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English Pages 830 Year 2004
HEAT AND MASS TRANSFER (SI UNITS) (For B.E./ B. Tech., I.E.S. and I.A.S. Examinations) (Fifth Edition)
R.YADAV
B.Sc. Engg., M.E., Ph.D., F.I.E., M.I.S.T.E. Emeritus Professor of Mechanical Engineering MNNIT Allahabad  211004 Former Professor and Head, Mechanical Engineering Department M.N.R.Engineering College (N.ow MNNIT) Allahabad Former Principal, RlT (now NIT) Jamshedpur AND
SANJAY
B.E. (MECH.) M.E. Assistant Professor of Mechanical Engineering NIT Jamshedpur AND
RAJAY
B.TECH. (MECH.), Senior Engineer, B.H.E.L., HARDWAR
CENTRAL PUBLISHING HOUSE 18 C, SAROJJNI NAIDU MARG ALLAlfABAD  2/1001

r

© COPYRIGHT 1998 Pubilisher and Author ISBN 8285444382
Second Revised Edition 1992 Third Revised Edition I 996 Fourth Revised Edition, 2001 Fifth Revised Edition, 2004
Published by : Central Publishing House, Allahabad Figures Designed by: Sun Graphics, Allahabad; 05322545767, 9415443743 and Printed by: Halcyon Press, 18C, Sarojini Naid� Marg, Altahabad
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NOMENCLATURE A
=
Ae
=
Area Cross Sectional Area
=
Surface Area
=
Constant
= = = =
Width Constant, Concentration
= =
Diameter Energy, Emmissive Power
=
Shape Factor
f
=
h
=
Frequency· Heat Transfer Coefficient
I
=
Intensity of Radiation
]
=
Total Radiation Energy
= = =
Thermal Conductivity Length, Thickness Mass, Mass Flow Rate
N, n
=
Normal
p
=
Pressure
q
= =
Heat Transfer Rate Heat Transfer
=
Thermal Resistance
s
=
Length
u
= =
Absolute Temperature Temperature
=
Overall Heat Transfer Co�fficient
u
=
Velocity
V
= =
Velocity, Voltage, Volume Width
=
Directions
As B b C C
D a
E F
k
L m
Q R
T
w X,
y,
Z
Sound Velocity, Specific Heat Diameter, Diffusion Coefficient
PREFACE TO FIFfH REVISED (S.I.) EDITION In the present edition, a revisi�n has bean made and printing mistakes have been removed. Many figures havl;been replaced for better understanding. I hope this edition will serve better to students appearing for B.E./B.Tech, I.E.S. and i.A.S. Examinations. .

R.YADAV
Allahabad
SANJAY
Godess Swarswati Puja Day, '2004
RAJAY
/
•
I
PREFACE TO FIRST (SI) EDITION Heat and mass transfer processes are the integral parts of the technology and environment. During the recent past, its importance has been recognised and a full course on this topic was introduced in almost all the engineering and technological institutions of the world. Most of the engineering institutions in India and abroad . have switched over from MKS or FPS to SI (System Internationale) system of units and the rest are in the transition stage. Though there are many books availabie on this topic in the international market but there is a scarcity of good books written i:1 SI units covering all the aspects of the subject matter. This book has been written to fill this gap. This book will suit the courses followed in the engineering colleges leading to B.E. degrees. It will also suit the courses of A.M.I.E. and U.P.S.C. examinations. I have attempted to present the matter iri a simple, lucid and precise manner. A large number of good practical and typical problems have been solved in all the nineteen chapters of the book and I hope the book will meet the requirements of the students, teachers and practical engineers. Sincere acknowledgement are due to the authors listed in the references whose books/papers have been consulted during the preparation of this work. Finally, the author wishes to thank his wife Meera and sons Sanjay and Rajay for the hardships which they had to face due to the authors engagement in the v.enture.
R. YADAV
CONTENTS Chapters 1. INTRODUCTION: Importance of Heat Transfer; Basic Modes of Heat Transfer; Conduction Mechanism; Convection Mechanism; Radiation Mechanism; Steady and Unsteady Heat Transfer; Conservation or Energy; Basic Law of Heat Conduction; General Heat Conduction Mechanism and Law; Basic Law of Convection; Basic Law of Radiation; Thermal Resistance and Conductance; Combined Heat Transfer Mechanisms; Thermal Conductivity.
129
2. GENERAL HEAT CONDUCTION EQUATION : General Heat Conduction Equation in Rectangular Coordinates; General Heat Conduction Equation in Cylindrical Coordinates; General Heat Conduction Equation in Spherical Coordinates; Solutions of Equations
3041
3. ONED I MENS ION AL S TE ADYST A TE HEAT CONDUCTION WITHOUT HEAT GENERATION:  Plane Wall or Slab; Hollow Cylinder or Tube; Spherical Shells; Conduction Shape Factor; Heat Flow through Composite Structures; Series Composite Wall; Critical Thickness of Insulation for Pipes; Composite Spherical Shells; Critical Thickness of Insulation for Spheres; Effect of Variable Thermal Conductivity; Heat Flow Through Plain Wall with NonUpiform Thermal Conductivity; Heat Flow Through Cylinder at Non Uniform Conductivity; Heat Flow Through Sphere with Non Uniform Conductivity; Heat Flow From a Long Tube Through_ which the Hot Fluid is Flowing.
42104
4. ONE DIMENSION AL STEADY STATE HEAT CONDUCTION WITH HEAT GENERATION: ' Plane Wall with Uniform Heat Generation; Hollow Cylinder with Uniform Heat Generation; Solid Cylinder with Uniform Heat Generation; Electric Wire Carrying Electric current; Solid Sphere with Uniform Heat Generation; Heat Flow Through Piston Crown; The Buried Cable; Variable Distributed Generation as a Function of Temperature; Dielectric Heating 5. HEAT TRANSFER FROM EXTENDED SURFACES : Straight Fin of Rectangular and Circular profile; Conduction Cooling of Turbine Bladings; Efficiency of Straight Rectangular Fins; Effectiveness of the Fin; Optimum Dimensions for Straight Rectangular Fins; Errors of Measurement of Temperature of Thermometer Well; Heat Flow Form a Bar Maintained at Different Temperatures; Gene:;alised Equations for Fios or Spines; Straight Fin of Triangular Profile; Optimum Dimensions for Straight Triangular Fin; Straight Fin of Concave Parabolic Profile; Optimum Dimension of Concave Parabolic
105152
� V 111)
Profii.e; Straight Fin of Convex Parabolic Profile; Straight Fin of Least Material; Selection and Design of Straight Fins
. . . 153205
6. T\VO AND THREE DIMENSIONAL STEADY S'{'ATE HEAT CONDUCTION : Conduction in a Semiinfinite Strip; Coi1ducuon in a Long Rectangular Rod; Graphical Method; Analogical Solution; ?inite Difference Method; The Nodal Network and Finite Difference Form; Finite Difference method; The Relaxation Method; The GaussSeidel Iteraction; The Matrix Inversion Method; Three Dimensional Heat Conduction.
. . . 206232
7. UNSTEADY STATE (TRANSIENT) HEAT CONDUCTION : Nonperiodic Heating and Cooling; Heating or Cooiings with Known Temperature Distribution; Newtonian Heating or Cooling with negligible Internal Resistance; TemperatureTime Response of Thermocouple; Heating or Cooling with negligible Surface Resistance; Transient Heat Conduction in Semiinfinite Solids; Heating or Cooling of Infinite Plate with Finite Internal and Surface Resistance; Infinitely Long Cylinder and Sphere with Finite Internal and Surface Resistance; Graphical Method (Temperature Time Charts) for Finite Internal and Surface Resistance; Periodic Heat Flow; Periodic Heat Flow in Semi infinite Thick Solid; Transient Conduction with Given Temperature Distribution; Multidimensional Effect.
. . . 233289
8. CONVECTION AND FLUID FLOW PROCESSES : Mean Bulk Temperature and Mean Film Temperature; The Ideal Fluid; Viscous Fluid The Stream Function; The Velocity Potential; Newtonian and NonNewtoniav Fluids;.Laminar Flow; Turbulent Flow; The Hydrodynamic ( Velocity) Boundary Layer Concept; Methods of Boundary Layer Control; Thermal Boundary Layer; Convection Boundary Conditions and Their Applications; General Momentum Conservation Equations; NavierStoke� Equation; Flow Regimes and Simple.Flow Solutions of Navier Stokes Equation; Velocity Distribution in Fully Developed Laminar Flow in Tube: Flow Loss in Passages.
... 290318
9. DIMENSIONAL ANALYSIS : Criteria for Similitude; Dimensions; Dimensionless Numbers and Their Physical Significance; Characteristics Length or Equivalent Diameter; Buckingham Theorem Frictional Loss in Pipes; Forced Convection Heat Transfer; Naturai or Free Convection; Forced Convection in High Speed Flow; Model Similitude for Heat Transfer Equipment.
... 319334
iO. LAMINAR FLOW FORCED CONVECTIVE HEAT TRANSFER : Equation of Motion for Hydrodynamic Boundary Layer Over a Flat Plate; Velocity Distrib'ution in the Boundary Layer (Exact Solution); Energy Equatibn of Thermal Boundary Layer Over a Flat Plate; The Prandtl Number; Temperatme Distribution in Boundary Layer (Exact Solution);
1;
,
Thickness of Thermal Boundary Layer; The Local and A veragc Heat Transfer Coefficient; The Nusselt Number; Approximate Hydrodynamic Boundary Layer Analysis (Integral Method); Approximate Method for Temperature Distribution (Integral Method); Velocity Distribution in the Entrance Region of a Tube; Momentum Equation for Hydrodynamic Boundary Layer in a Circular Tube (Cylindrical Coordinates); Energy Equation for Thermal Boundary Layer in Circular Tube (Cylindrical Co ordinates); Local and Average Heat Transfer Coefficients for Laminar Flow in Tubes; Heat Transfer in Tube (Solution of Energy Equation).
... 335391
11. TURBULENT FLOW FORCED CONVECTIVE HEAT TRANSFER : Turbulence; Turbulent Shearing Stress and Heat Exchange; Semiempitj.cal Theorie� of Turbulence; Analogy between Heat and Momentum Transfer; Reynold's Analogy; General Form of Reynold's Analogy; Colburn Analogy; Prandtl Analogy; Momentum Integral Method for Turbulent Boundary Layer over Flat Plate; Turbulent Heat Flow.
... 392421
12. EMPIRICAL CORELATIONS FOR CONVECTION : Laminar Flow Over Flat Plate; Laminar Flow Inside Tubes; Turbulent Flow Over Flat Plate; Mixed Boundary Conditions O\.er Flat Plate; Fully Developed Turbulent Flow in Tubes; Turbulent Flow in Noncircular Passage; Turbulent Flow Over Cylinders; Turbulent Flow Over Spheres; Turbulent Flow Across Banks of Tubes; Packed Beds; Horizontal Plates; Cylinders· and Wires; Inclined Plate; Spheres
. : . 422453
13. FREE CONVECTION : Temperature and Velocity Distribution in a Free Convection Boundary Layer Over Vertical Plate; Similarity Relations in Free Convection; Momentum and Energy Equation in Free Convection BoundaryLayer Over a Verli.cal Plate; Velocity and Temperature Distribution in Laminar Free Convection over Vertical Flat Plate (Exact Solution); Approximate Integral Method in Laminar Flow Over Vertical Plate; Turbulent Free Convection over a Vertical Plate (Integral Method); Free Convection from Rotating Bodies.
. .. 454479
14. BOILING AND CONDENSATION: Physical Mechanisms of Boiling; Boiling Regimes of Saturated Pool Boiling; Bubble Shape and Size Consideration; Bubble Growth and Collapse; Critical Diameter of Bubble; Factors effecting Nucleate Boiling; Forced Convection Boiling; Boiling Correlations; Physical Mechanisms of Condensation; Laminar Film Condensation on a Vertical Plate; Turbulent Filin Condensation; Film Condensation in Horizontal Tubes; Influence of the Presence of Noncondensable Gases; Enhancement of Film Condensation; Dropwise Condensation
... 480519
15. SPECIAL CONVECTIVE HEAt TRANSFER PROCESSES : Heat Transfer in Magnetohydrodynamic (MHD) Systems; Heat Transfer in High Speed Flow; Transpiration Cooling; Low Density Heat Transfer; Ablation
. . . 520537
16. HEAT EXCHANGERS : Heat Exchanger Types; Heat Exchanger Analysis; Log Mean Temperature Difference; LMTD Correction Factors; Heat Exchanger Effectiveness; Effectiveness by Using Number of Transfer Units; Heat Exchanger Design Considerations; Some Constructional and Thermal Aspects of . . . . 538602 Heat Exchanger Design 17. T HER MAL RADIATION PHENOMENA : Physical Mechanism; Surface Emission Properties; Surface Irradiation Properties; Concept of Black Body; Kirchoffs Law of Radiation; Intensity of Radiation and its Relations; Radiosity; Lambert's Cosine Law; Radiation Density; Radiation Pressure; Derivation. of StefanBoltzman Law from Thermodynamics; Planck's Distribution Law (Monochromatic Blackbody Emissive Power); Derivation of StefanBoltzman Law from Planck's Law;. The Wein's Displacement Law; Band Emission; Emissivity; Real Surface Behaviour; Grey Surface Behaviour; Solar Radiation
. . . 603663
18. RADIATION EXCHANGE BETWEEN SURFACES : Radiation Exchange Between Two Black Surfaces Separated by a Nonabsorbing Medium; The GeomcLrical or Shape Factor; Relationship Between Shape Factors; Radiant Heat Exchange Between Nonblack Bodies or Grey Bodies Separated by a Non absorbing Medium; Radiation Exchange from a Surface (Cavity); Radiadon Heat Exchange Between· Black Surfaces Connected by Refractory Surfaces; System·consisting of Two Grey Surfaces; System consisting of Two Black Surfaces; System consisting of Three Grey Surfaces; System consisting of Two Black Surfaces Connected by a Single Refractory Surface; System consisting of Two Grey Surfaces connected by Single Refractory Surface; System consisting of Four Grey Surfaces; Radiation Shields; Radiation Heat Exchange through an Absorbing and Transmitting Medium; Radiation Exchange with Specular Surfaces.
. .. 664752
19. MASS TRANSFER: Modes of Mass Transfer; Fick's Law of Diffusion; Diffusion of Gases and Liquids Through Solids; Equimolar Counter Diffusion; Conservation of Species; General Mass Diffusion Equation; Bom�dary and Initial Conditions; Isothermal Evaporation of Water into Air; Mass Transfer Coefficient; Correlation for Convective Mass Transfer.
753782
References Appendix
783788 789815
Index
8 17
.1 Intn;>duction Heat transfer is that science which predicts the rate of energy transfer taking place between material bodies as a result of temperature difference between them. According to thermodynamics, this energy transfer is defined as heat. The science of heat transfer not only explains how heat energy may be transferred but alsp predicts the rate at which the exchange will take place under certain specified conditions. There is a difference between thermodynamics and heat tra,!fer and it is the heat transfer rate which forms the basisfor difference. Thermodynamics deals with a system in equilibrium and predic� the amount of energy required to change a system from one equilibrium state to another. Moreover, from thermodynamics point of view, the amount of heat transferred during a process simply ·equals the difference between the energy change of the system and the work done. Evidently, this type of analysis neither considers the mechanism of heat flow nor the time required to transfer the heat. H simply suggests how much heat to supply to, or reject from, a system during a process between the prescribed end states without taking , care of whether or how this could be accomplished. The reason for this lack of information obtainable from thermodynamics analysis is the absence of time as variable. As an example, consider the cooling of a hot steel bar placed in a pail of water. Thermodynamical analysis will predict the final equilibrium.temperature of the steel bar_ water combination but it will not predict how long it _takes to reach this equilibrium condition or what the temperature of the bar will be after a certain length of time before the equilibrium condition is attained. But heat transfer study will predict the temperature of both the bar a�d water as a function of time. 1.1 Importance of Heat Transfer Study
The study of heat transfer has become an increasingly inten�_e concern in modem technology, in the earth sciences, in organic metabolism and in environmental : engineering. The importance of study· of heat transfer can be. understood from the following areas in which the knowledge of hea� transfer is required: (i) Energy production and conversion.There is not a single ,ipplication in this area that does not involve heat transfer effects in some way. For thermal design of boilers, steam turbines, condensers, feed heaters, other heat exchangers, cooling towers, internal combustion engines, gas turbines, jet propulsions, nuclear reactors, magneto hydrodynamic devices, engineers have to make a detailed heat transfer analysis. Solar energy conversion for space heating and electric power generation is full of heat transfer problems. For the efficient use of geothermal energy sources, there are numerous heat transfer problems that must be solved. (ii) Ref!:igeration and airconditioning.The ·thermal design of coml)ressors, lndensers. eva_pora�ors, incinerators· and cryogenic _storage equipments and other cooling and heating equipments involves an intensive stuay of heat transfer. (iii) Electric machines.Cooling systems for_ electrical motors, generators and transformers involve heat transfer effects as heat generated during the flow of current through the windings of these machines has to be dissipated effectively. Insulations provided on electric wires Reed a heat transfer analysis to avoid conditio�s which will . cause overheating. Electronic equipments need cooling systems and that involve heat transfer analysis for design.
Heat and Mass Tr.:µ1sfer
2
(iv) Chemical and petrochemical operations.Heating, cooling, evaporation and condensation in chemical operations are nothing but heat transfer problems. A lot of heat lransfer analysis is required in designing the fractional distillation plants for crude petroleum.
(v) Civil engineering.Construction of dams, and heavy structures, design of buildings for the minimisation of heat losses need heat transfer study.
(vi) Space vehicles.The rapid removal of heat from the outer surface of space vehicle is a must for its durability. Maintenance of an optimum temperature is necessary for proper function of the instruments mounted on the space vehicles. (vii) Manufacturing process.The casting of metals, extrusion, metal cuttings and heal treatment of metals involve heat transfer study.
(viii) Environmental engineering.Heat transfer processes are relevant to air and water pollution and strongly influence local and global climate. Now, the world is confronted with thermal pollution problem associated with the discharge of large amounts of waste heat from a power plant to the environment. Numerous heat transfer considerations relate to the design of cooling towers to alleviate the environmental problems associated with tl1is discharge. (ix) Earth sciences.Numerous heat transfer considerations relate to the study of meteorology, geophysics and oceanography. (x) Astronomy....:._The study of astronomy involves heat transfer problems.
(xi) Agriculture and food processing.To develop better variety of seeds one is confronted with heat transfer prob�ems. There is a lot of heat transfer study in food processing.
The various engineering problems involving heat transfer study can be classified in two groups ; (a) heat flow situations where maximum heat transfer is desirable with minimum possible heat exchange area (e.g. walls of LC. enginr.s combustion chambers, heat exchanger, outer space vehicles, gas turbine blades, etc.) (b) heat flow situations where heat transfer is undersirable and its flow is to be prevented (e.g. walls of building, steam pipes, etc.)
From the above examples, it follows that, in almost every branch of engineering, heat transfer problems are encountered which are not capable of solution by thermodynamic reasonings alone, but �uires a heat transfer analysis. Heat transfer studies mainly require the knowledge of thermodynamics:rfluid mecIJanics, physics and mathematics. 1.2
Basic Modes of Heat Transfer
There are thr� basic modes of heat 1rn;nsfei They are :
(i) Conduction. (ii) Convection. (iii) Radiation.
They have in common that temperature differences must exist and that heat IS always t.ransferred in the direction of decreasing temperature. But they differ entirely in the
Introduction
3
physical mechanism and laws by which they are governed. The following examples will distinguish clearly between the three modes of heat transfer. A class teacher wishes to distribute apple to a row of students. He may cause the apple to reach the farthest most student by any one of the following methods :(a) He could deliver the apple to the student nearest to him who delivers the same to the next and so on till finally it reacJ:tes the last student. (b) He could go to the last student and deliver the apple to him. (c) He could throw the apple so that it reaches the last student
Here we. assume that the apple .corresponds to heat energy while the student corresponds to the molecules. Conduction, convection anJ r.tdiation corresponds to first, second and third methods of delivering the apple to the last student. In conduction, heat transfer takes place due to direct molecular communication without any appreciable displacement of the system. In convection, heat transfer takes place due to molecular movement. In radiation, there is no need of any medium while in conduction and convection material medium is essential.
It will be worthwhile here to discuss the basic mechanism of various modes of heat transfer separately. 1. 3. Conduction Mechanism
Heat conduction is a mechanism of heat transfer from a region of high temperature to a region of low te:uperature within a medium (solid, liquid or gas) or between different mediums in direct physical contact. The mechanism of heat conduction may be explained by two conceptsmolecular activity and electron drift.
(i) Molecular activity./n conduction heat flow, the energy is transmitted by direct molecular communication without apprecia�le displacement of molecules. Alternatively, conduction may be viewed as the transfer of energy from the more energetic to the less energetic particles of a substance due to interactions between the particles. From the kinetic theory viewpoint, the temperature of an element of matter is proportional to the mean kinetic .energy of its constitutent of the velocity and relative position of the molecules is called internal energy. When molecules in one region acquire a mean kinetic energy greater thao that of the molecules in an adjacent region, as manifested by temperature difference, the molecules possessing the greater energy will transmit part of their energy to the molecules in the lower temperature region. For example, if the molecules of conducting material at one end of a rod are heated they are set into rapid motion and these, in communicating by elastic impact with their neighbours of lower kinetic energy, set the latter into more violent motion, and so on throughout the length of the rod. Thus, the phenomenon of heat conduction in solids is as a result of transfer of internal energy from one molecule to other. Alternatively, in a solid, conduction may be attributed to molecular activity in theform of lattice vibration.
(ii) Electron drift.An alternative picture of heat conduction is derived from the concept of electron drift. As we know, good conductors of heat are also good conductors of electricity, and since the conduction of electricity is postulated on the theory of free electron drift, so it appears rational to ascribe heat conduction primarily to the mobility offree or valence electrons.
Conduction is the only mechanism by which heat can flow in opaque solids. The best example of conduction is heating of a metal rod from one end.
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' 4
Heat and Mass Tnmsfer
Conduction is also important in fluids and gases but in nonsolid medium it is accompanied with convection and radiation. Conduction in liquids is due to elastic impact while in gase:, and vapors, it is due to molecular diffusion of kinetic energy.
Irrespective of the exact mechanism, which is no� yet fully known, the observable effect of heat conduction is an equaliz.ation of temperature. 1.4. Convection Mechanjsm
Convection is a process involving mass movement of fluids. The mechanism of convection is the transfer of heat energy by actual physical movement of fluid molecule from one place to another in which there exists a temperature gradient. Basically, this mode of heat transfer is comprised of two mechanismenergy transfer due to random moleculw motion (diffusion) and bulk or macro.scopic motion offluid.
When a temperature difference produces a density difference (buoyancy effects) which results in the mass movement of the fluid, the process is called free or natural convection. Some examples of free convection are heat flow from a hot place to atmosphere, chilling e;ffect of a cold wind on a warm body, and heating of room by a stove, etc.
When an external source such as pump or fan causes the fluid motion to take place, the process is called forced convection. Some examples of forced convection are heat exchange in condensers, nuclear reactors, all types of heat ex�hangers, air conditioning equipments, cooling system of internal combustion engines, etc.
A major concern of the heat transfer engineer is the prediction of the rate of heat transfer between a fluid and a solid boundary surface. This takes place in various steps. Suppose the solid surface is at higher temperature than the fluid. First, heat will flow by conduction from the surface to adjacent particles of fluid. The energy thus transferred will serve to increase the temperature and the internal �nergy of these fluid particles. Thus the fluid particles will move to a region of cold fluid (lower t.mperature in the fluid) where they will mix with and transfer a part of their energy to, other fluid particles. Note that this flow is of fluid as well as of energy. The energy is actually stored in the fluid particles and due to mass motion of fluid particles, it is also €arried away. Thus the heat transfer between a solid surface and fluid is due to conduction as well convection. In a fluid flow over a heated surface, a hydrodynamic and thermal boundary layers are formed. The random molecular motion generally dominates near the surface where the fluid motion is low (or zero). The bulk fluid motion d,ominates away from the surface and within the boundary layer. Hence the extensive study of heat transfer in the boundary layer is a must
Convection process involving phase changes leads to the two important fields boiling and condensation.
1.5. Radiation Mechanism / '· Radiation heat transfer is due to property of a matter to emit and absorb thermal energy (or different kind of rays) in the form of electromagnetic waves and to the fact that no material medium is required for the transfer of heat. In other words, radiation is a process by which heat flows from a high temperature body to a body at lower temperature· in the form of electromagnetic waves when the bodies are separated in space even when a vacuum exists between them. In fact, radiation heat transfer occurs most efficiently in a vacuum. the best example of the radiation is the.heat received by the earth from the sun.
5
Introduction
The radiant energy waves coming from the sun towards earth passes through almost vacuum (generally called ether) and also throqgh the atmospheric layes smrounding the earth. Radiant energy travels at a speed of light (3 x 108 m/s). In fact, according to electromagnetic theory all types of rays and thermal radiation differ only in their respective wave lengths. Although, we focus primarily on radiation from solid surfaces, emission may also occur from liquids or gases. 1. 6 . Steady and Unsteady Heat Transrer
Like other physical phenomenon, heat exchange is also accompanied by spacetime variation of temperature. Mathematically, t =f (X, y, Z, 'r)
(1.1)
dt =Om .t =f(x,y,z ) dr
(1.2)
However, for the analysis of heat transfer problems, two types of heat transfer are consideredsteady and unsteady. In the case of steady state, the temperature at each point of the system remains constant in the course of time, and it is a function only of space coordinates. Mathematically, During such a process, heat exchange rate is constant and there is no change in the internal energy of the system. Some examples of steady state arerheat exchange from combustion products to water in the boiler, heat exchange process in all type of heat exchangers, etc.
In the case of unsteady state heat transfer, the temperature at each point in the system changes with time. This results in the change in the heat exchange rate with time. The internal energy of the system also varies. Mathematically,
(1.3)
dt  =I: 0, t = (x, y, z, r) dr
Examples of unsteady state heat transfer are the temperature variations of walls of building in a day, water cooling system of internal combustion engines, etc.
Besides this, the heat transfer may be in one, two or three crrections. This depends upon the configuration of the system considered. Mathematically, the above phenomenon may be expressed as follows:
One dimensional heat transfer Two dimensional heat transfer Three dimensional heat transfer
Steady
Unsteady
t =f(x) t = f (x, y)
t =f (x, i) t = f (x, y, r)
t = f (x, y, z)
t = f(x, y, z; r)
It is worthwhile to mention that in the beginning of any heat exchange process, it is unsteady but after certain time it becomes steady if all the governing parameters remain the samy: If governinip�ramcters vary, it remains unsteady. •
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I
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/
6
Heat and Mass Transfer
1. 7. Conservation of Energy
Since the subject of heat tramsfer is an extension of thermodynamics in that it considers the rate at which energy_is transported, hence it is useful to consider this law. This law may be applied in heat transfer by two approaches. They are as follows :
(a) Conservation of Energy for a Control Volume.ln a control volume approach (Fig. 1.1), it may be stated as follows:
The rate (lt which energy (thermal and mechanical) enters a control volume minus the rate at which this energy leaves the control volume must be equal to the rate at which this energy is stored. This may be written mathematically as follows:
(1.4)
Eout
''
Est
 
   
Fig. 1.1 Control Volume for Conservation of Energy.
(b) The Surface Energy Balance.In many situations, one has to apply the conservation of energy requirement at the surface of a medium. fa such cases, the control surface does not include mass or volume (e. g. wall exposed to air) hence generation E g and storage item, Es, do not exist. Mathematically, The above relation is valid for bothsteady and unsteady cases.
(1.5)
1. 8. Basic Law of Heat Conduction
The fundamen� law of heat conduction was proposed by the French scientist, J.BJ. Fourier in 1822. With reference to Fig. 1.2 one dimensional form of Fourier's law for solids states that the rate of heat flow by conduction in material is equal to the product of · three quantities namely:I;
(a) k, a property of the material known as thermal conductivity. (b) area A, normal to the direction of flow. (c) Jtl Jx
the temperature gradient at the section i.e., the rate of change of temperature t with respect to the distance x in the direcLion of heat flow.
Mathematically, the conduction rate equation dQ = kA � ax. di
(1.6 a)
Introduction
7
In the above equation, negative sign has been arbitrarily affixed to make Q positive because of the fact that flow of heat always occurs in the direction of decreasing temperature, ie, the temperature gradient dtldx will always be negative. In the case of steady state heat conduction, we have simply dQ = Q = q and (� ) = !!:!... r ox dr dx
, hence A
q
dt q =  kA dx
Fig. 1 .2. Linear Heat Conduction.
w
(1.6 b)
Here, it is assumed that the·heat flow in y and z directions is zero and the temperature in any plane prependicular in the x axis is uniform throughout the plane. The heat transfer per unit area, q " known as heatflux is given by dt q = kq* =A dx
W /m2
(1.6 c)
To find the steady state linear heat conducted through a uniform slab is which area A does not vary with x as shown in Fig. 1 .2, equation ( 1 .6b) may be intergraed with the help of separation of variables. Assuming k uniform q
f
.x2
xi
dx =  kA
or q (x2  xi ) = kA (t1  tz)
f
' '1
2
dt
kA (t1  l z ) q =  Xz  X1
If (x2  x1 ) = L = thickness of the slab, then
w
(1.6 d)
Heat and Mass Transfer
8
Note that f or this case the temperature function t(x) in Fig. 1 .2. is linear.
In the transient state of heat conduction, the quantity of heat entering the leaving a volume element of the body is not the same at any given instant According to First law of thermodynamics this difference is used to increase the internal energy E of the element. Thus the internal energy change is given by
ar ar
....... ( 1 .7)
dE = c p V  dr
From equation (1.6 d), it is possible to find the unit of thermal conauctivity. If q is in W, L __is in m and A is m 2 and (t 1 t 2) is in °C or K as it is temperature di,fference, the unit of thermal conductivity is given by '
k = 1
qL
A {t1  t2 )
expressed in (W ) (m)
ie! k is expressed in WlmK or Wlm °C
{m 2) IK )
w
mK
In MKS unit, it is in kcal/hm °C
Note'that k is not necessarily uniform for any one material but may vary depending upon the temperature, pressure and humidity. For moderate pressure level, the effect of pressure is small. The structural characteristics of the materials lead to expect that the thermai conductivity will decease in the following orders. (i) (ii) (iii) (iv) (v)
Pure metals Alloys Nonmetallic crystalline and amorphous substance Liquids Gases.
1. 9 General Heat Conduction Mechanism and Law
If we are interested to consider the conduction process in general, including conduction in very dense gases as well as in liquids and solids, we have to take into account the additional and differe!lt microscopic transport mechanisms. A classification of common mechanisms together with the materials in. which each mechanism is important is listed below :Degenerate electron gas diffusion (metals): Elastic phonon diffusion (all solids). , Classical theory for common liquids. Classical gas, molecular diffusi on · (gases, porous solids, electrons in semiconductors). (v) Radiation, electromagnetic photons (internal emission and reabsorption, perhaps continuous spectra).
(i) (ii) (iii) (iv)
It is worthwhile �ention that the �acr�scopically observed conduction characteristics of any material is the sum of all above effects (pl� others) which may be in operation.
Introduction
9
It is possible to discuss the transport of these microscopic mechanisms in terms of single general model, as follows.
Let us consider the x direction in a material in which an energy gradient (dtldx) exists. As shown in Fig. 1 .3 , this model assumes a number of points in the x direction, A apart (small in size), where A is equal to the distance over which an exchange by any mechanism is essentially complete. A would be the mean free path in a gas or absorption length for radiation. Let e 1 and e2 be the energies per unit volume at two adjacent points. Let v be the velocity of energy transport The net rate of energy flux in the positive x direction is given by de qx = ve1  ve2 =  v (e 2 =e 1 ) =  v H
dx
1
11,
dt = c 11,1 v dt 1 de v =  11, dx dx dx
where c = relevant specific heat of the material. The local heat flux is given by "
qx =  k dt dx
Comparing the above two equations, one gets k= C AV
(1.8)
Equation ( 1.8) shows that k is the product of c, i.. and v. Table I.I shows these quantities for mechanisms I to 5. x
Fig. l .TModel for Conduction
It is worth mentioning that theproduct of the three columns in each case, indicates how transport of energy by each ·mechanism de�nds upon physical constants and upon temperature. The net temperature dependence of tbermal conductivity is indicated by the last column.
1.1 O. Basic Law of Convection The rate of heat transfer b1,,convection between a solid surface and a fluid may be computed by Newton's law of cooling (Fig. 1 .4) as follows. • (1.9) q = h A (t5  t00) or q" = h (t.  too)
where
q h A ts too
= rate of heat transfer by convection, W = average �onvecti.ve heat transfer coefficient, Wlm2 K = heat transfer surface m2 = temperature of the solid surface (wall), °C = temperature offluid fi:ee stream, °C
area,
q' = heat flux rate, ·WJm2·
J.'.
Heat and Mass Transfer
10
Table 1.1
Case
A,
I.
2.
cT 3
(A VsoTh a)
f ume only convective heat transfer through the fluid layer and no contact resisThe overall heat transfer coefficient is given by 1 I 1 = U= = A 'f. R A R R R, R R L2 L3 I l [ s+ 1 + 2 + 3 + ] R ' +T+h hs + A I + k I I 2 1 267 W!m 2K = I __ 1_ .0.2x102 0.7x102 0.0017 + 580 + 384 + 58 5i65 +
(a)q
= UA(t.\· 1) I
c�
Ans.
267x3200x 10 (11175) '=3075 W 4
(b) The temperature drop across the scale is given by q = or
M dt r= R. scale
flt q/A=AR
I
or (c)
Where
I
=
t!..t 0 •0017
flt = 3�75 X 0.0017 = 16330C 3200 X 10 4 q = u1 AV.  t3) 0.2 x 102 00017 + 384 4565 + l
Ans. =529.6 Wlm 2
or
307f" 529.6 x 3200 ;x 10 4 (116 t3) Ans. or t3 = 97.85°C Problem 3.4. A �•;di is constructed of several layers. The first layer consists of masonary brick (k = 0.66 Uti'mK) 25 cm thick, the second layer of 2.5 cm thick mortar (k = 0.7 W/mK). tile third layer of 10 cm thick limestone (k = 0.66 W/mKJ and the outer layer consists of 1.25 cm thick plaster (k = 0.7 WlmK). The heat transfer coefficient on the ir:'~ ·'er and exterior of the wall fluid layer are 5.8 Wlm2K and l l .6 W/m2K respectively. ; ,nd (a) the overall heat transfer coefficient from the air on the interior to the air at the exterior on the wall, (b) the overall thermal resistance per m2, (c) the rate of heat transfer per m2 if the interior of the room is 26°C while the outside air is at a temperature.of70°C and (d) the temperature at the junction between the mortar and the limestone. Solution. Refer to Fig. 3.10. Assume only convective heat transfer through fluid film layer itnd no thennal contact resistance. The overall hPat transfer coefficient is (a) V=•=I LI L2 L3 L4 l A r, R +++++ hi kl k2 k3 k4 ho
oneUimentional SteadyState Heat Conduction without Heat Generation  =I 1 25 x 102 2.5 x 102 IO x 102 l .25 x 102 0.7 + 1 l .6 5.8 + 0.66 + 0.7 + 0.66 +
= 1.1871 Wlm2K
I AU =1 X l.]87}
(b)
The o�erall thennal resistance = t R =
(c)
= 0.8423 m2K/W q = UA (t;  t) = 1.1871 x 1((26 ( 7))] = 39.174
+ 25 + 2.5 �10+1.25+
�
t,
R,
1.z
Ri
l.J
�
t4
�
ls
R.,
W!m2
57
Ans.
Ans.
1c
q + O,NV\1'Q.JV\NV\r.M,0.....AA�.J\NVV'O..,I\N,rO + q
· Fig. 3.10
or
I +1 (1 (7)  __l__ 39.174 = _ _ _ _ _ _2  3 +  _ _ __2____ 0.2555 l Ox l0 1.25 x l 0  0.66 x l + 0.7 x l + 11.6 x ·]
Ans. or t3 = 3 °C Problem 3.S. In a manufacturing operation, a large sheet of plastic, 1.25 cm thick, its to be glued to a 2.5 cm thick of cork board (k'= 0.046 WlmK). In order to_ affect
Heat and Mass Transter
58
a bond, the glue is to be maintamed at a temperature of 43°C for a considerable period of time. This is accomplished by a radiant heat source, applied uniformly over the surface of the plastic (k= 2.27 WlmK). The exposed sides of the cork and of the plastic have a heat transfer coefficient by convection of 11.6 Wlm2K, and the room temperature during the operation is 20°C. Neglecting heat losses by radiation from the sheets, calculate the rate at which heat must be supplied to the surface of the plastic to obtain the required temperature at the interface. The thermal resistance of the glue may be neglected. Also draw the thermal circuit for the system. Solution. Refer to Fig. 3.11.
Fig. 3.11. It is to be noted that both sidescork and plac;tic are exposed to room air. The plastic side is exposed to radiant heat source. A part of this heat supplied say q1 will transmit, through the plastic, glue and cork and finally it will be convected to air. Another part of this heat supplied s_�y q2 will directly convect to room air in the opposite direction of q 1 because of high temperature of plastic layer and low temperature of room air. t  ta t2  t3 t,  t2 =3 ql . = = R1 R2 Ra t2t3 t, t2 = 13 ,a or q = = , 1/h0A L/k1A L/k.zA
43  t3 13 20 or q1 = 0.025/0.046 X 1 1/11.6 X 1
(43 t3 ) x 0.046 = (t3  20 ) x 11.6 x 0.025 _ or '2 = 23.14°C or
OneDimensional SteadyState Heat Conduction without Heat Generation
59
Using the value of t3• the value of t1 can be calculated.
t i  43 43  23. 14 = 0.0125/2.27 X 1 0.025/0.046 X 1 or t1 = 43.43°C
Now q 1 = (23.14  20) x 1 1.6 = 36.424 W!m2K
Ans .
q2 = hA (t1  ta ) = 1 1.6 x 1 (43.43  20} = 27 1.7 88 W/m 'l/(
Thus the total heat supplied by radiant heat source
2 Ans. q = q1 +q2 = 36.424 + 27 1.7 88 = 308.212 W/m K Problem 3.6. The walls of a refrigerated van consist of 2.5 mm of steel sheet at the outer surface, 16 mm plywood at the inner surface and 12 cm of glass w90I in between. Estimate tAe heat loss and the capcity of the refrigerator in tons of refrigeration if the inside temperature is maintained at  16°C. The outer side surface temperature i., 32°C. The total surface area of the van is 25 m2. Assume k (steelsheet) = 23 WlmK, k (plywood) = 0.06 WlmK, k(glasswool) = 0.06 WlmK Solution . Refer to Fig. 3.12.
q
=
Where
t;  t,,
�
(l)
l
L L2 L3 "i:,R = A [ k,_ + k + k3 2 1
2 _!_ 0.25 X 10 + }2 X }9 ? + } .6 X 10l} = Q.0986 _  (25) [ 0.6 0.06 23
Fig. 3.12
Heat and Mass Transter
60
q
=
32  ( l O} 0.0906
= 463.57
W
Ans.
Since 1 ton of refrigeration = 3.48 kW
!·;�489
9
6 Capacity = � 1
Ans.
= 0.132 ton
8
Problem 3.7. A composite insulating wall has three layers of material held together by a 3 cm diameter aluminium rivet per 0.1 m 2 of surface. The layers of materials consist of 10 cm thick brick with hot surface at 200°C, 1 cm thick timber with cold surface at 10°C. These two layers are interposed by a third layer of. insulating material 25 cm thick. The conductivities of the materials are : k (brick) = 0. 3 W!mK k (insulations) = 0.12 W!mK k (wood) = 0.175 W!mK k {aluminium) = 204 WlmK Assuming one dimensional heat flow, calculate the percentage increase in heat (B.U. 1 77) transfer rate due to rivets. Solution. Refer to Fig. 3.13. Asswne no thermal contact resistance at the interface.
9
9
Brick
I n sulation
t, ° =200 C
Without rivets :
Wood
t4 • =10 C
Fig. 3.13
· L 1 [10 x 10 2 25 x 10 2 I x 1 0 2 'f. R = (l.) [ I + L2 + L3] __ + ] = 22.48 KIW A k1 k2 k3 = 0. 1 0.93 0. 12 . 0.175 The heat transfer rate is given by ti  t4 200.10 = =8.45 q = 22 .48 LR
w
OneDimensional SteadyState Heat Conduction without Heat Generation
61
With rivets :
Rivet area = (,r /4 ) (0.03)2 = 7 .0685 x 104 m 2
L rivet (10 + 25 + 1 1 102 = 2.496 R rivet = �   = k riva • Anver 204 X 7 .0685 X 1 Q4
K/W
Area of the layers with rivets area excluded = 0:1  0.0007 0685 = 0.099293 m2 The thermal resistance due to rivets is in parallel with the resistance of the composite wall. The conductivity of the materiaf for rivets is very high thus it provides high conductivity path. Thus the equivalent or effective.thermal resistance is 1_ 1_ + _ _1_ = _ Rrivet RMC111 Requiv 1 x Area with rivet . + _1_ ='1:,R Area without rivet Rrivet
1_ 1 _ X 0.099293 + _ =_ 22.48 0.1 2.496 or R equiv = 1/0.4448 = 2.248 K/W
t1 t The heat flow rate = q = 4 = 200 .  10 = 84.5 19 2.248 Req...,
W
The percentage increase in heat flow rate due to the presence of rivet
= �4.519  8.45 X 100 = 900 per cent Ans. 8.45 This example illustrates that where to reduce the loss of heat is the major problem. the use of rivet to hold the various insulating material should be avoided. Promlem 3.8. A furnace wall is made of inside silica brick (k = 1.86 W/mK) of thickness 10 cm and outside magnesite brick (k= 5.8 W/mK) of thickness 20 cm. The temperature on the inside surface of the wall of silica brick is 900°C and outside magnesite brick surface temperature is 150°C. Calculate the heat flow through this composite wall. If the contact resistance between the two walls is 0.00257 m2KIW, calculate the temperatures of the surfaces at the interface. Solution. Refer to Fig. 3.14 The thermal resistance of silica brick for unit area 2 =R, = �= 10 x l0 = 0.0537 m2 K /W k1A 1 .86 x 1 The thermal resistance of magnesite. brick
= R2 = !::!:._ = 0.2 = 0.0344 m2 K/W k2 A 2 5.8 X 1
The total resistance including contact resistance = I:.R = 0.0537 + 0.00257 + 0.0344 = 0.09067 m2 K/W.
Heat and Mass Transfer
62 The heat transfer = q = Also or
750 900  150 = 8271.75 W/m2 = 0_09067 :E R 900  t1 t2  150 = = 8271.75 W/m2 q= � RI ti = 900  q RI = 900  8271 .75 x 0.0537 = 85�.8°C
And t2 = 150 + q R2 = 150 + 8271.75 x 0.0344 � 434.5 C _ The temperature drop at the interface = t1  t2 = 455.8'  434.5 = 2 l .3°C °
Am.
Ans. Ans.
900•c
1 so·c
Fig. 3.14 Problem : 3.9. A furnace wall consists ·of steel plate 15 mm thick (k = 17.5 W/mK) lined on inside with silica bricks 160 mm ·thick (k = 2 W/mk) and on the outside with magnesia brick 210 mm thick (k = 5.3 W/mK). The inside and outside loss from the wall are maintained at 700°C and 130°C respectively. Calculate the heat loss from the wall for unit area. If the heat loss � reduced to 2900 W/m2 by providing an air gap between steel and silica bricks, calculate the necess!ll'Y width of air gap if the thermal conductivity of air may be taken as 0.035 W/mK. Solution. Refer to Fig. 3.15. The total resistance excluding the thermal resistance due to air gap is . 1 LI £ L3 t R = R1 + � + � =  [ + 2 + ] A k1 � k 3 .
=
.
.
· l I .5 x l0 2 l6 x l o 2 2l x l0 2 2 [ + 5 .3 ] = o· l204 m K/W 1 11.5 + 2
Heat loss from the wall exfluding air gap is q=
At 700  130 _ =  4734.2 W/m2 0. 1204 tR
Ans.
OueDimensional SteadyState He:at Conduction without Heat Generation
63
In order to reduce the heat loss to 2900 W/m2• the thermal resistances to be increased by providing air gap. The new thermal resistance would be }:.R new
700  130 29()()
q,,_
=�=
= 0.1965
rri2 K /W
Hence the thermal resistance of the air gap is
Rair
=
Lf?n�w  LR
=
or L = thickness air gap
0. 1965  0.1204
=
0.07 61 x 0.035
= =
0.07 6 1
2.663 x
L L =  = kA
103 =
0.035 x 1 2.663 mm Ans.
St�e l
� Li .f U Lr+ L 3 j
�,__ �� t2 t2 t t4
f1
3
Fig. 3.15.
Problem 3.10. A proposed selfcleaning oven ,.lesign involves use of a · composite window separting the oven cavity from the room air. The composite consists of two high temperature plastics. The thickness of plastic exposed to the interior of the oven is twice than that of the face exposed to the room air. The thermal conductivity of interior plastic is 0.15 W/mK whilst that of the other is 0.08 W/mK. during the self cleamng process, the oven wall and enclosed air temperature are maintained at 4300C, while the room air temperature is 30°C. The inside convection and radiation and outside convection heat transfer coefticients are the same and equal to 25 W/m2 K. Calculate the minimum window thickness required to ensure a maximmn temperature of 55°C at the outer surface of the window. Solution. Refer to Fig. 3.1 6. Assume thermal contact resistance to be zero. The inside surface of the plastic is heated by convection as well as by radiation. The heat transfer rate � ta  too
'3  t .,.,
 q = R0 LR
Heat and:� Transfer
64
But Ard
a
D
= R• +R 1 + R2 + Ro
_1_ = 1 + 1 = 1 + 11/h,. A 1/h; A R ..,..;v R ,. R 
R
l]
. = _!__ [_!_ + ...., A h r h;
l
L2 + 1 L t + 1 1 [ +  + ho A k A k h; hr 1 2A
1 l:R A
i,, . =430 C ·
0
,::=:177';'j7"T""'�
.. .t.a. ;:41o·c · ·:1. ; · h · · · . . · f'.. ·• . . ' . . . . ,�·: hr ·.· ·. · '1r · �: . . • · . • ' . . • . i::·
1//'. tI1//,,
· . · ·','• q. conv. :+ .:" . 1. Oven cavity · f:' · • · . · . . • · 1·: ·,......
6 3 or .!1. x 0255 = [ (0.002 �  10 t ) = 0.001 A 2 , 3 !> or
boo2 5(>2)  1036 {3oo3  50 )
.!1. = 308 W/m 2 Au. A The average value of conductivity may be calculated based on the value of q/A. =  k dt Now, dx A .!L ) or 308 = k (300  50 0.255 Jc = 0.31 4 W/mK.
Ans. Problem 3.29. A wall is made cf material whose thennal conductivity varies with temperature as follows:or
k = /co ,2 If the thickness of the wall is L. find the expression for steady state heat conduction through a wall per ttnit area if the two surfaces are maintained at different temperamres t1 and t2 •
If the heat transfer rate is expressed as the product of temperature difference and mean thermal conductivity divided by L, at what temperature must this conductivity be · calculated so that such equation gives the right result. Solution. Ref. to Fig. 4.28 • We have
.!1. A
=
 1' !!_ dx
dx
=  1co , 2 !!£.
3
OneDimensional SteadyState Heat Conduction without Heat Gcneratioo !L A
JL dx
A
3L
o
!I = �
or
J'2 t
=  ko
1
2
1
89'
dt .
(tr  ti )
Ans
x =O
t1
Fig. 3.28 If
q =
k,,. (tI L

lz )
then 3
ko t1 { k,,. = 
3
or
t,,, =
3  tz
t1  tz
) = k,,
2 t,.
t?  tf
3 {t1
 _ tz )
Problem 3.30. The wall of a steam boiler furnace is constructed with 1 20 mm of foamed fireclay and 490 mm of red brick. The wall temperature inside the boiler furnace is 1050°C. and the wall outside temperature is 45"C. The thermal conductivities of thewall materials are k (fire clay) = 0.3 (1 + 0.0008 t)
W!mK
k (red brick) ' = 0.7 WlmK. Find the heat lost per unit of the furnace wall and the temperature at the layer interface.
90
Sr,/ution. Refer to Fi.g. 3..29
We bave,
t 12 But k1 = k 0 {1 + at ,,. ) = 0.3 [ 1 + 0.0008 { i ; }]
\ 3 X 10120 _ Ri = 105 + t2 � 0.3 [ 1 + 0.0008
l
3 3 R z = !::..!:_ = 490 X 10 = 490 x 100.7 0.7 x 1 k2 A
+120mm•I·
49o m m � .
Fig. 3.29 Now, putting these values in the expression of heat conduction through the wall and brick layer, we get {t2  45) (1050  45) q _;=     = 490 X 103 1 20 _ 10_ __ 3 _ + 490 X lo3 _x ____ 0.7 0.7 ( 1050 + lz 0."3 [ 1 + 0.0008 2
)l
This results in a quardatic equation and solution gives tz = 820°C
OneDimentional Sttady•State Heat Condi.lction without Heat Generation q
=
820  45 490 x 10 3/0.7
=
1 107 Wlm2
91 A,rs.
Problem. J.31. A furnace wall of 25 mm thick is buik up of a layer of frreclay brick whose thermal conductivity may be taken as k = 0.84 ( l + 0.695 x 10 31) where k is in Wlm K and I is in °C. _ Calculate the loss of heat per unit area and the wall surface temperatures, if the temperature of the gas in the furnace is 1200 °C and the room temperature is 30°C. The local coefficient of heat transfer from the hot gas to the wall is 30 W/m2K and the local heat transfer coefficient from the wall to surroundings is 10 W!m2 K. Solution. The above problem may be solved easily by the method of successive approximations (also known as iteration method). Assume a mean wall temperature of t = 65 0°C. At this temperature, the mean thermal conductivity of fire•clay is k,,, = 0.84 (I + 0.695 x 103 x 65 0) = 1.219 WlmK. The overall heat transfer coefficient is given by
The heat transfer rate is given by q
=
Uo A ('I·  t�\ = 2.95 5 x 1 (1200  30) = 345 7.35 W/m2
1200 � ,, I/hi
t  1, Ri
i q =  = �
Now
=
3457.3 5 Wlm2
or or
t2
= 10 + q �
0
= 30 +
34;�35
= 375 .73 5 °C
83 7 5 375 73 5 = 730.24"C Now, the new mean temperature of the wall = 1111 = l0 · ; ·
At this 1111, the new thermal conductivity is
ic,,. = 0.84 (1 + 0.695 x 10 3 x 730.24) = 1 .266 W/mK The new overall hxat transfer coefficient is .
l U= · 1 0.25 1 30 + 1 .266 +
w
3.03 5 Wlm2 K
j; ·1
�. :
Heat and Mass Transfer
92
The. heat transfer rate is ' q = UA (t;  ·ti) = 3.0235 x 1 (1200  30) = 3537.49 W/m2 Now using the new value ofq, the wail temperatures may be given as
,, = 1200  35�49 = 1082.080(: 35 49 12 = 30 + ��. = 383.749°C 383.749 1 . The new mean temperature ofthe wall = tm = 0 82 + 732.91°C 2 And,
km = 0.84 (1 + 0.695 x 10 3 X 732.9) = 1.2678 W.'m K
And,
.1 = 3.0263 Wlm2 K. U= l l 0_25 + + 30 1.2678 .10
And,
'
,;
.
· q = 3.0263 x 1 (1200  30) = 3540.77 W!.l
and
3 1, = 1200  �·77 1081.97°C
and
12 = 30 + 3 �·77 = 384.01°c
Since the variation in k with previous value is much less and hence the new values may give ·correct value of k111 = 1.2678 WlmK and there is no need of further calculations. Problem 3.32. The inner and outer surface temperatures of a plain brick wall of dimension 6 m long x 4m high x 300 mm thick are 900°C and 30°C. Calculate (a) the average thermal conductivity (b) thermal resistance and (c) the heat loss from the wall. (d) also, calculate the temperature at 150 mm distance from the hot surface. Assume k (brick) = 0.9 ( 1 + 0.001 t); Wlm K where t is in °C Soiution. (a) The average thermal condn�vity is 1ft
0 1 k., = ko (1 + at. ) = 6.9 [1 + � (900 + (b)
R=
f� = 1.31'850.3(6 IVI
X
X 4)
3e)] = 1.3 135
= 0.009,4.8 KIW
Wlm K
AIIS.
Ans. 900  30 · 3 = 0.00948 = 91.77.2 X. 10 W 'f = (C) (d) The average thennal conduc�if¥ based on hot surface temperature 900°C and 1;  1o
r
OneDimensional StfadyS1me Heat Conduction without Heat Generation
93
temperature t at a distance of 150 mm is km
= ko
(1 +at m )
=
0.9 [ 1 + O.�l (900 + t )]
The heat conduction through this section is q
4 t = 
R
900  t
0. 15/k,,.A Since under steady state conditions, heat passing through each section of the wall is the same, hence =
900  t_ __ } 91 . 772 X 1()' = ___(_ (0. 1 5) 0.9 [1 + 0.0005 (900 + t )]
or 9 1 .772 x 103 x 0. 1 5 = (900  t) [1 + 0.0005 (900 + t)] x 0.9 or 0.0005 t2 + t  1334.7 = 0 or t = 291 °C
3 . 1 2 . Heat Flow through Cylinder at NonUniform Conductivity.
Ans.
3 . 1 2 . I Inner and Outer Temperatures Known.
Consider a long tube whose inner and outer surface r1 and r2 are maintained at uniform temperature t1 and t2• We are interested to examine the temperature pmfilP. in this tube when the thermal conductivity of the wall varies linearly with the temperatqre as k = k0 (1 + at) k0 = thermal conductivity at 0°C
Where
a = a constant which is negative for metallic conductors and positive· for insulating materials
The governing equation for unidirectional heat flow through a tube without heat generation is given by !!_ [r k dt ] = 0 dr dr
Intergrating the above equation, we have r.k dt dr
or
=Ci
Ci dt = k0 (1 + at ) 
dr
r
Intergrating the above equation again, we get t 2 ) = C1 In r + Cz /co
(,\ +; �
 ,
� 
(353)
Heat and Mass Transfer
94 The boundary conditions are·:At
and
r = r1 , t = t1 and r = r2,
t "" t2
(3.54)
� (ti . + � tf ) = C1 In r + C2
}
(3.55)
� (12 + � ti = C1 In 72 + C2
S ubtracting the equation (3.55) from (3.�). we get
c, In ,,,1,, 1 = ko [(,, + � ,,,
where
1 1,, ,q l +
�
= � (t2  t2 ) [ 1 + � (t1 + tz) ] =· /co {t1  tz ) [1 + t,,. =
a
Gn ]
(t1 + r�2
_ ko (t1  t2 ) ( l + a tm} . .. Ci In (r1fr2 )
:. C2 = � (t1 + � ti2)  C1 In r1
_ ( � ) k
0 (t1  1i) \1 + y ,,, _
,,, [ 1
qg  qgO

r (
ro
)
2]
where r0= outside radius q;o = heat generation rate at the centre line.
Find the drop in temperature from the centre line to the surface of 3 cm diameter rod having k (cylinder)= 22 W/mK, q';, = 12 x 1 05 W/m3• Derive the formula used.
4.7. A copper .wire 3 mm in diameter in insulated with plastic with 3mm thickness. Calculate the maximum current carrying capacity of the wire without heating any pa.rt of the plastic above l 20°C. Take k (plastic)= 0.35 W/mK, k (copper)= 330 W/mK Electrical conductivity of copper = 5.1 x I 06 ohm1 cm1 Assume the temperature of plastic is 50°C and h (plastic to atmospheric) = IO W/m 2 K.
..
150
Heat and Mass Transfer
4.8 A coil 4 cm thick and 40 cm inside diameter is placed in an oil both. At a current of 0.�1 amp and voltage of 0.15 volt ·the surface temperature remained unchanged. Calculate the maximuin temperature in the coil when the current was 3 amp. 4.9. Prove that the steady state temperature distribution in asphere in which heat is produced at a uniform rate per unit volume q/' is given. t.  �
= q�j [ I 
(f. r]
· •Where rO and t� ai:e outer radius and outer surface temperatures respectively.
4 .10. · The average heat production during ripening of oranges is estimated to be 1300 W/m2 . If the orange is assumed to be homogeneous having k = 0.6W/m2K calculate the temperature of the centre of the orange and the heat flow from the outer surface. Assume the diameter is 9 cm and · that the external surface temperature is 12°c. . 4.11. The inside dimension of oven made of brick (k = 6 W/mK) arc 2m x 1.5 m x 0.5 µi. The wall is of 12 cm thick. Calculate the heat loss per hour if the inner and outer surface temperatures are 500°C and HX)°C. 4.i2. ·The outer surfaces of a flat plate electric resistance heater 1.5 cm thick, of effective k = 16 W/mK ;ire at 350°C. Determine the temperature at the centre of the heater if the electrical power input is 1200k W/m3 • Neglect the heat flow through the edges. 4.13. A heating element is constructed from carbon in the shape of bar, 8 cm wide and 1.5 cm thick and 1 m long. A potential difference of 12 volts applied at the end of bar maintains a uniform surface temperature of 800°C. Calculate the temperature of the _centre of the bar. The electri, the heat transfer from the fin in actual case reduces to the form of the equation when the end is insulated. The error which results kom this approximation will be less than 8% when
(h/)
½
=½
S . 2 . Conduction Cooling of Turbine Bladings
· The gas turbine blades are subjected to high temperature, high centrifugal stress, gas bending, vibratory and thermal stresses. It is established fact that the cycle efficiency increases with the increase of maximum cycle temperature. But the highest cycle
Heat Transfer from Extended Surfaces
163
temperatur� is limited ?Y the quality of turbine blade materials and it is possible that due to overheatmg the turbme blades may fail. Thus it is essential to cool the turbine blades to prevent the failure of blades due to overheating.
.� I I
Tip
I
Root
Fig. 5.7. Conduction Cooled Turbine Blades. One of the method of cooling the turbine blades is the conduction cooling. It is indirect method of cooling of blades by conducting heat from the blades to an internally cooled turbine rotor. The cooling of turbine blades problem comes under the the category of extended surfaces. Consider a turbine blade of length L shown in Fig. 5.7. Assume that the blade cross section area A and periphery P are the same at all spanwise stations x, the unit surface conductance h and effective gas temperature t are both uniform along P and and the temperature gradients in all blade sections are negligible. These assumptions make the problem one dimensional and it is the case of an insulated end (i.e. case ii) Following the method of solution for the insulated end, the temperature distribution is given by t  too cosh m (h : X ) 0 cosh m L t,  t oo 0, Where
=
   mL__
✓ (k �� �Lr � {{:: f
This equation shows that for a given gas and root temperature, the blade temperature itself depends upon the dimensionl�ss parameter;s mL i.e. kjki. Fig. 5.8 shows. the typical blade temperature profile.
It is obvious from the figure that the value of mL should be low. Low value of ml required to increase the effectiveness. of conducti9n cooling � be oh1 .. i ncd by either decreasing the blade surface conductance ks = h PL or in�reasing the blade internal conductance ki = k AJL.
Problem 5.1. One end of a long rod is placed into a furnace while' the o�er end is projected into atmospheric air. Once the steady state condition is reached, tfle temperature of the rod is measured at two points 7 Cllf,'11fJar.liand found to be 120° C and 85°C when the alJ!1ospheric temperature is at 20°C. Calculate. llie)hennal conductivity of the rod if the diameter of the rod is 2.5 cm in diameter and heariran sfer coefficient Ii is 20 W/m2 K.
Heat and Mass Transfer
1 64
Solution. Assume the end of fin insulated. For insulated end, we have
t = .!!._ = cosh m (L  x ) _t__ cosh mL l?. 00
_
750
I
450 3 5 0 12 0.0
0. 4
0.2
x/L
0.6
0.8
1.0
Fig. 5.8 Longitudinal Temperature Distribution in a Conduction Cooled Turbine Blades. ,.,,
.
Assume that the first point where the measured temperature is 120° as the starting point. Hence the point w� ere the temperature is 85°C is at x = L. Hence l£,  t oo
or or
or
or or or
=
ts  t oo COSh mL
85 _ 20 = 120  20 cosh mL
l UU cosh mL = = 1 .537 65 mL = I
1 h.tr d =2
L
k. (tr /4) d 2 2 _ k _ 4h L d
_
r
4 x 20 x xHr = 1 5.68 W/mK 4 2.5 X 10
Ans.
Heat Transfer from Extended Surfaces
1 65
Problem . 5.2. An aluminium alloy fin 3.5 mm thick and 2.5 cm long protrudes from a wall. The base is at 420°C and ambient air temperature is 30°C. The heat transfer coefficient may be taken 11 W/m2 K and thermal conductivity k is 200 W/mK. Find the heat loss from the fin per unit width (depth) of mate.rial.
Solution. Refer to Fig. 5.2 (a) For insulated end, For rectangular fin
../(h P Ac k )
=k
Ac m = k 2 8 bm
qfi,. = k.2 8.b. m 0s tanh mL
m=
where
_
h = rrn )
'V \�/
f ( 11 x 2 x 1 000)
·v A
200 X 3.5
5.6 X 2.5 = 0.14 mL = 100
= 5.6 if 4 f, > r L >> H
2TCL
C o sh1 ( H/r )
2TCL
 l n(l/2Hl } I n ( I /r > {'1 l n ( L/ r )
J__ 
214
Heat and Mass Transfer
,� T2_

Vartical Cylind or ·more transport phenomena permits the analysis of each process by Analogus mathematical methods. We know that twodimensional Laplace equation is used to describe a potential field �or several phenom�na. This distribution in
Heat and Mass Transfer
218
tbe electrostatic and temperature field is given by
iE
ax 2
 +
and
t,
ax 2
+
iE
ay 2 2
0
a, 
ay 2
0
The analog field plotter is an instrument which is easily used to obtain field distribution described by the twodimensional Laplace equation.
This plotter uses a thin sheet of electrically conducting paper and to the scaled shape of the heat conducting object (Fig. 6.8). An electrical current flow pattern can be established in the paper by means of a suitably attached and energized electrodes. Boundary conditions corresponding to isotherms are obtained in the electrical field by attaching copper wires to the paper or by painting the areas with highly conductive silver paint and then attaching an e.m.f. source. The plain edges of the conducting paper correspond to insulated surfaces in the temperature field. Nutt d e t ec t o r
Po w e r s up p ly
Power unit
      , S l i d e r of I Vo ltage d i v iden ____p Leods
 1
i
1
I
I
I
I
l.
I
l
Fig. 6.8. Plotter
Lines of constant voltage are obtained by moving the stylus (pointer) across the paper, making small perforations in the paper wherever the null detector indicates that the stylus is at a specified voltage. The voltage level of the particular potential being evaluated is established by choosing a slider position on the voltagedividing potentiometer of the null detector. Selecting equal increments in voltage makes adjacent perforation lines ana1ogus to isotherms separated by the same temperature difference. As the constantenergy flow lines are orthogonal to the potential lines the free hand sketch •of this is possible and the result will be a network of curvilinear squares as obtained in flux plotting. The constant flow lines can also be traced constant by simply reversing the conducting and insulating portions of the boundary. The other types of analogus are (i) Fluid flow analogy
(ii) Membrane analogy
/
Two and Three.Dimensional Steady State Heat Conduction
219
6 . 5 . Finite Difference Method.
In multidimensional steady state circumstances, considerable mathematical difficulty arises and with complicated boundary conditions a useful solution by analytical methods are not possible. In order to overcome the mathematical difficulty various nu'merical methods have been cfeveloped which may be used to obtain approximate solutions. Among the numerical methods finite difference is one of them. Like all other numerical methods, this method is also particularly well suited for use with high speed digital computers. 6 . 6 . The Nodal Network and Finite Difference Form
In the analytical solution, the detennination of temperature is made at any point of i!}terest in a medium, whereas in a numerical solution the_detennination of temperature is possible at only discrete points. This suggests that the first step in any numerical analysis is to select these points. Suppose the medium of interest is a twodimensional heat conduction (Fig. 6.9). To select the discrete points, the medium of interest is subdivided into a number of small regions and a reference point of each is assigned at its center. The reference point is generally termed a nodal point or simply a node and. the aggregate of points is tenned as nodal network or mesh. For the general expression of a numerical approach to a problem, it is often convenient to adopt an indexing system which attaches a unit subscript to each point say i, j, k (tr, ;, , ) denoting x, y and z directions. For two dimensional system z tenn will be absent.
i+l,i i,j  1 (a}
t(x)
i+t I x +tt.x
j +1
I
/
L.,
'''' X
(b)_
Fig. 6.9. Two Dimensional Heat Conduction (a) Nodal �etwork (b) Finite Difference Approximation.
In this numerical method, an appropriate conservaton equation is written for each of the points in the nodal network. For any interior node of a twodimensional system with no generation and unifonn thermal conductivity, the exact form of governing equation (6. 1) is approximated in t,he finite difference form as follows.
220
Heat and Mass Transfer
Rectangular Coordinates : at I ax i + 1/2,j
;; Ii  1/l .j
_a2,
ax 2
i'j
=
=
t I + 1, J  t i,j
(6.10)
Ax
= t; + 1,j  ti1 ,j
a , 1ax L ia .j
(6.1 1)
Ax
Ax
a, 1ax L lll . j
(6.12)
Substituting equations (6.10) and (6.11) into (6.12), we get
Similarly,
it
ay 2
i,j
i ,j
=
= I; + !,j + t i  1,j  2t;,j {Ax f
ar I ay
li,j + w  a rtay l i,j  1a = t i, j + 1 + t i, j  1 2 Ay (Ay)
(6.13) 
2t i, j
If Ax = Ay and substituting equations (6.13) and (6.14) into (6.1) we get t;, j + 1 + t i,j  1 + t i + 1,j + ' i 1,j  4t ;, j = 0 or {t ;, i + 1  t;,A + (t i,j  1  t ;,i) + (t; + 1, i  t;,i ) + (t i  t,i  t ;, i } = 0
(6.14) (6.15) (6.16)
It is worth to note that each node represents a certain region and its temperature is a ml:!asure of the average temperature of the region, e.g., the temperature of node i.j. Fig. 6.9 may be viewed as the average temperature of the surounding shaded area. It is to be noted that the selection of the nodal points is rarely arbitrary depending often on matters such as geometric convenience and the desired accuracy. For higher accurary a large number of points are chosen.
Equation (6.15) is the approximate algebric equation a finite difference form of exact differential equation. This approximate, finite difference form of the heat equation may be applied to any interior node which is equidistant from its four neighbouring nodes. This equation reveals that sum of the temperatures associated with the neighbouring nodes should be equal to four times the temperature of the node of interest. Equation (6.16) is really a summation of the condution heat entering the node whose temperature is t;, i' This reveals that either equation (6.15) or (6.16) is derived from one of Kirchhoffs electrical laws written In its thermal analog. In other words the algebric sum of all heat flowing to a point must be equal to zero. Thus. UJ = O.
The equation (6.15) may also be obtained by performing an energy balance on an interior node. With the help of nodal energy balance, the finite difference form for node at a plane surface with convections (e.g. nodes 7 and 8 of problem 6.4) is given by Table . 6.2.
Two and Three Dimensional Steady State Heat Conduction
2h �x h. �x 2 t · . R  . 2 t 1·  l, J. + t 1, J. + I + t i,j  1 + t_  2 (+ = t, J ' k k ) •I
22 1 (6. 1 7)
It is to be noted that equations (6.15) are set of algebric simultaneous equations and it may be solved by many old and new methods. The truncation error can be pointed out by Taylor series about the pointes 'u
Cylindrical Coordinates : The finite difference equation (6.15) derived for two dimensional rectangular coordinate system cannot be easily used for cylindrical co ordinate system. A change of variable, however will reduce the cylindrical coordinate system to a rectaQgular system in the new variable. In cylindical coordinate (r, q,), the conduction equation i$
;·p,
dr 2
I i1r + I it = 0 t 
Let ( = In r, 1J = q, thus
r or
r2 d ,2
iJ dq, 2
(6.18)
I ot r2 d '
Substituting above equations in equation (6.18), we get = C2 cos A x + C3 sin h
Putting the values of I/> and 1/f in equation (7. 10), we get or
0 = "' 0
= Cr
1/f
= e A
2
2
e A a r .
cos A x + C3 s in A x
a r {B 1 cos A x +B 2 sin A x ) 2
or
(c2
B 1 e A a
r
=0
) (7.13 )
The second boundary condition is 0 = 0 at x = L for all r > 0 This gives B 2 sin A. L = 0
This is possible only if B 2 = 0 or sin AL = 0. If B 2 = 0, the solution does not give any meaning, hence sin A L = 0 or A. L = ntr where n = 1, 2, 3, 4, 5, The equaton (7.13) talces the form of 00
0 =
I, B
n= I
n
e  (mr JL fa r_ sin (n tr !L ) x
(7.14)
The above equation must satisfy the initial condition 0 = f (l) _at r = 0, which gives 0; (x) = f (x) =
I, B n
n= I
sin (ntr /L) x
(7. 15)
The right hand .side of eq'ii,ition (7.15) is a Fourier sineseries expansion of the function 0; (x) for which B ,. •s �xpressed as Bn
=
(t}
I:
0; (x) . sin n; x dx
Putting the value of B� in equation (7 .14), we get
(7.16)
248
• Heat and· Mass Transfer
(2)
oo  (n In L ) 2 a r . n 1t . sm  x 0 =  L e L L n=l
JL 6; o
(x ). sin n n x dx L
(7. 1 7)
If the plate is initially at a uniform temperature t; throughtout, then t ; (x )= t ; or 6; (x) 0; . Thus B,. is given by
=
L 26 n 1t n n •· J:.._ [  cos x ] x dx = sin L L nn L o
26 . ( = '  1  cos n tr )
(7.18)
ntr
For all even values of n, RHS of equation (7.18) beco�es zero and for all odd values of n,
( 1  cos nn ) = 2. B,. = 0 where n = 2, 4, 6, 8, ...........
Now equation (7.14) talces the form of ts ) = '_ � � 0, t , ts
= (i)1t ,.f= (!n ) e (nn/L ) ar . sin 2
or
(
where
n = I, 2, 3, ............odd number
1
n
,r x L
(7.19)
oO q =  k A � =  kA
Since
a:x
q (x, r )
Where
ax
= 4 (k: ) (ts  t; ) �/  (nn / L f a r __ cos n; x
(7.20)
n = 1 , 3, 5, ............ .
40 ' where n = I , 3, 5, 7 ....... . B ,. = nn
Intergrating equation (7 .20) we get the total heat flow across area A of the plane. L Q (x, < ) = (:,w1 ) (t, � t ; ) � (.1,) [ I  , (utL i' a . ,
•]
x cos "; x
(7.21) where n = 1, 2, 3, ........ . Fig. 7.6 shows the temperature history given by equation 7.19 and instantaneous heat rate (7 .20) for various stations in the plate.
Problem 7.8. In an industrial heatng process, a lar_ge plate having thickness 5 cm intially at a temperature of 40° C and suddenly its both surfaces are raised to and maintained at 500° C. Calculate (a) temperature at a plan_e 2 cm from the surface after 30 seconds of the sudden change in its surface temperature, (b) instantaneous heat flow rate
Unsteady State (Transient) Heat Conduction
249
across the surface after 30 seconds per m2, (c) total heat flow rate across the above plane from r = 0 to r = 30 seconds per m2• Assume a = 2 x 1 03 m2 /h, le = 40 W/mK.
'�
ut
� L/ = Fo 1 .0
1 .2
0.6
0.8
0.4
0.0
0.2
!:.� l+ 5, t1 t: .
0.8
4
1x +
0.6
5
3
� = 61
I 2
0.4
� I
'
cfo
0.0
l
0.2
0.4
F0 = � �2
0.6
�
0.8 .
�
I
� r:r
f l;:
0 1 .2
1 .0
Fig. 7 .6. Temperature History and Instantaneous Heat Rate for Infinite Plate with Negligible Surface Resistance.
Solution. Assuming negligible surface resistance (h ➔ oo ), the temperature distributions at a point x is given by t  ts  (4) � . n tr 0 _    L..  ei111tlL >2 a r . szn x t i  ts 1r n= l. 3 ___ : n L 0i
(1)
Here,
Now,
( }2 a r n;
nn  x L
=
nrc2 5
n 2 t?
( 51100 l
X
2 X IO 3 3600
X
30 = 0.0657 n2
(4) [e 0
Case 1111 t(_x,OJ = t; 11 k:1t1h�xco =Q o
(7.22)
· •�rm.._ X
·� �
C tlx,O) )=fi kit/hloo =h[taoltlo,t)J Case III I l
I+ X
l{��K_
!
·�
•• � ••• �.
0
,; ..._____ JI
Fig. 7.8. Transient Temperature distribution in a Semiinfinite Solid for Three Surface Conditions t(oo,r)=t8 forr>O
With the above boundary conditions of equation (7.22) is given in the form of Gaussian error function
'
Unsteady State (Transient) Heat Conduction 0 0;
 = elf (u ) = erf ( X I;  Is 2� t  ls
U
Where
=
253 (7.23)
)
X
2vTa�
The Gaussion error function is defined as erf (u) = or
/:,, (
eif (u) =
h f:
2 e u du
f 1\ •; + ;! \'
+ .... )
The values of error function is given in Appendix.
The instantaneous heat flow rate at any point of the semiinfinite solid is give by Ak(tst;) u2 e Qx = v(na 'f}
(7.24)
�W
The instantaneous heat flow rate at the surface (x = OJ is q 0• 'f (
) _kA (ts t; )
 ;===,.... v(na 'r)
w
(7.25)
Cumulative heat flow is given by
Q(O,'r )=2kAM (ts t;)
J
(7.26)
The general criterion for the semiinfinite solution to apply to a body of finite thickness subjected to one dimensional transient heat conduction is given by 2 0 :::?: 0.5 2�
Where 2 o is the thickness of the body.
Cylinder and Sphere. If cylinder and sphere are subjected to above type of heating and cooling, the temperature at the centre of cylinder or sphere of radius r O is given by.
(7.27) The values of functionerf (a r /ra2) for cylinder and sphere can be seen from Fig. 7.9. · Penetration depth and penetration time
. _ Penett�JiQ1f_qepth is the· locati�·.of a point where the change in temperature is within 1% diang�in���surface te!JlPerature, i:e.,
of the
Heat and Mass Transfer
254
t  ts = 0.9 Ii  Is
From the error function table, this corresponds to x /2✓(a f) = 1.8.
0.3
1>l"'
0.2
c:,
0.15
�
0 .,
�
0.08 0. 06 0.04 0.03 0.02 0.315 0 .01
I 0.05 0.2
0.3 0.4
0,6 �8 1.0
(lt
�
Fig. 7.9. Error integral for Cylinder and Sphere Penetration time is the time rP taken for a surface perturbation to be felt at that depth
in th{' range of 1% and is given by
d = 1.8 2Y{ai;f
or
fp
2
d. =
.(7.28)
13 a Case (b) Constant Surface Heat Flux t (x, r)ti =
'211 0" (a r I n )½ k
exp
Case (c) Surface Convection
(4
2)
:
t
q0" x  erfc k
(2 �)
(7.29)
(7.30)
or
t

Unsteady State (Transient) Heat Conduction
t;
(2 vra,rl[exp(� + h:�•)]
= erfc
f 00  ( i
Where erfc (u)
=
[erfa (,«,i,J)
I  erf (u) = complementary error function.
255
+ h
�l(7.J
Th� te'!lperature history rate in a semiinfinite solid with surface convection is shown m Fig. 7.10.

J
:? 0.05
�1.=?5l=�=" +�i�>,_l'.µ,..�,.__p
1.0
1S
Fig. 7.10. Temperature history with Surface Convection.
Problem 7.10. In an iron and steel industry, a large cast iron ingot ai 700°C is taken out from a furnace and its one surface is suddenly exposed and maintained at 50°C. Calculate (a) the time required to reach the temperature 300° C at a depth of 4 cm from the surface (b) instantaneous heat flow rate at a depth of 4 cm and on surface after half an hour. For ingot; take k = 48.5 W/mK, a= 0.06 m2 /h.
Solution. (a) Since there is a sudden temperature change in the solid hence we can use error function method. The tem�rature distribution is given by
!!_ = � = e,f (u) = 300  50 = 250 = 0_3846 700  50 650 0; t;  t s From table, for erf(u) = 0.3846, But or
or
U
=
2✓(a. ,) X
u = 0.36
= 0.36 =
0.06 � = �
=
(�f
16 x 60
4
}00 X 2 ✓(0.06 't)
· .(72)2 X 0.06
=
3.086 minutes
..
256
Heat and Mass Transfer
(b) The instantaneous heat flow rate at the surface is given by
kA (t..  ti }
(.!J..) =
or
 k A 9;
===  ===  l'br a i I  l'l,r a : }
q
A
48.5 X(  650 ) =  102687.96 W /m 2 Y(1r x 0.06 x 0.5 )
Ans.·
Instantaneous heat flow rate at the point of 4 cm is given by  k 8; A u_2 q, = Yl1r a i } e  =
vi
�8.5 x 650 e  (0.26) 2 =  95974.67 W/m2 Ans. n x 0.06 x 0.5)
The negative sign shows the heat lost from the ingot.
Problem 7.11. In a cooling process, a large mass at 500°C is suddenly subjected to 40 °C to one of its surface and maintained at that temperature. Calculate (a)· the temperature at a depth of 3 cm after 5 minutes, (b) after howmuch time, the temperature at the depth of 3 cm will reach to 300 °C, (c) the quantity of cumulative heat passed through the plane at a depth of 3 cm within first 2 hours. For mass, take, k = 1.2 WIm K, a = 0.003 m2/h Solution : Since, there is sudden temperature change in the object, hence we can use error function solution (a) we have 8/8i = erf (u)
3/100 = 0.948 u = =x== = 2 f{ii:i) 2 Y (0.003 X 5/60 )
where
t�rom table, for u = 0.948, erf �u) = 0.817
I  ts  t  40 = 0.817 500  40 t i  ts
or
(b)
(t)
t = 415.82°C
From table, for erf (u) = 0.565. or (c)
U = 0.572 =
300  40 = 0.565 = erf (u )
/ l'00 X = e3::_ ...,. :::_ _ '::_::_::._ 2 X ✓(a "C) "C 2 X ✓(0.003 X )
· .. . .. i = 0.229 hrs = 13.75 minutes
The cumulative heat rate is given by
Q) =2 A
Or (
t  ts
X
}.2 X
✓(.\
1t
2
,_
Ans.
Ans.
J/m� )(40  500) X 3600 =  5789.52 X } 0� Ans.
x 0.003
Unsteady State (Transient) Heat Conduction
257
Problem 7.12. A large mass of material initially at 20°C is subjected suddenly to 300 °C to its one surface and maintained thereafter. Calculate (a) the position of the maximum heating rate and its value after 10 minutes (b) maximum heat absorbed by the body during 10 minutes. Take k = 2.5 W/m K, a = 0.4 m2 I h (a )
Solution. The position of maximum heating rate is given by
x = • YI 2 a r ) =
Y( 2 x 0.4 x 10 /60) =
The heating rate is given by the expression o8  
where
ae Ot (b)
= _
=
3.65cm
Ans.
2 8; .x e _,, X 2 X l'1naJ f ½ { }
= � = (0.365) X 60 = 0.499 4a'f 4 X 0.4 X 10 2
U
8; = t;  ts
0.365m
= 20
 300 =  280°C
l ( 280)x0.365x eo 499 . 312 2 X ✓ (1t X 0.4) (IQ/60)
The maximum heat absorbed is Q /A

2 k 8; � fl'rT
= 
'V \,ia}
= 2 x 2.5 x 280 ,. / (10 x 60 x 3600
·v
tr X 0.4
15.855 K/h
Ans.
6 m2 ) = l .835x 10 J / ,
Ans.
Problem 7.13 A large steel mass initially at 20°C is suddenly subjected to 600 °C to its one surface and maintained thereafter. Calculate the time at which the temperature gradient will be maximum at 3 cm below the surface and its value. Take a = 0.05 m2 /h, k = 40 Wlm Kfor steel mass. Solution : The temperature gradient will be maximum when
= ;:
=
X
(1! r
a0 
where
a0 dX
dX
u =
✓( ,r X
1
Y(n a ii x2 n a i
1
9 = h :: 32.4 seconds 13
2X l
The temperature gradient"is given by f
e U
2
(3/1 00)2 X 103 tr X 0.05 X 9
0.05 X 9/13
)
= 0.636
:{0616) 2 = 17.74 K/m e
Ans.
258
_Heat and Mass Transfer
Problem 7.14. Water pipes are to be buried undergound in a wet soil (a=2. 78 x 103 m 2 /h) which is initially at 5°C. Due to sudden change in the environmental condition, the·soil surface ·temperature suddenly drops to 4°C and remains at this value for 8 hours. Determine the minimum depth at which the pipes be laid if the surrounding soil temperature is to remain above 0°C to avoid the freezing of water. Soil may be considered as semiinfinite soild. 'Solution. At the·critical depth of the earth, the temperature will just reach 0°C after 8 hour. We have
00
t t 8 = X  = e,f (u) = e,f 8; l;  oo t 2Var
).
(
0  (  4) i_ 0.444 e,f = = = ( 2 5 .,.. ( 4) 9
X
)
fai"
From table of error function, we have :;:;::x== =0.42 2 v(a r ) or
= 0.42 X 2
✓( 2.78
10 3
X
8 ) = 0.125 m
Ans.
Problem 7.15. A copper cylinder, 600 mm in diameter and 750 mm. in lengtt., is initially at a uniform temperature of 30°C. In a certain situation, the cylinqer 'is exposed to hot flue gases which result in a sudden increase in surface temperature amounting to 500°C. Calculate (a ) the temperature at the centre of cylinder 4 minutes after 3 minutes of the new operation. Also, calculate (b) the time required to attain a temperature of 400°C at the centre of the cylinder. Take a =1.12 x 104 m� /s. Assume the cylinder as semiinfinite solid. Solution, (a) The temperature distribution at the centre of cylinder is expressed as .. � 1 t,,, = erf(a T) 1.12 x 1 o 4 ; 3 x 60 = ] ·i' =[ X
8,
I;  t,,,
X
�
t· �
(0.3)
.•
\ �� \
= erf (0.224) = 0.35 from Fig. 7 .9
The temperature at the centre of cylinder is t � = 500 + 0.35 (30 SOO)=335.5°C t = too + 0.35 {t;
Ans.
(b) Since t = 335.5°C, hence 335.5 500 = 0_35 = e,f ar ( ) 30  500 rl or or
...· �=0.29 ro2 'r =
2 0.29 X ra2 = 0.29 X (0.3) a 1.12 X 10 4
233 seconds
Ans.
Unsteady Slate (Transient) Heat Conduction
259
Problem 7.16. A motor car travelling at 80 km/h is brought to rest within a five seconds period when the brakes are applied. The braking system of the motor car c:nsists of 4 brakes with each brake band of 350 cm2 area; these press against the steel drums of equivalent area. The brake lining and the drum surfaces are at the same temperature and th� heat generated during the stoppage action dissipates by flowing across the drums. If the drum surface is treated as semiinfinite plane, calculate the maximum temperature, rise. Assume that the mass of the car is 1 500 kg and the properties of drum surface are : k = 55 W/m K. and a = 1 24 x I05 m2 /s. ( Solution. Once the car is stopped, its kinetic energy is converted into heat energy which is dissipated through the drum.
Kinetic energy = ; m if = ½ X 1 500 ( 80 ;� OOO r = 3.7 X la5 J
The heat flow rate = q = 37 x 105  0.74 x 105 W 5
The instantanec us heat flow rate at the surface (x = 0) is given by. I\0. (a) Cylinder : The temperature history in this case is given by .
264
Heat and Mass Transfer
(b) Sphere::
t  t oo sin M"  M" cos M" .,  e  M,.2 a ,../ • 'i' x s in M,. = 4 1) ( t i  t oo r1 r ,c s l 2 M,.  sin 2M,.
Where
(r
r)
00
L
r M,. = ).,. 1
(7.42)
fa
7 . 9 . Grapliical. Method (TemperatureTime Charts) for Finite Internal and Surface Resistance (0 < B i< 100). · Solutions of unsteady heat conduction problems by analytical methods as discussed earlier are very time consuming jobs. Therefore, graphical evaluations or charts in non dimensonal forms have been plotted for practical use. We will consider here the charts prepared by Heisler. This is suitable for situations having finite internal and surface resistance (d < Bi < 100) The variables involved i n the transient state heat conduction problems may be correlated in the form of 9
 t oo
r
a  =  =! ( , t ;  t 00 Oi . Lez ' t
k h Le °
X)=
Le
/.
(Fo
X)
1 , Bi ' Le
(7.43)
This suggest that the nondimensional temperature distributions is the functions of three dimensionless parameters as discussed below :(i) Fourier number, F0 (a r /L�) : It indicates how rapidly a solid is cooled or
heated. For small values of a/L � , a large time r is required to obtain a significant "' temperature change. (ii) Biot number, B; (h Lc fk): It indicates the ratio of the resistance to_ heat transfer· at the surface of a solid to that within the internal body. It is similar to Nu (Nusselt number) used in convective heat transfer. (iii) Dimensionless length parameter, x/Le : It indicates the distance of any location from an arbitrary origine as compared to the characteristics length (plate thickness). For. circular and sphere , this parameter will be (r/ro) , Using the above nondimensional numbers , Heisler have prepared many charts (Figs. 7.13 to 7.18) for the determination of temperatwe history.
Figs. 7.137.15. These charts have l)een been plotted between dimensionless temperature distribution (t0t_) !(t; t_) d: tJ0/0i (the suffix 0 refers to x = O} and FO for • ·· various (1/Bi ) < 100.
· 1.'hese' fi��).giv< .__.
0 0 N
" ,.,,
,
i.,1,,
t.O
I ....
� ,, '\i �.., � I � L, � ,"?�
,, I,� �11 1, I.I.,��
II
,....._ .
M
IO
I
,
,;s

0
0 0
0
., ,., ,.,,
"
I
�C ' .,
,
....,
....
[/
 'gv
0 0
0
ci
.!¾ '"

�
r..:
co It
� 1 .0 0.7 0.5 0.4 0.3 0.2
l:lo
ei
0. 1 0.07 0.05 0.04 0.03 0.02
i
f
0.01 0.007 0.005 0.004 0.003 0.002 0.001
�
..,
2
3
(X't
Fo = � ro ♦
\
4 6 8 1 0 1 2 14 1 6 1 8 20 22 24 26 28 30 40 50 60 70 80 90 100 110 120 130 140 150 260 300 350
Fig. 7. 1 4. Heisler Chart for Central Temperature Histoty in an Infinite Cylinder
\
.� :,:�::�f: �;��.� ,, , �,�;,·, � �.::s �.,:� � 1 .0
0.4 0.3 0.2
F=
8i
Ii
),'
I::
I"
��> 0 .
0.4
,
,I,1,
u
V
11
I/
�
0 0. 0 1 0 . 020. 050 .1 0.2 0 .5 1 . 0 2 3 5 1 0 20 50 1 00 k 1 Bi = h.6
Fig. 7.16. Heisler PositionCorrection Factor Chart for Temperature History in Plate. 1 .0 r0rts  0 . 2 0.9
.4
0.8 0.7
�
...' 0 �
0 .2 ...'0. 1
....
'j ,
[/
8 80 0. 5 0. 3
..I
J
0. 6 I... (.)� 0 .4
,... �
_Ll�rttTl7
I 1,1
III
�
17
:/
I ll
'!H11
I/
"' r ro
I I
:
,
11111
,1 i•· 11111 0 50 1 00 0 . 0 1 0 .020 . 050. 1 0 2 0 .5 1 . 0 2 3 5 1 0 20
.·o
��1
 
k
Bi = h.r0
.,, , , .· Fig. 7 . 1 7. He_isler PositionCorrection Fact or Chart for Temperature History in Cylinder. . .�. 
Unsteady State (Transient) Heat Conduction 1 .0
� �Is
0.9
._ d.,
0.8 >
.....
0.6
90 0.5
�, o.6
0.2 >>0.1 >
o.r>i II 9
'.
I
v, I/ V
[\
0.4 0.3 ,
I/
II
0.7
a
;;:;""
T
0. ;
269
r ro
/
/·
I/
� 0 0.01 0.020.05 0.1 0.2 0.5 1 .0 2 3 5 k 1 = Bi h.r0
10 20 50 1 00
Fig. 7. 18. Heisler PositionCorrection Factor Chart for Temperature History in Sphere. Prob lem. 7. 18. A missile reenters the dense layers of the atmosphere at a very high velocity. The nose section is formed of 6 mm thick stainless plate and is not cooled on the inside surface. As a first aooroxim:ition, the eff�tive temperature o.f the air surrounding the nose region is considered to be unifonnly 2000°C. Assuming that the steel nose section is at a uniform temperature at :,uvc calculate me maximum permissible time in these surrounding if the maximum metal temperature is not to exceed 1000°C. Radiation effect will be neglected. This problem may be idealised as a 6 mm thick plate with one surface adiabatic and the other surface subjected to the surrounding conditions. Take p = 7300 kg/m 3, cP = 1.0718 kl/kg K, k = 150 W/mK, h = 3000 W/m'K Solution .
1  k = 150 x 1000 = 8.33 or B = 0.12 3000 x 6 B l·  h 8 Since B ; is grater than 0.1 , hence we can use Heisler chart for the solution of problem. 0i = l;  loo = 50  2000 =  1950°C The maximum temperature is at the surface. 0s = ls  t.,, = 1000  2000 =  1000°C = _I!:_ = 150 x3600 = 0_069 m2/h a· p Cp 1300 X 1.07} 8 X J03
270
Heat and·Mas.s Transfer
Froin chart, (Fig. 7 .16) the correction factor for plate at the surface (i.e. x/Lc = 1) is II' =Os = 0.94 80 60,
80 80 8:... I ( 1000 ) = 0.545 = 8; 8/ 8; =0.94  1950
With 80 /8;= 0.545 and 1/B; = 8.33, from chart (Fig. 7. 13) , for plate at the centre or
F 0 = ar = 6.0 f,2
T
6 X 62 X 3600 = 1 1 •26 S = ( 1 000)2 X 0 .069
Ans.
When the exposed surface temperature is 1000°C, the inside surface temperatun. is 8 0 = to  t_ = 8,10.94 =  1000/0.94 =  1063.8 °C t0 = 2000
 1063.8 = 936.2°C
Ans
. Pro�lem 7.19. A long piece of cylindrical iron bar 12 cm diameter is heated in a furnaCf1/ to a uniform temperature of 200 °C. It is then left to cool in a forced air cooler. If the heat transfer coefficient between its surfac� and ambient air at 40°C is 156 W/m2K; (a) bow long would it take for the centre of the bar to cool down to 50°C ? (b) what would be the temperature of surface and mid radius at this instant (c) what would be the error in the answer to (a) If Newtonian cooling is assumed ? Take k = 50/Wm2K, a = 0.0725 m2/it. Solution : J_Q__ _ t __ t_ oo = 50  40 = = 0.0625 t;  t 00 200  40 160
_!_ = _!_ = 50 X }00 = 5.34 l hO 156 x 6 B;
Since B; is greater than 0.1 , we can use Heisler chart. For central temperature. (a) Using Heisler chart Fig. 7.14, we get F0 = 8.2 = a r/o2
f = 8.2 x 82 = 8.2 x 6 x 60 = 24.43 minu tes Ans. or a 2 0.0725 X (100) (b) Using Heisler chart, for 1/B; = 5.34 1 and r IrO = 0.5, the correction factor is 0.98. Thus the temperature distribution at midpoint is 2
t  40 = 0.98 X 0.0625 200  40 ° or t = 49.80 C The correction. factor for r/rO = 1 is· O.92, thus the surface temperature is given by ts = _40 + 0.92 X 0.0625·x 16() =· 49.2°C_
Ans.
Unsteady State (Transient) Heat Conduction
(c) For Newtonian cooling
or
271
t  t.,, = e  (hA/p c V ) r t,  t.,, 0.0625
Talcing log, we have
V = e (h a /k ) (A/ ) r smce
a = k /p c
o.o725 ) ( 2 x l OO ) r since A IV = 2ko for cylinder. In (0.0625) =  ( 1 56 x 50 X 3600 6 In (0.0625) =  2.0944 x 1 03 r
or
r = · In (0.0625) = 1323.81 s = 22.06 minutes  2.0944 X 10 3
or
Thus the error in calculation is = ( 24.43  22  06 x 1 00 = 9.7% 24.43 Problem 7.20. In order tci prevent the freezing of oranges during cold nights, it is necessary to heat the surrounding atmosphere. For this, it is necessary to find out the temperature at the centre of an orange. Assume that the diameter of the orange is 8 c_m and initially it is at 1 5°C. It is suddenly exposed to the atmosphere at 5 °C in night during 1 pm to 6 pm. If the heat transfer coefficient is 10 W/m2K, state whether there will be a formation of frost in the centre of orange or not. The properties of the orange juice may be taken as k = 0.5 W/mK, a = 0.54 x I03 m2/h, tr, freezing temperature of the orange is 2°C. Solution
1
= 0.5 X 100 = 1 . 25 10 X 4
k
S ince B, is greater than .0. 1, Heisler chart may be used. at Fo = �
3 = 0.54 x lo x .S = l.6875
(0.04)2
Assume the orange as cylinder. Using Heisler chart for cylinder central temperature, Fig. 7.4, we get (!) = 0. 1 2
Or
f  t 00 t i  t 00
=
( _ (5) = _!____±_2_ 15  (  5 ) 1 5 + 5
I = (15 + 5) X 0. 12  5 = .:.. 2 . 6°C
As the centre temperature of the orange is below the freezing temperature ( 2 °C) there will be a formation of frost in the orange.
Problem 7.21. In a research project, a concrete wall 30 cm thick and originally at 40°C is suddenly exposed on air side of the hot gas at 800°C. The heat transfer coefficient on the hot side is 25 W/m2 K. The other face is insulated. Calculate (a) the time required to raise the temperature at the insulated face of the slab to 3 00°C. (b) Heat transfer to ,vall
Heat and Mass Transfer
272
per square metre of surface area. For concrete, take, k = 0.8 W/mK, c = 0.837 kJ/kgK, p = 200 kg/m3• Solution : The insulated face of the slab corresponds to the centre plan in a slab of thickness 2L so that 00/ox = 0 for x = 0. Thus 6 = 30 cm. 80 ot  ..,t 300 800 500 = 0 _ 657 = = [ ) = 760 8.I 40  830 tI. t«>
(l.) (.!.)
0·8 = 0.1066 = = Bi ho 25 x o.3 Since BI. is greater than 0.1 and less and than 100, we can use Heisler chart. a
=
(_!_) pC
=
0·8 2400 X 0.837 X 103
.=
3.982 x 107 m'ls
From Chart I, for 80/8.I = 0.558 and 1/BI. = 0.1065, we get a F0 = 0.4 = 't
c52 0.4 x (0.2)2 't or = 401 80.8s = I 1.16 h J.982 X 107 From Heisler chart, 8/80 = 0.5 Q� = pcL (ti t,,)
=
2400 x 0.837 x 0.2 (40  800) =  305337.6 kJ/m2
Q =  305337.6 X 0.5 =  }52668.8 kJ/m2 The negative sign shows that heat flows into the wall. Problem 7.22. A large steel plate 5 cm thick is initially at a uniform temperature of 450°C. It 1s suddenly exposed on both sides to a surroundings with convective heat transfer coefficient 285 W/m2K and temperature 60°C. Calculate the centreline tempera ture, and the temperature inside the plate. 1 .25 cm from the mid plane after 3 minutes. Take k (steel) = 42.5 W/mK, a = o:043 m2/h. Solution. The characteristics length = LC = o = 5/2 = 2.5 cm. •
� _ 0.043 x (3/60) _ F0 _ 2  3 · 44  2 LC (2.s x 1 02) 425 = 5.964 l/B.I = klhlC = klh o = 285 X 2.5 X 10 2 Since B.' is greater than 0. 1, hence solution by lumped technique would be inappro.
priate.
For F0 = 3.44 and 1/B.I = 5.964 and at x/LC = 0 (mid plane), from chart (Fig. 7. 13) for central temperature eo 'o  t"' = 0.6 = 0I
I
1.I  1«>
or
Unsteady State (Transient) Heat Conduction
t o =t oo + 0.6 {t i  toe} = 60 + 0:6 (450  60) =294 °C 1
273
Ans.
2 5 1.25 = 0 5 . The distance of 1.25 cm from the mid plane means x/Lc =x /f, = · � 25
. From chart (Fig. 7.16) at x/f, = 0.5 and 1/B, = 5.988, 8
t  t 00
t o  t oo

= 0.96
t =t oo + 0.96 {t o  t oo ) :: 60 + 0.96 (298  60) = 288.48 °C
or
Ans.
Problem 7.23. During the manufacture of 100 mm thick plastic sheets, they are brought to a uniform temperature of 180 ° C and then allowed to cool to a surface temperature of 50°C in air at 3.5°C befor� further processing. Calculate the cooling period required if natural convection cooling is employed with average surface coefficient of 9.8 W/mZJ.
.o
8
� � 8 LI o.
i ..,.
,
"'
q,
�
 "' "'
I/ I
I.
F/
I
''
I I
,
"'
�
�
.,�
�
!
I
II
, I'
I
I"
: t!
�It
.: i
.. , ........ . �I,
II
2.4
2.0
1 .6
1 .2
o.s
0.4 0.2
0.4 0.6
0.8
1 .0
0(ri) = (t (Tl)  tJl(t.  t_)
.
Fig. 1 0.5 (b) : Temperature Distribution in the Boundary Layer over Flat Plate (tan lj> = at/av at the swface and it is positive). .
Laminar Flow Forced Convective Heat Transfer Also, Vl dx
= ( t oo t s )
� = .' t dy ,
00
Uoo) d0 � = (t oo  t s ) d 0 d 1/ =  L A rr2x dry. dx dry dx
 t s) d0 = (t dy
y l�J
00
) d0  t s) d0 d ry = ( t oo  t s) A rr d ry dy 11
·v t�Jd
349
(10.37). (10.38)
(10.39)
Substituting equations (1 0.1 1), (10.12), (10.37), (10.38) and (10.39) in equation • (10.34), we get
\
. u oo d 20 d0 = (t t s) a oo ux d 2 d 11 1/
(l 00 t s)
Uoo d 20 · I U oo d0 =a  or    f vx d11 2 2 X d 1/
n d 0 f or   + v dry 2 2 2
Bue � = Pr a
d 1' + l f Pr d0 = 0 2 dry d712
(10.40;
Thus the partial differential equation (10.34) has been converted into ordinary differential equation. The boundary conditions to be satisfied are at 11 = 0 (i) f = 0
(ii) 0 = 0 at 71 = 0 (iii) 0 = I at 71 = oo The solution to equation (10.40) was first given by pohlhausen (1921). For solving this equation let d0 = A d1/ Hence the equation (10.40J becomes I Pr . f . A = O dA + d1/ 2
(10.41)
_ ..i:...
.
 
Heat and !vi.ts 5 or
Or > . 5 > o X
�
X
This equation shows that for Pr < I, 01 >O
(10.47,
(c) Case 3: For Pr > I (i.e. oil in which viscosity in higher but conducting nature
s lower).
8 = 0.99 a;: 17 < 5 or
0r X
o 5 < < �
X
(10.48)
This equation shows that for Pr > I, 01< o. For example; an oil with Pr = 1 000, has
o, .,, 011 0.
According to Pohlhausen, the following relation, in general may be assumed between the thermal and hydrodynamic boundary layers: 01 = O/(Pr)I/3 (10.49)
1 0 . 7 The Local and Average Heat Transfer Coefficient Once the temperature distribution in the bundary layer over a fl�t plate is known, we can find the heat transfer coefficient, local as well as average. The rate of heat flow per unit area, at any loc�on x along the plate is given by qx =  k dt = h ,, (t s  t oo) ( ) A ay z . y  o
Heat and � Transfer
· 352
Where (at{ 0.5.
(10.61)
Problem 10.1. Air at 16°C and a pressure of 1 bar is flowing over a plate ;,•
velocity of 3m!s., If the plate is 30 cm wide and at 60°C, find following quantities us., exact method at x = 30 cm and at the distance corresponding to the transition point. T,lkt> properties at the bulk mean temperature {16+ 60)/2 = 38°C. (a) boundary layer thickness (b) local friction coefficient (c) average friction coefficient (cl) shearing stress due to friction (e) thickness of the thermal boundary layer (t) local convectiveheat transfer coefficient (g) average convec�ve heat transfer coefficient (h) rate of heat transfer by convection (i) total drag force on the plate (J) toial mass flow rate through the boundary. Solution : Properties of air at 38°C_ are p = 1. 1374 kg/m3 k = 2.732 x 10 2 W/mK 1.005 kJ/kgK
(i) At x = 30 cm Cp =
(Fig. 10.1)
v = 16.768 x l Q6 m2/s
First of all it is necessary to check whether 1i1e flow is laminar or turbulent.
Laminar Flow Fared Convc•,:;tive Heat Transfer
Uw X =  = 3
X
0.3 XI 10 6 16.768
= 5.367 x 10
3:'i5
4
S ince Re < 5 x 105, hence the flow is laminar. V
(a) The boundary layer thickness at x = 30 cm is
5 x 0.3 = 6.474 x 10 3 m 8= 5x = 4 i{Rex f Y (5.367 X 10 ) or & = 6.474 mm
(b) Loral friction coefficient is 0.664 = C fx =  1/(Rex f
0.664
{(5.367 X 10 } 4
Ans.
= 2.866 x 10 3
Ans.
(c) Average friction coefficient is Cf = 2 X Cfx = 2 X 2.866 X 103 = 5.732 X 1() 2
Ans.
(d) Local shear stress due to friction =
2.866 X 103 X 1.137 4 X 9 = 14 _ 669 X 103 N/rrf 2
�e) Thickne�s of thermal boundary layer is given by Pohlhansen's equation 3 = 6.47 6 x 10 = 7.29 x 10 3 m = 7.29 mm (0.7)½ (Pf: ) ½ ,.,., · :r coefficient is given by �• (f) Local convecti. & =
o_ _
hx.X Nu x =  = 0.332 k =
=
r.
)½
Ans.
(Pr} ½
½ 0.332 x k (Re x )½ (Pr )
Ans.
X
0.332 X 2.732 X 10 
A :.\ .
2 (5.367' X 10 4)½ (0.7) ½ 0.3
= 6.219 W/rrt K
. (g) Average convective heat tran sfer coefficient is h = 2hx = 2 x 6.219 = 12.438 W/m2K (h) Rate of heat transfer is given by q =h A s (ts  t � = 12.834 X (0.3
X
0.3) (60  16 ) = 50.82 W
(i) Totai on the plate , . force i D "' ( ro)x � Area of plate on one side up to 30 cm !} = 14.669 X l Q3 X (() 1 X 0. 3) = 1 . 12 X J n3 N
mat
Ans.
336
Heat and Mass Transfer
(J) Total mass flow rate through the boundary = m = (5/8) p. u_ (82  8 1 ) Here, 8 1 = 0 at x = 0 and 82 = 8 at x = 30 cm :. m = (5/8) x 1.1374 x 3 x 6.44 x 103 = 0.0138 kg/s
Ans.
The transition point begins beyond Re = 5 x 105 . Hence the distance from the leading point to transition is given by u oo . x,, 3x 1 5 ' 5 X 10 =  = V 16.768 X 1�
16.768 X 106 = 2 794 m = 279.4 cm . 3 For x = 279.4 cm, the same above procedure is followed. The students are advised to solve. Problem 10.2. In a certain process, castor oil at 30°C flows past a flat plate. The velocity of the oil is 0 .0 8 m /s . The length of the plate is 5 m . The plate is heated unifonnly and maintained at_90°C. Calculate the following:or x 1 ,
=
5 X. 105
X
(a) hydrodynamic and thermal boundary layer thickness at the trailing edge of the plate. (b) total drag force per unit width on one side of the plate. (c) local heat transfer coefficient at the trailing edge. (d) heat transfer rate. At the qiean film temperature. tf = (90 + 30)/2 = 60 °C, th� thermophysical properties are taken as folllows:p = 956.8 kg/m3
a = 7.2 x 108 m2/s v = 0.65 x I Q4 m2/s
k = 0.213 W/m K
Use exact solution. Solution. First of all it is essential to establish the nature of the flow. The Reynolds number at the end of the plate is u oo L ReL = ___ = 0.08 X 5 = 0.615 X 10 4 4 V 0.65 X 10 • Since the Re is less than 5 x 105, hence the flow is laminar over the entire plate. (a) The hydrodynamic boundary layer thickness is given by 5 X5 A ns 0 _ 5X = 5 XL = 0 3 l 87 m = 3 l.8 7 cm . , 4 Re ., o ·  v{Re J J( L) ( .615 X 10 ) According to Pohlhausen the thermal boundary layer thickness is
,Y
8 1,
=
,
8

tPr1½
Laminar Flow Forced Convective Heat Transfer
Pr = V /a = 065 X 10 8 7.2 X 10
But
= 902.77
4
o, = o.3 l 87
{902.77)½
= 0.03298 m = 3.298 cm
The average skin friction coefficient
c, = 1.328 = fife
357
1.328
✓0.6 15 X 10
4
Ans.
0.01693
The drag force = D = CI x ( l/2)pu.,,2 x area of plate for one side = 0.01693 x ( 1/2) x 956.8 x (0.08)2 x (5 x 1) = 0.2591 Nim (c) The local Nusselt number at x = L is given by
NuL = 0.332 (ReL)½ (Prf3 .= 0.332 (0.615 X 104)½ (902.77f3 = 251.57 or hL = 251.57 x 0.213 = 10.7 1 6 W/m2 K
5
Ans.
(d) The average heat transfer coefficient = h = 2hL = 2 x 10.716 = 21.432 W/m2K The heat transfer rate = q = h As Lit = 21.432 x (5 x 1) x 60 = 6429.6 W Ans.
Problem 10.3. A flat plate is kept in a air stream at temperature 20°C. The velocity of the air is 3mls. The plate measures 50 x 20 cm and is maintained at a uniform temperature of l OLl',C. Compare the heat loss from the plate when the· air flows parallel to SO cm and (b) parallel to 20 cm side. Also, calculate the percentage heat loss increase. Solution. The mean film temperature = t1 = (20+ 100)/2 = 60°C
At 60°C, the thermophysical propertit?S of air are
p = 1.06 kg/m3, V = 18.97 X 106, k = 2.894 X 102 W/mK, Pr = 0.696 (a) When air flows parallel to 50cm side U= X 4 3 X 0.5 . Re x =  =  = 7.907 x 10 (lammar flow) 6 V 18.97 X 10
Now, Nux = 0.332 (ReJ/2 (Prf3 or
Or
3 h � = 0.332 7 .907 x 10 4)½ (0.696)½ = 82.734 k
(
82.734 x 2.894 x 10 hX _ 0.6 
2
= 3.99 W/m2 K
The average heat transfer coefficient 2 h = 2hx= 2 x 3.99 = 7.98 W/m K
The heat loss from one side of the plate ql =_ hAs Lit = 7.98 X (0.5 X 0.3) X 80 = 95.76 W
A r!S.

_\
Heat and Mass Transfer
358
(b) Whe_n air flows parallel to 20 cm side,
3 X 0.2
Re :,: =  = 
v
18.97 X 10o
= 3. 1 6 X 1 0�
h x Nu:,: = x = 0.332 (Rex )½ !Pr I½ "
or or
1 4 ½ hx. (0.696) ½ = 52.3 X = 0 .332 X (3.16 X 10
)
k
hx ;:: 52.3 X 2.894 X 10 0.2
2
= 7.567 W/m2 K
2 or h = 2 x 7.567 = 15. 1 34 W/m K The heat transfer rate = q2 = hAs Lit = 15.134 x (0.5 x 0.2) x 80 = 121.07 W
Ans.
Comparing the above two results, we find that more heat loss occurs when the air flows parallel to the shorter sides. The percentage heat loss increase is q2 � % X 100 = 121.07  63.84 X 100 = 89.64% 63.84 qi
Ans.
Problem 10.4. A flat plate is kept in a stream of air at 30°C. The velocity of air is 2 mis. The flat surface has a sharp leading edge and its total length equals 0.8 m. Calculate (a) the average shear stress and overall drag coefficient (b) compare the average shear stress with the shear stress at the trailing edge. Air properties at 30°C are p = 1.165 kg/m3 , µ = 6.717 x 102 kg/hm, , v = 70 x 1 0 0 11i 2 1� Solution. To check the nature of the flow over entire plate, the Reynolds number is calculated at the end of the plate. 11,, L 2 X 0. 1 , Re =   =6  1 05
1
V
20 X 1 0
Since the Reynolds number is less than 5 x 1 05 , the boundary layer over the entire plate is _laminar in nature. The average skin friction or drag coefficient is C1 = 1.328 = 1.328 Y{ReL)
MJ
= 0.4 1 99 X 102
The average shear stress is given by 2 'rw = :,.I PUoo2.C1 = 2I X l . 1 65 x 2 X 0.4199 x 102 .= 9.783 x 103 Nim 2 (b, ';. ne skin friction coefficient at the trailing edge is 2 C fx = 0.664 = 0.664 = 0.20995 X 10
. Y{Red
y( 10s)
An.s
Laminar Flow Forcer. Convective Heat Transfer
Shear � at the trailing edge is :,. ,. = , �'ii: . L�, = '
l
l kn�e.
1 .. ,
I
.._
359
< 1 . 1 65 , .1 x U.'.20995 x 1 u = 4 . 89 1 8x 1 0 3 N/11,2
4 . 89 1 8 " i O I 9_7g·· )' 1 0·· .l
'
2
The· above result is obtained by tht: fonnula itself. Problem 10.S. Assume a plate 0.5 m long is placed at zero angle of incidence in a stream of water at 20°C moving with 1.5 m/s. Detennine (a) the stream wise velocity component at the midpoint of the boundary layer (b) the maximum boundary layer thickness and (c) the maximum value of the normal component of velocity at thr trailing edge of the plate. Take properties of water at 20°C as p = 1.205 kg/m3, µ = 6.533 x 102 kg/h m. Use exact solution. Solution : (a) The boundary layer thickness by exact solution is given by o = _E_ = 5  (v;"
ilReJ
'V �
Evidently, the midpoint of the boundary layer (y = 0/2) occurs at 1J = y g = 2.5
The steam wise velocity component is obtained from the B lasius Solution in tablular fonn. At we get
1J = y
Vii
� = 0.736
U oo
 = 2.5
vx
u_
or u = 0.736 x 1.5 = 1.104 mis (b) The maximum value of boundary layer thickness occurs at x = 0.4. Thus
pu_ L = 1.205 X 1.5 X 0.5 X 3600 R eL = µ 6.553 X 102 The boundary layer thickness at trailing edge is
oL _
5L
=
5 x o.5
� ✓(0.498 X 10
5
)
= 0.498 X 105 . hence flow is laminar.
3 = 11.2 x 10  m = 11.2 m m
Ans.
(c) The maximum value of the nonnal component of velocity occurs at the outer edge of the boundary layer where u = u_ . Hence at ulu.. = 1, we have
360 or
l1
=
0.86uoo
Heat and Mass Transfer =
0.86 X 1.5
� ✓ (0.498
X
10
5
= )
5.78 X 103 m/s
Ans.
Problem 10.6. Air at atmospheric pressure and temperature 20°C flow over a plate at a velocity of 2 mis. The plate is maintained at 100°C. The length and the width of the plate are 80 x 40 cm. Calculate the heat transfer rate from (a) the first half of the plate, (b) full plate and (c) neat half of the plate. Use exact solution. Solution. The mean bulk temperature = tb = (100 + 20)/2 = 60°C.
The properties of air at 60°C. P = 1.06 kg/m3 Cp = 1.005 kJ/kg K
µ = 7.211 kg/mh v = 18.97 x lo6 m2/s
k = 2.894 x 102 W/mK . Pr = 0.696 (a) For first half of the plate, x = 40 cm u
x
x 0_.4__ ·= oo • _ = _2__ Re X = __ V
18.97 X lo6
0.421 X 10
5
Since Re 60, 1 x 104 < Re < 1 x 105 and 0.5 < Pr < 100. and
The fluid properties are evaluated at film temperature. 1 1 . 1 1 Summary of Formulae (2)
St = C1 / 2 St (Pr)2!3 = C112 = I /8
(4)
St =
(1)
(3)
St =
For Flat Plate :(5) (6)
(7) (8)
Ct / 2 1 + (um /u..J (Pr  I)
q, 1 2
1 + 5 J(q, 12) (Pr  1)
7 (o/x ) = 0.3 ¼ (Rex )
Cf:z = 0.0576 ¼ (Re x ) _ 0.072 Cf (ReL }¼
C t
 [(
lo&:·;: rs•  :,L ]
(9)
Nu" = 0.0288 (Rex )O.s (PrY3
( 1 1}
For laminar and turbulent flow
(10)
(12)
Nu = 0.036 (Re�o.s (PrY3
Nu = (Pr/3 [0.036 (ReL)0·8  836] For tubes
. Nu=0.023 (Re)08 (Pr/3
( 1 1 .59) (1 1.60)
Turbulent Flow Forced Convective Heat Transfer
411
Problem 11.1. The average drag coefficient for a turbulent flow past a flat plate is
expressed by
_ CI 
0.455
( 1 o g10 ReL j8
W�ere ReL is the Reynolds number. A flat plate 4 m longh and 60 cm wide is kept parallel to the flow of water which is flowing at a velocity of 4 mis. Find the drag force on both sides of the plate. Take v = 0.01 x lo4 m2/s Solution. The Reynolds numbers at the end of the plate is given by ReL
4 x 4 = = Lu.,,  = · V
0.01 X 10
4
16 x 10 6
This confirms that the flow is turbulent The average drag coefficient is expressed as 0.455
[] o g10 (16 X 10 6 )]
258
=
3 2.789 X 10
The drag force on both side of the plate is F = 2 x CI x (1/2) u.,,2 x area of one side = 2 x 2.789 x 103 x (l/2) x 1000 x 16(4 x 0.6) = 107.097N Ans. Problem 11.2. Ambient air at 30°C flows at 52 mis past a flat plate 0.7 f!'l long. The plate surface is maintained at 290°C. Determine the heat transfered from tbe entire plate length to air taking into consideration both laminar and turbulent portion of the boundary layer. Assume unit width of the plate and critical Reynolds number to be 5 x 105. Also, calculate the percentage error if the _ boundary layer is assumed to be of turbulent nature from the very leading edge of the plate. ts + t.,, = 290 + 30 = 160°C 2 2 ° At 160 C, the thermophysical properties of air are k = 3.638 x 102 W/m K, v = 30.08 x lo6, Pr = 0.682 For the entire plate, the Reynolds number is . Solution._The film temperature = t 1
=
Lu.,, 0.1 X 52 6 ReL = = 1.21 X 10 V 30.08 X 10 6
The critical distance xcfrom the leading edge at which the transition occurs is Xe
Rec V 5 X 10 X 30.08 X 10 = =  = 5
U .,,

Laminar boundary layer r�gion
52
6
0.2892 m
For the laminar boundary layer region, the average heat transfer coefficient is
412
Heat and Mass Transfer
h
= 0.664
.!._ {Rec Xe
)°5 (Pr }0333
X l0 = . 0.664 X 3:638 0.2892
2
X
5
(5 X 10
}°5 (0.682)°333 = 5 1 .9Q5 W/m2 K
The heat transfer from the laminar portion is q, = hAi:1t = 5 1 .995 X (0.2892 X 1) X (290  30) = 3909.6 W ,
Turbulent boundary layer region
For the turbulent boundary layer region, the average heat transfer coefficient is
h
=
t]
F
(pr R L )  (Rec (L �Xe ) [{ e o.8 o.8 ] I 1 0 2 [ = 0.036 3o:63 (5 x 10 5) 1 .2 1 x 10 (0.682)0 .333 = 1 .4.54 W/m2 K ; ( 6) 7 _ 2892 Heat b:ansfer from turbulent portion is given by = q, = h A L1t = 104.58 x (0.7  0.2892) x 1 (290  20)= 1 1599.95 W 0.036
0.8
Hence total heat transfer from the plate is q = qi + q, = 3909.6 +. 1 1599.95
=
Ans.
15508.1 5 W
Altier : The above solution could also be obtained by using the.direct formula for overall heat transfer coefficient h = (k/L) [0.036 ReL o.s  836] (Pr)0·333 82.799 W/m2 K
=
Ans, q = h A Lit = 82.799 x (0.7 x 1) (290  20) = 15649 W If the boundary layer is turbulent from the very begining, then heat transfer coeff. h = 0.036 (k/L) (ReL )0.8 (Pr)0.333
= q
0.036
X
= hAL1t =
rs (0.682)
3 ·638 X lO� (1.21 X 10 6 0.7 121'.04 x (0.7 xl) (290  20)
=
0
33
.3
=
121.04 W /m2 K
22876.56 W
Percentage error = 2287656  1 5649 x 100 = 46. 18% Increase 15649
Ans.
Problem 11.3. In a gas turbine system, the hot gas flows with the velocity of @ m/s over the surface of'a combustion · chamber. The temperature of the hot gas is 90b0c and the wall of the combustion chamber is at 300°C. Assume the combustion chamber to be a flat plate measuring 90 �m x 60 cm. The flow is parallel to 90 cm side. Taking transition Reynolds number equal to 5 x 105, calculate the heat loss from the gases to the combustion chamber. iake the following properties of gas p 0.495 kg/m3 , k = 0.074 W/mK, v = 94 x 1(}om2/s, Pr = 0.625.
=
Solution. The average Nusselt number _for a flow over a flate plate when both laminar and turbulent boundary layers are present is given by
413
Twbulent Flow Forced Convective Heat Transfer Nu = [ o.664 (ReL )°5 + 0.036 {(ReL
)°'8

.s ½ (Rec )° }] (Pr )
If transition occurs at Rec=· 5 xl OS, then
= [ 0.03$eL )°"8 
Nu
836] (Pr ) ½
The Reynolds number at the plate end is Lu_ = 0.90 X 80 (ReL ) = V 94 X 106
=
0.7659 X 10
6
The value of ReL = 0.1659 x 106 shows that both laminar and turbulent flow are present. Hence
=
[o.036 (o.7659 X 10 6
=
854.28
Nu or
hL 'h
°
) .s  836 ] (0.625)
½
845.28 x 0.074 = 70.24 W/m 2 K 0.9 The heat loss is given by q = h A Lit = 70.24 x (0.9 x 0.6) x (900  300) = 22757.76 W = 22.757 KW Ans. Problem 11.4. An aeroplane wing idealised as a flat plate is 100 cm wide and 5.5 m in length and maintained at 20°C. The aeroplane is flying at i speed of 500 km/h in an air at 0°C and 60 cm of Hg pressure. Find the heat loss from the wing and drag force on the wing if the flow is made parallel to the 100 cm width. or
=
Solution. The mean film temperature = t
= 20 + O = 10°C 2
1
°
The properties of air at 10 C are k = 2.51 1 x 102 W/mK, V = 14. 16 x lo6 m2/s. Pr = 0.7rl5 The pressure of the air at the flight altitude = p = 60 � x 1.013 = 0.7997 bar 76 From perfect gas law, the density of the air is 2
J!._ = 0.7997 X 10 p ·= RT.. 0.27 x (273 + 10) The flight speed = u _
=
5oo x lOOO
3600
=
=
l.0465 kg/m 3
138.88  mis
The Reynolds number for the entire wing is
Lu_ Rei_ =  = 1 X 1 38.88 14.16 x 106
= 9.8
X 106 .
414
Heat and M�s Transfer
Assume that the �ritical Reynolds number is 5 x I 05 • Hence . X •U X X 138.88 Rec = 5 x IQ 5 = � = _c____6 14.}6 X 1 0V or xc = 0.0509 m = 5.09 cm This shows that in the small portion of width the flow is laminar and in the rest of the portion is turbulent. So we can ssume the flow either a combination of laminar and turbulent or only turbulent. Let us consider the former. Hence, or or
° 333 Nu = [0.036 ((Re/ ·8)  8.36] (,Yr) 
° 333  836 �0.705) 2 h = 1 1 808} 6 x k = 1 1 808.36 x r5 1 1 x 1 0 = 296.5 W/m 2 K
hllk = 0.036( 9.8 x 1 06) [
0.8
]
Hence, heat lost = q = hA.M = 296.5 x (5.5 x (5.5 x I ) x (20  0) = 326 1 5 W Ans. That average friction coefficient is given by 1670 _ 0.072 C _ [ 0.072 0.2 Re ] � f  (Re )0.2 0.8 X 106) L L \
1670 = 2 . 708 X 1 0 J (0.8 X 106)
Drag force from one side of the wing c x ( 1 12)_p u:, x one side area of the wing 1

= 2 . 708 x 10 3 x ( 1 /2) x 1.0465 x 138.882 x (5.5 x 1) = 1 54.3 N
Ans.
Problem 1 1.5. Forced 2ir at 30 mis flows over a square flat plate maintained at 1 10°C. The drag force experienced by the plate is 1 2 N. Using Colburn analogy calculate the heat transfer coefficient and heat loss from the plate surface. Assume the flow to be turbulent and take following properties of air p = 1.029 kg/m3 , v = 2 l .09 x 1 0 6 m 2/s, Pr = 0.694, cP = 1.009 kJ/kgK. Solution. The drag force for turbulent flow is given by
I
2
0.072
I
x  x 1 .029 (30) x (L x L) F = C x  p u"' x area = 1 2 (Rel)0·2 2
' .
= 0.072
(L:J
x ½ x 1 .029 x 900 L2
6 0.2 1 2 1 .09 x 1 0 ') or x  � 1 .029 x 900 L2 12 = 0.072 ( 2 l x 30 ) ' 1 ' 8 12 L = ( or = 2.735 ) 1.9604 The Reynolds number at the plate end is
?
Turbulent Flow Forced Convective Heat Transfer Re =
415
Lu"" = 2.735 x 30 ·= 3 .89 x 1 0 6 6
2 1 .09 X 1 0The skin friction coefficient is l
V
0.072 = 3 .462 X 1 0 3 CJ, = (3 .89 X } 06)°'2
From Colburn analogy St . (Pr)213 = C/2 or or
h __ (Pr)213 = C /2 f pC U p ""
C
pC
U
"" h = f x P 1 2 (Pr)2 3
=
3.462 1 0 3 ;
x
1 .029 x 1 .009 x 3 0 x 1 03 = 68_77 W/m2 K (0.694)213
The heat loss from the plate is q = h A !1t = 68.77 x (2.735
x
Ans.
2.735) x (1 1 0  30) = 4 1 1 53 .2 W Ans.
Problem I 1.6. A refrigerated truck carrying food stuff is speeding on a high way at 96 km/h in a desert area where the ambient lfir temperature is 59°C. The body of the truck may be considered as a rectangular box measuring I O m long., 4 m wide and 3 m high. Consider the boundary layer on the four walls to' be turbulent and the heat transfer only from the four surface. If the wall surface of the truck is maintained at 1 0°C, calculate the following neglecting heat transfer· from front and back :(a) �eat loss from the four surfaces. (b) tonnage of refrigeration. (c) power required to overcome the resistance acting as four surfaces. Assume the flow to be parallel to I O m long side. . 50 + Hi So/ut10n. The mean film temperature = � =  = 30° C 2 The thermophysical of air at the mean film temperature of 30°C are :p = 1 . 1 65 kg/ni3 , c = 1 .005 kJ/kg K, k = 2.673 x 1 0 2 W!m2K, v = 1 6 x 1 0 6 m2/s,
Pr = 0.70 1
P
The Reynolds number at the end cf the side is · L · u"" IO x 95 x 1 000 ReL =  = ,6 = 16.49 x 1 0 6 V 3(i00 X J 6 X 1 0
For turbulent flow,
416
Heat and Mass Transfer
or
Nu = 0.036 (Re)°8 (Pr) 113
h L/k = 0.036 (16.49x 106) °8x (0.701) 113 = 18995.56
1899 5.56x 2.673x 10 2 10 (a) Heat loss from the four surface are
or
h =
50 _775 W/m2 K
q = hAM = 50.775 x [2(4 + 3)10](50  10) = 284340 W = 284.34 kW Ans.
(b) The cooling capcity required = 284340 W = 284.34 kW =
. 284340 = 81 .496 tons of refr"1gerat1on . . 3.489 X 1000
(c) The coeficient of friction C1
=
0.072
(Re)0·2
The drag force = F
=
0.072
02
(16.49x 10 )'" 6
=
2.593 X 1o 3
= c1 x ½ p u:, x area
95 X 1000 = 2.593 x 10 3 x 2l · x 1.165 x ( 3600 ) Power require� 0
X
I
I II
1 .8 .��...., .,_
2 p = 0.614 kg/m ; cp = 1.046kJ/kgK. k = 4.593 x 10 W/mK. Pr = 0.615·. µ = 29.7 X 6 10 kg/ms p u .. L · 0.614 x 8 x 0.3 = = 49616.16 Re = 6 µ 29.7 X 10 The Nusselt number is given by 117 NUx = 0.332 (Prt3 (Ret2 (I's/I..)°" + 00 0.l17 Nu,, = 0.332 (0.675) ¼3 (49616. 16)0s (273 6 ) 273 + 20 or h,, L or = 73.71 k
3
73.71 X 4.593 X 102  = 1 1 .28 W / m 2 K. h x = 0.3 The average heat transfer coefficient is given by
or
h = �
r h x dx = 2 h = 2 x 1 1.28 = 22.56 W/m2 K :i:
Heat transfer from both sides of the plate. Ans. q =) 2 (hA.,,1 t) = 2 x 22.56 x (0.3 x 1) x (600  20) = 7850.88 W. , j Problem 12.S. Hot air at the mass flow rate of 0.08 kg/s flows through an un,nsulated sheet metal duct of20 cm diameter. The inlet temperature of the air"is 1 00°C. The air gets cooled in its passage due to cold outside air and at a distance of 4 m. the inside air temperature is 80°C. The heat transfer coefficient between the outer surface of the duct and the cold ambient air at 6°C is 6 W/m2 K. Calculate the. following :(a) the heat loss from the duct over its 4 m length•. (b) the heat flux and the duct surface temperature at a"length of 4m. Solution : Refer to Fig. 12.6 for the system and its equivalent electrical netwo� • ho =6 W/m1 k Cold ::£;:�:1_
:��:
====:
tx=eo·c
      , ·

L: 4m
'• = eo·c
Ri
_h·,._,
Yh
I
• t
Fig. 12.6
Ro
YA
.1.
ho
•
fo = &·c
Empirical Correlations For Convection
437
100 + 80 = 'X)o C ' The me.an bulktemperature = '• = 2 ° At t6 =: 90 C, the thermophysical properties are :
p = 0.912 kg/m , Cp= 1.009 kJ/kgK, k= 3.�27 X 10 W/mK , � 2 � m Is. Pr = 0.69, µ = 22.14 x 10 kg/ms v = 22.1 x 10 . . � Re = U..; . Div ,.; 4m /,cDµ = 4 x 0.08/(,c x 0.2 x 22.14 x 10 Hence the flow is turbulent (a) The heat loss from the duct over its 4 m is 3 q = mcp ..1t = 0.08 x 1.009 x 10 x (1 00  20) = 6457.6 W (b) Th� heat flux is ..1t = q" = q /A _ · 4t � . {1/h; } + {I/ho) R; + Ro .
or
�
3
Ans.
·
h' D = 63.5 k
h · = 63.5 '
X
q = � = 80  6 = 276.749 W/ m 2 l: R 0.26739 ,, IA = 80  t = 80  t = 276.749 Also• q = q h, 9.928 or
23003.4
ho = 6 W/m2K.. already given h; is calcuiated from the correlation 03 o.a Nu = 0.023 (Rer (Pr)°"' = 0.023 (23003A) (0.69) = 63.5
3.127 X 10 2 = 9.928 W/ m 2 K 0.2 The thermal resistance =I: R = R ; + Ro = (1/h;) + (I/ho); m2 /WK. 1 "t" 0 or + (1/6) = 0.26739 m2 /WK. �  9.928 Hence, heat flux or
=
° t = 80  276·749 = 52.12 C
Ans.
Ans.
9.928 Problem 12.6. Following data are obtained from a metallic cylinder of 15 mm diameter and 1 00 mm in length heated internally by an electrical heater and subjected to cross flow of air in a low speed wind tunnel : Velocity of free stream air = M ·mis Temperature of free steam,I air = 25°C • I . Average temperature of ctlinder surface = 1300C Power dissipatioq by heater = 70 W
438
Heat and M� i'ransfer
If 10% of power dissipation is lost through the insulated end portions of the metallic cylinder, calcuate the experimental value of convective heat transfer coefficient for such a system. Compare this value with that obtained from the correlation ·suitable for this arrangement
Nu = 0.26 (Re) .r. (Pr)
036
0
3j
(Pr/Prs)°
where·all tl)ennophysical properties except Pr1 are evaluated at the mean bulk temperature (free stream) of air. Pr is evaluated at the average temperature of cylinder.
So_lution. The thennophysical properties of air at the 111ean bulk temperature (free stream) at 25°C are ...,
k = 2.6325 x 10 W/m K, 2
v
2
= 15.53 x 10 m /s, Pr = 0.702, Pr , at 130° C= 0.685 4
2
nie heat transfer rate from the cylinder to the air is given by or or
q = h A1 L1t
70 X 0.9 = h (,r
X
0.015 X 0.1)°(130  25)
x_ 0._ 9 ____ = 121.53 W/ m 2 K. 10 _ h = _____ 1r X 0.015 X 0.1 X (135  25)
(b) Reynolds number is given by Re� =
u_ . D = V
Using the correlation given, we have
15.53
Nu = 0.26 (14488) .r. (0.702) 0
or or
h. D = 72.263
k
72.263 X 2.6325 h = 0.015
X
15 X 0.015
036
X
10
= 14488
6
(0.702'/0.685)
0.25
2 10� = 126 .82 W/ m K
From the above, we conclude that the predicted results from correlation is higher than the experimen� value•
. Problem 12.7. In the transmission system of electrical energy, a copper bus bar of circular crosssecµon having 20 mm diameter is coole3'
4
(D )¾
h oc
or
Also,
h1 h2
Ans.
= 1 .4 14
q = hAs At = h (tr DL) tit
h1 D1 10 = 0.3535 q tfq2 =  = 1 .4 14 X Ans. 4Q h2D 2 Problem 12.13. To heat the room, the fire is initiated at the fireplace. In order to reduce exfilteration of room air through a chimney. a glassdoor. firescreen of height 0.7 m and width of 0.95 m is used. The surface temperature attained by the glass is 220°C. If the temperature is 20°C, calculate the free convection heat transfer rate from the fireplace to the room.
¼ ]2
Churchill and Chu correlation is valid for this case which is given as Nu =
[
0.825 +
Solut ion : The film temperature = (f
0.387 (Gr . Pr )
6
s 111 ] [ 1 + (0.492/Pr )½ 6
= (220 +
20)/2 = 120°C
The thermophysical properties of air at 120.°C are :
1c = 3.208 x Mf W/mK, v = 21.13 x .10 m2/s, Pr = 0.688 2
 6
Heat and Mass Transfer 3 p = l fl't = 1/ (120 + 273) = 2.544 X 10 /c1
Gr = L3 /Jg L! t v2
= (0.7 )
3
X
2.544
Gr. Pr =Ra = 3.83 X 10
9
X
10
3
X
(21.13 X 10 6
X
r
9.91 X (220  20) = 3.83 X 109
0.688 = 2.635 X 10,
9 2
¼6
0.387 h.635 x 10 9 ) Nu = 0.825 + hL = 165 k
or or
[1
Q/16 . (0.492/0.688>7
+
] 3/'21 _
2 = 5.571 W/m 2K h = 165 x 3.208 x 10 0.95 q = hA/J L!t = 5.571 x (0.7 x 0.95) (220  20) = 740.94 W
.·)
Ans
Problem 12.14. A vertical plate is heated from one side and maintained at 90°( when _ it is subjected to free convection with atmospheric air at'30°C. Calculate the loca and average heat transfer coefficient at 40 cm from the leading edge of the plate: For the above, following correlation is valid :Nux = 0.52 ( Pr . 0.95 + Pr
) ¼ .(Gr.Pr }
¼
The properties are evalulated at film temperature. Solution." The mean film temperature = tJ = (90 + 30)/2 = 60°C The thermophysical properties of air at 60°C are
k == 2.894 x 10: W/mK, V = 18.97 x 10 m2/s, Pr = 0.696 2
6
3 l l = = 3 X 10  .pt'l' K • P  273 + tf 273 + 60
Gr = L /Jg Llt = ( 0.4 ) V2 3
3
X
3 3 X 10
X
9.81 X (90  30) = 3 1 4 x 1 06
( 18.97 X 106 r
Gr.Pr = 3 1 4 x 1 06 x 0.696 = 2 1 8.544 x 1 06
Using the correlation
.
)1/4
0·696 Nu_. = 0.52 ( 0.95+ 0.696
(2 i 8.544 x 1 06) 1 14
449
Empirical Correlations For Convection or
h .L T = 50.98
·or h = 1 /L
AllS. 6
h dx = (4/3) hX = (4/3) x 3.688 = 2.766 W/m2 K
Ans. X Problem 12.15. rind the heat transfer from a 60 W incandescent bulb at 100°C to ambient air at 20.C. As,:ume the bulb as a sphere of diameter 5 cm. Also, find the percentage of power l0s1. by free convection. The correlation is given by Nu . = 0.6
f
O
(Gr.Pr) 114 • Solution. The mean filL l temperature = t = ( 1 00 + 20)/2 = 60°C. i
The thermophysical properties of air at 60°C are k = 2 . 894 x 102 W/mK, v = 1 8.97x 10o m2/s 1 1 Pr = 0.696, 13 =  = = 3 x 10 3 per K. 60 + 273 £! + 273 Gr =
D3 l3 M (fl.05)3 x 3 ?< : o 3 x 9.8 l oo 20) p } = = 8.17 x 105 v· (1 8.9'.7 X 10 6)
Gr.Pr = 8.17 x 105 x 0.696 = 5.686 x i 05 From the co. i ·,lation Nu = 3.6 (Gr.Pr) I14 = 0 . 6 (5.686 x 105) 114 = 16.47 or
h = 16 .47 x � �;4 x 1 0 ' = 953 W/m2 K �
The heat transfer = h A M = 9.53 (1t x 0.052) x 80 = 5.98 W g
Ans. Problem 12.16. A thin walled.vertical duct in the form of circular crosssection having diameter of 50 cm c.arries gas at 200°C. The duct is surrounded by air which may be considered still at 25 °C. Find the heat transfer rate by using the simplifie{i relation for air for laminar flow. h = l .42 . (MJL) 114 Solution.
h =. 1.42 (f)
Heat flow rate = h As M
1/4
= 1.42
(
200 25 ;
)0.25
= 5.164 W/m2 K
5.164 x (1t x 0.5 x l ) {200 25)
=
1419.52 W.
Ans. Problem 12.17. For cooling purpose; a transformer is immersed in an oil bath. The whole combination is housed in a cylinderical tank which is 70 cm in diameter and 1.25 m long. If the electrical loss is 1 . 25 kW, calculate the surface temperature of the
=
Heat and Mass Transfer
450
tank. The entire loss of electrical energy is assumed to be due to natural from the bOttom of the tank. The simplified relation for laminar boundary layer is applicable for the above Cast" which is given as h = l .>2 (A{
)\or a cylinderical plane. h = 1:42 (A�
(
fa
.vertical plane
Solution. The lteat transfer from the tank is due to th� sum of heat transfer from the side and top surface Area of side = A .side = TC D L = TC x 0.7 x 1.25 = 2.748 m2 Area of top = A1op = TC (O.7) 2 = 0.384 m 2 4
Qsitk == hsitk . A .ri,u..1t = 1.32 (M
)¼ X
L
2.748 X (t  25)
1 :: 1.32 ('  2 5) X 2.748 (t  25) = 3.835 (t  25) .25 1. 25 Q top = htop A top• Li t = 0.25
= i.42 (� )
VJ
X 0.384 (t  25) = 1.42 (\;}5)
025
X 0.384 (t  25) = 0.424 fr  25)
0.15
Q,_ = Qllido + Qlop
o··
3 25 1.25 X 10 = 3.835 (t 25{�. + 0.424 (t  25{
1250 = 4.259 (t 25{
25
(t  25/ = 293.49 or (t 25) = (293.49) 1 • = 94.2 Ans. t = 25+ 94.2 • = 119.2° C 0r Problem 12.18. A two stroke motor cycle •petrol engine cylinder consists of 12 fins. Each fin is of 2Q cm ou�r diameter and 10 cm inside diameter. The fins may be idealised as single horizootal flat plate of the same area. The average fin surface wmperature is 480°C and the atmospheric air is 20°C. Calculate the heat transfer rate from the fins when (a) the motor cycle is stationary and (b) the motor cycle in running at a speed of 70 km/h. Solution. Case (a) when motor cycle is stationary the heat transfer will be due to free convection. The average film temperature is t1 = (480 + W)/2 = 250°C. 25
The thermophysical properties of air at 250°C are:
K,
1 1 25
k = 4�266 x 10 W/m v = 40.61 x IO� m /s, 1 3 Pr = 0.661, /3 = 1/(273 + t1) = 1/(273 + 250) . = 1.912 x 10 K 2
2
Gr = ___.L /Jg ''"A t where the significant kngt'1 L = 0.9 D · I
3
vZ
45 1
Empirical Correlations For Convection (0.9 X 0.2)3 X 1 .912 X 10 3 X 9.81 X (480  20) = 3 _051 (40.61 X 10 6)
or
Gr �
Since
Gr.Pr = 3.05 1 x 107 x 0.661 =. 2.035 x 107 Gr.Pr < 109, he�ce the flow is laminar.
X
l07
Nu = 0.54 (Gr.Pr)114 = 0.54 (2.035 x 107)°25 = 36.26
36.26 x 4.266 x 10 2 _ _ W/ 2 K m = 8 51 0.9 x 0.2 The heat transfer is from both sides of fins. Hence. h=
q = h AS !lt = 8.5 1 [2 x 12 x ¾ (o.22  o.1 2)} 1, o, < O. The heat transfer rate at the wall can be expressed as follows.
or
(g_A )o =  k (i:Jy� ) =  k It ¼ h . =  (Gr x ) d8 (0) Nux = x 4 y=O
k
x
4
dr;
s
_ t _ ) ___!;_ (d 0 4 Vx dr;
!Qt )
(1 3.21 )
462 where
Heat and Mass Transfer g /3 its  t_ ) x 3 Gr" = local Grashof number = "·______ 1 .6
1 .0
0.8
0.5
" 0.4 � � o.3 X �
0.6 ol
v2
::, 1 N
a,"' 0.2 2
3
t:x f
�
0.2 0.1
4
2
3
( :x G
(b)
(a)
r� 4
5
6
7
Fig. 13.4. Velocity and Temperature Profiles in Free Convection on a Vertical Plate .
The variation of Nu" with Pr i's shown in Fig. 13.5 by so.lid line. An empirical e.1uation approximating the calculated results is as tollows:N"x
=
4 ,JjGrx 14 }
0.676 Pr ½
(13.23 )
( 0.861+ Pr )¼
The curve representing the above empirical equation follows the exact solution very closely.
T �x
i /
I
I
lI
l l_________ / 2
/
0
4
/
/
/
3
Fig. 13.6. Approximate Intergral Method in Free Convection
F, .:r C. •n·, ..t:(ion
� i"ce " "' v.)rie5 as x 1 .1 · )>  • II)

·�.
��� . ,_._....
\ �
s, ., ., n
C ..__� ��

h,ing the variation of upper limit with x cancels out an9 hence ,,e!oc ily ,u 1 0 . 0
E4uJ 1 i ng th1�. to the integral 01 R'. H. S. eq. (13 . 1 6) reduces to integral form of :nom:ntum cquau(,n . S 11 2 (13.25) = !}_ f u dy =  V g /3 J (t  t� dy o dy o dx o
Them1al boundary layer eq. (13.17) is
(du,) +
a, = a i, a, + u u ax oy oy .2
Integration of L.H.S. gives
J
6
o
=
d ('
U
[·
f
 too)
'�
s
o
[u
ox
0 ( t . ;,., + V · '� oy
t )] . •.
o (t  t� + (t  t� avl ox
ayJ
Since v = d at y = 0 and y == 6
and
dy
dy
{!;} = :: , the above equation reduces to
t
s
L u (t t � dy
Equating the integrated result of L. H. S. to integrated result R. H. S., we get the integral form of energy equation as
d dx
fi; O
u (t  t � dy =  a
[d (1  , �1 dy
O
(13.26)
Solutions of eqs. . (13.25) and (13.26) by momentum intergral method depend upon the �uitable function of velpcity and temperature distribution. Let us assume.
,
t  t 00 to  t
=
( i _ l. )2 6
which fulfills the condition t = to at y = 0 and t = t. at y = 6.
.�
(13.27)
466
Heat and Mass Transfer
The velocity distribution may be assumed as
Here
� = } (1
61 = 6.
/
 fr
(13.28)
Su!Jstituting eqs. (13.27) and (13. 28) into eqs. (13.25) and (13.26), we get
s:)
1 d (u.x2 u = g 6 /J (t,. t  )' 3 . 105 dx
and
d
t1x

V • u.x 6
(13.30)
6  2a
30 )  T
("x
(13.29)
Eqs. (13.29) and (13.30) involve u.x and 6 as de�ndent variables and x as an independent variable. It can be solved only when additional relationship for U,xand 6 may be introduced. We assume solutions of the following form. u.x = C 1 x"' and 6 = C2 x/" Substituting these relations in the above equation, we get
C 2 n Ci "' 2 m + n c/c x 2 m + 11 l = g x  vx ...,. f3 (t o  t .J 2 3
105
m + n C C x m + 11 l = 2a 30 2 . Cz
x
C3
 11
In order to see that these equations must be dimensionally .correct the exponents of x in ·· · all terms must be the same. Thus 2m + n  I = m n. m+n  1 =n 1 1 aid n = 2 4 Solving for C1 and C2 gives
This gives m =

C 1 = 5.17 V (: + 20 21
Since v /a= Pr hence
) ½ [g /3 (t
 t .J ] ½
v2
g f3 (t t ) ½ 20)½ [  _ ] x½ v ( 21
Ux = 5.17 V Pr +
2
(13.32)
Free Convection
467·
.
g /J {to t .J x [ O = 3.93 lPr ) ½ (Pr + '20) ] 21 ¼ . ·v 2 3
or
� = 3.93(Pr X
f¼
2
¼
4
(0.952 + Pr ) ¼4 ( Grx )¼4
The local heat transfer rate is
(A!I...). x =  k [o {t ayt..)],  o = 2 k {too t..) = hx {to  t..)
Thus the local Nusselt number is given by · Nu x
h .x ) = 2x . =0.508 =(x
o
k
[ Pr2 . Gr
(13.33)
]¼
0.952+ Pr x.
(13.34)
The average h over x = 0 to x =L is 4/3 times hx at x =L. Thus hL ) = (3/4) Nu = 0.667 (Nu) = (x. k .
For air having Pr = 0.715, eq. (13.34) reduces to
[
Rr 2 .GrL Pr + 0.952
]
¼
(13.36)
(NuJ = 0.378 (GrJ¼
by
· (13.35)'
Here all the properties are evaluated at.film temperature. �ass Flow Through The Bounday The mass flow through a particular crosssection for unit. width of the plate is given
s:
m = Um • (OX 1) p
Where Um =mean velocity =
¼
u dy =
¼
s:
U x.
i (1 fr dy
= �
Ux (13.37) :. m = . op 2 The mass flow rate between the two crasssections at x = x1 and x =Xz is given by .
If x1= 0 and x2 =L, then the total mass flow rate through the boundary
(13.28)
(13.39)
•
468
Heat and Mass Transfer
Putting the value of u., and � from eqs. (13.32) and (13.33), we get m1 =
[_p_ ] 12
5.17 v
(rr · + 1
m,
=
1 3 . 7 . Turbulent Method)
1.7 p . V
 ½ pg (t ) [ v
¼  t .. } 4 ] 2
 t..
}]½x ½
o
v
[
o
2
[g /3 (t
20 ¼4 x 3 .93 ( P r ) ½ (Pr + ) 21 or
20 11
G + � ]¼ rL
(Pr )2 r'Pr
21
)
(13.40)
Free Convection Over a Vertical Plate (Integral
Eckert and jackson (1950) solved the turbulent free convection heat transfer problem by integral method assuming thikness �. of the velocity and temperature bom1dary layers to be equal. Hydrodynamic and energy equations in integral forms are .
6
6
!!_ J u 2 dy =  i o + . g p J (t o  t .. } dy 6 0 0 p d J (t 0 6 o 6
 t ..
(q/Alo } u dy = p ep
. tto _t,__ [1  ( Y� )�]
(13.4i) (13.42)
These equations were solved assuming suitable velocity and te�perature distribution as
=
": = (
y.;
I )7 [ f] 1
(13.43)
4
The expressions for io and (q/A)0 at the wall might be expected to be the same for the forced convection and free convections boundary layers. Thus i 0
Cp
=
2 (" . �) ¼
(13.44)
V 0.0225 p u .. 
hX Pr 7213 . pu_
=
(}.0296 �
½5  · x) " ( . 1 3 . v �  0
(
U .. . �
)½
(13.45) . (t3.46)
Ii
.r'rcc Convcclion
In the resulting expression, we may replace u_ by ux, since u_ ancl u x arc s1mqar as ylo➔0. The following equation results.
)¼ ,Pr· )½
(q/A)o · .(t ) ( V  = . 0.0?.25 o  t _ u x p ep
·
u,. . o
(13 .47)
1
Substituting all these expressions into the integral equation, we get d Ux2 O ) = 0.125 g jJ \l  ) 6 0.0225 U. 2 0.0523 ( o t x · dx 0.0306 .!!.._ {ux . o) = 0.0225 u; dx
(v)0 ½ (p,f½ Ux
� )¼ Ux
0
(13.48)
•
The solution of the above equations have the form as Ux = C1 x"' and 0 = C2 x"
Substituting this results in eqs. ( 13.49) and equating the coefficients and exponent of x in the two equations we get m = 1/2, n = 3.7  8/ Thus C 1 = 0.0689 V C 25 (Pr ) /3 10 C2 = 0.00338
Finally, we get
aoo
Uz =
v2 g /3 (t O  ,_)
[ 1 +. 0.494Pr
Jz [1 + 0.494 Pr½ ] ½
1. 185 (v /x) (Gr
0 °1 o/x= 0.565 (GrJ .i (Pr) 1/15 [1 + 0.494 Pr ½] '
Nusselt number is given by
½] (P,)1½
(13.50)
(13.52)
Nux = 0.0295 (GrJ 1/s (Prf15 [1 + 0.494 Pr ½ ¾
and the average value is
f
Nu = 0.0246 (Grv 1/s (Prf15 [1 + 0.494 Pr ½ ]¾
(13.53)
This equation agrees well with experimental data.for water and for air. 13 . 8 . Free Convection From Rotating Bodies In rotating bodies, centrifugal and Coriolis forces produce free convection process. The knowledge of heat transfer from the rotating body to the surrounding atmosphere by free convection is considerably important as it is needed in the thermal analysis of the shafts, flywheel, turbine rotor, etc. (a) Rotating Disc. Kreith suggested that the bounary layer formed on the surface of a rotating disc remains laminar and uniform thickness for the rotational
i
Heat and Mass Transfer
470
ro
ro
number, Re less than 2.5 x 10.S defined as Re = r'J./v where is the angular velocity of the disc. Wagur suggested that the average heat transfer coefficient from a rotating disc of radius R in air below the critical Reynolds number may be well expressed by
(ro
1 R 2 ·) hR =1 0.335 Nu = k V ·
½
Here the rotational critical R:eynold��umber is given by the expression.
(13.54)
,\
; = 2.5 X 10 Rec = [email protected]
s
V
(13.55)
where Re stands for the critical radius. If the rotational Reynolds number is greatar than critical value, then the average heat transfer coefficient is given by. o 2 .s R hR Nu = K = 0.015 ( @ R)

R 100 ( e ·)
T
2
(13.56)
\� ig. 13.7 shows the formation of laminar and turbulent boundary layer over a rota��g disc in whi�h Re may be calculated by eq. (13.55). ' \ Laminar boundary layer
(b)
Fig. 13.7. Boundary Layer Over a Rotating Disc.
Rotating Gas Turbine Bladis
Fig. 13.8 shows the cooling of gas turbine blades by free convection proposed by Prof. Schmidt. Holes are drilled into the blades from the root and are closed at the ends near the tip by leaving some material during drilling. The coolant is supplied from the root of the blade and the coolant is subject¢ to a centrifugal acceleration Rm 0} where Rm is the'mean radius of the blade measured from the centre of the shaft and ro is the angular velocity of the rotating blades. The coolent passing through the drilled holes · is heated due to the IJl3at transfer from the hot gases flowing through the blade channel. The heated ::oolani rises upward. This motion of c.oolant is similar to the flow over a vertical plate �xcept 1!1e absence of gravitational fi�ld and presence._of e�ntrifqgaJ,.action. Thus the Gr number IS replaced Gr =
(R m ro2 ) /3 At L 3 v2
47 1
Free Convection
Where L is th� length o'f the cooling passage.
fo fact, Gr' is alway·s greater than 101 2 so the flow is turbulent and the relation for turbulent tlow over plate may be used.
•
(c)
Rotating Cylinders
The average heat transfer coefficient from a cylinder rotating in air above the critical velocity is expressed as where l3.9.
Nu = \P = 0. 1 1 [ { 0.5 ( R e(i)
)
Re ro = roir D2/v
2
For Correlation, see chapter 12. List of Formulae
(z) Flow Over Vertical Plates (Laminar Flow) (a) Exact solution
(1)
Nux = 0.676 [
(2)
Nu = 0.902
[
2
I Gri 14 ) Pr 1
¼
0.861 + Pr�
l
2 (GrL /4 } . Pr
0.861 + Pr2
.,
¼
35
+ Gr }Pr r
(13.57)
Heal and Mass Transfer
472
(b) Approximal solution
(3) .!±.. = � Ux
(4)
6
U max
=
(1 ...:. � )2 6
4
27 27
U:z
(5) u ,,. =  U"""' 48 = 5.17 .
(Pr
+
1Q__ )½
211
(6)
U :z
(T)
0.952 + Pr � = 3.93 [ ] X Gr:x . Pr �
V,
(10) m , = 1 .7 pv [
2 (Pr )
(11) Nu x
=
· (12) Nu L
3 hL = = 
hx . X k
k
] (p,, + 0.952) GrL
= 0.508 4
¼4
2 · /3 g (t  too ) ] [ 2 V
Nux
[
=
�X =
(14) Nux (15) · NuL
0.565 Pr1/15
[
4
.
4 Pr 2 Grx ] = 2� Pr + 0.952 ¼ · 6
Pr 2 G rL ] 0.667 [ Pr + 0.952
¾ ]0.1
1.0494 Pr 3
G�
=
0.0295 Pr1/15 [
=
0.0246 Pr % [
G rx
1 + 0.494 Pr GrL
1 + 0.494
7°'4
½J 1
½.
¼
(iz) Flow Over Vertical Plates (Turbulentflow)
(13)
½:
,,,. ½
10.4
..
Free Convection
473
Problem 13.1. A hot plate 30 cm in height and 1 . 1 wide at 120°C is exposed to the ambient still air at 20°C. Using the approximate solution, calculate the following :(a) maximum velocity at 15 cm from the leading edge of the plate. (b) boundary layer thickness at 15 cm from the leadinge edge of the plate. (d) local heat transfer coefficient at 15 cm from leading edge of plate (e) average heat transfer coefficient over the surface of the plate. (f) total moss flow through the boundary . (g) heat loss from the plate _ (h) rise is temperature of the air passing through the boundary . Soluti9n. The mean film temperature = t1 = (120+ 20)/?, = 70°C
At 70°C, the thermophysical properties of air are: p = . i.029 kglm3; k = 2.964 x 102 W/mK; V = 20.02 X 106 m2/s, 3 Pr = 0.694, /J = �1(273+ t ) = 11343 = 2.915 x 10 K 1 1
GrX = x3 13 g 2 v
M
=
GrX = 19.24 x 107 •• = 5.17 � = 5.17 X 20.02 X 10
(a)
or
6
(0.15)3 x 2.915 x 10 3 x 9.8 x 100 = 2.40S x 101 (20.02 X lQf 6
(Pr + ;� i½ (/J g �,,  r, ½ 1
(0.694 + 20 21
)½ X [ 2.915 X 10
(b)
Um
(c)
�=
x
X 9.81 X
(20.02 X 10
= 2.633 mis U max
3
6
r
Ans.
4 X 2.635 = 0.39 mis 4 U == x 27
27
27 X 0.39 = 0.2193 mis 27 Umux = =48
48
3.93 [ 0.952+
�r]¼
Grx .Pr
o = 0.0762 x 0.15
=
4
= 3.93 [
0.0114 m
=
0.952 7+ 0.694
2.405 x 10 x (0.694)
11.4 mm
. ] 2
¼= ' 4
Ans.
0.0762 Ans.
Heat and Mass Transfer
474
or
h" =
¼4 2 " . Pr .Gr ] = 2x I 'o =  = 0.508 [ Pr + 0.952
2 x 0· 1 5 x 2 · 964 x 10 2 = 5. 199 W Im 1( 0.0 1 14 0.15
3 x 5.199 = 3.899 W/m2K 3 Nu = (e) h = "
4
4
(f) m1 = 1.7 p .
V
]¼
· 
7• \�·24 x 1 0 = I .7 x. 1 .029 � 20 .02 x 1 0 6 [ (0.694) (0.694 + 0.952)
(g)
Ans .
GrL [ 2 (Pr ) (p,. + 0.952)
.. � ..
or
Ans.
]1/4
= 4.372 x 10 3 kg/s
H_eat lost from the plate will be trom tne oom s10es, nence q = 2h As (ts  t_ ) = 2 X 3.899 X (0.3 X 1 . 1) (120  20) = 257.33 W
(h) Heat lost = q = 257.33 = m Cp At M =
4.37}
X
257.3 3 = 585 70C 10 3 X }.005 X 103 /
Ans.
Ans.
Ans.
Problem �.2. Develop the relation between Reynolds and Grashof number assuming the heat transfer coefficients· over vertical plates for pure forced and free convection are equal in laminar flow. Solution. In the case of pure forced convection.
.,
Nu = 0.664 (Re)½ (Pr/3
and in the case of pure natural convection Nu = 0.667
Equating these two, we get
·r
. 0.664 (Re}½ (Pr }½ = 0.667 or or
(R e )
]¼
2 Gr (Pr } + 0.952 [Pr
2 2{Pr )4/ 13 = Gr (Pr ) Pr + 0.952
2 Pr + �.952 Gr = (R e ) [ ] (Pr }½
Gr (Pr 1 1 Pr + 0.952 2
l/
74
Ans.
A ns. Ans.
Free Convection
475
Problem 13.3. Develop an expression for the total shear force acting on the hot vertical plate when it is exposed to ambient air if the temperature of the plate is maintained constant. Solution. The local shear stress is
'l"x
=µ

OU I oy y o
=
µ l_ [u x l'._ ay o
(1  l'._o
)j
y =o
= �
o
The average shear stress over the whole surface is given by 'l" = _!_ L 'l"x dx L o
f
=
fo µ o dx L
_!_ L
U
x
After substituting the expressions for Ux and o in the above, we get.
11
where
(')½
20 72 f3glltm A 1 = 5.17 v (Pr +.) 21 y2 Bl,
= 3_93
Pr + 0.952 [ 2] 2 (Pr) /3 g tit mlv
¼
4
Problem 13.4. Two vertical plates each at 80°C are placed in a tank of water at 20°C if the height of the plates are 10 cm, calculate the minimum spacing which will prevent interference of the free convection boundary layers. Solution. Refer to Fig. 13.9 •
..L.'4'
10cm
L
Fig. 13.9
Boundary � Layers
476
Heat and Mass Transfer
Let 6 be boundary layer thickness at the trailing edge of the plate. Toe minimum ·spacing required = L = 2 6. Toe film temperature = t,,, = (80 + 20)/2 = 50°C
The thermophysical properties of water at 50°C are k = 64.74 x 102 W/m�, v = 0.556 x 106 m2/s

.
�
3.54, 13 = 11(1 + 273) = 1.323 = 3.095 x 103 K I x313g At (0.10)3 X 3.095 X 10= 3 X 81 X 60 i°= 5_892 X l09 Grx = 6 v2 (0.556 X 10 ) Pr
=
Gr"Pr
£x =
=
5.892 x 109 x 3.54
=
20.85 x 109
Since Gr Pr > 109, hence the flow is turbulent. For turbulent flow
1
·213 ·h 1.4094 Pr 1.0494 x (3.54)213r· r = 0.565 (3.54f 8/15 [ 0.565(Prf 8/15 [ = 0.0331 Grx 5_.892 x 109
6
=
0.0331 x 0.1
=
0.00331 m
=
3.31 mm
2 6 = 2 x 3.31 = 6.62 mm Problem i3.S. A vertical plate 15 cm x 15 cm in size at 40°C is exposed tc atr.,osphere at 20°C. Compare the free convection heat transfer from this plate with tha1 which would result due to forced convection over the plate at a velocity equal to twice th( maximum velocity which occurs in free convection boundary fayer. Solution. Refer to Fig. 13.10. Toe film temperature = = (40 + 20)/2 = 30°C. The thermophysisical properties 01 air at 30°C are k = 2.673 x 102 WlmK. v = 16 x 106m2/s., Pr = 0.70l, 13 = 1/('f+ 273) = 1/30 + 273) = 3.3 x 103 K I
'l
15cm
l
Boundary layer
Fig. 13.10.
•
Free Convection
Gr
=
L 3 {3gM v2
477
3 3 6 = (0.15) x 3.3 X 10 X 9.81 X 20 = 8.535 X 10
(i6 X 10�)
2
8.535 X 106 X 0.701 = 5.983 X 106 As Gr Pr< 109 , hence the flow is laminar. For laminar flow Gr Pr
Nu
or or
=
=
2
0.667 [ (Pr ) Gr ] Pr + 0.952
¼
0.667[(0.701 ) X 8.535 X 10 0.701 + 0.952 2
4 =
hL/K= 26.62
6 ]
¼
,;
26.62
26.62 x 2.673 X 10 i = _743 W/m 2K 4 0.15 The heat lost from one side of the plate qfree = h A s !!..t � 4.743 X (0.15 X 0.15) X 20 = 2.13 W h
=
Forced Convection :
The maximum velocity=
=
U max.
4
27
u ,.
·] ½
u ,. = 5.17 r {3g lH X {(Pr + 0.952)
2
½
3
= 5_17[3.3x'10 x 9.8lx 20x 0.15] =1.253 mis (0.70 + 0.952) umax = 4/27 x l.253 = 0.1856 mis
The average heat transfer coefficient with forced convection if velocity is assumed equal to 2umax then Nu= 0.664 (Re/2 (Prt3
Re b'.
Luoo v
= 2x 0.15x 0.1856 16x IO�
= 3.48 x 103
12
Nu·= h{: = 0.664 (3.48 x I03y (0.701) 113 = 34.79 or
2
h =34.79 X 2.673 X 10
0.15
Heat lost due to forced convection
= 6.199
qforced = h A,. At = 6.199 X (0.15
X
0.15) X 20 = 2.7898 W
478 Hence
Heat and Mass Transfer
� = 2.7898 = 1.3097 2.13 �arced
Ans.
Note: Many problems concerning free convection have been solved in chapter 12 using empirical equations.
EXERCISES Note : The thennophysical properties of fluids are given in Appendix. 13.1. Deduce the expression for boundary layer thickness ancl heat transfer coefficient in free convection over vertical plate for laminar flow by momentum integral �ethod. Calculate the rate of heat transfer by free convection from an 20 by 30 cm 13.2. horizontal flat plate at I 50°C facing upward to air at 40°C. 13.3. Calculate the heat transfer coefficient for free convection from vertical heated plate 40 cm high at 95°C in 15°C air. 13.4. The temperature of the surface of a horizontal cylinder in quiet air at 50°C is 250°C. What must be its diameter if the Grashoff number is 2 x 103 ? 13.5. Calculate the heat loss per metre of length of a nominal 10 cm horizontal steam pipe at 95°C to the still air in a room at 18°C. 13.6. Calculate laminar free convection heat transfer along a flat plate inclinded at an angle of a to the vertical direction using the integrated boundary layer equation. 13.7. (a), A hot plate of 30 cm in height and 100 cm wide is exposed to the ambient air at 25°C. The plate surface is maintained at 125°C. Calculate the heat loss from both surface of the plate. (b) If the height of the plate changed to 60 cm, calculate the percentage change in heat transfer coefficient 13.8. A hot plate of 40 cm in height and 70 cm wide at 150°C is exposed to the air at 30°C. Calculate the followings: (a) maximum velocity at 30 cm from the leading edge of the plate. (b) mean velocity at 30 cm from the leading edge .. (c) boundary layer thickness at 30 cm fonn the leading edge (cl) local heat transfer coefficient at 30 cm from the leadng edge. (e) average heat transfer coeficient (f) total mass flow through the boundal)' (g) total heat loss from the side of the plate (h) rise in temperature of the air passing the boundary (i) total drag force on the plate. 13.9. A vertical plate 50 cm in height and 60 cm wide at 100°C is exposed to ambi�nt air at 30 C. G"ve a comparison between the heat loss by the free convection from this plate with �at which would result from forcing air over
Free Convection
13.10. 13.11. 13.12.
13.13. 13.14.
.. � :: 13.15.
4,, , ,
the plate parallel to 50 cm side at a velocity equal to the maximum vdocity wh.ich occurs in free convection boundary layer.
(a) A steel plate 80 cm x 30 cm and 6 mm thick is heated uniformly to 350 �C and then exposed to atmosphere air at 20°C, If the plate is kept vertical calculate the approximate time required for the plate to cool to 100°C. (b) If the plate is dipped in the water at 20°C vertically find the approximate time to cool the plate to 100°C.
A rod in 10 cm diameter and 60 cm height is suspended in a room. The temperature of the rod surface is 120°C and room air temperature is 30°C. Calculate the heat loss from the rod by natural convection and radiation. Also calculate the heat transfer coefficient
The plain wall of a container kept at a temperature of 95°C is insulated with the the other at a distance of 12 mm each producing two air spaces. The outside wall of the insulation is cooled to 50°C by the surrounding air. What temperature does the medium sheet of foil assume ? What is the heat flow per square metre through the insulation ?
Neglect heat exchange by radiation. A steam pipe 30 cm nominal (standard weight) passes though a factory space maintned at l 7°C. If the pipe is insulated and has outside surface temperature of 100°C find the rate of energy loss by natural convection per metre of pipe length.
A cylinder 8 cm in diameter and 12 cm high having a surface temperature of 60°C is placed vertically in water at 15°C. Calculate the average surface coefficient and the heat transfer rate.
A turbine blade is cooled by free convection with water as coolant. The cooling passage is 10 mm in diameter and 12 cm long. The mean peripheral velocity at mean radius of 30 cm is 250 mis. The hole surface temperature is maintained at 250 °C and the cooling water temperature is 65 °C. Calculate the average heat transfer coefficient and heat loss per hour from the blade.
14 Boiling and Condensation
(Two Phase Flow Heat Transfer)
In the preceeding chapters of convective heat transfer, we have considered the fluid as a homogeneous single phase system. But in many situations, the fluid changes its phase during convective heat transfer processes. Boiling and condensation are such convective heat transfer processes that are associated· with change in the phase of fluid. In this chapter, we focus our attention on boiling and condensation. During boiling, there is a heat transfer from solid surface to liquid which results in change of phase form liquid to vapor state. Conversely, condensation of a vapor to liquid state results in heat transfer to the solid surface. There are many unique features of these.processes. Since there is a phase change in these processes hence the heat transfer to or from the fluid can occure without influencing the fluid temperature. Due to latent heat effect associated with the phase change heat transfer coefficients and rates are g_enerally much higher than those of convective heat transfer processes without phase change. The most unique feature is the achievement of high heat transfer rate with small temperature difference. Boiling and condensation processes find a large number of applications in many engineering problems. These are as follows: (a) boilers and condensers used in thermal and nuclear power plants and other industrial applications. (b) nuclear reactors, spacecrafts and rockets (c) evaporators and condensers used in refrigerating and airconditloning systems. (d) process heating and cooling (e) melting of metal in furnaces (f) heat exchangers used in refineries and sugar mills (g) concentration, dehydration and drying of foods and materials. Due to two phase flow phenomenon, the analysis of these processes is quite difficult (A) Boiling 14.1. Physical Mechanisms of Boiling Evaporation at a solidliquid surface is termep boiling. This is possible only when the temperature of the surface, t5 exceeds the saturation temperature corresponding to the liquid pressure tsat. Newton's law of cooiing is applied in this case as
q =h A. ( t5 tsat ) = h A. Me where 6.te = t5 t sat and is termed as excess temperature. The boiling process involves the formation of vapor bubbles, which grow and subsequently detach form the surface. The growth of vapor bubble and its dynamics depend in a complex manner on the excess temperature (AtJ, the nature of the surface and other thermophysical properties such as its surface ten:;ion. Literatures [14.1 to 14.3) discuss in details about this field of heat transfer. 

•'
Boiling and Condensation
48!
Boiling may occure in the folliwing forms:
·~·,
(i) Pool boiling : This refers to a situation in 'Yhich the liquid above the hot surface is essentially quiescent and its motion near the surface is due to free convection and to mixing induced by bubble growth and (jetachmenL This type of boiling occurs in steam boilers involving natural convection. (ii) Forced convection boiling : In this case, the fluid motion is induced by external means, as well as by free convection and bubble induced mixing. The liquid is pumped and forced to flow. Boiling of water in water tube boilers inv�ving forced convection is an example of this type of boiling. (iii) Subcooled or local boiling : In this type of boiling, the temperature of liquid is below the saturation temperature and bubbles form in the vicinity of the heated surface. These bubbles travel a short path and eventually condense in the liquid which is at a temperature less than boiling point. {iv) Saturated boiling; The temperature of the liquid exceeds the saturation temperature. Bubbles formed at the solid surface (liquidsolid interface) are then propelled through the liquid by buoyancy effects, eventually escaping from a free surface (liquidvapor interface). The physical mechanisfll� �ill be more clear in the subsequent article " boiling regimes". 1 4.2: Boiling Regimes of Saturated Pool Boiling There exists a definite boiling regimes assoc�aced with saturated pool, boiling or forced circulation boiling. Fig. 14.1 shows the tem�rature distribution in saturated pool boiling with liquidvapor interface. It is abvious from the_ figure that altho� th�re is a sharp decline _in _the liq�id temperature close to the sohd surface, the tempera1ute through most of the hqmd remams slightly above saturation. Bubbles generated at the liquidsolid interface therefore rises to and are transported across the liquidvapor interface. To understand this physical mechanism, let us examine the boiling curve as shown in Fig. 14.2 (a). . .. ·. : ... . ·. . \ : ...· . ......· : ::
I I
y
I
I
I
I
I I . I I
Vapor bubbles
I
tsa t
.
ts
Fig. 14.1 Temperature Distribution in Pool Boiling at LiquidVapor Interface
482
Heat and Mass Transfer
The boiling curve shown in Fig. 14.2 (a) pertains to . water and can be obtained by considering an electrically . �eated horizoJ!tal wire submerged in a pool of water at saturation temperture. Similar · ' ·;trend is found with other fluids. Basically three different regims are clearly seen. ' (i)_ Interface evaporation or free convection boiling : The boiling starts in a thin layer of liquid which is in the vicinity of heated surface without bubble formation. This happens when flte ::;; flte , A where flte , A . "' 5°C.The liquid in the vicinity of solid surface becomes superheated (i.e. liquid temperature higher than saturated temperature at a given pressure ) and rises to, liquid vaporinterface where evaporation takes place. The fluid motion is determined principally br free convection effects. It is to be noted that the value of convective heat transfer . coefficient is much larger than that associated with free convection without phase change. This termed as.first stage of boiling; (ii) Nucleate boiling: It exists in the range, Me, A ::;; Me s Me c where Me c = 50°C and comprises of second and third stage. Wheri liquid is over heated the vapor bubbles are formed at certain favourable spots termed nucleation or active sites, these may be wall surface irregularities, air bubbles and foreign particles. The bubbles grow to a certain size influenced by pressure, temperature and surface tension at the liquidvapor interface. Depending upon the temperature excess, the bubbles may collapse on the surface, may expand and detatch from the surface to be dissipated in the_ OO?Y �f the liq�id or_ at _sufficientir high tempe�ture may ri� to the surface of bqu1d before bemg d1ss1pated. This whole process results.m increased value of h and 'q;. During local boiling (i.e. second stage), the primary mec_hanism of heat transfer is due to igtense agitation at the heat transfer is surface which creates a high heat transfer rate. In saturated or bulk boiling (Le.third stage}, the bubbles may break away from the surface because of buoyancy action and,, . move into the body of the liquid. Theref9re, it is both the agitation caused by �·� bubbles and vapor transport of energy into the body of the liquid influence the heat . transfer rate. The maximum heat flux, q;, max or q;: c generally termed as the critical heat flux and is in excess of / MW/ m2 corresponds to point C. As the name implies, this point. is very critical and n(}ed.'further discussion. We have seen that the high heat transfer rates and convective heat transfer coefficients are associated with small value·s of excess temperature, it is desirable to operate many enginerring devices in the regime of nucleate boiling. The value of h in excess of 104 W/m2 K for water is the characteristics of this regime . which is much higher than those corresponding to convection without phase change. (iii) Film boiling: This comprises of fourth, fifth and sixth state of boiling'. Transition boiling, unstable film boiling or partial flim boiling (i.e., fourth stage) is found for M e, c � flte , � flte , o , where l!.te , D = 150°C.ln this zone, the bubble formation is so rapid that a vapor film or blanket begins to form on the surface. Insulating.effect of vapor film (its low thermal conductivity) ignores the beneficial effect of liquid agitation and consequently the heat transfer coefficient and heat flux decrease with the increase in temperature excess, flte. At any point on the surface, conditions may oscillate between film and nu�leate boiling. Film boiling (i.e. fifth stage for M e �. M e , D. In this regime, the surface is completely covered by a vapor blanki!t, and therefore, heat transfer from the surface to the
Boiling and Condensation liquid occurs by conduction and radiation through vapor. From the minimum associated with point D. the heat flux. q1" increases with the increase of llte . In the sixth stage. radiation through vapor dominates and q/ increases rapidly. Interfac e __..,_ B ub bles .... F i l m '1 . evapora t ion or. or . Nucle ate . . free co nvec t ion VI V IV Ill II
�
..
N
. E
I
..
� � x j
.... 
0
,,=
.., 0 �
j
l/l
      "'• �
!
Pure convection ., 0 .. heat t ransferred � 1 111 by sup er h eated � � S • • • • Q. .,. liqui d r t �n9 d t qu 1; to th e .: c vapor i n terface !·5 · :g where evaporatiu,.c :l: c
�=
takes place
.,. .. ...., =· � ., 0
· t::
C VI 0
0 .o _! c
.!! .0 .&J · CI .. j .0
..,
z
j
.,. C
....D 
.0
E
....  e �.. C C:
E
.&:. 0
Cl 0 ::,
..
.&:.
� 'o  Q. � Cl
"O > 0 0::
.. c .... '! "c; �
g.
..! �J � .., z
Fig. 14.2 (a) nii Bbi!ing Curve For Water (Saturated Pool Boiling) Temp eroture e x cess , h. t e = ts  tsat
Critical heat flux or Burnout Point : It is worthwhile to discuss in detail about the critical heat flux. The point of maxim� heat flux on the boiling curve initiates the transition from nucleate to film boiling and is termed as critical heat flux or burnout point. The whole boiling process remains in the unstable state beycz_nd the burnount point C, until situation corresponds to point F on the boiling curve. Any increase in q /' beyond the burnout point will induce a departure form the boiling curve in which t,he surface condition changes form ilte . c to llte . F• i.e. a large increase surface temperature . In general, t_. • F exceeds the melting point of solid, hence destruction or failure of the wall surface may occure. For this r�n. point C is often termed as boiling crisis or burnount point. We may be interested to operate. a heat transfer surface close to this value but it is not wise to exceed it. Fig. 14.2 (b)' shows the boiling of methanol on a horizontal tube.
484
Heat and Mass Transfer
. • �. 14.2 (b) Boiling df Methanol on Horizontal Tube (a) Nucleate Boiling, (b) Transition boiling, (c) Film boiling [Courtsey of Prof. J .W. Westwater, Illinois l Jniversitvl.
14.3
Boiling and Condensation
Bubble Shape and Size ConsidP.ration
It is interesting to note that the shape, size or inelination of bubbles do not influence much the heat transfer rate in nucleate boiling. On the other hand, the heat transfer rate is greatly influenced by the nature and condition of the heating surface and the surface tension at the solidliquid interface. The surface tension is a measure of wetting capability (i.e low surface tension, highly wetted surface) of the surface with the liquid. Fig. 14.3 shows the three typical shape of vapor bubbles signifiying different wetting characteristics. On the unwetted surface. (Fig. 14.3 a), the bubbles spread out, forming a wedge between the water and l,eating surface, thereby allowing hydrostatic forces to resist the action of buoyancy. In the case of partial wetted surface, (Fig. 14.3b) the angle of contact is smaller than that for unwetted surface which reduces the area occupied by vapor bubbles thereby increasing the heat transfer area. If the surface tension is !ow (or roughness counteracts the surfac� tension). the surface becomes fully wetted. as shown in Fig. 14.3 (c). The liquid tends to push away and shear of the bubbles causing them to become globular or oval and leave the surface while still very small. This situation is the most desirable aspect for large heat transfer rate because of minimum area covered by the insulating vapor film. Addition of agents to reduce the surface tension has been found · have the same effect as providing a wettable surface to give large heat transfer xate. Vapor a
Part i a l l y w•tt•d
To ta lly w•tt•d
Fig. 14.3. Typical Shapes of Steam Bubbles. � 14.4. Bubble Growth and Collapse Experimental evidences in nucleate boiling show that vapor bubbles are not always in thermodynamic equilibrium with the surrounding liquid. In other words, the vapor inside the bubble is not necessarily at the same temperature as that liquid. Consider the forces acting on a spherical vapor bubble. The pressure forces of the liquid and vapor must be balanced by the surfacetension force at the vaporliquid interface. The pressure force acts on an area of n r2 and the surface tension acts on the interface length of 2nr: Making force balance, we get tr r2 (p,.  Pi ) = 2 tr r (1
or
20' pv  pI = 
r where Pv =vapor pressure inside bubble.
( 14.1)
p1 =liquid pressure over the surface of bubble
(T11 _ T,at ), the bubble of radius r will grow otherwise it wi!l collapse. Here, T, is the temperature of liquid surrounding the bubble. (1
.. :f�,. 4'\ ;. ,.
a Fig. 14.4 Force Balance on a Vapor Bubble In the light of above expression, the physical explanation of vapor growth and collapse may be given as follows : Consider a vapor bubble in pressure equilibruim (i.e. ·neither growing nor collapsing). Assume the temperature of the vapor inside the bubble is the saturation temperature corresponding to tl,e pressure p (Fig. 14.4). If the liquid is. at the saturation 11 temperature corresponding to the pressure p1 , · it is below the. temperature inside the bubble. As a result of this heat must be conducte! out of the bubble, the vapor inside must condense, and the bubble must collapse. This type of collapsing of bubble occurs on the heating surface or in the body Qf liquid. In order for the bubble to grow and escape to the surface thev must receive heat from the liauid. This condition is satisfied only
Boiling and Condensation
487
when the liquid is in superheated condition so that the temperature·of the liquid is gi�ter than the. vapor temperature inside the bubble. This corresponds to the metastable condition of nucleate boiling. Though the process of bubble growth is complex in nature, however a simple qualitative explanation of the physical mechanism may be put forward in the following way. The growth of vapor bubble starts when heat is conducted to the liquidvapor interface from the liquid. Consequently evaparation then takes place at the interface, thereby increasing the total vapor volume. If the liquid pressure; p1 remains contant, eq. (14.1) suggests that the pressure inside the bubble must be reduced. The reduction in pressure inside the bubble will result in a reduction in the vapor temperature and larger temperature difference between the liquid and vapor if the bubble stay at its same spatial position in the liquid. However, the bubble will likely rise from the heated surface and the farther away it moves, the lower the liquid temperature will be. There is always � chance for the bubbles to move into a region where the liquid temperature is below tha1. of the vapor. Once this happensheat will be conducted out and the bubbles will collapse. From the above, it follows that the bubble gowth process may reach a balance at some ,ccation in the liquid or if the liquid is superheated enough, the bubbles may rise to the surface before being dissipated. 14.5 Critical Diameter of Bubble The following paramteres are responsible for the maximum diameter of the bubble fonned on the heating surface. (j 1v = tension between liquid and vapor 0"1s = tension between liquid and solid surface
..
: �... ,
,,.: . �.;
Thus
ov s = tension between vapor and solid surface a = angle formd by the bubble as shown in Fig. 14.5. Dc = maximum or critical diameter of bubble. g (p1

D c =f
Pv ) = buoyancy force
fa ,
O'zv · g
P, ) Oiv ] (p, O"zs
Using the dimensional analysis technique, we get (14.5) where C � a constant = 0.0148 .for water bubbles.
Fig.14.5 Critical Diameter of Bubble
HeatandMass Transfer
488
14.6. Factors Affecting Nucleate Boiling Following factors affect the nucleate boiling : (a ) Pressure : The pressure affects the rate ?f bubble growth and in t� it also affects the temperature difference (t.,  t.,, ) causmg the heat flow. The maxtmum allowable heat flux for a boiling liquid first increases with pressure until critical pressure is reached and thereafter it decreases.
t
1 8x1 0 3
13.S x1Cf
"'E � ... .c �
9 x1 0
3
4 . 5 x 10l 4 6 + A t , C
2
8
,o
Fig. 14.6 Effect of Metal on Nucleate Boiling (b ) Material, shape and condition of (he heating surface : The material of the hcar.1 r\f  . surface affects greatly the boiling heat transfer coefficient. As shown in Fig. 14.1.1 f under identical conditions the boiling heat transfer coefficient is different fc,; : �•· different metals, copper has high value than steel, zinc and chromium. The condition of the heating surface also influencs the heat transfer rates. A rough surface yields a better heat transfer rates as compared to smoth or coated surface. This i� because of the fact the smothness weaknes the metal tendency to. get wetted. Similarly, the shape of the heating surface also affect the heat transmission. (c ) Liquid properties ; It has been confirmed by experiments that the bubble size increase with the dynamic viscosity of the liquid. The increase in bubble size reduces the frequency of bubble formations which in tum reduces the heat transfer rate. In other words, the heat reduces the heat transfer rate. In other words, the heat transfer coefficient increases with the initial temperature of liquid. Further, high thermal coductivity of the liquid induces the heat transfer rates. (d ) Mechanical agitation : Experiments have cofinned that the rate of heat transfer increases as the degree of agitation increases. 14.7. Forced Convection Boiling
When a liqu.d is forced to flow through a heated tube the boiling from the inner surface of the tube is termed as forced convection boiling. The flow velocity strongly influences the bubble growth and the hydrodynamic effects are significantly different tha:1 those corresponding to pool boiling.
Boiling and Condensation
489
Fig. 14.7 shows the flow regimes for forced convection boiling inside a tube. When the liquid enters heated tube, initially the liquid remains in subcooled fonn and the heat transfer to this subcooled liquid is initialy by forced convection and may be predicted from the convection correlations. However, boiling is soon initiated with bubbles appearing at the surface growing and being carried into the mainstream of the liquid. In the bubble flow regime, there is a·sharp increase in the convection heat transfer coefficient. With the increase of the volume fraction of the vapor individual bubbles coalesce to fonn plugs or slugs of vapor designated as slug flow regime. This is followed by an annula r flow regime in which the liquid fonns a film. This film moves along the inner surface while vapor moves at a faster velocity through the core of the tube. Starting fonn the forced convection. the heat transfer coefficient continues to increase through bubbly flow and . annular flow regimes. Immediately after the annular flow regimes dry spots start appearing on the inner surface and the heat transfer coefficient begins to decrease. The transition regime is characterised by the growth of the spots until the surface is completely dry and all remaining liquid is in the fonn of droplets appearing in the vapor core. The heat transfer coefficient continues to decrease through this regime. This regime is followed by the mistflow regime which persists until all the droplets are converted into vapor which is then superheated by forced convection from the surface, There is a slight increase in the heat transfer coefficient during this regime. A vast literature [2022] • is available on this topic.
{> Flow ,: ., ·
,,;i
Flow regime
l
1: Ql ·o Ii=
Ql
* 8 �
C:
� iii
r
C:
..fl 0
C:
0
�
0 LL
Ql
J: L'''...._:��
Saturated vapor
Saturated liquid
Distance from inlet
Fig. 14.7 Flow Regimes for Forced Convection Boiling Inside a Tube. 14.8. Boiling Correlations Free Convection Boiling. As discussed earli�r. when evaporation takes place at the liquidvapor interface the heat transfer is solely due to free convection and it is governed by. Nu =/1 (Gr ) /2 (Pr )
where functions/1 and/2 depend on the geometry of the heating surface.
Heat and Mass Transfer
490
Fritz established a correlation for water boiling at atmospheric pressure in free convection in a vertical tube heated from outside. h =1.973
(Aq )¾
= 1 5.I I.!Me }3 W /.m2 K
[g(P
P,)] ½ [ , ,. ]
( 1 4.6)
Nucleate Pool Bciling (i) Rosenhow [ 14.3] recommended the following correlation for nucleate pool boiling. qs• _  µ, · h fg where
µ, = h1, = P1 = Pv = (J = = CIX
est
=
1
At,
CJ
(14.7)
1 C ,1 . h fr Pr, 1 .
liquid viscosity, kg/ m.s
enthalpy of vaporisation, J/kg.
density of saturated fluid, kg/ m3
density of dry saturated vapor, kg/m3 surface tension, N/m specific heat of saturated liquid, J/kg K surface fluid constant
Pr, = Prandtl number of saturated liquid
(t,  tsat ) = excess temperature, K ,, = surface heat flux, W/m2 qi Here the �ripts f and g refer to (liquid ) and vapor (gas) states repectively. The coefficient C,1 depends on the surfaceliquid combination and may vary form 0.0027 to 0.014. Taole 14.1 gives the values of Cs, for various water surface combinations Tables 14.l [!{eference 14.4 ] l!,.tc
=
Surface
Copper
polished scored Stainless Steel ground and polished mechanically polished Platinum Brass
cs,
0.013 0.0068
0.008 0.013 0.013 0.006
., ": .
... ... '..
,;
Boiling and Condensation
49 i
(ii) Jacob [24] recommended a simple expression for the convective heat transfer coefficient associated with free convection and nucleate boiling ir. water h = C (A tS (1 � )0.4
(14.8)
Where p and Pa are the system and standard atmospheric pressure respectively. Table 14.2 gives the values of C and exponent n for h in units of W/m2K
Swface .
Horizontal
Vertical
Table 14.2 for C and n [ 1 4.4] q/ f!6.W/M2) q; 3
(5)
509
(r: T:,, )
+ 0.4 Cpv Ate )
]¼
{ts  tsa1 }
h a2
(14)
_ P,  P., u,,.  {__ .....:...;;__ 3µ
(15)
m = P, (P, 
(16)
{t.at  t8 ) x ] 0 = [4 k µ g P, {Pz  P., ) h1r
P., } g b
3µ
o3
•
Heat and Mass Transfer
.) l V
(17)
hx = k/o
(18)
h
(1 9)
z= [
h
(21)
h _. 1 . 1 3
P, (P,  Pv ) k g ht, µ L {tsa1  t.,3 ) [
]¼
m = .!L h t,
(22)
hind = hw:rtiaal x (sin 8 ) ¼
(23)
h,.. =
(24)
g h 1, = 0.728 [
(25)
P  Pv ) k g h t, h v = 1 . 1 3 [g ( , L µ {t,..,  t1 }
(26)
(27) (28) 1
]¼
= 4 hL 3
(20)
 )
P, (P, Pv k g ht, . 4 µ X (t_..  3t., )
McAdams
o.0077 [
g , P p { � 11, )k ' ] (Re J" �
�
(�  � ) k 3 h t, ] M D (t,..  t1 ) 3
]
¼4
ht ho.rt, = 0.728 Pi (Pi  Pv ) k g , [
D (t,.., 3 t., ) µN . 3
P (P  Pv ) k h t, h = 0_555 [ , 1 µ D (t,..,  t1 )
ht, =
h t, + ¾ Cpl (,..  t., )
]¼.
¼
fer single tube
fer vertical tube
]¼
fer bank of horiwntal tubes
Problem 14.S. The outer surface of a vertical tube of 1 .5m length and outer ,tiameter of 10 cm is exposed to saturated steam at atmospheric pressure and is maintained at 50°C by the flow of cool water through the tube. Calculate the rate of hea� transfer to the coolant and the rate of condensation of steam.
Boiling and Condensation
511
Solution. Assume laminar film condensation on a vertical surface and absence of on condensable gases in the steam Thennophysical properties of saturated vapor at atmospheric pressure are:
tsai = 100°C. P,, = 0.596 kg/rrr, h18 = 2257 kJ/kg
Thennophysical properties of water at film temperature t1 = (100 + 50)/2 = 75°C are: p1 = 975 kg/m3, µ =_375 x I06 N.sJm2, k = 0.668 W/mK.
For laminar condensation on a vertical surface h = · 1. 1 3
or
[
3
P, (P,  P, ) g k htg µ L (tsai  ts )
]¼
3 3 975(975  0.596) x 9. 8 1 x (0.668) x 2257 x 1 0 ] = 1 1 3 . h [ 375 X 1 0 6 X l .5 (100  50)
1/4
= 4365 _3 Wlm2K
The heat transfer rate = q = h (tr DL) (tsa1  ts ) = 4365.3 {tr x 0. 1 x 1.5) (100  50) = 102854.95 W = 102.854 kW The condeµsation rate is giv�n by
Ans.
Ans. 3 2'257 x 10 The assumption 1,f :am:nai film condensatiQJl may be checked by calculating Re. m = ..!L = _! 02854.95 = 0. 0455 kg/s
h fa
5 __ = 1544_8 5_ .04 _ 4m _ "" _ 4�_ = _ __4_x _0_ Re = _ 6 µ b µ ,cD 3...,5 x 10 x tr x 0. 1
Re = 1544.8 < 1800. · l�TJt:e the flow is laminar Problem 14.6. A steam crn,denser installed at the exhaust hood of a steam turbine consists of a squarP array of 900 horizontal tu"bes each 8mm in diameter. The tubes are exposed to saturated steam at a pressure of 0. 15 bar and the tube surface temperature is maintained at 25°C. Calculate the heat transfer coefficient and the rate at which the steam is condensed per ,.mit length of the tubes. ,S.o/ution. Assume film condensation on the tubes and absence of noncondensable
gases. Thermc,physical properties of vapor from table are : At p = 0. 15 bar, lsai = 54°C, p. = 0.098 kg/m3, hfg = 2 373 kJ/kg Thermophysical properties of saturated water at film temperature, t1 = (54. + 25)/2= 39.5°C are: P1 = 992 kg/m3 , µ= 663 x io6 N.s/m2, k = 0.631 W/mK. With 900 tubes. a"'.30 830 tubes of.square array could be formed.
512
Heat and Mass Transfer The heat tr..nsfer coefficient for a square array is given by
2 2 0.0 8 .81 x (0.63 1)3 x 2373 x l 03] With N = 30' h = 0 _728 [99 (99  9 )3 9 30 X 8 X 10 X 663 X 10 6 (54  25)
114
= 4326.5 W/m2K The rate of condensation for the single tube of the array i'> m = .!L = 1
"1g
h(1tD} (tsa1  t) 4326.5 (1t � 0.008) 4  25) ? = 1 .328 x 10 3 kg/s.m = hfg . 2373 x 10
For complete array, the rate of condensation is
m = N2 • m 1 = 900 x 1 .328 x 10 3 = 1 . 1 9.52 kg/s.m Problem 14.7. Compare the performance of horizontal and the vertical tube for the heat transfer coefficient, hat transfer rate and of condensation if in e_ach case the length of the tube is 1 .5m and outer surface of the tube is maintained at 80°C by the flow of cooling water through it. Comment on the results. Solution.
h h:
)114
1 0.768 ( ; 0_ 6
Ans. This shows that the heat transfer coefficient with horizontal tube is 7 1 .7% more than with vertical. =
0.768(L/D) 114
qh
hh
�
V
 = ,; =
=
=
1.717
1.1 1 1
Ans. A�
We observe that the performance of horizontal tubes for condensation purpose is much superior to vertical tube the therefore �orizontal tube is prefered. Problem 14.8. Being a condenser designer you are entrusted to design a steam condenser to condense 5000 kg/h of dry and saturated steam at a pr�ssure of 0. 1 bar. A square array of 400 tubes, each of 6mm in diameter is to be used. The tube surface temperature is to be maintained at 24°C. Calculate the heat transfer coefficient and the length of each tube assuming single pass. Solution. At 0.1 bar, the pr.operties of dry and saturated vapor are : 3 tsot = 45.74°C, Pv = 0.0676 kg/m , hfg = 23 93 kJ/kg _
.,
513
Boiling and Condensation
°
The properties of saturated water at the mean film temperature, 'l = (45.74 + 24)/2
= 30 C are
p1 = 993.95 kg/m3, k = 62.53 1 x 102 W/m.K, µ = 728.15 x 10 6 kg/m.s
Since the tube is to be arranged in square array, hence of horizontal · the number · tubes in vertical column is N = ✓(400) = 20 The average heat transfer coefficient for steam condensing on bank of horizontal tubes is
[
]l
P, CP,  P)/? g h[g h = 0.728 ND µ (tsa1  t)
/4
3
99. .95 993.95  ·0.0676) X (62.53 1 X 10 2) X 9.81 X 2393 = 0.728 x [ J ( 20 X 6 X 1 0 3 X 728. 15 X 1 0 6 (45.74  24) =
1/4
X
1 03]
5389.1 W/m2 K
The heat flow rate = q = h As (tsa1  t) = h (N 2 1t DL) (t501  t) = m h1g or
5389.1 (400 X 1t X 0.006£) (45.74  20) =
or
L
=
5000 3600
X
2393
X
3 10
332361 . 1 = 3 17m · 1045889.7
Problem 14.9. Steam at 0.08132 bar is arranged to condense over a 60 cm square vertical plate. The surface temperature is maintained at 28°C. Calculate the following : (a) film thickness, local heat transfer coefficient and mean flow velocity of condensate at 30 c_m from the top of plate. (b) average heat transfer coefficient and total heat transfer from the entireplate. (c) total steam condensation rate. What would be the heat transfer coefficient if the plate is inclined at 30 degree with the horizontal plane ? Comment on the results. Solution. Assume laminar flow film condensation on the vertical plate. Properties of saturated vapor at 0.08132 bar are : 3 ,SOI = 42°C, Pv = 0.0561 kg/m , hfg = 2402 kJ/kg
The properties of saturated water at the mean film temperature, 't= (42 + 28)/2 = 35°C are : p1 = 993.95 kg/m2, k = 63.53 l x l0 2 W/mK, µ = 728. 1 5 x l 0 6 kglm.s
Heat and Mass Transfer
5 1�
The film thickness at a distance x from the top edge of the plate is given by
6
4 x 62.53 l x l0 2 x 728. 1 5 x l0· (42  28) x = [ 9.8 1 X 993.95 (993.95  0.056 1) X 2402 X 103 (a)
]
1 14
= l .8 l9 x l0 4 (x)l/4
At x = 0.3m from the top edge of plate,
c5 = 1 .8 1 9 x 10 4 (0.3) 114 = 1 .346 x 10 4 m = 0. 1 346 mm
Ans.
The heat transfer coefficient 62 53 x 10 2 = 4645 W/m K .6 h = !£ = . 2 :r O J .346 X 10 4
Ans.
The mean flow velocity of condensate =
o2
�
3µ
\? = . 0.8 1 x \(,0.4256 x 10 4,' = 993 95 x 0.8 08 x 1 0 2 6 3 x 728.15 x 1 0
mis
At the bottom of the plate.
Ans.
o, = 1 .8 1:) x 1 0� 4 (x)114 = l .8 1 9 x 1 0 4(0.6) 114 = l .6 x 10 4 m Average heat transfer coefficient h
=
ii
2 = i x 6253 1 x 10 = 3908. 1 8 W/m2 K 3 30 J .6 X 1 0 4
Using McAdam's correction h = 1 .2 x 3908. 1 8 = 4689.8 W/m2 K The total heat flow rate is
q =· hA� (tsa,  t) == 4689.8 {0.6 X 0.6) {42  2.8) = 23636.59 W
Ans.
(c) The rate of steam condensation is
m = ..!L = 2363 65 93 = 0.00984 kg/s hfg 2402 x 10
Ans.
If the plate is iDclinded at 0 with the horizontal, then g is changed to g sin 0, hence hinc/ined = h,ertical X (sin 0)1 14 = 4689.8 X (sin 30)1 14 = 3943.63 W/m2 K
Ans.
Boiling and Condensati�n
5 15
It is worthile to check the type of flow 4 4 x 0.00984 Re = m = = 90.09 < 1 800 728. 1 5 X 1 0 6 X 0.6 µb Hence the assumption is correct. Problem 14.10. A vertical flat plate in the from of fin is 50 cm in height and is exposed to steam at atmospheric pressure. The plate surface is maintained at 60°C. Calcu late (a) the film thickness at the trailing edege of the film, (b) overall heat transfer coefficient, (c) heat transfer rate and (d) co·ndensate mass flow rate. Assume laminar flow conditions and unit width of the plate. Solution. At atmospheric pressure, the properties of vapor are : tsa, = 100°C, hfg = 2257 kJ/kg, Pv = 0.596 kg/m3 The properties of saturated water at the mean filni temperature 'l = (100 + 60)/2 = ° 80 C are :
,,.
6 2 2 3 p1 = 971 .8 kg/m , k = 67.4 13 x 1 0 W/mK, µ = 355°.3 x 10 N s/m or kg/m.s (a) The film thickness at a distance x from the top edge o�the plate is given by i/4 4 k µ (tsat  l.) X 0 =[ ] . Pi (p,  p hfg ) g 1/4 6 = [4 X 67.413 _X 1 0 z X 355.3 X 1 0 ('00  60) X 0.5] 9.81 X 97 1 .8 (97 1 .8  0.596) X 2257 X 103
=
i.14
x 1 0 4 m = 0. 174 mm
(b) The overall heat transfer coefficient
2
= !!3 ! = i3 x 67.413 x 1 04 =
5 1 65.7 W/m2K 1 .74 >.< 1 0Using McAdam's correlation which is 20% higher than the Nusselt's result. h = 1 .2 x 5 1 65.7 = 6 198.84 W/mK (c) The heat transfer rate is q = hA5 (tsa,  t) = 6 198.84 (0.5 x 1) (100  60) = 123976.8 W (d) Condensate mass flow rate h
m
6
= _g_ =
hfg
123976· 8 = 0.0549 kg/s 2257 x 103
It is essential to check whether the flow is laminar or not. 4 4 x 0.0549 Re = m = = 6 1 8 < 1 800 355.3 X 1 0 6 x 1 µb I
Ans.
Ans.
Heat and Mass Transfer
' ) 6
l
This shows that the assumption of laminar flow is correct Problem 14.11. An outer surface of a 'Ylinderical drum having 30 cm diameter is exposed to saturated steam at 1.5 bar for condensation. The surface temperature of the drum is maintained at 89°C. Calculate (a) length of the drum and (b) thickness of condensate layer to condense 60 kg/h of steam. Solution. Assume film condensation and laminar flow. The properties of saturated vapor at 1 .5 bar are : tsa1= 1 U.34°C, Pv = 0.862 kg/m3, h1, = 2226 kj/kg
The properties of saiurated water at th� mean filrri temperature, 11 = ( 1 1 1 .344 + 89)/2 = 100°C are : p1 = 958.3 kg/m3 , k= 68.227x 102 W/mK, µ = 283 x lfr6kg/m.s (a)
The film thickness at the bottom edge of the drum is
¼
_ 4 k µ It_  t., } x " 5 [P, (P,  Pv ) g h 1, ] 6
2
= 4 X 68.227 X 10 X 283 X 10 ( 1 1 1 .33  89) / [ ] 958.4 (958.4  0.862) X 9.81 X 2226 X 10
The average heat transfer coefficient is 4 5 k ,=3 4 h ·= 3
Using McAdam's correlation.
X
68.227x 102
1 .712 X 104 (L )
¼
¼
4
=1.712 X 10 (L)
¼
= 531 3.6 (L)¼
531 3 ·6 = 6376.32 (L)¼ h = 1 .2 X (L)¼ The heat transfer rate = q=hA., (t.,.,,  t.,) = m h11
or 6376.32 (L) ¼ (1t'x 0.3L) ( 1 1 1 .34  89) = (50/3600) x 2226x 103 I
•
' '·
._
...
 . .· ¾3
or 134253.12 (L)314 = 30916.66
or L = ( 30916.66 } 134253.12
= 0.1412m
Hence the. length of drum '= 0.1412m /I '
1/
Ans.
(b) 5 = 1.712 X 104 (L)74 = 1 .712 X 104 (0.1412)74 = 1.0494x lo4 m = 0.10494 mm 1/
Ans.
Boiling and Condensation
517
It is worthwhile to check the validity of assumption. 4 x (60/3600) = 249.9 < 1800 Re = 4m = 6 b _µ 283 X 10 X tr X 0.3 :fence assumption is correct
EXERCISES F'or thermophysical properties, see the tables given in Appendix (a ) Boiling
·. ,
..
14.1. List the application of boiling heat transfer. 14 .2. Discuss the physical mechanism of boiling. 14.3. What do you mean by pool boiling ? How does it differ from forced convection boiling? 14.4. Discuss the phenomenon of nucleate boiling: Mention the factors that affect nucleate boiling. 14.5. Discuss the various regimes of saturated pool boiling. . 14.6. What do you mean by burnout point ? 14. 7. , Discuss the bubble shape and sire. 14.8. Discuss the fl,ow regimes of forced convection boiling inside a tube. 14.9. From experience house wives know that the risk of burning foods with siliconcoated pans is less than with ordinary uncoated utensils. Explain this. 14.10. A long electric wire of Imm diameter earring electric current dissipates 4000 W/m and attains a surface temperature of l 25 °C when submerged in water at atmospheric pressure. Calculate the boiling heat transfer coefficient \ 14 . 1 1 Calculate the nucleate pool boiling heat transfer for water under atmospheric pressure in contact with ground/ polished stainless steel when the excess temperature is 20°C.
14.12. Estimate the heat flux and boiling heat transfer coefficient for water boiling on a horizontal flat plate taking excess temperature to be 10°C. Fritz correlatiom may be used. h = 1.54 (q /A )¾ = 5.58 (L\t• )3 W/m2 K.
Ans. 54150 W/m2, 5580 W/m3 k. 14. 13. An electrical horizontal conductor, 2 mm diameter with emissivity 0.5 carries electrical current which cau�s its surface temperature to rea� 550°C when immersed in water under atmospheric pressure. Calculate the power disripation per unit length of the conductor. I4. I4. A wire having Imm diameter and I 5cm long carrying electri�_, current is submerged horizontally in ')Yater at 7, bar. The voltage drop in·the wire is 2.I 5V and a current of 131.5 Amp flows. Determine the heat flux and boiling
518
Heat and Mass Transfer
heat transfer coefficient if the surface temperature is to be maintained at 180°C. Ans. 6 x 105 W/m2, 39885 W/m2 K. 14. 15. Calculate the current at which Imm diameter nickel wire will bum out when submerged horizontally in water at atmospheric pressure. The electirc resistance of the wire is 0.13 0./m. 14.16. Calculate the nucleate pool boiling heat transfer coefficient for water boiling under atmospheric pressure on the outer surface of a vertical, 15mm diameter tube maintained at 10°C' above the· saturation temperature. What is the effect when the tube is_positioned horizontally? (b ) Condensation
14.17. Discuss the mechanism of condensation. 14.18. Piffetentiate between the mechanisM of filmwise and drop wise condensation. 14.19. Which type of condensation is used in design of condenser ? Give the argument 1420. Derive the Nusselt theory o_f laminar flow film condensation on a vertical plate. 14.21. For laminar film condensation on vertical plate prove the following. ..:_ 4k . µ (t a  ts ) x O[ P1 (P1  Pv ) g . h fg
l
¼
r ,r 4
g • h fg and hr. = 0.943 A IA Pv ) k µL (tsai  ts )
¼
14.22. Dry and saturated steam at atmospheric pressure condenses on a 40mm diameter tube. The surface temperature, is maintained at 60°(::. Calculate the heat transfer coefficient if the tube is lm long and is oriented (a) horiwntally, (b) vertically Also, calculate the rate of steam condensation in each case. Ans. 555� W/m2 K, 7500 W/m2 K, 0.01207 kg/s, 0.01621 kg/s
•
14.23. A 60cm square vertical plate is exposed to saturated steam at one atmospheric. The surface teinperature is maintained at 80°C. Calculate the following (a) 'local convection coefficient at the middle and at the bottom of the plate. (b) average convection coefficient for the entire plate. (c) t0tal condensation rate and total heat transfer rate to the plate. 14.24. Dry and saturated steam at one atmospheric pressure condenses on the outer surface of¾ a vertical 15cm diameter pipe of length l .2m. The surface temperature is 90°C. Calculate the total condensation rate and the heat transfer rate to the pipe. . 4.25. A vertical plate 3 m high, is exposed to saturated _steam at atmospheric pressure. The surface temperature is 55°C. Calculate the condensation rate and the heat tranfer rate per un,it width ol th,� 1 late.
..
Boiling and Condensation
519
If the plate height were halved, could turbulent flow condition still exist ? What effect would reducing the height have on the heat transfer rate per unit width of the plate. 14.26. A horizontal tube of 6 cm diameter, is exposed to steam at 0.15 bar. The surface temperature is maintained at 35°C. Calculate the condensation rate and heat tranfer rate per unit length of tube. 14.27. Dry and saturated steam at a pressure of 0.08 bar is to be condensed on a · square array of 100 tubes each of diameter of mm. The tube surfaces are maintained at 26°C. Calculate the condensation rate per unit length·of the tube.
15
Special Convective H_e at Tr_a nsfer Processes
Mode� t�chnology has �reated large number of special convective heat transfer processes. In this chapter, a' brief,description of fivesuch processe� will be presented. These are (a) · Magnetohydroclynamic sy�tem
(b) High � flow
(c) T.ranspiration cooling
· (d) Lowdensity fl�w (e) Ablation
Here, the treatment of these · topics is not intended to be comprehensive. Only the basic p�ysical mechanisms will be discussed. 15.
Heat Transfer in Magn�tohydrodynamic
(MHD) Systems
· g to Faraday's law of electromagneti� induction an electric conductor moving gnetic field �riences a a retarding force as well as induced electric field and cwrenL s· effects are found when a conductive fluid moves through a magnetic field A conductive uid may be obtained by ioni!ling the gas which may occur as the result of an elevation o mperature or by suitable seeding process. \
Our man objective is .to find out the heat tqtnsfer rate· to a conducting flµid under the influence of magnetic field. The applkation of this field of heat transfer is �00 =0, 1J= 0 d1J
8 ='0 of y = oo. Tl= .,...
(15.55)
Heat and Mass Transfer
528
Using the above boundary conditions, Pohlhausen has given the· solution of equation (15.55) as Pr
d 00 {7] , Pr}= 2Pr f ( 'i) 00
I
�
d 2 1J
[I� (dJ) o d 2 1J
2
Pr d7J] d7J
(15.56)
where 00 is used to indicate the adiabatfo wall solution. Fig. 15.6 shows a graphical plot of the solution. 10 �:""lli!!!lliS;;:!:::.�r��.., 0.8
JI.J :_ l, 0.4 0.6
The recovery factor is given by r = 00 (0, Pr)
•·y . •.
For Pr = 1 (i.e. for air), the recovery factor is given by Laminar flow ;
r r
Turbulent flow ;
=
(Pr) 1fl
= (Pr
(15.57)
,
�.�',· . 0
(15.58) ''•' :·
) 1/3
•••
These recovery factors along with equation (15.43) may used to calculate the adiabatic wall temperanue. As there is a large property variations across the high speed boundary layers the I thermophysical properties are evaluated at a reference temperature given by � f 1 (f5.5� t • = t_ + 0.5 {ts  t_) + 0.22 {tas  t .. ) Following correlations arc. used.for highspeed heat transfer calculations. 5
(i) Laminar boundary layer (Rez L 5 x 10 ) 7/
¼ St • (Pr·)73 = 0332 (Re•) 2 x
x
(ii) Turbulent boundary Layer (5 x HP L Rez L 10 )
• ( ·)½ = 0.0288 (Re·)½
Stx Pr.
x
7
.\
(15.60)
(15.61)
· 529
Special Convective Heat TransferProcesses 10 > C c
r t
o r condensing vapor ( Y100) th l
i
=t 2 i,
At 1
Area
➔
Fig. 16. 15. LMTD for Condenser t
'u f
I
I
C h«Cc·
or e v a pora t ing l i qui d ( c,.... ao) Hot
M1
l
t112
tc,
= tc2
Area 
2
tcz
Fig. 16.6. LMTD for Evaporators
fl"l:!at Exchanger:,;i
553
without phase change and Ch l d(At ) dA o 61 1 Ada +b At ) q
(3) (4)
(5)
(6)
J�
f
or
The constant a may be expressed by solving U1 = a + b At1 and U2 = a +b At2 which results in U1 M2  U 2M 1 a = At 2  M 1
(8)
Substituting the value of a from eq. (8) into eq. (7), we get u l A t z  Uz At 1 ProVed q = ,� In IU 1 At 2 / U2 At i )
Problem 16.8. For experimental purpose, a steam surface condenser is to be designed to transfer 300 kW of thermal energy at a condensing temperature of45°C. The cooling water enters the condenser at 20°C with a flow rate of 25,000 kg/h. Calculate the surface area required to handle this load if the overall heat transfer coefficient for the condenser surface 1400 W/m2K. If the outside tube diameter is 30mm and the tube length is I m, calculate the number of tubes. Compare the result with the use of arithmatic mean temperature difference.
Heat Exchangers
563
Solution. Refer to Fig. 16.22 We know that during the condensation of steam, temperature remains constant th,roughout the cOiidenser, i. e. lh l = lh 2 = 45°C
Using the heat bamnce on the'water side, we get q = me Cpc (tc2  tc1 ) 300 = 2SQOO 3600
or
X
4. 18 {tc 2  20}
300 X 3600 . + 20 = 30.33 oC 25()()() X 4.18 At 1  At 2 L,MTD = At,,. = In (At i /At 2 ) Here' At I = thi  Icl = 45  20 = 25 °C and At2 = th2  Ic2 = 45  30.33 = 14.6 7°C tc 2 
or
14.67 = 1937°c :. At ,,. = 2s In (25/14.67} The heat exchange rate is given by q = UA At,,. or 300 x 103= 1400 x A x 19.37 or
A
=
2 300 x 10 = ,.... i 1.06 m  1400 X 19.37 3
4 = ntrdoL = n x n 0.03 x 1 = 1 1 .06 m2 __::· l l.06 = 1 17.3 = 1 18. or n = 1r X 0.03 The arithmatic mean temperature difference is At + + Ala = 1 M2 = 25 14.67 = 19_83oc 2 2 Now q = U A Ala or 300 x 103 = 1400 x A x 19:83 or
A
=
300 x 103
1400
X
19.83
Ans.
Ans.
= I O. S m2
Enor = 1 1 .06  10.8 X 100 = 2.3 % 1 1 .06 Ans. Problem 16.9. A counterflow tubular oil cooler is to be designed to cool 1500 kg/h of oii from temperature 90°C to 30°C by means of water entering the cooler at 20°C .id leaving the cooler at 50°C. Calculate the amount of water flow rate required and L'le hea• transter area. Take cP of oil as 3 kJ/kgK and overall heat transfer coefficient equal to eo,J W/m2K.
Heal and Mass Tiansfcr
Solution. R�fer_ Lo Fig. 16.20. The heat exchange raLe is given by
 x 3 x 10 3 (90  30) = 74999.99 . q = mo cpo (l02  to1 ) = 1500 3600 Also, q =_ 14999.99 = Tnw C ( tw2  t1 ) or m w =
pw
74999.99
4.18 (50  20) X 10
3
w A ns.
= 0.598 kg/s
The logmean temperaLure difference is M 1  M2 A m = ,�,u.l In (�11 /� t2 )
Here, M l = �(h i  tc2
J(
=
M
90  50 = 40°C and �/ 2 = th2
m
=
40  10 = 21.64cc In (40/10)
 te l
=
30  20 = 10°c
or 74999.99 = 1200 x A x 21.64 or A =
2 74999.99 = 2.88 m 1200 X 21.64
A ns.
Problem 16.10. An industrial gas turbine plants employs a counterflow concentric tube heat exchanger which cools the lubricating oil coming out from its bearings. The oil mass flow rate is 0.2 kg/s and its specific heat is 2.2 kJ/kgK. The mass flow rate of the coolant (water) which flows in the opposite direction is 0.16 kg/s and its specific heat is \ 4.18 kJ/kgK.. The lubricating oil enters the cooler at 110°C and leaves at 40 °C. The cooling water enters the cooler at 20°C. The heat transfer coefficient from the oil to the I tube surface is 2300 W/m2K and from tube surface to water is 5700 W/m2K. The , resistance of the tube wall 'may be neglected. If the mean diameter of the tube is 15mm, . calculate the length of the tube. Neglect the resistance of the scale if any formed.
Solution. Refer to Fig. 16.20 Let subscripts o and w refer to the oil and water respectively. Using the heat balance m 0 Cpo · (to1  to2 ) = 'fnw . Cpw(t..,2  t .., 1 )
or 0.2 X 2.2 (110  40) = 0.16 X 4.18 (tw2  20) or t..,2 :: 66.05°C.
The �c transfer rate is q = ..._m0 cp0 (t01  t02) = 0.2 x 2.2 (I IO  40) = 30.8 kW
The LMTD is given by
.,
Heat Exchangers
565
Here, fl/1 = lh l  lc2 = 101  lw2= 1 1 0  66.05= 43.95°C fl t 2 =
lh2  le 1=lo 2
°  I,., 1 = 40  20 C
flt = 43.95  20 = 30.41oc m ln (43.95/20) Since the resistance due to tube wall and scale formation it any are negligible, hence
•
h_ h_ w . o _ _I__= _ = 2300 x 5 700 =1638_75 W/m2 K ., U= _ h h + 2300 + 5700 1 w 0 __!_ + _ hw ho
The heat exchange rate is given by q = VA fltm or or
30.8 x 103= 1638.75 x A x 30.41 A = 0.6 I 8 m2 = 1t d L
_or
L =
0·618 = 13.11 m 1t X 0.0 }5
Ans.
Problem 16.11. A single pass surface condenser is to handle 5100 kg/h of dry and saturated steam at 50°C. The outer and inner diameters of the tubes are 20mm and 17 mm respectively. The length of the tube is 3m. The cooling water enters the tube at 20°C. The velocity of water though the tube is 2 m/s and the temperature rise in 1he cooling water is 10°C. Calculate number of tubes. Take the following data Heat transfer coefficient of steam side=11,700 W/m2 K. Fouling factor of water side=0.0002 m2 K/W. Thermal conductivity of tube material=93 W/mK. Fouling factor of steam side=0.0009 m2/W. · Solution. Refer to Fig. 16.22 The mean bulk temperature of water=tb=(20 + 30)/2=25°C The thermophysical ,properties of water at 25°C from table are p= 996.95 kg/m3, k= 60.7.385 x I 02 W/mK, v = 0.9055 x l � m2/s, Pr= 6.22 flt1  M z � LMTD = fl tm = In (At1 / M 2 ) �ere,
Ai1=50  20= 30°C and At2=50  30=20°C
A tm = 30  20 = 24.66oC In {30/20} The heat transfer coefficient of water side is not known· hence it has to be found out Nu = 0.023 (Re)08 • (Pr)0.3 U oo D But Re = = V
2 X 0.011
0 C)O'i 'i X 106
=
0.375 X 10 5
Heat and Mass Transfer
566
Nu =0.02J (0.375 X 105)08 (6.22)03 = 181.589
10 2 = 64 93.2 W/m2 K or h = h; = 181.589 x 60.7885 x 0.017
The over
Cc, then (tc 2  fc; ) > (thl  th2 )
From the above inequality it follows that tc2 will real;h tc i as seen from Fig. 16.14 and the condition (i) will be satisfied. If Ch< Cc, then (th l  thi) > (tc2  tc1 ) The above inequality shows that tc2 will w.ach the value of tc1 as seen from Fig. 16. 14 and the condition (ii) will be satisfied. The minimum fluid may be either the hot or cold fluid depending upon the mass flow rates and specific heats. Hence if Cc < ch, then Cc (tc2  tc 1 ) t tc2  qacOlal  = ,....,. E == c1 qmax Cc (th )  te l ) th l  tc 1
If
(16.37)
ch < C c , then
t h 1  th 2 Ch (t h 1  th 2 ) qactnal  = ,....,. E == th !  tc 1 C h (th l  te l ) qmax
(16.38)
It i$ possible to put the above expression for effectiveness in different forms . If C,,.;n is .the lower value of two capacities Cc and C,., then (16.39) q= =C,,,;n (th! tc1 ) Using the above value of qmax, effectivness is C (t 2  te l. ) E = c,cCmin (t1i 1  te l }
(16.40)
c,. (t1,1  t1, 2 )
(16.41) Cmin (th I  tc 1 .l Once the �ffectiveness is known it is very easy to calcuiate the rate of heat transfer by the equation
or
=
q = E C min (t1, 1  le 1 )
(16.42)
because the equation (16.42) does not involve the outlet temperature. Equation (16.42) is of course also suitable for design purpose instead of equation involving LMTD. 1 6 . 7 . Effectiveness by Using Number of Transfer Units (NTU) In the preceding section, we have discussed the definition of effectiveness of a heat exchanger in terms of inlet and outlet temperatures. It is worthwhile to derive expressions for effectiveness which may be free from outlet temperature terms and hence be useful for the design of heat exchangers. In the heat exchanger design, heat capacity terms and NTU are important parameters and here effort is made to express effectiveness in these terms. For this, let us consider two type of heat exchangersparallel flow a• · • counter flow.
582
Heat and Mass Transfer (i) Parallel _ Flow Heat Exchanger Consider a parallel flow heat exb), we have
(L
d (t h  tc } + ...!._ ) d A  U C,. Cc (t,.  •c } In�grating the above equation from A = 0 to A = A we get lo (t1i2  tc2 ) =  UA C,. {t1,1  tc i } or
( l + C,. ) Cc
1  tc2 � = exp [  {UA/C,. } (1 +C,. I Cc )] � "2 t 1,1  te l
(16.43)
From equations (16.40) and (16.41), we have the expressions for effecti"eness
Hence,
c,. (r,.1  , ,.2e) •cc (tc2  rc d = ,E = ,,Cmin {t 1,1  tci } Cmi,, (t 1,1  tci } · E Cnun (t,.1  tc1 ) t1,2 = t1,1 
(16.44)
c,.
(16.45)
Eliminating t,.2 and tc2 from equations (16.43) with the help equation (16.44) _and (16.45), we get
If Cc > c,. then Cmin = c,. and Cmaz = Cc ; hence equation (16.46) becomes E=
1  exp [ ( UA/Cmin ) {I +Cmin /Cmaz
I]
I + (C""" /Cmaz }
(16.47)
If Cc < C,.then Cuan = c0and Cmaz = C,.; hence equation (16.46) becomes E=
1  exp [ (U A IC'""" ) jl +Cmax / Cmi,, )] 1 +· (Cmin IC'""" )
(16.48)
583
Heat Exchangers After rearranging equations {16.47) and (16.48), we get a common equation E = 0
1  exp [ ( U A /Cmin } (1 +Cmiil /Cmax )] 1 + (Cmm /Cmlll }
(16.49)
where Cmi,. and Cmlll represent the smaller and larger of the two heat capacities Cc and
c,. .
The dimensionless ratio, AU /Cmin = NTU, here NTU means number of transfer u nits. Thus the effectiveness of a parallel flow heatexchanger is given by E=
1  exp [_· NTU {1 + (cmin / C,nax } } ] I + (cmm / C= }
...
(16.50)
(ii)
CounterFlow Heat Exchanger Consider a counterflow heat exchanger. From the equation (16.13c), w� get d (t,.  tc ) {t,.  tc )
=
U
[.!__.!_] t1i le
dA
!:!_ • C,.
=
[1 _ C,.C ] dA c
(16.51)
Integrating above equation from inlet condition to outlet condition 2, i.e. from A == 0 to A = A, we get
lhz  tc I t,. 1  lcz
or
 =
exp [· (UA JC,. ) (1 C,. I Cc )� 
(16.52)
From equations (16.44) and (16.45), we have E Cmi,i {t1i1  tc i ) thz = l1t1  3Jki lc z = tc 1 C,.
+
E Cmin lt1i 1  tc i ) Cc
Substituting the values of th2 and tcz in equation (16.52), we get !hi 
E Cmin {t1i1  tc i ) .c · ,.
tel
= exp [ (U A JC,.) (1 C,. I Cc )] (16.53) (t 1 _ tc 1 ) ,. t,. 1  tc 1  ______...;._ Cc If c,. < Cc, then Cmi,. = c,. and Cmax = Cc hence the above equation becomes th 1  E (,,. !  tc1 )  tc1 . = exp (UA I cmi,. } (1  C,,.;,. / Cm ... ) / C (t t ) ) ,,..J  tc 1  E (c i 1 ,. 1 c n mar m . E Cmin
,•'
or
,
;
1 E 1  e (Cmm /C ma:c )
.
.
�� =
exp
[ ( U A /Cmi,.) (1  Cmin /Cmlll )] ·
Heat and � Transfer
584
After solving the above equations, we get 1  exp [ (u A .'Cmin } (1  CIIWI /Cmax
E
l]
( 1 6.54)
If Cc < C11, LJien Cmin =Cc and Cmax =C11 ; hence equation ( 16.53) becomes . C . (t11 1  t c d  t c1 ti.1  E Cmax � = exp [ (u A /Cmin ) 1 ) Cmin � \ {t111  tc 1)  E {t11 1  t c i }
cf::
n·_
Solving the above equation, we get 1  exp [ UA/Cmin {l Cmin /Cmax }] . E _ l  (Cmin !Cmax ) exp [ (UA/Cmin ) (1  Cmin /Cmax }J
( 1 6.55)
From equations ( 1 6.54) and ( 1 6.55), it is found that there is a same expression of effectiveness of counterflow heat exchanger for both cases. Writing the equation ( 16.55), in terms of NTU, we get .
.
_ e 
max )]
1  exp [ NTU ( l  C min I C
I  (Cmin / Cmax ) exp [ NTU (1  C min !Cmax )]
( 16.56)
We find thai: effectiveness for parallel flow and counterflow heat exchange is give by different expr�sions
+
1  exp  NTU (1 Cmin !Cmax ) (e )pa,r,11e1 flow = ___...__.________,._ 1 + Cmin !Cmax 1  exp [ NTU (I  Cmin / Cmax )] 1  (Cmill / Cmax ) exp [  NTU ( i  Cmin ICmax )]
( 16.57) ( 16.58)
Here it is worthwhile to discuss two limiting cases of equations ( 16.57) and ( 16.58). Case (i) : When (C,n;,. /Cmax) 0.
=
Using the above case, we get a commo:1 e�pression for parallel and counter flow heat exchanger. E =1  eNTU '( 1 6.59) Such cases are found in evaporators and condensers in which one fluid remains at constant temperature throughout the exchanger. Hence Cma:c =oo and thus C,,.;,. /ma:c == 0. Case fii) : When (Cmin /Cmax) =1 (a) In the case of parallel flow heat excha'lgerusing (Cmin ICmaJ =1, we get · 2 (NTU ) 1 e E =
( 16.60)
Heat Exchangers
j . :s
(b) Using the condition (C mi,JC ma.,) = 1 for the case of counter flow heat exchanger, the expression for effectiveness "becomer indeterminate. However, it is possible to find the value of E by applying the.calculus of limits. 1  (C mi,JC ma:r.) = 0, NTU (1  C mi,. /C,,.,.J C,.. hence the effectiveness of heat exchanger is by E =
qactual qrw:c
=
t h l  t1, 2 = 110  35 110  25 t h l  te l
=
0.882
(c) Here, C,,.;,.=Ch = 1.5 k W/K and Cmax = C, = 2.045 kW/I{, hence CwuJCmtll = 1.5/2.045 = 0.733
Ans.
Heat Exchan�er:s The effectiveness is given by E =
I
l
After rearrangement.
exp [NTU
589
(1tJ
Cmin � �Cnun\,, [ xp NTU ( I  max.� ) \C::f C
Cmin E  1  exp [ NTU (1 )� = E (Cmin I Cmax ): 1 Cmax
or or or
__0_·88 2  1  =exp [_.:_NTU ( 1 0.732)] 0.882 x 0.733  l
0.333=exp [NTU x 0.267)  0.267 NTU = In 0.333
NTU = 4.118 or This value of NTU may also be obtained from chart for CmiJCmax= 0.733. E = 0.882 and We know thatNTU = UA /Cmin
or
4.118= 1500 A/1.5 X 103 A =
3 4.118 X 10 = 2.74 m 2 1500
Ans. Problem 16.21. Steam at atmospheric pressure enters the shell of a surface condenser in which the water flows through a bundle of tubes of diamater 30 mm at the :ate of 0.06 kg/s. The water enters the condenser at 20°C and leaves at 75°C. Steam condenses on the outside surface of the tube. The overall heat transfer coefficient is expected to be 250 W/m2K. Using NTU method, calculate the effectiveness, length of the tube and steam condensation rat.e. Assume that only latent heat of steam is lost during condensation. Take the latent heat of vaporisation at 100°C, hfg= 2261 kJ/kg. Solution. Throughout condenser the hot fluid (i.e. steam), remains at constant temperature. Hence Cmax is infinity and thus Cmin is obviously for cold fluid. Thus Cmin / Cmax "'0. When Ch > Cc, then effectiveness is given by 1c2  le I qacrual 75 20 E = = = =0.6875 100  20 thl  le i q,_
For or
Cmin =me. Cpc= 0.06 X 4.18 = 0.2508 kJ/K
Cmin !Cm= "' 0. t= 1eNTU 1.'' . 0.6875= 1 .:..'. emu
Ans.
590
or or or But
Heat an� ivfass Transfer
eNTU = 1  0.6875= 0.3125  NTU = In 0.3125 NTU= l.l63 NTU = UA /Cmi,. = U.TCDL/Cmi,.
3 (NTU).Cmi,. = 1.163 x 0.2508 x 10 U TC D 250 TC X 0.030 Using the 0·1erall energy balance
or
L =
m h . hfg
=
=
12.37 m
Ans.
m e .Cpc (t c2 t c 1)
m1, X 2261 = 0.06 X 4.18 (75 · 20) _ mh _ 0.06 X 4.18 (75  20 ) X 3600 = 22.96 kg/h or Ans. 2261 Problem 16.22. A finneC: tube, crossflow heat exchanger is to be designed to heat pressurized water by means of hot exhaust gases. The water flowing at the rate of 1.5 kg/s enters the exchanger at 35°C and leaves at 125°C. The hot exhaust gas heat transfer coefficient based on the gas side is 110 W/m2K. Using NTU method, calculate the effectiveness and required gas side surface area. Take the following properties. Exhaust gas : Cp = 1 kJ/kgK, Water (at tb = 80°C): Cp = 4.197 kJ/kgK. or
Solution. Refer to Fig. 16.36.
Fig. 16.36
The heat capacity of cold fluid= Cc= mr Cpc= 1.5 x 4.197 = 6.2955 kW/K The heat capacity of hot fluid may be obtained by writing·overall energy balance as m,. is not given. Hence or,
m,. .Cph{t h l  t h 2)
ch =
=
me .cpc (t c2  t e l )
Cc {tc 2  t c d = 6.2955 ( 125  35) = 2.832 kW/K {300  100·} t{ 1,1  t h2)
Hence, Cmax = Cc = 6.2955 kW/K and Cmi,. = Ch = 2.832 kW/K If Cc > Ch• then effectiveness is t  1, 300  100 E _ hl ! 2 _ {30035)  ,,., t,,)
=
0.754
Ans.
C,n;,. Cmax
=
2·832 6.2955
Heat Exchangers = 0.4498
591
"' 0.45
With Cmi,. /Cmai "' 0.45 and E = 0.754, it results in form Fig. 16.33. NTU
=
uh Ah "' 2 .. 1 c"""
2 Ah = 2.1 X 2.832 X 10 = 54.06 m Ans. . 110 Problem 16.23. A shell and tube type steam condenser imployed in a large steam power plant affects a heat exchange rate of 2200 MW. The condenser consists of a single pass shell and 32,000 tubes, each executing two passes. The water at the rate of 3.2 x 104 kg/s passes through the tubes which are of thinwalled and of diameter 30 mm. The steam condenses on the outer surface of the tubes. The heat transfer coefficient on the steam side may be taken as 11500 W/m2K. Steam condenses at 50°C w)lile water enters the condenser 20°C. Using LMTD correction factor method and NTU method, calculate the following:(a) Outlet temperature of cooling water and (b). lengµi of the tube per pass. :rake tht! following properties :Wai ;r (at tb "'27°C): cP = 4.18 kJ/kg, µ = 855 x lo6N.s/,r;2 , k = 0.6.13 W/mK and Pr = 5.83. Neglect thennal resistance of tube material and fouling effects L Solution. Refer to Fig. 16.37.
or
3
1c, ++tc2
h:32000
A_
Fig. 16.37
Using the overall energy balance, th,� cooling water outlet temperature may be obtained. q = me. cpc(te z lc1 ) or or
2200 X 106 = 3.2 X 104 ?( 4t8() (te2  20)
Ans. le 2 = 36.44°C (i) LMTD correction Factor Method:• q = FU A ..1tm where A = 2 n nDL
To find out overall heat. transfer coefficient, h; has to be calculated.
Heat and Mass Transfer 4
Here, mass flow rate of water through each tube = m = 3·2 x 10 = 1 kg /s 32000 4 4X 1 Re = !±...!!.J!. = 4 m = = 4.96 X 10 6 µ trDµ Tr X 0.03 X 855 X 10 Since Re > 2500, hence the flow is turbulent and Nu= 0.023 (Re)0·8 (Pr)0·4 = 0.023 (4.% x 104)0·8 (5.83)04 = 265.69 h = h; = 265.69 x 0.613 = 5428.93 W/m 2 K or 0.03 If thermal resistance of tube material and fouling factors are m:;glected, then U =
l = l = 3687.92 W/nf K (1/hd + (1/ho) {1/5428.92) + (1/11500)
•
LMTD is given by !!.t I  M 2 In {!!.t i / M 2)
l!.tm = 
..1t2 = 50  20 = 30°C and ..1t2 = 50  36.44 = 13.56°C
Hae
!!.rm = 30  13.56 = 20_7oc In (30/13.56)
io find correction factor, we have to find P and R
p =
fez  tel thl  tc1
= 36.44  20 = 0_548 ax! R = 50  20
thl  th2 lc2  tel
=
50  50 = 0 �6.44  20
According P = 0.548 and R = 0, we get F = 1. q = FU (2ntr DL) L1tm Hence 2200 x 106 = 1 x 3687.92 (2 x 32000 x tr x 0.03L) 20.7 or L = 4.77 m or (ii) NIU Method :Since the heat exchanger is a condenser, hence Ch= Cmai = and Cmua = Cc = mc.Cpc = 3.2 X 104 X 4180 = 13376 X 104 W /K Since c,. > Cc, hence 00
E
Hae or or nr
=
fc2  le I
t,.1  tc1
= 36.44  20 = 0.548 50  20
"Cmi,.!Cmax. = 0 E = 1 eNTU 0.548 = 1  eNTU
,... NTU = In 0.452 NTU = 0.794 = UA!Cmi,.
Ans.
Heat Exchangers
or
or
593
0.794 = U (2 ntrDL) Cmin· �
.,L =
0.794 X 13376 X 10 = 4.77 m _ 3687.92 X Z X 32000 1r X 0.03 4
.1
Ans.
16.8 Heat Exchanger ;Design .Con�iderations In the selection or design of a heat exchanger, following considerations are almost always made (a) Heat transfer requirements (b) Cost (c) Physical size (d) Pres.suredrop characteristics The way that the requirements are met depends on the relative merits placed on items. (b) to (d) above. In order to increas� the overall heat transfer coefficient, the fluids may be forced at higher velocities but tllis higher velocity results in a larger pressure drop through the exchanger, meaning larger pumping costs. If the pressure drop is to keep minimum, the surface area of the exchanger will be larger due to lower value of overall heat transfer coefficient but there is a limit of physical size that can be accommodated, and mor.eo\er, a larger physical sizeresults in higher costs. This suggests that a correct compromise between all the these conflicting factors will lead to the proper design of a heat exchanger. 16. 9. Som e Constn,1cti'onal and Thermal Aspects of. Heat Exchanger Design. (A) Shell and Tube Type (a) Tubes : Heat exchanger tubes are available in a variety of metals and sizes, The materials include steel, copper, admiralty, Muntz metal, brass, 7030 copper nickel, aluminum bronze, aluminum and stainless steels. The wall thickness of a tube is usually· defined by Brirmingham wire gauge (BWG) and is available in variety of forms. A large number of sizes of tubes are available which are listed in Appendix. The most common sizes include 1.905 cm O.D. arid l.54 cm O.D. The flow velocity through tubes may · ·' range from 1 to 2·m/s. (b) Tube Pitch : The tubes are hold by the two tubes sheet placed at both ends. ,The tube holes in the tube sheet should not be very close to each other otherwise it will weaken the tube sheet. In general, tubes are laid out on either square or triangular patterns as shown in Fig. 16.38. The square pitch offers the advantages of accessibility of tubes �
•
(0)
•
I
(bl
Fig. 16.38 Common Tube Layout for Exchanger (a) Square Pitch (b) Triangular Pitch.
594
Heat and Mass Transfer
for external deaning and lower pressure drop when fluid fl�s in the direction indicated. In square p:tch, the velocity of the fluid undergoes continuous fluctuation because of the constructed area between adjacent tubes compared with the flow area between successive rows. Triangular pitch officers ·even greater turbulence because the fluid flowing between adjacent tubes at high velocity impinges directly on the succeeding row. This means that ignoring the pressure ctr.op and cleanability, the triangular pitch is superior for attainment of high shell side heat transfer coefficient and the in&ase in heat transfer coefficient under comparative conditions may be 25 per cent greater than for square pitch.
The values of tube pitch, PT for square and triangulat arrangements of 1.905 cm OD and 2.54 cm. OD tubes are 2.54 cm. and 3.175 cm respectively. The square arrangement may be rotated by 45° , It is possible to have the triangle arrangement with cleaning lanes by spreading the tubes wide enough. (c) Shells. Shells of heat exchangesbasically fall in the category of.pressure vessels, and therefore designed as a thick or thin cylinders. The smaller size of shell (up to 30 .cm) is fabricated from steel pipe. Above 30 cm and including 61 cm. the actual side diameter and the nominal pipe diameter are the same. For shells operating under pressure up to 22 bar, the standard wall thickness with inside diameters from 30 cm to 61 cm inclusive _is 0.952 cm. For operating pressure above 22 bar, the greater wall thickness may be taken. Shells above 61 cm in diameter are fabricated by rolling steel plate. Generally, shells are of cylindrical shapes but the big size condenser of power plant may be of rectangular shapes.
(d) Tube Sheet. Due to temperature difference between tubes and shell thermal stress sets up between them. These are various methods to reduce this thermal stresses and accordingly the exchangers are generally classified. In general, two basic fo�s of tube sheets are used to reduce thennal stress. They are � (i) Fixed tube sheet
(ii) Floating tube sheet Each of the tube sh�et arrangement has various designs. In one of fixed tube sheet arrangement, admiralty metal tubes are rolled into an integral sheet and shell hub which is silversoldered to the shell. The tube bundle is not removable, hence this construction is only for those �ases in which the fluid in the shell does not seriously foul the tubes. In the cases, where there is a change of severe thermal stresses due to large difference in temperature between the shell and tubes, afloating head (or sheet) design is used. Fig. 16.39 shows such a construction in which the different parts of an exchanger are listed.
Fig. 16.39 Floating Head Exchanger with Two Tube Passes and oneCrossbaffled '.ihell Pass (Adapted trom TEMA)
595
Heat Exchangers �I . Shell
2. Shell cover 3. Shel! channel end flange
12. Channel cover 13. Chann�l nozzle
14. Tie rods and s�cers
4. Shell cover end flage
15. Transverse baffles or support
5. Shell nozzle
f6. Impil)geme� t b;,ur1c 1 7. Longitudinal 18. Vent connection ·
6. Floating tuq� sheeL 7. Floating head
8. Floating .head flange
plates
I
Mffie
, 1�. Drain connection
9. Floating head backing device 20� ,Test connection 10. Stationary tube sheet 21. Support saddles 11. Channel 22. Liftina ring This design allows cleaning of tubes by simply removing the channel covers. The tube bundles may also be removed for cleaning the outer sµrfaces of th� tubes. In most of shell and tube heat exchangers baffJes are used to produce turbulence. The passes in tubes or shell side may be more than one � (e) Baffles. it is a well k9own fact that with the presence of tw;bulence in ijquid the heat transfer coefficient increases. In order to induce turbulence ·outside the tubes, baffles are employed which cause the liquid to flo\\ through the shell at right angles•to !he axis of tubes. This results in the creation of considerable turbulence · even wnen u . small quantity,::of liquiq flows through th� tube. Numbe� of baffles to be provided. in the exchanger depends upon many factors such as length ·of the shell, degree of turbulence, etc: The centre to center distance between the b'affles is called the baffle pitch. The baffle size varies from onefifth of the inside.shell diameter to the inside snell diameter. In order to hold securely the baffles, baffle spacers are used which consists of throughbolts screwed into the tube sheet and a number of smaller lengths of pipe which form shoulders . between adjacent baffles. Thoug.h, tl)ere are severa( types of baffles but the most common are segment al baffles which are drilled.plates with height equivalent to threeforth of the inside diameter of the shell, Such baffles are known as 25 percent cut off baffles. Other 1 types of baffles include disc and doughtnut and orifice plate. ' . (f) Tubesheet Layout s and Tube Count s. · Fig. 16.40 shows actual tubesheet fayout with partition arrangement for a 33.65 cm. I. D. shell with 2.54 c..m 0. D. tubes
TIie 11..,,., linN lnolica portltlens IKatH 111 Ille floatiftt lleael ce.•
Fig. 16.40 Tube Sheet Layout
. Heat and Mass Transfer
596
on 3.175 cm triangular pitch arranged for six tube passes. It is worthwhile to note that tubes are not usually laid out symmetrically in the tube sheet. In order to minimise the contraction effect of the fluid entering the shell, extra space is usually allowed in the shell by omitting tubes ,directly under the inJet nozzle. When tubes are laid out with minimum space allowances bet:ween partitions and adjoining tubes and. within a diameter of free obstruction called ilie out er t ube limit , the number of tubes in the layout is the tube count . It is desirable to hav� _equal number, of tubes in each passe but it is not always possible ; however the unabi,nce should not be more than 5 percent.
(g) Shell Side Heat Transfer Coefficient . The shell side heat transfer coefficient depends on baffle spacing type of pitch, tubes size, clearance and fluid flow characteristics. To find the Reynolds number of flow in the shell, it is essential to know the expression for hydraulic or equivalent diameter. The equivalent diameter for square pitch is D = e
4 x �P/  ir D ; I 4 ) 4 x free area = Wetted perimeter K D0
( 16.62)
where PT is the tube pitch and D0'is the tube outside diameter. For triangular pitch, the equivalent diameter of shell . is
2 4 x [ 0.5 PT x 0.86 PT  0.5 ir Do /4 ] _ De  "�...._ (l/2) K D0 ,
( 16.63)
For values of Re number from 2000 to 106, the shell side heat transfer coeffici�nt is given by
hD · = 0.36 (D......!._!_ G o s c µ)½ ( )0. (_p__ ) .J!:_ 5
.. k
_ _e
m
k
where Gs = mass velocity, kg/s. m2 = m!As
µ "'
1 4
( 16.14)
As = shell side or bundle crossflow area. McAdmes equation may be also used.
(b) Pre�sure Drop : (z) Shell side : The pressure drop on the shell side is given by !).ps =
p fGs 2 D ( N + 1) · � bar 10 De ef>s
Where Ds=inner shell diameter p=density
N = number of baffles The tube side p�ssure drop is given by
p JG/ L n 6p,° = 
( 16.65)
ef>s = viscosity ratio = (µ/µw)s°·14
f = friction factor
,
(16.66)
Heat Exchangers
597
where n = number of tube passes aria L = tube length The loss due·to change in direction of fluid in multipas·s tubes is given by Apr = 4 np V ( N + 1) bar 2
2 x 10
5
where V = velocity, m/s and � density, kg/m3 0.040 0.030 0.020
0
f""'
r'..2(1  a ).J) Eu Plate 2 reflects energy=( 1  cxu) ( 1  cx).1 )Eu Pfute 1 absorbs energy=CX>..1 ( 1  cx'2) (1  CX).1)Eu Plate 1 reflects energy=( 1  CX).1 )2 (1  cxu)Eu Plate 2 absorbs energy =cxu ( 1  CX).1 )2 ( 1  cxu)Eu This process is continued until all of Eu must be absorbed. : Consider the energy emitted by plate 1 : Plate 1 emits energy=E>..1 Plate 2 absorbs energy= . �u E>..1 Plare 2 reflects energy=(1 · cxu) EA.1 Plate 1 absorbs energy =CX).1 (1  cxA.2)E).1 Plate 1 re.tlects ener� =( 1  CX).1 ) ( 1 aA.2)E).1 Plate 2 ab�orbs energy =cxu (l  CX>..1) (1  cxu)E>..1 Plate 2 reflects energy = (1  cx.,1 ) ( 1  cxu)2 E).1 Plate 1 absorbs energy = cx.,1 ( 1  cx>..1 ) ( 1  a:u)2E>..1 Plate 1 reflects energy = ( 1  cxA.j)2 ( 1  cx'>,2)2 EA.1 Plate 2 absorbs energy = cx;u (1  CX).1 )2 (1  cxu)2EA.1 This process is continued until all EA.1 must be absorbed :
( 17.25)
( 17.27)
1
(17.28)
( 17.29)
Considering equations (17.25) to ( 17.29),,net energy lost by plate 2 is given by q=energy emitted  energy absorbed = Eu [EA.2 ( 1  CX'A1 ) CX).2 + EA.2 ( I  CX).. 1 )2 ( 1  cx).2) au + ......]  [EA.I CX)..2 + E)..1 ( 1  Cl).; ) ( 1  cx)..2) CX)..2 + EA. I ( 1  .b
u == 1  E ')J, aA.1
a A.1
615
( 17.30)
Hence the statement of the Kirchoffs is law is proved. Note that the same method may be applied for Kirchoffs law based on total values. 1 7 . 6 . Intensity of Radiation an d its Relations In the previous articles, we have observed that the radiation emitted by a surface propagates in all possible directions. In addition, radiation incident upon a surface may come from different directions. We are often interested in knowing the directional , distribution of the emission and irradiation and that may be studied by introducing the concept of radiation intensity. 17.6.1. Solid Angle. Before defining intensity of radiation, let us first define a new term called solid angle. Consider emission of radiation in a particular direction from an elemental area dA I as shown in Fig. 11. 10 which also depicts the spherical coordinate system. The zenith angle, 8 and azimuthal angle, IP of the spherical coordinates system specify the direction of emission. Assume that this radiation passes through a differential �mall surface area in space, dA ,. (i.e .. area normal to 8 , IP direction) which subtends a solid angle dm, when viewed from a point on dA 1 • The differential solid angle d(J) (Fig. 17.11) is defined by a region between the 'rays of a sphere and is measured as the ratio of the element ofarea d4,. on the sphere to the square of the sphere's radius. Mathematically,
d4,. ( 17.31) r2 It is worthwhile to note.that the differential area dA,. represents an area that is no.mat to the ( 8,IP) direction. dro
= 
E mitt•d Ra diation
n
dA n

�__... _ _ _ z
I
  .J
X
Fig. 17.10. Directional Nature of Radiation.
616
Heat and � Transfer
dAn
Fig. 17.1 1. Definition of Solid Angles. nj ,
I
l I
\ d A ,fll sint ded♦
"""' . ·""
r s in e
>:·
,., �

�"'
0o
(Gray)
1
4 3 Waver length, a,m 2
5
Fig. 17.34 (a) Gray Body Emission
1 7 .19. Solar Radiation. The fear that the energy sources (fossile fuels) on the earth may be exhausted in the coming years has directed the attention of the world to use solar radiation which is an inexhaustible energy source. Today world is passing through an energy crisis. Hence it is essential that we should· go for the solar energy. Solar radiation is at present not used directly for industrial purposes because its concentration is lo":'.. However, in recent years, researches are going on extensively to use solar radiation
644
· Heat and Mass Transfer
as an alternative energy source. Now it is possible to use solar radiation for heating houses, distilling fresh water from sea, cooling homes, etc. Electricity is also being generated by the use of solar thennal energy but its capacity is low. Various types of solar collectors and absorbers have been developed. Solar radiation is also an important factor in the evaluation of heating or cooling requirement for aircraft, missiles and buildings.
(i) The Solar Resource. The sun is nearly spherical radiation source whose radius is r0 = 0.695 x 109 m and located 1.50 x 102 m from the earth. Within its inner core (r �.23 r0), there is a continuous fusion process in which mass is converted into energy resulting in the release of tremendous heat energy and thereby maintaining the temperatures of the inner core from 8 x 106 K to 40 x 106 K. The fusion process produces X ray and Y ray electromagnetic radiatlon which emanate from the high density core. In the low density gaseous region called convective zone (0.7r0 � �r0) energy is also transported by convection. It is believed that temperature drops from about 1.3 x 106 K to SOOK through this convective zone. The outer iayer of convection zone which is called photosphere is essentially opaque and well defined. Three gaseous layers lie outside the photosphere, with temperature ranging from about SOOOK within the inner layer to as high as 106 K in the outer layer. Solar radiation consists of energy that is emitted by various layers of the sun, with major contribution being provided by the photosphere. It is to be noted that although the radiation emitted by the sun originates from the various temperature zones for practical purposes, the sun can be regarded as a black body emitter at an effective temperarure of about 5800K. Using this value of the sun temperature the effecti.ve total emissive power of the sun may be calculated as ESU/1 = CJ'T:,, = 5.67 x I 08 (5800)4 = 6.416 x 107 W/m2
Out of this enormous solar radiation flux, only a small fraction reaches the outer fringes of the earth's atmosphere, and an even 'smaller portion reaches the surface of the earth itself. Fora horizontal surface outside the earth's atmosphere, solar radiation appears as a beam of nearly parallel rays that form an angle 0, the zenith angle, relative to the surface normal. The .extraterrestrial solar irradiation, Gs0 depends on the geographic latitude, as well as the time of the day and year. It is expressed as G.s0= G., .f. cos0 (17.98) where G� called the solar constant, is defined as the flux of solar energy incident on a �urface oriented normal to the sun's rays, when the earth is at its mean distance from the sun. The value of solar constant, G., = 1353 W/m2. The . term f is a correction factor to account for the eccentricity of the earth's orbjt about the sun and its value ranges from 0.97 to 1.03.
Fig. 17.35 shows distribution of the monochromatic extraterrestrial solar irra_diation ( Gs0h and monochromatic solar irradiation on the earth surface, G e,.• The radiation is concentrated in the !ow, wavelength region (0.2��3µm) of the thermal spectrum with the peak occurring at about 0.5 µm.
The solar irradiation flux experiences a significant changes while passing through atmosphere before it reaches the  surface of the earth mainly due to absorption and scattering. The atmospheric gases such as ozone (03), oxygen (Oi), carbon dioxide (COi) and water vapor (H20) absorb the .radiation and that lowers down the radiation reaching the
Thermal Radiation Phenomena
645
earth �urface. The ab�orption by ozone is strong in UV region. In the visible region, ther� 1s some abs_orp�on by ozone and oxygen, and in the near and far infr•� the maximum absorption 1s by water vapor.
In addition, there is also continuous absorption of radiation by the dust and aerosal content of the atmosphere throughout the solar spectrum.
There are two types of atmospheric scattering namelyRayleigh and Mie, as shown in Fig. 17.36. Atmospheric gas molecules is responsible for Rayleigh scattering which is also known as molecular scattering. This scattering of radiation is uniform in all directions and hence approximately half of the scattered radiation is redirected back to space, while the remaining portion strikes the surface of the earth. The scattering of radiation due to dust and aerosol particles of the atmosphere is known as Mie or particles scattering. This is concentrate,.• (2 n''f f 9
sin0 d0 dqJ \ ·i
d). = 0.5
r
J
4
2
n IJ..,1, dJ.
n
B l ac k b o dy 200 0 K
at
In order to have spectral integration multiply and divide by Eb and using the relation for intensity of radiation with emission, we get E = 0.5 Eb
From Table
J
4E
2
J..,b dJ. = 0.5 Eb [F (CH4) Eb
 F (0+ZJ]
A 1 T == 2 X 2000 = 4000 µ m. K; F(0+2 ) = 0.480877 A2T
=
4 x 2000 . = 8000 µ m. K; F(o+4)
=
0.856288
E = 0.5 Eb(0.856288  0.480877) = 0.5 Eb (0.375411)
A ns. Comments:(i) due to directional restriction the total hemispherical emissive power is reduced by 50 percent. = 0.5 x 5.67 x 108 (2000)4 x (0.3754.11) = l.7 x 105 W/m2
(ii) Spectral restriction reduces the total hemispherical emissive power by 62.4585 percent
Problem 17.11. A large isothermal enclosure is maintained at a uniform temperature of 2500 K. Calculate the following :(a) the en1issive power of the radiation that emerges from a small aperture on the enclosure surface. (b) the wavelength A.1 below which 10 per cent of the emission is concentrated and the wavelength Ai above which 10 per cent of the emission is concentrated.
656
Heat and Mass Transfer
(c) the maximum spectral emissive power and the wavelength at which the emission occurs. the irradiation incident on a small object placed within the enclosure. (d) Solution. Refer to Fig. 17.4 1
lnclosur• · 1 , 2sooK
Fig. 17.4 1
Assume that the rueas of aperture and object are very small relative to enclosure surface. (a) The emission from the aperture of any isothermal enclosure will exhibit the characteristics of a black body. Hence E = EbCD = a T4 = 5.67 x I 08 ( 2500)4 = 22. 14 x 1()5 W/m2
Ans.
(b) The wavelength A. 1 corresponds to .the upper limit of the spectral band containing 10 per cent of emitted radiation. Hence for fraction Fco+ 'A.) = 0. 1, the Table yields A 1 T= 2200 µm . K
Ans. A I == 2200 = 0.88 µm . 2500 The wavelength Ai corresponds to the lower limit of the spectral band containing 10% of emitted radiation. Thus or
or
F{i..2+l ) = 1  F(0+i..2) = 0. 1 F{ O+i..2} = 1  0.1 = 0.9.
From Table, we get J..iT == 938S µm. K or
or
A. 2 == 9385 == 3.754 µm 2500 � (c) From Wein's displacement law Ama:c T = 2898 µm . K. A
max
== 2898 2500
=
1. 1592 µm
Ans.
Ans.
6S7
Thennal Radiation Phenomena
·The maximum emissive power is
C1 ,.., '\  S _ 3.142• x 108 x (1 . 1s92r S E . = 'J.. ,b exp (C2/A. 7)  1 exp ( l.439 x l �4/�.1592 x 2500)  l
= 12.5 x 105 W/m2 • µ�
Ans.
(d) The irradiation incident on a small object placed within the enclosure may be treated as equal to emission from a black body at the encloure surface temperature. Thus Ans.
Problem 17.12. A diffuse fire brick wall is exposed to a bed of burning coals at 2000 K. The surface temperature of the brick wall is maintained at liOO K. The spectral emissivity of the wall is shown in Fig. 17.42. Calculate the following : : (a) total emissivity of the fire brick wall (b) total emissive power ofthe fire brick wall (c) total absorptivity of the fire brick wall for irradiation due to emisfion from the bed of coals. 0.8
 t
0.5 �
E ).3
,...1,' E u '' ''
0.1 f, Eu
Solution. Refer to Fig. 1 7 43.
Fig. 17.42.
' 10
1, µm
T1 = 600K Brick wall
Bed of coal
Fig. 1 7.43.
658
Heat and Mass Transfer
(a) The total emissiviLy is given by
f 00�
,. ,b dA E = Eb
Writing the integrals in parts with reference to Fig. 17.42
J
"' l
0
E ,.,1, di,,, Eb
Using the black body functions, we get
E = 'E "' I F(O+.U ) + E 2 [F(0>A2)  F(O>Al i] + E i 3 [1  F(O>U ) ]
Using Table, we get For A1 Ts = 2 x 600 = 1200 µm. K ; Fco, Al ) =0.002134
For
A2Ts = 10 x 600 E=
0.1
X
=
6000 µm. K; Fco,,.2)
=
0.737818
0.002134 + 0.5 � [0.737818  0.002134) + 0.8 [1  0.737818)
= 0.578 (b) The total emissive power of a brick wall is given by E= E ..i> = 0.480877 For tv2_Tc = 1 0 X 2000 = 20,000 µm. K;.F(O➔AZ) = 0.985602
Substituting the values of.fractions, we get a = 0.1 x 0.480877 + 0.5 [0.985602  0.480877] + 0.8 [l  0.985602] = 0.3 1 1 9 A ns. Problem 17.13. A flat plate solar collector with no cover plate is used to collect the solar radiation to heat water. The surface emissivity of the absorber is 0. 1 while its solar absorptivity is 0.95. At a given time of the pay, the absorber surface temperature is 125°C when the solar irradiation is 800 W/m2, the effective sky temperature is .:..5°C, and the ambient air temperature is 30°C. Calculate the useful heat removal rate in W/m2 from the collector and the efficiency of the collector. Assume that the heat transfer convection c.oefficient for the calm dry conditions is given by h = 0.22 (ts  t..
W/m2K
Y3
Also, assume the following Steady state conditions, bottom of the surface is well insulated and absorber surface is diffuse. Comment on the results. Soiotion. Refer to Fig. 17.44. t s ky:  s• c G =" 800 \ W/m 2
.
. .::
;
. .
�. .. • . .. . . . . . · .
I1 G sky
_
•
ts  125 C A b sorber T S u r f a c e �  (X's E 01 � � "', ·�
�5
o' L,
?,
CB753/� I
= U s efu l H e a u s e R e m ov e d t
Gs
cx s k y G s k y
[
q, "c o n v
\__ I ___ / /4E,.,
1 Tu � es ..___.''       ::'.J C ar r y i n g wa t er Q
S o l a,· C o i i e ctor
Fig. 1 7.44 Using the energy balance on the absorber Ein = Eau,
or
where Erad = ·energy lost due to radiation, W/m2 q!e = useful heat removal rate, W/m2
�'·· ·  ·] 11{
1, use
C ontro l Volume
660
qconv
Hem and Mass Transfer
.
= energy !os,l due to conveclion, Wfm2 = h(t8  t_) = 0.22 (t8  t_)½ 4
Gsty= earth's irradialion, W/m2 Gs = �lar irradiation: w;m2
a =: absorptivity. Since the sky radiation �� concentrated in appro_ximately the same spectral region as that of the surface emission, we may reasona:�ly assume : 103 kg mole/m3 bar and the binary diffusion coefficient of hydrogen in 20°C calculate the following::(a) the molar concentration of hydrogen at the opposite faces of the membrane. (b) · the molar and mass diffusion flux of hydrogen through the membrane. 19.12. Under steady state condition, oxygen diffuses through stagnant carbon monoxide at 5°C and 76 cm of Hg and 30 mm of Hg respectively. If the diffusivity of oxygen in carbon monoxide is 0.19 cm2/s, calculate the rate of dif1usion of oxygen. 19.13. In order to separate helium from a gas stream, a thin .Plastic membrane is used. · Under steadystate condition, the concentration of helium in the membrane is 0.022 kg mole/m3 and 0.005 kg mole/m3 at the inner and outer surface respectively. If the membrane is 1.5 mm thick and the binary diffusion coefficient of helium in the plastic is 1.02 x 109 m2/s, calculate the diffusion flux rate. 19.14. Compare the mass diffuston coefficient for the binary mixtures of following gases at 360 K and 1 atm:(a) Ammonia and air (b) Hywogen and air 19.15. A rubber membrane, 0.6 mm thi
kq/s.m
8 .02 8.09
8 .29 8.49
8 .69 9 .09 9.29 Y.49 9 .69 9 .89 10.09 10.29 10.49 ,0.69 1 0. 89 1 1 .09 1 1 .29 1 1 .49 1 1 .69 1 1 . 89 12.02 12.09 12.29 12.49 1 2.69 1 3 . 05 1 3 .42 1 3.79 14. 14 1450 1 4. 85 1 5.. 19 1 5.54
1 5 . 88 1623 1 6.59 1 6.59 17.33 17.72 1 8. 1 1 8. 6 19.1 1 9.7 20.4 2 1 .5 24. 1 25.9 27.0 28.0 300 32.0 37.0 45.0
kgxl