كواستنةوةى كةرمى Heat Transfer گواستنهوهى گهرمى Gwastnaway Garmi نقل الحرارة
2. Note: All questions are solved according to the mass and heat
transfer databook.www.ferhang.com 2 3. Chapter 1Introduction Heat
Energy and Heat Transfer What is Heat energy?Heat is a form of
energy in transition and it flows from one system toanother,
without transfer of mass, whenever there is a temperaturedifference
between the systems. Heat always moves from a warmerplace to a
cooler place.3 .What is Heat transfer?The science of how heat flows
is called heat transfer. .Importance of Heat Transfer What is
Importance of Heat Transfer?Heat transfer processes involve the
transfer and conversion of energyand therefore, it is essential to
determine the specified rate of heat transfer ata specified
temperature difference. The design of Equationuipments likeboilers,
refrigerators and other heat exchangers rEquationuire a detailed 4.
analysis of transferring a given amount of heat energy within a
specified time.Components like gas/steam turbine blades, combustion
chamber walls,electrical machines, electronic gadgets,
transformers, bearings, etcrEquationuire continuous removal of heat
energy at a rapid rate in order toavoid their overheating. . .
.Thermal Equilibrium 4What is Thermal Equilibrium?Two bodies are in
thermal Equilibrium with each other when they havethe same
temperature. In nature, heat always flows from hot to colduntil
thermal Equilibrium is reached. Hot objects in a cooler room
willcool to room temperature. Cold objects in a warmer room will
heat up toroom temperature. 5. . . . .If a cup of coffee and a red
popsickle were left on the table in a roomwhat would happen to
them? Why?The cup of coffee will cool until it reaches room
temperature. Thepopsickle will melt and then the liquid will warm
to room temperature. . .What are the ways of heat transfer? The
heat transfer processes have been categorized into three basic
modes: :51. Conduction 2. Convection 3. Radiation 6. Conduction 1.6
Mechanism of Heat Transfer by ConductionWhat is
conduction?Conduction is the transfer of heat through materials by
the directcontact of matter. Particles that are very close together
can transferheat energy as they vibrate. This type of heat transfer
is calledconduction. Dense metals like copper and aluminum are very
goodthermal conductors.6 . . . .How are the particles arranged in a
solid, a liquid and a gas? 7. Solids usually are better heat
conductors than liquids, and liquids arebetter conductors than
gases. .The ability to conduct heat often depends more on the
structure of amaterial than on the material itself. For Example,
Solid glass is a thermalconductor when it is formed into a beaker
or cup. But When glass is spuninto fine fibers, the trapped air
makes a thermal insulator. . .When you heat a metal strip at one
end, the heat travels to the otherend. As you heat the metal, the
particles vibrate, these vibrations makethe adjacent particles
vibrate, and so on, the vibrations are passed alongthe metal and so
is the heat.7 8. . .Metals are different, The outer electrons of
metal atoms drift, and arefree to move. When the metal is heated,
this sea of electrons gainkinetic energy and transfer it throughout
the metal. But Insulators, suchas wood and plastic, do not have
this sea of electrons because of thisthey do not conduct heat as
well as metals. . . .8 9. Why does metal feel colder than wood, if
they are both at the sametemperature?Metal is a conductor, wood is
an insulator. Metal conducts the heataway from your hands. But Wood
does not conduct the heat away fromyour hands as well as the metal,
so the wood feels warmer than themetal. . . .9 10. 10( = )WhereQ is
heat transferred rate with Watt.K is thermal conductivity of the
material.A is area of the material with 2 . is change in the
temperature. is the distance with m. . Q . K 2 A . T . xThermal
Conductivity The thermal conductivity of a material describes how
well the materialconducts heat. . 11. Example 1-1. One face of a
copper plate 3 cm thick is maintained at 400,and the other face is
maintained at 100. How much heat is transferredthrough the plate?3(
) 400 ( cm( ) 1.1 ( 00 (. ) 1Solution. From Appendix A the thermal
conductivity for copper is 370 W/Mat 250. From Fouriers law370 ) )
250 (. W/M) )A( : 11 = = =(370)(100 400)3 10= 3.7 MW/m22.
Convection Convection is the transfer of heat, occurs by the motion
of liquids andgases due to random molecular motion a long with the
macroscopicmotion of the fluid particles. Convection in a gas
occurs because of gasexpands when heated, hot gas rises and cool
gas sink. But Convection inliquids also occurs because of
differences in density. . . . . . 12. What happens to the particles
in a liquid or a gas when you heat them?The particles spread out
and become less dense. 12 . 13. Fluid movement:When the flow of gas
or liquid comes due to differences in density andtemperature zone,
it is called natural convection. When the flow of gasor liquid is
circulated by pumps or fans it is called forced
convection.Convection depends on speed. Motion increases heat
transfer byconvection in all fluids.13 : . . ( = )WhereQ is heat
transferred rate with Watt.h is Heat transfer coefficient.A is
surface area of the material with2. is change in the temperature. .
Q . h.) ) 2 A . T 14. Convection depends on surface area. If the
surface contacting of the fluid isincreased, the rate of heat
transfer increases. Almost all devices made forconvection have fins
for this purpose. . . .Example 1-2 Air at 20 blows over a hot plate
50cm by 75cm maintained at250. The convection heat transfer
coefficient is 25 W/m2. Calculate theheat transfer.14 15. )75cm( 50
( cm( ) 1.2 ( ) 20 ( 25 (. W/m2( ) 250 (. .Solution. From Newtons
law of cooling15: q = hA(Tw-T)= (25)(0.50)(0.75)(250-20) = 2.156
kWRadiation Radiation - It is the energy emitted by matter which is
at finitetemperature. All forms of matter emit radiation to changes
the electronconfigureuration of the constituent atoms or molecules.
The transfer ofenergy by conduction and convection requires the
presence of amaterial medium whereas radiation does not. In fact
radiation transfer ismost efficient in vacuum. - . . . .How does
heat energy get from the Sun to the Earth? 16. There are no
particles between the Sun and the Earth so it CAN NOT travel
byconduction or by convection. .16 17. Radiation is heat transfer
by electromagnetic waves. Thermal radiation iselectromagnetic waves
(including light) produced by objects because oftheir temperature.
The higher the temperature of an object, the morethermal radiation
it gives off.17 18. . ) ( . . 4 2 = (1184 )WhereQ is heat
transferred rate with Watt. is Boltzmann constant and is Equal to
5.669 108 is emissivity . . Q(5.669 10 ( 8 () Example 1-5. Two
infinite black plates at 800 and 300 exchange heat byradiation.
Calculate the heat transfer per unit area. ) 1. ( ) 800 ( ) 300 ( .
.Solution. Equation (1-10) may be employed for this problem, so we
findimmediately: ) ) 1.1 ( ( 4 T2q/A = (T14) 19. =
(5.66910-8)(10734-5734) = 69.03 kW/m3Thermal radiation We do not
see the thermal radiation because it occurs at infraredwavelengths
invisible to the human eye. Objects glow different colors
atdifferent temperatures. A rock at room temperature does not
glow.The curve for 20C does not extend into visible wavelengths. As
objectsheat up they start to give off visible light, or glow. At
600C objects glowdull red, like the burner on an electric stove. As
the temperature rises,thermal radiation produces shorter -
wavelength, higher energy light. . . 20 ( C( . . 600 ( C( . . .At
1,000C the color is yellow and orange, turning to white at 1,500C.
Ifyou carefully watch a bulb on a dimmer switch, you see its color
changeas the filament gets hotter.19 20. 500 ,( . C 000 ,( ) 1 C )
1 .The bright white light from a bulb is thermal radiation from an
extremelyhot filament, near 2,600C. A perfect blackbody is a
surface that reflectsnothing and emits pure thermal radiation.
2,600 (. C( .The white - hot filament of a bulb is a good blackbody
because all lightfrom the filament is thermal radiation. The curve
for 2,600C shows thatradiation is emitted over the whole range of
visible light. 2,600 ( C( . .20 21. Thermodynamics and Heat
Transfer-Basic Difference21 22. - Thermodynamics is mainly
concerned with the conversion of heatenergy into other useful forms
of energy and IS based on (i) the concept ofthermal Equilibrium
(Zeroth Law), (ii) the First Law (the principle ofconservation of
energy) and (iii) the Second Law (the direction in which
aparticular process can take place). Thermodynamics is silent about
the heatenergy exchange mechanism. )ii( ) ) )i( ) )iii( ) ( ( .
.The transfer of heat energy between systems can only take
placewhenever there is a temperature gradient and thus. Heat
transfer is basically anon-Equilibrium phenomenon. The Science of
heat transfer tells us the rate atwhich the heat energy can be
transferred when there is a thermal non-Equilibrium. That is, the
science of heat transfer seeks to do whatthermodynamics is
inherently unable to do. . . .22 23. However, the subjects of heat
transfer and thermodynamics are highlycomplimentary. Many heat
transfer problems can be solved by applying theprinciples of
conservation of energy (the First Law) . ) (.Dimension and Unit
Table 1.1 Dimensions and units of various parameters1( . )
1Parameters Units Mass Kilogram, kg Length metre, m Time seconds, s
Temperature Kelvin, K, Celcius oC Velocity metre/second,m/s Density
kg/m 3Force Newton, N = 1 kg m/s 2Pressure N/m2, Pascal, Pa Energy
N-m, = Joule, J Work N-m, = Joule, J Power J/s, Watt, W Thermal
Conductivity W/mK, W/moC Heat Transfer Coefficient W/m2K, W/m2oC
Specific Heat J/kg K, J/kgoC Heat Flux W/m 223 24. Viscosity
N-s/m2, Pa-s Kinematic Viscosity m2/s review Example: A red brick
wall of length 5m, height 4m and thickness 0.25m. Theinner surface
110 and the outer surface is 40 . The thermal conductivity ofthe
red brick, k=0.7 W/mK. Calculate the temperature at a point 20m
distancefrom the inner surface.0.25 (. m( 4( m( )5m( : .)k=0.7
W/mK( 011 ( ) 40 (. (20 ( . m( Given: length =L=5m, Height =h=4m,
thickness = =0.25m, 1 = 110 ,2 = 40 .24 25. Solution: = 25 = (1 2 )
= 0.7(5 4)(110 40)0.25= 3920 = but x=20cm=0.2m3920 = 0.7(5 4)(110
40)0.2 3920 = 14 (110 2 )0.256 = 110 2 2 = 110 56 = 54 () 26.
Example: Calculate the heat transfer by convection over a surface
of (1.5 2)area if the surface is at 190 and the fluid is at 40 .
The Value of convectionheat transfer coefficient is25 W/m2 . K .
Also Estimate the temperaturechange with plate thickness if thermal
conductivity of the plate is1 W/m. K.5 : ) 2 126( 90 ( ) 40 ( . )
125 ( . W/m2 . K(1( . W/m. K( Given: Area=A=1.5 2 , = 190 , = 40
.Heat transfer coefficient = h = 25 W/m2 . Kthermal conductivity of
the material = K = 1 W/m. KSolution: = = ( ) = ( ) 27. = 25 1.5
(190 40) = 5 625 () = 275 625 = 1 1.5 = 3 750()Example: Two
infinite black plates at 800 and 300 exchange heat byradiation.
Calculate the heat transfer per unit area.: ) 800 ( ) 300 ( .
.Given: black plates means() = 100% = 1.00,1 = 800 = 1073, 2 = 300
= 573.Solution:4 2 = (14 )4 2= (14 )= (5.669 108 ) 1 (10734 5734 )=
69.03 /2 (Answer) . Radiation : 28. Example: a reactor wall, 320mm
thick is made up of an inner layer offiber brick(k=0.84 W/m.K),
covered with a layer of insulation (K=0.16 W/m.K).the reactor
operates at a temperature of 1325 and the ambint temperatureof 25.
If the insulation outer surface not exceeds 1200 . convection
heattransfer coefficient between outter surface and surrounding is
0.12 /2 . . determine the thickness of the fiber brick and
insulating material.320 ( mm( : . )K=0.16 W/m.K ( )k=0.84 W/m.K(325
( ) 25 (. ) 1200 ( . ) 10.12 (. /2. ( .Given: reactor wall
thickness=320mm=0.32m, Thermal conductivity ofinner layer (fiber
brick) = k = 0.84 W/m.K , Thermal conductivity of outter
layer(insulation) = k = 0.84 W/m.K , reactor temperature = 1 =1325,
ambinttemperature = = 25, = =28 29. 290.122 . , = 2 =
1200.Solution: = = (1 ) = (1 ) = 0.12 1 (1200 25) = 141 For fiber
brick: = = (12 )141 = 0.84 1 (1598 )167.85 =(1598 )167.85 = 1598 =
1598 167.85 . (1) 30. 30For insulation material: = = (1 2 )141 =
0.16 1 ( 1473)0.32 881.25 = 14730.32 (2)Substituting Equation (1)
into Equation (2)881.25 =(1598 167.85 ) 14730.32 881.25 =125 167.85
0.32 881.25 (0.32 ) = 125 167.85 282 881.25 = 125 167.85 282 125 =
881.25 167.85 157 = 713.4 =157713.4= 0.22 the thickness of the
fiber brick= = 0.22 (Answer)insulating material= 0.32 = 0.32 0.22 =
0.10 (Answer) 31. 31:Data Book . properties value thermal specific
heat(C) density() ( 06 ) conductivity(k) ( ) . (properties) 21 :
(properties) ( 06 ) thermalconductivity(k)specificheat(
C)PrandtlNumber(Pr)ThermaldiffusivityKinematicviscosity(v)density
() 32. 0.1553 3.020 4183 0.6513 106320.478 106985 ( 06 ) ( 06 ) (
06 ) (properties) (properties) . 06 ) . ) ( 04 ) ( 00 ) ( 06 )
(properties) ( 06 ) 06 ) ( 04 ) ( 00 ) ) . . properties value :Data
Book . heat transfer .Convection questions Why does hot air rise
and cold air sink?Cool air is more dense than warm air. . 33. Why
are boilers placed beneath hot water tanks in peoples homes?Hot
water rises.So when the boiler heats the water, and the hot
waterrises, the water tank is filled with hot water. . .Radiation
questions Why are houses painted white in hot countries?White
reflects heat radiation and keeps the house cooler. Why are shiny
foil blankets wrapped around marathon runners at theend of a
race?The shiny metal reflects the heat radiation from the runner
back in,this stops the runner getting cold.. .33 34. 1. Which of
the following is not a method of heat transfer?A. Radiation B.
Insulation C. ConductionAnswer : B. .C .B .A34B : 2. In which of
the following are the particles closest together?A. Solid B. Liquid
C. Gas D. FluidAnswer : A. .D .C .B .AA : 3. How does heat energy
reach the Earth from the Sun?A. Radiation B. Conduction C.
Convection D. InsulationAnswer : A. .D .C .B .AA : 4. Which is the
best surface for reflecting heat radiation?A. Shiny white B. Dull
white C. Shiny black D. Dull black 35. 35Answer : A. D. . C . B
.A.: 5. Which is the best surface for absorbing heat radiation?A.
Shiny white B. Dull white C. Shiny black D. Dull blackAnswer : A. .
D . C . B .A: 36. Chapter 2STEADY STATE CONDUCTION - ONE DIMENSION
2 - Temperature in a system remains constant with time.
Temperaturevaries with location. . .36 37. Thermal Diffusivity and
its Significance Thermal diffusivity is a physical property of the
material, and is the ratio of thematerial's ability to transport
energy to its capacity to store energy. It is anessential parameter
for transient processes of heat flow and defines the rateof change
in temperature. In general, metallic solids have higher
Thermaldiffusivity, while nonmetallics, like paraffin, have a lower
value of Thermaldiffusivity. Materials having large Thermal
diffusivity respond quickly tochanges in their thermal environment,
while materials having lower a respondvery slowly, take a longer
time to reach a new Equilibrium condition. . . () . .1. Several
plane Walls together (Thermal resistance).2. Heat generation.3.
Types of fin. ) ( .37 .. . 38. Several plane Walls together
(Thermal resistance).A plane wall is considered to be made out of a
constant thermalconductivity material and extends to infinity in
the Y- and Z- direction.The wall is assumed to be homogeneous and
isotropic, heat flow isone - dimensional ) ( . . (Z-( (Y-) .Thermal
Resistances and Thermal Circuits: : From page 43 in the Data
book:38 : =Conduction in a plane wall: = Convection: =1 Consider a
plane wall is between two fluids of different temperature: 39. = 1
+ + 2Overall Heat Transfer Coefficient (U) : A modified form of
Newtons Law ofCooling to encompass multiple resistances to heat
transfer. : )U( . = 39 40. Example: An exterior wall of a house is
constructed by a 0.1m layer ofcommon brick (k=0.7 W/m.K) and a
0.04m layer of gypsum (k=0.48 W/m.K)then followed by 0.058m loosely
packed rock with wool insulation (k=0.065W/m.K). If the temperatue
difference through composite wall is 20 . Findthe amount of heat is
transfering.k= 0.1 ( ) 0.7 m( : 0.058 ( m( )k=0.48 W/m.K( 0.04 ( m(
.)W/m.K . )k=0.065 W/m.K( ) 20 ( . .Given: Thermal conductivity of
brick = 0.7/. , thermal conductivity ofPlaster = 0.48/. , thermal
conductivity of wool= = 0.065/. ,Temperature difference= = 20=293
K40Solution:From page 43 in the Data book: 41. =41 = + + ==0.10.7
=0.142 ==0.040.48 =0.083 ==0.0580.065 =0.892 = + +
=0.142+0.083+0.892 =0.142 + 0.083 + 0.892 =1.117 = =2931.117 = 293
1.117 42. kF kG42=2931.117= 262.232 ( )Series Composite Wall and
Parallel Composite Wall:F G kkNote: departure from one -
dimensional conditions for .Circuits based on assumption of
isothermal surfaces normal to xdirection or adiabatic surfaces
parallel to x direction provideapproximations for . : : . X X x q
43. 43 44. A Cylindrical Shell - Expression for Temperature
Distribution - 44 45. rqWhat does the form of the heat Equation
tell us about the variation ofwith in the wall? Is the foregoing
conclusion consistent with the energyconservation requirement? How
does vary with ? Temperature Distribution for Constant K :45: K rrq
rr q rr q r s s,1 ,2 ,2lnln / 1 2 2sT T rT r Tr r r 46. Heat Flux
and Heat Rate: dT k q k T Tr s sdr r r rk q rq T Tr r s sr rLkq rLq
T Tr r s sConduction Resistance: 2 1,1 ,2ln /22 1,1 ,22 1r r
Example 2-2 A thick walled trbe of stainless steel (18%Cr, 8%Ni, k
= 19W/m) with 2cm inner diameter and 4cm outer diameter is covered
with a3cm layer of asbestos insulation ( k = 0.2 W/m). If the
inside walltemperature of the pipe is maintained at 600, calculate
the heat loss permeter of length. Also calculate the tube
insulation interface temperature.8% ( Cr, 8%Ni, k = 19 W/m ) 2.2 (
) 13( cm( 4( cm( 2( cm( .)k = 0.2 W/m( 600 ( . ( .46 ,2 1,ln /Units
K/W2ln /Units m K/W2t condt condr rRLkr rRk ,1 ,22 12ln /22ln / 47.
Solution. The accompanying Figure shows the thermal network for
thisproblem. The heat flow is given by: . 47=2(12
)ln(21)+32ln()=2(600100)219+52ln()0.2=680 W/mThis heat flow may be
used to calculate the interface temperature betweenthe outside tube
wall and the insulation. We have . =2ln (3/2)2=680 W/mwhere Ta is
the interface temperature, which may be obtained as . Ta Ta = 595.8
48. The largest thermal resistance clearly results from the
insulation, and thus themajor portion of the temperature drop is
through that material. .2-6 Critical thickness of insulation L T T
2 ( )ir rln( ) 1k r hdq ,0.17r 0.0567 5.67o 48qoo i0 odrkhro
Example 2-5 Calculate the critical radius of insulation for
asbestos (k = 0.17W/m) surrounding a pipe and exposed to room air
at 20 with h=3.0 W/m2. Calculate the heat loss from 5cm diameter of
the pipe at 200, whencovered with the critical radius of insulation
and without insulation. )k = 0.17 W/m( : .)h=3.0 W/ m2( ) 20 ( 5 )
200 ( cm .Solution. We calculate ro asm cmkh3.0The inside radius of
the insulatyion is 5.0/2 = 2.5 cm, so 49. 2 (200 20)0.04r 0.0133
1.33o 49W mqL105.7 /1(0.0567 )(3.0)ln(5.67 / 2.5)0.17h r T T W mqLi
o (2 )( ) (3.0)(2 )(0.025)(200 20) 84.8 /So, the addition of 3.17
cm (5.67-2.5) of insulation actually increases the heattransfer by
25 percent.As an alternative, fiberglass having a thermal
conductivity of 0.04 W/mmight be employed as the insulation
material. Then, the critical radius wouldbe5.67 ) 2 - 3.17 ( 2.5 cm
.0.04 ( W/m( . m cmkh3.0Heat generation Page 47 in the Data
BookHeat generation : . Tw T Tw 50. = +50. Tw .x= 0 maximum
temperature(To) maximum temperature = To = Tw +q.2kL2 X . . : ( 0 )
0= (2) 51. 51Amount of heat generation=q. =qvolume(W/m3 )Example: a
plate of 24mm thickness is shaped to nuclear fuel element exposedon
the sides to heat convection at 200 with a convection heat
transfercoefficient of 900 /2 . Generates heat at 20 MW/m3.If
thermalconductivity of the plate is 25.4 /. . Determine the surface
temperatureand the maximum temperature.24 ( mm( : 900 /( ) 200 ( 20
(. MW/m ) 3 .)2 .25.4 ) . . /.)Given: = 2 = 24 = 0.024 =0.0242=
0.012 , =200 , convection heat transfer coefficient of = = 900 /2 ,
Amountof heat generation=q. = 20 MW/m3 = 20 106 W/m3, thermal
conductivityof plate = = 25.4 /. Solution: from page 47 in the Data
Book:surface temperature = = +. 52. 52 = 200 +20 106 0.012900 =
466.66 surface temperature = = 466.66 ()maximum temperature = To =
Tw +q.2kL2maximum temperature = To = 466.66 +20 1062
25.4(0.012)2maximum temperature = To = 523.35 (Answer) insulation :
. x L 2 L= . 53. . x Data Book : .53 54. Example: a plate wall is
1m thick and it has one surface (x=0) insulated whilethe other
surface (X=L) is maintained at a constant temperature of 350 ,
K=25w/mK, and uniform heat generation per unit volume of 500 W/m3
existsthroughout the wall. Determine the maximum temperature in the
wall andthe location of the plane where it occurs. )x= ) 0 )m : ) 1
(K =25w/mK) , ) 350 ( )X=L( 500 ) . W/m ( 3 .Given: thick = L = 1m
, = 350 = 623 .Thermalconductivity= K =25w/mK, = . =500 /3Solution:
from page 47 in the Data Book:maximum temperature = To = Tw
+q.2k54L2 To = 623 +5002 25(1)2 55. 55To = 623 +50050 maximum
temperature = To = 633 (Answer)The location of maximum temperature
is (x=0).(Answer) : . : =q.2k(L2 x2 ) +x2L(Tw1 Tw2 ) +12(Tw1 + Tw2
) :max =k2 q. L(Tw2 Tw1) : 56. 56 =q.L22k+k8 q.L2(Tw1 Tw2 )2
+12(Tw1 Tw2 )Example: The thickness of a steel plate is 25mm and (K
=48 W/m ), having auniform heat generation per unit volume of
30*106 W/m3 , the temperatureon the two surfaces of the steel
plate, are 180 and 120 . Determine thevalue and position of the
maximum temperature and heat flow from eachsurface of the plate. )K
=48 W/m( 25 ) mm) : 80 ( 30 ( ) 1 *106 W/m ) 3201 (. ( .Given:
thick = 2L = 25mm =0.025m = 0.0125, Thermal conductivity=K =48 W/m,
heat generation=q. =30*106 W/m3, maximumtemperature=Tw1=180,
minimum temperature=Tw2 =120 .Solution: =q.L22k+k8 q.L2(Tw1 Tw2 )2
+12(Tw1 Tw2 ) =30 106 (0.0125)22 48+488 30 106 (0.0125)2(180
120)2+12(180 120) = 48.828 + 4.608 + 30 = 83.43 . (Answer) 57.
Example: A wall 8cm thick has its surfaces maintained at 0 and 100
. Theheat generation rate is 3.25*105 W/m3 . If the Thermal
conductivity of thematerial is 4W/mK, determine the temperature at
the mid plane, the locationand the value of the maximum
temperature.00 (. 8( ) 0( ) 1 cm( : 4( W/mK( 3.25 ( . *105 W/m ) 3
.Given: thick = 2L = 8cm =0.08m = 0.04, , heatgeneration=q.
=3.25*105 W/m3 , Thermal conductivity= K =4 W/mKmaximum
temperature=Tw1=100=373K, minimum temperature=Tw2 =0=273K.Solution:
from page 47 in the Data Book: :57 =q.2k(L2 x2 ) +x2L(Tw1 Tw2 )
+12(Tw1 + Tw2 )=0 =3.25 1052 4(0.042 02 ) +00.04(373 273) +12(373 +
273)=0 = 65 + 323 = 388 () 58. :58max =k2 q. L(Tw2 Tw1)max =42 3.25
105 (0.04)(273 373)max = 0.0153m = 1.53 cm (Answer) : =q.L22k+k8
q.L2(Tw1 Tw2 )2 +12(Tw1 Tw2 ) =3.25 105 (0.04)22 4+48 3.25 105
(0.04)2(373 273)2+12(373 273) = 65 + 9.615 + 25 = 99.61 () 59.
Example: a 6cm thick slab of insulating material (K =0.38 W/mK) is
placedbetween two parallel electrodes and is subjected to high
electric powerproducing 39000w/3 . Under steady state condition,
the convection heattransfer coefficient is 11.8W/2 .K at left side
while the right side is at 30 , ifambient air temperature is 25 at
left side. (1)Calculate The left side surfacetemperature. (2)
Determine Location and magnitude of maximumtemperature. )K =0.38
W/mK( 6( cm( : 39000 ( w/ ) 3. 8.(. ) 30 ( W/2.K11( ) 25 ( . ) 1( .
) 2( .Given: thick = 2L = 6cm =0.06m = 0.03, Thermal conductivity=
K=0.38 W/mK, heat generation=q. = 3900059Wm3, heat transfer
coefficient=11.8W/2. ambient air temperature = 25 = 298K, minimum
temperature=Tw2=30=303K. 60. Solution: from page 47 in the Data
Book: = +60. 1 = 298 +39000 0.0311.81 = 397.15 ()max =k2 q. L(Tw2
Tw1)max =0.382 39000 (0.03)(303 397.15)max = 0.0152m = 1.52cm = =
0.0152= 1.52cm (Answer) : =q.L22k+k8 q.L2(Tw1 Tw2 )2 +12(Tw1 Tw2 )
=39000 (0.03)22 0.38+0.388 39000 (0.03)2(397.15 303)2+12(397.15
303) = 46.18 + 11.995 + 47.075 = 105.25 = = 105.25 () 61.
61FinsTypes of the fin: :1. Long Fin 2. Short Fin (end
insulated)=Thin Fin 3. Short Fin (end insulated) 4. Circumferential
fin 5. Rectangular fin 6. Triangular fin short fin Long fin : short
fin . fin : Long fin . fin 62. Example: Find out the amount of heat
transfer through an thin iron fin oflength 50mm, 100mm width and
thickness of 5mm. Assume k=210W/m. forthe material of the fin and
h=42W/2 . , if the atmospheric temperature is20 . Also determine
the temperature at the tip of the fin if basetemperature is
80.)50mm( : )k=210W/m.( 5(. mm( 00 ( mm1( ) 20 ( . )h=42W/2.( ) 80
(.Given: Type of the fin is thin iron fin i.e. Short Fin (end
insulated), length=L=50mm=0.05m, width=w= 100mm=0.1m thickness
=t=5mm=0.005m, thermalconductivity=k=210W/m., heat transfer
coefficient= h=42W/2 . ,atmospheric temperature = =20. Base
temperature = 80.Determine the temperature at the tip of the fin.62
63. 63Solution:From page 49 in the Data Book:Perimeter = p =2
(width) + 2(thickness)p =2 (0.01) + 2(0.005)=0.02 +
0.010=0.03m.Area =A= width * thicknessA=0.01 * 0.005 = 0.00005 2. =
= 42 0.03210 0.00005= 10.954 64. = ( )tanh (mL) = 42 210 0.03
0.00005 (80 20)tanh (10.954 0.05) = (0.115)(60) 498 = 3.436 64
=cosh ( )cosh() 2080 20=cosh 10.954( )cosh( 10.954 0.005 ) 2080
20=11.153 2080 20= 0.866 2060= 0.866 20 = 52 = 72 () 65. Example: a
rod of carbon steel (k=54 W/m.K) with a cross section of
anEquilateral triangular (Each side is 5mm) is 80mm long. It is
attached to aplane wall which is maintained at a temperature of
400. The surroundingenvironment is at 50 and convection heat
transfer coefficient is 90W/2.K.Calculate the heat dissipated by
the rod. )k=54 W/m.K( : 80 ( . mm( 5( mm( ) 400 (. ) 50 ( 90 ( . .
W/2.K(65Given:short fin fin : Type of the fin is short fin (end not
insulated), thermalconductivity=k=54W/m.K, length=L= 80mm=0.08m,
Base temperature= =400, surrounding temperature = =50 , heat
transfer coefficient=h=90W/2 . K, Calculate the heat dissipated by
the rod. 66. hLmk66Solution:(0.005)2 = 2 + (0.0025) () 2 = 4.33 103
From page 49 in the Data Book:Perimeter = p =3 * 0.005=0.015m = =12
= =12 ( ) = =12 0.005 ( 4.33 103 )A=1.08 105 2 . = = 90 0.01554
1.08 105= 48.112 = ( ) (hLmktanh(mL) + ()1 + () tanh(mL)) ( )0.5 =
(400 50) ((48.112 0.08) + (90 0.0848.112 54)1 + (90 0.0848.112 54)
(48.112 0.08)) (90 0.015 54 1.08 105 )0.5 67. (0.99) + (2.77 103 )1
+ (2.77 103 )(0.99)67 = (350) (tanh(3.84) + (2.77 103 )1 + (2.77
103 ) tanh(3.84)) (7.87 104 )0.5 = (350) () (0.028) = (350)
(0.992771 + (2.76 103 )) (0.028) = (350)(0.99)(0.028) = 9.72
()Example: Circumferential aluminum fins (K=200W/mk) is rectangular
1.5cmwide and 1mm thick are fitted onto a 2.5cm diameter tube. The
fin basetemperature is 170 and the ambient fluid temperature is 25
. Estimatethe heat lost per fin, If h=130W/2k.)mm 5.( ) 1 cm ) 1
)K=200W/mk( : 70 ( 2.5 (. ) 1 cm( ) 25 ( . .)h=130W/2k(Given: the
type of the fin is circumferential fin From page 50 at the Data
Book:Circumferential rectangular fin = +2= 0.0.15 +0.0012= 0.01552
= 1 + = 0.0125 + 0.0155 = 0.028 = (2 1 ) = 0.01(0.028 0.0125) =
1.55 105 2 68. 682 1 = 2(22 ) = 2(0.0282 0.01252 ) = 1.5 103 21.5 (
= )0.5= 0.01551.5 (130200 1.55 1050.5)= 0.395 0.4 =21=0.0280.0125=
2.24 curve curve number 2 X-axis% 40 fin efficiency Y-axis = 85% =
0.85 = h(Tb T ) = 0.85 1.5 103 130(170 25) = 24.033W
(Answer)Example: A heating unit is made in the form of a vertical
tube fitted withrectangular section steel fins. The tube height is
1.2m and its outer diameter is60mm. the fins are 50mm height and
3mm thickness along the tube. Thenumbers of fins used are 20. The
base and air surrounding temperature are80 and 18 respectively,
h=9.3W/2k. for fin material (k=55.7W/mK).Calculate the total heat
transferred from the tube with fins.: 60 (. mm( 2.( m . ) 13( . mm(
50 ( mm(8( ) 20 (. ) 80 ( ) 1 . )k=55.7W/mK( .)h=9.3W/2k( .Given:
the type of the fin is circumferential fin 69. 69Solution: =
Calculation: for heat transfer with one fin :From page 50 at the
Data Book:Circumferential rectangular fin = +2= 0.05 +0.32= 0.0652
= 1 + = 0.06 + 0.065 = 0.125 = (2 1 ) = 0.03(0.125 0.06) =
0.03(0.065) = 1.95 103 22 1 = 2(22 ) = 2(0.1252 0.062 )0.02621 .5 (
= 0.5)= 0.0651.5 (9.355.7 1.95 1030.5)= 0.153 =21=0.1250.06= 2.08 2
70. curve curve number 2 X-axis% 99 fin efficiency Y-axis = 99% =
0.99 = h(Tb T ) = 0.99 0.026 9.3(80 18) = 0.239(62) = 14.841W = =
14.841 20 = 296.833 W (Answer)Example 2-6 A current of 200 A is
passed through a stainless steel wire (k = 19W/m). it is 3mm in
diameter. The resistivity of the steel may be taken as 70cm, and
the length of the wire is 1m. The wire is submerged in a liquid
at110 and experiences a convection heat transfer coefficient of
4kW/m2.Calculate the center temperature of the wire.k = 200 ( ) 19
A( ) 2. ( )76 cm( 3(. mm( .)W/m0( . ) 11 )m ) 14( . kW/m2(
.Solution. All the power generated in the wire must be dissipated
byconvection to the liquid:70 71. P = I2R = q = hA(Tw-T)The
resistance of the wire is calculated from(70 10 )(100) q MW mT 71 L
0.099(0.15)26ARwhere is the resistivity of the wire. The surface
area of the wire is dL, sofrom Equation ) dL( . )( T W w (200)
(0.099) 4000 (3 10 )(1)( 110) 3960 2 3 and Tw = 215 (419)The heat
generated per unit volumeq is calculated from )q ) P qV q r 2 L so
that33 2560.2 /3960(1.5 10 ) (1)(5.41*107 Btu/hft3)215 231.6(5.602
10 )(1.5 10 )(4)(19)q r42 8 3 2 woo Tk 72. Chapter 3 Steady state
conduction multiple dimension72 - Nodal Equation: nodal Equation :
.nodal Equation )Node( (x) node . . (y) Example: write the nodal
Equation for node 1 and node 6 as shown in theFigure below. 73. : 1
6 .73Answer: )Node( )x( node . . )y( For node 1:: 74. .
(conduction) ( 2) ( 1) . (conduction) ( 3) ( 1) . (conduction) ( 0)
( 1) . (conduction) ( 0) ( 1) () + . = 0(2 + 3 + 4 + 5 ) + . = 074(
2+ 3+ 4+ 5) + . ( ) = 0( 2+ 3+ 4+ 5) + . ( 1) = 0(( 1)2 1+ ( 1)3 1+
( 1)4 1+ ( 1)5 1) + . ( 1) = 0 ()For node 6::. (conduction) ( 7) (
1) . (conduction) ( 4) ( 1) . (conduction) ( 9) ( 1) 75. .
(conduction) ( 16 ) ( 1) () + . = 0(7 + 8 + 9 + 10 ) + . = 075( 7+
8+ 9+ 10) + . ( ) = 0(+ + + ) + . ( 1) = 0( ( 1)7 6+ ( 1)8 6+ ( 1)9
6+ ( 1)10 6) + . ( 1) = 0 ()Example: write the nodal Equation for
points 1,2,4 as shown in the Figurebelow.2,4 , . : 1 76.
76Answer:For node 1: )Node( (x) node . . (y) 77. 77:. (conduction)
( 2) ( 1) . (conduction) ( 0) ( 1) (Conduction) . (Convection) () +
. = 0(2 + 4 + ) + . = 0( 2+ 4+ ) + . ( 1) = 0[ (2 1)2 12+ ( 1)4 1+
(2 1) ( 1 )]+. (22 1) = 0 () 78. insulation : 1 78For node 2:
)Node( (x) node . . (y) :. (conduction) ( 1) ( 2) . (conduction) (
3) ( 2) . (conduction) ( 0) ( 2) (Conduction) . (Convection)
(Convection) (Node 2).() + . = 0 79. (1 + 3 + 5 + ) + . = 0279( 1+
3+ 5+ ) + . ( 1) = 0[(2 1)1 2+ (2 1)3 2+ ( 1)5 2+ ( 1)( 2 )] + . (
1) = 0 ()For node 4: )Node( (x) node . . (y) :. (conduction) ( 1) (
0) . (conduction) ( 0) ( 0) . (conduction) ( 7) ( 0) 80.
(Conduction) . (Convection) () + . = 0(1 + 5 + 7 + ) + . = 080( 1+
5+ 7+ ) + . ( 1) = 0[ (2 1)+ ( 1)+ (2 1)+ ( 1)( 1 ) ]+ . ( ) = 02[
( 1)1 4+ ( 1)5 42+ ( 1)7 4+ ( 1)( 1 ) ] + . ( ) = 0 ()Example:
write the nodal Equation for node 1 as shown in the Figure below.:
1 . 81. 81Answer:For node 1: )Node( (x) node . . (y) 82. () + . =
0(2 + 3 + + ) + . = 082( 2+ 3+ + ) + . ( 1)= 0[ (2 1)2 1+ (2 1)3 1+
(2 1) ( 1 )2+ (2 1) ( 1 )] + . (2 1) = 0 ()Example: write the nodal
Equation for node 5 as shown in the Figure below.: 5 . 83.
83Answer:For node 5: )Node( (x) node . . (y) 84. () + . = 0(2 + 6 +
4 + 7 ) + . = 084( 2+ 6+ 4+ 7) + . ( 1) = 0(( 1)2 1+ (2 1)6 1+ (
1)4 1+ (2 1)7 1) + . ( 1) = 0 ()Example: write the nodal Equation
for node 2 as shown in the Figure below.: 2 . 85. 85Answer:For node
2: )Node( (x) node . . (y) :. (conduction) ( 0) ( 2) . (conduction)
( 0) ( 2) (Conduction) . (Convection) . (Convection) (AB) (AB). ()2
= ()2 + () : 2 = ()2 + ()2() + . = 0 86. (6 + 4 + ) + . = 086( 6+
4+ ) + . ( 1) = 0[( 1)6 2+ ( 1)4 2+ ( 1) ( 2 )]12+. ( 1) = 0 =12 [(
1)6 2+ ( 1)4 2+ (()2 + ()2 1) ( 2 )]12+. ( 1) = 0 ()Example: write
the nodal Equation for node 1 as shown in the Figure below.: 1 .
87. 87Answer:For node 1: )Node( (x) node . . (y) : 88. .
(conduction) ( 0) ( 1) (Conduction) . (Convection) . (Convection)
(AB) .(AB) ()288()2 =()22+()22 : = 2+()22() + . = 0(6 + + ) + . =
0( 6+ + ) + . ( 1) = 0[( 1)6 1+ (2 1) ( 1 ) + ( 1) ( 1 )]+. (1222
1) = 0 =12 89. 289[( 1)6 1+ ( 1) ( 1 )()2+ (2+()22 1) ( 1 )]18+. (
1) = 0 ()Chapter 4 Unsteady state conduction 90. The process of
heat transfer by conduction where the temperature varies withtime
and with space coordinates, is called unsteady or transient.
.Lumped Capacity : Page 57 in the Data Book90 91. 91 :1. . Area
volume 2. . = [] 57 Example: a thermocouple junction, which may be
approximated as a sphere,is to be used for temperature measurement
in a gas. The h between the 92. junction surface and the gas is 400
/2 . With the junction properties areK=20W/mK,: 400 (. /2. ( (h) .
92 )K=20W/mK( = 400 /. and density 8500kg/3 .If thermocouple
junctiondiameter assumed to be 0.7mm, how long will it take for the
junction to reach199 , If junction temperature is at 25 and placed
in a gas steam that is at200.Given: h = 400 /2 . , Thermal
conductivity= K=20W/mK, =400 /. , density = = 8500kg/3 junction
diameter = 0.7mm = = 0.35 = 3.5 104, , Initaial temperature =25 ,
fluidtemperature==20.Solution:Lumped Capacity From page 57 in the
Data Book: = [] = = 4 2 = 4 (3.5 104 )2 = 1.53 106 2 = =4 33 93. 93
=4 (3.5 104 )33= 1.75 1010 3 = []199 25200 25=
(4001.531064001.7510108500) 0.99 = (1) Taking logarithms for both
sides 0.99 = 1 0.01 = = 0.041 () 94. Semi Infinite with error
function: 94 : . . q^./A flux . fluid :xoio= 2at page 58 in the
Data BookExample: A thick concrete wall fairly large in size
initially at 30 suddenly hasits surface temperature increased to
600 . Determine the depth at whichthe temperature become 400 after
25 minutes. Thermal diffusivity is4.92* 107m2 /s and K=1.28 W/mK.:
) 30 ( ) 600 (. . K=1.28 W/mK 4.92 = * 107m2 /s. 400 ( 25 (Given:
Initaial temperature = =30 , surface temperature= =600. T x=400 ,
Thermal diffusivity ==4.92* 107m2 /s,Thermal conductivity=K=1.28
W/mKSolution: 95. fluid dimension Semi Infinite with error
function95From page 58 in the Data Book:x oi o= 2 400 60030 600=
24.92 107 15000.35 = 0.0540.35 = () =0.054 . . (1)According to the
table from page 59 in the Data Book: 6.30 erf(z) column 6.30924
6.30 (z) 6.33 . =0.054 . . (1)0.33 =0.054 = 0.01782 = 1.782 The
depth= x=0.01782 m=1.782 cm (Answer) 96. Semi Infinite with
Convection Boundary condition: : . . q^./A flux . _T h . fluid :
Chart 96 97. . (Fluid) (Fluid) . Fluid (h) Fluid 06 : chart .
(time) . (time) : (time) x x Y-axis (T) (T) 97 curve Y- curve curve
axisX-axis . X-axis x2t x . curve X-axis htk curve X-axis . Y-axis
curveY-axis Y-axis . (time) X-axis assume curve htk X-axis 98.
Y-axis curve curve Y-axis . .Y-axis Example: a steel ingot (large
in size) heated uniformly to 415 is hardened byquenching it in an
engine oil maintain at 20 with = 58/2.Determine the time required
for the temperature to reach 595 at a depth of12mm. Steel has those
properties = 78333 , = 465 /. , = 48/.983 , = 465 /. , =48/. . The
ingot may be approximated as a large flat plate.: ) ( ) 415 ( .) =
58/ ) 20 ( ) 212 (. mm( ) 595 ( = 7833 .Given: Initaial temperature
= =415 , fluid temperature= =20. T=595 , depth x= 12mm= 0.012m. =
78333 , = 465 /. , =48/. Solution: fluid (dimension) . 99. Semi
infinite with convection boundary condition.From page 60 in the
Data Book: . :Infinite Solid : Heisler chart . .1. Infinite plate2.
Long cylinder3. Sphere1. Infinite plate . (x) (L ( 2. (Fluid)
convection .Data 00 chart Bookchart 99 100. Data Book 00 . Data
Book 00 chart : (x) . (surface) .Heat transfer ) : . Data Book 07
chart (rate100 101. Example: A plane wall is 0.12m thick ,made of
material of density 7800kg/3 ,thermal conductivity 45W/mK, and
specific heat 465 J/kg.K. It is initially at auniform temperature
of 310 , the wall is exposed suddenly to convection onboth sides at
30 with a convection heat transfer coefficient of 450W /m2 .K
.Determine the temperature after 8 minute at (1)mid plane (2) 0.03m
fromcenter. )7800kg/ 0.12 ( ) 3 m( : 465 (. J/kg.K( 45 ( W/mK( )
310 ) 30450 (. W /m2 .K( 0.03 ( m( ) ) 4( ) 1( ) 2Given: density =
= 7800kg/3 , Thermal conductivity= K=20W/mK, C =400J/kg.K
Thickness=2L=0.12m = 0.06 ,initial temperature= = 310,fluid
temperature==30. h = 450 /2 . , time =t
=8minute=8*60sec=480sec.101Solution: convection Infinite Solid
(Fluid) infinite plate 102. From page 63 and page 65 in the Data
Book:102= =207800 400= 6.41 106 2 /2 =X-axis=6.41106 4800.062 =
0.85 ==450 0.0620= 1.35 curve curve X-axis . 6.02 Y-axis = 0.52 =
0.52 = 30310 30 103. 145.6 = 30145.6 + 30 = = 175.6103Temperature
at the mid of the plate= = 175.6 ()From page 63 and page 66 in the
Data Book:X-axis==4500.0620= 1.35 ==0.030.06= 0.5 curve curve
X-axis . 6.49 Y-axis = 0.89 =/ 0.89 =/ 30175.6 30129.58 = / 30/ =
159.58 () 104. Example: a plate of 20cm thickness at temperature of
500 ; k=57W/m,suddenly air at 25 is blown over the surface with
h=200 /2 . , =11.85 1052 /. (1) At what time the mid temperature
will be 240.(2)At what time the temperature at 1cm from the surface
will be 240. )k=57W/m( 20 ( ) 500 ( cm( : .)= 11.85 1052/( )h=200 /
25 ( ). 2 (2( ( .) 1( ) 240 (104.) 1( ) 240 cm(Given:
Thickness=2L=20cm = 10 = 0.1, initial temperature= =500 ,Thermal
conductivity= K=20W/mK, fluid temperature==25. , = 200/2 . , =
11.85 1052 /Solution: convection Infinite Solid (Fluid) infinite
plate 105. 1051- Mid temperature:From page 63 and page 65 in the
Data Book: = =240 25500 25= 0.45 ==200 0.157= 0.35 = 2.7 = 2.7 =
22.7 =11.85 105 0. 12 106. 0.027 = 11.85 105 106 =0.02711.85 105 =
227.84 ()2- At what time the mid temperature at 1cm from the
surface will be240. ) 1( ) 2 ( : (x) . (surface) .L=10cm but
x=10-1=9cm =0.09mFrom the chart of page 66 in the Data
Book:=0.090.1= 0.9 ==200 0.157= 0.35 = 0.91 = 0.91 =/ 0.91 =240 25
250.91( 25) = 215( 25) = 236.26 = 261.26 107. Temperature at the
mid of the plate=261.26 when the surface will be240. )= 21.2 ( ) 2
( .107From page 6 3in the Data Book: = =261.26 25500 25= 0.49
0.5X-axis=2.35 = 2.35 = 22.35 = 22.35 =11.85 105 0. 120.0235 =
11.85 105 =0.023511.85 105 = 207.04 ()2. Long cylinder . (r) .
(Fluid) convection . 04 chart Data Book 108. Data Book 04 chart .
Data Book 09 chart : (r) . (surface) .Heat transfer ) : Data Book
76 chart (rate.Example: a long cylindrical bar of radius 80mm comes
out off oven at 830and is cooled by quenching in a large bath of 40
coolant. The heat transfercoefficient between the bar and coolant
is 180/2 . . Determine the timerequired the shaft center to reach
120 . If (K=17.4 W/mK and = 5.28 106 2 /)80 ( mm( : 830 ( ) 40 ( .
(180 (. /2. ( (K=17.4 W/mK and = ) 120 ( . 5.28108106 2/) 109.
Given: Thermal conductivity= K=17.4W/mK, = 5.28 106 2 / , 0 =80 =
0.08, initial temperature= = 830 ,fluidtemperature==40. h = 180 /2
. , = 120 , time =t =?Solution: convection Infinite Solid (Fluid)
Long cylinder From page 63 and page 68 in the Data Book:: =109 =120
40830 40= 0.1 =0=180 0.0817.4= 0.82X-axis=2 =02 =5.281060 .082
=5.281066.41030.0128 = 5.28 106 110. 110 =0.01285.28 106 = 2424.24
() curve curve Y-axis : . 2 X-axis 3. Sphere . (r) . (Fluid)
convection . chart Data Book 04 Data Book 04 chart . Data Book 09
chart : (r) . (surface) . 111. Heat transfer ) : . Data Book 76
chart (rateMulti dimension : 111From page 74 in the Data book: :
112. (semi-infinite plate) (a) .() (1) =112 infinite plate () ( ( 1
Semi-infinite solid with convection boundary condition (infinite
rectangular bar) (b) .(X1 ) (X2 ) = infinite plate (X ( 1 infinite
plate (X ( 2 (Semi-infinite rectangular bar) (c) . 113. () (X1 )
(X2 ) =113 infinite plate (X ( 1 infinite plate (X ( 2 ()
Semi-infinite solid with convection boundary condition (rectangular
parallelepiped) (d) .(X1 )(X2 ) (X3 ) = infinite plate (X ( 1
infinite plate (X ( 2 infinite plate (X ( 3 (Semi-infinite
cylinder) (e) .() () = () Infinite solid (Long cylinder) 114. 114
() Semi-infinite solid with convection boundary condition (short
cylinder) (f) .() () = () Infinite solid (Long cylinder). infinite
plate () Example: a cube of aluminum 10cm on each side is initially
at a temperatureof 300 and is immersed in a fluid at 100 . The heat
transfer coefficient is900W/2 .. Calculate the temperature in the
center of one face after 1min.Take = 8.418 1052 /, = 220/. 10 ( )
300 ( cm( : 900 ( . W/ ) 100 (. ). 2 115. = 1 . 8.4181151052/, =
220/.Given: 2L=10cm=0. 1m =0.05m, initial temperature == 300,
fluidtemperature = =100, heat transfer coefficient =h=900W/2 .
,time==1 min=60sec. = 8.418 1052 / = 220/. Solution:From the chart
of page 65 in theData Book: : = 2=8.418 105 600.052= 20.02 20 ==900
0.05220= 0.2 = 0.02 = 0.02 = (X1 ) = 0.02 = 100300 1000.02 200 =
1004 = 100 = 100 + 4 = 104 () 116. Example: a cube of aluminum 12cm
on each side is initially at a temperatureof 400 and is suddenly
immersed in a tank of oil maintained at 85 . Theheat transfer
coefficient is 1100W/2 .. Calculate the temperature at 1cmfar from
the three faces after 2 minutes. Take = 8.418 1052 / =220/. 12 ( )
400 ( cm( : ) 85 (. 1( 2 cm( 1100 (. W/2.(.Given: 2L=12cm=0. 12m
=0.06m, initial temperature == 400, fluidtemperature = =85, heat
transfer coefficient =h=1100W/2 . ,x=0.01m from the surfaces=0.11m
from the center, time==2 min=120sec. =8.418 1052 / = 220/. : (x) .
(surface) .116Solution:From the chart of page 65 in the
DataBook:(X1 ) = (X2 ) = (X3 ) = 2 =8.418 105 1200.062= 2.85 ==1100
0.06220= 0.3 = 0.48 117. = 0.48 =117 1cm from the surface: : (x) .
(surface) .L=0.06cm but x=0.06-0.01=0.05cmFrom the chart of page 66
in the Data Book: ==0.050.06= 0.8 ==1100 0.06220= 0.3 = 0.94 = 0.94
=/ (X1 ) = ( )=0 ( )=0.05 = 0.94 0.48=0.45(X1 ) =0.45(X1 ) = (X2 )
= (X3 ) =0.45(X1 ) (X2 ) (X3 ) = 0.45 0.45 0.45 = 85400 85 118.
0.091 = 8531528.7 = 85 = 28.7 + 85 = 113.7(
)MODULECONVECTIONConvection Heat Transfer-Requirements The heat
transfer by convection requires a solid - fluid interface,
atemperature difference between the solid surface and the
surrounding fluid118 119. and a motion of the fluid. The process of
heat transfer by convection wouldoccur when there is a movement of
macro-particles of the fluid in space froma region of higher
temperature to lower temperature. ) - ( . .Convection Heat Transfer
Mechanism Let us imagine a heated solid surface, say a plane wall
at a temperatureTw placed in an atmosphere at temperature T ,
Figure 2.1 Since all real fluidsare viscous, the fluid particles
adjacent to the solid surface will stick to thesurface. The fluid
particle at A, which is at a lower temperature, will receiveheat
energy from the plate by conduction. w T T ( 1 . 2 . ) . A .The
internal energy of the particle would Increase and when theparticle
moves away from the solid surface (wall or plate) and collides
with119 120. another fluid particle at B which is at the ambient
temperature, it will transfera part of its stored energy to B. And,
the temperature of the fluid particle at Bwould increase. In this
way, the heat energy is transferred from the heatedplate to the
surrounding fluid. Therefore in the process of heat transfer
byconvection involves a combined action of heat conduction, energy
storage andtransfer of energy by mixing motion of fluid particles.
) B ( B B . . .Figure Principle of heat transfer by convection Free
and Forced Convection When the mixing motion of the fluid particles
is the result of the density120 121. difference caused by a
temperature gradient, the process of heat transfer iscalled natural
or free convection. When the mixing motion is created by
anartificial means (by some external agent), the process of heat
transfer is calledforced convection, it is essential to have
knowledge of the characteristics offluid flow. Since the
effectiveness of heat transfer by convection dependslargely on the
mixing . ) ( . .Basic Difference between Laminar and Turbulent Flow
In laminar or streamline flow, the fluid particles move in layers
suchthat each fluid particle follows a smooth and continuous path.
There is nomacroscopic mixing of fluid particles between successive
layers, untill the fluidwill turn around a comer or an obstacle is
to be crossed. 121 122. . .If a line dependent fluctuating motion
is observed in directions whichare parallel and transverse to the
main flow, i.e., there is a randommacroscopic mixing of fluid
particles across successive layers of fluid flow, themotion of the
fluid is called' turbulent flow'. The path of a fluid particle
wouldthen be zigzag and irregular, but on a statistical basis, the
overall motion ofthe macro-particles would be regular and
predictable. . .Formation of a Boundary Layer When a fluid flows
over a surface, irrespective of whether the flow islaminar or
turbulent, the fluid particles adjacent to the solid surface
willalways stick to it and their velocity at the solid surface will
be zero, because ofthe viscosity of the fluid. Due to the shearing
action of one fluid layer over theadjacent layer moving at the
faster rate, there would be a velocity gradient ina direction
normal to the flow. . . .122 123. Chapter 5Principales of
ConvectionForced convection (flow over flat plates)Page 111 in the
Data Book:2 Laminar Flow Forced Convection Heat Transfer2.1Forced
Convection Heat Transfer PrinciplesThe mechanism of heat transfer
by convection requires mixing of oneportion of fluid with another
portion due to gross movement of the mass of thefluid. The transfer
of heat energy from one fluid particle or a molecule toanother one
is still by conduction but the energy is transported from one
point123 124. in space to another by the displacement of fluid.When
the motion of fluid is created by the imposition of external
forcesin the form of pressure differences, the process of heat
transfer is called forcedconvection. And, the motion of fluid
particles may be either laminar orturbulent and that depends upon
the relative magnitude of inertia and viscousforces, determined by
the dimensionless parameter Reynolds number. In freeconvection, the
velocity of fluid particle is very small in comparison with
thevelocity of fluid particles in forced convection, whether
laminar or turbulent. Inforced convection heat transfer,
Gr/Re2>1 and we have combined free and forced convection when
Gr/Re2 1 .2.2. Methods for Determining Heat Transfer CoefficientThe
convective heat transfer coefficient in forced flow can be
evaluatedby: (a) Dimensional Analysis combined with experiments;(b)
Reynolds Analogy an analogy between heat and momentumtransfer; (c)
Analytical Methods exact and approximate analyses ofboundary
layer124equations.2.3. Method of Dimensional AnalysisAs pointed out
in Chapter 5, dimensional analysis does not yi eldequations which
can be solved. It simply combines the pertinent variables
intonon-dimensional numbers which facilitate the interpretation and
extend therange of application of experimental data. The relevant
variables for forcedconvection heat transfer phenomenon whether
laminar or turbulent, are 125. (b) the properties of the fluid
density p, specific heat capacity Cp,dynamic or absolute viscosity
, thermal conductivity k.(ii) the properties of flow flow velocity
Y, and the characteristic125dimension of the system L.As such, the
convective heat transfer coefficient, h, is written as h = f ( ,V,
L, , Cp, k) = 0 (5.14)Since there are seven variables and four
primary dimensions, we wouldexpect three dimensionless numbers. As
before, we choose four independent orcore variables as ,V, L, k,
and calculate the dimensionless numbers byapplying Buckingham s
method:1 = a b d c a b c d 3 1 3 1 3 1 V L K h ML LT L MLT MT = o o
o o M L T Equating the powers of M, L, T and on both sides, we getM
: a + d + 1 = OL : - 3a + b + c + d = 0T : - b 3d 3 = 0 By solving
them, we have : - d 1= 0. D = -1, a = 0, b = 0, c = 1.Therefore, 1
= hL/k is the Nusselt number.2 = a b c d a b c d 3 1 3 1 1 1 V L K
ML LT L MLT ML T = o o o o M L T 126. Equating the powers of M, L,
T and on both sides, we get126M : a + d +1 = 0L : - 3a + b + c + d
= 1 = 0T : - b 3d 1 = 0 : - d = 0.By solving them, d = 0, b = - 1,
a = - 1, c = - 11 VLand 2 32/ VL;or, (Reynolds number is a flow
parameter of greatest significance. It is theratio of inertia
forces to viscous forces and is of prime importance to ascertainthe
conditions under which a flow is laminar or turbulent. It also
compares oneflow with another provided the corresponding length and
velocities arecomparable in two flows. There would be a similarity
in flow between twoflows when the Reynolds numbers are equal and
the geometrical similaritiesare taken into consideration.)a b c d 3
a 1 b c 3 1 d 2 2 1 4 p V L k C ML LT L MLT L T o o o o M L T
Equating the powers of M, L, T, on both Sides, we getM : a + d = 0;
L : - 3a + b + c + d + 2 = 0T : - b 3d 2 = 0; : - d 1 = 0By solving
them, 127. 127d =- 1,a = l, b = l, c = l,VLC ;4 p 5 4 2k VL C C
=ppk VL k 5 is Prandtl number.Therefore, the functional
relationship is expressed as:Nu = f (Re, Pr); or Nu = C Rem Prn
(5.15)where the values of c, m and n are determined
experimentally.Example: in aprocess water at30 flows over a plate
makntained at 10with a free stream velocity of 0.3m/s. Determine
the hydrodynamicboundary layer thicknedd, thermal boundary
thicknedd, convection heattransfer coefficient, heat trandfer rate
. Consider the plate of 1m*1m. Takethe following properties of air:
v=1.006*1062 / sec , k=0.5978W/m.,Pr=7.02Given:Fluid temperature=
=30 , plate temperature= =10stream velocity =u= 0.3m/s. plate
1m*1m. Kinematicviscosity=v=1.006*1062 / sec ,
thermalconductivity=k=0.5978W/m., Pr=7.02. Determine the
hydrodynamic 128. boundary layer thicknedd, thermal boundary
thicknedd, convection heattransfer coefficient, heat trandfer
rate?128Solution : )Fluid( . )Chapter ) 5From Page 111 in the Data
Book: = = = =0.3 11.006 106 = 298210.735 < 5 105 From Page 112
in the Data Book:0.5Hydroulic boundery thickness = hx = 5x
RexHydroulic boundery thickness = hx = 5 1(298210.735)0.5hx = 9.156
103 m (Answer)0.333 = = PrTx = (9.156 103 ) (7.02)0.333 = 4.784
103m (Answer) 129. 0.333( 0.333( . = = 0.664 129From Page 112 in
the Data Book: )Data Book( )Nusselt Number( : 0 .5 Pr( Nux = 0.332
Rex0.5 ( = 0.664 0.3330.5 = 0.664 (298210.735)0.5 (7.02)0.333 =
0.664 (546.086) (1.913) = 693.84 . (1)From Page 111 in the Data
Book: = = = 10.5978=0.5978 . . (2)Equation (1)= Equation (2)693.84
=0.5978 = 693.84 0.5978 = 414.7772 . Convection heat transfer
coefficient=h=207.388W/2. (Answer)Heat transfer rate =Q=hAT= hA( )
= 414.777(1 1)(30 10) = 414.777 (20) = 8 295.551 () 130. Example:
water flows over a flat plate having a uniform surface temperature
of80 , the plate is 15mm * 15mm side. Water is at 20 and the flow
velocity is3m/sec. Determine the heat carried away by the water.
Use the followingproperties, Pr=3.68, v=0.5675*1062 / sec ,
k=0.6395W/m.KGiven: plate temperature= =80 , plate 0.015m*0.015m.
Fluidtemperature= =20 , stream velocity =u= 3m/s.
Kinematicviscosity=v=0.5675*1062 / sec Pr=3.68,
thermalconductivity=k=0.6395W/m.K. Determine heat trandfer
rate?130Solution : )Fluid( . )Chapter ) 5From Page 111 in the Data
Book: = = = =3 0.0150.5675 106 = 79295.154 < 5 105 From Page 112
in the Data Book: 131. )Data Book( )Nusselt Number( : 0.333( ( Nux
= 0.332 Rex0.333( . = = 0.664 1310 .5 Pr0.5 ( = 0.664 0.3330.5 =
0.664 (79295.154)0.5 (3.68)0.333 = 0.664 (281.593) (1.543) =
288.546 . (1)From Page 111 in the Data Book: = = = 0.0150.5978 =
0.023 . . (2)Equation (1)= Equation (2)288.546 = 0.023 =288.5460
.023 =12 301.676 /2 . Convection heat transfer
coefficient=h=6150.838/2 . Heat transfer rate =Q=hAT= hA( ) = 12
301.676(0.015 0.015)(80 20) = 2.76 (60) = 165.6 ()Example: A plate
of 100cm*50cm and 2cm thick is placed in a horizontalplane. The top
surface is maintained at 100 . If the air is flowing over the 132.
plate at 3m/sec and 20 . Find the heat lost by the plate per hour.
Whatshould be the bottom temperature of the plate at steady state
condition?Take thermal conductivity of the plate 20W/m.k , 100cm
side of the plate isparallel to the air flow. Take properties of
air at mean temperature of 60.Given: plate 1m*0.5m,plate
temperature= =100 , stream velocity=u= 3m/s Fluid temperature= =20
, thermal conductivity of theplate=k=20W/m., Determine heat
trandfer rate?132Solution : )Fluid( . )Chapter ) 5From Page 33 in
the Data Book we will find properties of air at meantemperature of
60Kinematic viscosity=v=18.97*1062 / sec Pr=0.696,
thermalconductivity of the air=k=0.02896W/m.,From Page 111 in the
Data Book: 133. = =133 = =3 118.97 106 = 158144.438 < 5 105 From
Page 112 in the Data Book:0.5 = = 0.664 0.333 = 0.664
(158144.438)0.5 (0.696)0.333 = 0.664 (397.673) (0.886) = 234.036 .
(1)From Page 111 in the Data Book: = = = 10.02896 =0.02896 . .
(2)thermal ( )thermal conductivity( thermal ( )Nu( )conductivity of
the plate. )Convection( . )conductivity of the airEquation (1)=
Equation (2)234.036 =0.02896 = 234.036 0.02896 =6.776 /2 .
Convection heat transfer coefficient=h=3.388/2 . Heat transfer rate
=Q=hAT= hA( ) = 6.776 (1 0.5)(100 20) 134. = 6.776 0.5 (80) = 271.1
134Heat lost per second= = 271.1 Heat lost per hour= 3600 = 975.986
()Bottom temperature: = = (1 2 ) )Q( )thermal conductivity of the
air( . )Conduction( . )thermal conductivity of the plate(271.1 =
20(0.5 1)(1 100)0.02271.1 = 10 (1 100)0.02271.1 =(1 100)0.020.54 =
1 100 1 = 100.54 ()Example: a flat plate 1m wide and 1.5m long is
to be maintained at 90 in airwith temperature of 10. Determine the
velocity with which air must flowover the plate so that heat
dissipated from plate is 3.75 Kw. Take air propertyat 50. Take the
flow laminar.Given: plate 1m*1.5m,plate temperature= =90 , Fluid
temperature= =10, heat dissipated from plate =Q= 3.75 Kw=3750W.
135. Solution: From Page 33 in the Data Book we will find
properties of air atmean temperature of 50Kinematic
viscosity=v=17.95*1062 / sec , Prandtl Number=Pr=0.6968,thermal
conductivity of the air=k=0.02826W/m.,Heat transfer rate =Q=hAT=
hA( )3750= h(1.5*1)(90 10)3750= 120*h135 =3750120= 31.25/2 . From
Page 111 in the Data Book: = = =31.25 1.50.02826= 1658.704From Page
112 in the Data Book: 0.5 = = 0.664 0.333 136. 1658.704 = 0.664 2=
(2 820.925)21360.5 (0.698)0.3330 .5 (0.887)1658.704 = 0.664 0
.51658.704 = 0.588 0.5 =1658.7040.588= 2 820.925= 2 820.925()= 7
957 618.81From Page 111 in the Data Book: = =7 957 618.81 =
1.517.95 106142.83 = 1.5 =142.831.5= 95.22 / sec ()Example: air at
30 flows with a velocity of 2.8m/s over a plate 1000mmlength 600mm
width 25 mm thickness. The top surface of the plate ismaintained at
90. If the thermal conductivity of the material is 25W/m.
,calculate (i) heat lost by the plate (ii) bottom temperature of
the plate.Use the following properties for air: = 1.06/3 ,
C=1.005Kj/kg.K,k=0.02894W/m. ,v=18.97*1062 /, Pr=0.696.Solution:
137. 137Solution : )Fluid( . )Chapter ) 5From Page 111 in the Data
Book: = = = =2.8 118.97 106 = 147601.476 < 5 105 From Page 112
in the Data Book:0.5 = = 0.664 0.333 = 0.664 (147601.476)0.5
(0.696)0.333 = 0.664 (384.189) (0.886) = 226.02 . (1)From Page 111
in the Data Book: = = 138. =138 10.02896 =0.02894 . . (2)thermal (
)thermal conductivity( thermal ( )Nu( )conductivity of the plate.
)Convection( . )conductivity of the airEquation (1)= Equation
(2)226.02 =0.02894 = 226.02 0.02894 = 6.54 /2 . Convection heat
transfer coefficient=h=6.54 /2 . Heat transfer rate =Q=hAT= hA( ) =
6.54 (1 0.6)(90 30) = 235.44 Heat lost = 235.44 ()Bottom
temperature: = = (1 2 ) )Q( )thermal conductivity of the air( .
)Conduction( . )thermal conductivity of the plate(235.44 = 25(1
0.6)(1 90)0.025 139. 235.44 = 15 139(1 90)0.02515.696 =(1
90)0.0250.392 = 1 90 1 = 90.392 ()Example: a square plate 12 area
heated uniformly to constant temperatureof 90 after that is cooled
by air at 20 flowing over the plate at 2m/sec.Calculate heat lost
from the plate. Take the following properties of air , =1.075/3 ,
C=1.008Kj/kg.K, k=0.0286W/m.k, = 19.8 106 /2 .Given: plate
1m*1m,plate temperature= =90 , Fluid temperature= =20 , stream
velocity =u= 2m/s thermal conductivity of theplate=k=20W/m., =
1.075/3 , C=1.008Kj/kg.K,k=0.0286W/m. , = 19.8 106 /2 , Determine
heat trandfer rate?Solution : 140. )Fluid( . )Chapter ) 5140From
Page 111 in the Data Book: = = =1.008 19.8 1060.0286= 0.697 = = =
=2 1 1.07519.8 106 = 108585.858 < 5 105 From Page 112 in the
Data Book:0.5 = = 0.664 0.333 = 0.664 (108585.858)0.5 (0.697)0.333
= 0.664 (329.523) (0.887) = 194.07 . (1)From Page 111 in the Data
Book: = = = 10.0286 =0.0286 . . (2)Equation (1)= Equation (2)194.07
=0.0286 = 194.07 0.0286 141. = 5.55 /2 . Convection heat transfer
coefficient=h=5.55/2 . Heat transfer rate =Q=hAT= hA( ) = 5.55 (1
1)(90 20) = 388.544 Heat lost = = 388.544 ()Example: air at 20 is
flowing over a flat plate which is 200mm wide and500mm long. The
palte is maintained at 100 . Find the heat lost per hourfrom the
plate if the air is flowing parallel to 500mm side with
2m/secvelocity. if the flow is parallel to 200mm side, What will be
the effect onheat transfer? Properties of air at 60 are these:
v=18.97*1062 / sec ,Pr=0.7, thermal conductivity of the
air=k=0.025W/m. .Given: Fluid temperature= =20plate 0.5m*0.2m,plate
temperature= =100 , stream velocity =u= 2m/s ,
Kinematicviscosity=v=18.97*1062 / sec , Pr=0.7, thermal
conductivity of theair=k=0.025W/m.,Determine heat trandfer
rate?Solution : (A)If the air is floeing parallel to the 500mm
side:141 142. )Fluid( . )Chapter ) 5142From Page 111 in the Data
Book: = = = =2 0.518.97 106 = 52714.812 < 5 105 From Page 112 in
the Data Book:0.5 = = 0.664 0.333 = 0.664 (52714.812)0.5 (0.7)0.333
= 0.664 (229.597) (0.888) = 135.378 . (1)From Page 111 in the Data
Book: = = = 0.50.025 = 20 . . (2)Equation (1)= Equation (2)135.378
= 20 =135.37820 = 6.768 /2 . Convection heat transfer
coefficient=h=6.768 /2 . Heat transfer rate =Q=hAT= hA( ) = 6.768
(0.2 0.5)(100 20) 143. = 6.768 0.1 (80) = 54.15 143Heat lost per
second= = 54.15 Heat lost per hour= 3600 = 194 940 ()(B)If the air
is floeing parallel to the 200mm side:From Page 111 in the Data
Book: = = = =2 0.218.97 106 = 21085.9258 < 5 105 From Page 112
in the Data Book:0.5 = = 0.664 0.333 = 0.664 (21085.9258)0.5
(0.7)0.333 = 0.664 (145.209) (0.888) = 85.62 . (1) 144. 144From
Page 111 in the Data Book: = = = 0.20.025 = 8 . . (2)Equation (1)=
Equation (2)85.62 = 8 =85.628 = 10.7 /2 . Convection heat transfer
coefficient=h=10.7/2 . Heat transfer rate =Q=hAT= hA( ) = 10.7 (0.2
0.5)(100 20) = 10.7 0.1 (80) = 85.616 Heat lost per second= =
85.616 Heat lost per hour= 3600 = 308 232 ()The effect on the heat
transfer:heat transfer will increase from 54.15 Watt to 85.62 Watt.
145. Chapter 6Forced convection (flow across tubes or ducts)145
146. Flow over one tube or one duct page 115 in the Data
BookExample: air at 90 and 1 atmosphere, flows across a heated
1.5mmdiameter wire at a velocity of 6m/sec. The wire is a
temperature of 150 .Use the following data. Kinematic viscosity
25.6*1062 / sec , thermal146 147. conductivity of the air
0.03365W/m.K , Pr=0.689, Calculate heat lost perunit length.Given:
Fluid temperature= =90, wirediameter=d=0.5mm=1.5*103 , air velocity
=u= 6m/s , wiretemperature= =150 , Kinematic viscosity=v=25.6*1062
/ sec ,thermal conductivity of the air=k=0.03365W/m., Pr=0.689,
Determineheat lost per unit length=147?Solution : = = = =6 1.5
10325.6 106 = 351.562From Page 115 in the Data Book: = =
0.333Reynolds numbaer is between 40-4000 C=0.683 , m=0.466 from the
table at page 115 in the DataBook. = = 0.683 (351.562)0.466
(0.689)0.333 = 0.683 (15.361) (0.883) = 9.267 . (1)From Page 111 in
the Data Book: = = = 1.5 1030.03365 = 0.044 . . (2)Equation (1)=
Equation (2) 148. 9.267 = 0.044 =1489.2670.044 = 207.889 /2 .
Convection heat transfer coefficient=h=207.889/2 . = = = 1.5 103 =
4.712 103 Heat transfer rate =Q=h T= h ( ) = 207.889(4.712 103 )
(150 90) = 0.979 (60)= 58.779 == 58.779( )Example: air at 20 flows
with a velocity of 0.33m/s is around of 25mmdiameter horizontal
tube 400mm long, calculate heat transfer if the tube wallis
maintained at 180 .Given: Fluid temperature= =20, air velocity =u=
6m/s , tubediameter=d=25mm=0.025m, tube length=L 400mm=0.4m,
tubetemperature= =180 , Determine heat transfer rate .Solution : =
= + 2=180 + 202=2002= 100We will find properties for air at = 100
149. From page 33 in the Data Book: = 0.946/3Kinematic
viscosity=v=23.13*1062 / sec , Pr=0.688, C=1009j/kg.K,thermal
conductivity of the air=k=0.0321W/m.,149From page 111 in the Data
Book: = = = =0.33 0.02523.13 106 = 356.679From Page 115 in the Data
Book: = = 0.333Reynolds numbaer is between 40-4000 C=0.683 ,
m=0.466 from the table at page 115 in the DataBook. = = 0.683
(356.679)0.466 (0.688)0.333 = 0.683 (15.465) (0.882) = 9.325 .
(1)From Page 111 in the Data Book: = = = 0.0250.0321 = 0.778 . .
(2)Equation (1)= Equation (2)9.325 = 0.778 =9.3250.778 = 11.973/2 .
150. Convection heat transfer coefficient=h=11.973/2 . = = = 0.025
0.4 = 0.031 2Heat transfer rate =Q=h T= h ( ) = 11.973(0.031) (180
20) = 60.182 heat transfer rate= = 60.182 ()Flow across tube Banks
page 120 in the Data Book150 151. : ) ( )maximum velocity( = ( :
)in line( 1. St151StD) 152. . ) ( )Staggered( 2. .StSt D152(1) = ()
(2) =Sl2(Sd D) = Sl2 + (St2)2Example: a tube bank uses an in-line
arrangement of 30mm diameter tubeswith Cp=Cn=60mm a tube length of
2m. There are 10 tube rows in the flowdirection and seven tubes
high. Air at 27 and 15m/s flow in cross flow incross flow over the
tubes, while a tube wall temperature is maintained at100 .
Determine the temperature of air leaving the tube bank and the
rateof heat transfer.Given: tubes diameter=30mm=0.03m, pitch
transverse to flow=St=SP =Cp=60mm=0.06m, pitch along the flow=Sl=Sn
= Cn=60mm=0.06m,tubelength=L=2m, number of rows=n=10, = = = 10 7 =
70 , air velocity=15 m/s, airtemperature=1=27, tube wall
temperature =100.Solution: = = + 2=100 + 2702=1272= 63.5 60 We will
find properties for air at = 60From page 33 in the Data Book: =
1.06/3Kinematic viscosity=v=18.97*1062 / sec , Prandtl
Number=Pr=0.696,C=1005 j/kg.K, thermal conductivity of the
air=k=0.02896W/m..It is in-line arrangement.From page 120 in the
Data Book: 153. = = (153StSt D) = (0.060.06 0.03) 15 =0.060.03= 2
15 = 30 /Reynolds Number to be calculated on the basis of maximum
fluid velocity( ):From page 111 in the Data Book: = = = =30
0.0318.97 106 = 47 443.331From Page 122 in the Data Book:(FLOW
ACROSS BANKS TUBES) = = The values of C and n are given in table of
page 122:StD=0.060.03= 2SlD=0.060.03= 2 = 0.229 , = 0.632 = = =
0.229 (47 443.331)0.632 = 0.229 (902.282) = 206.622 . (1)From Page
111 in the Data Book: = = 154. =154 0.030.02896 = 1.035 . .
(2)Equation (1)= Equation (2)206.622 = 1.035 =206.6221.035 =
199.459 /2 . Convection heat transfer coefficient=h=199.459 /2 . =
= ( ) = 0.03 2 70 = 13.194 2Heat transfer rate =Q=h T= h ( ) =
199.459(13.194) (100 27) = 192 121.363 Total heat transfer rate= =
192 121.363 ()Mass flow rate= . = . = 1.06 15 10 0.06 = 9.45 / = .
(2 1)192 121.363 = 9.45 1005 (2 27)192 121.363 = 9 587.7 (2
27)20.038 = (2 27)20.038 + 27 = 22 = 47.038The exit air
temperature=2 = 47.038 (Answer) 155. Example: 20mm out diameter
copper tubes are arranged in in-line at 30mmpitch Sp and 25mm pitch
Sn. The entry velocity of air is 1 m/s at 20 the tubewall is at
40.determine the exit air temperature and total heat transfer
rateif the numbers of tube are arranged as 6 rows with 6
columns.Given: tubes diameter=20mm=0.02m, pitch transverse
toflow=St=SP=30mm=0.03m, pitch along the flow=Sl=Sn=25mm=0.025m,
airvelocity=1 m/s, air temperature=20, tube wall temperature =40of
is at 20 the tube wall is at 40 , number of rows =6 rows, number
ofcolumns =6 columns.155Solution: = = + 2=40 + 202=602= 30We will
find properties for air at = 60From page 33 in the Data Book: =
1.165/3Kinematic viscosity=v=16*1062 / sec , Pr=0.701, C=1005
j/kg.K,thermal conductivity of the air=k=0.02675W/m..It is in-line
arrangement.From page 120 in the Data Book:StSt D = = () =
(0.030.03 0.02) 1 =0.030.01= 3 /Reynolds Number to be calculated on
the basis of maximum fluid velocity( ): 156. 156From page 111 in
the Data Book: = = = =3 0.0216 106 = 3750From Page 122 in the Data
Book:(FLOW ACROSS BANKS TUBES) = = The values of C and n are given
in table of page 122:StD=0.030.02= 1.5SlD=0.0250.02= 1.25 = 0.367 ,
= 0.586 = = = 0.367 (3750)0.586 = 0.367(124.274) = 45.608)Nusselt
Number( : 16 . )C ) 1note : Rows deep= columnMultiply the above
Nusselt Number by C1. See tables 122:For in line tubes C1=0.94 =
45.608 0.94 = 42.871 . (1)From Page 111 in the Data Book: 157. =
=157 = 0.020.02675 = 0.747 . . (2)Equation (1)= Equation (2)42.871
= 0.747 =42.8710.747 = 57.391 /2 . Convection heat transfer
coefficient=h=57.391 /2 . Assuming the length of tubes =1m = = ( )
= 0.02 1 36 = 2.261 2Heat transfer rate =Q=h T= h ( ) =
57.391(2.261) (40 20) = 57.391(2.261) (20) = 2595.247 Total heat
transfer rate= = 2595.247 ()Mass flow rate= . = . = 1.165 1 6 0.025
= 0.174 / = . (2 1)2595.247 = 0.174 1005 (2 20) 158. 2595.247 =
174.87 (2 20)14.841 = (2 20)14.841 + 20 = 22 = 34.841The exit air
temperature=34.841 (Answer)Example: air 1 atmosphere and 10 flows
across a bank of tubes 15 rows highand 10 rows deep at a velocity
of 7m/s measured before entering the tubebank. The tubes are
maintained at 65. The diameter of the tubes is 2.54 cmthey are
arranged in an in-line manner so that the spacing in both normal
andparallel directions to the flow is 3.81 cm. Calculate the total
heat transfer perunit length for the tube bank and the exit
temperature. Air properties at27.5 are = 1.137/3 , = = 1.894 105/.
,Pr=0.706, C=1007 j/kg.K, thermal conductivity of the
air=k=0.027W/m..Given: air temperature==10 , air velocity==7 m/s,
number ofrows=n=15 rows, number of columns=rows deep =10 columns,
tube walltemperature =65, tubes diameter=2.54 cm=0.0254m, pitch
transverse toflow=St=SP=3.81cm=0.0381 m, pitch along the
flow=Sl=Sn=3.81cm=0.0381 m, = 1.137/3 , = = 1.894 105/. ,
Pr=0.706,C=1007 j/kg.K, thermal conductivity of the
air=k=0.027W/m..158Solution:note : Rows deep= columnIt is in-line
arrangement.From page 120 in the Data Book: 159. = = (159StSt D) =
(0.03810.0381 0.0254) 7 = 21 /Reynolds Number to be calculated on
the basis of maximum fluid velocity( ):From page 111 in the Data
Book: = = = =1.137 21 0.02541.894 105 = 32 020.897From Page 122 in
the Data Book:(FLOW ACROSS BANKS TUBES) = = The values of C and n
are given in table of page 122:StD=0.03810.0254=
1.5SlD=0.03810.0254= 1.5 = 0.250 , = 0.620 = = = 0.250 (32
020.897)0.620 = 0.250(621.396) = 155.349 . (1)From Page 111 in the
Data Book: = = 160. =160 0.02540.027 = 0.94 . . (2)Equation (1)=
Equation (2)155.349 = 0.94 =155.3490.94 = 165.134 /2 . Convection
heat transfer coefficient=h=165.134 /2 . = = ( = 10 15= 150) =
0.0254 150 = 11.969 = = = ( ) = 165.134(11.969 ) (65 10) = 108
711.137 = 108 711.137 /Total heat transfer per unit length== 108
711.137 / ()Mass flow rate= . = . = 1.137 7 15 0.0381 = 4.548 / = .
(2 1)108 711.137 = 4.548 10057 (2 10)108 711.137 = 4580.408 (2
10)23.733 = (2 10) 161. 23.733 + 10 = 22 = 33.733The exit air
temperature=33.733 (Answer)Internal flow at Page 123 in the Data
Book161 162. Example: In a straight tube of 60mm diameter, water is
flowing at a velocityof 12m/sec. The tube surface temperature is
maintained at 70 and theflowing water will be heated from the inlet
temperature 15 to an outlettemperature of 45 . Take properties of
water at its mean temperature.Calculate (1) heat transfer
coefficient. (2) Heat transferred. (3) The length ofthe
tube.162Given: tubes diameter=60mm=0.06m, water velocity=u=1m/s,
tube wall temperature =70,inlet temperature=1=15,outlet
temperature=2=45.Solution: = =1+22=15+452=602=30We will find
properties for water at = 30From page 21 in the Data Book:Water
properties at 30 are not possess, we will find Water properties
at20 and 40 after that we will find at 30 due to
interpolation.temperature density()Kinematicviscosity(v)Number(Pr)
specificheat( C)thermalconductivity(k)20 1000 1.006*106 7.020 4178
0.597840 995 0.657*106 4.34 4178 0.628Water properties at 30:30 =20
+ 402=1000 + 9952= 997.5/3 163. 16330 =20 + 402=1.006 106 + 0.657
1062= 0.8315 1062 /30 =20 + 402=7.02 + 4.342= 5.6830 =20 + 402=4178
+ 41782= 4178 /. 30 =20 + 402=0.5978 + 0.6282= 0.6129 /. From page
111 in the Data Book: = = = =0.06 120.8315 106 = 865 904.991 >
2300 From Page 125 in the Data Book: : = = 0.023 0.8 N=0.4 for
heating of fluidsN=0.3 for cooling of fluidsOutlet temperature (45
) > inlet temperature (15 ) heating of fluids and n = 0.4 =
0.023 (865 904.991)0.8 (5.68)0.4 = 0.683 (56231.05) (2) = 2586.628
. (1)From Page 111 in the Data Book: 164. = =164 = 0.060.06129 =
0.097 . . (2)Equation (1)= Equation (2)2586.628 = 0.097
=2586.6280.097 = 26422.405 /2 . Convection heat transfer
coefficient=h=26422.4052 . () = = 24= 0.0624= 2.827 103 2Mass flow
rate= . = . = 997.5 12 2.827 103 = 33.844 / = . (2 1 ) = 33.844
4178 (45 15) = 4242006.96 2. The heat transferred = = 4242006.96
(Answer)Heat transfer rate =Q=h T= h ( 1 +22)4242006.96 = 26422.405
(70 30)4242006.96 = 26422.405 40 =4242006.9626422.405 40= 4.013 2 =
= 165. 4.013 = 0.06 L165 =4.0130.188= 21.289 () )hydraulic
diameter( (Dh =4 AP. )Data Book( ( 120 )hydraulic diameter( )d( .
:Example: Water flows in a duct having a cross section 5mm*10mm
with amean bulk temperature of 20 . The duct wall temperature is
constant at60 , for fully laminar flow calculate the water flow
velocity and amountof heat transfer, if exit water temperature is30
. Take the followingproperties of water. =100kg/3 ,v=1.006*1062/sec
, Pr=7.02,C=4216J/kg.K, k=0.5978W/m.K .Given: = = 20, = 2 = 30,duct
wall temperature =60.Solution: = = 20 =1 + 22 20 =1 + 30240 = 1 +
30 1 = 10 166. The type of the flow is Laminar. From Page 127 in
the Data Book:166Because22=510=12 = 3.391 . (1) = = 0.005 0.01 = 5
105 2 = 2() + 2() = 2 0.005 + 2 0.01 = 0.03from page 126 in the
Data Book: =4 =4 =4 5 1050.03= 6.66 103Assuming duct length=1 m. =
= = 6.66 103 1 = 0.02 2From Page 111 in the Data Book: = =3.391 =
6.66 1030.59782.027 = 6.66 103 =2.0276.66 103 = 304.101 /2 . 2 >
1 167. = = . (2 1 ) = ( 1671 + 12). (2 1 ) = ( 1 + 12). 4216(30 10)
= 304.1 0.02(60 20). 4216(20) = 304.1 0.02(40). 84320 = 243.28.
=243.2884320= 2.885 103 / = . = 2.885 103 = 1000 5 1052.885 103 =
0.05 =2.885 1030.05= 0.0577sec()Amount of heat transfer: = ( 1 +
12) = 304.1 0.02(60 20) = 243.28 Heat transfer rate= = 243.28 ()
168. Example: n-butyl alcohol flows through a square duct of 0.1m
side with avelocity of 30mm/sec. the duct is 4m long. The walls are
at constanttemperature of 27 . The bulk mean temperature of n-butyl
is 20 .Determine Heat transferred. Take the following properties of
n-butyl. , ,=810kg/3 , =29.5*104Ns/2 , Pr=50.8, k=0.167W/m.K
.Given: duct wall temperature =27, fluid
velocity=u=30mm/sec=0.03m/sec, = =20.168Solution:The type of the
flow is Laminar.From Page 127 in the Data Book:Because22=0.10.1=1 =
2.976 . (1) = = = 0.1 0.1 = 0.01 2 = 2() + 2() = 2 0.1 + 2 0.1 =
0.4 from page 126 in the Data Book: =4 169. =4 169 =4
0.010.4=0.040.4= 0.1 From Page 111 in the Data Book: = =2.976 =
0.10.1672.976 = 0.598 =2.9760.598= 4.969 /2 . = = = 4 0.1 = 1.256 2
> = = ( 1 + 12) = ( 1 + 12) = 4.962 1.256(27 20) = 43.648 Heat
transfer rate= = 43.648 () 170. Example: Air at 20 flows through a
tube 8cm diameter with a velocity of9m/sec. The tube wall is 80 .
Determine the tube length required for exit airtemperature of 36
.Given: inlet temperature=1=20, tube diameter=d=8cm=0.08m,
airvelocity=u=9 m/s, tube wall temperature =80,
outlettemperature=2=45.Solution: = =1 + 21702=20 + 362=562= 28 30We
will find properties for air at = 30From page 33 in the Data Book:
= 1.165 /3 , = 16 1062 /, = 0.701, = 1005 J/kg. , =0.02675 /. From
page 111 in the Data Book: = = = =9 0.0816 106 = 45000 > 2300
From Page 125 in the Data Book:Dittus-Boelter equation: = = 0.023
0.8 171. N=0.4 for heating of fluidsN=0.3 for cooling of
fluidsOutlet temperature (36 ) > inlet temperature (20 ) heating
of fluids and n = 0.4 = 0.023 (45000)0.8 (0.701)0.4 = 0.683
(5279.223) (0.867) = 105.337 . (1)171From Page 111 in the Data
Book: = = = 0.080.02675 = 2.99 . . (2)Equation (1)= Equation
(2)105.337 = 2.99 =105.3372.99 = 35.22 /2 . = = 24= (0.08)24= 5 103
2Mass flow rate= . = . = 1.165 9 5 103 = 0.052 / = . (2 1 ) = 0.052
1005 (36 20) = 847.47 Heat transfer rate =Q=h T= h ( 1 +22) 172.
847.47 = 35.222 (80 28)847.47 = 1831.544 172 =847.471831.544= 0.462
2 = = 0.462 = 0.08 L0.462 = 0.25 L =0.4620.25= 1.838 ()Tube
length=L=1.838 m (Answer) 173. Chapter 7Free Convection)Reynolds
Number( )Free convection( : Constant wall ( )Grashof Number( .
)Modify Grashof Number( )temperature . )Constant heat flux(
Example: a flat electrical heater of 0. 4 m*0.4 m size is placed
vertically in stillair at 20 . The flux from the heater is 1200 W/2
. Determine the value ofconvection heat transfer coefficient and
the plate temperature.Given: Area=(0.4 m*0.4 m)air,
temperature==20, heat flux=q=1200W/2.Solution: From page 33 in the
Data Book:173We will find air properties at 20.temperature density
Kinematic Number(Pr) thermal thermal 174. () viscosity(v)
diffusivity() conductivity174(k)20 1.205 15.06*106 0.703 21.417*106
0.02593We will find coefficient of thermal expansion () from page
29 in the DataBook at20. = 3.42 103 1This question is constant heat
flux.Constant heat ( )Modify Grashof Number( . )fluxModify Grashof
Number = = 42 =(9.81)(3.42 103 )(1200)(0.4)4(0.02593)(15.06 106 )2
=1.035.88 1012 = 1.75 1011 = 1.75 1011 0.703 = 1.23 1011 > 109It
is turbulentWe will choose a suitable rule to find Nusselt Number
(Nu): from page 135 inthe Data Book: = 0.17( )0.25 = 0.17(1.23 1011
)0.25 = 0.17(592.35) = 100.7 . (1) = = 175. =175 0.40.02593 =
15.426 . . (2)Equation (1)= Equation (2)100.7 = 15.426 =100.715.426
= 6.52 /2 . Convection heat transfer coefficient= = 6.522 . ()Heat
transfer rate =Q=h T= h ( ) = ( )= ( )1200 = 6.52 ( 20)183.82 = (
20)183.82 + 20 = = 203.82 The plate temperature= = 203.82 () .
)Upper surface is heated( : 176. Example: A flat plate 1m by 1m is
inclined at 30 with the horizontal andexposed to atmospheric air at
30 . The plate receives a net radiant heatenergy flux from the sun
of 700W/2 which then is dissipated to thesurrounding by free
convection. What temperature will be attained by theplate?Given:
Length=L=1 m, width=w=1 m, angle with the horizontal=30 angle with
the Vertical = = 90 30 = 60o , air temperature= = 30,heat
flux=q=700 W/2 .Solution: From page 33 in the Data Book:176We will
find air properties at30.temperature
density()Kinematicviscosity(v)Number(Pr) Specificheat
(C)thermalconductivity(k)20 1.165 16*106 0.701 1005 0.02675We will
find coefficient of thermal expansion () from page 29 in the
DataBook at30. = 3.3 103 1 177. 177This question is constant heat
flux.Constant heat ( )Modify Grashof Number( . )fluxModify Grashof
Number = = 42 =(9.81)(3.3 103 )(700)(1)4(0.02675)(16 106 )2
=22.666.848 1012 = 3.3 1012 = (3.3 1012) 0.701 = 2.31 1012 >
109It is turbulentWe will choose a suitable rule to find Nusselt
Number (Nu): from page 136 inthe Data Book: 60 = (2.31 1012 ) (0.5)
= 1.155 1012 = 0.17( )0.25 = 0.17(2.31 1012 )0.25 = 209.58 .
(1)from page 111 in the Data Book: = = = 10.02675 = 37.38 . .
(2)Equation (1)= Equation (2) 178. 209.58 = 37.38 =178209.5837.38 =
5.6 /2 . Convection heat transfer coefficient= = 5.6W/2 . ()Heat
transfer rate =Q=h T= h ( ) = ( )= ( )700 = 5.6 ( 30)124.86 = (
30)124.86 + 30 = = 154.86 The wall plate temperature= = 154.86
()Example: consider a surface 0.5 m high kept in an angle of 45
from thehorizontal at a constant wall temperature of 40 , in air at
20. Determinethe value of convective heat transfer
coefficient.Given: Length=L=1 m, angle with the horizontal=45 angle
with the Vertical = = 90 45 = 45o , air temperature= = 20,wall
temperature = =40 . 179. 179Solution: = = + 2=40 + 202=602= 30From
page 33 in the Data Book:We will find air properties
at30.temperature density()Kinematicviscosity(v)Number(Pr)
Specificheat (C)thermalconductivity(k)20 1.165 16*106 0.701 1005
0.02675We will find coefficient of thermal expansion () from page
29 in the DataBook at30. = 3.3 103 1This question is constant wall
temperature.)Constant wall temperature( )Grashof Number( .Grashof
Number = = 32 ( 134 ) =(9.81)(3.3 103 )(0.5)3(40 20)(16 106 )2 180.
180 =0.082.56 1010 = 312 500 000 = (312 500 000) 0.701 = 219 062
500 < 109It is Laminar.We will choose a suitable rule to find
Nusselt Number (Nu): from page 134 inthe Data Book: = 0.59(
45)0.250.25 = 0.59((219 062 500) (0.707)) = 0.59(111.56) = 65.82 .
(1)From page 111 in the Data Book: = = = 0.50.02675 = 18.69 . .
(2)Equation (1)= Equation (2)65.82 = 18.69 =65.8218.69 = 3.52 /2 .
Convection heat transfer coefficient= = 3.52W/2 . () 181. Example:
a long rectangular duct of width and height of 0.75 m and 0.3
mrespectively. The outerSurface temperature of the duct is
maintained at 45 . If the duct is exposed to air at 15 in a aramp
space beneath a home, whatis the heat loss from the duct per meter
length?Given: width=0.75 m and height = 0.3 m, Surface temperature
== 45 . airtemperature= = 15181Solution: = = + 2=45 + 152=602=
30From page 33 in the Data Book:We will find air properties
at30.temperature density()Kinematicviscosity(v)Number(Pr)
Specificheat (C)thermalconductivity(k)20 1.165 16*106 0.701 1005
0.02675We will find coefficient of thermal expansion () from page
29 in the DataBook at30. = 3.3 103 1 = = 182. = 0.75 0.3 = 0.225 2
= 2() + 2() = 2 (0.75) + 2 (0.3) = 2.1 from page 126 in the Data
Book: =1824 =4 =4 0.2252.1=0.92.1= 0.42 This question is constant
wall temperature.)Constant wall temperature( )Grashof Number(
.Grashof Number = =32 ( 134 ) =(9.81)(3.3 103 )(0.42)3(45 15)(16
106 )2 =0.0712.56 1010 = 281 068 455.9 = (281 068 455.9) 0.701 =
197 028 987.6 < 109It is Laminar.We will choose a suitable rule
to find Nusselt Number (Nu): from page 137 inthe Data Book:3.1
Horizontal cylinders long cylinders: 183. = ( )107 < < 1012 =
0.125, = 0.333 = 0.125((281 068 455.9) (0.707))1830.333 =
0.125(587.2) = 72.27 . (1)From page 111 in the Data Book: = = =
0.420.02675 = 15.7 . . (2)Equation (1)= Equation (2)72.27 = 15.7
=72.2715.7 = 4.6 /2 . Convection heat transfer coefficient= = 4.6/2
. () = ( ) = 4.6(1.319 )(45 15)= 182.02 the heat loss from the duct
per meter length== 182.02 () 184. Example: A vertical slot of 20mm
thickness is formed by two 2m*2m squareplates. If the temperatures
of the plates are 115 and 25 respectively,calculate the following:
(1) the effective thermal conductivity. (2) The rate ofthe heat
flow through the slot.Given: Length=L=20mm=0.02 m, width=2 m and
height = 2 m , temperature ofplate 1 =1= 115 . temperature of plate
2 =2= 25184Solution: = =1 + 22=115 + 252=1402= 70From page 33 in
the Data Book:We will find air properties at70.temperature
density()Kinematicviscosity(v)Number(Pr)
thermaldiffusivity()thermalconductivity(k) 185. 70 1.029 20.02*106
0.694 28.55*106 0.02966We will find coefficient of thermal
expansion () from page 29 in the DataBook at70. = 2.91 103 1185 =3
( 137 ) =(9.81)(2.91 103 )(0.02)3 (115 25)(28.556 106 ) (20.02 106
) =2 1055.71 1010 = 34 983.92We will choose a suitable rule to find
Nusselt Number (Nu): from page 138 inthe Data Book: = 0.22 (0.2 +
0.23 )()0.25 = 0.22 (0.6940.2 + 0.6940.23 34 983.92)(20.020.25) =
0.22(10.46)(3.16) = 7.27 . . (1)From page 111 in the Data Book: = =
= 0.020.02966 = 0.67 . . (2)Equation (1)= Equation (2)7.27 = 0.67
=7.270.67 186. = 10.86 /2 . Convection heat transfer coefficient= =
10.86 W/2 . () = (1 2 ) = 10.86(2 2)(115 25) = 10.86(4)(90) = 3
909.6 ()Example: The horizontal air space over a solar collector
has a spacing of2.5cm. The lower plate is maintained at 70 while
the upper plate is at 30 .Calculate the free heat convection across
the space for air at 1 atm.Given: Length=L=2.5 cm=0.025 m,
temperature of plate 1 =1= 70 .temperature of plate 2 =2=
30186Solution: 187. 187 = =1 + 22=30 + 702=1002= 50From page 33 in
the Data Book:We will find air properties at50.temperature
density()Kinematicviscosity(v)Number(Pr)
thermaldiffusivity()thermalconductivity(k)50 17.95*106 0.698
25.722*106 0.02826We will find coefficient of thermal expansion ()
from page 29 in the DataBook at70. = 3.10 103 1 =3 ( 137 )
=(9.81)(3.10 103 )(0.025)3 (70 30)(25.722 106 ) (17.952 106 ) =1.9
1054.617 1010 = 41 151.38We will choose a suitable rule to find
Nusselt Number (Nu): from page 138 inthe Data Book: = 0.069 0.333
Pr0.074 = 0.069 (41 151.38)0.333 (0.698)0.074 = 0.22(34.402)(0.973)
= 2.311 . . (1)From page 111 in the Data Book: 188. = =188 =
0.0250.02826 = 0.88 . . (2)Equation (1)= Equation (2)2.311 = 0.88
=2.3110 .88 = 2.612 /2 . Convection heat transfer coefficient= =
10.86 W/2 . () = (1 2 ) = 2.612 (70 30)= 2.612 (40) = 104.494 ()
189. Chapter 8Radiation Heat
Transfer1898.1RADIATIONDefinition:Radiation is the energy transfer
across a system boundary due to a T, by themechanism of photon
emission or electromagnetic wave emission.Because the mechanism of
transmission is photon emission, unlike conductionand convection,
there need be no intermediate matter to enable transmission.The
significance of this is that radiation will be the only mechanism
for heattransfer whenever a vacuum is present.4.2Electromagnetic
Phenomena. 190. We are well acquainted with a wide range of
electromagnetic phenomena inmodern life. These phenomena are
sometimes thought of as wave phenomenaand are, consequently, often
described in terms of electromagnetic wave length,. Examples are
given in terms of the wave distribution shown below:One aspect of
electromagnetic radiation is that the related topics are
moreclosely associated with optics and electronics than with those
normally foundin mechanical engineering courses. Nevertheless,
these are widely encounteredtopics and the student is familiar with
them through every day life experiences.From a viewpoint of
previously studied topics students, particularly those witha
background in mechanical or chemical engineering, will
