field in a medium resulting from the conditions that are imposed on the boundaries of the medium. The determination of temperature distribution helps in— (a) Finding heat transfer rate (b) Finding the structural integrity of the medium through determination of thermal stresses, expansion and deflections (c) Optimising the thickness of an insulating material for a conductor. 10. What are the conditions on which the solution of heat conduction equation depends? The solution of heat conduction equation depends upon following conditions— (a) Boundary conditions (b) Initial conditions 11. Why do we require initial conditions? If the solution of heat conduction equation is time dependent, initial conditions are required. The initial conditions describe the temperature distribution in a medium at the initial moment of time and these are needed for the time dependent problems (transient or unsteady state). The initial conditions can be expressed as— (i) t = 0, T = T(x, y, z) (ii) t = 0, T= To = Constant 12. What are the boundary conditions? How many are needed?
40
Heat and Mass Transfer
Boundary conditions are the physical conditions existing at the boundaries of the medium. If heat conduction equation is of first order, only one initial condition is required. If heat equation is of second order, two boundary conditions are needed for each coordinate in determining temperature distribution. 13. What are the different kinds of boundary conditions? There are three kinds of boundary conditions which are(a) Constant surface temperatureT(0, t) = T, = Temperature at surface (b) Constant surface heat flux(i) Finite heat flux(k aT , ) = q, = fixed heat ax x=o (ii) Zero heat flux due to insulated surface( dT 0 ax)x=o (c) Equal conductive and convective heat flux at surface condition(_ k dT = h[T,  T(0, t)] dx )x=o 14. Using onedimension heat conduction equation with uniform thermal conductivity, steady state and no heat generation, find heat `Q' to be supplied to a wall of width `I,' to maintain temperature difference of (T1  T2) across it.
H L .I Onedimensional heat conduction equation isa 2T
0 as qg = 0, aT= 0 axe = Now integrating two times, we getT = Cix + C2 Using boundary conditions x = 0, T = T1 —> C2 = T1 X = L, T = T2 —> T2 = CiL + T1
General Heat Conduction Equation
or
C1 =
4I
T2 — Ti .
L Now putting the values of C1 and C2 , we get— T
—(T2
Ti )
x + Ti L The above is temperature distribution inside the wall along xdirection. Now to find heat, we have to find the change of temperature w.r.t. x coordinate. aT _ T2 — T1 L ax —
As per Fourier's law, heat flow is— Q = —kA
aT ax
= —kA ( T2 — 11 ) L ) — kA(Ti—T2) L 15. The temperature distribution across a wall of 2 m thick is given by— T(x) = 820 — 300x — 50x2 where T is in degree and x is meter. If the wall has surface area = 5 m2, internal heat generation qg = 10 kw/m3, k = 40W/mk, find— (a) Rate of heat entering and leaving the wall (b) Rate of change of internal energy
x= 0
x=2
T = 820 — 300x — 50x2
Given
aT = —300 — 100x
ax
Heat entering
aT Qx=0 = —kA () ax x=o
or
Qx=0 =
—40 x 5 x (300 — 1004=o = 200 x 300 = 60 kW
42
Heat and Mass Transfer
Heat leaving Qx=2 = —kA ( or
ax 1=2 —40 x 5 x (300 — 100x)x =2 Qx=2 = = —200(300 — 200) = +200 x 500 = 100 kW
Rate of change of internal energy (Au) is— Au = Rate of heat entering + Rate of heat generation — Rate of heat leaving = 60 + qg xA xL— 100 = 60 + 10 x 5 x 2 — 100 = 60 + 100 — 100 = 60 kW 16. The temperature distribution at a certain instant of time in a concrete slab during curing is given by— T = 3x2 + 3x + 16 Where x is in cm and T is in Kelvin. The rate of change of temperature with time is given by (assume diffusivity to be 0.0003 cm2/s) (a) 9 x 104 k/s (b) 48 x 104 k/s (c) 12 x 104 k/s (d) 18 x 104 k/s (IES — 94) T = 3x2 + 3x + 16
= 6x + 3 ax ax
a2 7 xx2 Now or
=
6
a2T = 1 aT ax2 a at a2T aT =a at ax2 = 0.0003 x 6 = 0.0018 k/s
Option (d) is correct. 17. Consider the following statements: The Fourier heat conduction equation Q = —kA 1. Steadystate conditions
dT , presume dx
General Heat Conduction Equation
43
2. Constant value of thermal conductivity 3. Uniform temperatures at the wall surfaces 4. Onedimensional heat flow of these statements (a) (b) (c) (d)
1, 2 and 3 are correct 1, 2 and 4 are correct 2, 3 and 4 are correct 1, 3 and 4 are correct (IES  98)
The statement (b) is correct. 18. A steam pipe is passed through a room in which air and wall temperature are at 30°C while surface temperature of the pipe is 400°C. If the diameter of the pipe is 40 mm and average heat transfer coefficient is 20 W/m2°C, what is the rate of heat loss from the pipe for one metre length of pipe.
i i i 1)
Ta = 30°C
T ., = 400°C steam pipe
\
i ,
'''
Q =hxAx(Ts  Too ) = 20 x (,rdL) x (400  30)
= 20 x ir x 40 x 103 xlx 370 = 0.93 kW 19. A copper pipe of diameter 5 cm is kept at a temperature of 50°C in a large room where air and wall are at 20°C. If the average heat transfer coefficient is 6.5 W/m2°C and emissivity of the body is 0.8, calculate the total heat lost by the pipe per unit length. Qconv = hA(T1  T2)
= h(ndL)(Ti  T2)
= 6.5 x 7r x 5 x 102 x 1 x (50  30) = 30.6 W/m QRad = aA E (T14 — T24) = 5.67 x 108 x (7r x 5 x 102 x 1) x 0.8 x (3234  2934) = 25 W/m Total heat lost
Qtotal = QConv + QRad
= 30.6 + 25 = 55.6 W/m
44
Heat and Mass Transfer
20. The radial temperature distribution at a certain instant of time in a cylindrical column during curing is given by T = 5r + 250 where r is in cm and T is in Kelvin. Find the rate of change of temperature with time if diffusivity is 0.002 cm2/s. Onedimensional steadystate heat conduction in radial direction is given by—
a ( aT) _ 1 aT ar C. ar ) — a at T = 5r + 250 aT 5 ar 
Now
Hence,
aar (r or or
• 5) =
at
a 1 at 5= 1 aT a aldT =ax5 at = 0.002 x 5 = 0.01 k/s
21. The radial temperature distribution in a sphere at a certain time is given by T = 6r2 + 200 where r is in cm and T is in Kelvin. Find the rate of change of temperature with time if diffusivity is 0.0004 cm2/s. Onedimensional steadystate radial heat distribution is1
a ( r 2 an = 1 aT
ar ) a at T = 6r2 F 200 aT _ 12r ar I a [r2(12r)] = 1 aT a at r2 ar 1 1 aT — x 12.3•r2 = r2 a at aT _ a x 36 = 0.0004 x 36 = 0.0144 k/s at r2 ar
Now
or or
z
General Heat Conduction Equation
45
22. In steadystate the heat generation rate per unit volume is qg at radius r of solid rod given by 2 X 106
qg = 500[1 —(i) 1
Find the maximum temperature of the rod having R = 40 cm when temperature at surface is 20°C. Take k = 100 kW/m°C. For cylindrical surface and radial heat flow, we have
t(r al a ;) = qk r or
aar (r aaT ) +
i.
x 500[1 (f )2 1 X
106 =
z
0
On integration, we getr4 = C (r aT) soo x io6 r2 a r ) 100 x io3 L 2 4R2 aT = 0 at r = 0, —
Now
=0
aT = 5000 ('  ' 3 ar
2
4R2
Now integrating again [r2
T=Now
r4
+
5000 4 16R2
C2
T = Tmwhen r = 0 Tmax = C2 2
4
r ,1 + Tmax 16R 2
T = —5000L 4
[r2
or Now when or
Tmax = T + 5000
4
4
16R 2 ] r = R, T = Ts = 20°C
Tmax = 20 + 500OUR2 ) = 20 + 5000 x 3 x 16 x 102 6 = 20 + 150 = 170°C
46
Heat and Mass Transfer
23. The temperature distribution in a long cylindrical to be is—
T(r) = 800 + 1200r — 3000r2 where T is in centrigrade and r is in metre. The cylindrical tube has inner radius = 250 mm and outer radius = 400 mm. Find the heat entering and leaving the cylinder.
Given ...
T(r) = 800 + 1200r — 3000r2
aT = 1200 — 6000r ar
Hence, heat flow per unit length is— aT —kA(— ar 1=0.25 = —50 X (27r X 0.25)(+1200 — 6000r)r=o.25 = —50(0.570(1200 — 1500) = 257r(300) = 23.55 kW Similarly, heat flowing out per unit length is— Qr=0.25 =
Q r=o.4 = —kA
( a Tr)
r = 0.4
—50 x (27r x 0.4)(1200 — 6000 • r)r=o.4 = —40 is • (1200 — 2400) = 40 • ir • 1200 = 150.72 kW
Qo.4 =
24. The temperature field in a body varies according to the equation T(x, y) = x 3 + 4xy. The direction of fastest variation in temperature at the point (1, 0) is given by (a) 3i + 8j (b) j (c) 0.6i + 0.8j (d) 0.5i + 0.866j (GATE — 1967)
T = x3 + 4y Now
aT = 3x 2 and aT = 4 ax
ay
General Heat Conduction Equation
Hence,
47
=3, (al =4
PT)
ax x=i ay y=o Hence, direction of fastest variation in temperature at (1, 0) is given by 3i + 4j or 0.6i + 0.8j. .. Option (c) is correct. 25. Two surfaces of a 1.1 m thick wall are maintained at temperature of 200°C and 60°C respectively. Determine the heat flux through the wall, if the thermal conductivity of the wall material varies with temperature as follows— k = 0.2(1 + 0.04T) where k is in W/mk and T is in °C. (UPTU — 200708)
or On integration
OT q = Q = —ky A x qax = —kaT L 200 q f dx = —f [0.2 (1 + 0.04T)]aT o 60 2 200
q x L = —0.2 [T + 0.04 xT 2
60
= —0.2 [(200 — 60) + 0.04 (200 2_ 602)] = —0.2[140 + 0.02[4 — 0.36] x 104] = —0.2[140 + 728] = 173.6 W/m q = 173.6 = 173.6 _ 157.8 W/m2 L 1.1 — 26. A 50 cm thick steel plate (ksteei = 36 W/mk) of large surface area generates heat at a uniform rate of 20,000 W/m3. Both the surfaces of the plate are exposed to a gas (hgas = 25 W/m2k) at temperature of 100°C. Find the temperature at: (i) centre, (ii) surface, and (iii) at 10 cm right 2 h = 25 W/m k
ksteel = 36 W/mk T., = 100°C x=0
x = 25 cm
.
X
48
Heat and Mass Transfer
2T
x2
n qg +— =v
k
a 2 r,
or
k
aT _
qg
axe
or
qg —
+ Cl ax — k x aT = 0 at x = 0, hence C1 = 0 ax qg T = —T x2 + C2
Now on integration,
(i)
At the surface, we have1c( axx=o.2,5 T) = h(Ts — T—) —k(—
or or
k
X
0.25) = 25 (Ts — 100)
20 x 103 x 0.35 — Ts — 100 25 Ts = 100 + 280 = 380°C
Putting the value of x = 0.25 in equation (i), we have— 25 x 103 x (0.25)2 + c2 36 C2 = 380 + 43.4 = 423.4 °C T = —694.4x2 + 423.4°C
380 = or .
The temperature is maximum at x = 0, Tx=0 = 423.4°C Temperature at x = 0.1 m Tx=0.1
= 423.4 — 694.4 x (0.1)2 = 416.5°C
27. A flat plate has thickness 5 cm, thermal conductivity 1 W/mk, convective heat transfer coefficient on its two flat faces are 10 W/m2k and 20 W/m2. The overall heat transfer coefficient for the plate is— (a) 5 W/m2K (b) 6.33 W/m2K (c) 20 W/m2K (d) 30 W/m2K (ESE — 2002)
General Heat Conduction Equation
49
k = 1 w/mk
2 = 20 w/m k
hi = 10 w/m2k
H— 5 cm
ER =
P
1 5 x 102 1 + + k hi h2
5 x 102 1 = 1 + + 10 1 20 = 0.1 + 0.05 + 0.05 = 0.2 U
=E
1 = 5 W/m2K R = 0.2
Option (a) is correct. 28. A 0.5 m thick plane wall has its two surfaces kept at 300°C and 200°C. Thermal conductivity of the wall varies linearly with temperature and its value at 300°C and 200°C and 25 W/mK and 15 W/mK respectively. Then steady heat flux through the wall is— (a) 8 kW/m2 (b) 5 kW/m2 (c) 4 kW/m2 (d) 3 kW/m2 (ESE = 2001) k1 +k1 25+15 k —— = 20 2
Q=k A
Option (c) is correct.
x
2
(300 — 200) _ 20 x 100 = 4 kW/m2 0.5 — 0.5
Chapter
3
STEADY STATE HEAT CONDUCTION
A A A A A A A
LOGMEAN AREA CONCEPT OF THERMAL RESISTANCE THERMAL RESISTANCE OF SLAB THERMAL RESISTANCE OF SPHERICAL WALL COMPOSITE WALL
A A A A A A
COMPLEX COMPOSITE WALL OVERALL HEAT TRANSFER COEFFICIENT
A A A
THERMAL CONTACT RESISTANCE COMPOSITE CYLINDER COMPOSITE SPHERE CRITICAL THICKNESS FOR WIRE CRITICAL THICKNESS FOR SPHERE LAPLACE EQUATION 2D HEAT CONDUCTION SHAPE FACTOR FINITE DIFFERENCE EQUATION RELAXATION METHOD
INTRODUCTION Onedimensional heat flow means that there is a single predominant direction in which temperature gradient is existing. Hence, heat flow takes place in one direction while heat flow in other directions is neglected. Therefore, only one space coordinate is required to describe the distribution of temperature within the heat conducting body. The term of steady state means that the temperature distribution within the body does not change with time. Fairly good estimates of heat conduction can be made in many cases by making the assumption of one dimensional heat conduction. Heat flow through: (i) a plane wall, (ii) a very long hollow cylinder or pipe, and (iii) a very thin wire or rod can be found out easily using this method. Multidimensional heat conduction problems are generally complex. Laplace and Fourier obtained analytical solution of partial differential equations for different heat conduction problems. The analytic approach can be applied to such problems which involve simple geometrical shapes and boundary conditions. Finite difference forms of Laplace equations are numerical methods which give good approximation to complex heat conduction problems. Finite difference equations can be easily solved with the help of high speed computers.
Steady State Heat Conduction
5I
1. Derive an expression under onedimensional steady state heat conduction for temperature distribution for plane wall or slab.
T2 L —d x=L
x
x=0 Heat conduction through slab
Consider a plane wall with thickness `L'. Heat is flowing under steady state condition in such a way that temperatures of two faces be maintained at T1 and T2. If thermal conductivity 'lc' is uniform, then general heat conduction equation is
a 2 r,
a2 r,
a2 r,
+qg
1 al,
K a
ax2+ a y2+ z2
at
For onedimensional steady state heat conduction without internal heat generation a2 , a2 , a 2 7, we have = 0, = 0, y2 z2 t2 = 0 and qg 0. =
The equation becomes a2T
=0
axe
On integration, we get aT = C1 = constant ax On integration again, we get T = C1 x + C2 Applying boundary conditions, we get and or
x = 0, T = Ti or Ti = C2 x = L, T = T2 or T2 = CiL + T2 Cl
T2 — T1 L
The temperature distribution becomes T= or
T2 — Ti x + Ti L
T2  T1 _ x T2 — T1 — L
52
Heat and Mass Transfer
Now heat flowing from the plane wall is— q=
aT —m a x
=kA4[711 (T1 T2)1 kA(T — T2 ) L 2. Derive an expression for temperature distribution under one dimensional steady state heat conduction and without internal heat generation for a hollow cylinder or tube. Consider a hollow cylinder of length 'I,' with outer and inner radius as ro and r1. For steady state onedimensional heat conduction in radial direction with no heat generation, the equation is
a2T 1 aT are
r ar =u
ar
aT ) _° r
Or On integration, we get
Or
raT = C1 = constant ar aT Cl ar = r
Radial heat flow in a hollow cylinder
Integrating again, we get T = C1 log r+ C2 Applying boundary conditions, we get At
r =ri,T=Ti or
and
r = ro, T = To or
Ti = Ci log ri+ C2 To = Ci log ro
F
C2
Steady State Heat Conduction
53
From there two equations, we get To — Ci =
log Ti log
C2 
ro + (Ti — To) log ri ri
log r 0 The temperature distribution equation becomes T—
To — Ti
tog r +
Ti log r° + (Ti — To) log ri ri
log 11) rt
log i i) r
or
T —Ti
log — ri
To — Ti
log /11
The rate of heat flow is Q = —M
OT
ar log r
= —kA
a
[ +(To Ti ) r Tt
ri
log — r° ri
1 = —kA[(To Ti ) r
log i il ri =—
kA (To — Ti ) r
But area
log ° r
A = 27r r • L Q—
2 irkL(Ti— To) log( 713
3. Find logmean area for a hollow cylinder. What is its purpose? Consider a hollow cylinder and plane wall made of same material with same wall thickness (ro — r) and temperatures at two sides as Ti and To. In order to have the same heat conduction from the hollow cylinder and the plane wall, the problem is to find logmean area
54
Heat and Mass Transfer
(Am) for the plane wall which will give the same heat transfer from the plane wall as it is from the cylinder.
o — r1Cylinder and plain wall with same thickness and temperatures
Heat flow through hollow cylinder is Q=
27r Lk(Ti — To )
(i)
loge 115 Heat flow through the plane wall is Q  k Am (Ti —To) (r  ri) Now on equating Equation (i) and (ii), we get 2irLk(T —To ) log 0r or
Am =
kAm(Ti — To) r0 — r, 2 7rL(ro — log — r0
L(27rro — 2 7rri) 2 7rr0 • L log 27rri •L
Ao — Ai log A — ° Ai Here Ao and Ai are outer and inner surface areas of the cylinder. The purpose of logmean area of the cylinder is that the problem of heat transfer from the cylinder can be transformed into that of from the plane wall so that the solution of the problem becomes simpler. 4. Derive an expression for temperature distribution under onedimensional steady state heat conduction for a hollow sphere. Consider a hollow sphere with radius r0 and ri with steady temperatures at To and Ti. The radial heat flow in the hollow sphere is
Steady State Heat Conduction
a2T 2 aT + are r ar
or
ar
55
=0
(r2 aT) = o
ar
Now on integration, we get— aT r
r2
Integrating again, we get— T=—C l + C2
(i)
r
Applying boundary conditions, we get— at
r = ri, T =
at
r = ro, T = To
= To =
ri ro
On solving, we get— Ti — To = 1 Cl 1 (to rij
and
C2 = T
1 (Ti — To) — ri
ro Putting the values of C1 and C2, in equation (i), we getT—
1 1 Cr — rij
To — Now rate of heat flow is— Q = kA
aT ar
= —kA 1(T—To) r2 1) ro ) — To) 1 = —k • 4ir r2 • — r2 (1 1) To ri 4,rk(T —To) ri ro (ro —
+ C2 + C2
56
Heat and Mass Transfer
5. Find logmean area for a hollow sphere. zink(Ti — T0 ) Pi • To • • • (i) fro — n) k Am (Ti — To) • • •00 Qslab (ro — ri) where ro — ri = thickness of slab and Am is logmean area of the slab. Now equating both equations, we get— Qsphere = Qslab 4 k (Ti — T0 )ri • r0 kAm(Ti — T0) (ro — ri) (ro — ri) Am or = 47cri ro or Amt = (47r ri r0)2 = (47r ri2 • 4ir ro2) Am = Ai • Ao where Ai = inner surface area of sphere Ao = outer surface area of sphere. Qsphere =
6. Explain electrical analogy for heat transfer. Find thermal resistance for conduction and convection heat transfer. Consider heat flow through a plain wall or slab, then heat flow is— (T1 — T2)Ak Q— L — T2 — AT _ Thermal Potential L L LI Ak Ak Ak Comparing this equation with equation for electric circuit— AV Electric Potential Current = 1 = Hence, we can have following electrical analogy— (a) The rate of heat transfer (Q) is analogus to current (I) (b) Thermal potential (T1 — T2) is analogus to electric potential (V1 — V2 = AV) (c) Thermal resistance (Rth) is analogus to electric resistance (R) 0MAF0
V2 T1 0Wr0 T2
R
Rth th
Electric circuit
L
k
Thermal circuit
Concept of resistance
Steady State Heat Conduction
57
Similarly, for convective heat transfer, we have Q = hA(Ti — T2) or
where Rth =
Q=
AT T1 —T2 Rth 1 hA
1 which can be shown as— hA Q
T1 0I \ AAJ0 T2 I Rth = hA Thermal circuit
7. Derive an expression for thermal resistance of cylindrical wall. Guidance:
First derive the heat transfer through the cylindrical wall as done in question 2
Q
2 • K • L(Ti — To) log /11 ri
or
Q=
T —To
 To
0 log r
Rth
27rK • L
Here Rol =
log r0 r 27r KiL
which can be shown as—
Q 0—"AA/0 T, ro log Rth
2irkL
Thermal circuit
8. Derive an expression for thermal resistance of spherical wall. (UPTU 2003 — 4, 2008 — 9) Guidance: First derive the equation of heat transfer of spherical wall as done in question 4 and then find thermal resistance as explained below
Q=
4irk(Ti — To) (ro—
ri • ro
58
Heat and Mass Transfer
T — To ro — 47rk• ri •ro
T — To Rth where Rth —
10 — k • ri • ro
which can be shown as— Q 0—AAA,0 To
ro —
Rth — 4.1* .
ro
Thermal circuit
9. What is a composite wall? When a wall is consisting of a number of slabs of different materials and thicknesses as shown in the figure, it is called composite wall. T1
T2 T3 T4
®
Q
1 IT
Q
® ®
2 3 LA —+ LB
4._
4
Lc
H
Composite wall
Here
RthA =
LA kAA
, RthB =
LB . kB A
, AthC =
Lc kc A
In case there is a perfect contact between the different layers and there is no temperature drop occurring at interface (as shown in the figure), we can draw the analogous electric circuit as shown below—
Q
RthA
RthB
RthC
Since all resistances are in series, the heat transfer is—
Steady State Heat Conduction
Q =
59
T1—T4
LA
LB
+— +
kAA
LC
kB A kc il
10. What is an overall heat transfer coefficient (u)? The overall heat transfer coefficient is defined as the heat transmitted per unit area, per unit time and per degree of temperature difference between hot and cold fluids on either side of a slab. Therefore thermal resistance to heat flowing through any body consists of sum of convective and conduction resistance. Hence heat transfer is— Q = U • A • (Th1 — Th 2)
where
U=
1 1 L 1 —+—+—
h1 k h2
Th1
Th2 F. L ] Overall heat transfer coefficient
11. Derive an expression for overall heat transfer coefficient for a plane wall. (UPTU — 2008 — 9) Consider a plain wall with thermal conductivity l' and it has hot fluid with heat transfer coefficient `1/1' at one side and cold fluid with heat transfer coefficient '122' as shown in the figure
Hot fluid (h1)
Cold fluid (h2) h2
Heat transfer coefficient for plane wall
60
Heat and Mass Transfer
The equivalent electrical circuit is as given below— T hl 1 h1 A
L kA
1
h2 A
The total thermal resistance is—
1
L 1 + + h1A kA h 2A
R= Hence heat transfer is—
Th.,— The Q— 1 L 1 hiA
I
kA h2 h2 A
Ax (Thi — Th2 ) 1 L 1 —+ — + — h1 k h2 = Ux A x (Thi — Th2) where
u=
1 1 L 1 —+ — + — h1 k h2
12. Draw the analogus electric circuit of heat flow for a complex composite wall as shown below:
10
®
Q
®
®
®
Q
DO
The analogus electric circuit is— RthB AVVVQ —.0—i\N\F9
RthA
RthE
_iv\
p
01NA/0
JW\r
RthC
RthD RthF
RthG
Steady State Heat Conduction
6I
13. What is thermal contact resistance? How can it be reduced? (UPTU — 2008 — 9)
At the interface of two slabs, there is a discontinuity in temperature gradient due to surface roughness and void space in between the interface of two slabs. There is a large resistance to heat flow at the interface. This resistance is called thermal contact resistance. The drop of temperature from T2 to T3 is due to the thermal contact resistance. T1 T2
Q
T3 T4
Temperature drop at the interface
The contact resistance is given by— Rcontact
T2 — T3 QIA
The contact resistance can be reduced by— (a) fine surface finish of the slabs (b) continuous contact of slabs without any void space (c) high contact pressure (d) similar metals. 14. The temperature distribution at certain instant of time in a plane wall of 50 cm thick, is given by the relation— T = 450 — 500x + 100x2 + 150x3 where temperature T is in degree Celsius and x in metres measured from the hot surface at 450°C. The thermal conductivity of the wall material is 10 W/mK. Calculate the rate of heat energy stored per unit area of the wall at that instant of time. (UPTU — 2002 — 3)
Given
T = 400 — 500 x + 100 x2 + 150 x3
dT = —500 + 200 x + 450 x2 dx Heat entering the wall from the face being heated i.e. x = 0 is— = —kA
(dT d xj x=o
= —10 x 1 x (500 + 200 x + 400 x2),=0
62
Heat and Mass Transfer
= —10(500) = 5000 W Heat leaving the wall i.e. at x = 0.5 m is— Qout = —kA
X
(dT
dxj x.0.5
(500 + 200 x + 450 x2)x=o.5 = —10 x (500 + 100 + 112.5) = 2875 W = —10 X 1 X
Qstored = Qin — Qout
= 5000 — 2875 = 2125 W 15. The wall of a building is a composite consisting of 250 mm layer of common brick (k = 0.72 W/mK) and 10 mm layers of Gypsum plaster (k = 0.12 W/mK) on both the sides of the bricks. During a hot day (at steady state), the temperature of outside plaster (exposed to ambient air) is 40°C and the temperature of inside plaster (exposed to inside air) is 25°C. Find (i) the heat flow rate through the wall per unit area of the wall (ii) the temperature of the interface of brick and outside plaster. (UPTU — 2003) Gypsum 1 12
3 Bricks
Q Gypsum
40°C
Q
25
—H 10
250
The equivalent electric circuit is—
40
25
R1 =
Total thermal resistance
ki A
R2 =
k2 A 9
Rth
=
R3 =
+ 4 + 4 ki A k2 A k3A
L3
k3 A
Steady State Heat Conduction
63
1 31 A ki23 +k +k = 1 F 10 + 250 + 10 1 x A L0.12 0.72 0.12]
Now or
or
Similarly,
or or
=1 (83.3 + 347.2 + 83.3) x 103 1 = 0.514°C/W = Ti  T4 _ 40  25 Q Rth 0.514 = 29.18 W Q = Q12 = Q23 = Q34 k2 A(T2  40) Q12 = 42 0.12(T2  40 29.18 = 10 x103 40 = 29.18x 102 = 2.43 0.12 T2 = 40  2.43 = 37.57 °C. k3 A(T3  T2) Q23 = L23 29.18  +0.72 x 1(37.57  T3) 250 x 103
T2 
2.18 9 3757  T3 = 0 x 250 x 103 .72 = 10.13 T3 = 37.57  10.13 = 27.44 °C
16. One side of a plane wall is maintained at 100°C while other side is exposed to a convection environment having T = 10°C and h = 10 W/m2K. The wall has the dimensions of 3 x 5 m, has k = 1.6 W/mk and is 40 cm thick. Calculate heat transfer rate through the wall. (UPTU  2003  4) Area = A = 3 x 5 = 15 m2 aT Q = kA = hA(Toc  T2) ax or
+1.6 x (ii  T 2) = 10(T2  10) 0. 4
64 Heat and Mass Transfer
2 h=1.0W/M k
H— 40 cm —0 T. = 10°C T.1 = 100°C
or
or
T2
10 (T2 — 10) 4 100 — T2 = 2.5 T2 — 25 125 T2 = = 35.7°C 3.5 1.6 x (3 x 5)(100 — 35.7) Q= 0.4 = 3.858 kW Ti — T2 =
17. The total thickness of a furnace wall which is made of an inner layer of fire brick covered with a layer of insulation is 32 cm. Thermal conductivities of fire brick and the insulation are 0.84 and 0.16 W/m°C respectively. The furnace temperature is 1325°C and the temperature of surrounding is 25°C. Calculate the thickness of the fire brick and that of the insulation for minimum heat loss through the wall. Assume the maximum heat loss through the wall. Assume that the maximum temperature in the insulating material should not exceed 1200°C. (UPTU — 2004 — 5) k= 0.84
k— 0.16 T.i = 1325°C
T.,= 25°C
T2 = 1200
1
2
3
32 cm —'I
Steady State Heat Conduction
65
The equivalent electric circuit isTa
Rw
Rin
Let thickness of insulation = x Hence, thickness of wall = 0.32  x Now, thermal resistance of wall isR,, =
L„, k„, A
0.32  x 0.84 x 1
 1.19(8.32
 x)
Thermal resistance of insulation is =
Lin = kin • A
x
0.16 x 1
=
6.25 x
Therefore, heat transfer isT1  T Q1.3 
R„,+ Rin
1325  25 1.19(0.32  x) + 6.25 x
1300 0.38 + 5.06 x
Similarly, Q12 
Tl  T2 R„,
1325 1200 1.19(0.32  x) Now or or or or or
125 0.38 1.19 x
Q1.3 = Q1.2 = Q2.3
1300 125 0.3 + 5.06 x 0.35  1.19 x 10.4 (0.35  1.19 x) = 0.38 + 5.06 x 3.64  12.38 x = 0.38 + 5.06 x 17.44 x = 3.26 x = 0.186 m Thickness of insulating material = 18.6 cm Thickness of wall = 32  18.6 = 13.4 cm
18. A board is composed of three equal layers, middle being of packed saw dust. (k = 0.02 W/me) with the side layers made of plywood each of 2 cm thickness (k = 0.12 W/ m°C). The three layers joined together by bolting using four steel bolts of 1 cm diameter at the corners (k = 40W/m°c). Calculate the heat flow per m2 area if one surface is at 35°C and the other at 20°C. (UPTU  2004  5)
66
Heat and Mass Transfer k= 0.12
k= 0.02
Bolts (4 Nos) having k = 40 T.i = 35°C T2 = 20°C
Equivalent electrical circuit is—
Rp1 = Rp2 =
RSD =
L
= 0.02 = 0.167 1.12 x 1 kp A
L kspx A
0.02 = 1 0.02 x 1
Area of bolt = LT x (0.01)2 4 = 0.785 x 104 Rb =
L = kb • Ab
0.06 40 x 0.785 x 104
= 19.11 Resistance of four bolts is1
= 1+ E E 1+ E —
Rbt
or Now or
Rb
Rb
Rb
19.11 4 = 4 = 4.77 2 Rp + RsD = 0.167 x 2 + 1 R2p+SD = 1.33 Rbt =
Rb
Rb
Steady State Heat Conduction
67
Above is based on area = 1 m2, but actual area is in fact without bolt area. 1 Actual R2p±sp = 1.33 x 1 — 4 x 0.785 x 10 4 1.33 1— 0.0003 1.33 — 1.33 0.9997 Now equivalent circuit is— R 2P+SD T1
o T2
Rbt
Rbt
R2P + SD
R = eR2P + SD X Rbt
4.77 + 1.33 Reg' — 4.77 x 1.33
6.10 4.77 x 1.33 = 0.96 — T2 35.20 Q= Rego — 0.96 15 = 15.625 W/m2 0.96 19. A mild steel tank of wall thickness 10 mm contains water at 90°C where atmospheric temperature is 15°C. The thermal conductivity of mild steel is 50 W/mk and heat transfer coefficients for the inside and outside of the tank are 2800 and 11W/m2K respectively. Calculate: (i) rate of heat loss per unit area of tank surface (ii) temperature of the outside surface of the tank. (UPTU — 2006 — 7) Heat transfer from each material is equal, Q12 = Q23 = Q34 = Q14 = heat flow per unit area hi (T — T2) =
k (T2 — T3)
= h2(T3 — T4)
Tl — T4
Q14 = 1
— + 10 X 103 + hl
k
1 17,2,
68
Heat and Mass Transfer
~ 10 mmH
90 —15 1 102 +1 2800 + 50 11 75 3.57 x104 + 0.2 x103 + 0.09 75 (3.57 + 2 + 900) x104 75 x 104 905.57 = 0.08 x 104 = 800 W/m2 Now or or
Q14 = Q34
800 = h2(T3 — 15) = 11(1'3 — 15) T3 — 15 = 72.72 T3 = 87.72°C
20. Two aluminium plates 5 mm thick with ground roughness of 2.54 mm are bolted together with a contact pressure of 20 bar. The overall temperature difference across the plate is 80°C. Calculate the temperature drop across the contact joint. The contact resistance of the joint is estimated to be 0.88 x le m2 K/m and the thermal conductivity of aluminium is 202 W/mK. (UPTU — 2002 — 3) Total thermal resistance per unit area is— R=
k AL x A
x 2+ kontact
5x10 3 x 2 + 0.88 x 104 202
Steady State Heat Conduction
69
T2
T1
AT= Ti — T2 = 80°C
AL
AL
I
H 5 —I
5
= 0.49 x + 0.88 x 104 = 1.37 x 104 AT 80 Q R 1.37 x104 = 58.2 x 104 W/m2 Same heat (Q) will be flowing through the contact. Hence, we have— Temperature drop across contanct (ATc)
Q
•••
Rcontact
ATc = Q X Rcontact = 58.2 x 104 x 0.88 x 104 = 51.2°C
21. A conical cylinder of length with radius `r1' and `r2' (where r2 > r1) is insulated along its outer surface. `T1' & `T2' are the temperatures maintained at two ends. If heat flow is along its axis only, calculate heat flow through the cylinder.
r2
Consider heat flow through distance 'x' from the end of the cylinder where radius is rx Now Hence,
rx = r +
r2 —
Ax = 7r ( ri +
r2 —
)2
70 Heat and Mass Transfer Therefore, heat flow is—
aT
Q = — kAx a x = —kir (n Qa x
or C
ri. I
L
or
Q ri+
or
Q [
x x )2 Xa T
ax
= —k x iv x )T
T2
1
1
x iL
= i rk (T 1 — T2)
L

o
1
ri
=
r2 — ri L
irk (Ti — T2)
1 1 = irk (Ti — T2) r2 11 Q•L=
or
aT
Ti
r2 — ri x r2 — 11
L
kxxf
)2 =
r2 — 11 ri. + L (
—Q x L r2 — ri
or
L
L X)2
0
1 r2 — n 7'2 L
r2 — ri
r2— ri.
Q • ax
or
+
ri r2
or
7rk (Ti — T2) irk (Ti — T2) ri • r2 L
Q=
22. A composite wall is made of different slabs as shown in the figure. If the heat flow is one dimensional, find the heat transfer per unit area. A
B
D
Ab = 1 m2 kb = 100 W/mk 600°C
A, = 2 m2 ka = 50 W/mk
Ad = 2 m2 kr) . 10 W/mk C
A, = 1 m2 k, = 25 W/mk H 10 cm—'H— 10 cm —.H— 10 cm—.1
100°C
Steady State Heat Conduction
The equivalent electric circuit of the composite wall is— Rb AAAr600 0—WV0
Ra
Now
100
0—AAN0 Rd —JV\Ar— Rc
Ra =
Xa = 0 .1 = 0.1 x 102 2 x 50 Aa X ka
Rb =
4 = 0.1 = 0.1 x 102 1X100 Ab X kb
Rc =
xc = 0.1 = 0.4 x 102 lx 25 cc A xk
Rd =
Xd = 0.1 = 0.5 x 102 2x10 Ad xkd
Equivalent resistance (Rbc) of Rb and Rc is1 = 1 + 1 Rb Rc Rbc 1
=
01)(102
+
1
0.4)(102
= 1000 + 250
or
Rbc =
1250
= 8 x 104
Now the electric circuit reduces to600
500
Ra
Equivalent resistance
Rbc
Rd
R = Ra + Rbc + Rd = 0.1 x 102 + 0.08 x 102 + 0.5 x 102
= 0.68 x 102 Now heat flow Q is— Q=
T1 —T2
R
= 600 —100 0.68 x 1012 = 500 x 102 0.68
7I
72
Heat and Mass Transfer
= 73.5 x 103 W. = 73.5 kW 23. A plate made of material of thermal conductivity 10 W/m°C is heated from bottom surface at constant rate such that the upper surface, which is exposed to the surroundings, is maintained at constant temperature of 250°C. The upper surface convects and radiates heat to the surroundings. The surroundings' temperature is 110°C. The convection coefficient and radiation factor are respectively 75 W/m2°C and unity. Calculate the temperature gradient between upper and lower surface of the state. (UPTU — 2004 — 5) QR + Qconv 7:c =110°0
F12 =1
'Ok
h = 75
1'
T,,,, = 250°C
t
_1._ k = 10
I Heat
Heat flow through conduction is the sum of convection and radiation heat transfer Q con = Qconv + QR
To j = 75 x 1 x (250 — 110) = 75 x 140 = 10.5 kW QR = A • Fi2 • a(T: — TM = 1 x 1 x 5.67 x 108(5224 — 3834) = 5.67 (742.5 — 215.2) QR = 5.67 x 527.3 = 2.99 kW Qconv + QR = 10.5 + 2.99 = 13.49 kW aT Qcond = — kA ax Qconv = h • A • (Tw —
Now
= —10 x 1 x dT = 13.49 x 103 dx or
dT = —1349°C/m dx = —1.35°C/mm
Steady State Heat Conduction
73
24. A square plate heater of surface area 200 cm2 is inserted between two slabs of same surface area. SlabI is 1.0 cm thick with 'lc' = 50 W/m°C and slabII is 2.0 cm thick with 'lc' = 100 W/m°C. If outside heat transfer coefficients on side I and II are 50 and 100 W/m2°C respectively and surroundings' temperature of air is 30°C, find maximum temperature of the system. The rating of heater = 1 kW.
Suppose Tna, is temperature at heater. Q = heat through slabI + heat through slabII =
T — T_ T — T_ + L1 + 1 L2 1 ki k Iii Al k2A2 h2 A2
= (TLi. x—T1
1X10 2
50 x 200 x104 +
1 +
1
50 x200 x104
1 2 x 102 1 + 100 x 200 x 104 100 x 200 x 104
1 1 T — 30) [ —., + —(m 1 x 10 +1 0.5 x 102 + 0.5
1000 = (Tmax  30) [ 1 + 1 ] 1.01 0.505 = (Tmax  30) (0.99 + 1.98) or
Tmax  30 = 1000 336.7 2.97 Tmax = 366.7°C.
25. A refrigerator having inside dimension of 0.5 m x 0.5 m and 1.0 m height is maintained at 6°C. The walls of the refrigerator are constructed from two mild steel
74
Heat and Mass Transfer
sheets 3 mm thick (k = 46.5 W/m°C) with 50 mm glass wool insulation (k = 0.460 W/ m°C) between them. The average heat transfer coefficient at the inner and outer surface are 11.6 and 14.5 W/m2°C respectively. The surroundings' temperature is 25°C. Find the rate at which heat must be removed from the interior, and the temperature at outer surface of the refrigerator. (UPTU — 2004 — 5) kw = 0.046 W/m °C
ks = 46.5 W/m °C T,= 25 °C ks
= 6 °C
ho = 14.5 hi = 11.6 3 m+— 50 m —+ 3 md T, = 25°
T,= 6°C
TS 1 hoA
LW
Ls ksA
kA
Ls keA
1 h jA
The refrigerator has four walls through which heat is coming in. Hence, area is Area = 2(LxW+Lxh+hxW) = 2(0.5 x 0.5 + 0.5 x 1.0 + 1 x 0.5) = 2(0.25 + 1) = 2.5 m2 The thermal resistance is— 1 + Ls L„, + Ls + 1 ho A ks A k„,A lc, A h1 A 1[1 A 14.5 =
2 x3x103 50 x103 + 1 11.6 46.5 0.046
[6.8 x 102 + 0.013 + 1.087 + 8.6 x 102] 2.5 = 2.1 x 1.254 = 0.5 °C/W 5 The heat transfer isQ=
To — ER
Steady State Heat Conduction
256
75
19 0.5
0.5 = 38 W
Now if Ts is surface temperature, then heat transfer is— Q = hoA(To — Ts) 38 = 14.5 x 2.5(25 — Ts) 38 25 — Ts = = 1.05 2.5 x 14.5 Ts = 25 — 1.05 = 23.95°C
or or
26. A wall is 30 cm thick and area of 15 sq. m is made of bricks (k = 0.35 W/m°C) having plasters of both sides of thickness 2 cm (k = 0.6 W/m°C). The rate of heat flow is 1.8 kW and temperature at inside and outside of wall are 140°C and 40°C. A glass window is provided for observation from outside to inside, find the area of glass window (kg = 1 W m° C). Brick Plaster
Plaster
To = 20°C
= 100°C Glass window
0.02
0.02
The equivalent electric circuit is— To = 20
= 100
R„ = = 0.02 = 2.22 x 103 0.6 x 15 k p A ' LP
Rb
=
4 kb • A
=
0.3 = 57 x 103 0.35 x 15
76
Heat and Mass Transfer
R
g

Lg
0.34 1 x Ag
kg x Ag
1
1
Rtotai
2Rp + Rb
+
0.34 Ag
1 Rg
Ag 1 + 2 x 2.22 x 103 + 57 x 10— 0.34 = 16.27 + 2.94 Ag
As Now or or or or
 To Q Krotai 100 10 = 1800 Ribtai 90(16.27 + 2.94 x = 1800 1464.3 + 264.6 x Ag = 1800 264.6 x Ag = 335.7 Ag = 1.27 m2
27. Find heat transfer through a composite cylinder. Or
Derive an expression for the steady state overall heat transfer coefficient for a composite hollow cylinder whose inner surface is exposed to hot fluid and outside surface is exposed to a cold fluid. (UPTU  2009  10) A composite cylinder consists of one inner hollow cylinder and one or more outer hollow cylinders. Inner hollow cylinder has radius ri and r1, and outer hollow cylinder has radius r1 and ro. In case, the material of the inner and 'outer cylinder has conductivity of k1 and k2, then total thermal resistance can be found out.
T,
To
1 hp4,
log
log
ri 2irk1L
2jrk2L
r.
1 h0A,
Steady State Heat Conduction
Q
Q=
77
Temperature difference Thermal resistance — To r r0 +log ri rt 1 + log + 1 hi Ai 27rk1 . 2irk2L ho Ao
28. Find heat transfer through a composite sphere. Consider a composite sphere consisting of inner and outer hollow spheres having conductivities of materials as k1 and k2 as shown in the figure.
To
Equivalent electrical network of the thermal system is— To
Ti
1
ho • 4 ird Q
1 r0 — r1r1— k2 ro n 4 g k1 ri • ri hi 4 nri2 Temperature difference Thermal resistance To — T — 1 + + ho • 4nro2 4nk2ro • ri
—
+
1 hrr r2
29. A 50 cm diameter pipeline in the Arctic carries hot oil at 30°C and it is exposed to surroundings temperature of —20°C. A special powder insolation 50 mm thick having a thermal conductivity of 7 x 103 W/mK surrounds pipe. The film heat transfer coefficient on the outside of the pipe is 12 W/m2K. Calculate the rate of heat loss from the pipe per meter of length, neglecting thermal resistance of the pipeline. (UPTU — 2002 — 03)
78
Heat and Mass Transfer
ri = 25 cm To = —20°C f
ro =25 +5 =30 cm
k— 7 x
le W/mk
2 h = 12 W/m k
The equivalent thermal resistance To
Ti
log r°
hAo
2irkL
Q=
—T° r log 0 1 I + 27aL hx2nro • L 30 — (20) log 0.3 1 0.25 + x1 12x 27rx0.3x1 27rx7x10—
27rx 50 = 26.05 + 0.28 = 11.92 W/m 30. A standard iron pipe having 5 cm inner diameter and 2.5 mm of thickness is insulated with magnesium insulation (k = 0.02 W/m°C). Temperatures at the interface between the pipe and the insulation is 300°C. The permissible heat loss through the pipe is 600 W/m and the temperature of the outer surface of the insulation is not allowed to exceed 100°C. If the thermal conductivity of the pipe material is 20 W/m°C. Calculate the minimum thickness of insulation required and the temperature of inside surface of the pipe. (UPTU — 2004 — 5) = 20 W/m°C
2.5 cm 2.75 cm k2 — 0.02 W/m°C
Steady State Heat Conduction
79
The equivalent thermal resistance circuit is— T2 = 300°C
Q
T3 = 100°C
r2 log — ri
r3 log — r2
2jrk1 L
27dc2L
_ Temperature Difference Thermal Resistance
600 =
or
600 =
or
600 =
Ti — T2 =
T2 — T3
r log —2 r1
P3 log — r2
27rk1 L
27rk2 L
T1 — 300 300 —100 2.75 = log /.3 2.5 log 2.75 27rx10 xl 27rx 0.02 xL 200 x 2 x x 0.02 x 1 r log 2.75
or
log 2r335
8x 600 = 0.04186
r3 = e 0.04186 = 1.0427 2.75 or r3 = 23 .867r2 — .. Thickness of the insulation = r = 2.867 — 2.75 = 0.117 cm or
Now
600 =
300 log2.75 2.5 2 irki L
T1 —
600 x log 2.75 2.5 2n x 20 x 1 T1 — 300 = 0454 T1 = 300.454°C T1 — 300 —
or or
31. For the situation below, what would happen to the average temperature at face 'C' if thermal conductivity of the solidII was increased?
80
Heat and Mass Transfer Insulated
Face 'B' Solid II
Face 'A Solid
T = 20°C
q0
Solid Ill Face 'C' /////////////////// Insulated
(a) No change (c) increase
(b) becomes 20°C (d) decrease (GRE)
The equivalent circuit is— Rb TB
x2 k ..2A2
To 0/VVV0 Xi
1
V—
ki Ai
Ta = 20°C
Tc
hA1
x2 k3A3
1
1
ER= KL x + i iii
1
A
+
1
+ 1 hili
x2/ k2 ri2 X2 /k3 A3
When k2 increases, it is apparent that ER will decrease. Now AT Q= ER Hence, Q will increase Now Q is also Q
— T — 10 e1 hA1
Since Q has increased, Te will also increase. Hence, option (c) is correct. 32. The temperature drop through a twolayer furnace wall is shown in the figure. Assume that the external temperatures T1 and T3 are maintained constant and that T1 > T3. If the thickness of the layers x1 and x2 are the same, which one of the following statement is correct? (A) k1 > k2
(b) k1 < k2
(c) k1 = k2 (GRE and GATE — 1997)
Steady State Heat Conduction
Material I k1
8I
Material II k2
H xi
Equivalent thermal circuit is—
x k1A Q.
Q
x k2A
Ti — T2 . T3 — T2 x x ki A k2 A
. Ti  T2 .c T3 — T2 x ki A
(as x1 = x2 = x)
x k2 A
( T1— T T —T but 2 = slope and 3 3 = slope l x x
Slope of T1 — T2 line = Slope of T3 — T2 line 1 1 k1 k2 or
Slope of (T1 — T2 ) line = k2 Slope of (T3 — T2 ) line k1
It is apparent that the slope of (T1 — T2) line > the slope of (T3 — T2) line. Hence, k2 > ki Hence, option (b) is correct. 33. A 20 cm dia, 1.2 m long cylinder loses heat from its periphery surface by convection. Surface temperature is constant at 100°C and the field temperature is constant at 20°C. The average convection heat transfer coefficient over the surface of the cylinder is 20W/m2K. The heat transfer. (a) 120 ir W (c) 320 ir W
(b) 240K W (d) 480K W (IES — 1989)
82 Heat and Mass Transfer
10 cm
T. = 20°C h = 25 W/m2k = 100°C
Heat transfer through convection at surface 100 — 20 _ T1 — 1: — Q— 1 1 hxA 25 x 2 x x 0.1 x 1.2 = 80 x 25 x 2ir x 0.1 x 1.2 = 480x W Option (d) is correct.
34. A composite slab has two layers of different materials with thermal conductivity `k1' and `k2'. If each layer has same thickness, the equivalent thermal conductivity of the slab would be— (a) kJ. k2
(b) k1 + k2
(c) (k1 + k2) (k1 k2)
(d)
2 ki k2
ki + k2 (IES)
The slabs can be shown as below:
k1
k
k2
H x 4 x I x ki
F'— 2x —d
x k2
2x k
x x ER1 = + k1 k 2 =
ki + k2 ki .k2
ERi =ER2 ki + k2 = 2 ki •k2
k
ER2 = x
2x k
Steady State Heat Conduction
or
k—
83
2 k1• k2 ki+ k2
Option (d) is correct. 35. A hollow stainless steel spherical container having 20 cm inside diameter and 30 cm outer diameter (with k = 50 W/mK) holds certain liquid. Its inner and outer surfaces are at 250 and 100°C. Find radial heat flow and temperature at r = 12.5 cm from the centre. Heat transfer is— T — To ro — ri 4gro ri x 250 —100 0.15 — 0.10 47r x 0.15x 0.10 x 50 = 28.26 kW
Q
r,  10 cm r,,,= 15 cm r  12.5 cm
To = 100°C
Now T
—
250 100 — 250 T—
(r — ri ) fro — ri)
0.15 (0.125 — 0.10) 0.125 (0.15 — 0.10)
T — 250 = —150 x 0.15 x 0.025 0.125 x 0.05 = 250 — 90 = 160°C 36. A hollow cylinder has inside radius 2.5 cm and outside radius 5 cm. Inside temperature is 300°C and outside temperature is 110°C. Find the temperature at
84
Heat and Mass Transfer
3.75 cm from centre if k = 70 W/mk. Also find heat flow through the cylinder per length. Heat transfer is— Q=
T —To ivb —
2irkL 300 —100 27r x 70 x 1 = 126.8 kW/m Temperature distribution is— r T—T To —
1,15
1
T —300 100 — 300
3.75 "'5
2.5 3.75
or
T — 300 = —200 x "'5
or
2.5
= —117 T = 300 — 117 = 183°C
37. A hollow sphere is made up of two materials. First one with k1 = 70 W/mk with ID = 10 cm and OD = 30 cm. The second sphere has k2 = 15 W/mk and with OD = 40 cm. Inside temperature is 300°C and outside temperature is 30°C. Find the rate of heat flow through the sphere.
Steady State Heat Conduction
85
The equivalent thermal circuit is— = 300°C
T3 = 30°C r2 —
r3 — r2
47rk1rir2
Q=
4,rk2r2r3
Tl — T3
r3 — r2
r2 —
47rkirir2
41rk2r2 r3
300 — 30 0.15 — 0.05 0.2 — 0.15 + 47r x 70 x 0.15 x 0.05 47r x 15 x 0.2 x 0.15 = 11.24 kW 38. What do you understand from critical thickness of insulation? Or
Derive an expression for critical radius of insulation for cylindrical body. Give practical example to explain the concept of critical radius. (UPTU — 2008 — 9) In case of plane walls, the heat transfer rate decreases as thickness of the wall increases since thermal resistance R is equal to the thickness of the wall divided by kA i.e. thermal resistance keeps on increasing. This is not true always for pipes and spheres with the addition of insulation as in their cases, thermal resistance firstly decreases and after certain extent of addition only thermal resistance increases. The thermal resistance of pipes and spheres therefore firstly decreases with the addition of insulation and heat transfer reaches to a maximum value at certain value of insulation thickness. The outer radius with insulation when heat transfer is maximum is called critical radius of insulation. If further insulation is added after the critical radius the thermal resistance increases resulting in the decrease of heat transfer. The adding of insulation after critical radius is undesirable as heat transfer from electric wire to surroundings will decrease, resulting in heating and damaging of wire.
86
Heat and Mass Transfer
Consider a cylinder of length with inner radius `r;' and outer radius with `le as thermal conductivity. The thermal resistance consists of conductive and convective resistance. log L. ER —
+
1
2 xkL h • (27r r • L)
Now as insulation is increased, the radius increases which increases conductive resistance but it decreases convective resistance. Initially the net result is the decrease of total thermal resistance with the addition of insulation as the rate of increase of conductive resistance is less than the decrease of convective resistance due to larger surface area. After reaching certain radius rc, the increase of conductive resistance is higher than the decrease of convective resistance. This radius is called critical radius and net result is the decrease of total thermal resistance and the increase of heat transfer when insulation is added beyond rc.
n rc
Radius of insulation
39. Where is this principle of the critical thickness of insulation used? The principle of the critical thickness of insulation is used in electrical industry as we want to provide such a thickness of insulation to wires and cables carrying current which permits an increase in the heat transfer from these currents carrying conductors but at the same time insures adequate insulation. The temperature of the conductors does not increase with increased heat transfer when provided with critical thickness of insulation and they remain safe otherwise the insulation of the conductors will burst whenever heat transfer is less and temperature rises in the conductors. 40. Derive an expression for the critical radius of insulation for a cylinder. (UPTU —2003)
Steady State Heat Conduction
87
Consider a cylinder with radius 'ri' and it has insulation upto radius 'r' as shown in the figure. The thermal conductivity is `le and heat transfer coefficient is 'h'. The equivalent electric circuit of the heat transfer is—
log f:
1
h(2nrL) 2irkL
Total thermal resistance is log r ER — Differentiate ER and put
+
1 h(27rrL)
asR = 0 in order to find minimum total resistance. a
a or
DrkL
ri 1 r 1 ar 2irkL h(2irrL)r2
ER
r=
o
k
This radius is called critical radius of insulation re. k — h 41. Derive an expression for critical radius of insulation for a sphere. (UPTU — 2007 — 8)
Consider a sphere with radius `ri, and outside radius 'r' with insulation. The thermal conductivity is `le and heat transfer coefficient is 'h' with the surroundings. The equivalent electrical circuit for the heat transfer is—
88
Heat and Mass Transfer
r — ri
1
4irkr x
h(47u)
The total thermal resistance isER =
azR r
= 0 for minimum thermal resistance
azR _ ar or
r— 1 47rk rxri h(47rr2 )
rc =
+ri
2 4gkr2 xri 4ghr3
0
2k h
42. It is desired to increase the heat dissipation rate over the surface of an electronic device of spherical shape of 5 mm radius exposed to convection surroundings with h = 10 W/m2K by encasing it in a spherical sheath of conductivity 0.04 W/mK. For maximum heat flow, the diameter of the sheath should be (a) 18 mm (c) 12 mm
(b) 16 mm (d) 8 mm (IES — 1996)
For a sphere, critical insulation is— rc =
2k h
2 x 0.04 10 = 0.008 m = 8 mm Option (d) is correct. 43. A cable of 10 mm diameter is insulated with a material for which k = 0.2 W/mk and it is exposed to surroundings at 20°C. Find the critical insulation if heat transfer coefficient is 9.3 W/m2K. _k _ c h 0.2 9.3 = 0.021 m = 21 mm
r
Steady State Heat Conduction
89
44. A pipe covered with insulating material has inner diameter 0.1 m and outer diameter 0.11 m. Thermal conductivity of insulating material is 1.0W/mK. The temperature of the fluid inside the pipe is 100°C and surroundings temperature is 20°C. Find the critical radius of insulation and heat loss per metre length of pipe for critical radius of insulation. Take 'h' = 8W Im2K. The critical radius is— 1 r =— k = — = 0.125 m h 8 Equivalent thermal circuit is—
log
1
h(2nr,L)
2rckL
Q=
rc 1 ri + 27rkL 27rrc hL log
100 — 20 0.125 log 1 0.05 + 27rx1x1 27rx 0.125x8x1 = 620 W/m 45. Two insulating materials of thermal conductivity 'Ic' and '2 k' are available for lagging a pipe carrying a hot fluid. If the radial thickness of each material is the same, then— (a) Material with higher thermal conductivity should be used for the inner layer and one with lower thermal conductivity for the outer. (b) Materials with lower thermal conductivity should be used for the inner layer and one with high thermal conductivity for the outer. (c) It is immaterial in which sequence the insulating materials are used. (d) It is not possible to judge unless numerical values of materials are given. (GATE — 1994) The critical radius— (a) first material
(rci) =
(b) second material
2k (rc2) = h
90
Heat and Mass Transfer
It is apparent that
rc2 = 2rci Hence, the first material is to be used inside and second material is to be used for outside as it has bigger critical radius. Option (b) is correct. 46. An electric wire of 2 mm diameter is given a 25 mm thick insulation (k = 0.5 W/m°C). The surroundings has temperature of 25°C and heat transfer coefficient h = 10 W/ m2°C. The wire has surface temperature of 120°C. Compare the heat dissipation with and without insulation. Find the thickness of insulation when heat dissipation is maximum and find its value.
The equivalent electric circuit is— Ti
log
1 h(2772L)
J_ 27d 1.5 D) 7T2
1
i 0±
0 L
2.
2irL
S
log (t
1
Vertical isothermal cylinder of length L buried in a semiinfinite medium (L » D)
S
2,rL
49 log (ff
ItDH Contd.
Steady State Heat Conduction
3.
123
Large plain wall
S=L
,11_4.
Disk buried paralld to the surface in a semiinfinite medium (z » D) rT2
S = 4D or
5.
S = 2D if z = 0
Isothermal sphere buried in a semiinfinite medium
Iz .4.
S= 1
6.
2,W 1 — 0.25 x
D
z
Isothermal sphere buriad in a semiinfinite medium (surface insulated and medium at T2)
Conk!.
I 24
Heat and Mass Transfer
7.1
S=
2irD 1+ 0.25 x D
z
7.
The edge of two adjoining walls of equal thickness
S = 0.54 W
8.
Corner of three walls of equal thickness
S = 0.15 L
76. What is finite difference method? It is a numerical analysis method which is used to obtain approximate solution in multidimensional steady state condition in a body. Numerical methods are preferred over analytical methods as: (i) useful solution is difficult to be obtained in multidirectional steady state condition with complicated boundary conditions, and (ii) numerical methods are well suited to be used with high spreed digital computers. In the analytical solution, the determination of temperature is made at any point of interest in a medium, whereas in a numerical solution the determination of temperature is possible at discrete point. 77. How is Laplace equation for two dimensional heat conduction approximated to the finite difference forms or equations?
Steady State Heat Conduction
I 25
Or
Explain: (i) node, (ii) nodal network, and (iii) finite difference form for 2D heat conduction. In numerical method, the temperatures are determined at discrete points. To select the discrete prints, the body is subdivided into a number of small regions and a reference point to each is assigned at its centre. The reference point is termed as node or nodal point. The aggregate of nodal points is called as nodal network or mesh. The body with nodes and nodal network is as shown in the figure. Consider five modes as shown in the figure which are assigned reference numbers with respect to the centre node which is assigned as (i, j). Material between nodal points may be replaced with fictitious rods having the same conductivity (k). Now 2 D Laplace equation isj
+1
I
AMENEUEllb• Ay j
+ 1, j
MENNII=EMMI gNIENIMMENF valIMMENEr "1411111MIP"
i
— 1, j
j
j
Two dimensional nodal network and finite difference approximation
a2 T
a2T ax2
But
and
=0
CaT ax),+
Ax
2'
a TI
J — T1, j
_
Ax
C ax iv2,j
Now
ax)i+1.;
—
2'
l.ax2 Ca )0
Ax Ti
,j +
 , j 2 ,j
(Ax) 2 r
Similarly
a2 T' ay2
.aY)i,j+1/2
Ay
a T)
ay )i,J _1 2
I 26
Heat and Mass Transfer
Ti,j+1+
 2T, 2
(Ay) If Ax = Ay, then Laplace e uation from equation (i) and (ii) — a 2 T a2 T _ ax2 or
Ti+1,
Ti, j+1
ay — 4
Ti , j_i 
Ti, =
0 0.
Hence, each node represents a certain region and its temperature is a measure of the average temperature of the region i.e, the temperature of node (i, j). The above equation is the approximate algebraic equation in finite difference form of the exact Laplace differential equation in two dimensional. This approximate finite difference form of the heat conduction equation may be applied to any interior node which is equidistant from its four neighbouring nodes. This rquation therefore reveals that the sum of the temperatures associated with the neighbouring nodes should be equal to four times the temperature of the node of interest. In other words, the algebraic sum of all heat flowing to a point must be equal to zero. (Eq = 0) 78. What are the nodal finite difference equations for: (i) interior node, (ii) node at an internal corner with convection, (iii) node at plane surface with convection, (iv) node at an external corner with convection, and (v) node at insulated boundary. The nodal finite difference equations are as per Table given below. Nodal finite difference equations Finite difference equation (Ax = Ay = Az)
Configuration 4 •
Interior Node Qio + Q20 + Q30 + Q40 = 0
.0
•3
[k(T, — To ) + h(T2 —
+ k(T3 — TO+ k (T4 —To)]
Alxb Al
or T1 + T2 + T3 + T — 4 To = 0 Node at an internal corner with convection 2
kx
A/xb Al
4
(T3 — 0
h1,T. 2
Mxb
( T4 — To) + k x
Al
+ kx
Mxb 2x A/
A/xb ( T2 — To) + hx Aix b(T_ — To ) = 0 2M
•3 2(
•
( T1 — To) + k
+ T4) +(T2 + T3) +
2.h•A/
2(3+
h•
)To = 0
—0
Steady State Heat Conduction
I 27
Node at plane surface with convection
4 •
kx 0
1•
h1 T,
Alxb
Al
(T1 — To) + k x
2,6,/
(T2— To)+ k
A/ b
2,6,/
(T4 — To) + h x Al x b(T_ — To) = 0 (271 + T2 + T4 ) +
•
Aixb
2.h.A1To. (h•Al 2 +2) To = 0 k k
2 Node at an external corner with convection
1 —0 hi Ta 2
kx
Alxb
EIxb
(Ti To) + k x (T2 — To) + h x Al x b 2A/ 201 x (T,, — To) = 0 ( Ti + T2)
+2hAx
k
x 70, — 2 (
h. AI
k
+1) To = 0
Insulated Node at insulated boundary Aix b Aix b Aix b kx x (71 To) + k x (T2 — To) + k x 20 / A/ 20/
(T3 — To) = 0 Ti + 2 72 + T3 — 4T0 = 0
79. How can finite difference equations be solved? The finite difference equations can be solved by— (a) The relaxation method (b) the GaussSeidel iteration method (c) the matrix inversion method 80. Describe relaxation method of solving a system of finite difference equations. The steps involved are— (a) Stepl. Sundivide the system into a number of small regions and assign a reference number to each node. (b) Step2. Assume temperatures for all nodes. (c) Step3. Using the assumed temperatures, calculate the residuals at each node. (d) Step4. Release the largest residual to zero by changing the corresponding nodal temperature by an appropriate amount. (e) Step5. Change the residuals of the surrounding nodes to correspond with temperature change in step4. (f) Continue to relax residuals and obtain temperatures until all residuals are close to zero.
I 28
Heat and Mass Transfer
81. The boundary temperatures of a thin plate are as shown in the Figure. Ditermine the temperature at the centre of the plate. y/ 4
200 °C 400 °C
T
2 0 =— = 1 2 100 °C
I
100 °C I 1.—w= 2 m—.I
Now or
•
H= 2 m
1
0
3
2 +Az 1
x
4x To = l'i + T2 ÷ T3 ÷ Tet as Ax = Ay =1 To = 400 +100 +100 + 200 4 = 200°C
82. The temperature distribution and boundary condition in various parts of a solid is as shown below. Determine the temperature at nodes marked as A, B and C. Also find the heat convected over the surface exposed to convection. Assume k = 1.5 W/mK and h = 500 W/m2°C. Insulated T. = 30 °C Convection face h= 500 W/m2°C
For node 'A', we have TA =
T1+ T2+ T3+ T4
4
= 172.9 + 200 +132.8+137 4 = 160.7°C For node 'if , we have TB =
T1 + 2 x T2+ T3 4
Steady State Heat Conduction
I 29
129.4 + 2 x 103.5 + 45.8 4 = 95.6°C.
= For node 'C', we have
hxAx 1 xl: (27'1+1'2 + TO + 7, _ 2 , hx Ax k 1
=2
+2
(2 x 103.5 + 67 + 45.8) +
500 5 0.1 x 30 1
500 x 0.1 +2 1.5 = 37.4°C. Now the heat convected out by the exposed surface is— Qconv
=hxAxAT = (Az x Ay) x E(T — To ) but Az = thickeness = 1 = 500 x 1 x 0.1 x [(45.8 — 30) + (37.4 — 30) + (67 — 30) + 2 x (200 — 30)] = 7258 W
83. The figure as shown below indicates temperatures in a region of a solid with boundary conditions. Find: (i) thermal conductivity, and (ii) the heat flow from surface 1. (Annamalai University — Nov 2005, May 2006) Insulated
1 ,— 0.1 m—).
Dr„,
Ta = 300 °C, h = 10 Whn
Surface 1
(Q)surfacel = h • A • I AT 1 = h • (Az x Ax) x [— x (500 — 300) + (356 — 300) + (337 — 300)1 2
Also
= 10 x 0.1 x [100 + 56 + 37) as Az = 1 = 193 W Q= k • A • EAT
I 30
Heat and Mass Transfer
193 = k • (Ax x Az)[(TE — TF) + (TA — TD) + (TB — Tc)] = k x 0.1 x 1 x [(500 — 500) + (455 — 356) + (454 337)] = k x 0.1 x (97 + 117) or
193 0.1 x 196 = 9.85 W/mK
k=
84. The boundary temperatures of a thin plate are shown in the Figure. Determine the temperature at the centre of the plate. (Annamalai University — April 2007) YA •4
200 °C
t
Ay 400 °C
100 °C
•
4 = 1.5 m
1
100 °C
X
2
Here
Ax # Ay
and
2 Ax = — = lm and Ay = 15 = 0.75 m 2 2
Now
To =
3
•
3 + T4 when Ax = Ay 4T
T1 + T2 +
(T1+ T3) To =
•1
—Ax
v=2m1
and
0
Ay Ax + (T2+ T4) Ax Ay when Ax # Ay 4
1 0.5 (400 + 100) + 0. 7 (100 + 200 1 5 4 0.75 x 500 +
1 x 300 0.75
4 = 375+400 = 775 4 4 = 193.75°C
Steady State Heat Conduction
I3I
85. Consider two dimensional steady state heat conduction in a square region of side 12 subjected to the boundary conditions as shown in the Figure. A
Y
T= 400 °C
L
P 0 0 N
T2.
•
T3.
1
0
T4
T=600 °C
T= 300 °C
L Calculate Ti., T2, T3 and Ti considering Ax = Ay = — . Calculate the heat transfer rate 3 through the boundary surface at x = L per 1 m length perpendicular to the plane of figure for L = 0.1 m, k = 20 W/mK. (Annamalai University — May 2004)
For node T1, we have— T1 =
T2 + T4 + 600 + 400 4
...(i)
T2 =
200 + T3 + T1 + 400 4
•• •00
T3 =
200 + 800 + T4 + T2 4
•• .(iii)
For node T2, we have—
For node 7'3, we have—
For node T4, we have— T3 + 800 + 600 + Ti 4 Rearranging the equations (i), (ii), (iii) and (iv), we haveT4 =
1000+ T2 + T4 — 4 T1 = 0 600 + T3 + T1 — 4 T2 = 0 1000 + T2 + T4 — 4 T3 = 0 1400 + T1 + T3 — 4 T4 = 0 Now using the relaxation method—
. . .(iv)
I 32
Heat and Mass Transfer
S.No. Step 1. Initial assumed value 2. change AT2 = —50 3. change AT4 = +50
Ti 500 500 500
R1
0 —50 0
T2 450 400 400
7'3 500 500 500
R2
—200 0 0
R3
0 —50 0
T4 550 550 600
R4
200 +200 0
As all residuals are zero on assigned temperatures, hence temperatures are— Ti = 500°C T2 = 400°C T3 = 500°C T4 = 600°C Now heat transfer rate at x = L is Q=k•b•Ax•
dT L but Ax = Ay = and b = 1 dy 3
= 20 x dT = 20 x [(T3 — 800) + (T4 — 800)] = 20 x [(500 — 800) + (600 — 800)] = 20 x [300 — 200] = —10 kW 86. Steady twodimensional heat conduction takes place in the body as shown in the figure below. The normal temperature gradients over surface P and Q can be considered to be uniform. The temperature gradient
at surface Q is equal to 10 ax ax K/m. Surfaces P and Q are maintained at constant temperatures as shown in the figure, while the remaining part of the boundary is insulated. The body has a constant conductivity of 0.1 W/m • K. The values of
(a) (c)
ay
and aT — at surface P are: ax
ax ax
aT = 0 Kim = 20 K/m, ay
(b) aT 20 Kim, aT = 10 Kim
ax
= 10 K/m, aT = 10 K/m ay
(d)
ay
ax
ax ax
= 0 K/m, aT = 20 K/m ay
Y Surface Q at 0°C
Surface Pat 100°C X
Figure
Steady State Heat Conduction
I 33
The surface P is horizontal (Parallel to xaxis). Surface P has uniforms temperature. Hence
(
T
ax
0 at surfaceP.Now the surface Q has only temperature gradient —
ax),
p
10 K/m. Now balance of heat is— (Q)P = ( Q)Q
or
or
k • Ap
(1
aT
x ay
k•A x ( a x)Q Q
p
X b) x (aT) p ( aT ay 1
(2 x b) x 10
= 20 K/m
The option (d) is correct. 87. A long pipe of 0.6 m outside diameter is buried in earth with it axis at a depth of 1.8 m. The surface temperature of pipe and earth are 95°C and 25°C respectively. Determine the loss of heat per unit length of the pipe. Assume conductivity of earth = 0.51 W/mK. / / / / / / / / / / / / / /
Earth Z = 1.8m D = 0.6 m
Pipe buried in earth
= k • S (T
pipe
where
Tearth)
S = shape factor as given in Answer 75.
2ir x L 7 log 41 D where D = diameter and Z = depth 2.7rx1
4 x 1.8 0.6 = 2.53 log
I 34
Heat and Mass Transfer
2 = 0.51 x 2.53 (95 — 25) L
= 90.8 W/m 88. A sphere (diameter = 1.6 m) is buried in earth at a depth = 5.5 m. Heat in sphere is generated at a rate of 580 W. The earth conductivity = 0.51 W/mK and earth surface is at 6°C. Determine the temperature of the sphere under steady state condition.
Earth
f Z = 5.5 m
D = 1.6 m
Sphere buried in earth
Shape factor
S=
2irD
D
1— 0 . 25.— z
Now
= 2x7rx1.6 1.6 10.25x — 55 = 10.8 m / Q = k • S • (Tsphere Tearth) 580 = 0.51 x 10.8 (Tsphere — 6) Tsphere
580 6 + 051 x 10.8 = 6 + 104.9 = 110.9°C. _
Chapter
4
HEAT TRANSFER FROM EXTENDED SURFACES
KEYWORDS AND TOPICS A A A A
FINS LONG FINS INSULATED TIP FINS FINITE LENGTH FINS
A A A A
FIN EI,PECTIVENESS EFFICIENCY OF FIN ERROR IN MEASUREMENT HYPERBOLIC FUNCTIONS
INTRODUCTION The rapid heat dissipation from a surface to surroundings is required in many applications. The heat dissipation depends upon: (i) convective heat transfer coefficient (h), (ii) the temperature difference between surface and surroundings (ts — L), and (iii) surface area (A). The surface area exposed to the surroundings is frequently increased by the attachment of protrusions to a surface. This method of increasing surface area provides a easier method of increasing heat transfer rate from a surface. The protrusions are called fins. The fins can be rectangular, circular, trapezoidal and parabolic. The selection of a fin for a particular application depends upon factors which will give: (i) maximum efficiency, (ii) minimum material for cost, weight and space consideration, (iii) minimum resistance to the heat flow, (iv) adequate strength, and (v) ease of manufacturing. The heat dissipation from extended surface is found out using onedimensional heat flow method. 1. What are fins? Why are they used? Fins are basically extended surfaces. They are used to increase the heat transfer rate from a surface to the adjoining fluid. The rate of heat transfer increases as the effective surface area of the surface increases with the attachment of fins.
I 36
Heat and Mass Transfer
2. How does heat transfer take place from a fin? The heat transfer takes place from the fin to the adjoining fluid by conduction and convection. 3. Mention different types of fins? Fins are of two types: (a) fins of uniform crosssectional area, they can have rectangular or circular profiles (b) fins of nonuniform crosssectional area: they have tapered shapes with crosssection as rectangular or circular.
Fins of uniform crosssection
Fins of nonuniform crosssection
Annular fin Fins of uniform crosssection
4. What are various applications of fins in industry? Fins are generally used in industry in following applications: 1. Cooling of IC engines
Splines
Heat Transfer from Extended Surfaces
2. 3. 4. 5.
I 37
Cooling of electric motors Cooling of transformers Cooling of refrigerators Cooling of electronic components.
5. Derive an expression in onedimensional differential equation of temperature for a fin of uniform crosssection. For the analysis of heat flow through the fin, following assumptions are made: 1. 2. 3. 4. 5. 6. 7.
Steady state heat conduction along xaxis No heat generation in the fin Uniform heat transfer coefficient over the entire surface of the fin The material of the fin is homogenous and isotropic. The fin and base material has no contact resistance. Negligible radiation from the fin. The fin temperature at base is uniform (T0).
Fin
Base
Fin of uniform crosssection
Consider an elemental crosssection `dx' thick at distance 'x' from the base of the bin. On basis of heat balance for this element, we get: Conduction heat entering at 'x' = Conduction heat leaving at `x + dx' + Heat leaving by convection Qx =
Qx+dx + Q.,
But heat leaving at 'x' is aT Qx = —kA— x Heat leaving at x + dx is Qx+dx = —kA
aT ax
a
+ ax(Qx)
I 38
Heat and Mass Transfer
= kA kA— aT + aT (kAA) • dx (kAA) dx ax ax ax = kA aT ax
kA a2T dx ax2
Also convective heat transfer is Qconv =
h (P •ddx)(T x)(T 
where P = perimeter, Ta = temperature of surroundings On substituting these values, we get T kA" d xx  h(P h(P• dx)(T ax2 a2T m` (T ax2 where Now take
=0 =0
(i)
m = hP kA a 29 a2 7, 0 = T  Tc, and = ax2 ax2
Hence, equation (i) becomes
a 29
m2 0 0 ax2 Equation (ii) is a differential equation of second order which has solution as
0 = cie" + c2emx Two boundary conditions are required for solving c1 and c2. 6. What are different conditions of fins which can yield a set of two boundary cases for solving temperature distribution equation for the fins? There can be three conditions of the fins: (a) Fin is infinitely long resulting temperature at end of the fin is same as that of the surrounding fluid. The boundary conditions possible with such fins are: (i) At x = cc, T=T i.e. 0 = 0 (ii) At x = 0, T = To i.e. 0 = 190 where To is temperature at base. (b) Fin has its end insulated. The boundary conditions possible with such fins are: (i) a T = o at x = L a T = To at x = 0 (c) Fin has finite length. The boundary conditions possible with such fins are:
Heat Transfer from Extended Surfaces
I 39
.
aT = L, k— h(T  71,0 ax (ii) At x = 0, T = To i.e. 0 = 00 (i) At
x
7. Derive expressions for temperature distribution and rate of heat flow for infinitely long fin. Also explain why fins are made of high conducting metals. The general equation for temperature distribution for a fin is 0 = c le" + c2emx Now boundary conditions are (i) x = 0, T = To or 0 = Oo (ii) x = cc, T = 7'. or 0 = 0 Applying second boundary condition, we get 0 = c le — + c2eor
or
ci + c2ee ,,,, ci = c2e as _ 3 0 e c2 = 0 and 0 = ciemx 0=
from equation (i) (ii) Using first boundary condition in equation, we get 00 = ci Hence, temperature distribution in the fin is
or
0 = Ooemx 0 — = emx 0o
T T = emx To  To The rate of heat dissipation from the fin is or
kA( aT ) ax x=o = kA(00 x m x emx)x=0 = +kAm00
ems, = 
= +kAliii) x 00 kA = VkAhP (T0  Tc.c)
(i)
I 40
Heat and Mass Transfer
x Temperature distribution of infinitely long fin
Temperature distribution in an infinite long fin depends upon the value of 'x' as shown in the figure. From the above diagram, it is apparent that temperature of the fin falls from base temperature (T0) as we move away from the base. As temperature falls along the length of the fin, heat transfer reduces along the length of the fin. If `m' is higher, the temperature falls much faster along the length of the fin which is undesirable. The ideal case is that base temperature remains constant throughout the length of the fin which is possible when 'Ic' is high. It is the reason why fins are made of high conducting metals such as aluminium, copper and brass, etc.
8. Derive expressions for temperature distribution and heat flow for a uniform crosssectional fin which has insulated end. The general equation for temperature distribution of a fin is given by 0 = c le" + c2emx On differentiating, we get ae = cime" + c2memx ax Now boundary condition that end is insulated gives
at
ae n aT ax = ax= x= L
(i)
Heat Transfer from Extended Surfaces 1
0= + c2memL 2mL c1 = c2e x = 0, 0 = 00 we get,
or Now when From equation (i)
Oo From equation (ii) and
41
= CI + C2
we get 00
C2 =
and ci =
1+ e2mL
00 1+ e
2mL
Putting the values of c1 and c2 in equation (iii), we get 0 = 00[ 1+emx e2mL +
= ryo
emx 1+ e2mL
emL mx e [ e' x xe 1+ e2mL emL 1+ e2mL
e m(Lx) = e0 e mL
m(xL) e mL emLemL + +e
em(Lx) =
Oo
em(Lx)
emL + e
cosh m(L — x) Oo cosh mL Now temperature distribution is —
cosh m(L — x) cosh mL
0
0o
T — Ta cosh m(L — x) To — T — cosh mL The rate of heat flow is given by 0 = —kA
"
C ax )x=o
= —kA[Qo x
rm sin h m(L — x))1 cosh mL
x=0
Q = kAm00 tanh(mL) =
hP kA Ootanh mL
9. Derive expressions for temperature distribution and heat flow for a fin which has finite length and loss of heat from fin tip is by convection.
I 42
Heat and Mass Transfer
The general equation of temperature distribution for a fin is 0 = ciemx + c2emx At the tip of the fin, we have Heat loss by conduction = heat loss by convection at x = L
kA a0 — = hAO at x = L ax or or
a =0 hx0+kx— ax c2emL) ÷ cimem.L., + c2memL) =0 _ h x (ciemL + k(
ciena, h il +c2eniqh +11 =0 mk ) mk ) Now we know at x = 0, 0 = 00 and applying it in general equation, we get .
00 = c1 + c2 or c1 = 00  c2 From equations (i) and (ii), we have
emq h 1) mk Cl = 00 + )ci +ng,(h e +1 mk _niLi h _ 1)mk ) ci [1 = 00 eFrnqh + i j mk e
or
or
00 [11+ h lemL 1 R mk ) i Cl = (emL + eniL)± h (emL — emL ) mk
and
C2= 00 
041+ h jemL mk mL mL (e + e )+ h (emL —mL e) mk
00(emL [ + Cm L)+ h (e mL emL ) (1+ h j emL1 mk mk = (emL ± emL) ± h (em L _ e _mL ) mk
Heat Transfer from Extended Surfaces
I 43
00 (1 — h jet' mk (emL nil.
h mi. mi. + enil. ) + (e — e ) mk
Substituting the values of c1 and c2, we get (e m(Lx) + em(Lx)) + h (em(Lx) — em(Lx))
mk
0= 00
(emL + emL) + h (emL — emL) mk
cosh m(L — x) + 0 = 00 x
h
sin m(L — x) mk h cosh (mL) + — sinh (mL) mk
Now to find heat transfer Q from the fin, we use Q = —kA( 09 ) x_ o
—m sinh (mL) — = —kA x 00 x
h mk
cosh (mL) +
sinh (mL) + = kArrlOo cosh (mL) +
= V hP kA 00
h mk h mk
h mk
x m cosh (mL)
sinh (mL)
cosh (mL) sinh (mL)
tanh (mL) + h mk 1+ li tanh (mL) mk
10. What is the significance of fin effectiveness? (UPTU — 20067) The ratio of heat transfer rate from a surface with fin and heat transfer from the surface as such without fin is called fin effectiveness (E). The value of fin effectiveness should be greater than one.
I44
Heat and Mass Transfer
with fin
Fin effectiveness E —
()without fin
In case we take infinitely long fin, then fin effectiveness is given by E=
VhPkA00 — hA00
00 hA kP
For E > 1, we have to have M < 1. kP 11. What do you understand from the efficiency of a fin? (UPTU — 20067) The efficiency of a fin is defined as the ratio of actual heat transfer from fin surface to the maximum heat transferred which could be achieved from the fin. It is apparent that maximum heat transfer would occur when the temperature of entire fin surface becomes uniform and equal to the base temperature To nfin =
Qfin max
12. The temperature distribution in a stainless fin (k = 0.17 W/cm°C) of constant crosssectional area of 2 cm2, and length of 1 cm, when exposed to an ambient of 40°C (h = 0.0025 W/cm2°C) is given by
(T — Ta) = 3x2 — 5x + 6 where T is in °C and x is in cm. If the base temperature is 100°C, then the heat dissipated by the fin surface will be: (a) 6.8 W (b) 3.4 W (c) 1.7 W (d) 0.17 W (IES — 1994) We know It is given—
ao Qfil, = —k11()
x=0 T — Ta = 0 = 3x2 — 5x + 6
ao ax
= 6x — 5
or —kA(5) = 5 x 0.17 x 2 = 1.7 W
Qfm =
Option (c) is correct.
Heat Transfer from Extended Surfaces
I 45
13. Addition of fin to the surface increases the heat transfer if 1 hA is: kP
(a) (b) (c) (d)
equal to one greater than one less than one greater than one but less then two. (IES  1996)
The fin can transfer additional heat in case effective (E) is move than one. Qfin
E=
hP kA 9 0
Qwithout fin
1
hA
hAeo
>1
>1
il kP hA < 1 kP Option (c) is correct. 14. A long rod 20 mm in diameter has one end maintained at 110°C. The surface of the rod is exposed to ambient air at 20°C with convection coefficient of 5 W/m2K. Calculate the temperature distribution equation and heat loss from the rod having thermal conductivity of 1.5 W/mK. (UPTU  2002  03) Guidance: A long rod applies that it is a infinitely long fin therefore 0 = emx 00 where
m=
hP
kA P = perimeter = 7cd =
A = area =
m
=
7r x
0.02 = 6.3 x 102 m
X42x (2 x 102)2 4 4
5 x 6.3 x 102 , = 8.16 15x 3.14 x10—
T  Ta  e8.16x T
To —
Now heat loss from fin is Q = jhP kA (To  To.)
= 3.14 x 104 m2
I 46
Heat and Mass Transfer
= \ 5 x 6.3 x 102 x 15 x 3.14 x 104 x (100  20) = 3.07 W. 15. One end of a long, 2 cm diameter rod is inserted inside a furnace while the other end projects into the ambient air at 28°C. The steady state temperatures of the rod at two points 10 cm apart are measured as 125°C and 91°C. Find the thermal conductivity of the material of rod where h = 15 W/m2K. (UPTU  2002  2003) Guidance: The fin is infinitely long and temperature distribution is given by: 0 = Ooemx Fin Base
Now and
Ox = 00e" = ooem(x+Sx) Ox+Sx Now dividing equation (i) and equation (ii)
ex = 0oe'
= e m•Sx m(xFox) Ope
exFox Tx  7: = em.Sx TxFox  To.
125  28 m. sx =e 91  28 97 =emxo.i as 8x = 10 cm 63 or or But
and
0.1 m = log 97 63 = 0.43 m = 4.3 m2 = hP where kA P = perimeter = A = Area = ird2 4
Heat Transfer from Extended Surfaces
I 47
2 hP m = = 15xxxd kA rd2 kx 4 (4.3)2 = 15 x 4 k•d 18.63 = or
k=
15x4 = 30x100 kx 2x102 30 x 100 = 161 W/mK 18.63
16. A fin of circular xsection, diameter 2.5 cm is placed in a furnace with large portion of it is projecting in a room where temperature is 28°C. After steady state conditions prevail, the temperature at two points 10 cm apart are found to be 110°C and 85°C. The value of h = 28 W/m2K. Find value of k. (UPTU  2006  7) Guidance: It is like last problem where fin is infinitely long. Also ex
= em•Sx
e x+ox
Tx  Ts° Tx+ox  Tso
= emxSx
where Sx = 0.1 m (given)
110  28  eo.ixm 85  28
or
Now
82  eo.im 57 82 0.1 m = log — 57 = 0.36 m = 3.6 28 x nd hP 2 m  kA k x ird 2 28 k x 2.5 x 102
or
k=
28 x 102  2800 m2 x 2.5 (3.6)2 x 2.5 2800 = 86.42 W/mK 12.96 x 2.5
I 48
Heat and Mass Transfer
17. It is required to heat the oil to 300°C for frying purpose. A long laddle is used in frying pan. The section of the laddle is 5 mm x 18 mm. The surrounding air is at 30°C and the thermal conductivity is 205 W/mK. If temperature at a distance of 380 mm from the oil should not exceed 40°C, determine convective heat transfer coefficient. (NMU — 2002)
= 30°C
Now as laddle is long, we have temperature distribution as T—T ,x —e To —T Now it is given that T = 40 at x = 0.38 m 40 — 30 — e_0.38m 300 — 30 or or But
e—0• 38m = 10 270 m = 8.67 h x 2(0.005 + 0.018) m2 — hP — kA 205 x 0.005 x 0.018 (8.67)2 x 205 x 90 x 106 0.046 = 30.17 W/m2K
h—
18. Consider a stainless steel spoon (k = 15.1 W/mK) partially immersed in boiling water at 95°C in a kitchen at 25°C. The handle of the spoon has a crosssection 0.2 cm x 1 cm and it extends 18 cm in air from the free surface of the water. If h = 15 W/m2K, find temperature difference across the exposed surface of the spoon handle. State your assumptions. (Shivaji University — 1997)
Heat Transfer from Extended Surfaces
I 49
Assumptions: The handle of spoon is thin and heat loss from its free end is negligible i.e. the free end of the spoon is insulated. For insulated tip fin, temperature distribution is T — To, cosh m(L — x) To — T cosh mL Let TL is temperature at the tip of the handle i.e. at x = L = 0.18 m 71— 7' 1 cosh mL To — Tos
Now
m2 —hp— 15 x 2(0.2 + 1) x 102 kA 15.1 x 0.2 x 1 x 104 = 34.52 m1 mL = 34.52 x 0.18 = 6.21 cosh (mL) = cosh (6.21) = 250 TL —7' 1 1 To — To, cosh 6.21 250
95 — 25 = 0.28 250 TL = 25.28°C Now the drop of temperature from base temperature i.e. (To = 95) to TL (25.28)
or
— 25 =
To —
= 95 — 25.28 = 69.72°C.
19. Two long rods of same diameter, one made of brass (k = 85 W/mK) and other made of copper (k = 375 W/mK) have one of their ends inserted into a furnace. Both the rods are exposed to same environment. At a distance of 105 mm away from the furnace, the temperature of brass rod is 120°C. At what distance from the furnace, the same temperature would be reached in the copper rod? = 120°C 105 mm Brass Furnace Copper x = 120°C
Since both rods are long, we can apply temperature distribution as in the infinite long fin,
I 50
Heat and Mass Transfer
T pmx To  T — For brass rod at x = 0.105, TL = 120°C 120  T  emb xo.los To T For copper rod TL = 120°C at x 120  T x = e _mcx To T Equating equations 1 and 2 mb x 0.105 = mc x x TL
or
x=
x 0.105
me
hP
til kb `4 x 0.105 hP
\i' kc A x=
k kb
x 0.105
;i 375 x 0.105 85 = 0.221 m = 221 mm
20. A very long rod 25 mm in diameter has one end maintained at 100°C. The surface of the rod is exposed to ambient air at 25°C with convective coefficient of 10 W/m2K: 1. What are the heat losses from the rods, constructed of pure copper (k = 398 W/ mk) and stainless steel (k = 14 W/mK)? 2. Estimate how long the rods must be to be considered infinite. (PU  2003) For infinitely long fin, the heat transfer isQ = NihPkA (To  To.)
For copper fin, k = 398 Qc„=
x
= 29.37 W
x 25 x 103) x 398 x x (25 x 103)2 x (100  25)
Heat Transfer from Extended Surfaces
I5I
For stainless steel fin, k = 14 W/mk k Iccu
Q ss =A; " x Q.
14 — ..\ii 398 x 29.37
= 5.51 W Now to find length of infinite long fin, we equate the heat loss of infinite long fin to the fin of finite length with tip insulated
. or
VhPkA (To — Ta) = VhPkA (To — Ta)tanh (mL) tanh mL = 1 mL = 2.6
For copper rod, we have m = 1 hP
10 x 7.854 x 102
1 kcu A
Now
398 x 4.90 x 104
=2 mLci, = 2.6 6 = 1.3 m
L. = 2 For stainless steel rod, we have m= Now
10 x 7.854 x 102
14 x 4.9087 x 104
— 10.69
mL„ = 2.6 L„ =
2.6
10.69
= 0.247 m
21. The aluminuim square fins (0.5 m x 0.5 m) of 1.0 cm length provided on the surface of an electronic semiconductor device to carry 46 x le W of energy generated by an electronic device and the temperature at the surface of the device should not exceed 60°C. The temperature of the surrounding medium is 20°C. Take k = 190 W/ mk & h = 12 W/m2K. Find the number of fins required to carry out the heat transfer neglecting heat loss from the end of the fins. The fins can be considered insulated tip fins. Hence, Q = IhP kA (To — Ta) tank mL
m
hP _  / kA —
12.5 x 2(0.5 + 0.5) x 103 190 x 05 x 05 x 106
I 52
Heat and Mass Transfer
= 23 m1 mL = 23 x 1 x 102 = 0.23 Q = V125 x 2(05 + 03) x 103 x 190 x 05 x 0.5 x 106 x tanh(0.23) x (60  20) = 11.5 x 103 W Number of fins required = Qt"4.al Qfin
x 103 ,=4 11.5 x 10— 22. Determine the heat lost per hour by a mild steel rod of 2 cm diameter and 10 cm length. The temperature at one face of the rod is 220°C and ambient temperature is 20°C. Take k = 50 W/mK and h = 20 W/m2K. Neglect the heat lost by the free end of the rod. The fin is equivalent to the fin with insulated tip. Hence, Q = VhP kA tanh mL(To  Too ) m
hP _ _ Al kA
20 xxxd 50x 7xd2 4
180 = v80 = 8.95  50 x 2 x 102 = \ 1 mL = 8.95 x 0.1 = 0.895 Q = •NlhP kA tanh mL(To  To.) = V20 x Ir x 0.02 x 50 x Ir X (0.0D2 x tanh (0.895) x (220  20) = 20 W = 20 Joules/s Heat loss per hour = 20 x 60 x 60 = 72 x 103 Joules = 72 kJ 23. A circular fin of 100 m length and 5 mm diameter extends horizontally from a casting at 200°C. The fin is in an environment with T., = 20°C and h = 20 W/m2K. What is the temperature at the free end? Take k = 133 W/mK. (UPTU  2003) The fin has finite length and temperature distribution is
Heat Transfer from Extended Surfaces
T—T To — T
I 53
h . Ruh m(L — x) mk h cosh mL + — sinhmL mL
cosh m(L — x) + _
Temperature is TL at x = L TL — 7' = To — T
Now
h sinh mL mk
cosh mL +
20 xxxd
m _ lhP _ i kA
=
Now
1
133 x
20 x 4 133 x d
x x d2
4
i 80 i 133 x 0.005
= V120.3 = 10.96 mL = 10.96 x 0.1 = 1.096 TL — 20 _ 1 200 — 20 — cosh (1.096) + 10.9620x 133 x sinh (1.096) 1
= 1.66 +
20
10.96 x 133
x 1.32
= 0.595 TL — 20 = 0.595(180)
= 107.26 TL = 127.26°C
24. What do you understand by fin effectiveness and fin efficiency? Derive an expression for these for a fin of insulated tip using standard expression for the heat flow rate through the fin. (UPTU — 2003) The fin effectiveness is defined as the ratio of fin heat transfer rate to the heat transfer rate that is possible from the surface without the fin. E—
Qfin
hA(To — L)
Fin efficiency is the ratio of actual to ideal heat transfer from the fin. It is a parameter which evaluates the thermal performance of a fin.
I 54
Heat and Mass Transfer
Actual heat transfer
elfin = Ideal heat transfer through fin if entire fin surface is at base temperature = 9fin where Qideal Qideal =
hP x L(To — To.)
The above means that fin has constant temperature of 70' throughout its length which is possible in case fin material has very high 'V .
T,
T,
Ideal fin
The differential equation for temperature distribution is— a 29 or
m0 axe 0=c
+ c2emx
For insulated tip fin, we have—
x = L ae = 0 and x = 0, 0 = 00 '
We get— C1 =
e
9 oemL ng, + em L & C2
9 oe — niL
e
ra
+e
0 = 0 cosh [m(L — x)]
°
cosh mL where 0 = To —
Q = —kA (t)x=o = —kA
rm sinh (mL)1 (To T) cosh (mL)
= VhP kA (To — Too) tanh (mL)
For the above fin, we have— Qfin
hA(To — hP kA x (T — Too) tanh (mL) hA (To —
Heat Transfer from Extended Surfaces
I 55
_ hA A tanh (mL) —
Gm hPL(To —
11hP kA • (To — T_) tanh (mL) hPL(T — 1:) A x tanh (mL) = l_ A hp
25. A straight fin of rectangular profile is constructed of stainless 188 steel and has a width of 5 cm and thickness of 2.5 cm. The base temperature is maintained at 100°C and the fin is exposed to a convection environment at 20°C with h = 47 W/m2K. Calculate the heat lost by the fin of one meter length. (UPTU — 20034)
The fin is of finite length with end insulated. `k' for 188 steel = 14 W/mk Q = 11hP kA (To — T) tanh (mL)
m_
hP kA
47 x 2(5 + 25) x 102 14 x5x 25x104
47 x 15 x 102 14 x 5 x 25 = V402.8 = 20.07 Now
mL = 20.07 Q = X147 x 0.15 x 14 x 125 x 104 (100 — 20) tanh 20.7
= 0.35 x 80 x 1 = 28 W 26. One end of a long rod 3.5 cm in diameter is inserted into a furnace with other end projected outside the furnace in air. After steady state is reached, the temperature of the rod is measured at two points 180 mm apart and found to be 180°C and 145°C. The atmospheric temperature is 25°C. If the heat transfer coefficient is 65 W/m2°C, calculate the thermal conductivity of rod. Assume that the end of the fin is insulated.
Temperature distribution is 0 = 19 0
cosh (m (L — x)) cosh mL
I 56
Heat and Mass Transfer
1 x = L, Or, — 00 cosh mL OL = 180 — 145 = 35 00 = 180 — 25 = 155 35 _ 1 155 cosh mL
at
cosh (mL) = 155 = 4.42 35 mL = 2.166 m — 2.166 — 2.166 L 0.180 = 12.03 65 x rd 65 x 4 hP 2 x gd 2 = m = k kA = k xd 4 k=
=
260 kx 35x102
260 = 51.33 3.5 x 102 x (12.03)2
27. The both ends of a 6 mm diameter copper rod (U — shaped) having k = 330 W/mK are rigidly connected to a vertical wall as shown in the figure. The wall temperature is constant at 110°C. The developed length of the rod is 50 cm and is exposed to air at 30°C. The value of h = 30 W/m2K. Find: (i) temperature at centre of the rod, and (ii) heat transfer by the rod. (PU — 1996)
T, = 100°C T,. = 30°C
Guidance: At the centre of the fin, temperature would be constant due to symmetry. Ushaped fin can be considered to be two fins of length 25 cm each (total length = 50 cm) which are insulted at tip. TL — T 1 — cosh (mL) To — Too TL — 30 _ 1 100 — 30 cosh (mL) m2
= hP — kA
30 x ird — 30 x 4 330 x d 2rd2 330 x 4
Heat Transfer from Extended Surfaces
I 57
120 330x6x103 m= 7.78 m1 mL = 7.78 x 25 x 102 = 1.945
Now
70 cosh (1.945) TL, = 30 + 19.6 = 49.6
TL,— 30 =
Q = 211hPkA (To — Ta)tanh (m • L) = 2 1130 xxx6x 103 x 330 x x (6 x102)2 x 70 x tank 1.945 = 9.76 W 28. How can fin effectiveness be increased? Fin effectiveness is Qfin
E=
Qwithout fin
kP
A/hP kA (To —
Ta) hA(To — Tot)
)1/2
— (hhA
Hence, effectiveness can be increased by: (a) Using material of higher thermal conductivity (k) (b) Increasing the ratio of perimeter (P) to crosssectional area (A) (c) Using lower 'h' i.e. natural convection instead of forced convection. 29. What is the reason for the error in the measurement of temperature of fluid flowing through a pipe when measured by a thermometer placed in the thermometer well of the pipe? How can error be reduced? `T0' at pipe wall
ii
11
Thermometer well
Liquid
.V1h.
1.%
1121.
I 58
Heat and Mass Transfer
A thermometer well is a small tube welded radically to a pipe through which a fluid whose temperature is to be measured is flowing. The well is filled with little of other liquid and then thermometer is dipped in this liquid. The liquid receives the heat from the fluid flowing whose temperature is being measured. The heat flows from the fluid to liquid through the wall of the well. The temperature (TL) recorded by the thermometer is that of the liquid at the bottom of the well and it will be less than flowing fluid temperature (Ta). The temperature in the well will reduce as we move towards the top of the well or towards the wall of tube in which the well is located where temperature is 70'. Hence, heat will flow from the bottom of the well to the wall of the tube in which the well is located. Therefore, the well can be considered a fin but with a difference that the heat is flowing from the surroundings to the base (wall of the tube). Ta Area =irxdxt t= thickness of well wall Perimeter = irxd
Ta
Well acts as a fin
The temperature distribution on the wall of the thermometer well is same as that in a fin of finite length IL 1 TL —TOO _ h To — Ts., cosh (mL) + — sinh (mL) mk As h sinh mL is small and it can be neglected. Hence, we have mK TL — Tso . 1 ..:. 1 ..:__. 1 1 kA To — T cosh mL mL L 1 hP L
kxjrdxt — 1 1k•t hxrd LN h
Now the thermometer is measuring II' instead of To.and quantity (TL — T) is a error in the measurement. The error in the measurement can be reduced by following ways: (a) (b) (c) (d)
Well material should have moderate thermal conductivity. Keeping thickness of wall (t) of the well as small as possible. Keeping length of well as large. Using this method for measuring temperature of the fluid having low heat transfer coefficient.
30. Temperature of the air flowing in a tube is measured with the help of a thermometer placed into a thermometer well filled with oil. The well is made of stainless steel
Heat Transfer from Extended Surfaces
I 59
(k = 55.5 W/mK), 12 cm long and 0.15 cm thick. The surface heat transfer for air and well is 23.5 W/mK and temperature shown by the thermometer is 360°K. Evaluate the measurement error if the temperature at the well of tube is 315°K. Temperature distribution is TL  Too = 1 To  Too cosh mL m = lhP i kA
1 23.5 x Ird A 555x7rdxt _
or
235 A; 55.5 x 0.15 x 102
= V282.28 = 16.8 mL = 16.8 x 12 x 102 =2 360 x T 1 1 315 x Tso cosh (2)  3.76 = 0.266 360  Too = 83.79  0.266 Too 0.734 Too = 360  83.79 = 276.21 T  276.21  0.734 = 376.3°K Measurement error = 376.3  360 = 16.3°C
31. A mercury thermometer is placed in a oil well for the measurement of temperature in a pipe. The well is made of steel (k = 50 W/m°C) and is 14 cm in length and 1.0 mm in thickness. The temperature recorded by well is 100°C while pipe wall temperature is 50°C. If the heat transfer coefficient between the air and well wall is 30 W/m2°C, calculate the true temperature of air. (UPTU  2004  5) Temperature distribution is TL  T = 1 (mL) To  Too cosh mL =
h x rcd r Ph kA = k x rcd x t
I 60
Heat and Mass Transfer
= Li
kxt
— 0.14 J
30 ., v50x1x10—
= 3.42 Now we have 100 —T ___ 1 _ 1 50 — T cosh 3.42 15.3 1530 — 15.3 To. = 50 — To. 14.3T_ = 1480 To. = 103.5°C 32. A thermometric well is placed in a pipe having diameter of 55 mm. Pipe well temperature is 100°C and 'h' inside is 300 W/m2K. Thickness of thermowell is 1.2 mm and its length is 50 mm. Thermal conductivity of thermowell material is 30 W/ mK. If the temperature of gas flowing through the pipe is recorded by thermometer as 210°C, determine the true temperature of gas. If the error in gas temperature is to be reduced by 60% by increasing the length of thermowell, find the new length of thermowell. Temperature distribution is
TL —Ts. To — T.,
1 cosh mL
m .I h = l
I k•t
300 3 30x1.2x10 i
= 91.29
mL = 91.29 x 50 x 103 = 456.45 x 102 = 4.56 210— T. _ 1 __ 1 100 — T — cosh 4.56 48 10080 — 48T_ = 100 — To. 47T = 9980 T.. . 9980 47 = 233.6°C Error = 233.6 — 210 = 23.6°C Allowable error 40% of present error = 0.4 x 23.6 = 9.44 71— Too = 9.44 Now 9.44 1 __ cosh (91.29 x L) 233.36 —100
Heat Transfer from Extended Surfaces
I6I
cosh (91.29L) = 133.36 9.44 = 14.13 91.29L = 3.32 or
L
91.29 = 0.357 m
33. A thermometer well 2 cm in diameter 1 mm thick is made of steel (k = 30 W/mK and it is used to measure the temperature of air flowing through a pipe. Calculate the minimum length of pocket so that the error is less than 1% of applied temperature difference. Air temperature is 300°C and h = 100 W/m2K.
Temperature distribution is TL —7' 1 To — Tso cosh mL Now error = TL — Too = 1% of applied temperature 1 x (To — 7',,,) = 10 0 1 _ 1 cosh mL 100 or cosh mL = 100 or mL = 5.3 m= 1h 1 k•t 100 = 1 30 xlx103 1
or
= V33.33 x 102 = 57.7 mL = 57.7 x L = 5.3 L = 5.3 = 9.19 x 102 57.7 = 91.9 mm
34. Determine the energy input required to solder together two very long copper wire pieces of diameter 1.7 mm. The melting point of solder is 200°C and environment temperature is 20°C. Take 'h' for environment as 16 W/m2K and k for wire material as 330 W/m°C.
Guidance: The problem is nothing but a heat transfer by two fins with the solder thickness acting as hot surface at base temperature at 200°C and two pieces of wire are acting as long
I 62
Heat and Mass Transfer
fins transferring heat to the environment. The heat transfer for one fin is calculated which is later doubled to find total heat transfer from the solder to the environment through two long pieces of wire being soldered as shown below: Solder acts as base plate at 200°C
Wire
Wire Ta = 20°C Soldering of wires: two fins
For long fin, heat transfer by it is Q = . \ I kA Ph (T,  E,‘) ir x (1.7) 2 rd2 x 106 = 2.27 x 104 m2 4 4 P = ird = x x 1.7 x le = 5.34 x 103 m
A=
Q = \1330 x 2.27 x 104 x 5.34 x 103 x 16 x (200  20) = 1.44 W Heat lost by both wire type fins is Qtotai = 2 x 1.44 = 2.88 watt In order to solder and maintain the temperature of solder at 200°C, we have to supply 2.88 watt of energy to the melted solder. 35. A cutting tool of 60 mm diameter is to be grounded for sharpening. The tool has length of 1 meter. During grinding 40 watt of energy is dissipated to the tool which is convected to the air at 20°C. If h = 7.1 W/m2°C and 'le of tool material = 42.4 W/ m°K, find the temperature of the rod when grinding is taking place. Guidance: The problem can be visualized as a fin consisting of tool which is convecting heat out through its surface while tip of the tool is at a constant temperature To tanh (mL) + h km Qtool = V hP kA (7's  Tc.c) [ h 1+ — tanh (mL) m
Grinder
Ta = 20°C
Grinder and cutting tool: fin
Heat Transfer from Extended Surfaces
I 63
(60 x 103)2 A = rd2 = ir x = 2.83 x le m2 4 4 P = ird = ir x 60 x le . 188.5 x 103 m m
Now
liP — 17.1 x 1885 x 103 1 42.4 x 2.83 x 103
_ Il kA
= 3.34 m1 mL = 3.34 x 1 = 3.34 7.1 h = = 5 x 102 km 42.4 x 3.34
Putting these values in the equation of heat transfer, we get i [ tanh 3.34 + 5 x 102 Qtooi = 0.1 x 1885 x 103 x 42.4 x 2.83 x 103 x (To — 20) x 1 + 5 x 102 tank 3.34 40 = 0.4 x (To — 20) [1] or
40 = 100 To — 20. _ 04
or
To = 120°C
36. An experimental arrangement for measuring the thermal conductivity of solid material involves the use of two very long rods that are equivalent in every aspect (dimension, shape, etc.) except that one is fabricated from standard material of known conductivity kA while the other is fabricated from the material whose thermal conductivity kp is desired. Both rods are attached at one end to a heat source of fixed temperature Tb, are exposed to a fluid of temperature T.. and are instrumented with thermocouple to measure the temperature at fixed distance 'x', from the heat source. If the standard material has kA = 200 W/mK and measurements reveal the value of TA = 75°C and TB = 60°C at x for Tb = 100°C and T..= 25°C, what is thermal conductivity kp of the test material? (UPTU — 20078) T,.= 250°C kA = 200 W/mk A
..
Tb = 100°C kB B x
Temperature distribution is T — Tc., = e`"X Too
Tb—
where m = ithP \i' kA
I 64
Heat and Mass Transfer
For material 'A' 75  25 emix 100  25 =
where m1 =
hP kA x A
75 = 0.405 mix = log 50
or For material fi
60  25  2x em 100  25
where m2 =
hP kp x A
m2x = log 75 = 0.762 35 Now dividing equation (i) to equation (ii), we get
or
Pit = 0.762 _ 1.88 m1 0.405 .
kfl = 1.88 kA kp = 1.88 x kA = 1.88 x 200 = 376 W/mK
or
37. Define hyperbolic functions. Hyperbolic functions are: 1. sinh x = 2. cosh x = 3. tank x =
ex  ex z ex ± ex z ex _ ex ex + ex
38. What are the derivatives of the hyperbolic functions? The derivatives are: (sinh x) = cosh x 1. dx d (cosh x) = sinh x — 2. dx 3. dx
x) =
1 cosh2 x
Heat Transfer from Extended Surfaces
Hyperbolic Functions
x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.5 4.0 4.5 5.0 6.0 7.0 8.0 9.0 10.0
sink x
cosh x
tank y
0.000 0.100 0.201 0.305 0.411 0.521 0.637 0.759 0.888 1.027 1.175 1.337 1.510 1.700 1.904 2.136 2.376 2.646 2.942 3.268 3.627 4.022 4.457 4.937 5.466 6.050 6.695 7.406 8.192 9.060 10.018 16.543 27.290 45.003 74.203 201.71 548.32 1490.50 4051.50 11013.20
1.000 1.005 1.020 1.045 1.081 1.128 1.186 1.255 1.337 1.433 1.543 1.670 1.811 1.971 2.151 2.352 2.577 2.828 3.108 3.418 3.762 4.144 4.568 5.037 5.557 6.132 6.770 7.474 8.253 9.115 10.068 16.573 27.308 45.014 74.210 201.72 548.32 1490.50 4051.50 11013.20
0.000 0.100 0.197 0.291 0.380 0.462 0.537 0.604 0.664 0.716 0.762 0.801 0.834 0.862 0.885 0.905 0.927 0.935 0.947 0.956 0.964 0.970 0.976 0.980 0.984 0.987 0.990 0.991 0.993 0.994 0.997 0.998 0.999 0.999 0.999 0.999 1.000 1.000 1.000 1.000
I 65
I 66
Heat and Mass Transfer
39. A cylinder of 1 m long and 50 mm in diameter is placed in an atmosphere at 45°C. It is provided with 10 longitudinal straight fins of a material having k = 120 W/mK. The height of 0.76 mm thick fins is 1.27 cm from the cylindrical surface. The heat transfer coefficient between cylinder and atmosphere air is 17 W/m2K. Calculate the rate of heat transfer and the temperature at the end of fins if the surface temperature of cylinder is 150°C. (Annamalai University, 2004 — 5) 5= 0.76 mm
h= 1.27 cm L (a) Front
(b) Side view
Considering fin having short length and its end insulated, we have— T—T cosh m(L— x) = To _ T cosh (mL) = ihP N kA p = 2 x (length of cylinder + 8) . 2 x 1 = 2 A = Lx S = 1 x 0.76 x 103 = 76 x 105 m2
Now
m
m=
1
17 x 2 120 x 76 x105
= V3.728 x 102 = 19.3 m1 Temperature at end when x = L is— cosh 19.3 (L — L) T — 318 = cosh (19.31x 12.7 x 102) 423 — 318 or
T —318 = 1 105 1.03
Heat Transfer from Extended Surfaces
or
I 67
T= 102 + 318 = 420 K
Q = I f hPkA (To — 11,) tanh (mL) = (17 x 2 x 120 x 76 x 105) (423 — 318) tanh (19.3 x 12.7 x 102) = 3.1 x 105 x tanh (2.451) = 3.1 x 105 x 0.983 = 320 W Now for 10 fins, heat transmitted is 10 x 320 = 3.2 kW
Qtotal =
40. A circumferential rectangular profile fin on a pipe of 50 mm outer diameter is 3 mm thick and 20 mm long. The thermal conductivity is 45 W/mK. The convection coefficient is 100 W/m2K. Base temperature is 120°C and the surrounding temperature is 35°C. Determine: (i) heat flow rate, (ii) fin efficiency and (iii) fin effectiveness. (Annamalai University 2007 — 8)
✓
A
r/A
✓
A
II
%/A
r/.
i_ T
s i'
nc127—
t k—
L —.1 (c) Equivalent fin
(a) Side view
(b) Top view Circumferential rectangular fin
. ihP l kA p = 2(7rd + 5) . 27rd A = 7rd x 8 = 7rd8 m
P . 27rd . 2 A ,rdx5 8
I 68
Heat and Mass Transfer
.
IN
=
=
100 2 11 45 x ( 5 I 100x 2
I 45 x 0.003
= 1/14815 = 38.49 m1 As fin is short, we can assume that the end is insulated. Q= VhPkA x (To — Too) tanh (mL) = V100 x (27r x d) x 45 x (IrchS) x (120 — 35) tanh (mL)
But
mL = 38.49 x 0.02
= 0.78 Q = 11100x 2x 45x7r 2 x d 2 x (5 x (85) x (tanh 0.78)
= 119000 x 7r2 x (0.05)2 x 0.003 x 85 x 0.65 = 815 x 103 x 85 x 0.65 = 45 W Fin efficiency Thin =
tanh (mL) mL
= tanh0.78 = 0.65 0.78 0.78 = 83.33% Fin effectiveness (E) =
—
tanh(mL) / hA \" kP
0.65 _ 100x8 1 N 45 x 2
0.65 _ 0.65 0577 100 x 0.003 90
= 1.13 41. A brass rod (k = 133 W/mK), 100 mm long and 5 mm in diameter extends horizontally from a casting at 200°C. The rod is in air at T = 20°C and h = 30 W/m2K. What is the rate of heat transfer to air and temperature of the free end of the rod? (UPTU — 2008 — 9)
Heat Transfer from Extended Surfaces
I 69
I hP 1 il — = 1 kA p = ird = iv x 0.005 m 0.0157 m
m
A= 5 d 2 = 1.96 x 105 m2 4 P ird 4 4 = =800 A 7r 2d 0.005 —xd 4 m=
V300 x 800 133
= V1804.5 = 42.49 m1 mL = 42.49 x .01 = 0.425 As fin is finite length and we can assume that the end is insulated, thenQ = ,i 1hPkA x (To  Tos)tanh(mL) =
f
ir
AI 300 x (iv x 0.005) x 133 x (4 x 0.0052) x (200  20) x tanh (0.425)
= 0.11 x 180 x 0.405 = 8.036 W T To, cosh m(L  x) = To  T cosh (mL) Now at x = L, we have T  20 = 1 cosh 0.425 200  20
Now
T = 20 + 180 1.093 = 184.72°C. 42. Calculate the amount of anergy required to solder together two long pieces of base copper wire 1.625 mm diameter with solder that melts at 195°C. The wires are positioned vertically in air at 24°C. Assume that the heat transfer coefficient on the wire surface is 17 W/m2K and thermal conductivity of wire alloy is 335 W/mK. (Annamalai University 2002  3)
I 70
Heat and Mass Transfer
Assume wire as long fin, then— Now
Q = ‘IhPkA x (To — To0 p = r d = iv x 0.001625 = 5.11 x 103 m gc/ 2 = x (0.001625)2 4 4 = 2.07 x 106 m2
A=
.
Q = \,I17 x 335 x 5.11x 103 x 2.07 x 106 x (195 — 24) = 7.76 x 103 x 171 = 1.3272 W
43. Two long metal rods of same diameter are placed on a surface with the end temperature being 300°C. The rods extend to the room where the temperature is 30°C. The temperature at 25 cm from the hot end in one rod is found to be 200°C while in other rod, this temperature is attained at 400 cm from the hot end. Find the ratio of conductivities of these rods. (Annamalai University — 1999 — 2000)
x1 = 25 cm .1 T, = 300° C
T = 200° C
T,. = 30° C
.. x2 = 400 cm T = 200° C
Two Fins For rod1, we have T —T = mi xi e To — Tc, or or
200— 30 = e 0.25.m1 300— 30 m1 = 1.85m 1
For rod2, we have
or
200— 30 . eo4 m2 300— 30 m2 = 1.567m1
Heat Transfer from Extended Surfaces
I7I
Now we have Q = V hPkA (To  Tj I hP — xkA(ToTso) kA = m x kA(To  TO Qi = m1 x ki x A(To  Too) =
Q2 = M2 Q1 = Q2
Now
X
k2
X
AM  TO
= m2 = 1.567 k2 m1 1.85 = 0.625 k1
44. Calculate the amount of energy required to solder together two very long pieces of bare copper wire 1.5 mm in diameter with solder that melts at 190°C. The wires are positioned vertically in air at 20°C. Assume that heat transfer coefficient on wire surface is 20 W/m2°C and thermal conductivity of wire alloy is 330 W/m°C. (UPTU  2009  10) Assume wire as long fin, then Q = V liPkA x (To  TO P = ird = ir x 1.5 x 103 = 4.71 x 103 m A= .
7r x (15 x 103)2 Ord 2 = = 1.766 x 106 m2 4 4
Q = V20 x 4.71x 103 x 330 x 1.766 x 106 x (190  20)  N15489 •8 x 108 x 170 = 74.1 x le x 170 = 1.26 W
Chapter
5
TRANSIENT HEAT CONDUCTION
A A A A A A
LUMPED BODY BIOT NUMBER FOURIER NUMBER GEOMETRIC FACTOR CHACTERISTIC LENGTH SEMI INFINITE SOLIDS
A A A A A A
TIME CONSTANT RESPONSE OF THERMOCOUPLE SENSITIVITY OF THERMOCOUPLE LUMPED ANALYSIS HEISLER CHARTS GAUSSIAN ERROR FUNCTION
INTRODUCTION Transient heat conduction is nothing but the unsteady state heat conduction. In unsteady state, the heat flow as well as the interior temperature distribution of the system changes continuously
dT dt
dT dt
with time. It means that — # 0 during transient heat conduction while — = 0 for steady state heat conduction. The transient heat conduction occurs while: (i) heating or cooling of metal billets, (ii) cooling of ICengine cylinder, (iii) cooling and freezing of food in canning industry, (iv) brick manufacturing, (v) vulcanisation of rubber, and (vii) starting and stopping of various heat exchange units in power installation. If solids have smaller thermal conductivity, then there is interior temperature distribution while heating or cooling. In case solids have high heat L
conduction or low internal heat conduction resistance ( while their surface convection
kA
resistance is high I R , then the temperature throughout the solids can be considered to be AJ uniform and they form lumped heat capacity systems.
Transient Heat Conduction
I 73
1. What is transient or unsteady heat conduction? When heat energy is either removed from or added to a body, its internal energy changes, resulting a change in its temperature at each point within the body. This temperature at any point within the body is a function of time and location of the point in the body in the direction of heat flow during the transient period. The conduction occurring during the transient period is called transient conduction or unsteady state conduction. Hence, in unsteady state— T = f (x, t) = function of direction & time 2. Explain the energy balance in a body during transient heat conduction. During transient heat conduction, the energy balance is— The net rate of heat transfer from the body to surroundings = the net rate of internal energy change of the body .
hA(T — T) = —mCp
aT at
where, h = coefficient of convective heat transfer A = surface area of body T = temperature of body
aT
= change of temperature of body per unit time
at C, = sp. heat of material of body m = mass of body.
3. Explain the main characteristics of unsteady or transient heat flow. The unsteady state of heat flow has always variation in heat flow and temperature. The variation can be continuously increasing/decreasing (nonperiodic) or periodic. The heat treatment processes like quenching, annealing and normalizing are such nonperiodic unsteady state of conduction in which heat flow and temperature of the body increases or decreases continuously till steady state reaches. The heat flow in a building between day and night is periodic unsteady state involving periodic variation in heat flow and temperature. 4. What is the analytical method of solving transient heat conduction? If temperature distribution of the body is given at certain instant of time, then we can find out (i) heat flow inside or outside the system: (ii) rate of change of temperature with time, and (iii) rate of change of energy within the system. Consider a plain wall with temperature distribution at any instant is— T = a + bx + cx2 + dx3
I 74
Heat and Mass Transfer
where a, b, c and d are constants. The general differential equation for onedimensional heat conduction is—
a2T qg _ 1 aT ax2
a at
k
where, a = thermal diffusivity of the system qg = internal heat generation per unit volume. k = thermal conductivity of the system.
a
q1
q2
k
qg
x= 0
x=L L
We can now find out the following— (a) The rate at which heat is entering and leaving the wall is— qin = —kA
gout
=—
(a
a x Jx=0
in k
= —kA
a
+ a x (gin)
x d xi
(aT
a x )x=L (b) Rate of change of temperature with time is given by— T [a2 T qg] =a + at ax2 k
(c) Rate of change of energy within the wall can be given by— Estored = (q1
where
q2)
qg(A • L)
A = area and L = length of the wall.
(d) Maximum temperature within the wall is found out by— a (aT ) =0 ax at 5. The temperature distribution at certain instant of time in a plane wall, 50 cm thick is given by the relation— T = 450 — 500 x +100x2 + 150 x3
Transient Heat Conduction
I 75
where temperature T is in degree Celsius and x is in metres measured from hot surface. The thermal conductivity of the wall material is 10 W/mK. Calculate the rate of heat energy stored per unit area of the wall at that instant of time. (UPTU — 2002 — 3) Temperature distribution is— T = 450 — 500 x + 100 x2 + 150 x3 a T = —500 + 200 x + 450x2 ax (aT axl.o = —10 x 1 (500 + 200 x + 450 x2 )x.o = 5000 W
Qin = —kA
Qout = —kA( a T ax x=0.5 = —10 X 1 (500 + 200 x + 450 x2 )x=o.5 = —10(500 + 200 x 0.5 + 450 x 0.52) = 2875 W Heat stored = Qin — Qout = 5000 — 2875 = 2125 W 6. In previous problem, find the time rate of temperature change at x = 0 and x = 0.5 if a uniform heat generation (qg) of 500 W/m3 is present in the wall and emissivity of wall a( — k p Cp
is 6 x 103 m25.
Time rate of temperature change is—
aT _ a [a2T _E gg ] k
at — axe T = 450 — 500 x + 100 x2 + 150 x3
On differentiating, we get—
a T = —500 + 200 x + 450x2
Now
ax
Again differentiating, we get— a2 T
axe
= 200 + 900 x
Put the values in equation (i)
a T = 6 x 103[200 + 900 x + 500 1 10 at = 6 x 103 [250 + 900 x]
(i)
I 76
Heat and Mass Transfer
(a_T ) 6 x 103 x 250 d tx=0 = = 1.5°C = 6 x le [250 + 450] Cat/ x=03
= 6 x 700 x 103 = 4.2°C 7. What is a lumped heat capacity system? If the internal conductive thermal resistance of a body or system is negligible as compared to the surface convective thermal resistance, then the system is called a lumped heat capacity system. For example, if the size of a body is very small, the temperature gradient exists in the body is negligible, then body can be assumed at constant temperature. 8. Find the expression of cooling of a lumped body. Or
Find the temperaturetime history of a lumped body. Or
Show that for lumped heat capacity system
T

—e
hA pC V xT
P
(UPTU  2002  3, 2007 8, 2009  10)
Lumped system analysis means that the temperature gradient in the body is negligible and body can be assumed to have uniform temperature everywhere at an instant of time. The analysis of unsteady heat transfer with negligible temperature gradient is called lumped system analysis. Consider a lumped body (volume V', surface `A', density 'p', thermal conductivity IC' and sp heat `Cp') at initial uniform temperature 'Ti' and it is suddenly immersed in a fluid having temperature To; . The heat is dissipated by convection into the fluid from the surface of the body. The rate of heat flow out from the body is equal to the rate of decrease of internal energy of the body.
hA(T — Too) = —MC
aT paz
= —pVC
aT
Transient Heat Conduction
Now take
=T—T
ae aT at = at —hA ae = pVCp
at
On integration, we get— or
—hA
log 0 =
We know
p
t + Cl
T = Ti at t = 0 or Ot=o = = log 0i = C
—
Putting the value of C1, we get— hA t + log Oi pVC p
log 0 = or
hA
log = Oi
pVCp hA
or
Oi
e
pvcp t hA
T— pVCp t Ti —T = e The temperature of the body decreases with time as shown belowor
2 E T lime (t) —.Temperature — time history of the lumped body
9. Find the expression for heat transfer from a lumped body. We know the temperature distribution any instant is given by— hA
t T —T pVC =e P Ti—T
I 77
I 78
Heat and Mass Transfer
The heat transfer at that instant is given by— qt = hA(T — Too ) Heat transfer for time at is— qt clt = hA(T — Too)a t hA
But
t
T — Too = ( Ti — T )e PVCP hA
qt at
T
hA(Ti —T )e Pvc P at
Total heat flow from initial instant to time t is— hA „
qt at = hAf (Ti —
0
Too)
t
e P' 'P at
0 hA
„t Qt = pVCp(Ti — T) 1— eP"
10. What do you understand from Biot number (Be), Fourier number (Fo) and geometrical factor (GF)? What is the value of Biot number for lumped system analysis? Biot number is the ratio of internal conductive thermal resistance to surface convective resistance Internal conductive thermal resistance B= Surface convective resistance (V A) _ kA —1 hA h • Is
where — = characteristic length of the body (ls) A The lumped system analysis can only be applied when Biot number is equal or less than 0.1 i.e. 0.1. This implies that internal resistance to heat flow is very small as compared to convective resistance to the heat flow. The Fourier number is the ratio of the rate of heat conduction to the rate of thermal energy storage in the solid. Rate of heat conduction Fo = Rate of thermal energy storage in the solid kA
AT
Fo = pVCp
IcAt (AT) t
p (A x Is)Cp is
Transient Heat Conduction
2
I 79
at 2 (4)
P Cp (4) Fourier number signifies the degree of penetration of heating or cooling effect through a body. Geometric factor (GF) is the ratio of volume to area. It is also called characteristic length of the body. The characteristic length (1s) for certain common body shapes are—
(a) For infinite (large) plate of thickness 'I,' exposed to convection environment on both sides— Volume AL L _ = = is Area 2A 2 (b) For a solid cylinder of radius R and length 'L— K R2 L RL 4= 2K R2 + 2K RL 2(R+ L) (c) For a long cylinder of radius 'IV where 'I,' » 'R'— KR2 L R Is = 2 2 rRL (d) For a solid sphere of radius 'R'— KR 4=
2KR
3
2
3
(e) For a cube of side 'L'— / =
L3

6L2
L 6
11. What is a time constant? For a lumped body, the temperature difference of a body and fluid decays exponentially as T —T —e Ti —T
The exponent quantity I term
hA pV C
hA pVC x t P
p xt in the above equation is a dimensionless quantity as
pV Cp has unit of time. In case time period (t) becomes equal in magnitude to hA
(pVCp , then exponential temperature decay equation becomes— hA T —Too _1 =e — Too
where
= 0.368
pV C p hA
=t=
I 80
Heat and Mass Transfer
or
1
T—
Tso
Ti — Tso
— 1 — 0.368
—T — 0.632 Ti — Tso — T = 0.632(Ti — To0
or or
Hence, time constant is time required by a temperature measuring instrument to record a temperature difference equal to 63.2% of initial temperature difference. 12. What do you understand from the response of thermocouple and the sensitivity of thermocouple?
pvC
e r where 1
hA p
i 8
0.632
T1 T2
23
T4
A Time constant (t) Transient temperature response to different time constants (r)
Thermocouple is a temperature measuring instrument. The response time of a thermocouple is defined as the time taken by the thermocouple to indicate the source time. The sensitivity of thermocouple is defined as the time required by the thermocouple to indicate 63.2% of the initial temperature difference. The temperature difference is reduced by 63.2% after one time constant. Hence, a thermocouple is sensitive in case it can have low time constant and thermocouple can achieve 63.2% of the initial temperature difference quickly. The low value ( pVCp of the time constant is desirable and it can be achieved by— hA V (a) Decreasing characteristic length ( — of wire of thermocouple i.e. radius of wire is A small (b) Using light metal for wire of thermocouple having low density and low specific heat (c) Increasing the heat transfer coefficient.
13. Find the expression for temperature of plane wall at any instant using lumped heat analysis.
Transient Heat Conduction
I8I
Consider a plane wall of thickness having one surface subjected to constant heat flux `q' and other surface is exposed to convective environment. The wall has Biot number (Bi) 0.1 h
E h(T— Teo) q
x=0
x=L
and lumped analysis can be applied. The wall has initial temperature Ti and convective environment has temperature T._ Now applying the energy balance for the wall, we have—
dT P at
A x q — hA(T — Too) =mC Put
0 =T
ae aT at = at d0 P a ad Axq hA0 at mcp mcp ae = a — h0 at
A x q — hAO =mC or or or where We know Now
or or
at = ae a —b0 Axq hA a= &b= mCp mCp 0 = Oi at t = 0 e f at = o e, a—b• 0 t— 1
log
a—bOi l a+b0
b
a +bOi _bt —e a —b0
14. A copper plate 2 mm thick is heated upto 400°C and then quenched in water at 25°C. Find the Biot number and verify if the lumped heat capacity analysis is applicable.
I 82
Heat and Mass Transfer
Also find the: (i) time constant, and (ii) time required for the plate to reach the temperature of 40°C. The data given: h = 100 W/m2K, C,, = 400 J/kgK, k = 385 W/ mK, plate size = 25 m x 25 m, p = 8800 kg/m3. (UPTU  2003  4) Characteristic length is = V
A =
L • B • (5 (5 2 L•B = 2
where
= thickness
= = 1 mm 2 100 x 1 x 103 hls Biot number Bi = k = 385 = 2.5 x
< 0.1
Hence, lumped heat capacity analysis can be applied as Bo < 0.1 pVCp pCp x is Time constant (r) = hA  h 8.800 x (400) x 1 x 103 100 = 35.2 sec Now exponential temperature decaying equation isT  Tso _dr =e 40  25 400  25
or
et/35.2
t = log 15 35.2 375 t = 35.2 log 25 = 113.3 sec
15. Balls of 12 mm diameter are anneded by heating to 877°C and then slowly cooling to 127°C in air at a temperature of 52°C and h = 20 W/m2K. Calculate time required for cooling process for the balls. Take p = 7800 kg/m3, k = 40W/mK and C,, = 600 J/kg K (UPTU  20056) 4 irR3 3 _ Characteristic length is = — A  4r R2 V
= 6 = 2 mm 3
R 3
Transient Heat Conduction
I 83
20 x 2 x103 his k 40 = 1 x 103 < 0.1
Biot No (Bi) =
Since Bi < 0.1, lumped heat capacity analysis can be applied hA t h T T pCpV = pCp x1, xt e e  Tso
or
xt
127  52 75 877  52 825 x t = log 825 pCp is 75 = 2.39 2.39 x p x Cp x ls t= h pCp l,
2.39 x 7800 x 600 x 2 x 103 20 = 1122.17 sec = 18.7 min 16. A steel ball 5 cm in diameter is heated to a temperature of 900°C and placed in still surrounding atmosphere for cooling, if atmospheric temperature is 30°C, calculate the initial cooling rate of the ball in °C/min. For steel take p = 7800 kg/m3, CP = 2 kW/kg • K Assume h = 30 W/m2°C. (UPTU  2004  5) Equating conduction and convection heat transfer
mC a P
or
atT
aT
= hA(Ti  T) hA
(Ti Toc) = (Ti Lc) pcpis at VC is = characteristic length V=R 5 =— A 3 = 6 cm aT 30 (900  30) at 7800 x 2 x103x 5 x102 6 30 x 870 x 6 7800 x 100 = 0.2 °C/sec
I 84
Heat and Mass Transfer
= 0.2 x 60 °C/min = 30°C/min 17. A solid steel ball 5 cm in diameter and initially at 450°C is quenched in a controlled environment with convective coefficient of 115 W/m2K. Determine the time taken by centre to reach a temperature of 150°C. Take C,, = 420 J/kg K, p = 8000 kg/m3 and k = 46 W/mK. (PU 2003) Guidance: For lumped system, temperature is uniform throughout the body. However, lumped system analysis is possible in case Biot number is less than or equal to 0.1. V = Characteristic length = — = 5 cm A 3 6
115 x 5 x102 6 x 46 = 0.021 < 0.1
. hl B1 = s k
Hence, lumped system analysis is possible. NowT T T,

_ e mP"
; /1 xt

150  90 450  90 
e
—115 2 xt 8000x420x5/6x10
8000 x 400 x 5 x 102 x log 360 115 x 6 60 = 440.4 sec = 7.34 min
t=
18. A person is found dead at 5 pm in a room whose temperature is 20°C. The temperature of the body is measured to be 25°C when found, and the heat transfer coefficient is estimated to be 8 W/m2K. Modelling the human body a 20 cm diameter and 1.7 m long cylinder, calculate actual time of death of the person. Take thermophysical properties of the body.
k = 6.08 W/mK, p = 900 kg/m3, C,, = 4000 J/kg • K (NMU  2002)
Human body can be modelled to
Transient Heat Conduction
I 85
Characteristic length is = A V 2 rR L
RL
2 ir(R2 + RL)
2 (R + L)
15 x102 x1.7
2(15 x10 2 +1.7) = 0.069 m The body temperature can be considered uniform in case Biot Number 0.1 . hl 8 x 0.069 B1 = = ks 6.08 = 0.092 Now we can apply lumped system analysis T —Tpcpisxt
e Ti — T 25 — 20 =e 37 — 20
900 x 0.069 x 4000 x t
17 x 3.23 x 10+5 t = log — 5 = 37943 sec = 10.54 hour 19. An egg with mean diameter of 4 cm is initially at 25°C. It is placed in boiling water for 4 minutes and found to be consumer's taste. For how long should a similar egg for same customer be boiled when taken from refrigerator at 2°C. Use lumped system analysis and take thermophysical properties of egg as: k = 12 W/mK, h = 125 W/m2K C,, = 2000 J/kg K, p = 1250 kg/m3. (PU — 1997)
Consider an egg being boiled as shown— 2 h = 125 W/m k
 100°C
3 = 0.02 3 Characteristic length /s = V=R
I 86
Heat and Mass Transfer
Applying lumped system analysis, we have— T—T
epcpisxt
Ti — —125
0 02 x(4 x60)
1250 x2000 x T x 100 3 =e 25 —100 T= 100 — 75 x 0.165 = 87.5°C
The consumer requires egg with uniform temperature of 87.5°C. Applying again lumped system analysis for the egg with initial temperature as 2°C 375 xt T T = e moo
Ti — T —375 xt 87.5 —100 = e 5000 2 —100 98 x 5000 12.5 375 = 275.7 sec = 4.6 min
or
t = 10g
20. A ball of 20 cm diameter with initial temperature 400°C is exposed to air of temperature 25°C. If h = 80 W/m2K, calculate: (i) time required to cool the ball to 85°C, (ii) initial rate of cooling, (iii) instantaneous rate of heat transfer at the end of one min. Take k = 40 W/mK, p = 7800 kg/m3, C,, = 450 J/kg K. R = 1 x 103 Characteristic length Vs' = V = 3 3
= 3.33 x
m
B. = hls k
80 x 3.33 x103 40 = 6.66 x 103 < 0.1
Since Biot number is 0.1, lumped system analysis can be applied T—
—h Too
—T
= e
85 — 25 400 — 25 — t=
e
pC 1 xt P' 80 xt 7800 x 450 x3.33x10'
7800 x 450 x 3.33 x 103 x log 375 80 60 78 x 450 x 3.33 x 101 x 4.094 8
Transient Heat Conduction
I 87
= 5981.5 sec = 1.66 hour Now for initial rate of cooling, we have— —mC P
dT
at dT
or
at
= hA(Ti — hA (pV)Cp(Ti —
V R Characteristic length `/,' = A 3 = 3.33 x dT
m
80
at — 7800 x 450 x 3.33 x10
3
(400 — 25)
80 x 375 — 2.567°C/s 78 x 45 x 333 Instantaneous heat transfer rate is— qin = hA (T — To j h
But
T—
xt
= (Ti — To je PCP4 h
Here
xt
qin = hA(Ti — To )e PCPis t = 60 sec qin = 80 x 47r x (1 x 102)2 x (400 — 25)7800 —80 x 60
x 450 X e 3'33 x1°3 x 7800 x 450 = 37.68 x e13.41 = 25 W Total energy transferred during first minute is— Q = pVCp (Ti — Toc)(1—e
P
ht Cp l,
80x60
4 7r x (102)3 x 450 x (400 — 25) x [1 _ e7800x450x3.33x103 = 7800 x — 3 4 x 7v X 450 x 106 [1 com] x 375 = 7800 x — 3 = 7.8 x 0.15 x4 X7r(1 — 0.664) x 375 = 7.8 x 0.15 x 4 x 7V (0.336) x 375 = 1851 W 21. A thermocouple junction is in the form of a 4 mm diameter sphere. The properties of the material are Cp = 420 J/kg K, p = 8000 kg/m3, k = 40W/mK, h = 40 W/m2K. The junction is initially at 40°C is inserted in a stream of hot air at 300°C. Find: (i) time constant, (ii) thermocouple is taken out from the hot air after 10 sec and is kept
I 88
Heat and Mass Transfer
in still air at 80°C. Assuming heat transfer coefficient in still air is 10 W/m2K, find temperature attained by junction 20 sec after removing it from hot air stream. (PU — 1955) — 2 x103 Characteristic length is — V — R A 3 3 40 x3 x103 h s B = i = k 40 = 6.67 x let 0.1 Hence, lumped system analysis is possible. h xt T—T pcp is T—T = e Time constant r =
pC„Is
8000 x 420 x x 103 3 40
= 56 secs Temperature (T1) attained by junction after 10 secs when exposed to hot air stream (Toc = 300°C) Ti 300 5t6 40 — 300 — e _15(6) — 300
here t = 10 secs
e 0.84 40 — 300 — Tl — 300 = —0.84 x 260 = 300 — 218.4 = 81.6°C Now temperature attained by the junction in 20 sec when exposed to air at 30°C having initial temperature of 81.6°C. h2 T2 — 30 pc 4 xt 81.6 — 30 — e 10x20 8000x420xx103 3 =e = 0.915 T2 = 30 + 51.6 x 0.915 = 30 + 47.21 = 77.21°C 22. A mercury thermometer bulb idealized as a sphere of 0.3 mm radius is used for measuring the temperature of fluid whose temperature is fast changing. Taking the
Transient Heat Conduction
I 89
following data: k = 120 W/mK a = 5 x 105 m2/s, h = 10 W/m2K, specify whether thermometer is able to read the temperature faithfully if time for temperature change of the fluid is 3 secs. If not, what should be the diameter of a thermocouple to read the temperature of the fluid taking following data: k = 100 W/mK, a = 120 m2/s, h = 10 W/m2K.
0.1 x 103 Characteristic length is — R 3 3 = 0.333 x 104 m 10 x 0.333 x104 hi s= 100 k = 0.333 x le 0.1
B•=
Hence, lumped system analysis is applicable Time constant
=
p Cp h
( PC p ) k k h
1 s
1 k a h xis 1 x 100 x 0.333 x 104 5 x105 10 = 2 x 10 x 0.333 = 6.66 sec > 3 sec
= — X—
As time constant is more than the time of fluctuation of fluid, the temperature cannot be recorded faithfully by the thermometer. It is possible with thermocouple with some other radius having time constant as 3 sec. 1 x— k xis =3=— a h i 3x120 x100 or — 100 = 36 Now or
is = 11 =36 3 R = 36 x 3 = 108 mm = 0.108 m
23. A household iron has a surface area of 0.05 m2 and is made of stainless steel having weight of 1.4 kg and 500 W capacity. If h = 17 W/m2K on the surface and air at 30°C, how long it takes iron to reach 110°C. Take p = 8000 kg/m3, C,, = 500 J/kg°C & k = 20 W/m2C. Find out the Biot number.
I 90
Heat and Mass Transfer
Bd—
m h V hi pA A— 20x
(8001x.50.05) 20
= 0.0375 0.1 Now
where
a — b 0i _ e —bt a —b0
a=
Aq 8c. b _ hA mCp mCp
0.05 x 500 17 x 0.05 b= 1.4 x 500 1.4 x 300 a = 0.0357, b = 1.2 x 103 01 =T; —T = 0 0=T—T = 110 — 30 = 80 a=
0.0357 = 0.0357 — 80 x1.2 x103 1.2 x 103t = log 0.0603 0.0357 = 0.524 t — 0.524 — 436.67 sec 1.2 x1024. A thermocouple junction of spherical form is to be used to measure the temperature of the gas stream for which h = 400 W/m2K and k = 20 W/mK, C,, = 400 J/kg K, p = 8500 kg/m3. Find: (i) junction diameter needed for the thermocouple to have time
Transient Heat Conduction
I9I
constant of one second, (ii) time required for the thermocouple junction to reach 198°C if junction is initially at 25°C and is placed in a gas stream which is at 200°C. Characteristic length = A V = 4/3 x R23 = 1/3 R 4 7r R Time constant r = or
pClg P = 1 sec (given)
8500 x 400 x RI3 =1 400 400 x 3 — 3.53 x 104 m 8500 x 400 d = 2R = 2 x 3.53 x 104 m = 0.706 mm
or
R=
Hence, junction diameter
Now temperature distribution is— T—T
Ti — T
=e
198 — 200 I = e as time constant r = 1 sec 25 — 200 t = log 175 2 = 4.47 sec 25. How the transient heat conduction for a plain wall can be analysed with the help of Heisler charts. (UPTU — 2006 — 7) Or How is transient heat conduction in solids having finite conduction and convective resistance (0 < Bi < 100) determined? All practical heating and cooling of metal objects involve both internal and external resistance. In lumped system analysis, we consider body at uniform temperature, thereby neglecting the internal resistance to heat transfer. Heisler charts are used to solve the transient heat transfer problems taking into consideration both internal and external resistance to heat transfer. The charts are used when Bi > 0.1. Such chart solutions for transient analysis are available for standard objects like rectangular, cylindrical and spherical shapes. Heisler charts are provided at the end of this chapter.
x=—L
x=L x=0
I 92
Heat and Mass Transfer
Consider a plane wall of thickness `2L'. It has initial temperature 70' and it is suddenly exposed to environment at temperature ' . Analytically we have seen that the temperature of the object depends on Biot number, Fourier number and location of section
0.1
—8
0.01
I
I
I
1 2 3 4 8 12 F0 =
as
L2 Centreline temperature for a plane wall of thickness (2L)
Heisler had drawn the graphs for different values of
hL
Bi =
hL)taking Ton k
To —1:
on
yaxis and af on xaxis. Centreline temperature can be obtained from this chart. To determine the temperature at a location other than centreline, other chart is used which have graphs for different values of
T'
t
taking "; on yaxis and on xaxis where hL Tat) — T(x, t) = temperature at x at time t, T(0, t) = centreline temperature and x = the distance measured from centreline. L
0.2
I N
H C
0.5
L
0.001
0.1
1
10 1 Bi
100
Temperature as a function of centreline for plane wall of thickness 2L
The heat gained or lost by the plane wall can be determined using chart which is plotted with
Transient Heat Conduction
I9I
constant of one second, (ii) time required for the thermocouple junction to reach 198°C if junction is initially at 25°C and is placed in a gas stream which is at 200°C. Characteristic length = A V = 4/3 x R23 = 1/3 R 4 7r R Time constant r = or
pClg P = 1 sec (given)
8500 x 400 x RI3 =1 400 400 x 3 — 3.53 x 104 m 8500 x 400 d = 2R = 2 x 3.53 x 104 m = 0.706 mm
or
R=
Hence, junction diameter
Now temperature distribution is— T—T
Ti — T
=e
198 — 200 I = e as time constant r = 1 sec 25 — 200 t = log 175 2 = 4.47 sec 25. How the transient heat conduction for a plain wall can be analysed with the help of Heisler charts. (UPTU — 2006 — 7) Or How is transient heat conduction in solids having finite conduction and convective resistance (0 < Bi < 100) determined? All practical heating and cooling of metal objects involve both internal and external resistance. In lumped system analysis, we consider body at uniform temperature, thereby neglecting the internal resistance to heat transfer. Heisler charts are used to solve the transient heat transfer problems taking into consideration both internal and external resistance to heat transfer. The charts are used when Bi > 0.1. Such chart solutions for transient analysis are available for standard objects like rectangular, cylindrical and spherical shapes. Heisler charts are provided at the end of this chapter.
x=—L
x=L x=0
I 94
Heat and Mass Transfer
Similar solutions of heat conduction can be obtained for a sphere of radius `r0'. Heisler's charts can be used for finding temperature at centre of the sphere (radius `r0') at different instants of time. The charts are drawn with central temperature on yaxis with respect to a 1 Fourier number ( F0 x t on xaxis for various . The charts are also available for Bi ro JJ finding temperatures at other positions as a function of r and also for finding Po Q0
Heating and cooling of a sphere 27. A solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in same surroundings. If the temperature difference between the body and that of the surroundings is T, then— (a) (b) (c) (d)
both spheres will cool at the same rate for small values of T. both spheres will cool at the same rate for all values of T the hollow sphere will cool at a faster rate for all the values of T. the solid sphere will cool at faster rate for all the values of T (IES 92)
pC„V hA Volume of solid sphere is higher than the volume of hollow sphere. Time constant Z =
Hence,
rsolid > Thollow
Therefore, hollow sphere will cool faster than solid sphere. Option (c) is correct. 28. A large metal plate of thickness 5 cm is initially at 460°C. It is suddenly exposed to a fluid at 100°C with a convection coefficient of 142.5 W/m2K. Find the time needed for its mid plane to reach a temperature of 316°C and surface temperature at the same instant of time. Take k = 21.25 W/mK and a = 1.2 x 105 m2/s. (UPTU — 2007 — 8) Guidance: Here
Use Heisler charts. 2L = 5 cm or L = 2.5 cm
Transient Heat Conduction
1 Bi
21.25 k hL — 142.5 x 2.5 x 102 = 5.96 = 6 —
The centreline temperature is— Tg
_ 316100 216 _ 06 — 460 —100 — 360 — .

Now for
— 1 = 6 and T°'t = Ts —1: Bi
We get the value of
at F, = 2 = 2.2 t= =
0.6,
2.2 (2.5 x 102)2
2.2 x L2 a
1.2 x 105
45.83 sec
0.6
1
2.2
=6
F = at o
2
L
Centreline temperature of infinite plate
Now
x = 1 for surface of the wall, L Bi
=
6, we get
=
—
0.92
6
0.92
I 95
I 96
Heat and Mass Transfer
T2.5,45.83 — 100
0 92 460 100  . T2.5,45.83 = 100 + 360 x 0.92 = 100 + 331.2 = 431.2°C
29. Two identical balls made of pure iron and copper having diameter of 6 cm and initial temperature of 500°C are being cooled in oil having temperature of 100°C and heat transfer coefficient of 10 W/m2K. It is desired that both balls should reach a temperature of 150°C at same time. Which ball should be inserted into oil first? After how much time second ball should be inserted into oil? Justify the answer and formula used. Properties are— (a) Pure copper: p = 8954 kg/m3, k = 386 W/mK Cp = 383 J/kgK (b) Iron: p = 7897 kg/m3, k = 73 W/mK Cp = 452 J/kgK Find Biot number for copper and iron balls. hls (Bd copper = is = characteristic length of ball = where
R = radius 1, = 3 = 1 cm
Now
(B d copper =
hR 3k
10 x3x102 = 2.5 x 3 x 386 (lid iron =
hR 3k
=
< 0.1
10x 3 x102 = 13.6 x 104 < 0.1 3 x 73
Hence, lumped system analysis is possible in both the cases. _ h xt To Tcc pcp4 Now T Tcc e T = 150 And For iron ball 10
or
500 100 = e 7897x452x1x102 xis 150 100 = 7897 x 452 x102 log 400 t, 10 50 = 3569.44 x 2.079 = 7421 sec
3
Transient Heat Conduction
I 97
For copper ball 8954 x 383 x 102 400 x log = 7223 sec 10 50 is  tc = 7421  7223 = 298 sec t, =
Now
Hence, copper ball has cooling faster and it has to be inserted after the iron ball. 30. A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of gas stream. The convective heat transfer coefficient of the bead surface is 400 W/m2K. Thermophysical properties of thermocouple material are k = 20 WmK, C,, = 400 J/kg K and p = 8500 kg/m3. If the thermocouple initially at 30°C is placed in a hot stream of 300°C, the time taken by the bead to reach 298°C is (a) 2.35 sec (c) 14.7 sec
(b) 4.9 sec (d) 29.4 sec (GATE 2004)
Temperature of thermocouple will vary exponentially.  Thermocouple
—..To,= 300°C . _,..
c: To = 30° d = 0706 mm h
T 1: pCpl' xt = eBiF0 = e_mt = e To — 1: V —R = = 0.706 = 0.1176 mm 1c = A 3 6
7: T =e T_ To or
400 3 xt 3500x400x0.1176x10
1x103 = e
103 xt xt =
log 300  30 300  298 = 4.9 sec.
t=
Option (b) is correct. 31. A steel ball of 5 cm diameter at 500°C is suddenly placed in a controlled environment maintained at 100°C. Taking following data, find the time required to maintain centre point temperature of 150°C in the ball. C,, = 450 J/kg°C, k = 35 W/m°C, h = 10 W/m2K, p = 8000 kg/m3. (UPTU  2009  10)
I 98
Heat and Mass Transfer
Biot number Bi =
his k
R /, for steel ball = — = 25 = 0.833 3 3 10 x 0.833 = 0.238 35 As Bi > 0.1, hence the lumped heatsystem analysis can not be used. Heisler charts are to be used. 1 1 = 4.2 Bi 0.238 The centreline temperature isB1 =

_ 150  100 _ 50 = 0.125  500 100  400
From Heisler chart, for centreline temperature = 0.125 and 1 = 4.2, we haveBi Fourier Number F0 = tNow •••
at 2 Po
=2
2. rj a a=
p.Cp
35 = 9.7 x 106 m2/s 8000 x 450
2 x (25 x 102)2 9.7 x 106 = 1.29 x 102 S. = 129 s
t=
32. A large metal plate 10 cm thick is initially maintained at a temperature of 500°C. It is suddenly exposed to a surrounding at 140°C with a heat transfer coefficient of 570 W/m2K. At a later instant, its centre plane reaches a temperature of 350°C. Find the temperature of a plane at a distance of 2.5 cm from the mid plane at the same instant. Take k = 170 W/mK, a = 4.78 x 105m2/s. (UPTU  2008  9) Here
2L = 10 cm L = 5 cm = 5 x 102m
Transient Heat Conduction 1
—
Bi
k
170
hL
570 x 5 x 102
I 99
—6
The centreline temperature is— To,t — Ts. = 350 —140 Ts — T 500 — 140
=
210 360
= 0.583
1 Now from Heisler chart for centreline temperature = 0.583 and — = 6, we get— Bi
Fourier Number Fo = a2t = L
or
t= =
3.8 x
3.8
L2
a 3.8 x (5 x 102 )2 4.78 x 105
= 1998 Now
to find temperature at a plane at distance x = 2.5 from the midplane. x 2.5 = = 0.5 L 5
x Using position correction chart for — = 0.5 and 1 = 6, we get— L Bi Tx — T Ts —1:
or or
= 0.95
Tx —140
= 0.95 500 —140 Tx = 140 + 360 x
0.95 a 482°C.
33. What are infinite and semi infinite solids or bodies? Give the examples of semi infinite body. What is the general criteria for any body to be considered as semi infinite body? An infinite body is a body which extends itself infinitely in all directions of space. If an infinite solid is split in the middle by a plane, each half is known as semi infinite solid. A semi infinite solid is a body that has a single boundary surface. Which extends to infinity in one direction as shown in the figure. The earth and any thick slab can be considered to be a semi infinite body which helps in obtaining the temperature variation in it nearer to its surface. The general criteria for an infinite body to be considered semi infinite body is—
200
Heat and Mass Transfer
8
21,1 at
0.5
where 8. thickness, a = thermal diffusivity and t = time. The heat conduction is a semi infinite body occurs is one dimensional direction.
A semiinfinite solid
34. How is temperature distribution in semiinfinite solid found out using Gaussian error function? Find the expression for: (i) instantaneous heat flow rate at given xlocation, (ii) heat flow at surface, and (iii) total heat flow for a time interval. Consider a semi infinite solid, initially at uniform temperature T,. The surface of the solid is now (t > 0) subjected to some boundary condition. The temperature distribution and heat flow at any position 'x' in the solid with time can be obtained by using the equation—
a 2T 1 aT — — where 0 ax` a at
x
Transient heat flow in a semi infinite plate
The body is initially at uniform temperature Ti including the surface at x = 0. The surface temperature at x = 0 is instantaneously changed to and held at T, for all times greater than
Transient Heat Conduction
20 I
t = 0. Thus the boundary conditions are— T(x, 0) = T(0, t) = Ts for t > 0 t) = Ti for t > 0
The solution of the differential equation for transient heat conduction as given with the above three boundary conditions gives the following temperature distribution at any time 't' at a plane parallel to and at a distance 'x' from the surface740 — Ts
Ti — Ts
where
_erf
=
x, 2V at
= erf (0)
and erf (0) = Gaussian error function. 2Ia•t
The values of Gaussian error function for various values of '0' are tabulated at the end of this chapter. Also we have— erf O =
where
2 reedn x
tl = similarity variable =
2,\Ia•t
Now
x 2\la•t
T(x, t) — Ts = (Ti — Ts) exf
On differentiation, we get—
aT
[Ti — Ts j
ax
Vir•a•t
X
exp
—x2 [4.a•ti
As per fourier's law, the instantaneous heat flow rate at a given xlocation within the semiinfinite solid at any specified time 't' is— (x, t)
= —k • A.
a T. ax
[ X2 = —k • A (T, —Ts) x exp 4. a • t a•t
At the surface we have x = 0, hence surface heat flow is— kA(Ti — Ts) Q0,0 =
g • a•t
It is apparent from the above that the surface heat flow diminishes with time. The total heat flow in time interval from t = 0 to t = t is—
202
Heat and Mass Transfer
—kA(Ti —TS ) ft 1 Qtotal —
v r•a
.s1t
dt
—2 k • A • (Ti — Ts)
•a
35. What is the expression for the temperature at the centre of cylinder or sphere of radius `R' in Gaussian error function? Under similar conditions of heating or cooling, the temperature at the centre of infinite cylinder or sphere of radius 'IV can be given as— T — Ts Ti — Ts The values of function erf
[ a
= erf
[ a.2t
for the cylindrical and spherical surfaces can be obtained
/V' J from the figure below: 1.0 0.8 0.6 0.4 0.3 0.2 0.15 0.1 r
1 0.08 0.06 Infinite cylinder
0.04 0.03 0.02 0.015 0.01 0.1 0.15 0.2 0.3 0.4 0.6 0.8 a.t R2 —
Error integral for cylinders and spheres
Transient Heat Conduction
203
36. What do you understand from penetration depth and penetration time? The penetration depth is the depth from the surface (x = 0) where the temperature changes is within 1% of the applied change in temperature (7; — Ti). Hence T
= 0.99 = erf (1.8)
Ti — Ts
Penetration depth x= 1.8 x 2 x
a• t
x = 3.6V a•t
Or
Therefore the penetration time at a given depth indicates the time taken by the heat to get the penetration of 1% from the surface. Hence—
__ 1 x = a
x )2
36
37. The ground at a particular location is covered with snow pack at —10°C for a continuous period of 90 days. The average soil properties at that location are: (i) k = 0.4 W/mK, and (ii) a = 0.15 x 106 m2/s. Assuming an initial uniform temperature of 15°C for the ground, determine the maximum depth to bury the water pipes from the surface to avoid freezing. Ts = —10° C /
/
/
/
/
/
/
_ earth surface /
/
/
Soil
b
Water pipe buried in soil
The temperature distribution in the soil can be given as74,0 — 7; = erf Ti  Ts
2Vot.t
Here T(x, t) = 0 when water starts freezing Ts = —10°C Tl = 150°C Hence
0 — (10) 15 — (10)
= erf
x 2 ' a .t
= 0.4
From the table for erf (0) = 0.4, the value of 0 = 0.37
204
Heat and Mass Transfer
tfr = or
21/ a •t
= 0.37
x = 2 x V0.15 x106 x (90 x 24 x 60 x 60) x 0.37 = 0.8 m
38. A large block of steel is initially at 35°C. The surface temperature is suddenly raised and maintained at 250°C. Calculate the temperature at a depth of 2.5 cm after a time of 30 seconds. The thermal diffusivity and thermal conductivity of steel are 1.4 x 105m2/s and 45 W/mK respectively. (Annamalai University — 2008 — 9) For semi infinite solid, we have— T —Ts = erf [ Ti — Ts 2 11(37 xt x 30 x 102 — 21,1a•t 241.4x 105 x 30
Now
= 0.61 From Gaussian error function table, the value of— erf [0.61] = 0.612 T250 = 0.612 35— 250 T = 118.5°C. 39. A large concrete high way initially at a temperature of 70°C and stream water is directed on the highway so that the surface temperature is suddenly lowered to 40°C. Determine the time required to reach 55°C at a depth of 4 cm from the surface. Annamalai University — (2007 — 8) T1 x = erf [ 21ji c Ti — Ts For concrete, a = 8 x 107 m2/s from data book. Now
erf[
5540 15 = ix ]= = 0.5 70 —40 30 2V a • t
Transient Heat Conduction
From the table for erf
[2 \,/ xa • t 2 cc•
205
= 0.5, we have= 0.48
X
2
1 x — a
2 x 0.48) 4x102 )2 (`2x0.48
1 8x107
17.36 x 104 8x107 = 2170 seconds or 36.2 min 40. A thick steel slab is initially at a uniform temperature of 25°C. When the slab is exposed to hot flue gases, the surface temperature suddenly changes to 450°C. Make calculations for the temperature in a plane 250 mm from the slab surface 5 hours after the operation of change in surface temperature. Find also the heat flowing into 2 square metres of this plane and the total energy flowing through the surface during 5 hours period. Assume h = 160 kJ/m • hr • degree, p = 8000 kg/m3 and C = 0.48 kJ/ kgdegree.
a= Now
p• C
T Ts = erf TTi — Ts
160 = 4.17 x 102 m2/hr 8000 x 0.48 ,x
a•t
0.25
j= erf
2 V4.17 x 102 x 5
= erf (0.274) From the table of Gaussian error function, it is found out
Hence or or
erf (0.274) = 0.30 T T S = 0.30 Ti  Ts T  450 = 0.30 25400 T = 322.5°C
206
Heat and Mass Transfer
The instantaneous heat flow rate of this plane at t = 5 hr is k • A•(TT,)•exp Q(0.25,5)
=
x2 (. 4 a• t
V tr•a•t
= 160 x 2 (25  450). erf
[
0.252 4 x 4.17 x 102 x 5
= 1.56 x 105 kJ/hr The total heat flow from this plane in 5 hr isQtotal = 1/7r
=
x k • A • (Ti  Ts ) x V a
f 5 2 x 160 x 2 x (25  450) x '\,1 4.17 x 102 e tc
= 1.68 x 106 kJ. 41. A thick copper slab (a = 1.1 x 104 m2/hr, k = 380 W/mK) is initially at a uniform temperature at 10°C. Suddenly the surface is raised to 100°C. Calculate the heat flux at the surface 5 and 10 min after raising the surface temperature. How long with it take the temperature at the depth 5 cm from the surface to reach 90°C. Heat flux at t = 5 min isqt
=5 min =
k • (T  Ts) V r.a• t 380 x (10 100)
=
Vtrx1.1x104 x5x 60 = 106.2 kW/m2 Similarly heat flux at t = 10 min isqt=lo mm =
380 x (10 100) ~tc x1.1x104
x 10x60
= 75 kW/m2 Now temperature distribution isTX TS
Ts
= erf
x 211 a • t
Transient Heat Conduction
90 100 = erf 10100 or
erf
x 2,0H 1t.t
207
a•t
= 0.11
From the table of Gaussian error function, we have= 0.10
2fa•t 5x102
 0 10
2111.1x lex t or
t = 568 hrs
42. The surface temperature of a large aluminium slab initially at 200°C is suddenly lowered to 70°C. What is the total heat removed from the slab per m2 area when the temperature at a depth of 4 cm has dropped to 120°C. Assume a = 8.4 x 103 m2/s and k = 204.2 W/mK. Tx  T, = erf  Ts
x 2v a•t
120  70 x = 0.385 200 70 Va•t From the table of Gaussian error function, we have
or
erf ,
x = 0.36 21 a•t (0.04) 2 = (0.36)2 4 x 8.4 x 10' x t 16 x104 = 0.37 sec 4 x 8.4 x 103 x 0.1296 The total heat removed per unit area is2 x k x (Ti  Ts) x , t qtoto = V a or
t=
= 2 x 204.2 x (70 200) x ; 1/ Jr
= 1988 kW/m2
0.37
A; 8.4 X 103
Heat and Mass Transfer
208
HEISLER CHART (PLANE WALL)
1.0 0.7 0.5 0.4 0.3 0.2 0.1 0.07 0.05 0.04 0.03 I 2 0.02
0.01 0.007 0.005 0.004 0.003 0.002 0.001 0
•
...am I. In.
,Ite4
kLIL_
Mb r
II
416.110111. III
1) kh
X
0 cf3
13
1,„
90
100
80 6'0 6:9 'PS
v0 v
4 6 8 10 12 14 16 18 20 22 24 26 28 30 40 50 60 7080 90 100110 130 150 200 300 400 500 600
9 IIMIN1INIM1
3
a
111MMILIM11.1.9MIL91=1611=1.
NwiLal NoleL Lm misamomhm.
'A4
2
0
44 0 'tti%\ 0%
IIMINIMEWNICUOS.
1
Heishler chart for temperature history at the centre of a plane wall
Transient Heat Conduction
CORRECTION FACTOR 1.0
XL = 0.2
0.9 0.4 0.8 0.7 0.6 0.6
L
0.5 0.4
0.8
0.3 0.9 0.2
0.1
1.0
0 0.01 0.02 0 05 0.1 0.2
05
1.0 2 3 5
10
20
50 100
1_k B1 hL
Correction factor chart for temperature history in a plate plane wall
209
2 I 0 Heat and Mass Transfer
P14
>4 C.)
14
I
"Z5
1.0 0.7 0.5 0.4 0.3 0.2 0.1 0.07 0.05 0.04 0.03 0.02 0.01 0.007 0.005 0.004 0.003 0.002 0.001
It
,.Z.4.,,N ''
,
9
la \
,'z4 .41:1 ,4 .fir 1,21."SISr.V.4 — 4,......z.v....,.4.44, 14141\ 1:111441". NS %%hiN\._ .. 011\litiMA\SZN\
*0
01
voimanualmw Nommommunimummi.
4
sx%
70 + 0
1
• MN••MENNEMMENNEN■
ASso AO
6 8 10 12 14 16 18 20 22 24 26 28 30 40 50 60 70 80 90 100 110 120130 140 150 200 300 350
\\
mmanominmeugusw.
mh.5001 :,Z,`" ■m1.
limelipL
N, M11i'1111U1:118
nil fl
0 26,
0
\ .... IawlM\ RIM LW"
it
3
I
• a .a NVOOMMILIMMW& NOOMMAIIIIMM >‘ IMEEIOMUNM k
2
usImmthuLwitainsam I INNII=NEMIAILIM
1
at
F0  112
Heishler chart for temperature history in a cy linder
0
0.7
1.0 0.5 0.4 0.3 0.2 0.1 0.07 0.05 0.04 0.03 0.02
0.01 0.007 0.005 0.004 0.003 0.002 0.001
taIN •
A
V.
1.0
1.5
2
\
'11101MIn,
2.5
\\
S va ,90 \ AS
(90
k
3 4 5 6 7 8 9 10 15 20 25 30 35 40 45 50 70 90 110 130 170
\\\ \\\
c,)0
kitMIN "N`1.411\11111.11L '116. 716. \\WV \11... \\ I MUM Nikh. ' 11\\NW
0.5
NEW itilLISIMIMInkfl.Wink 'MAWR
ELFWERMWW
M141M1 MIOIM,M,MCM "1110/0/MOMMI. MIIMMILIMM13.1060.1Welh. .NIZIMMOMMOMMI. . 10611111MILIMIln. MittalLINEL LMLW‘k
'EP
0
at F0 = — L2
210 250
\\\
Heishler chart for central temperature history in a sphere
2II Transient Heat Conduction
2I2
Heat and Mass Transfer
HEISLER CORRECTION FACTOR 1.0
= 0.2
0.9 04 0.8 0.7 06 r/R
0.6 0.5 0.4
08
0.3 09 0.2 0.1
1.0
0 0.01 0.02 0 05 0.1 0.2
0.5 1.0 2 3 5 1_k B1 hR
10
20
50 100
Correction factor chart for temperature history in a Cylinder. 1.0
r
=6:2
0.9 4 0.8 0.7 06 r/R
0.6 0.5 113
0.4
0.8
FX H 0.3 0.9 0.2 0.1
1.0
0 0.01 0.02 0 05 0.1 0.2
0 5 1.0 2 3 5 1=k B1 hR
10
20
50 100
Correction Factor Chart for Temperature History in a Sphere.
Transient Heat Conduction 2 I 3
GAUSSIAN ERROR FUNCTION erf 0 = ,2 id' en2dri 1 gJ0
0
erf 0
0
erf 0
0
erf 0
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40
0.0 0.0225 0.0451 0.0676 0.0901 0.1125 0.1348 0.1569 0.1709 0.2009 0.2227 0.2443 0.2657 0.2869 0.3079 0.3286 0.3491 0.3694 0.3893 0.4090 0.4284
0.42 0.44 0.46 0.48 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.0 1.05 1.10 1.15 1.20 1.25 1.30
0.4475 0.4662 0.4847 0.5027 0.5205 0.5633 0.6039 0.6420 0.6778 0.7112 0.7421 0.7707 0.7970 0.8270 0.8427 0.8614 0.8802 0.8952 0.9103 0.9221 0.9340
1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95 2.0 2.10 2.20 2.30 2.40 2.50 2.60 2.80
0.9431 0.9523 0.9592 0.9661 0.9712 0.9763 0.9800 0.9838 0.9864 0.9891 0.9909 0.9928 0.9940 0.9953 0.9967 0.9981 0.9987 0.9993 0.9995 0.9998 0.9999
Chapter
6
FORCED CONVECTION HEAT TRANSFER
A REYNOLDS NUMBER A VISCOSITY A PRANDTL NUMBER A SHEAR STRESS A NUSSELT NUMBER A FREE CONVECTION A STANTON NUMBER A FORCED CONVECTION A A LAMINAR FLOW RAYLEIGH NUMBER A CRASHOFF NUMBER A TURBULENT FLOW A HYDRODYNAMIC BOUNDARY LAYER A SKIN FRICTION COEFFICIENT A FILM TEMPERATURE A THERMAL BOUNDARY LAYER A BULK MEAN TEMPERATURE A COEFFICIENT OF FRICTION A LOCAL HEAT TRANSFER COEFFICIENT A EQUIVALENT DIAMETER A AVERAGE HEAT TRANSFER COEFFICIENT A DIMENSIONAL ANALYSIS
INTRODUCTION Thermal convection occurs when a temperature difference exists between a solid surface and a fluid flowing past it. The convection process is essentially a process of heat energy transport achieved by the circulation of fluid particles. As heat energy is transported by the transport of fluid itself, hence transported heat energy depends upon: (i) conduction, (ii) fluid flow, and (iii) mixing of fluid medium. The flow can be laminar or turbulent. In turbulent flow, irregular velocity fluctuations are always superimposed on the main stream flow and this provides additional mixing effect on the flowing fluid particles. Hence, these fluctuations are responsible for the higher transfer of heat and momentum in turbulent flow. The rate of heat and momentum transfer in turbulent flow is many times larger than what is possible in laminar flow. The heat transfer by convection is determined by studying the analogy between heat and momentum transfer. The convective heat transfer problems may be solved using the empirical correlations of dimensionless numbers such Nusselt number, Prandtle number and Reynolds number.
Forced Convection Heat Transfer
2I5
1. What do you understand by the fluids? Fluid is a substance which cannot acquire any static equilibrium under the action of any shear fore of even a small magnitude. In other words, fluids cannot acquire a static equivalent deformation (0) in order to achieve an equilibrium with external applied torque. However, a fluid resists the applied torque by attaining the constant value of the rate of change of deformation ( d° — and then the acceleration of fluid becomes zero. Hence, a fluid dt is a substance which can: (i) attain the shape of the container as it has no definite shape of its own, (ii) deform continuously under the action of shear force, (iii) attain the equilibrium rate of deformation, and (iv) the deformation does not disappear on the removal of torque. 2. What is the viscosity of a fluid? The property which characterizes the resistance which a fluid offers to applied shear force is called viscosity. The resistance does not depend upon the deformation (0) but on the rate of deformation d°). When there is a relative motion between different layers of the fluid, t then there is a tangential friction force in between the layers. Viscosity is less in gases but larger in liquids. When a solid mass slides over a surface, a friction force is developed to oppose the motion. Similarly, when a layer of a fluid slides over another layer of the same fluid, a friction force is developed between them opposing the relative motion. The tangential force is called viscous force. Consider a fluid moving in streamlined manner on a fixed horizontal surface as shown in figure. The layer of the fluid in the contact of the surface remains stationary while the velocity of other layers increases with distance from the fixed surface. There is therefore an internal tangential resistive force acting between two layers which is opposing their relative motion.
a)
N _ 0
Velocity Velocity gradient due to viscosity
This resistive force is called viscous force. In order to maintain the flow of the fluid, we have to apply an external force to overcome the tangential resistive or viscous force. According to the Newton's equation of viscosity, the tangential shear stress (resistive force per unit area) existing between two adjacent fluid layers is directly proportional to the velocity gradient existing in the fluid in a direction perpendicular to the fluid layers.
2I6
Heat and Mass Transfer
Shear stress (r) a or where
du y
du = du = Velocity gradient dy = Viscosity
Idu r
U U„,
u0 Velocity gradient
Fluids following Newton's law of viscosity are called Newtonian fluids while other fluids are called nonNewtonian fluids. 3. How is heat transferred in convection? Heat is transferred in convection by two mechanisms. One is heat transfer from a hot surface to adjacent fluid by random molecular motion which is called diffusion. The other one is the transport of heat by bulk movement of the fluid from higher temperature region to lower temperature region. 4. What is the classification of convective heat transfer? The convective heat transfer can be classified as natural and forced convection depending upon how the fluid motion is initiated. The natural convection is a process in which the fluid motion results due to density changing on heat transfer. Natural circulation of the fluid is caused by the rise of warmer fluid and the fall of colder fluid.
Hot plate Natural convection
In forced convection, the fluid is forced to flow over a hot surface. The heat transfer rate is much higher in forced convection.
Forced Convection Heat Transfer
2I7
Q Relative velocity of fluid layers Fan
Forced convection 5.
Hot plate
What is a laminar flow (viscous flow)? Laminar flow is a flow in which flow takes place in layers. There is no mixing of fluid particles between any two adjacent layers. The shape of lamina (layer) depends upon the shape of the boundary through which flow is taking place. In laminar flow, the fluid particles move in unmixing layers or streams and follow smooth continuous paths. The fluid particles retain their relative positions at successive crosssections of the flow passage. There is no transverse displacement of fluid particles. Soldiers marching in orderly manner is an analogy to laminar flow. The shape of laminae if flow takes place between two parallel flat plates are plane sheets parallel to each other as shown below. In case the flow takes place through a circular pipe, the laminae become concentric sheets as shown in figure.
////////////////////
//////////////////// Layers in plane sheets
Concentric cylindrical layers
6. What are the conditions which help the flow to be laminar? The flow will be laminar when1. 2. 3. 4. 5.
Velocity of flow is low Diameter of pipe is small Viscosity of the fluid is high Density of the fluid is less Reynolds number is less than 2300 for flow in pipes and less than 3.5 x 105 for flow over plates.
7. What is turbulent flow? The turbulent flow is when the velocity of flow reaches a certain limit such that the fluid particles no longer move in layers or laminae. Violent mixing of fluid particles now takes place due to which they move in random manner. As a result the velocity at any point varies both in magnitude and direction from instant to instant. In turbulent flow, the motion of fluid particles is irregular. The fluid particles move along erratic and unpredictable path. The
2I8
Heat and Mass Transfer
velocity of fluid particles fluctuates both along the direction of flow and also perpendicular to the flow. A crowd of commuters on a railway station rushing for boarding a train is an analogy of turbulent flow. The flow in pipes having Reynolds number above 4000 is always turbulent. /////////////////////////////
///////////////////////////// Turbulent flow
8. What is Reynolds number? How is it useful? (UPTU — 2002 — 3) Reynolds number is the ratio of inertia force to the viscous force of the flowing fluid. It is a dimensionless number and it is given for plate and pipe as— (a) Re — (b) Re =
pVL pVd
for plate where L is the length of plates for pipes where d is the diameter of pipe
Reynolds number is very useful for1. Predicting whether flow is laminar or turbulent. 2. Finding out the coefficient of friction (f) in order to determine the loss of heat due to viscosity. 9. What are the parameters on which convective heat transfer depends? Convective heat transfer strongly depends on: (i) fluid properties such as viscosity, thermal conductivity (k), density (e) and specific heat (Cr) (ii) geometry and roughness of the solid surface, and (iii) type of flow (laminar or turbulent). 10. Explain the concept of velocity boundary layer development over a flat plate.
U
h— Laminar
Transition to— Turbulent —I U u = 0.99 U (5= thickness of
boundary layer •••
= laminar sublayer
\%\ Leading edge
5 Re=3.5x10
5 Re=5x10
Growth of boundary layer on a flat plate
Forced Convection Heat Transfer
2I9
The figure above shows the development of a boundary layer on a flat plate held parallel to the flow of a fluid having free stream velocity as U. The retardation of the fluid increases as more and more of the flat plate is exposed to the fluid flow. The boundary layer having velocity gradient and shear stress thickens as the distance of the fluid from the leading edge of the plate increases upto certain distance from the leading edge, where the Reynolds (pux number where x is distance from the leading edge is less than 3.5 x 105. The flow in this part of the boundary layer is laminar. In the laminar boundary layer, velocity distribution is parabolic and shear stress can be given by Newton's law of viscosity. The laminar boundary layer can grow to certain thickness beyond which it becomes unstable. Hence, transition from laminar to the turbulent boundary layer occurs in the transition region. As the name suggests, the flow in the transition region is neither fully laminar nor fully turbulent. It fluctuates between the two types of flow. In turbulent boundary layer, there exists always a thin laminar sublayer close to the flat plate surface. In this sublayer, the flow is always laminar. The velocity distribution in turbulent boundary layer is logarithmic but it varies similar to laminar boundary layer from zero to free stream velocity. 11. What is hydrodynamic boundary layer? Or Describe briefly the hydraulic boundary layer over a flat plate when a free stream of fluid flows longitudinally over it. (UPTU — 2003 — 4) Hydrodynamic boundary layer is a thin layer of fluid in immediate vicinity of boundary surface in which large velocity gradients and shear stresses exist. When a fluid flows past a solid surface, the fluid particles on the surface have the same velocity as that of the surface due to viscosity of the fluid. In case the boundary surface is static, the fluid velocity du (u is fluid velocity in xdirection and y is gradually increases. The velocity gradient dy perpendicular distance from the boundary surface) varies with y and it has high values close du to the boundary. This high velocity gradient gives rise to large shear stress (/ T= I at dy ) the boundary surface. The term hydrodynamic layer is used to describe the thin layer of fluid flow on the boundary surface within which the velocity gradient is appreciable magnitude. Outside the hydrodynamic boundary layer, the velocity of fluid is same as the main stream velocity. Hydrodynamic boundary layer is defined as the distance away from the surface where local fluid velocity is 99% of the free stream velocity (u_). u  99% uo, Distance u
°Hydrodynamic
1111110"" Velocity .
Hydrodynamic boundary layer
220
Heat and Mass Transfer
12. Explain thermal boundary layer for flow of a cold fluid on a hot plate. Or Describe briefly the thermal boundary over a flat plate with flow of fluid. (UPTU — 2002 — 3) If a fluid flows on a surface having different temperature than the surface, the thermal boundary layer is developed similar to velocity boundary layer. Consider a fluid at temperature 'Too' flows over a surface at temperature 70', the fluid particles adjacent to the surface get the same temperature that of the surface. These fluid particles exchange heat energy with particles in adjoining layers and this continues till the adjoining layers have no temperature difference which happens at large distance away from the surface. As a result, a temperature gradient is developed in the fluid layers and temperature variation occurs in the fluid flow from `To' to 'Too' . The flow region over the surface in which the temperature variation in the direction normal to the surface is observed is called thermal boundary layer. The thickness of the thermal boundary layer (8th) is defined as a distance from the surface at which the temperature difference is 0.99 (To — Too). It can be appreciated that the heat transfer cannot penetrate further into the free stream as the distance from the leading edge increases. The convection heat transfer rate anywhere along the surface is directly related to the temperature gradient at that location. Therefore, the shape of the temperature profile in the thermal boundary layer leads to the local convection heat transfer between surface and flowing fluid. The boundary layer thickness is 8th =
T —T
— 0.99.
To — Too
Y
00
Thermal boundary layer
To.
Uo.
OC
T
T Hot plate at To
13. What do you understand from the boundary layer concept? In the boundary layer concept, the fluid flow field over a body is divided into two regions as under: (1) A thin region near the surface where velocity and temperature gradient are large and it is called boundary layer. (2) The region outside the boundary layer where velocity and temperature of the fluid is nearly equal to the free stream values. The hydrodynamic and thermal boundary layer thicknesses are defined as the distance from the surface at which the local velocity and temperature of the fluid reach 99% of free
Forced Convection Heat Transfer
22 I
stream velocity and temperature. The heat transfer between a fluid and a surface can take place when both the velocity and thermal boundary layer exist simultaneously. 14. Explain the formation of simultaneous velocity and thermal boundary layers. During flow of fluid over a heated surface, both velocity and thermal boundary layers are developed simultaneously. If the effects of fluid viscosity is stronger than thermal effects, then the velocity boundary layer is thicker than thermal boundary layer and vice versa. For liquid metals thermal effects are much stronger than viscous effects and hence Soh > 8. For oil and greases, the viscous effects are much stronger than thermal effects and hence 8> 8th. For gases, 8th = S.
T mmik VAowiwiwzbviwiwiwit,,w////
To —
Liquid metals
—4— =1,1 eth
Oils and Greases
15. Explain the variation of coefficient of friction (Ctx) and local heat transfer coefficient (hi) with velocity and thermal boundary layer. At surface, heat transfer is—
aTI qo = —1c( = hx(To — Tco) C '0' y=0 )
= k C a) Y To — Tso
or
n
Now as boundary layer thickness (5th) increases with increasing `x', the temperature (a T gradient must decrease as x increases. Hence, the value of h, will decrease as 'x' Y )31= 0 increases. There is a very significant increase in the heat transfer coefficient in the transition region due to fast intermixing of the fluid particles. In turbulent region, heat transfer coefficient (hi) decreases due to added resistance to heat transfer from laminar sublayer and much thicker boundary layer thickness (5th).
Laminar
Transition, Turbulent
Variation of heat transfer coefficient
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Heat and Mass Transfer
16. What is Prandtl Number (Pr)? What is its significance? (UPTU — 2003 — 3) Prandtl number (Pr) is defined as the ratio of molecular diffusivity of momentum to the molecular diffusivity of heat Pr =
Molecular diffusivity of momentum Molecular diffusivity of heat
µCp k
pip k
a
Kinematic viscosity Thermal diffusivity of heat It means that Prandtl number is a measure of relative value of momentum transfer and heat energy transfer in the velocity and thermal boundary layers. In other words, Prandtl number measures relative thickness of velocity and thermal boundary layers. Therefore we have (a) Pr >> 1 then 8» 8th for oil. It means heat diffuses very slowly. (b) Pr = 1 then S = 8th for gases. It means that momentum and heat diffusion take place through fluid at the same rate. (c) Pr 1
///////,//// //,/,//// Thermal boundary layers
17. What is Nusselt number (Nu)? (UPTU — 2002 — 3) Nusselt number (Nu) is the ratio of rate of heat flow by convection process under a unit temperature gradient to the rate of heat flow by conduction process under a unit temperature gradient through a stationary thickness of metres Heat flux through convection Nu = Heat flux through conduction in fluid
Forced Convection Heat Transfer
For convection: For conduction:
223
Qconv — h x (T — Too) h when T — Too = 1 A k Qcond =dT kx — when dT 1 A L L hL Nu = —
Nusselt number is a measure of the convection heat flow occurring at the surface. Nusselt number in thermal boundary layer is analogus to the friction coefficient (Cf) in the velocity boundary layer. Bigger the Nusselt number, more is the heat transfer coefficient (hx) and higher is the convection heat transfer from the surface. It represents the slope of the temperature gradient curve in the thermal boundary layer thickness. If 'Nu' is unity, it indicates that there is no convection and heat transfer is by pure conduction in the boundary layer. Larger value of 'Nu' indicates large convection in the fluid. 18. What is Grashoff number (Gr)? (UPTU — 2002 — 3) Grashoff number is used in natural convection and its role is same as that of Reynolds number in forced convection. It is defined as the ratio of the product of inertia force and buoyancy force to the square of viscous force Gr = Inertia force x bouyancy force (Viscous force)2 (p V2 L2) x (p •
• g • AT • L3c )
(tiV L)2 _ p2 • —
• g • AT • L3 /I
2
1 , A T = temperature difference Tf + 273 + Too between surface and fluid, Tf = mean film temperature = To 2 , = significant length where /3 = coefficient of volumetric expansion =
of the body (height for vertical plates and cylinders, diameter for horizontal cylinders and spheres). In natural convection, the fluid flow is produced by bouyant effects resulting from a temperature difference. For free convection, the transition from laminar to turbulent occures when Gr = 109. 19. What is Stanton number (St)? It is the ratio of the heat transfer at the surface to that transported by fluid by its thermal capacity. Heat flux to the fluid St = Heat transfer capacity of fluid
224
Heat and Mass Transfer
h x AT
h
pxCp xu_xAT pCpu_ h•L
pu_L ti • Cp k Nu Re • Pr Hence, Stanton number can be defined as the ratio of Nusselt number to the product of Reynolds number and Prandtl number. 20. What is Rayleigh number (Ra)? Rayleigh number is the product of Grashof number and Prandtl number. Ra = Gr x Pr g•/3•L3 AT
v•a Rayleigh number is useful in natural convection heat transfer. The flow is laminar in natural convection if Rayleigh number is less than 109. 21. What is skin drag? When fluid flows past a solid surface, it exerts a tangential shear stress or friction drag on the surface parallel to the direction of flow resulting due to the velocity gradient in the boundary layer. This is called skin drag. Skin drag is the force exerted by the fluid on the surface in the direction of the flow. 22. What is skin friction coefficient or friction drag coefficient? How is skin friction drag force found out? Skin friction coefficient is the ratio of shear stress at the surface in the direction of flow to the product of density and velocity head. (T) y=0 Cfx =
2
P 2
The average skin friction coefficient can be obtained by integrating skin friction coefficient for entire surface and dividing by surface length.
cf
Cfx • dx
Forced Convection Heat Transfer
225
Skin friction drag force (F) is found out from average skin friction coefficient, U
2
F = Cf (p =)A 2 where A = area of contact between fluid and surface and tt,,, = free stream velocity of fluid. 23. Find average heat transfer coefficient if hx = C • xm where C = constant and 'x' is distance from the leading edge of the plate. We know— h • dx
h=
Cx _01 = Jo
c [ x0.9 ix x
0.9 0
C x 0.9 = 1.11 hx
cxc" 0.9
24. Heat transfer coefficient for laminar natural convection over a hot vertical plate of length is given by— hx = C X X114 Find average heat transfer coefficient. The average heat transfer coefficient is L J hx • dx h= o
L
IL
=
C•
c.ximax [X 3/4 t •L
C • L3/4 3 71. . •L
226
Heat and Mass Transfer
4CL 4 3 25. Tabulate forced convection correlation for flow over flat surface (both laminar and turbulent flow). All fluid properties are calculated at film temperature (Tf) where— T° +T Tf = 2 S.N.
Correlation
Laminar flow Re < 3.5 x 105
1. Boundary layer thickness
5x Re
=
2. Local skin friction coefficient
cfx
=
_ 0.664 VRex
r. 1.328 2 .• fx = Al Rex Nux = 0.332 Re, Pr1/3
3. Average skin friction coefficient
Cf =
4. Local Nusselt number 5. Average Nusselt number
Turbulent flow Re > 5 x 105
Nux = 2 Nux = 0.664 x Rex. Pr1/3
0.381 (Re)"
c = 0.0592 fx
(Re)1/5
074 C f = 0. 1/5 (Re) Nux = 0.0296 x Re2.8 x Pr1/3 Nu = 0.037 x Reo.8 x pr1/3
In case fluid is initially laminar and thereafter turbulent over remaining part of the surface,
then correlations are1. Boundary layer thickness ( 0.381 1.0256 )
Re1/2
Rex
for 5 x 105 < Re < 107
2. Average Nusselt Number Nu = Pr15 [0.037 Re" — 872] where
Re =
p •uL
and Nu =
hL
In case of constant heat flux condition, we can have Nuz = 0.453 Rex/2 • Pr1/2 Average Nu = 0.6795 Rem • Pr 1/2
Forced Convection Heat Transfer
227
26. What do you understand from film temperature? (UPTU — 2003) In the boundary layer thickness, the temperature of the fluid is maximum at the surface and it decreases as the distance of the fluid from the surface increases. At the surface, heat transfer from hot surface to the fluid takes place purely due to conduction. However, away from the surface and deep into the fluid, heat transfer occurs due to convection. The temperature of the fluid varies from surface to fluid upto the thermal boundary layer thickness. In order to perform convection analysis to determine the heat transfer coefficient, an average temperature of the fluid and the surface is taken. The temperature is known as mean film temperature To + T_ T = f 2 Note: In case heat transfer between the tube wall and a fluid, the bulk mean temperature is taken which is mean of inlet and outlet temperature of fluid from the tube. 27. The temperature profile in a thermal boundary layer for flow over a flat plate is given by:
T —Ts _ 3 y 1 ( Y j To. — Ts — 2 6th 2 (5th x 6thx = 4.53 x
Revs P,"3
Find hx and average h. The temperature y profile is— T —Ts = 3/2 Y 1 ( Y )3 T — Ts 8th 2 Sth (
c, Ts ) = (T _ To 3 x 1 1 x 3Y2 _ 2 oth 2 (51 25th uY/y=o — y=o 3(T
aj 1
k
Now Hence But
hx =
raT) a y )y=0 T,.— Ts
—
3k k • Re1/2 Pr1/3 — 0.332 — 26th x
,x Nu., — hk = 0.332 Rein Pr15 h= 1f L = 2(hx)x=i, o Nu = 2 Nux
228
Heat and Mass Transfer
28. Explain the velocity profile and boundary layer for fluid flow through a tube pipe. Boundary layer
Fluid flow fully developed H— Entry length —)1 Parabolic velocity profile for laminar flow Laminar sublayer
Velocity profile for turbulent flow
Consider a fluid entering a tube. Due to viscosity, fluid layer adjacent to tube surface immediately becomes stationary and velocity of the subsequent layers increases as distance from the surface increases. The boundary layer develops from the entrance of tube similar to the flow over a plane surface. The velocity of fluid in the boundary layer is zero at the wall of the pipe and it is maximum value at the centre of the pipe after covering a certain distance which is called entry length. The fluid flow after entry length is fully developed. The initial uniform region of velocity profile is called core and it shrinks throughout the entry length. Finally, at last point of entry length of centre at tube, boundary layers from the two sides merge at core top. Now velocity profile is fully developed inside the tube as flow is fully pervaded by the viscous action. In case of laminar flow a parabolic velocity is developed. When flow is turbulent, a much flatter velocity profile is observed. 29. Explain the development of the thermal boundary layer in a tube/pipe. (UPTU — 2003) A thermal boundary layer is developed similar to velocity boundary layer when a fluid having different temperature than that of the tube flows through the tube. The fluid particles in the layer in contact with the surface of the tube assume the wall surface temperature (To). This initiates convection heat transfer in the tube and the development of the thermal boundary layer along the tube. Now temperature gradient exists in the thermal boundary layer. The thickness of the boundary layer increases along the flow until boundary layer reaches the tube centre, thereby filling the entire tube. The region of flow in which the thermal boundary layer develops and reaches the tube centre is called thermal entry region. The thermal entry length (xth) is related to hydrodynamic entry length (xv) as under— xth = Pr for laminar flow xv
After thermal entrance region, we have thermally developed region.
Forced Convection Heat Transfer
229
30. What criteria is used to identify types of flow in a tube (pipe)? The Reynolds number for a fluid flow in a tube is given by— Thermal boundary layer Thermally developed region
6th = thickness of boundary layer [4 Thermal entrance region Thermal boundary layer
Re =
pVd
ii where d = diameter and V = velocity of flow. The flow is considered to be laminar when Re < 2300 and it is considered to be turbulent for Re > 2300. Reynolds number can also be given in mass flow rate— Re =
pVd
ii M = Flow volume x density = (Lr d2 V) x p V=
Re=
4M
rd 2 p p( 4Mj d , x µ ird`p
. 4M irdit 31. What is the equivalent diameter pipe (hydraulic diameter)? Find equivalent diameter for rectangular duct and hollow crosssection pipe. The concept of equivalent diameter pipe or hydraulic diameter is used to simplify the analysis when the flow is taking place in the ducts of square or rectangular crosssectional area. Equivalent diameter is a circular pipe which would present the same resistance to flow and heat transfer rate which is being presented by the actual duct.
230
Heat and Mass Transfer
Equivalent diameter (de) — Li 4 x flow crosssectional area Wetted perimeter Rectangular duct has equivalent diameter as worked out below.
T
Depth = D
IH— Breadth = B
4(B x D) 4A = P 2(B + D) B=D=a
de = In case of square duct,
4 x a2 2 x 2a =a Hollow tube has equivalent diameter as worked out below. de=
4 x 1r4 (D4' — di2) 4A = P r (Do + di) = (Do — di)
de =
Equivalent diameter for a rectangular duct with a pipe inside can be worked out as under
I
I B
de =
4(B x D d2) 4A 4 = P 2(B+ D)F ird
Forced Convection Heat Transfer
23 I
32. Tabulate forced convection correlations for flow inside a pipe. 1. Laminar fluid flow (Re < 2300) (a) Friction factor
f = 64 Re h_ f
(b) Frictional heat loss
LV 2
2. g • d
f
(c) Frictional pressure drop (AP) p 2 2 d (d) Nusselt number for surface having constant temperature is = P .g. h f = f
Nu = 3.66 (e) Nusselt number for constant heat flux condition is Nu = 4.364 All fluid properties are calculated at mean bulk temperature (Tb) which is Tb =
Tinier + Toutlet
2 2. Turbulent flow inside a pipe (Re > 2300). The Nusselt number has same value for both constant surface temperature and constant heat flux condition. Nu = 0.023 Re" x Pr" n = 0.4 if fluid is getting heated = 0.3 if fluid is getting cooled
where 3. Fluid is liquid metals (Pr 3 x 105 16.8 x 102 Hence flow is turbulent. Let 'x' is length for transition Reynolds number. Re = 3 x 10 — or
x=
V •x
3 x105 x16.8 x106
15
= 0.336 m For laminar flow, we haveNu =
hx = 0.664 Rem Pr15 T
Forced Convection Heat Transfer
26 I
x 0.664 Re f/2 Pr1/3 x 0.026 x 0.664 x (3 x 105)1/2 (0.708)1/3 0.336 = 9.99 W/m2K
k h=—
or
For entire surface including laminar and turbulent region, we have Nu = (0.037 Re"  872) Pr1/3 = [0.037 (17.86)0.8  872] (0.708)1/3 h•L
k h
 Nu 0.026 x [0.037(17.86)08  872] x (0.708)1/3 = 32.95 W/m2K
Now heat transfer isQ = hA(T, 
= 32.95 x (2 x 1) (40  10) = 1976 W 64. For a forced flow over a flat plate of length '1,', the local heat transfer coefficient 'hi' is known to vary as x0.5 where 'x' is the distance from the leading edge of the plate. Determine the ratio of the average Nusselt number for the entire plate (NuL) to the local Nusselt number at x = L(NuL). (UPTU  2007  8) hx = x0.5 hL, = — 1 f L h dx L x 1 f L x°• 5 • dx =L o [x +°5 1L L
Now
0.5 jo
1 L°5  2 L0.5 L 0.5 (hx)x.1, = L0.5 Nu (Nux)x=i,
h•L k (hx)x=z, L
k
Nu (Nux)x=L
2L°5 (hx)x=z,
L °.5
=2
262
Heat and Mass Transfer
65. Air at 34°C is flowing over a flat plate maintained at 120°C with a free stream velocity of 5 m/s. The plate is 10 m long and 1 m wide. Calculate the local heat transfer coefficient at 0.5 m and 5 m from the leading edge of the plate. Also determine the heat transferred from the first 50 cm of the plate to air. For air take Pr = 0.697, v = 20.76 x 106 m2/s, k = 0.03 W/mK. (UPTU  2007  8) 1. At
x = 0.5
Rex =
p•V•x

V•x
v /I, 5 x 0.5 = 20.76 x 106 = 1.2 x 105 < 3.5 x 105
Hence, flow is laminar. Hence, we haveNux = 0.332 Re1/2 Pr1/3 = 0.332 x (12 x 104)1/2 x (0.697)15 = 0.332 x 3.464 x 0.886 = 1.02 xx Nux = hk = 1.02
or 2. At
x 1.02 = 0.03 x 1.02 0.5 = 0.0612 W/m2K =x =5m 5x5 Rex = 20.76 x 106 = 1.2 x 10+6 > 5 x 105
(hx)x=o.5 = xk
Hence, flow is turbulent. Hence, we have (Nu)x = 0.0296 x Re°.8 x Pr1/3 = 0.0296 x (1.2 x 10+6)0.8 x (0.697)15 = 0.0296 x 73003.72 x 0.8866 = 1916 Nu = x
h xx x k
Heat transfer coefficient at x = 5 m or
k
• Nux (hx),5 = x
= 0.03 x 1916 5
Forced Convection Heat Transfer
263
= 11.496 W/m2K Now heat transfer coefficient at x = 0.5 (hx)x.0.5 = 0.0612 hay = 2 x 0.0612 = 0.1224 Q = 0.1224 (0.5 x 1) (120 — 34) = 5.2632 W 66. Water at 40°C with a flow rate of 1 kg/s is heated in a tube to 60°C. Surface of the tube is maintained at 100°C. Find the length if the diameter of tube is 25 m. For water (at 50°C) — Cp = 4.81 J/kg.K, µ = 5.48 x 104 kg/ms, k = 0.643 W/mK, Pr = 3.56 For water (at 100°C) — µ = 2.79 x 104 kg/ms (UPTU — 20078) Properties of water at bulk temperature 40 + 60 — 50°C 2 4m 4x1 Velocity V= p • d2 1x103 x x (25 x103)2 = 2.038 m/s 1x103 x 2.033x 25x103 5.48 x 104 = 9.33 x 104 > 2300
Re =
p•V•d
Hence, flow is turbulent. Hence we have—
But
Nu = 0.023 x Re" x pro.4 = 0.023 x (9.33 x 104)0.8 x (3.56)0.4 = 0.023 x 9460 x 1.66 = 361.6 hd Nu = k
or
h= k x Nu
Perimeter
0.643 x 361.6 25x 10 3 = 9.3 x 103 P=7cd T, — _ T, — To —
h.P.L e
mC P
264
Heat and Mass Transfer
or
L=
m CP 10g Ts— h•P Ts — To 1 x 4181 lo 100 — 40 xxx 25x10— g 10060 9.3x103
4181 x 0.405 9.3xivx25 = 2.322 m 67. Match ListI and ListII and select the correct answer. ListI
ListII
(Analogy between) A. Momentum and Heat transfer B. Momentum and mass transfer C. Mass and heat transfer
(Number) 1. Stanton number 2. Lewis number 3. Sherwood number
Codes: (a) (b) (c) (d)
A 1 1 3 3
B 3 2 1 2
C 2 3 4 1
Option (a) is correct 68. Match ListI and ListII and select the correct answer. ListI A. B. C. D.
ListII
(Number) Biot number Reynold number Prandtle number Grashof number
Codes: (a) (b) (c) (d)
A 4 4 2 2
Option (b) is correct
1. 2. 3. 4. B
C
D
1 3 1 3
2 2 4 4
3 1 3 1
(Ratio of) Buoyance force to thermal diffusivity Momentum diffusivity to thermal diffusivity Inertia force to viscous force Conductive to convective resistance of slab
Forced Convection Heat Transfer
265
69. Match ListI and ListII and select the correct answer.
A. B. C. D.
ListI Reynold number of pipe Prandtle number Nusselt number Mach number of pipe
Codes
A (a) 4 (b) 4 (c) 2 (d) 2
Option (b) is correct
B 1 3 3 1
C 3 1 1 3
1. 2. 3. 4.
ListII Film coefficient, diameter and thermal conductivity Flow velocity, acoustic velocity Heat capacity, dynamic viscosity, thermal conducivity Flow velocity, diameter and kinematic viscosity
D 2 2 4 4
Chapter
7
FREE CONVECTION
KEYWORDS AND TOPICS A COEFFICIENT OF THERMAL EXPANSION A BUOYANCY FORCE A BOUNDARY THICKNESS A GRASHOF NUMBER A PRANDTL NUMBER A SIGNIFICANT LENGTH A CONTINUITY EQUATION A MOMENTUM EQUATION
A ENERGY EQUATION A DIMENSIONAL ANALYSIS A RAYLEIGH NUMBER A NUSSELT NUMBER A FILM TEMPERATURE A VON KORMAN INTEGRAL TECHNIQUE A CONVECTION CORRELATION A LAMINAR & TURBULENT
INTRODUCTION The heat flows from hot body to the surrounding fluid due to the temperature difference. The fluid in vicinity of hot body gains temperature and its density decreases. Due to the difference in densities of the warm fluid in vicinity of the hot body and the cold fluid away from the hot body, the warm fluid begins to flow in upward direction. The force making the warm fluid to flow upward is called buoyancy force. Whenever heat transfer is caused due to the motion of fluid by densities difference only, it is called free or natural convection heat transfer. The circulation pattern consisting of upward movement of the warm fluid and downward movement of cold fluid is called convection currents. These currents are set up naturally due to gravity alone and these are responsible for heat convection. Free convection would be nonexistent if there is no gravitational force. Heat transfer with free convection is insured in many applications such as: (i) transmission lines, (ii) cooling of transformers, (iii) heating of houses, and (iv) cooling of core of the nuclear reactor. 1. What is free convection? How does it differ from forced convection? In forced convection, heat transfer by the fluid motion is induced by a pump, fan or a blower, whereas in free convection, fluid motion is induced by a charge of density of the
Free Convection
267
fluid resulting from heating or cooling. It is similar to hot air balloon which rises as hot air inside the balloon has lower density as compared to surrounding air.
Pair
Hot air balloon
The fluid motion is set up as a result of the buoyancy force caused by density difference which takes place due to temperature difference. The heat transfer by convective motion as a result of buoyancy forces is called free convection.
Fluid motion on heating
Vertical hot plate (Ts )
iffr Laminar flow
//////////////////////// Horizontal hot plate fluid motion
T
A
Turbulent Flow
I
Fluid at To
I
T+T 37+13 — T= s = 25°C 2 2 The properties of air from data book at 25°C v = 15.71 x 106 m2/s, k = 2.614 x 102 W/mK Pr = 0.707 Now = 1 = 1 Tf. 25 +373 = h = 1.7 5 m Gr =
g 13 • AT x 1.75 V
2
1 = 3.35 x 103 K1 298 9.81 x 3.35 x 103 x (37 —13) x 1.75) (15.71 x 106)2
= 1.72 x 1010 The Rayleigh number is— Ra = Gr • Pr = 1.72 x 1010 x 0.707 = 1.2 x 1010 > 109
Free Convection
293
Flow is turbulent Using correlation for above Ra, we get
Now
Now
Nu = 0.13 (Ra)1/3 = 0.13 x (1.2 x 1010) = 0.13 x 2.29 x 103 = 298 k x Nu and x = 1.75 h=— x 2.614 x 102 x 298 1.75 = 4.45 W/m2K Q = hA(T,  Toj = 4.45 x (ivDL) (37  13) = 4.45 x (iv x 0.25 x 1.75) (24) = 146.7 W
26. Two concentric spheres of diameter 20 cm and 30 cm separated by air. The surface temperature of two spheres are 47°C and 7°C respectively. Find heat transfer to outer sphere. Using relation Nu = 0.228 Ra0.226. The properties of air at 27°C are: v = 1.57 x 105 m2/s, Pr = 0.712 and k = 2.6 x 102 W/mK Film temperature isTo F T 47+7 = Tf =  27°C 2 2 f4 =
1
=
1
=
1
T1 27 + 273 300 = 3.3 x 103 K1 x = characteristic length = Gr =
d0
2
g • (3 • ATx3 V2
9.81 x 3.3 x 103 x 40 x (0.05)3 (1.57 x 105)2 9.81 x 3.3 x 40 x 125 x104 2.46 x 1010 = 65.8 x 10+4 = 6.58 x 105 Ra = Gr x Pr = 6.58 x 105 x0.712 = 4.7 x 105
=
30 2 20 = 5 cm
294
Heat and Mass Transfer
From the relation given
Now
Nu = 0.228 x (Ra)o.226 = 0.228 x (4.7 x 105)0.226 = 4.37 h = —k x Nu x  2.6 x 102 x 4.37 = 2.27 W/m2K 0.05 Q = hAAT = 2.27 x (47 r r?) x 40 = 2.27 x 4ir x (0.1)2 x 40 = 11.4 W.
27. Determine the coefficient of heat transfer by free convection and maximum current density for a nichron wire 0.5 mm in diameter. The surface of the wire is maintained at 300°C. The wire is exposed to still air at 20°C and resistance per metre length of the wire is 6 ohm/m. Use relation Nu = 1.18 (Gr Pr)1"8. Properties of air at 160°C arek = 3.61 x 102 W/mK, Pr = 0.687 v = 30.35 x 106 m2/s (Annamalai University 2003) Film temperature isTf Now
Ts F 1._ 300+20 =160°C 2 2
fi =
1 = Q • 31 x 103 k1 160 + 273 Characteristic length x = 5 x 104 m g•i3x ATd3 = 9.81x 2.31x103 x 280 x(5 x104)2 v2 (30.35 x 106)2 = 0.86 Ra = Gr • Pr = 0.86 x 0.687 = 0.591 Gr =
Now
Given relation is
Now
Nu = 1.18 (Re = 1.18 x (0.591)1/8 = 1.105 h = —k x Nu x  3.61 x 10:2 x 1.105 5x
Free Convection
= 79.77 W/m2K Q = hA(T, — Lc) = 79.77 x (iv x 5 x = 35 W/m
295
x 1) x (300 — 20)
Heat generated by wire— Q = I2R but R = 6 ohm/m or or
/2 = Q = 35 = 5.833 R 6 I = V5.833 = 2.42 ampere
28. The water in a tank at 30°C is heated by passing the steam through a pipe of 50 cm long and 5 cm in diameter. If the pipe surface temperature is maintained at 80°C, find the heat loss from the pipe per hour. If the pipe is kept vertical, what will be the heat loss per hour from the pipe. Film temperature is— T=
T+T
2 = 50°C
=
80 + 20 2
Properties of water at 50°C are— p = 988.1 kg/m3, v = 0.556 x 106 m2/s = 0.55 x 103 kg/ms, C, = 4200 J/kgK k = 0.64 W/mK, 13 = 5.1 x 104 K1 Pr= 11CP = k = 3.60
0.55 x 103 x 4200 0.64
Case 1: Pipe is horizontal. Hence, x = d = 0.05 g•PATxx 3 = 5.1 x104 x 9.81x 60 x (0.05)3 v2 (0.556 x 106)2 = 1.21 x 108 Ra = Gr Pr = 3.60 x 1.21 x 1011 = 4.26 x 1011 Gr =
From the correlation, we get— Nu = 0.53 (Ra)1/4 = 0.5 x (4.26 x 1011)1/4 = 75.7
296
Heat and Mass Transfer
h= —k x Nu  0.64 i x75.7 x 5 x 102 = 968 W/m2K Q = h(ivdL) (T,  TO = 968 Or x 5 x 102 x 0.5)(80  20) = 4.55 kW Case 2: Pipe is vertical Now characteristic length x = L = 0.5 instead of d = 0.05 Hence,
Ra = 4.26 x 1014 Nu = 0.13 (4.26 x 1014)1/3 = 978.2 h = —k x Nu = 0'64 x978.2 x 0.5 = 1250 W/m2K Q = h(ivdL) (T,  TO = 1250 (iv x 0.05 x 0.5) (60) = 5.89 kW
Note: The convective motion is more when tube is vertical resulting in higher heat transfer. 29. Match ListI and ListII and select the correct answer using the codes given below the lists. ListI (Flow Pattern)
ListII (Situation)
1. Heated horizontal plate
2. Cooled horizontal plate
Free Convection
(C)
297
3. Heated vertical plate
///////////////////
(D)
4. Cooled vertical plate ///////////////////
Codes: (a) (b) (c) (d)
A 4 3 3 4
B 3 4 4 3
C 2 1 2 1
D 1 2 1 2 (IES — 1995)
In flow pattern (A), air is moving up which is possible when air gets heat from plate and its density reduces to facilitate upward motion. The plate has to be hot. In flow pattern (B), air has to be cooled down and gain density to move down. Plate has to be cold. In flow pattern (C), air has to be heated up so as density reduces to facilitate upwards motion. Hence, plate has to be hot. In flow pattern (D), air has to be cooled down so as move down to plate surface due to high density. Hence, plate has to be cold. Option (C) is correct. 30. Assertion (A): A slab of finite thickness heated on one side and held horizontal will lose more heat per unit time to cooler air if the hot surface faces upwards when compared with the case where hot surface faces downwards. Reason (R): When the hot surface faces upwards, convection takes place easily whereas when hot surface faces downwards, heat transfer is mainly by conduction through air. Mark the answer as: (a) (b) (c) (d)
If both (A) and (R) are true and (R) is the correct explanation of (A) If both (A) and (R) are true but (R) is not the correct explanation of (A) If (A) is true but (R) is false If (A) is false but (R) is true. (IES — 1996)
The convective motion for plate facing up and down as shown below:
298
Heat and Mass Transfer
Plate facing up
Plate facing down
The convective motion is possible when hot plate is facing up as air heated up with reduced density can move up facilitating higher rate of heat transfer. The upward motion of heated air is not permitted when hot plate is facing down and heat transfer is mainly by conduction. Both assertion (A) and reasoning are correct and reasoning is the correct explanation of assertion Option (a) is correct. 31. Two vertical plates each 120 mm high and maintained at 85°C are placed in water at 15°C. Find minimum spacing which will prevent interference of free vertical o.i /15 [ 1 + 0.494(Pr) 2/3 8 th boundary layer. Use relation = 0.565 (Pr)8 for turbulent Gr flow. Guidance: The problem is to find the thickness 8th of boundary layer at trailing edge of the vertical plate. The spacing between two plates has to be double of this thickness.
H 6th —bh— 3th H
85 +15 — 50 2 For water at 85°, the properties are— Tf =
Now
k = 0.674 W/m°C, v = 0.566 x 104 m2/s Pr = 3.54 1 1 Q = T f. + 273323 and x = 120 x 103 K1 Gr =
g [3 AT x 3 V
2
= 11.9 x 109
9.81x
323
x 70 x 120 x 103
(0.566 x 104)2
Free Convection
299
Ra = Gr x Pr = 11.9 x 109 x 3.54 = 42.1 x 109 > 109 Flow is turbulent and as per correlation givenSt,
[1 + 0.494 x Pr,,, = 0.565 x (Pr)8/15 Gr
0.1
,,, 0.1 1 + 0.494 x (3.54)— = 0.565 x (3.54)8/15 [ 42.1 x 109 S or Now spacing
t' = 3.06 x 102
8th = 120 x 103 x 3.06 x 102 = 3.67 x 103 m S = 28th = 2 x 3.67 x 103 = 7.34 x 103 m = 7.34 mm
32. A vertical 0.8 m high and 2 m wide double pane window consists of two sheets of glass separated by 2 cm air gap at atmospheric pressure. If temperature of outside and inside glasses of air gap are at 12°C and 2°C, find heat transfer through the double pane. Use relation Nu = 0.197 Ralm (f! 9 where H = height of plate. c Guidance: The problem is to be solved considering two vertical plates separated with a spacing 'x' which is the characteristic length to be used for calculating Grashof number Ti= 2°C
To = 12.5
T = f
Film temperature
To + Ti 2
12 + = 7°C 22
Properties of air at 7°C
Now
Pr = 0.717, k = 2.46 x 102 W/mK v = 1.4 x m2/s R 1 1 _ 1 K1 Tf 7 + 273  280
300
Heat and Mass Transfer
Gr =
g•
•
AT • x3
V
2
9.81 x 280 x 10 x (0.02)3
Now
(1.4 x 105)2 = 14.3 x 103 Ra = Gr x Pr = 14.3 x 103 x 0.717 = 10.25 x 103 Nu = 0.197 (Ra)1/4 x ( H ) 119 X
= 0.197 x (10.25 x 103)1/4 x
(0 202) 9
= 1.3 The effective thermal conductivity of double pane is keff = Nu x k = 1.3 x 2.46 x 102 = 3.2 x 102 W/mK keff A(To  Ti) x = 25.6 W
Q=
k x Nu, keff = k x Nu. Note: While h = —
3.2 x 102 x (0.8 x 2) x 10 0.02
Chapter
8
RADIATION HEAT TRANSFER
A A PHOTON A MAXWELL'S ELECTROMAGNETIC THEORY A A A QUANTUM THEORY A A STEFANBOLTZMANN'S LAW A A PROVOST THEORY OF EXCHANGE A A SPECTRAL EMISSIVE POWER A A BLACK BODY A A ABSORPTIVITY A A REFLECTIVITY A A TRANSMISSIVITY A A DIFFUSE SURFACE A A GREY SURFACE A A OPAQUE SURFACE A A KIRCHHOFF'S LAW A A WIEN'S DISPLACEMENT LAW
SPECULAR AND DIFFUSE REFLECTION IRRADIATION RADIOSITY MAX. PLANCK'S LAW SOLID ANGLE STERADIAN INTENSITY OF RADIATION SHAPE FACTOR RECIPROCITY THEORY SURFACE RESISTANCE SPACE RESISTANCE GREY BODY FACTOR GREENHOUSE EFFECT SOLAR RADIATION AND CONSTANT SKY TEMPERATURE
INTRODUCTION The radiation is the transmission of heat energy between two bodies having no physical contact between them. The transmission of heat energy as radiation does not affect or heat up the medium between the heat source and the receiver. The transmission of heat energy by radiation does not infact require any intervening medium and this transmission can take place even when there is vacuum between the bodies. Infact transmission of heat energy by radiation occurs most effectively in vacuum and it is reduced or eliminated in the presence of different medium materials. Energy transmitted by radiation is not continuous but it is in the form of discrete packets of energy called photons which are successively released by hot bodies. The photons propagate through space as electromagnetic waves. Whenever these photons in the waves strike
302
Heat and Mass Transfer
a receiver body, then the conversion of wave motion into heat energy takes place. The heat energy on conversion is partly absorbed, reflected or transmitted on the surface of the receiver body. 1. What is radiation? Radiation is the transmission of heat by a body due to its temperature. Every hot body of a temperature above absolute zero emits thermal radiation. Radiation unlike conduction and convection does not require any medium. In fact, radiation is most effective in vacuum. 2. What are the basic theories to explain the mechanism of radiation heat transfer? There are two basic theories to explain the mechanism of radiative heat transfer which are— (a) Maxwell's electromagnetic theory. According to this theory, heat transfer in the form of electromagnetic waves takes place from a body when its temperature is above absolute zero. (b) Max Planck's concept or quantum theory. According to this theory, heat transfer in the form of photon or quanta of energy takes place from a body when its temperature is above absolute zero. The photon has energy = hv, where h = Planck's constant and v = frequency of photon. 3. Explain types of radiation on the basis of wavelengths. The radiative heat transfer takes place in the form of electromagnetic waves emission by a body above absolute temperature. The wavelength of the electromagnetic emission may vary from zero to infinity. However, the bulk of thermal emission has wavelength varying from zero to 100 1.tm (thermal emission from sun has wavelength from 0.1 lam to 4 gm). Types of radiation on the basis of wavelength is as tabulated below. S.No.
1. 2. 3. 4. 5. 6. 7.
Type of Radiation
Wavelength
4x 107 to 1.4x 104 µm 1 x 105 to 2 x 102 pm 0.02 to 0.4 0.38 to 0.76 1.1,m 0.1 to 100 1.1,m 0.1 to 4µm 92 to 2 x 101° 1.1,m
Gamma Ray xrays Ultraviolet rays Visible radiation Thermal radiation Solar radiation Radio waves
4. What is the relation of speed, wavelength and frequency of electromagnetic radiation? The relation between speed of light (C = 3 x 108 m/s), frequency (f) and wavelength is C = f•A
Radiation Heat Transfer
303
5. What are the parameters deciding the type of thermal emission? The thermal radiation can have wavelength varying from 0.1 to 100 pm and wavelength of emission depends upon nature, temperature and state of the emitting surface. Emission of radiation in gases depends upon the thickness of gas layer and the pressure of the gas. 6. What is quantum theory? According of quantum theory, the thermal radiation propagates in the form of discrete quanta or photon. Each photon has energy which is— where
E = hv h = Planck's constant = 6.625 x 1034 Js v = frequency of photon
Each photon is a particle having energy, mass and momentum similar to the molecule of a gas. Hence, radiation can be considered a photon gas moving from one location to another. The mass and momentum of a photon are— E = mq? = hv hv mass m = CZ o
and
momentum = m • Co = hv2 x Co Co
hv Co 7. What is StefanBoltzmann equation? As per StefanBoltzmann equation, the flux of heat energy emitted by a black body is proportional to the fourth power of its absolute temperature where
qb = 0 74 qb = quantity of energy emitted per unit area and per unit time a = StefanBoltzman constant = 5.669 x 108 W/m2 K4 T = absolute temperature
8. What is Provost theory of exchange? According to Provost theory of exchange, all bodies radiate thermal radiation at all temperatures. The amount of thermal radiation radiated per unit time depends on the nature, area and temperature of the emitting body. The rate of heat transfer is faster at higher temperature. Besides emission, a body also absorbs a part of thermal radiation which is being emitted by other bodies and which falls on it. Three conditions may prevail—
304
Heat and Mass Transfer
(a) If a body radiates more than what it absorbs, its temperature falls. (b) If a body radiates less than what it absorbs, its temperature rises. (c) If a body radiates and absorbs same amount of thermal radiation, its temperature remains constant. 9. What is emissive power (E)? The emissive power of any body is defined as the energy emitted by the body per unit time and per unit area. It is expressed as W/m2. The emissive power of a black body is proportional to the fourth power of its absolute temperature as per StefanBoltzmann equation. 10. What are total emissive power (E) and monochromatic (spectral) emissive power (Ea)? How are they related to each other? (UPTU — 2006 — 7) The amount of radiation emitted per unit wavelength varies at different wavelengths. Hence, monochromatic emissive power is defined as the rate of energy radiated per unit area of body and per unit wavelength. Total emissive power is— E = f EbAAA, o where A = wavelength of emission Eb = monochromatic emission power of black body at wavelength A. 11. What do you understand from a black body? A black body is an ideal hypothetical body which can— (a) absorb all incident radiation from all directions at all wavelengths. Hence, there is no reflection, transmission and scattering of incident radiation (b) emit maximum amount of radiation as compared to a real body at any temperature. 12. Why term black is used for black body? (UPTU — 2003) Or
What do you understand by coloured surface? (UPTU — 2005 — 6) As black body absorbs all incident radiation including visible light spectrum therefore, the black body appears black to human eyes. 13. Is ice a black body? Most of black coloured surfaces have high value of absorptivity and they are classified as black body. However, ice appears white but ice also absorbs all incident radiation. Hence, ice is also a black body.
Radiation Heat Transfer
305
14. How can an artificial black body be created? An artificial black body can be created by making radiation to enter through a small aperture in a hollow sphere. The radiator undergoes many reflections within the sphere till complete radiation is absorbed inside. Here aperture can be considered as black body.
Small aperture
Black body
15. Explain absorptivity, reflectivity and transmissivity. Or
Explain why radiation is usually treated as a surface phenomenon. (UPTU — 20067) When a radiation is incident on a body, three things happen which are: (i) a part is reflected back, (ii) a part is transmitted through, and (iii) the remainder is absorbed. In other words, we have Incident
Reflected Q
Absorbed Q,
Transmitted Q.,
Incident radiation = Absorption + Reflection + Transmission Q = Qa+
Qp
Q,
Absorptivity (a) is the fraction of incident radiation which is absorbed by the body a =
Qa Q
Reflectivity (p) is the fraction of incident radiation which is reflected by the body Qp
P=
Q
Transmissivity ('r) is the fraction of incident radiation which is transmitted by the body
306
Heat and Mass Transfer
16. Prove p + + a = 1. Now incident radiation is— Q=Qp+ar+Qa Now dividing by Q, we get—
QP ar + Q. = Q Q Q Q Q p++a=1 17. What do you understand from: (i) white body, (ii) opaque body, (iii) diffuse grey body, and (iv) transparent body? (UPTU — 20023, 2003, 20078) 1. A body is called a white body if it reflects all incident radiations. Hence, for a white body, we have— p = 1 and
a=
=0
2. A body is called an opaque body if it does not transmit any part of incident radiation. Hence, for an opaque body, we have— r = 0 and
p+a=1
3. Diffuse surface: For real surfaces, the spectral emissivity EA is a complex function of wavelength 'A,' and emissivity calculation for all wavelengths become very difficult. Hence, grey and diffuse approximations are commonly used in finding emissivity. A surface is called to be diffuse surface, if its properties are independent of direction of radiation. 4. A body is called grey body if its monochromatic emissivity is independent of wavelength. The emissivity of the grey body does not change with temperature. Thus, a grey body has a characteristic emissivity value which does not vary with temperature and the value of emissivity of a grey surface is less than one. 5. A body is called transparent if radiation passes through it fully i.e. 1" = 1 and a = p = 0. 18. What do you understand from emissivity and emission from real body? Emissivity (E) is the ratio of the emissive power of any real body to the emissive power of a black body at the same temperature. It is the ability of a real body to radiate heat in comparison to a black body at the same temperature Emissivity = E = E Eb
Emissive power of the real body Emissive power of black body
Radiation Heat Transfer
307
The emission power of a real body is given by using emissivity of the body and emission power of a black body as per StefanBoltzmann equation E = E [6A T4]
.1\ Black body
Grey body Real body Wavelength (X) .Spectral emissivity versus wavelength
Note: Spectral emissivity is for a particular wavelength i.e. E A =
while emissivity
(total hemispherical emissivity) is average emissivity in all directions and wavelength i.e. E . The average emissivity of a black body, grey body and real body is as shown = — Eb below. 19. What is Kirchhoff' s law? (UPTU — 2003, 2005 — 6)
According to Kirchhoffs law, the ratio of total emissive power to absorptivity is constant for all bodies which are in thermal equilibrium with the surroundings. Therefore, it means that emissivity of a body is equal to its absorptivity. When body attains thermal equilibrium with its surroundings, it means— E1 E2 = E3 Eb al a2 a3 ab But a black body has absorptivity ab = 1, hence we _ E2 E3 Eb 1 al a2 a3 or
El
Eb
=
E2
Eb
= a2 and
E3
Eb
= a3
But as per definition: E2
Eb  E 19 al = E 1,
Eb = E 29
E3
= E3
Eb Hence, a2 = E 2 and a3 = E 3 Kirchhoffs law can be also stated that the emissivity (E) and absorptivity (a) of a real surface are equal for radiation with identical temperature and wavelength.
308
Heat and Mass Transfer
20. What is Wien's displacement law? (UPTU — 2003, 2003 — 4) According to Wien's displacement law, the product of maximum wavelength of a radiation from the body and the absolute temperature of the body is constant. Amax x T = Constant The above is the relationship between the temperature and the wavelength at which the maximum value of monochromatic emissive power occurs. 21. Explain the types of reflection of radiation. Reflection can be: (1) specular reflection or (2) diffuse reflection. In specular reflection, the incident radiation and reflected radiation make equal angle with the normal to the surface at the point of incident. Polished and smooth surfaces give specular reflection. Incident radiation
Normal
Reflected radiation
/////////// //////////// Surface Specular reflection
In diffuse reflection, the incident radiation is reflected in all directions. The rough surfaces give diffuse reflection. Incident radiation Reflected radiation
//////////// ////_,///////////// Surface Diffuse reflection
22. What is irradiation (G)? It is total amount of radiant energy which is incident on a surface per unit area and per unit time.
Radiation Heat Transfer
309
23. What is radiosity (I)? (UPTU — 2005 — 6) It is the total amount of radiant energy leaving a surface per unit area and per unit time. Radiosity (J) comprises: (1) emission (E) from the surface, and (2) reflected part of irradiation (p x G) as shown below.
pxG
Emission and reflection
Radiosity J = E p x G where E = emission, p = reflectivity, G = irradiation. For black body, For nonblack body
p = 0 and J = E J = E Eb p x G.
24. Explain MaxPlanck's law for spectral emissive power (EbA). The energy emitted by a black body varies with wavelength and temperature. For a particular wavelength, a body tends to radiate more heat at a particular temperature. According to MaxPlanck's law, the spectral emissive power (EbA) at any temperature and wavelength is— EbA =
1AT
— 1) where Cl and C2 are constants, A = wavelength in p,m and T is temperature in absolute. The variation of spectral emissive power with wavelength and temperature is as shown below— (ec2
17 1.1
gt 'N
Amaxx T = constant Ti> T2 > T3 > T4
E to T2 a
Band emission
15 ai o.
r
=i EbAdA. 0 0 Ai A2
Wavelength (A.)
Spectral black body emissive power
Following are the observations from the graph— (a) Emissive power at any temperature increases with increase of wavelength upto some value and after this it decreases.
3I0
Heat and Mass Transfer
(b) At any wavelength, emissive power increases as temperature is increased. (c) At elevated temperatures, maximum emission takes place at shorter wavelengths. Hence, most of emission is taking place at elevated temperatures in a narrow band on both sides of wavelengths at which emission is maximum. (d) Connecting 'A ' points on graph for various temperatures gives Wien displacement law i.e. A x T = constant Ai E (e) Total radiation EB is r —Eb,d,. while band emission is Ebo, Ao = f bAdA. o o 25. Prove StefanBoltzmann law using MaxPlanck's equation. The emissive power of a black body is as per MaxPlanck's equation is Eb =
f o
EbA dA
 cl A,5
=f 10 ec2 /AT _ 1 clA
A1 =x
Now substitute or
A2 dA = dx on differentiation 1 dx dA =  — x2
or
Eb =
r
Ci X
5
( 1 )
c` ec2xiT 1 3
= Ci J«X X o
[ c2 x x
dx
1
e T —1] dx
c` 3 =C1 f X X [e o
x2
—c2 x
—2(c2 x)
T +e T
—c2 xnxx 3 = c1 f X X [e T jdx
o
Now applying integral formula fo
in 1 • m+i
x e dx m —ax — a
Eb = Ci
6 c1 3! = 4 4 c2 x n n)3+1 (C2 X — T T4
=
6 c1 T4 4 4 C2 X n
=
6 1
4 C2
4
[1±1 14 2 4
—3(c2 x
+e T + ...1dx where n = 1, 2, 3
Radiation Heat Transfer
3II
6c1. T4 (7r 4
4
90
6ci x7r4 x T4 c2 x 90 = a x T4 —
where
a=
iir 4 = 5.67 x 108 W/m2K c2 6cx 90
26. Prove Wien's displacement law using MaxPlanck's equation. According to MaxPlanck's equation— Ebb = C2
Put
AT
A5 — 1) — ec2''' —1
(ec2
=x c2 xxT
or EbA =
x x5 x T5 cl (ex —1)
On differentiating and putting a (E1,) = 0, we can get maximum spectral emissive ax power for a particular wavelength.
a r clx x5x T5 ) o a ax (Eb), = 0, ax el, (ex —1) or or
ci x T5 5x4(ex _ 1) _ x5(ex) 5 C2 Ci X T5 ,5 E.,2
x
—
0
(e x —1)2
X4 , [5ex — 5 — x x (ex 1)2
eX]
=0
4
or
Ci
c3
X
(e x —1) [e 4
or
(ex —1)
X(5 — x) — 5] = 0 (ex (5
5)
3I2
Heat and Mass Transfer
The value of x is found out by trial error as— x = 4.96 or or or
C2
— 4.96
6 = 0.0029 — 4.9 Lax = constant Lax
27. What do you understand from solid angle? A solid angle is defined as a portion of the space inside a sphere enclosed by a conical surface when the vertex of the cone is located at the same location as the centre of sphere. Consider an area dAs on the surface of sphere (radius r) whose centre is at 'c'. The area dAs subtends an angle 'of at c as shown in the figure which is called solid angle. dAs w=
r
2
dAs r
Solid angle
Consider a sphere is located at 'r' which has radius as 'r' and centre at 'c'. Then solid angle is— CO
=
dils r2
=
4 nr 2 r2
= 4n.
In case we consider hemisphere with radius 'r' and centre at 'c', solid angle is CO =
dAs
The unit of solid angle is steradian (Sr).
r2
=
2 nr 2 r2
= 27C
Radiation Heat Transfer
3I3
28. Explain intensity of radiation (I). (UPTU — 2003, 2007 — 8) Intensity of radiation is the rate of heat radiation in a given direction from a surface per unit solid angle and per unit area of the projection of the surface on a plane normal to the direction of radiation. dA2 = area on sphere surface = r 2 sin 0. de • drp
dA1 de Intensity of radiation
Consider radiation from an elementary area dA, located at centre of sphere having radius `r' as shown in the figure. The radiation is absorbed by other elementary area dA2 located on the hemispherical surface. Now we have— (a) Projected area of dA1 on a plane perpendicular to the line joining dA1 and dA2 which is equal to dA1 cos 0 (b) Solid angle subtended by dA2 is— dA2 dA2 = r 2 As per definition, we have— d Q12
Intensity of radiation =
(dA1 cos 19) X
dA2
where dQ12 = rate of radiant heat transfer from dA1 to dA2. However,
dA2 = rde(rs: 0)d0 = r2 sin 0. de. deP I—
dQ x dAi x cos°
sin 0 • dO. d0 r2
dQ12 r2
dA1 sin 0 • cos 0 • dO d0
or
dA1 xsin0xcos0x dex dO dQl2 =Ix =
3I4
Heat and Mass Transfer
29. Find total radiation or emissive power from a black surface in terms of intensity of radiation. The radiation from a surface will take place through a hemisphere which has origin at the surface.
dQ = I x dAi sin 0 cos 0 dO x dtp For the hemisphere, we have 0=z12 0=27r f
dQ = IdA1
sin 0 • cos 0 • dO • thip e=o
(p=0
0=z12
or
Q = IdAi
2xxxsinecose• de 0=0 0=z12
=
X dAi X Ir
f sin2 0 • dO e=o
x I x dAi [cos2 r/2 2 0 =rx/xdAi =
or or
Q = x I dA1 E= x I
The total emissive power of a black surface is equal to
times its intensity of radiation.
30. How is emissive power defined? Emissive power denotes the energy radiated per unit area and per unit time and per unit solid angle along the normal to the area.
Normal
Radiation Heat Transfer
E—
3I5
dQ dAxdcoxdt
where E = emissive power dQ = heat energy radiated d w = solid angle dt = time dA = area. 31. What is shape factor? (UPTU — 2003, 2007 — 8) Shape factor is defined as the fraction of radiation energy leaving one surface that falls on the other surface. A fraction of radiation energy leaving one surface that reaches the other surface because of the following reasons— (a) orientation of emitting and absorbing surfaces (b) characteristic of emitting and absorbing surfaces. (c) medium in between emitting and absorbing surfaces n2 ni dA2
dA1 Radiant heat exchange
The surface A l will emit radiation Qi = 6A1 T4 Out of this Qi radiation, a part of radiation (Q12) will be falling on surface A2 depending upon orientation. The fraction reaching surface A2 depends upon shape factor (F12) as given below Q12 = Q1 and
X
F12
where F12 = shape factor
2 F12 = 1 f f cos 01 X cos 02 X dAl x 2dA irr A2 A2
3 I 6 Heat and Mass Transfer 32. What is reciprocity theorem? According to reciprocity theorem, net radiant heat exchange between two surfaces may be evaluated by computing shape factor from either of the two surfaces. In other words, the shape factor (F12) from surface A l to surface A2 has definite relation with shape factor (F21) from surface A2 to surface A1 which is given by— A1F12 = A2F21 33. Evaluate shape factor and prove reciprocity theorem. Consider two black bodies A1 and A2. Heat transfer is taking place between these two bodies in nonparticipating medium. Consider two elemental areas dA1 and dA2. Let `r' be the distance between the two areas and line joining the elemental areas make angle 01 and 02 with respective normals to the areas as shown in the figure. The projected area of dA1 in the direction of radiation is dA1 cos 01. The radiation energy leaving from area dA1 and falling on area dA2 is— dQ1_2 = Intensity x projected area x solid angle. dQ1 _2 = Il x (dA1 cos 01) x
dA2 cos 02 ) r2
nl
dA1
Radiant heat exchange
However, intensity
Il
=
Ebl
=
a 7i4
a 77 4 dQ1 _2
'1 x (dA1 cos 191)
7r
(cIA2 cos 02 ) r2
Now on integration, total radiation from body A l is— Q12 =
where
a it f
cos° i x cos 02 x dA x dA2 2 JA2 f
r
= F12 x (6A1T14) 1 f r cos OiX cos 02 F12 = A — .1A j2 Al AI r2 = Shape factor
X
dA1 X dA2 (i)
Radiation Heat Transfer
3I7
Similarly, radiation leaving A2 and absorbed by Al can be evaluated as under6T2 f f Q21 =
COS 91 X COS 92
JA,JA2
r
2
xdA1 xdA2
= F2_1 (6A2 T24) I r cos 91X cos 92 X dA1 F21 = A 2 2 JAIJA2 r
where
x
dA2
From equations (i) and (ii), it is apparent— A1F12 = A2F21 34. Explain shape factor of a surface with respect to itself (F1_1). What is F1_1 for a concave, flat and convex surface? (UPTU — 2003) Consider a concave surface. The fraction of radiant energy leaving the concave surface is intercepted by another part of the same concave surface. The shape factor of a surface with respect to itself is denoted by F1_1. The shape factor F1_1 for concave surface is not zero while it is zero for flat and convex surface.
Concave surface A
F11 Flat surface
F11 Convex surface surface
35. Show shape factor from a radiating surface to a subdivided receiving surface is simply the sum of individual shape factors. Consider A l radiating surface and A2 surface consisting A3 and A4 as receiving surfaces. A2 = A3 + A4
F1 Receiving surface 14
Radiant surface Relation between shape factors
3I8
Heat and Mass Transfer
For surface Al and A2, applying reciprocating theorem Al F12 = F2_1 A2 = Q124 = C1 X Q12 a Ti
Also
Al F13 = F31 A3 = Q134 = C1Q13
and
= „1 , y A l F14 = F41 A4 =Q14 a r, — Q14
a Ti
Total Q12 = Q13 + Q14 ...
Al F12 = Al F13 ÷ Al F14
or
F1.2 = F1.3 ÷ F1A
36. Find the relationship in shape factors in case radiating surface (A1) is subdivided in to subareas A3 and A4. Consider radiating area Al is subdivided to subareas A3 and A4.
F4_2
Receiving surface
Radiating surface
Q12 = Al F1_2 Q32 = A3 F3_2 Q42 = A4 F4_2
But
Q12 = Q32 + Q42
A l F12 = A3 F3_2 + A4 F4_2 37. Find relationship of shape factors for an enclosure consisting of n surfaces. The enclosure has n surfaces and heat leaving one surface is absorbed by all n surfaces Q = Qi_i + Q12 + Q13 • • • + Qln = Q x Fi _ i + Q X Fi2 Q X Fi3
or
F1 _1 + F12 + F13 ••• F1.n = 1
Similarly, we can write— F21 + F22 + F23 • • • F2n = 1 F3_1 + F3_2 + F3_3 ... F3n = 1 Fn_1 Fn_2 Fn _3
Fnn = 1
Qx
Radiation Heat Transfer
3I9
Enclosure with nsurfaces j=n
We can write
Fij = 1 where J = 1, 2, 3, ... n. It is called summation rule. j=i
38. Find various shape factors when a black body is inside a black enclosure. Consider a black body (A1) inside a black enclosure (A2)
All radiation emitting from black body Al is intercepted by the enclosure A2. Hence, shape factor— F12 = 1 Since two bodies are interacting radiation, we have from reciprocity rule— A1 F1 _ 2 = A2 F21 AlE, F2_1 = A r 12 2
Al A2 Also heat omitting from A2 is absorbed by surfaces Al and A2.
But
F21 + F22 = 1 Al F21 = ri2 F22 = 1
Al A2
320
Heat and Mass Transfer
39. Find shape factors of a tube with a crosssection of an equilateral triangle as shown below.
Using the relation of an enclosure with three surfaces, we have— F1.1 ÷ F12 ÷ F13 = 1 However, surface 1 is flat, hence— F11 = 0 Also from symmetry—
or Similarly, and
F12 = F13 F12 ÷ F13 = 1 F12 = F13 = 0.5 F21 = F23 = 0.5 F31 = F32 = 0.5
40. Find shape factors for hemispherical surface and a plane surface as shown below.
Using summation rule w.r.t. surface 1, we have— F11 + F12 = 1 But by reciprocity rule, we have— F1 _ 2 A1
F2 _1 x A2
Using summation rule w.r.t. surface 2, we have— But
F21 + F22 = 1 F22 = 0 as it is a flat surface. F21 = 1 A2 F12 = F21 x
Radiation Heat Transfer
32 I
= A2 Al .
F11 = 1 — F12 = 1 — A2 A1 A2 = iv r2 A 1 = 2ir r2 A2 1 — = — or F12 = 0.5 and F11 = 0.5 Ai 2
However, .'. Hence, shape factors are—
F11 = 0.5, F12 = 0.5, F22 = 0 and F21 = 1 Note: F2_1 = 1 means that radiation of flat surface is completely intercepted by hemispherical surface. F12 and F11 equal to 0.5 each means that radiation from hemisphere is equally intercepted by flat surface and hemisphere. 41. Find the shape factors of cylindrical cavity as shown below.
Using summation rule w.r.t. surface 1, we have— F11 + F12 = 1 Also summation value w.r.t. surface 2, we have— But
F21 + F22 = 1 F22 = 0 as surface 2 is flat surface F21 = 1
Now using reciprocating rule
or
Al F1 _ 2 = A2F21 A :x1 F12 = — A
(i)
322
Heat and Mass Transfer
Jr 4 , x1 — 7rd` 7rdh+ —
d 4h+ d
From equation (i) Fll = 1  F12
=1
d 4h+ d
4h = 4h+d 42. Find shape factors for conical cavity as shown below.
h
Using summation rule w.r.t. surface 2, we have— But .%
F22 ÷ F21 = 1 F21 = 0, as surface 2 is flat F21 = 1
Using reciprocity rule, we have— A1 x F12 = A2 F21 or
A2 F12 = — X F21 Ai 7rd2
=
4 x 1 where 1= slant height irdl d 4h2+(12 2
d %/4h2+d2
Radiation Heat Transfer
323
Using summation rule w.r.t. surface 1, we have— or
F11 ÷ F12 = 1 F11 = 1 — F12
=1
4h2 + d2 43. Find shape factors for a sphere (diameter = shown below.
d)
kept in a cubical box of side
Since the sphere is within the cube, hence— F12 =1 Now using reciprocity rule A1 F12 = A2 F21
d Sphere in box
or
Al F21 = A2 — X F12
2 = rd = ,r 6d2
6
Using summation rule— or Also or
F11 + F12 = 1 F11 = 1 — F12 =1—1=0 F21 + F22 = 1 F22 = 1 — F21 = 1 — it/6
'el'
as
324
Heat and Mass Transfer
44. Find shape factors for diagonal partition within a square duct as shown below.
Using summation rule— F11 ÷ F12 ÷ F13 = 1
But F11 = 0, as surface 1 is flat Due to symmetry
F12 ÷ F13 = 1 F12 = F13 = 0.5
Using reciprocating rule— F12 X A1 = F21 X A2 Al E, F21 = r 12 A2 — 1 x‘ . 12,a x 1
2
x 0.5 2
=
xaxl x 0.5 = 0.71.
45. Calculate the shape factor for a very small disc dAi and a large disc A2 kept parallel at a distance 'I?' directly above the smaller disc. A2
and dA I are discs at vertical separation of height `h'. Consider an elemental area
at radius 'x' and thickness `dx' on disc A2. The elemental area dA2 is— dA2 = 27rx • dx
Radiation Heat Transfer
325
The shape factor of area dAi w.r.t. dA2 is— 1 f COS 91 • COS 92 • dAl ' dA2 F1 _2 = 1 7r R 2 dAl jAl jA2 Here ••• However, ... But and
01 = 02 = 0 F12 =
1 f 1 COS2 9° dAi • dA2 7r R 2 dAl jAl jA2
i = dA1 as area is small fA, dA F1_2 =
C 0 S 2 9 dA2
7r R2 R 2 = X2 ÷ b2
c o s2 0 =
5A2
112
R2
F12 = IA h
=
h2 h2 +x 2
2xx•dx h2 2 + x2 X X x (b2 + x2 )
2x = f0r 11h2 2 + x2 h2 + x2 dx 2 = h 2r+ r2 46. Calculate shape factor between two opposite sides of a hollow cube if shape factor between two adjacent sides of it is 0.2, as shown below.
Applying summation rule w.r.t. surface 1, we have— Given
However,
F11 + F12 + F13 + F14 + F15 + F16 = 1 F12 = F16 = F14 = F15 = 0.2 Fll + 0.2 + F13 + 0.2 + 0.2 + 0.2 = 1 F11 + F13 = 0.2 F11 = 0 as surface is flat F13 = 0.2
Similarly, we can find out other shape factors F2_4 = 0.2 F5_6 = 0.2
326
Heat and Mass Transfer
47. Explain Hamilton and Morgan charts for evaluating shape factors for simple configurations. Hamilton and Morgan charts are available for finding shape factors of two surfaces in the following configurations— (a) Aligned Rectangular Parallel Plates
Plates 1 and 2 (size 'X' and 'Y') and spacing `L'. The graphs are drawn for F1_2 versus LC for various
L (b) Perpendicular rectangular plates with common edge
I Plates 1 and 2 of sizes (X x Y) and (X x Z) perpendicular to each other with one common edge. The graphs are drawn for shape factor (F1_2) versus
X
for various
X
.
(c) Coaxial parallel discs
Discs 1 and 2 with radius r1 and r2 and interspacing r2 L shape factor (F1_2) versus — for various . L
. The graphs are drawn for
Radiation Heat Transfer 0.9
I
I
I
I I Y/L = 20
—
0.8
WIC L: , .
— 0.7
10
_
A011 ..0 111111F, 011P110111
—
1.5 1.0
— _
0.3
P d."
0
10
20
— —
0.2 0.1
AIWW111"
0.1
_
0.8 0.6
iIIIIII al P"AiAiii
—
0.2
— —
AA 4
0.5 0.4
_ .
0.6
F12
I
30
4.0
50
60
X/L Shape factor for aligned parallel plates
0.5 Y/X = 0.02 0.0 0.4
0.1
0.2 I 0.3 F12
0.2 0 1 0.1
0 01
0.2
2
04 06 08
4 6 8 10
Z/X Shape factor for perpendicular rectangles with common edge
327
328
Heat and Mass Transfer 1.0 r2
0.8 ri
4 3 r2IL 0.6
2
15
1.25
F12 1.0 0.4 0.8 06
0.4
0.2
03
0 01
0.2
0.4 06 08 1
2
4 6 8 10
Shape factor for coaxial parallel discs
48. What do you understand by equivalent electrical network for radiative heat transfer? or Derive an expression for surface resistance of a grey body. (UPTU — 2003) Consider an opaque nonblack surface or diffused grey surface with radiosity (J) and irradiation (G). Let E be emissive power of surface and Eb the emissive power of black surface at same temperature. J = (E = p.G)
p.G.
////////////////////// Radiosity (J)
J=E+ = E Eb
p•
///////////////
G. p•
G
where
E
= emissivity
Radiation Heat Transfer
329
Now for opaque surface, transmissivity z = 0. Hence, p + a = 1 or p = (1 — a) J= EEb +(1— a) • G
According to Kirchhoff s law E = a J= EEb + (1 — E )G.
E Eb 1— E The net heat transfer from the surface is equal to the difference of radiosity and irradiation
or
G —
J—G= Q
or
Q = A (J — G) =
A(j J — E 1— E
—A
(J—JE—J+EEb)
1—E
A E (Eb
1—E
Comparing with electric circuit: Now Q is analogus to current (i), (Eb — J) is analogus to potential difference (AV) and is analogus to resistance (R) . The term 1A
E E
1— E AE
is called surface resistance. The equivalent
electrical circuit and surface resistance network are0
1\ A&
0
Eb
1—E AE
J
49. Derive the expression for heat exchange between two nonblack surfaces. Draw the equivalent electric circuit for this heat flow. or Find grey body factor.
Heat and Mass Transfer
Consider two nonblack surfaces with radiosity J1 and J2. Now out of all radiation emitted by first surface, a fraction J1 A1 F1.2 is received by second surface. Similarly, out of all radiation emitted by second surface, a fraction J2A2 F2.1 is received by first surface. Therefore, net heat transfer between first and second surface is— Applying reciprocity rule,
Q12 = Jlx A1 x F1.2 — J2 x A2 x F2.1 A l F12 = A2F21 Q12 = Al X Fi.2(Ji — J2) — —
1 Al X Fi_2
The equivalent electric circuit and space resistance are
1 Al x F12 Two grey surfaces
Electric circuit
Space resistance
1 is called space resistance which depends upon geometry and distance Al fi2 between the surfaces. Besides space resistance, both surfaces have their respective surface resistances as shown below. \\.\\\\\\\\\\\\\\\\\\\
N
The term
\\\\\\\\\\\\\\\\\\\\\
330
A1F12
[ 1 — £2] A2E2 Surface and space resistance
The equivalent electrical circuit is— Eb1
Eb2
Q12 —
Ebi — Eb2
Ebl — Eb2
— 1 — Et + 1 ± 1 — E2 X Al El
f12
E2
A2
Radiation Heat Transfer
A
4)
0.(T4 _
i
i
T2
1 — E2 X A2
= 1 — + 1
E2 F12 = Fg 1 _2 Al a ( Ti4 — T24) El
where
Grey body factor = Fg1_2 —
33 I
1—E1 El
Al
1 1 + 1— E2 x Al E2 ti2 F12
50. Find the grey body factor for: (i) two infinite parallel plates, (ii) two concentric spheres of radius r1 and r2, (iii) two black surfaces, and (iv) A very small body in a large enclosure.
N\N\\\\\.\\\\
\\\\\\\\\\\\\n)
Case 1: Infinite parallel grey plates
= A2
If two grey surfaces are infinite parallel plates, radiation emitted by first surface is fully absorbed by second surface and vice versa. Hence, F1.2 = F21 = 1 and A l = A2. 1
Fgi _2 = El
1
1— E2
F12
E2
1
El
—1±1±k1
E2
1
— 1 — —1—1
El E2
Case 2: Concentric grey spheres of radius r1 and r2
X Al A2
332
Heat and Mass Transfer
The radiation emitted by first sphere is completely absorbed by second sphere. Hence, F1_2 = 1 and Al = 42r r?, A2 = 4 ir d 1 Fgi _2 = 1 — el + 1 + 1 — G 712 2X ..2. El 1 E2 12 =
1 2 1 + (1 _1)11 El E2 r22
Case 3: Two black surfaces In this case
E1= E 2 = 1 1 1 11 =F1.2 + 1 + F12 1 = F12 Fg12
...
Fg 12 = 1 — 1
or
For black bodies, grey body factor is same as shape factor. Case 4: A very small body in a large enclosure 0A1
A2
A2 >>> Al
All radiation emitted by Al is intercepted by A2. Hence, F12 = 1. Also
Ai Al A
—> 0
1 1 Fg _2 = i _ 1 = 1 — el l 2 + 1 — E2 1 Ei + XA +1+1—G X0 E1 E2 A2 F12 el ez = E1
51. What is the purpose of radiation shield? Find the ratio of heat transfer between two infinite parallel surfaces with radiation shield and without radiation shield. (UPTU — 2002 — 3)
81
Without shield
// / // // / i82 i
,,. / T2
T1 ei
\\\\\\\\\\\\\\
T1
\\\\\\\\\\\\\\
Heat transfer between two surfaces can be reduced by placing a radiation shield between them. The heat transfer is reduced as placing a radiation shield between two surfaces offers additional resistance in the heat flow
e3 With shield
Radiation Heat Transfer
333
Case 1: Without shield. For infinite parallel surfaces, the grey body factor isFgi _2 =
1
i1 + 1 _ 1) E].
Q12 =
E2
Ai o (Ti4 — T24) ( 1 + 1 _ 1) E1 E2
Case 2: With shield Q13 = Q32 a ( 4 _ Al Ti T34 ) Al a (T34 — T24)
( 1 + 1 _ ij
(1+1_
El E2
E3 E2
ij
4 ( 1 + 1 _ ij + 4 ( 1 + 1 _1) T2
T1
Also
T 34 —
Q13 =
E3 E2
El E3
(1 + 1 _ 1)+ (1 + 1 _ E3 E2 El E3 1) Ai a (114 — T24) ( 1 + 1 _ ij+ ( 1 + 1 _ E3 E2 El E3 1)
( 1 + 1 1) Q12
Q13
=
Ej.
E2
( 1 + 1 —1)+( 1 + 1 _ 1) E3 E2 El E3
52. If two infinite parallel plates have equal emissivity, find heat transfer with and without shield having the same emissivity.
Q12 Q13
Now given
( 1 ± 1 _1) El E2 ( 1 + 1 _ ij+ ( 1 + 1 _ i) E3 E2 E1 E3
El = E2 = E3 = E ( 1 + 1  1) Q12 E E ) = a_3 (1 ± 1 +111 E E I k,E E )
_0+ p
334
Heat and Mass Transfer
__ 1
__1 1 +1 2 Hence, heat transfer is reduced to half with the use of shield of same emissivity. ..T 34 = T14 + T24 2
Also
53. Find total resistance in case 'n' radiation shields of same emissivity are used. Also find the ratio of heat transfer with 'n' shields and without shields. Assume temperatures of surfaces remain at T1 and T2. When 'n' shields are used, we will have— (a) (n + 1) space resistance but F1_3 = F3_4 = F4_5 = 1 (b) (2n + 2) surface resistance Total resistance is— 1 ER =[(2n+2)(1 — E )+(n+1)xlixE A Aa (Ti4 — T24)
Qwith'n'shields =
(2n +2)( —1)+n+1 1=
Aa(T 4 — r24,) 2(n +1)
(n + 1)
E
= Aa (Ti4 — T24) (n + 1) r2 —1) E without shield = Q(i.e. n=0)
Qwith shields
_
A CT (Ti4 — T24)
2 —1 E 1
n +1
Qwithout shields Note: The equation of heat flow for parallel plates by n shields with emissivity E si, E s2 ... E Sri as Q12 =
A a ( Ti4 — T24) 1 + 1+ 2L, ÷ 1 El E2 n=1 Esn
Radiation Heat Transfer
335
54. Show heat transfer by radiation from a cavity is— Q = AiE Ti4
[ — Fil 1— Fll(1 —Ei)
Consider a cavity as shown below which is at uniform temperature T1. Out of the emitted radiant heat energy, a part of this energy is intercepted by the surface itself. Certain portion of this intercepted energy by the surface is further absorbed by the surface and the remaining is reflected to the surroundings. Let us chalk down emission, absorption and reflection of the radiant energy. (a) Emission of radiation from the surface— Al E 1a Ti4
Cavity
(b) The amount of radiant energy intercepted by the surface = F11 x (A E l a T 14) (c) The amount of intercepted radiant energy absorbed by the surface— = E l x (Fil Ai x E l X aT14) as (al = E 1) (d) The amount of energy reflected— = (1 —E 1)(F11 XAi xE i xaTi4) (e) The amount of reflected energy which is again intercepted and absorbed— = (1 — E 1) (Fi? Ai E 1 X a Ti4) (f) The amount of reflected and intercepted energy which is again reflected— = (1 — E 1)2 (Fl ? A iE 1 X a T4) (g) Reflected energy which is again absorbed— = (1 — E 1 )2 (A iE ? F1? a T4) (h) The fraction of this intercepted energy which is again reflected = (1 —
F11 a T14) Keeping the above absorption and reflection in view, the net rate of emission from the surface is— E 1)3 (A 1E 1
Q = First emission — total absorption
336
Heat and Mass Transfer
= Ai E la Ti4  [E ? Fn. Ala Ti4 ÷ (1  E OE ? Fli Ala T4 +(1  E 1)2 E1 Fil Ala Ti4 ÷ ...1 = A1E 1 676"[1  E 1F11  E 1(1  E 1)Fii  E 1(1  E 1)2 F1?...] = AiE la T4[1  E iFidl ÷ (1  E OFii + (1  E 02 Fli] }
E1 F11
= AiE la Ti4{1
1 F11 (1 E1) 1F11 } = AlE la T4{ 1  Fn. (1 _ Ei) 55. For two infinite parallel planes with emissivity El and E 2, the interchange factor for radiation from surface 1 to surface 2 is given by(a)
1
E1E2
1
(b) — + —
El + E2  E1 E2
E1 E2
(c) E1 + E2
(d) E1E2 (IES 93) Fg12
1 but F12 = 1 = 1  E1 + 1 1  E2 + E2 El F12
1 — 1 1+1+ — 1 1 Ei E2 E1 E2 El ÷ E2  E1 E2
Option (a) is correct. 56. What is the net radiant interchange per square metre for two very large plates at temperatures 800 K and 500 K respectively? (The emissivity of the hot and cold plates are 0.8 and 0.6 respectively. StefanBoltzmann constant is 5.67 x 10_8 W/m2K4.) (IES 96) Fg1. 2 
El E2 El ± E2  E1 E2
0.8 x 0.6 0.48 = 0.8 + 0.6  0.8 x 0.6 0.8 + 0.6  0.48 = 0.52 =
Q — A = F8120(1'14  T24) = 0.52 x 5.67 x 108 (8004  5004)
Radiation Heat Transfer
337
= 29484 (4096  625) x 108 x 108 = 40260 W = 10.260 kW/m2 57. Match ListI with ListII and select the correct answer using the codes given below the lists. ListI (A) Infinite parallel plates (B) Completely enclosed body large compared to enclosing body (subscript for enclosed body)
ListII 1. E 2. E 1 E 2
(C) Two rectangles with common side perpendicular
1 1 — —1  1 E1 E2 1 4. 1+ 1 E1 A2 c2 3.
(D) Concentric cylinders
(a) (b) (c) (d)
Codes: A 1 3 2 3
For parallel plates,
B 2 1 1 1
C 4 4 3 2
1 Fg12 = 1 E1 + 1 + 1  e2 Al El F12 E2 A2 F12 = 1 and Al = A2 1 1 +11 E1 E2 F12 = 1
Fg 1 _ 2
For concentric cylinders,

Fgi _ 2 =
1 E1
1 Ai 1 A 112 E2
)
1
For complete enclosed body, Al 0 and F12 = 1 112 1 Fg _2 = 1 _ 1 e 1 — E2 1 +1+ x0 E1 E2 =c1
D 3 2 4 4 (IES  95)
338
Heat and Mass Transfer
For two rectangles, we have Fg1_2 = E l E 2 Option (d) is correct. 58. Consider two infinitely long black body concentric cylinders with the diameter ratio D2/Di = 3. The shape factor for the outer cylinder with itself will be—
(a) 0 (c) 2/3
(b) 1/3 (d) 1 (IES — 97) Fg12 =
Now
1 El — 1 1 E21 Al E1 + F12 + E2 A2
dl L d1 Al 1 F12 = 1, A2 = gc12 L — d2 = 3 &E 1 =E 2 = 1 for black bodies. 1 F812 = 0 +1+ 0 = 1
Option (d) is correct. 59. Two large parallel plates having emissivity of 0.5 each exist at temperature of 1200 K and 300 K: (a) Find the heat transfer rate per m2 of the plate. (b) Find heat transfer rate if a radiation shield with an emissivity of 0.5 on both sides is placed between the two plates. (UPTU 2002 — 3) (a) The grey body factor is— Fg1 _ 2 =
1 (E 1 = E 2 = 0.5 also A l = A2) El 1 E2 Al El + F12+E2 X A2
Radiation Heat Transfer
=
Q12
A
339
1 1 = = 1 +0.5+1+0 5 +1+1+1 3 05 0.5 0.5
= Fg1_2 x a (Til — T24) _1 3 x 5.67 x 108 x (12004 — 3004) — _1 3 x 5.67 x 108 (20736 — 81) x 108 — = 39.038 kW/m2
(b) The grey body factor with shield— 1 Fg1_2 = ( 1 + 1 _ i) + ( 1 + 1 _ 1) E1 e3 e3 e2 =
1 ) 1 (1 0.5 + 0.05 — 1)+(005 + 0.5 1 1 = 2(2 + 20 —1) = 1 42 Q12
A
= Fgi_2 x a x (Ti4 — T24) = 4 x 5.67 x 108 (12004 — 3004) 2 = 4 x 5.67 x (20655) 2 = 2.79 kW/m2
60. Calculate the energy emitted by a grey surface of emissivity 0.45 at a temperature 800°C. The surface measures 5 m x 6 m. (UPTU — 2003 — 4) Energy emitted by a grey surface is— Q = EA6T4 = 0.45 x (5 x 6) x 5.67 x 108 (800 + 273)4 = 1.01 x 103 kW
340
Heat and Mass Transfer
61. A flat plate 5 m2 receives normally radiant energy with an intensity of 660 W/m2. The absorptivity of the plate is 2 times its transmissivity and 3 times its reflectivity. Find the energy absorbed, transmitted and reflected in watts. (UPTU — 2003 — 4) Given (a) absorptivity 2 times its transmissivity a = 2r or r = a 2 (b) absorptivity 3 times its reflectivity a = 3p or P a Now a+p+ = 1 a or a + (±1 + — = 1 2 3 or
6a +3a +2a = 1
or
a= 6
6
11
r = a =3 2 11 =a=2 3 11 ' Radiant energy intercepted = 660 W/m2 (a) Absorbed energy = a x 660 x area 6 x 660 x 5 = 11 = 1.8 kW (b) Energy transmitted = 660 x p x area = 660 x 2 x5 11 = 600 W (c) Energy reflected = z x 660 x area 3 x 660 x 5 = 11 = 900 W 62. Two parallel rectangular plane black surfaces of 1 m x 2 m each, face each other and are located at a distance of 4 m. The temperature of one surface is 200°C and the other is 100°C. Calculate the net heat exchange between the two surfaces. (UPTU — 20023, 20045)
Radiation Heat Transfer
34 I
Guidance: The problem is to be solved with Hamilton and Morgan chart for aligned rectangular parallel plates by finding shape factor F12.
® 0
x
'N
L . 4, x = 1 and Y =2 X =— 1 = 0.25 L 4 Y 2 L = 4 = 0.5 From the graph for calculated X and Y L L Fg12 = 0.05 Q12 = Fg1_2 x A a (Ti4  Ti) = 0.05 x (1 x 2) x 5.67 x (4734  3734) x 108 = 0.05 x 2 x 5.67 x (500.5  193.5) = 174 W
Here
63. The outer surface of a tube of 100 mm diameter is maintained at 120°C by the passage of steam through its interiors. A radiation shield is installed around the tube, with an air gap of 10 mm between the tube and the shield. The shield is at a temperature of 35°C. The tube and shield are diffuse grey surfaces with emissivity of 0.8 and 0.1 respectively. What is the radiant heat transfer from the tube per unit length? (UPTU  2005  6) The arrangement of tube and shield isShield (d2 = 0.12 m) Tube (d1 = 0.1 m)
T1 = 120°C
T2 = 35°C
Fg1 _2 =
1 1+(1 ) Al 1 El e2 A2
342
Heat and Mass Transfer
=
=
1 x1 ) xd2 x 1
1 ) irdi
1 _, ( 1 0.8 ' 0.1 1
1.25 + (10 1) 0.1
0.12
=
1 1.25 + 7.5
= 1 = 0.1143 8.75 Q = Fgi _2 x A x ax (Ti4  Tit) = 0.1143 x 5.67 x 108 x (iv x 0.1 x 1) (3934  3084) = 0.203 x 108 x (238.5  89.99) x 108 = 0.203 x 148.51 = 30.14 W/m 64. A small convex object of area A1, temperature T1 and emissivity E l is enclosed within a larger enclosure at temperature T2 and emissivity E 2. Derive an expression for the net heat exchange between the two. (UPTU  2007  8) F g12 = 1El
E1
1 +
1 F12
+
1 E2 E2
X
Al
A2
T2 £2
For convex surface Also
F11 = 0, hence F12 = 1 Ai A2 >» A l and A > 0 Al 1 F$1_2 = 1 = c1 — 1+1 cl Q = Fgi _ 2 x Ai a (Ti4  T24) = E illi a (Ti4  Th
65. Two large, diffuse, grey, parallel surfaces (emissivity for both the surfaces is 0.8) are separated by a small distance. A thin radiation shield is used to reduce the radiation
Radiation Heat Transfer
343
heat transfer rate between the two surfaces to 10% of the original. Find the emissivity of the shield. (UPTU — 2007 — 8) Case 1: Without shield, shape factor is 1 1 1 —1 El E2
Fg12
1
= 1 + 1
_ 0.8 0.8 1 = 1 2.5 —1 1.5 = 0.67 Qws = Fg1.2 A a (T14 — = 0.67 x A x a(Ti4 — Case 2: With radiation shield, shape factor is 1
Fg1_2 =
1 + 1 _].)+( 1 + 1 —1)
E3
E3
E2
1 =
1
1 E3
1 + 1 E3
1 2.5 — 2 + E3
1 0.5+ Qwith shield
E3
= Fg12 xAx a (Ti4 — T24) 1 xA x a (Ti4 — T24) 0.5 +1 ' E3
Given
Qwith shield
x A x a (7,14
0 = 100 1 Qws
) = 0.1 x 0.67 xA x a (Ti4 — T24)
E3
or
1 = 0.067 0.5 + 2 E3
344
Heat and Mass Transfer
1 = 0.0335 + 0.134 E3
or
0.134 = 0.9665 E3
or
E 3 = 0.134 0.965 = 0.14
or
66. A thin aluminium sheet with an emissivity of 0.1 m on both sides is placed between two very large parallel plates that are maintained at uniform temperature T1 = 800 K and T2 = 500 K and have emissivities 0.2 and 0.7 respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result to that without the shield. (UPTU — 2006 — 7) The setup of plates and shield is— Aluminium shield = 0.1
T1 = 800 K
T2 = 500 K
= 0.2
E2 = 0.7
1. Case 1: Without shield, the shape factor is 1 1 Fg12 = 1 1 _ 1— 1 + 1 _ 1 0.2 0.7 Ei +E 2 1 1 5 +1.43 —1 5.43 = 0.184 Qws
A
= Fg12
x a x (71 — Tit)
= 0.184 x 5.67 x 108 (8004 — 500 = 1.04 x 108 (4096 — 625) x 108 = 3.61 kW/m2 2. Case 2: With shield, the shape factor is Fg12shield
=
+ E1 E3
1 1) +
+ E3 E2
1)
Radiation Heat Transfer
345
1 ( + _ i) + ( + 1. ) 0.2 0.1 0.1 0.7 1 (5 + 10  1) + (10 + 1.43  1) 1 14 + 10.43 1 24.43 = 0.041 Qshield
A
Qshield
Qws
Fg12shi.eid
a x (Ti4  T24)
= 0.041 x 5.67 x 108 (8004  5004) = 0.041 x 5.67 x 108 3471 x 108 = 807 W/m2  807  22.35% 3610
67. A 20 cm diameter spheric ball at 800 K is suspended in the air. Assuming that the ball closely approximates a black body, determine(i) the total black body emissive power (ii) the total amount of radiation emitted by the ball in 5 min. (UPTU  2006  7) Q =A x ax T 4 A = ird2 = x 0.22 = 0.1256 m2 Q = 0.1256 x 5.67 x 108 x 8004 = 2.917 kW
Heat dissipated in 5 min. Qtote = 2917 x 5 x 60 = 875.1 kJ 68. What is greenhouse effect? Why is it a matter of great concern among environmental scientists? (UPTU  2006  7) Earth has atmosphere around it for many kilometres which is mainly air with pollutants like carbon dioxide, water vapour and methane, etc. These pollutants are transparent for shorter wavelength radiation but these pollutants are opaque for longer wavelength radiation. The earth receives solar radiation in day time and solar radiation has shorter wavelength facilitating solar radiation to pass through the atmosphere. Solar radiation reaches the earth and it warms up the earth on absorption.
346
Heat and Mass Transfer
As sun has very high temperature, hence solar radiation has short wavelength. During night earth cools down by radiating its heat energy into atmosphere. As earth has relatively low temperature, hence earth emits radiation at longer wavelength in the infrared region. The pollutants do not permit longer wavelength radiation emitted from the earth to pass through the atmosphere and this radiation is absorbed in the atmosphere. This absorbed radiation is again reradiated to the earth surface. On account of this interception of energy in lower atmosphere and reradiation towards earth, the temperature of lower atmosphere and earth increases. Such warming of lower atmosphere and earth due to interception of radiation from earth is called greenhouse effect. Due to greenhouse effect, there is no significant change between day and night temperatures. Increase of pollutants in atmosphere results from (1) excessive deforestation (more carbon dioxide in atmosphere), and (2) industrialization and burning of fuels (concentration of CO2, in ethane and CFCs in atmosphere).
Solar radiation can pass through due to short wave length Infrared radiation is Atmosphere reflected by pollutants with pollutants in atmosphere Greenhouse effect
Greenhouse effect is of great concern to environment scientists due to the following adverse consequences from it— (a) It heats up the earth surface and evaporates the surface water. (b) Global warming affects food production. (c) Global warming melts ice of polar regions and glaciers, thereby increasing sea level. Increase of sea level leads to flooding of low coastal areas of the world. (d) Carbon dioxide undergoes photochemical reactions at higher atmosphere to form carbon monoxide. Carbon monoxide is a toxic gas. 69. What is solar radiation? What is the effective sky temperature? (UPTU — 20023, 20067) The fraction of energy emitted by sun which reaches earth is called solar energy. Sun can be considered a nuclear reactor in which heat is being generated continuously by fusion reaction i.e. hydrogen atoms forming helium atoms. The reason for a fraction of solar energy reaching earth is that the sun subtends only at an angle of 32 min at the earth. The earth is located at a mean distance of 1.496 x 108 km from the sun. The intensity of solar radiation reaching the earth is almost constant and it is called solar constant (G5). The solar constant (G5) is the rate at which solar radiation flux is received on earth's surface normal to sun rays. Considering sun as black body and solar radiation is evenly distributed on the
Radiation Heat Transfer
347
surface area of the sphere formed by earth's orbit around the sun, we have solar constant as r )2 Solar constant G, = ( '4'1 a Ts = 1353 rorbit
Due to eccentricity of earth's orbit, the distance of earth from sun is more in winter season as compared to summer season. Hence, the value of solar constant `GS' varies as per earth's orbit around sun. Maximum quantity of solar energy reaches the earth surface when: (1) rays are normally incident on earth surface, (2) rays travel a smaller distance (seasons of year), and (3) there is less absorption. The solar radiation reaching the earth surface is the sum of direct solar radiation (Gd) and diffuse solar radiation (Gdf) as shown in the figure. The scattered radiation is received by earth from all directions. Diffuse solar radiation varies from low value on clear day to maximum on totally cloudy day. Diffuse solar radiation (Gdf) Direct solar radiation (Gd)
Atmosphere ,
Earth
The solar radiation passes through atmosphere and it undergoes absorption and scattering in the atmosphere. The atmosphere also contains pollutants like carbon dioxide, water vapour and other suspended particles. The emission of heat energy from the atmosphere to the earth surface is called atmospheric emission. The atmospheric emission depends upon effective atmospheric absolute temperature. The atmospheric emission constant `Gsky' is given by G sky = aTs2ky W/m2K
Effective atmospheric temperature (Tay) ranges from 285 K in warm and cloudy day to 230 K for cold and clear day. 70. If solar constant on earth surface is 1353 W/m2, find the temperature of sun considering it is a black body. Sun's diameter = 1.4 x 106 km and mean distance of the sun from the earth = 1.5 x 108 km. The solar radiation is spread over a sphere of radius (r) = 1.5 x 108 km ..
,,
,
t I
L
0 Earth 1
l
,‘
Sun , ,,,
i I
348
Heat and Mass Transfer
Qsun = Asun
x a x 71un
G = Qs
r grorbit
or
1353 =
= Asun ) Tx Ts4un 2
grorbit
2 4 rxds xax Tsw,
4 x — rorbit (1.4 x 109)2 x 5.67 x 108 x Tstn 4 x (1.5 x 1011)2 4 x 1353 x 2.25 x 1022
or
Q = 1096 x 1012 sun T = 1.96 x 10" x 5.67 x 10—
or
Tsun = 5762 K.
71. A furnace has temperature of 2000°C and it emits radiation. Find: (1) monochromatic emissive power at 2 pm wavelength, (2) wavelength having maximum emission, (3) maximum emissive power, and (4) total emissive power. Take C1= 0.374 x 1015 and C2 = 1.4388 x 102 1. Using MaxPlanck's law, for black body Eb A —
Eb b =
c 1 c2 e —1
6.374 x 1015 x (2 x 106)5 1.4888 x10 2
e 2 x106 x2273 —1 = 5.2 x 1010 W/m2 2. As per Wein's displacement law 2898 x 106 Amax
= 7 = 1.28 pin
2898 x 106
 2273
3. Maximum spectral emissive power 0.374 x 1015 x (1.28 x 106)5
(FbA,)max
1.4888x102
e 12,8x106 x 2273 
1
3.74 x 105 x 0.29 x 10" = 6.5 x 1012 W/m2 167 1
Radiation Heat Transfer
349
4. Total emissive power per m2
E = a T4 A = 5.67 x 108 x (2273)4 = 1.513 x 106 W/m2 72. The radiative heat transfer rate per unit area (W/m2) between two plane parallel grey surfaces (E = 0.9) maintained at 400 K and 300 K is(a) 992 (c) 464
(b) 812 (d) 567
(StefanBoltzmann constant a = 5.67 x 108 W/m2K4) (GATE  93) Shape factor
Fg12
Q
A
1 1 _ = 1 + 1 _1 1 + 19 1 09 0 El E2 = 0.818 = F g 12 a (ril  Ti)
Q = 0.818
A
x 5.67 x 108 (4004  3004)
= 812 W/m2 Option (b) is correct. 73. For the circular tube of equal length and diameter as shown below, the view factor F13 is 0.17. The view factor F12 in this case will be
(b) 0.21 (d) 0.83
(a) 0.17 (c) 0.79
(GATE  2001) Using the summation rule F11 +
F12 + F13 = 1
350
Heat and Mass Transfer
Since surface A l is flat, hence F11 = 0 F12 ÷ F13 = 1 F12 = 1 — F13 = 1 — 0.17 = 0.83
or Option (d) is correct.
74. Match List1 (type of radiation) with ListII (characteristic) and select the correct answer using the codes given below the lists. ListI (Type of radiation) A. Black body B. Grey body C. Specular D. Diffuse
1. 2. 3. 4.
ListII (Characteristic) Emissivity does not depend on wavelength Mirrorlike reflection Zero reflectivity Intensity same in all directions
Codes A 2 3 2 3
(a) (b) (c) (d) A. B. C. D.
B 1 4 4 1
C 3 2 3 2
D 4 1 1 4 (GATE)
Black body has a = E = 1 and p = 0 (option 3 in ListII) Grey body has emissivity which is independent of wavelength (option 1 in ListII) Specular has mirrorlike reflection (option 2 in ListII) Diffuser has emissivity independent of direction (option 4 in ListII) Hence, code is — 3, 1, 2, 4 Option (d) is correct.
75. Sun emits maximum radiation at A = 0.52 IL Assuming sun as black body, find the surface temperature of the sun and emissive power at that temperature. According to Wien's displacement law, we have As
Now
Amax x T = 2898 Amax = 0.52 henceT = 2898 0.52 = 5580 K
Q = a 74 = 5.67 x 104 X (5580)4 = 5.67 x 107 W/m2
Radiation Heat Transfer
35 I
76. A spaceship launched from the Earth to the Venus. The distance of sun to Venus is 108 x 106 km while to Earth it is 150 x 106 km. If temperature of spaceship at earth is 300 K, find the temperature at Venus. Guidance: The surface area of spaceship remains same and the solar radiation intercepted depends upon the distance from the Sun to Earth and Venus respectively. Venus
Earth
If Q is total emission from Sun, then
= Solar constant at Earth = crAspace X Ti 2 47r rsun_earth
Similarly
 Solar constant at Venus = 6 Aspace
4 7trs2unvenus
rsunvenus r h
2 =
1 T2
X
Ti
4
(los x 106
300 )4 150 X 106  T2
or or
1
(150)2 300  108) = 1.18 T2 = 1.18 x 300 = 354 K T2
77. Sun's surface at 5800 K emits radiation at wavelength of 0.5 A furnace at 300°C will emit through a small opening a radiation at a wavelength of nearly(a) 10 p. (c) 0.25 p.
(b) 5 p. (d) 0.025 p. (IES 97)
Using Wein's displacement lawx Ts = A mf 
X
Tf
X Ts
T1
 0.5 x 106 x 5800 = 5.06 x 106 573
352
Heat and Mass Transfer
= 5.06 gm Option (b) is correct. 78. Assertion (A). In a furnace, reradiation from the walls has the same wavelength as the incident radiation from the heat source. Reason (B). Surfaces at the same temperature radiate at the same wavelength. Mark the answer as— (a) (b) (c) (d)
If both (A) and (R) are true and (R) is correct explanation of (A) If both (A) and (R) are true but (R) is not the correct explanation of (A) If (A) is true but (R) is false If (A) is false but (R) is true (IES — 98) As per Wein's displacement law—
An, x T = constant or
1
T
m The wavelength of radiation depends upon temperature of surface. Assertion is true as all walls of furnace are at same time. Reason is true as surfaces at same temperature radiate at same wavelength and it is a correct explanation of assertion. Hence, option (a) is correct. 79. Two long parallel surfaces each of emissivity 0.7 are maintained at different temperatures and accordingly have radiation heat exchange between them. If is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of same emissivity on both sides. The number of shields should be— (a) one (c) three
(b) two (d) four (IES — 92) 1 Qwith shield = n+1 Qwithout shield Qwith shield = 25% Qwithout shield 1 = 0.25 n+1
or or Option (c) is correct.
n+1=
0.25 = 4
n=3
Radiation Heat Transfer
353
80. A source of radiation has an intensity of 800 watts/m2. Find the number of photons per sec per sq metre represented by this intensity if the wavelength is 500 n.m (use speed of light = 3 x 104 m/s and Planck's constant = h = 7 x 10r34 J/s) (a) 10.4 x 1021 (c) 4.4 x 1021
(b) 6.8 x 1021 (d) 2.0 x 1021 (IES 88)
Energy of a photon is E=h•v = 7 x 1034 x c A = 7 x 1034 x
3 x 10 4 500 x 1012
= 21 x r.20 1U = 4.2 x 1020 5 If there are n photons per m2 n x E = 840 840 4.2 x 1020 = 2 x 1021
or
n—
Option (d) is correct. 81. Mark the answer as— (a) (b) (c) (d)
If both (A) and (B) are true and (R) is the correct explanation of (A). If both (A) and (R) are true but (R) is not the correct explanation of (A). If (A) is true but (R) is false. If (A) is false but (R) is true.
(1) Assertion (A): The nose of aeroplanes is painted black. Reason (R): Black body absorbs maximum heat which is generated by aerodynamic heating when plane is flying. (IES — 96) (2) Assertion (A): In an airconditioned room, the reflective coating should be on the inside of the window. Reason (R): Window pane glass is transparent to solar radiation (IES — 96) (3) Assertion (A): Solar radiation is mainly scattered or transmitted but not absorbed by the atmosphere. Reason (R): Absorptivity of atmosphere is low. 1. The nose of aeroplane: The nose of aeroplane houses the radar equipment and any colour not hampering electromagnetic transmission form radar can be given to nose.
354
Heat and Mass Transfer
The assertion is false. The reason that black body absorbs maximum heat is true. Hence, option (d) is correct. 2. Airconditioned room: In airconditioned room, the temperature inside the room is maintained at low temperature. The glass is transparent to solar radiation and solar radiation heats up the cooled space inside the air. To avoid this, reflective coating is provided on the inside of the window. Hence, both assertion and reasoning are true and reasoning is the correct explanation of assertion. Hence, option (a) is correct. 3. Solar radiation: Atmosphere is transparent for solar radiation and radiation consists of both scattered and transmitted radiation. Assertion is therefore true. The reasoning is also correct and reasoning is the correct explanation of the assertion. Option (a) is true. 82. Match ListI with ListII and select the correct answer using the codes given below the lists. (A) (B) (C) (D)
ListI Window glass Grey surface Carbon dioxide Radiosity
1. 2. 3. 4.
ListII Emissivity independent of wavelength. Emission and absorption limited of wavelength Rate at which radiation leaves a surface Transparency to shortwave radiation
Codes: (a) (b) (c) (d)
A 1 4 4 1
B 4 1 1 4
C 2 3 2 3
D 3 2 1 2 (IES — 96)
(a) Window glass: They are transparent to short wave radiation—(option 4 of ListII) (b) Grey surface: Their emissivity is independent of wavelength—(option 1 of ListII) (c) Carbon dioxide: their emission and absorption limited of wavelengths (option 2 of ListII) (d) Radiosity: It is the rate at which radiation leaves a surface (option 3 of ListII) Hence option (c) is correct. 83. A square room 3 m x 3 m has a floor heated to 27°C and has a ceiling at 10°C. The walls are assumed to be perfectly insulated. The height of the room is 2.5 m and the emissivity of all surfaces is 0.8. Find: (i) the net heat exchange between the floor and ceiling, (ii) wall temperature. Assume shape factor of ceiling to floor as 0.25. The room is depicted as below.
Radiation Heat Transfer
355
0Surface = ceiling 0Surface = floor 0Surface = wall
T 2.5
Room
Applying summation rule But ••• Similarly,
F11 + F12 ÷ F13 = 1 F11 = 0 as ceiling is flat surface F13 = 1 — F12 = 1 — 0.25 = 0.75 F22 ÷ F21 ÷ F23 = 1 F22 = 0, F21 = F12 = 0.25 F23 = 1 — 0.25 = 0.75
The equivalent circuit diagram for space resistance when two interacting surfaces are connected with third surface which acts as refractory wall is—
The total space resistance between surfaces 1 and 2 is 1 ) 1, ▪ , 1 1 1 Rspace Al F12 Al F13 ▪ A2 F23 or
Rspace =
1 1
± A2 2A1 fi2 A2 Al F122
Ai a(Ti4  T24) Q12 = 1—E1 + 1—E2 Al + ± A2 — 2A1 F12 but Al = A2 A2 A2 —Al F2 El E2 Ai a (Ti4 — T24) e)+ 2(1— F12) E J) (1 F122)
Ai a(Ti4 — T24) 2[(1
E
J)+ 11F12
356
Heat and Mass Transfer
3x3x 5.67 x 108 (2.884 x3004) —_296 W 1— 0.88 1 2 + [( 0.8 ) 1+0.25 Heat will flow from floor (surface2) to ceiling (surface1) as sign is negative. And wall temperature is— —
T3 =
+ V2I) =
(2884 + 3004)
= 294 K 84. Liquid oxygen at —153°C is in a spherical vessel (diameter = 20 cm). A spherical container (diameter = 30 cm) enclosed the vessel concentrically and space in between is evacuated. Vessel surface and container surface have same emissivity of 0.04 and temperatue of container surface is 27°C. Find: (i) heat transfer rate, and (ii) rate of evaporation of liquid oxygen if latent heat = 209 kJ/kg. The vessel and container are as shown below.
Liquid oxygen
Vessel Container
Fgi _2 =
1E1 El
1 1 + (1 E2 x Al r12 E2 A2
But F1_2 = 1 as container is completely enclosing the vessel. 1 1 Fgi _ 2 = 1 1  1) 1 +( 1 1) d2 El e2 A2 0.04 0.04 )4, 1 25 + (25 — 1) x 022 0.32 1 1 25+10.67 25+24x4 9 1 = 0.028 35.67 Q12 = Fgi_2 x A x cr(Ti4 — T24) = 0.028 x (n x 0.22) x 5.67 x 104(1204 — 3004) = —1.58 W
Radiation Heat Transfer
357
Negative sign shows that heat is flowing into liquid oxygen and liquid oxygen will evaporate. Rate of evaporation per min is1.58 x 60 Mevap 209 x 103 = 0.45 x 103 kg/min 85. Two parallel plates are radiating to each other having temperature of 500 K and 300 K. Find heat transfer rate: (i) assuming the plates as black body, (ii) plates having emissivity of 0.4. Each plate has surface area = 2 m2. Case 1: Plates are black bodies Q12 = F12 x a x A(Ti4 — Ti) F12 = 1 Q12 = Fi2, X a X A(Ti4 — T24) = 1 x 5.67 x x 2 (5004 — 3004) = 6.17 x 103 W = 6.17 kW Case 2: Plates are grey bodies. As plates are black
1 + 1 1 — e2 X Al e1 A2 F12 e2 A1 = A2 and F12 = 1 1 1 = Fgi _2 = 1 + 1 _1 1 + 1 1 El E2 0.4 0.4 = 0.25 Q1_2 = 0.25 x 2 x 5.67 x 108 (5004 — 3004) = 1.54 kW Fg12 = 1— e
Now
86. Three concentric spheres of diameter 20, 30 and 40 cm are as shown. Innermost and outermost spheres have temperatures of 200 and 800 K respectively. Find temperature of intermediate sphere. Assume perfect vacuum for annular spaces and emissivity = 0.2 for all spheres. The set up is
358
Heat and Mass Transfer
1 1 2 + Al 1 + 0.2 ( 1 _ 1) 1 1 0.2 10.2 1/ El 0.32 2 2 = 0.148 1 1 Fg2.3 = 1 + 0.32 ( 1 1) 1 + A2 1 1J 0.2 0.42 10.2 ) E2 A3 e3 = 0.138 Q12 = Q23 Fg12 X Ai x a(T4 T24) = Fg2.3 X A2 x a (T24 
Case 1:
Fg12 =
Tl

= Fg2 3 X A2 X (71 Al Fg12 0 138 X 0.32 'pet 74\ ki 2  31 0.148 0.22
2.1(T24  2004) 3.1 T24 = 8004  2004 = (4096  16) x 108 = 4086 x 108
8004  T24 =
or
4 4086 x 108 T 3.1 = 1318 x 108 Tl = 603 K
or
87. Two long parallel plates of same emissivity 0.5 are maintained at different temperatures and have radiation heat exchange between them. The radiation shield of emissivity 0.25 placed in middle will reduce radiation heat exchange to(a) 1/2 (c) 3/10
(b) 1/4 (d) 3/5 (ESE  2002) A a (Ti4
Q
Qshield
_ T24)
1 —1  1 — El E2 A a (7,4 _ 7,4) 1 2 (1 ± 1 _ 1) ± ( 1 ± 1 _ 1) El E3 e2 E 3
Radiation Heat Transfer
(1 + 1 _ i.) + ( 1 + 1 —1) E1
=
E3
E2
1 1 —+——1 E2
=
E2
(0.5 =
359
1 —1) + ( 1 —1) +0.25 0.5 +0.25 (01.5 + 0.5
E2
1)
(2+41)+(2+41) (2+21)
_5+5__10 3 3 _ 3 Q 10 Option (c) is correct. Qshield
88. Two large parallel plates with a small gap, exchange radiation at the rate of 1000 W/ m2 when their emissivities are 0.5 each. By coating one plate, its emissivity is reduced to 0.25. Temperatures remain unchanged. The new rate of heat exchange shall become— (a) 500 W/m2 (c) 700 W/m2
(b) 600 W/m2 (d) 800 W/m2 (ESE — 2002) Qi a (T4  T24, ) A — 1+— 1 —1 E1 E2
or
1000 = a (T14  T24) 1 1 1 0.5 ' 0.5 — a (T i4 — T24) = 1000(2 + 2 — 1) = 3000 W/m2
Now
G A
 Th 1 +1 1
_ a (7i4
El E2
3000 3000 _ — 1 ± 1 — 1 — 2 + 4 —1 0.5 0.25 = 600 W/m2 Option (b) is correct. 89. Match ListI and ListII and select answer using the codes—
360
Heat and Mass Transfer
(A) (B) (C) (D)
ListI Black body Grey body Specular Diffuse
ListII Emissivity does not depend on wavelength Mirrorlike reflection Zero reflectivity Intensity same in all directions
1. 2. 3. 4.
Codes A 2 2 3 3
(a) (b) (c) (d) A. B. C. D.
B 1 4 4 1
C 3 3 2 2
D 4 1 1 4 (IES — 2002)
Black body: It has zero reflectivity (option 3 of ListII) Grey body: Its emissivity does not depend in wavelength (option 1 of ListII) Specular: It has mirror like reflection (option 2 of ListII) Diffuse: In diffuse radiation, intensity is same in all directions (option 4 of ListII) Option (d) is correct.
90. An enclosure consists of the four surfaces 1, 2, 3 and 4. The view factors for radiation are F11 = 0.1, F12 = 0.4 and F13 = 0.25. The surface areas Al and A4 are 4 m2 and 2 m2 respectively. The view factor F4_1 is— (a) 0.75 (c) 0.25
(b) 0.5 (d) 0.1 (ESE — 2001)
Applying summation rule
or
F11 + F12 + F13 + F14 = 1 0.1 ÷ 0.4 + 0.25 + F14 = 1 F14 = 1 — 0.75 = 0.25
Using reciprocity law A1F14 = A21 F41
or
4 x 0.25 = 2 X F41 F41 = 0.5
Option (b) is correct. 91. Solar radiation of 1200 W/m2 falls perpendicularly on a grey opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power 600 W/m2, the radiosity of that surface will be— (a) 600 W/m2 (c) 1200 W/m2
(b) 1000 W/m2 (d) 1800 W/m2
Radiation Heat Transfer
36 I
Equivalent electric circuit is— — Q Eb 0V\A—ci j
I—E AE
Q
Eb — J
A — 1— E E
J
Q
= 600 = 1200 — 1 — 0.5 0.5 J = 600 W/m2
A or
— 1200 J
Option (a) is correct. 92. A spherical aluminium shell of inside diameter 2 m is evacuated and used as a radiation test chamber. If the inner surface is coated with carbon black and maintained at 600° K. The irradiation on small test surface placed inside the chamber is (a = 5.67 x 10_8 W/m2 K4) (a) 1000 W/m2 (c) 5680 W/m2
(b) 3400 W/m2 (d) 7348 W/m2 (ESE — 1999)
The set up is—
As per summation rule F11 But
+ F12 = 1 F11 F12
= 0 for black body =1
Q = F12 X T4
A
1 x 5.67 x 108 x 6004 = 7348 W/m2
=
Option (d) is correct. 93. A...body reflects entire radiation incident on it— (a) transparent
(b) black
362
Heat and Mass Transfer
(c) grey
(d) white
Option (d) is correct. Note: It is beneficial to wear white clothes in summer. 94. The monochromatic emissivity of a white body at all wavelengths and temperatures is equal— (a) zero (c) 0.6
(b) 0.1 to 0.4 (d) unity
Option (d) is correct. 95. "All bodies above absolute zero temperature emit radiation". This statement is based on— (a) Stefan's law (c) Provost theory
(b) Planck's law (d) Wien's law
Option (c) is correct. 96. Total emissive power E of a diffuse surface is related to radiation intensity I as E equal to— (a)
4x/
(c) 7r2 x i
(b) it x / (d) 47r x /
Option (b) is correct. 97. The absorptivity of a white washed wall is— (a) 0.1 (c) 0.5
(b) 0.3 (d) 0.9
White washed wall has maximum reflectivity and minimum absorptivity. Option (a) is correct. 98. A body at 1000°C in black surroundings at 500°C has an emissivity of 0.42 at 1000°C and emissivity of 0.72 at 500°C. Calculate the rate of heat loss by radation per m2— (i) When the body is assumed to be grey with e = 0.42 (ii) When the body is not grey. Assume that the absorptivity is independent of the surface temperature. (a) 20.6 kW, 18.5 kW (c) 54.393 kW, 47.962 kW
(b) 32.6 kW, 28.5 kW (d) 68.96 kW, 52.9 kW
Radiation Heat Transfer
363
Case 1: Body is grey with E = 0.42 The grey body emissivity remains constant and e = a Q = E a (VI  T24)
Net
A
= 0.42 x 5.67 x 108(12734  7734) = 54.893 kW/m2 Case 2: Body is not grey
Hence, body will emit with emissivity 0.42 at 1000°C and absorb with absorptivity 0.72(e = a) at 500°C Q
= 0.42 x 5.67 x 108 x 12734  0.72 x 5.67 x 108 x 7734 = 62538  14576 = 47962 W/m2 = 47.962 kW/m2
Option (c) is correct. 99. A small body at 100°F is placed in a large heating oven whose walls are maintained at 2000°F. The average absorptivity of the body varies with temperature of the emitter as follow: Temperature Absorptivity(a)
100°F 0.8
1000°F 0.6
2000°F 0.5
What is the rate at which radiant energy is absorbed by the body per unit surface area? (a = 0.1714 x 108 BTU/hr ft2 K4) (a) 2.11 x 10+4 BTU hrft2
(b) 3.38 x 10+4 BTU hrft2
(c) 13.7 x 104 BTU hrft2
(d) 3.16 x 10+4 BTU hrft2 (GRE)
— Q = Heat absorbed at 2000°F  Heat emitted at 100°F = a (a x T14  E Th = 0.1714 x 108(0.5 x 24604  5604 x 0.8) = 0.1714 x 108(18.31  0.08) x 1012 = 3.16 x 10+4 BTU/hrft2 Option (d) is correct. 100. The earth receives solar radiation at a rate of 3.2 J/m2 min. Assuming that earth radiates like a black body, calculate the surface temperature of the sun. Angle subtended by the sun on the earth is 0.53° and a = 5.67 x 108W/m2K.
364
Heat and Mass Transfer
Earth is rotating about sun as shown below.
Angle subtended w = or
= 0.53 x 180 = 9.25 x 103 radian Qs„,i = irD2 x a x T4
Qs„,, is distributed evenly on the sphere surface of radius R. Hence, radiation received at earth surface (distance R) per unit area is— 7rD2 X X T 4 aarth A 4irR2 0.7,4 ( 2 8.2 = (given) 4 R) = 60 5.67 x 108 a 8.2 x (9.25 x 103)2 x T = 4 60 T4 . 8.2 x 4 or 60 5.67 x 108 x (9.25 x 103)2 T = 5800 K 101. A black body of surface area 10 cm2 is placed inside an enclosure having constant temperature 27°C. The black body has constant temperature 327°C by maintaining by heating by electricity. Find electric power required to maintain temperature. a = 5.67 x 108 W/m2K. Guidance: The black body emits radiation at 327°C and absorbs radiation falling on it which is emitted at 27°C. Net heat flowing out is the difference of two. Q = Aa(rit — = 0.1 x 5.67 (6004 — 3004) x 108 = 69 W Hence, 69 W electric power is to be given. 102 A piece of charcoal and steel of same area are heated to same temperature and left to cool. Which will cool faster? Charcoal acts like a black body and emission is— Q b = Aa(Tia — 7,4
surrounding)
Radiation Heat Transfer
365
Steel if acts like a grey body with emissivity E„ the emission isQg = E s A a (Ti4  nunounding) Since E s < 1, hence charcoal will cool faster. 103. Find the shape factor for surface Al for the following configuration: Al and A2 are squares. All sides are 1.0 m each. Distance between Al and A2 is 1.0 m. F.— 1 m
Guidance: Take area A2 + A3 = A4 and find shape factor F41 from Hamilton and Morgan graphs. Similarly, find out shape factor F31. Now F41 x (A4) =  F21 x A2 ÷ F3_1 x A3 and F21 can be found out. 1. Shape factor F4_1
Hence, from graph
= = = = X 1 ' X 1 2 F4_1 = 0.1.
2. Shape factor F3_1
Z _
X
From graph
_ ' X
F31 = 0.2
—
366
Heat and Mass Transfer
Now
F4_1 x A4 = F31 x A2 + F21 A3 0.1 X (1 X 2) = 0.2 x (1 x 1) + F21 X (1 X 1) F21 = 0.2  0.2 = 0
104.Find F1_2 for configuration of two offset squares of area A as shown in figure
A5
Let
A5 = A2 ÷ A4, A6 = Al ÷ A3 A1 F12 = A6 F6_5  Al F1_4  A3 F32  A3 F3_4 A1 F12 = A3 F34 = A6 F6_5  A3 F32  Al F12 2 A1 F12 = A6 F6_5  A3 F32
Due to symmetry or or
F1_2
Now
=
1
2A1
A
E,
/ 65  A3 F32)
A i = A3 = A and A6 = 2A 1 F12 = 2 (2 x F6_5  F32)
From Hamilton and Morgan graph,
F6 _5 (for =
nd = =0.25ndF32 ( — =1, — =1 = 0.2
)
F12) =0.5(2x0.250.2)=0.15 105.A hollow enclosure is formed between two infinitely long concentric cylinders of radii 1 m and 2 m respectively. Radiative heat exchange takes place between the inner surface of the larger cylinder (surface2) and the outer surface of the smaller cylinder (surface1). The radiating surface are diffuse and the medium in the enclosure is nonparticipating. The fraction of the thermal radiation leaving the larger surface and striking itself is
Surface —1 Surface — 2
Radiation Heat Transfer
(a) 0.25
(b) 0.5
(c) 0.75
(d) 1
367
(GATE — 2008) = 1 =0 = 1 = A2 F21 = (iv • d2 • L) F21 F12 = 2 X F21
F11 ÷ F12 F11 F12 AiFi2 Or • di • L) F12
But Now or or
1 F21 = — X F12 = 0.5
or
2 =1 F22 ÷ F21 F22 ÷ 0.5 =1 F22 = 0.5
Now or or Option (b) is correct.
106. Two parallel plates are at temperature T1 and T2 and have emissivities € 1 = 0.8 and € 2 = 0.5. A radiation shield having the some emissivity E3 on both sides is placed between the plates. Calculate the emissivity E3 of the shield in order to reduce the radiation loss from the system to one tenth of that without the shield. (UPTU — 2009 — 10)
1 F g12 = 1
+
1
—1
=
1 1 0.8 + 05 —1
E2 1 1.25+21 As
1 = 0.444 2.25
1 Qshield = — Q, hence, (Fg12),hield =—1 x Fg12 = 0.0444 10 10 1 (F g 12)shield =
1 + 1 1)±i + 1) E1
E3 J E3 E2 ) 1
0.0444 =
0.8
1 E3
1
1
368
Heat and Mass Transfer
.
= 1.25+22+ or or or
1.25 +
2 E3
=
E3
1.25+
e3
1 = 22.52 0.0444
1 2 = 21.27 e3 = 0.094
E3
107. Two concentric spheres of diameter D1 = 0.8 m and D2 = 1.2 m are separated by an air space and have surface temperatures of T1 = 127°C and T2 = 27°C. Find the net rate of radiation exchange between the spheres if: (i) the surfaces are black, and (ii) the surfaces are diffuse and gray with E1 = 0.5 and E2 = 0.05. (UPTU — 2008 — 9) If surfaces are black, then Fg 12 = F12 = 1 Surface —1
Surface —2
Concentric spheres
If surfaces are gray, then Fg 12 =
—
1 1—E1 1 1—E2 di + + X A2 142 E1 F12 E2 1 105 1 10.05 1.22 ++ x 0.5 1 0.05 0.82 1 1+1+2.138
1 = 0.242 4.138 T1 = 127 + 273 = 400 K and T2 = 27 + 273 = 300 K =
Radiation Heat Transfer
369
A = 4 Kr? = 4 x x x 0.42 = 2.01 m2 Qblack = a A Fg 12 = (T14 — T24) = 5.67 x 104 x 2.01 x 1 x (4004 — 3004) = 5.67 x (256 — 81) x 2.01 = 1994.42 W Quay = 5.67 x 108 x 2.01 x 0.242 x (4004 — 3004) = 482.65 W 108. A cylindrical cavity of diameter 10 cm and depth 20 cm is maintained at 60°C. Find the heat transfer rate from this cavity to atmosphere at 30°C. Assume cavity and atmosphere to be black bodies. (UPTU — 2008 — 9)
d 10 10 = = 4h+ d 4x20+10 90 = 0.111 Q= Al X F1 _ 2 X a X (T14 — T24)
F1_2 =
= rdh+
7rd2 ) x 0.111 x 5.67 x 108 x (3334 — 3034) 4
= n x 0.1x 0.2 +
x (112 x 0.111 x 5.67 x 10_8 (123 — 84.3) x 108 4 = (0.063 + 0.008) x 0.63 x 38.7 = 1.73 W (
Chapter
9
HEAT EXCHANGERS
KEYWORDS AND TOPICS A A A A A A A A A
HEAT EXCHANGER DIRECT CONTACT HEAT EXCHANGER INDIRECT CONTACT HEAT EXCHANGER FOULING PARALLELFLOW COUNTERFLOW CROSSFLOW EVAPORATOR CONDENSER
A A A A A A A A A
DOUBLE PIPE HEAT EXCHANGER SHELL AND TUBE HEAT EXCHANGER OVERALL HEAT TRANSFER COEFFICIENT LOG MEAN TEMPERATURE DIFFERENCE HEAT CAPACITY EFFECTIVENESS NUMBER OF TRANSFER UNITS REGENERATOR CORRECTION FACTOR
INTRODUCTION Heat exchanger is a thermal device which is used for the heat transfer from hot fluid to cold fluid. Temperature of each fluid changes as they pass through a heat exchanger. The heat exchanger can be: (i) direct contact heat exchanger in which heat is exchanged between the fluids by coming into direct contact, (ii) recuperator heat exchanger in which fluids are separated by a wall, and (iii) regenerator heat exchanger in which each fluid is made to flow alternately or periodically. The heat exchange in the heat exchanger may be occurring in the form of latent heat or sensible heat or the combination of both. The heat exchangers are also classified as: (i) parallel flow, (ii) counter flow, and (iii) cross flow depending open the direction of motion of fluids. Heat exchangers are used in many applications such as: (i) boilers, (ii) automobile as radiators, (iii) pasteurising plant as milk chillers, (iv) refrigeration plant as condensers and evaporators, (v) water heaters, (vi) air heaters, and (vii) power plants as condensers. 1. What is a heat exchanger? Heat exchanger is a device which is used in transferring heat from one fluid to another. The objective of a heat exchanger is either to cool or heat a particular fluid by other fluid.
Heat Exchangers
371
Boilers, condensers, cooling towers, radiators and preheaters are common heat exchangers used in industry. Heat exchange may take place in most of applications from hot water to cold water or hot water to cold air or vice versa in a heat exchanger. 2. What do you understand from direct contact heat exchanger? In direct contact heat exchanger, hot and cold fluids are made to come in direct contact so that transfer of heat can take place. Such exchangers have to have fluids in different states such as one is liquid and other is gaseous. Cooling tower of airconditioning plant is commonly used direct contact heat exchanger in which hot water is cooled by surrounding air. Hot water
Pump Cold air
Cold air
Cold water Cooling tower: Direct contact heat exchanger
3. What do you understand by indirect contact heat exchanger? Hot and cold fluids do not come in contact in indirect contact heat exchanger. Hot and cold fluids through the exchanger and heat transfer takes place through the walls separating the flow of the fluids. Such type of indirect contact heat exchangers are called regenerators. There is other type of indirect contact heat exchangers which are called recuperators. In this type hot and cold fluids flow alternatively through the same space. The heat is accumulated during the flow of hot fluid in the walls of the exchanger is transferred to the cold fluid when it is made to flow through the exchanger after the flow of hot fluid. 4. What is fouling? How is it taken care of? (UPTU — 2007 — 8) The walls separating the fluids in a heat exchanger do not remain clean after the exchanger has been in use for some time. The walls become fouled with deposits resulting from impurities in fluids and rust formation. Due to fouling the thermal resistance in the transfer of heat from hot fluid to cold fluid increases and overall heat transfer coefficient of the exchanger decreases. In order to take care of this adverse effect of fouling, an additional thermal resistance called fouling resistance or fouling factor is used while designing the heat exchanger. 5. How can heat exchanger be classified according to flow arrangements of fluids in the heat exchanger?
372
Heat and Mass Transfer
or Show temperature distribution in counterflow and parallelflow heat exchangers. (UPTU — 2002 — 3) The flow arrangement of fluids in the heat exchanger can be— (a) Parallelflow (b) Counterflow (c) Crossflow Parallelflow: In parallelflow heat exchangers, hot fluid and cold fluid flow in the same direction inside the exchanger. The temperature difference of the fluids of the inlet of the exchanger is maximum and temperature difference reduces as fluids flow through the exchanger. The temperature difference is minimum at the outlet at the exchanger. Fluid1 => (hot)
c> Fluid1
Fluid2 => (cold)
1=> Fluid2
Th1
lot fluid The AT2
AT1
C2
Tc1
Cold fluid
Distance from inlet Parallel—flow
Counterflow: In counterflow heat exchanger, hot fluid and cold fluid are made to flow in opposite or counter directions. The inlet of hot fluid and outlet of cold fluid is at one end while the outlet of hot fluid and inlet of cold fluid is at other end of the exchanger. The difference of temperatures of hot and cold fluids remains almost constant throughout the length of the exchanger. Fluid1 E (hot)
Fluid1 (hot)
Fluid2 Cc,
then Cc = C • and Ch = C and effectiveness is 1— exp[—NTU(1+ C c," )1 min
E— c rni
1
+
n
1
Cmax
25. What is capacity ratio? Write the expression of effectiveness of a parallelflow heat exchanger in terms of NTU and capacity ratio. The dimensionless parameter giving the ratio of C • to C is called the capacity ratio. C • Capacity ratio = R = mm max
Now the effectiveness of a parallelflow heat exchanger is 1— eN(1+ R) E =
where
N=
1 +R
NTU and
R=
capacity ratio.
26. Find the expression for effectiveness of counterflow heat exchanger. The variation of temperature is4
Th —4—
z
ATi T C2 Th2 AT2 Tc1
dTc x —A dx Counterflow
Area —0
Heat Exchangers
dQ = u • dA(Th — Tc) dQ = —thhchdTh = me cc dTc = —Ch dTh = —Cc dTc —dQ —dQ dTh = and dT = G c Ch
Also
or
d(Th — Tc) = —dQ
1 (1 Ch Cc
= dQV — Now putting the value of dQ— d(Th — Tc) = u 2 j,„, akih — ic)
or
Th—Tc
or
log
Ti.,z —
X
dA x ( 1 1 Ch Cc
= u (1
1
(Th — Tc)
X dA "
1 Tc1 =uxAx ( cc
1
)
Ch
Tit, — Tr z
Do observe that AT2 is now Th2 — Tci and ATI = Th1 — Tc2 as flow is counter or
Tk — Tc1 = exp [uA (— ci — c— l, Th, — Tc2 c h Now effectiveness is— Ch (Th, — Cc (Tc2 E=
Th2
Cmin
= Thi
Tc2 = Tc1
— Tc,)
Cmin (4, — Tc1 )
E Cmin (Thi Tc1 )
Ch E Cmin (Thi
Cc
Substituting the values of Th2 and Tc2 Cmin ( Th, — Tc, T, Ch \ —Exp[uAH —  )1 E Cmin(Th, — Tc,)] Cc Ch Thi —[ Tc,+ Cc
[ Th,—
or
E
—
(T — Tc,) i
1
1
E X Cmin
E
Ch Cmin
Cc
= Exp [uAH Cc
Ch
)1
387
388
Heat and Mass Transfer
EX
1
or 1
1. Case 1: In case Ch
>
C
•
Ch E Galin
=Exp
[uA ri
Cc
Cc, then C, = C
cc,
_
cc )] Ch )
uA and Ch = C and „  NTU = N, and Cmin
• R = Cmin  heat capacity. The equation (i) becomesmax 1— E R  e N(1 R) 1— E 1 e N(1R) or E = R eN(1R) N(1R) _ 1 e N(1R) e R
2. Case 2: In case Cc >
Ch,
then
1 E X 1
Ch =
Cmin and C, = Cm. Equation (i) becomes
Cmin
Cmin
= Exp [N (1 Cmin j] Cmax
E Cmin
Cmax
or
E =
N(1R) _ 1 e N(1R) R e
27. What is the value of effectiveness of a condenser? In a condenser, the hot fluid condenses at constant temperature. Hence, we have— = oc = C max Ch = = Q Thi Th, 0
as
Thi = Th2 = condensing temperature. Capacity ratio R = Cmin
Cmax
Cmin = 0 oc
Th1
Condenser
Now Putting
1e N(l+R) for parallelflow 1+ R R = 0, E = 1 — e —N
E=
(i)
Heat Exchangers
e
Similarly,
E=
Putting
R = 0,
e
389
N (1— R)
R for counterflow
N (1— R)
E=1
— e —N
Hence, for both types of flow, we have effectiveness of a condenser as— E=1—
28. What is the value of effectiveness of an evaporator? In an evaporator, cold fluid evaporates at constant temperature. Thi
Th2
Tci
T 2
Evaporator
Q =cc(Tc2—Tc1) or
=C
C — c = Tc,—Tc, 0 R
as
Tc2 = Tcl
= Cinin = Cinin = 0 oc
E=1
— e —Isi
both for parallel and counterflow.
29. What is the value of effectiveness for a regenerator? In case of regenerator, C • = Cm
R = Cmin — 1
C„„,, Hence, for parallelflow regenerator1 — e 2N E= 2 For counterflow regenerator if we put R = 1 E=
1— e — N(11) 1—1xe
11 1—1
—N(11)
0
or it is indeterminate
To find the value of effectiveness, L. Hospital's rule has to be used for regenerator when
R —> 1 E = Limit R —> 1
aR
a aR
[i—eNGR1
[1—
Re(1R)] N
390 Heat and Mass Transfer NeN(1R) ReN(1R) x (—N)
= —N —1—N 1+N
30. How can effectiveness of heat exchangers be found out by graphical method? The effectiveness of heat exchangers can be found out from the graphs prepared by Kays and London. The graphs are plotted for NTUm versus effectiveness (E) for different heat capacities (R). Following graphs are included— Effectiveness for parallelflow heat exchanger Effectiveness for counterflow heat exchanger Effectiveness for 1 — 2 parallel counterflow heat exchanger Effectiveness for two shell passes and 4, 8, 12 tube passes Effectiveness for crossflow heat exchanger with both fluids unmixed Hot fluid 0 Cold fluid
100
1 80 Effectiveness
(a) (b) (c) (d) (e)
60
40
20
0 0
2
3
NTU [ UA ] cmm Parallelflow heat exchanger
4
5
Heat Exchangers Hot fluid Cold fluid ..
100
.
cii,inicmax 7 80
. 0.25 0.50 0.75
Effectiveness
1.00
60
40
20
1
2
3 NTU Counterflow heat exchanger
F
5
Shell fluid Tube fluid
One shell one shell pass 2, 4, 6 etc, tube passes
2 NTU
3
4
5
One shell and 2, 4, 6 etc, two passes exchanger
39 I
392
Heat and Mass Transfer
100 amidCmax = 0
!2
...........
80
1
60
Shell fluid Tube fluid
a)
I I
40 w
Ii
2 Shells 20
Two shell passes 4, 8, 12, etc., tube passes
1
2
3
4
5
NTU Two shell passes and 4, 8, 12, etc., tube passes heat exchanger
100
80 Cold fluid
Hot fluid 40 w
20
1
2
3
4
NTU Crossflow heat exchanger
5
Heat Exchangers
393
31. A counterflow heat exchanger is used to heat water from 20°C to 80°C by using hot exhaust gas entering at 140°C and leaving at 80°C. The log mean temperature difference for the heat exchanger is— (a) 80°C (c) 110°C
(b) 60°C (d) not determinable as zero/zero is involved (IES 96)
Temperature distribution is
80 AT2 20
AT1 = Thi — Tc2 = 140 — 80 = 60° A T2 = Th2 — Tc1 = 80 — 20 = 60
ATff, = LMTD =
ATI — AT2 AT log A 1 L1T2
60 — 60 0 0 log 60
Indeterminate and hence—
60
LMTD =
ATI + AT2 60 + 60 2 2
60
Option (b) is correct. 32. A counterflow shell and tube exchanger is used to heat water with hot exhaust gases. The water (c = 41 80J/kg°C flows at a rate of 2 kg/s while the exhaust gas (1030 J/ kg°C) flows at the rate of 5.25 kg/s. If the heat transfer surface area is 32.5 m2 and the overall heat transfer coefficient is 200 W/m2°C, what is the NTU for the heat exchanger? (a) 1.2 (c) 4.5
(b) 2.4 (d) 8.6 (IES 95)
NTU =
Cmin
thw X Cwater = 2 x 4180 = 8360 .1/°C Ch = rith x ch = 5.25 x 1030 = 5407.5 J/°C Ch < Cc Cmin = Ch = 54075 J/°C Cc =
Now
ux A
394
Heat and Mass Transfer
NTU =
200 x 32.5 = 1.2 5407.5
Hence, option (a) is correct. 33. Water at the rate of 1 kg/s is heated from 35°C to 75°C by oil in a double pipe counterflow heat exchanger. Oil has a specific heat of 1.9 kJ/kgk and it enters the heat exchanger at 110°C and 2.5 kg/s. The overall heat exchanger coefficient is 300 W/m2K. Calculate the area of heat exchanger tube. (UPTU — 2002 — 3) Temperature distributions —4—
Th1
AEI 7,2
The
AT2
Tci
Applying energy balance— Heat lost by hot oil = Heat gained by cold water inhCh(Thi — Th2) = me cc(Tc2 — Tc1) 2.5 x 1.9 (110 — Th2) = 1 x 4.8 (75 — 35) Th2 = 74.8°C Now A Ti = Thi — Tc2 = 110 — 75 = 35 and A T2 = Th2 — Tc1 = 748 — 35 = 39.8
Now Now or
2 — ATI _ 398.35 LMTD =Tm = AT AT2 — log 39.8 log 35 ATI = 37.4°C Q = 1 x 4.2 x (75 — 35) = 168 ld Q =uxAxAT,„
A—
u x AT„,
168x103 300 x 37.4 = 14.97 m2 34. In a counterflow heat exchanger, water is used as a coolant for hot oil from a temperature of 160° to 60°C. Water enters the heat exchanger at a temperature of 25°C. The mass flow rates of water and oil are each 2 kg/s. The diameter of the tube
Heat Exchangers
395
is 0.5 m and the overall heat transfer coefficient is 250 W/m2K. Calculate the length of the heat exchanger. Take specific heat for oil and water as 2.035 and 4.187 14/kgk. (UPTU — 2003 — 4) Temperature distribution is— rhi ATI
T
c2
2
AT2
ci t
Counterflow
First is to find the outlet temperature of the water Heat lost by oil = Heat gained by water 2 x 2.035 x (160 — 60) = 2 x 4.187 x (Tc2 — 25) Tc2 = 73.6°C AT, = Thi — Tc2 = 160 — 73.6 = 86.4°C AT2 = 60 — 25 = 35°C Now
Now
ATff, = LMTD —
AT, — AT2 AT log AT2
86.4 — 35 log 86.4 35 = 57°C Q= u • A • AT„, = u • Or • d • 1) • AT. _
u•ir • d •
AT„,
2 x 2.035 x 103 x (160 — 60) 250x x 0.5 x 57 = 18.2 m 35. 1000 kg/hr of oil (C,, = 2.09 kJ/kg°C) is to be cooled from 80°C to 400°C in oil cooler by using water flow of 1000 kg/hr at 30°C. Give your choice for a parallelflow or counterflow heat exchanger, if the overall heat transfer coefficient is 24 W/m2C. For water Cp = 4.18 kJ/kg°C. (UPTU — 2004 — 5)
396
Heat and Mass Transfer
First to find out outlet temperature of the water Heat lost by oil = Heat gained by water 1000 x 2.09 x (80 — 40) = 1000 3600 x 4.18 x (Tc2 — 30) 3600 Tc2 = 50°C 80
—1— 80 ATI 50
40
T 5°
AT2
40
30
I
30 Parallelflow
Counterflow
Since outlet temperature of cold fluid is higher that outlet temperature of the oil, hence parallel flow is impossible. Therefore counterflow cooling is possible ATI = 80 — 50 = 30 AT2 = 40 — 30 = 10 LMTD = ATff, =
ATI — AT2 AT log ' AT2
30 —10 log
10 = 18.2°C 1000 3600 x 2.09 x 103(80 — 40)
Now
Q=
Also
Q = u • A • AT. A—
u • AT„, 10
00 x 2.09 x 103 x 40 _ 3600 24 x 18.2 = 53.2 m2 36. A double pipe counterflow heat exchanger is used to cool 10,000 kg/h of an oil (C,, = 2095 Joules/kg k) from 80 to 50°C using 8000 kg/h water (Cp = 4180 J/kg k) entering at 25°C. Find the heat exchanger area required with an overall heat transfer coefficient is 300 W/m2. (UPTU — 2005 — 6)
Heat
Exchangers
397
First to find outlet temperature of water. Heat lost by oil = Heat gained by water 8000 10,000 x 2095 x (80 — 50) = 360o x 4180 x (Tc2 — 25) 3600 . Tc2 = 43.8°C k 80 AT1 1 43.8
50 AT2
25 Counterflow
Now
ATi = 80 — 43.8 = 36.2°C AT2 = 50 — 25 = 25°C ATI — AT2 LMTD = ATff, = AT log 1 AT2
Now
= 36.2 — 25 log36.2 25 = 30.3°C Q =uxAxAT,n
or
A—
Q
ux AT,n
10,000 x 2095 x (80 — 50) 3600 300 x 30.3 = 19.2 m2 37. Water is to be heated from 20°C to 90°C by using a single pass heat exchanger. For this purpose, saturated steam at 120°C is condensing on the outer tube surface of a single pass heat exchanger. The heat transfer coefficient is 1800 W/m2K. Calculate the surface area of a heat exchanger capable of heating 1000 kg/h of water from 20°C to 90°C. Also calculate the rate of condensation of steam. Take hfg = 2200 kJ/kg. Guidance: The temperature of steam (hot fluid) in the heat exchanger remains constant at 120°C and heat is passed to cold fluid i.e., water by condensing steam (i.e. latent heat of evaporation.)
398
Heat and Mass Transfer Hot fluid 120°C
120°C — AT2 90°C —
ATI
Cold fluid J— 20°C
Parallel flow
Now
AT1 = 120 — 20 = 100°C AT2 = 120 — 90 = 30°C LMTD =
—
ATI — AT2 ATI log A2
Now
100 — 30 log 100 30 = 50.14°C Q =uxAxAT,„
or
A=
u • AT„,
1000 x 4180 x (190 — 20) _ 3600 1800 x 50.14 = 0.776 m2 Heat is lost by steam by the way of condensing thereby losing heat of evaporation (hfg). If m, is steam which has condensed, then 1000 x 4180 x (90 — 20) m5 = Q = 3600 2200 hfr = 37 x 103 kg/s 38. In a certain double pipe heat exchanger, hot water flows at a rate of 5000 kg/hr and gets cooled from 96 to 65°C. At the same time 50,000 kg/hr of cooling water at 30°C enters the heat exchanger. The flow conditions are such that the overall heat transfer coefficient remains constant at 2270 W/m2K. Determine the heat transfer area required and the effectiveness, assuming two streams are in parallel flow. Assume for the both specific heat = 4.2 kJ/Kg k. (GATE — 97) First to find outlet temperature of cold fluid.
Heat Exchangers
399
Heat lost in cooling hot fluid = Heat gained in cold water 0 x 4.3 x (95 — 65) — 50,000 x 4.2 x (Tc2 — 30) 3600 3600 or
Tc2 = 3° + 10° = 33°C
65
TAT2 33 Parallelflow
6,T1 = 95 — 30 = 65 6,T2 = 65 — 33 = 32 65 — 32 LMTD = 6,7,„ = 65 = 46.6°C log 32
Now
5000 x 4.2 x (95 — 65) = 175 x 103 W Q 3600 Q= u • A • AT,„
or
A—
Q
u • AT„,
175 x 103 2270 x 46.6 = 1.65 m2 —
Now
Q. will be when hot fluid is cooled to the inlet temperature of cold fluid Q. = mh Ch(Thl — Tc1) =
3600
x 4.3 x 103(95 — 33)
= 361.7 x 103 W Qaman 175 x 103 E= 361.7 x 103 Qmax = 0.48 39. A very long, concentric tube heat exchanger is having hot and cold water inlet temperature of 85°C and 150°C. The flow rate of the hot water is twice than that of the cold water. Assuming equivalent hot and cold water specific heats, determine the
400
Heat and Mass Transfer
water outlet temperatures and the heat exchanger effectiveness for the following modes of operation: (i) counterflow, and (ii) parallelflow. (UPTU — 2007 — 8) Guidance: The heat exchanger is very long and two situations arise depending upon type of flow. In case of parallelflow, both liquids will come out at the same temperature `T'. In case of counterflow, the fluid having low heat capacity will have larger temperature difference from inlet to outlet. Since cold water in this problem has lower heat capacity, hence it will have larger temperature difference, resulting cold water to attain the inlet temperature of hot fluid. 85°C
85
rc
2
15°C Parallelflow
Counterflow
1. Case 1: Parallel flow
heat lost = heat gained m h x ch x (Th i = rhc cc (T Tc1) 2 x cw (85 — 7) = cw(T — 15) 3 T = 185 T = 61.67°C 2 Cc (85 — 61.67) Q E= G(85 —15) Qmax 2 x 23.33 = 0.67 70 2. Case 2: Counterflow heat lost = heat gained mh • Ch (1' hi — Th2) = me Cc(Tc2 Tc1) 2 cw(85 — Th2) = cw(85 — 15) or The = 85 — 35 = 50°C Q 2 Cc (25 — 50) 2 x 35 E= (85 —15) 70 G Qma. =1 40. If heat capacities of hot and cold fluids are equal (thhch = cc) in a counterflow heat exchanger, show that ATI. = AT2 = AT at any section and temperature profiles of the fluids are linear and parallel. (GATE — 96)
Heat Exchangers
40 I
The temperature profile of a counterflow heat exchanger is—
H—
x —01
dx H Counterflow
For differential element of heat exchanger, heat flow is— dQ = —MhChdTh= —Mc Cc dTc = u x dA x AT
But
thhCh = thcCc dTh = dTc
.%
or or . Now
d(Th — Tc) = 0 Th — Tc = constant
ATi = AT2 = AT = constant —mh x ch x dTh = u x dA x AT
or Similarly,
dTh —u AT = = constant dA nth ch dA dA
As ATi = AT2 = AT and
dT dA
=
—u AT thc cc
= constant
= constant for fluids temperature profiles, hence these lines
are linear and parallel. 41. If ATI. = AT2 in a counterflow heat exchanger, show AT„, = ATI. = AT2. i — AT2 AT. = LMTD — AT log
Assume
Now as
ATI
AT2
ATi = c i.e. a constant temperature, ATi — AT2 = x AT2 = c — x i.e. it is variable c —(c — x) x AT. = = log c log c c—x c—x ATi —> AT2, x —> 0
402
Heat and Mass Transfer
limit (ATm) = limit x—>o x—>° log
ATff, =
= C
0 Indeterminate and using L' Hospital rules we have— 0
c—x a (x) ax
lixn2>icl ax
(log
c c—x 1
= limit [ x —>0 C —
x
c
c
x [(c — x)2
= limit [c — x] c x30 ATm = ATI = AT2 42. In a certain exchanger, both the fluids have identical mass flow rate and specific heat product. The hot fluid enters at 76°C and leaves 47°C and cold fluid enters at 26°C and leaves at 55°C. The effectiveness of heat exchanger is— (a) 0.16 (c) 0.72
(b) 0.58 (d) 1.0 (GATE — 1997)
Temperature distribution is76°C 55°C
47°C 26°C
Counterflow
Heat transferred Q = rhh x ch(Thi — Th2) = mh x ch(76 — 47°) Qmax = Cmin x (Thi — Tc1) = Cmin(76 — 26) Cmin = mh ch
= the cc
_ rithch(76 47) nth ch (76 — 26) = Qmax Q
0 58 — —* — 50 — Option (b) is correct.
Heat Exchangers
403
43. A hot fluid at 200°C enters a heat exchanger at a mass flow rate of 104 kg/hr. Its specific heat is 2000J/kg k. It is to be cooled by another fluid entering at 25°C with a mass flow rate 2500 kg/hr and specific heat 400 J/kg k. The overall heat transfer coefficient based on outside area of 20 m2 is 250 W/m2K. Find the exit temperature of the hot fluid when the fluids are in parallelflow. (GATE — 1999) Temperature distribution is200°C
2
25°C
Guidance: As only inlet temperatures are given, NTU method is to be used. 104 = mhCh = 3600 x 2000 = 5.55 kJ Ch 2500 C = thccc = x 400 = 0.278 kJ 3600 C • = Cc = 0.278 kJ C = Ch = 5.55 kJ uA 250 x 20 N =NTU = — 17.98 Cmin 0.278 x 103 Cmin _ 0.278 — 0.05 555 C In parallelflow, effectiveness is— 17 98(1+0.05) 1— eN (1+ R) 1—e ' R
—
E=
E=
1+ R 1 — e1838 1.05
1 — 6.32 x 109 1.05 = 0.952 Ch(Th, Th2) E= Cmin (Th, Tcd 0.952 =
555 (200 — 0.278 (200 — 25)
1 + 0.05
404
Heat and Mass Transfer
Th2
= 200 0.952 x 175 x 2.278 555 = 200 — 8.35 = 191.65°C
44. In a certain heat exchanger, both the fluids have identical mass flow ratespecific heat product. The hot fluid enters at 76°C and leaves at 47°C and the cold fluid entering at 26°C and leaves at 55°C. The effectiveness of the heat exchanger is— (a) 0.16 (c) 0.72
(b) 0.58 (d) 1.0 Ch = Cc = Cmin C h (Thi — Th2 ) E
E=
Cmin (Th, — =
(T — Th2 )
(Th, —
76 — 47 — 76 — 26
58
50 0.
Option (b) is correct. 45. In a condenser, water enters at 30°C and flows at the rate 1500 kg/hr. The condensing steam is at a temperature of 120°C and cooling water leaves the condenser at 80°C. Specific heat of water is 4.187 kJ/kg k. If the overall heat transfer coefficient is 2000 W/m2K, the heat transfer area is— (a) 0.707 m2 (c) 70.7 m2
(b) 7.07 m2 (d) 141.4 m2 (GATE — 2004)
Temperature distribution is120
120°C
FT2 80
ATI 30
Parallelflow
ATI = 120 — 30 = 90°C AT2 = 120 — 80 = 40°C LMTD = Tm =
ATI — AT2 90 — 40 ATI 90 log 47) log A L1T2
Heat Exchangers
405
= 50 = 61.73 0.81 1500 x 4187 x 50 Q = me x c, x (80 — 30) — 3600 = 87229 W Now
Q =uxAxAT,„
or
A=
u x ATni
87229 2000 x 61.73 = 0.707 m2 Option (a) is correct. 46. In a counterflow heat exchanger for the hot fluid, the heat capacity = 214/kg k, mass flow rate = 5 kg/s, inlet temperature = 150°C, outlet temperature = 100°C. For the cold fluid, heat capacity = 4 kJ/kg k, mass flow rate = 10 kg/s, inlet temperature = 20°C. Neglecting heat transfer to the surroundings, the outlet temperature of cold fluid is— (a) 7.5 (c) 45.5
(b) 32.5 (d) 70 (GATE — 2003)
Temperature distribution is150 100°C 200°C
Counterflow
or
Heat lost by hot fluid = Heat gained by cold fluid 5 x 2 x (150 — 100) = 4 x 10 x (T,2 — 20) 5x 2x 50 Tc 2 — 20 = 4 x 10 = 12.5°C Tc2 = 20 + 12.5 = 32.5°C
Option (b) is correct. 47. Air enters a counterflow heat exchanger at 70°C and leaves at 40°C. Water enters at 30°C and leaves at 50°C. The LMTD in deg (°C) is—
406
Heat and Mass Transfer
(a) 5.65 (c) 19.52
(b) 14.43 (c) 20.17 (GATE — 2000)
Temperature distribution is1 70°C AEI
T 50°C
40°C I 472 30°C
T
Counterflow
ATI = 70 — 50 = 20°C AT2 = 40 — 30 = 10°C LMTD =
ATI — AT2 20 — 10 = ATI l0 20 log A 7, 10 CAA 2 10 = 14.43 0.693
Option (b) is correct. 48. Two fluids A and B exchange heat in a countercurrent heat exchange. Fluid A enters at 420°C and has a mass flow rate of 1 kg/s. Fluid B enters at 20°C and also has a mass flow rate of 1 kg/s. Effectiveness of heat exchanger is 75%. Determine the heat transfer rate and exit temperature of fluid B (specific heat of fluid A is 1 kJ/kg k and that of fluid B is 4 kJ/kg k). (Gate — 99) Guidamce: As only inlet temperatures of both fluid have been given, we have to use NTU method. Ch = thh ch = 1 x 1 = 1 kJ Cc = cc = 1 x 4 = 4 kJ • = Cam, = 1 Cc = Cmax = 4 E 0.75 =
Cc (Tc, — Cmin (4, — 4 x (Tc, — 20) 400
Tee = 20 +
400 4 x 0.75
4 X (Tc,— 20) 1 X (420 — 20)
Heat Exchangers
407
= 20 + 75 = 95°C Q= Mc X Cc X (95 — 20) = 1 x 4 x 75 = 300 kW 49. A heat exchanger is designed to cool hot water from 83°C to 45°C with flow rate of 5 kg/mm. Cold fluid is flowing at 9 kg/min has inlet temperature of 25°C and specific heat of cold liquid is 2.85 kJ/kg k. Specific heat of hot water is 4.18 kJ/kg k. Explain the type of heat exchanger required for the above purpose. Heat lost by hot fluid = Heat gained by cold fluid rhh ch(Thi — Th2) = rhc cc(Tc2 — Tc1)
60
x 4.18(83 — 45) = 60 9 x 2.85(Tc2 — 25)
.
Tc2 = 56°C
Since Tc2 (= 56°C) is higher than Th2(= 45), only counterflow arrangement is possible. 85
85 56° 45
56
45
25
25
Parallel flow
Counterflow
50. A heat exchanger is used to cool engine oil from 60°C to 40°C by cold liquid at inlet temperature of 10°C and outlet temperature of 30°C. The total heat is 100 kW and overall heat transfer coefficient based on outer surface area of tube is 75 W/m2C. Estimate the heat transfer area for single pass parallel flow and counter flow heat exchanger.
T
60°C
60 40A—c2 3
AT1
40°
0 OW
10°
1 10°C Parallelflow
Counterflow
1. Parallelflow ATi = 60 — 10 = 50°C AT2 = 40 — 30 = 10°C
408
Heat and Mass Transfer
I AT,n = LMTD = AT — AT2 AT2 log AT2 = 24.9°C Q = uA AT,n Q 100 x 103 A = x 24.9 — 53.6 m2 uAT,n 75
or 2. Counterflow
As
ATi = 60 — 30 = 30 AT2 = 40 — 10 = 30 ATi = AT2, them I + AT2 AT,r, = AT 2 30 + 30 = 30 2 A=
Q 100 x 103 u • AT„, — 75 x 30
= 44.44 m2 Note: Area for counterflow is less. Hence, counterflow is more effective. 51. A counterflow heat exchanger having heat transfer area 10 m2 is used to cool engine oil from 100°C by water available at 20°C. The mass flow rate of oil and water is 2.4 kg/s and 1 kg/s. The overall heat transfer coefficient is 300 W/m2°C. Calculate the exit temperature of water and total heat transfer rate. Specific heat of oil and water is 2050 J/kgk and 4180 J/kgk. Guidance: Since inlet temperatures of hot and cold fluids are given, we have to use NTU method C, = inc cc = 1 x 4180 = 1480 W/°C Ch = 'hitch = 2.4 x 2050 = 4920 W/°C Cc < Ch Cc = Gain Au 10 x 300 NTU = = 0.72 4180 Cmin Capacity ratio R = Cmin _ 4180 Cma, — 4920 = 0.85 Effectiveness for counterflow is1 — e N (1—R) E= 1— R e —N(1—R)
Heat Exchangers
409
1 e032(10.85) 1—
0.85 x e —032(10.85) 1_e0.108
1— 0.85 x
10.897 0.763
e0108 — 1—
0.103 — 0.43 0.237 Q = E Cmin(Th — Tc1 )
= 0.43 x 4180 x (100 — 20) = 143.8 kW Exit temperature of water =
Tc2
Tc2 = Tci ER(Th1 Tc1)
= 20 + 0.43 x 0.85 (100 — 20) = 20 + 29.24 = 49.24°C. 52. In a tubular counterflow heat exchanger, 1080 kg/hr of water is heated from 40°C to 80°C by hot flue gas (sp. heat = 10 kJ/kg k). Flue gas enters at 200°C and leaves at 100°C. The overall heat transfer coefficient is 200 W/m2K. Find the area of exchanger: (i) by LMTDmethod, and (ii) NTUmethod. Case 1: LMTD method 4 200°C
AT1
180°C
100°C I AT2 40°C T Counterflow
ATi = 200 — 80 = 120°C AT2 = 100 — 40 = 60°C OTm
—
AT2 — AT2 120 — 60 AT to 120 log 1 60 AT2
= 86.6°C Q = A IcCc(Tc2 — Ti) = 1:0 x 1 x 103 x (80 — 40) = 50.2 kW
4I0
Heat and Mass Transfer
Q
=tixAx 6T,„
A =
50.2% x 103 2 200 x 86.6 — 2.9 m
Case 2: NTU method.
Now
C = 1080 x 4.18 x 103 = 1255 W/°C c 3600 Ch x (200 — 100) = Cc x (80 — 40) Ch = Cc X 40 = 0.4 x Cc 100 = 502.0 W/°C G > Ch C•• = Ch = 502 E=
Q Qmax
Thi )
Crain (Thi—
= 0.625 R = capacity ratio = 502 = 0.4 1255 For counterflow, effectiveness is E
0.625 =
or or
Now
1— eN(1R) 1 — ReN(1R) 1 _ eN(10.4) N 1— 0.4e (10.4)
1 _ e0.6N 10.4eo.6 N 0.625 — 0.25 e0.6N = 1 _ e0.6N 0.75 x e"N = 0.375 e+0.6N = 0.75 = 2 0.375 0.6 N= log 2.0 = 0.693 N 0.693 0.6 = 1.155 u N —A Crain
Crain
cinax
200 —100 200 — 40
Heat Exchangers
or
A—
4II
N x Cmin
1.155 x 502 200 = 2.9 m2 53. A tube type exchanger is used to cool hot water from 80°C to 60°C. The task is accomplished by transferring heat to cold water that enters the heat exchanger at 20°C and leaves at 40°C. Should this exchanger operate under counterflow or parallelflow conditions? Also determine the exit temperatures if the flow rates of fluids are doubled. Guidance: The outlet temperature of cold fluid is less than the outlet temperature of hot fluid. Hence, exchanger operates as parallelflow heat exchanger. Ch = thh Ch and C, = th c, Heat lost = Heat gain thh ch (80 — 60) = th • c,(40 — 20) Ch = Cc = C • = Cmaac C Capacity ratio R = c,mm — 1 1..max Thz 80 — 60 = 1/3 Thl —Tc, 80 — 20 For parallelflow, effectiveness is— 1+ eN(l+R) 1— e2N E= 1+R 2 E—
or or
1 x 2 = 1 — e2N 3 e2N = 1 — 0.67 = 0.33 N = 0.55
Now the flow rate is doubled which will change the value of NTU without changing R. Cmin)new = = 2x th h C Now
(NTU)new
x C•
u uA — 2 X Cmin (cminInewh 1 1 =— x NTU = — N 2 2 1 = x 0.55 = 0.275 2
4I2
Heat and Mass Transfer
(E )new =
1 — e
N(1FR)
1 e0.275X2
1+ R — 1+1 E new = 0.211 (Th)new = Th1 E (Th1 Tc1) = 80 — 0.211 (80 — 20) = 67.3°C (Tc2)new = Tc1 + E (Th1 Tc1) = 20 + 12.67 = 32.67°C 54. A tubular heat exchanger consists of 195 tubes each of 2.2 cm outer diameter and 5 m length. The flow of fluids are in counterflow. If overall heat transfer coefficient is 325 W/m2K, find outlet temperatures of the fluid and total heat transfer rate. Given inlet temperatures of hot and cold fluids are 125 and 22°C and mass rate of hot and cold fluids are 21 kg/s and 5 kg/s. Also ch = 2100 J/kg k and cc = 4100 J/kg k.
Now
Ch = x ch = 21 x 2100 = 44.1 x 103 J/kg°C Cc =mc x cc = 5 x 4100 = 20.5 x 103 J/kg °C Cc < Ch C • = Cc = 20.5 X 103 20.5 = 0.47 44.1
R = Crain C NTU = N =
uA Crain
325 x x 2.2 x 102 x 5 x 195) 20500
N = 1.07 For counterflow, effectiveness is— E=
1— e
N(1R)
1— ReN(1R) 1.07(10.47) 1— C 0.47e1.07(10.47) 1—
= 0.59 Tc1 = Tc1 +
E X Crain (Th1 Tc1)
Crain 0.59 x (125 — 22) = 22 + 1 = 22 + 60.2 = 82.2°C
Th2 = Th — E
Crain
(Th i — Tci)
Cmax
= 125 — 0.5 x 0.47 (125 — 22)
Heat Exchangers
4I 3
= 96.7°C Q = E Cmin(Thi Tc1)
= 0.59 x 20500 (125 — 22) = 1245 kW 55. A heat exchanger cools a liquid flowing at 8 kg/s from 100 to 50°C. Specific heat is 3850 K/kg°C. The cooling water has flow rate 10 kg/sec in tube side and entry temperature of 10°C. Overall heat transfer coefficient is 400 W/m2C. Find the heat transfer area: (i) for parallelflow, one shell and tube pass, (ii) for counterflow, one shell and tube pass, (iii) for one shell pass and two tube pass, (iv) for crossflow, both fluids unmixed. Heat lost = Heat gained rhh ch(Thi — Th2) = rhc cc(Tc2 Tc1) 8 x 3850 (100 — 50) = 10 x 4180(Tc2 — 20) Tc2 = 46.8°C Q = thhch(Thi — Th2) = 8 x 3850 x 50 = 1540 kW
1. Case 1: Parallelflow
50°C I AT2 46.8°C
ATI = 100 — 10 = 90°C, ATI = 50 — 46.8 = 3.2 ATni = LMTD =
AT2
—
log A L1T2 A=
u • AT„,
=
90 —3.2 90 = 25.7°C log
1540 x 103 420 x 25.7
= 150 m2 2. Case 2: Counterflow —4— Ari
100°C 46.8°C
50°C I AT2 10°C
1
4I4
Heat and Mass Transfer
ATI = 100 — 46.8 = 53.2°C A T2 = 50 — 10 = 40°C AT 53.2 — 40 n., = = 46.3°C log ziic A=
Q 1540 x 103 u • ATni — 420 x 46.3
= 83.2 m2 3. Case 3: One shell pass and two tube pass. Here tube side temperatures are Tti = 10°C,
Tt2 =
46.8°C
Shell side temperatures are— Tsi = 100 & Ty2 = 50°C 12 — Th _ 46.810 — P—T 0 41 Ts, — Tt ,  100 —10 — . 100 —50 s,— Ts2 R=T = 46.810 = 1.36 Tt2  Ttl For P = 0.41 and R = 1.36, from graph correction factor (F) is— F = 0.88 A Tff, for counterflow = 46.3°C (as found out above) .. Corrected mean temperature for the exchanger is— A Tm = F X (A Tm)counter flow = 0.88 x 46.3 = 40.7°C A=
Q
u x ATm = 94.6 m2
=
1540 x 103 400 x 40.7
4. Case 4: For crossflow, both fluids unmixed. Correction factor for P = 0.41 and R = 1.36. The correction factor is— F = 0.9 Hence, corrected mean temperature is— ATm
ATwcross = (.1 flow X 0.9 = 46.3 x 0.9 = 41.6°C
Heat Exchangers
4I 5
1540 x 103 u x AT„, — 1500 x 41.6 = 92.5 m2
A=
56. In a parallel flow heat exchanger, oil is cooled by water from 135°C to 65°C. The cold water has inlet and outlet temperature as 20°C and 50°C. Find: (i) exit temperature of fluds for counterflow, (ii) minimum outlet temperature of oil by increasing tube length in parallelflow. Heat lost = Heat gained thhch(Thi — Th2) = thccc(Tc2 — Tc1) thhCh(135 — 65) = thccc(Tc2 — Tc1) thhCh x 70 = rilc cc x 30 Hence,
Ch < Cc
R= C = 3° = 0.43 C 70 The effectiveness of parallelflow heat exchanger is E=
But
E=
0.61 =
13565
— 0.61
135 — 20 1— e —N(l+R) 1+ R 1 _ e—N(1+1)
1+1 N = 1.43
Or
Case 1: Counterflow E counter =
Now
1 e N (1— R) N (1— R) 1— R
1 — e1.43(10.43) 1.43(10.43) 1 —1.43 x e= 0.69 Th2 = Thi — E (Th1 — Tc1) = 135 — 0.69 (135 — 20) = 55.9°C Tc2 = Tc1 F E R(Thi — Tc1)
= 20 + 0.69 x 0.43(135 — 20) = 54°C
4I6
Heat and Mass Transfer
Case 2: When heat exchanger is too long in parallelflow, the outlet temperature of both fluids is same Th1
Ch x (Th1 
= Cc(T  Tc1)
T  Tci
= Cc (Thi 
= 0.43 (Th1  T) 0.43 x Th1 + TC , T 1.43 T = 54.5°C
0.43 x135+ 20 1.43
57. Calculate the overall heat transfer coefficient (u0) based on outer surface of a pipe with outer diameter of 4 cm and inner diameter of 3 cm. Thermal conductivity (K) is 53 W/m°C. Heat transfer coefficient hi and 1/0 are 2000 W/m2°C and 1000 W/m2°C. Fouling factors are Fi = 14 x 105 m2°C/W and F0 = 15 x 105 m2. ovv. Overall heat transfer coefficient isuo =
1 d1 o do do 1 do = — +—log—+—+=x+F F o d ( hi ) 2k di ho I
1 4 ( 1 + 0.04 log 4 + 1 + 4 x 14 x 105 +15 x105 3 2000 2 x 53 3 1000 3 = 459 W/m2°C 58. Water at the rate of 3.783 kg/s is heated from 37.78 to 54.36°C in a shell and tube heat exchanger. On the shell side one pass is used with water as heating fluid, 1.892 kg/s entering the heat exchanger at 93.33°C. The overall heat transfer coefficient is 1419 W/m2K and the average water velocity in 1.905 cm diameter tubes is 0.366 m/s. Because of space limitations the tube length must not be longer than 2.438 m. Calculate the number of tube passes, the number of tube per pass and length of tubes. (UPTU 2006  07) Guidance: The surface for heat transfer is the product of perimeter, length of tube, number of tubes per pass and number of passes. Given:
Tci = 37.78,
Tc2 =
54.36°C, rh = 3.783 kg/s
Heat Exchangers
4I7
Thi = 93.33°C, rith = 1.892 kg/s, u = 1419 Wm2K d = 1.905 cm, V = 0.366 m/s, L = 2.438 m Heat gained = Heat lost inc x cc x (Tc2  Tc1) = rhh x Ch X (Th1  Th2) 3.783 x (54.36  37.78) = 1.892 x (93.33  Th2)  3.783 x 16.58 or 93.33 Th2 1.892 = 33.15 or Th2 = 93.33  33.15 = 59.18°C P .T2  Ti Here Tsi = Thi, Ts2 = Ts,  T11 54.36 37.78 P93.3337.78
.
=
Th2,
Tri = Tci and
Tt2 = Tc2
16.58 _ 0.3 55 55.55
93.3359.18 T R  si Ts2 Tt2  Tti 54.36 37.78 .34.15 _ 2 16.58 Using P = 0.3, R = 2 for one shell on 2, 4, 6... tube passes, correction factor isF = 0.975 Now OT,„ for a simple counterflow exchanger 1 93.33°C Ail 54.36°C
T
59.18°C1 AT2 37.78°CtCounter flow
ATi = 93.33  54.36 = 38.97°C AT2 = 59.18  37.78 = 21.40°C  AT2 38.97  21.4 OT,n A Ti = log 39.97 in = log A 21.4 LIT2
4I8
Heat and Mass Transfer
Now
= 17.57 = 28.2 0.624 (A Tift)c,,,rected = 28.2 X 0.975 = 27.5°C 3.783 x 4187 x (93.33 — 59.18) Q — A= U • (ATm)corrected 1419 x 27.5 A = 13.86 m2 th, = 3.783 kg/s
Crosssectional area of tubes to handle flow 3.783 me chow. — p x V — 103 x 0.366 = 10.3 x 103 m2 If there are 'n' tubes in tube side, then number of tubes is— n=
Take
atow 10.3 x 103 i— x x (1.905 x 102) re 4
.10.3 x 4 x 103 .10.3x40 x x 3.63 x 10" x x 3.63 = 36.146 n = 36
Now we require surface area of 13.86 m2 for ensuring heat transfer which has to be provided by 36 tubes of 1.905 cm dia and length `l'. Each tube can have 2 or more pass. First take 2 pass
or
13.89 =ir x dx Ixnx2 13.86 /= xx1.905 x102 x 36 x 2 = 3.218 m
But length has to be not more than 2.438 m. Hence, we take now 4 pass. Then length of tube is— l= 3.218 — 1.609 m < 2.438 m Therefore, the heat exchanger has— (a) number of tubes per pass = 36 (b) number of tube passes = 4 (c) length of tube = 1.609 m 59. In a counter flow heat exchanger, hot fluid enters at 60°C and cold fluid enters at 30°C. Mass flow rate of the hot fluid is 1 kg/s and that of the cold fluid is 2 kg/s. Sp
Heat Exchangers
4I 9
heat of hot fluid is 10 kJ/kg K and that of the cold fluid is 5 kJ/kg K. The LMTD for heat exchanger in °C is— (a) 15
(b) 30
(c) 35
(d) 45 (GATE — 2007)
60°C
Th y i ATI —1—Tc2
Th2
.4. LiT2
i
Tci= 30
Counter flow
Counter flow
Heat capacity of hot fluid = mh ch = 1 x 10 = 10 kJ/ks Heat capacity of cold fluid = me cc =2x5 = 10 kJ/ks As mh ch = me cc ••• ATI = AT2= x or 60 — x = 30 + x or x = 45 A T ff, = ATi + AT2 2 15+15 2 = 15 = Option (a) is correct 60. Hot oil is cooled from 80 to 50°C in an oil cooler which uses air as the coolant. The air temperature rises from 30 to 40°C. The designed heat exchanger uses a LMTD value of 26°C. The type of heat exchanger is— (a) parallel flow (c) counter flow
(b) double pipe (d) cross flow (GATE — 2005)
420
Heat and Mass Transfer
80
80 50
30
50
40
40
'
30
Parallel flow
Counter flow
For parallel flow— A Ti — AT2 AT log 1 AT2 = 24.85°C
A T ff, —
—
(80 — 30) — (50 — 40) log 8030 5040
For counter flowAT1 — AT2 — (80 — 40) — (50 — 30) AT log 8040 log 1 5030 AT2 = 28.85°C
A Tn., —
As ATff, in counter flow is greater than the designed LMTD, hence counter flow heat exchanger is suitable. Option (c) is correct. 61. The LMTD of a counter flow heat exchanger is 20°C. The cold fluid enters at 20°C and the hot fluid enters at 100°C. Mass flow rate of the cold fluid is twice that of the hot fluid. Sp heat at constant pressure of the hot fluid is twice that of the cold fluid. The exit temperature of the cold fluid is— (a) 40°C (c) 80°C
(b) 60°C (d) cannot be determined (Gate — 2008) —4— AT1 +
h1 = 100 Tc2
Th2 AT2 Tc1
As
mh
• ch = mc' cc
A Ti = AT2
t
Heat Exchangers
42 I
m = ATiFAT2 2 ATI = AT2 = OTm = 20
LMTD = or
Exit temperature of cold fluid (Tc2) = Th1 — AT,„ = 100 — 20 = 80°C Option 'c' is correct. 62. A concentric tube heat exchanger uses water, which is available at 15°C, to cool ethylene glycol from 100°C to 60°C. The water and glycol flow rates are same at 5 kg/s. Determine The effectiveness of heat exchanger. Take Cp (water) = 4178 J/kg K, Cp (ethylene glycol) = 2650 J/kg K. Can you comment, whether the heat exchanger is working in parallel flow or counter flow modes of operation? (UPTU 2008 — 9) Th1 Th2 TC1
Given:
Now
Th1
= 100, Th2 = 60 and Tci = 15°C
Ch = Mh Ch = 5 x 2650 = 13250 kJ Cc = mc Cc = 5 x 4178 = 20890 kJ Ch < Cc Cmin = Ch
Also heat given = heat gained Ch x (Th1 — Th2) = Cc x (Tc2— Tc1)
or
13250 x (100 — 60) = 20890 (Tc2 — 15) 13250 x 40 20890 = 15 + 25.37 = 40.37°C
Tc2 = 15 +
or
Effectiveness
E=
Q Qmax —
100 — 60 100 —15 = 0.47
Cmin
Th2 ) min (Thi Tc,)
40 85
422
Heat and Mass Transfer
Since Tc2 (outlet temperature of water) is greater than Th2 (outlet temperature of ethylene glycol), hence parallel flow is used in the heat exchanger. 63. In a cross flow, both fluids unmixed heat exchanger has water at 6°C flowing at rate of 1.25 kg/s. It is used to cool 1.2 kg/s of air that is initially at a temperature of 50°C. Calculate the exit temperature of air and exit temperature of water. Assume overall heat transfer coefficient is 130 W/m2K and area is 23 m2. (Annamalai University — 2007 — 8) C, = me Cpc = 1.25 x 4.18 = 5.225 kJ/sK Ch = 1.2 x 1.05 = 1.26 kJ/sK Here Ch < Cc . Ch = C • = 1.26 x 103 J/sK
Now
NTU =
U•A C. 130 x 23 1.26 x 103
= 2.37 1.26 R — Cmin — 5.225 C = 0.241
Now
Now for NTU = 2.37 and R = 0.241, effectiveness for counter flow heat exchanger can be determined from the graph E = 0.83
Now . Similarly
or or
E=
Th 2) TO
Ch (Th, — Cmin
SO — Th2 — 0.83 506
Th2 = 50 — 44 x 0.83 = 13.48°C E=
Cc (Tc2 — TO Cmin Th
TO
5.225 1.26
X
(Tc2 — 6) (50 — 6)
4.146 x (Tc2 — 6) 44 Tc2 = 6 + 8.81 = 14.81°C.
0.83 =
64. A counter flow concentric tube heat exchanger is used to cool engine oil (C = 2130 J/ kg K) from 160°C to 60°C with water available at 25°C as the cooling medium. The flow rate of cooling water through the inner tube of 0.5 m is 2 kg/s while the flow rate of oil through the outer annulus OD = 0.7 m is also 2 kg/s. If h is 250 W/m2K,
Heat Exchangers
423
how long must be the heat exchanger to meet its cooling requirement. (Annamalai University 2004 — 5) Heat lost by oil = Heat gained by water MhCh(Thi — Th2) = McCc (Tc2 — Tc1)
or or
Now
2 x 2130 x (160 — 60) = 2 x 4186 x (Tc2 — 25) 2130 x 100 Tc 2 = 25 + 4186 = 25 + 50.88 = 75.88°C Q = 2 x 2130 x 100 = 426 x 103 W 1
160
AT1
1
60
7508
25
Counter flow
Counter flow Now and
...
AT1 = 160 — 75.8 = 84.2°C AT2 = 60 — 25 = 35°C AT1 —AT2
Q. =_ togATi AT2
Now
= 49.2 0.878 = 56.05°C Q =hxAx Q, 426 x 103 = 250 x A x 56.05
Now
A — 426 x 103 250 x 56.05 = 30.4 m2 A= 7r xD1 x L
or
L=
or
84.2 — 35 84.2 log 35
30.4 nx0.5 = 19.36 m
A AT2
424
Heat and Mass Transfer
65. Match List I (Heat Exchanger Process) with List II (Temperature Area Diagram) and select the correct answer using the codes given below the Lists. List I (Heat Exchanger Process) (A) (B) (C) (D)
Counter flow sensible heating Parallel flow sensible heating Evaporating Condensing
List II (Temperature Area Diagram) 2
1
t T
1.
A
A —P.
A
A
5
T
A —3
Codes
AB 4 (a) 3 (b) 3 2 (c) 4 3 (d) 4 2
Option (c) is correct.
CD 1 2 5 1 2 5 1 5
Chapter10
CONDENSATION AND BOILING
A A A A A A A A
CONDENSATION FILMWISE CONDENSATION DROPWISE CONDENSATION VAPORIZATION LAMINAR FILM CONDENSATE NUSSELT' S THEORY OF FILM CONDENSATION FILM THICKNESS FILM HEAT TRANSFER COEFFICIENT
A A A A A
POOL BOILING FORCED CONVECTION BOILING SUBCOOLED BOILING SATURATED BOILING REGIMES OF SATURATED POOL BOILING A NUCLEAT BOILING A UNSTABLE FILM BOILING A STABLE FILM BOILING
INTRODUCTION Condensation and boiling are nothing but special convective heat transfer processes which are associated with phase change of a fluid. Condensation is a process which involves a change of phase from vapour to liquid while boiling involves a change of phase from liquid to vapour. These processes are widely used in many applications such as: (i) boilers and condensers of power plants, (ii) evaporators and condensers of refrigeration plants, (iii) heating and cooling, (iv) melting of metals in furnaces, and (v) heat exchangers used in refineries and sugar mill. Since there is a phase change in these processes, hence the heat transfer is large without much change in fluid temperature. During a phase change there is: (i) liberation of latent heat by the fluid in condensation process in condenser, or (ii) absorption of latent heat by the fluid in boiling process in boiler. 1. What is condensation? Discuss the physical mechanism of condensation on a vertical plate. (UPTU — 20056) The condensation is a process in which vapour phase of substance changes into liquid phase by releasing its latent heat of vaporization. Vapour changes into liquid phase when it comes
426
Heat and Mass Transfer
into contact with any surface having temperature lower than the saturation temperature corresponding to the vapour pressure. During condensation, temperature does not charge as vapour is converting into liquid by releasing its latent heat of vaporization. 2. What are types of condensation? The condensation process can be classified as1. Filmwise condensation 2. Dropwise condensation 3. What is filmwise condensation? Why is it less effective but still preferred? In the filmwise condensation, the vapour condensation takes place on the cold surface. The condensate wets the surface and tries to spread out due to gravitational force. Ultimately the condensate falls down from the surface on increasing thickness of the liquid film due to gravity.
Ts < Tsaturation
Vapour
Film of liquid formed by vapour condensation on cold surface
Cold surface (Ts)
Filmwise condensation
Filmwise condensation is not preferred as layer of liquid is deposited on the cold surface having high resistance to heat flow which hinders the transfer of heat from the vapour to the cold surface. The filmwise condensation leads to low transfer of heat from the vapour to the cold surface and hence formation of filmwise condensation is less effective but preferred in most of the applications as it can be easily achieved in applications. Filmwise condensation takes place on clean serface and when vapour is having no impurity like organic substance or fatty acid. 4. What do you understand by dropwise condensation? or Which type of condensation has a higher heat transfer film coefficient and point out the reason thereof. (UPTU — 20067) In dropwise condensation, condensate does not spread on the cold surface to form liquid film. Instead condensate forms liquid droplets which grow in size and ultimately they break away from the cold surface. The condensation of vapour takes place most effectively as condensate does not form liquid film on the cold surface. The cold surface remains continuously exposed to the vapour to facilitate effective condensation. Such type of
Condensation and Boiling
427
condensation takes place when the cold surface is highly polished or has coating of teflon or zinc. Impurities like organic substances or fatty acids in the vapour are also helpful for dropwise condensation.
0
0
0
0
0
0
0
0
0
0
0
0
Tsurface < Tsaturation
Dropwise condensation
5. What type of condensation is assumed in designing condensers? (UPTU — 20067) Although heat transfer rate in case of dropwise condensation is much higher than what is possible in case of filmwise condensation, but it is very difficult to achieve dropwise condensation. Therefore, in spite of low heat transfer rate possible in filmwise condensation, film condensation is assumed for designing condensers. 6. What do you understand by boiling and condensation? Boiling and condensation are also heat transfer processes like conduction, convection and radiation but main difference is that boiling and condensation involve a phase change of the substance. In boiling, liquid phase of the substance changes into vapour phase while in condensation, vapour phase changes into liquid phase. Boiling and condensation are heat transfer processes which take place at constant temperature. The heat transfer rate in boiling and condensation is very high as compared to other modes of heat transfer due to latent heat associated with phase change of the substance. 7. Differentiate between dropwise and filmwise condensation. (UPTU — 20034 and 20067) S.N.
Dropwise condensation
1. The vapour condenses into small liquid droplets of various sizes. These droplets fall down from the surface in random fashion. Hence, condensate as droplets tends to leave the surface quickly without forming a film due to gravity. 2. No additional thermal resistance to the heat flow from vapour to the surface. 3. Heat transfer is 5 to 10 times more than filmwise condensation. 4. Occurs either on highly polished surfaces or vapour contaminated with impurities like organic substances and fatty acids. 5. Difficult to achieve or maintain.
Filmwise condensation The vapour condenses as liquid film on surface and stays on the surface for a longer time.
Additional thermal resistance due to formation of liquid film. Heat transfer is hindered by liquid film. Clean surface and no impurity
Easy to achieve
428
Heat and Mass Transfer
8. Derive an expression for velocity distribution in the liquid condensate film over a vertical plate for laminar film condensation. (UPTU — 2002 — 3) Nusselt's theory of film condensation is applied for vapour condensation occurring on a vertical plate. Nusselt's analysis of film condensation is based on following simplifying assumptions— (a) Vapour is pure, dry and saturated (b) The flow of condensate is due to the action of gravity and it is laminar. (c) The liquid film has attained surface temperature at contact with the cold surface and saturation temperature exists at liquidvapour interface. (d) Viscous, shear and gravitational forces act on the fluid in the film along the surface. Normal viscous and inertia forces are neglected. au (e) No velocity gradient at liquidvapour interface i.e.( — = 0 where u = velocity, ay j y.5 y = distance from surface and S = film thickness. (f) Heat transfer in the film is by conduction and temperature distribution in the film is linear. (g) Heat transfer is taking place under steady state. Consider condensation of vapour on a vertical plate as shown in the figure. Take origin `o' at upper end of the plate, xaxis along the vertical surface and yaxis at perpendicular to it. Take height of plate = L, width = b, thickness of film = S at distance 'x' from the origin. The film thickness is zero at upper and of the plate and it increases gradually till lower end of the plate where thickness is maximum. 0
Viscous force du dy
Velocity profile
Filmwise condensation and element of volume in film
dy
x dx dx
µ [
Bouyant I force = pv.g. (b.dxdy)
du d2u +µ dy]b x dx dy dY
Viscous force I Body force = (b.dx.dy)
Force balance in element of volume
Now consider the equilibrium of gravity, buoyant and viscous forces acting on the element of volume (b • dx • dy) in the liquid film pi, •
du g • (b • dx • dy) — pv • g • (b • dx • dy) = µ— (b • dx) — dy
du
d2u 2 • dy)(b • dx)
dy dy
Condensation and Boiling
d 2u dy e
or
429
— (PL Pv)g ti
On integration, we get— u = —(PL — Pv) (Y2). g + co, + C2 2 il, Apply boundary conditions— At
y = 0, u = 0
and at
y = 8,
du y
=0
We get the value of c1 and c2 as— ci = (PL — Pv)• g • O and c2 = 0 µ The velocity profile on putting the values of c1 and c2 is— u = PL — Pv [ (5y Y 2 1 µ L 2J = PLPV (5 2 [y2
1 (y ')21
o 2
The mean velocity in the liquid film at a distance y from the plate is— umean = I J.° u • dy (5 0 y )1dy 2 = 1 I*6 PL — Pv g o(.2,[y — —1 ( — 8J o µ 8 2 8 (PL — Pv) • g • 52 3µ 9. Find the expression for the change of mass flow rate of condensate through the laminar liquid film formed on the vertical plate. The mass balance in the element of volume in the liquid film formed on a vertical plate is as shown in the figure— Now m = density x flow area x mean flow velocity (um)
m
, dy10dx 1
= PL x (b x 8) x
(PL — P g) • g • 32
3µ
1
m+ dm
Mass balance
430
Heat and Mass Transfer
As film thickness '8' increases as 'x' increases when vapour moves down condensing on the vertical plate, the mass flow rate of the condensate is therefore dependent or a function of distance 'x'. As the condensate flows from 'x' to `x + dx', the film thickness grows from `8' to '8+ d8'. The growth of film is resulting due to additional condensation of the vapour on the cold surface. The mass of condensate added to the film from distance 'x' to `x + dx' can be found out by differentiating mass flow (m)— dm = d [pL(pL Pv)• g • b • 8 3 1c18 •dx d8 3µ dx = [PL(PL — Pv)• g • b • 82 i ds
10. Find the expression for film thickness (3) in the liquid condensate forming laminar film on a vertical plate. The heat balance in the element of volume in the liquid film formed on a vertical plate is as shown below— dy
dQ = hA AT
dQ  hfe dm
dx
dQ = KA ST
The rate of heat flow (dQ) in the film element is equal to the rate of latent heat release on condensation of vapour at the surface. Hence, dQ = latent heat of vaporisation x change of mass flow rate = hfg x dm =
hfg
[PL(PL — Pv)• g • b • 82 1
d8
(i)
The mode of transfer of heat (dQ) after condensation is pure conductance across the film. Hence, dQ —
k(b • dx)
(Tsar — Ts)
Equating equation (i) and (ii), we have— hfg PL (PL Pv). g b 32 d6 = or
k • b dx (Tsar — Ts) c
83 x d8 = kxµx(Tsat—Ts)xdx p L (pL — pv)• g • h fg
Condensation and Boiling
43 I
Now on integration and considering `S' varies with `x', we get84 k x tix(Tsat — Ts) x i ci
4
PL (PL — Pv) • g • hfg
Now boundary conditions are8 = 0 at
x = 0, ci = 0
Hence, the film thickness is— 8=
[4xkxtix(Tscit—Ts)
xitm
PL(P L — Pv) x g x h fg
The film thickness (8) increases as the fourth root of the distance (x) from the top of the plate. 11. Find the expression for the heat transfer coefficient for the film on the vertical plate.
The heat flow from the vapour to the surface is assumed to be taking place by only conduction through the condensate film. Hence, dQ =
k • (b . dx) (Tsat — 7's) (5
(i)
Now heat transfer at the surface due to only convection is—
00
dQ = hx(b • dx) (Tsai — Ts)
where hx = local heat transfer coefficient at distance 'x' from the top of the plate Equating equations (i) and (ii), we have— k • (b • dx) ,,,,
(5
or
ki sat — Ts) = hx(b X dx) (Tsat — Ts) k hx = 75
1/4
But
S—
4 x k x µ(Tsa, — T s) x x P L(P L — p v) x g x hfg 1
1/4
hx
Now
_[ PL (P L — pv)X k 3 X g X hfg 1 4 11(7'sat — Ts)X x
h = 1 rhxdx Lo 1
IL PL (PL — PV) X k3 X g X hfg
L Jo
4 11(7'sat —
Ts ) X x
i 1/4
X dx
1/4 1 X PL (PL PV)k 3 ' g • hf g 1/4 dx Xf L X = o L L 4 au (Tsai — Ts ) JJ
432
Heat and Mass Transfer
4 [PL (PL Pv) x k 3 x g x h fg 11/4 3 4x ttxLx(Tsat —Ts) [PL(PL—
Also
hz, = (hx)x.L, —
ill4
Pv) x k 3 X g
h fg
4xµxLx (T„t — Ts)
h = —4 x hL 3 Hence, average heat transfer coefficient is 4/3 times the local heat transfer coefficient at the trailing edge (x = L) of the plate. The average heat transfer coefficient is generally given as.
h = 1.13 [
PL (PL — Pv)x k3 x g x hfg
1/4
µ x L x (Tsai — Ts)
In case the surface is inclined '0' with horizontal, then average heat transfer coefficient is1/4 h = 113 [PL (PL — Pv) x k3 xgxhf g x sin 0 ti x L x (Tsat — Ts) 12. Write the expression for average heat transfer coefficient for laminar film condensation on: (i) a horizontal cylinder, and (ii) horizontal tube banks. Average heat transfer coefficient for laminar film condensation on a horizontal cylinder is— h = 0.725
rr [ PL
(PL Pv) • k 3 g • hfg
i1/4
eilD(Tsat Ts)
where D = diameter of cylinder. Average heat transfer coefficient for laminar condensation on horizontal tube banks is— • 1/4 pv). k3. g hfg h = 0.725 [ PL (PL D '(Tsat — Ts) where D = diameter and N = number of horizontal tubes in a vertical tier so that condensate from one tube flows directly to next tube below it. For a square array, the total tube is a2 while N = a. 13. What is boiling heat transfer? Boiling is a convective heat transfer process and it differs from other modes of heat transfer as it involves a phase change from liquid to vapour. As the phase change from liquid to vapour requires a large quantity of heat as latent heat of vaporization, possible heat transfer by boiling is large as compared to what can be transferred by other modes of heat transfer. Boiling occurs when a liquid comes into contact with a hot surface whose temperature is
Condensation and Boiling
433
higher than the saturation temperature of the liquid. Heat transferred to the liquid is— Q=11)