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Department of Aeronautical engineering
School of Mechanical engineering
Vel Tech Dr RR & SR Technical University
Course Material
U6AEA !" #eat Transfer
1
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U6AEA ! #EAT TRA$S%ER U$ T #eat Con'uction (
Basic Modes of Heat Transfer – One dimensional steady state heat conduction: Composite Medium – Criticalthickness – Effect of variation of thermal Conductivity – Extended Surfaces – nsteady state! Heat Conduction:"umped System #nalysis – Heat Transfer in Semi infinite and infinite solids – se of Transient – Temperaturecharts – #pplication of numerical techni$ues!
U$ T Convective #eat Transfer (
%ntroduction – &ree convection in atmosphere free convection on a vertical flat plate – Empirical relation in freeconvection – &orced convection – "aminar and tur'ulent convective heat transfer analysis in flo(s 'et(een
parallel plates) over a flat plate and in a circular pipe! Empirical relations) application of numerical techni$ues in pro'lem solvin*!
U$ T Ra'iative #eat Transfer (
%ntroduction to +hysical mechanism – ,adiation properties – ,adiation shape factors – Heat exchan*e 'et(eennon – 'lack 'odies – ,adiation shields!
U$ T V #eat E)changers (
Classification – Temperature -istri'ution – Overall heat transfer coefficient) Heat Exchan*e #nalysis – "MT-Method and E./T Method) pro'lems usin* "MT- and E./T methds!
U$ T V #eat Transfer *ro+lems n Aerospace Engineering (
Hi*h.Speed flo( Heat Transfer) Heat Transfer pro'lems in *as tur'ine com'ustion cham'ers – ,ocket thrustcham'ers – #erodynamic heatin* – #'lative heat transfer) Heat transfer pro'lems in no00les!
TE,T -../S
1! 2unus #! Cen*el!) 3Heat Transfer – # practical approach4) Second Edition) Tata Mc5ra(.Hill) 6776!
6! %ncropera! &!+!and -e(itt!-!+! 3%ntroduction to Heat Transfer4) 8ohn 9iley and Sons – 6776!
RE%ERE$CE -../S
1! "ienhard) 8!H!) 3# Heat Transfer Text Book4) +rentice Hall %nc!) 1 ;1!
6! Holman) 8!+! 3Heat Transfer4) Mc5ra(.Hill Book Co!) %nc!) /e( 2ork) < th Edn!) 1 1!
=! Sachdeva) S!C!) 3&undamentals of En*ineerin* Heat > Mass Transfer4) 9iley Eastern "td!) /e( -elhi)1 ;1!
?!Mathur) M! and Sharma) ,!+! 35as Tur'ine and 8et and ,ocket +ropulsion4) Standard +u'lishers) /e(-elhi 1 ;;!
U$ T 0
Basic Modes of Heat Transfer
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One dimensional steady state heat conduction:
Composite MediumCritical thickness
Effect of variation of thermal Conductivity
Extended Surfaces
nsteady state! Heat Conduction: "umped System #nalysis
Heat Transfer in Semi infinite and infinite solids
se of Transient Temperature charts
#pplication of numerical techni$ues!
C.$DUCT .$
*ART 0 A
3
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12 Define #eat Transfer2
Heat transfer can 'e defined as the transmission of ener*y from one re*ion to another re*ion due to temperature difference!
2 3hat are the mo'es of #eat Transfer4
ConductionConvection,adiation
!2 Define Con'uction2
Heat conduction is a mechanism of heat transfer from a re*ion of hi*h temperature toa re*ion of lo( temperature (ithin a medium @solid) li$uid or *asesA or 'et(een differentmedium in direct physical contact!
%n condition ener*y exchan*e takes place 'y the kinematic motion or direct impact of molecules! +ure conduction is found only in solids!
52 Define Convection2
Convection is a process of heat transfer that (ill occur 'et(een a solid surface and afluid medium (hen they are at different temperatures!
Convection is possi'le only in the presence of fluid medium!
2 Define Ra'iation2
The heat transfer from one 'ody to another (ithout any transmittin* medium iskno(n as radiation! %t is an electroma*netic (ave phenomenon!62 State %ourier7s 8a9 of con'uction2
The rate of heat conduction is proportional to the area measured – normal to thedirection of heat flo( and to the temperature *radient in that direction!
α . # dT dx
dT . C#
dxQ =
(here # – are in m 6
dT dx
. Temperature *radient in Dm
– Thermal conductivity 9Dm !
:2 Define Thermal Con'uctivity2
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Thermal conductivity is defined as the a'ility of a su'stance to conduct heat!
;2 3rite 'o9n the e
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1 2 3rite 'o9n the e stea'y state con'uction e t9o 'imensional con'uction e
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1;2 3rite 'o9n the general e
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%f the temperature of a 'ody does not vary (ith time) it is said to 'e in a steady stateand that type of conduction is kno(n as steady state heat conduction!
2 3hat is meant +y Transient heat con'uction or unstea'y state con'uction4
%f the temperature of a 'ody varies (ith time) it is said to 'e in a transient state andthat type of conduction is kno(n as transient heat conduction or unsteady state conduction!
62 3hat is *erio'ic heat flo94
%n periodic heat flo() the temperature varies on a re*ular 'asis!
Example:
1! Cylinder of an %C en*ine!6! Surface of earth durin* a period of 6? hours!
:2 3hat is non perio'ic heat flo94
%n non periodic heat flo() the temperature at any point (ithin the system varies nonlinearly (ith time!
Examples :
1! Heatin* of an in*ot in a furnace!6! Coolin* of 'ars!
;2 3hat is meant +y $e9tonian heating or cooling process4
The process in (hich the internal resistance is assumed as ne*li*i'le in comparison(ith its surface resistance is kno(n as /e(tonian heatin* or coolin* process!
(2 3hat is meant +y 8umpe' heat analysis4
%n a /e(tonian heatin* or coolin* process the temperature throu*hout the solid isconsidered to 'e uniform at a *iven time! Such an analysis is called "umped heat capacityanalysis!
!=2 3hat is meant +y Semi"infinite soli's4
%n a semi infinite solid) at any instant of time) there is al(ays a point (here the effectof heatin* or coolin* at one of its 'oundaries is not felt at all! #t this point the temperatureremains unchan*ed! %n semi infinite solids) the 'iot num'er value is ∞!
!12 3hat is meant +y infinite soli'4
# solid (hich extends itself infinitely in all directions of space is kno(n as infinite solid!
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%n semi infinite solids) the 'iot num'er value is in 'et(een 7!1 and 177!7!1 B i 177!
! 2 Define -iot num+er2
%t is defined as the ratio of internal conductive resistance to the surface convectiveresistance!
B i Internal conductive resistanceSurface convective resistance
B i LhL K
!
!!2 3hat is the significance of -iot num+er4
Biot num'er is used to find "umped heat analysis) semi infinite solids and infinite solids
%f Bi 7!1 " → "umped heat analysisB i ∞ → Semi infinite solids7!1 B i 177 → %nfinite solids!
!52 E)plain the significance of %ourier num+er2
%t is defined as the ratio of characteristic 'ody dimension to temperature (ave penetration depth in time!
&ourier /um'er Characteristic body dimension
Temperature wave penetrationdepth in time
%t si*nifies the de*ree of penetration of heatin* or coolin* effect of a solid!
! 2 3hat are the factors affecting the thermal con'uctivity4
1! Moisture6! -ensity of material=! +ressure
?! TemperatureI! Structure of material
!62 E)plain the significance of thermal 'iffusivity2
The physical si*nificance of thermal diffusivity is that it tells us ho( fast heat is propa*ated or it diffuses throu*h a material durin* chan*es of temperature (ith time!
!:2 3hat are #eisler charts4
%n Heisler chart) the solutions for temperature distri'utions and heat flo(s in plane(alls) lon* cylinders and spheres (ith finite internal and surface resistance are presented!
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Heisler charts are nothin* 'ut a analytical solutions in the form of *raphs!*ART 0 -
12 A 9all of =26m thic ness having thermal con'uctivity of 12 9BM 2 The 9all is to +einsulate' 9ith a material having an average thermal con'uctivity of =2! 3Bm/2 nneran' outer surface temperatures are 1=== ° C an' 1= °C2 #eat transfer rate is 15== 3Bmcalculate the thic ness of insulation2
iven Data
Thickness of (all " 1 7!< mThermal conductivity of (all 1 1!6 9Dm !Thermal conductivity of insulation 6 7!= 9Dm ! %nner surface TemperatureT1 1777 °C J 6K= 16K= Outer surface TemperatureT= 17 °C J 6K= 6;= Heat transfer per unit area D# 1?77 9Dm 6!
Solution
"et the thickness of insulation 'e " 69e kno(
overall T Q R∆= L&rom e$uation @1=A @orA LHMT -ata 'ook pa*e /o! =?9here
∆T T a – T ' @orA T1 – T ==1 6
1 6 =
1 1
a b
L L L R
h A K A K A K A h A= + + + +
1 =
=1 6
1 6 =
L M1 1
a b
T T Q
L L Lh A K A K A K A h A
−=+ + + +
Heat transfer coefficient h a) h ' and thickness " = are not *iven! So ne*lect that terms!
[ ]1 =1 6
1 6
T
T L L
K A K A
−⇒
+
[ ]1 =1 6
1 6
6
6
T
#
16K= 6;=1?77
7!<1!6 7!=
7!7
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2 The 9all of a col' room is compose' of three layer2 The outer layer is +ric !=cmthic 2 The mi''le layer is cor = cm thic > the insi'e layer is cement 1 cm thic 2 The
temperatures of the outsi'e air is °C an' on the insi'e air is " = °C2 The film co"efficient for outsi'e air an' +ric is 25 3Bm /2 %ilm co"efficient for insi'e air an'cement is 1: 3Bm /2 %in' heat flo9 rate2
Ta e/ for +ric 2 3Bm/ / for cor =2= 3Bm/ / for cement =2 ; 3Bm/
iven Data
Thickness of 'rick " = =7 cm 7!= mThickness of cork " 6 67 cm 7!6 mThickness of cement " 1 1I cm 7!1I m%nside air temperature T a .67 °C J 6K= 6I= Outside air temperature T ' 6I °C J 6K= 6 ; &ilm co.efficient for inner side h a 1K 9Dm 6 &ilm co.efficient for outside h ' II!? 9Dm 6
'rick = 6!I 9Dm cork 6 7!7I 9Dm ! cement 1 7!7; 9Dm !
Solution
Heat flo( throu*h composite (all is *iven 'y
overall T Q R
∆= L&rom e$uation @1=A @orA LHMT -ata 'ook pa*e /o! =?
9here∆T T a – T '
=1 6
1 6 =
1 1
a b
L L L R
h A K A K A K A h A= + + + +
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[ ]
=1 6
1 6 =
=1 6
1 6 =
6
L M1 1
D1 1
6I= 6:;D
1 7!1I 7!6 7!= 11K 7!6; 7!7I 6!I II!?
D :!I D
a b
a b
a b
a b
T T Q
L L Lh A K A K A K A h A
T T Q A
L L Lh K K K h
Q A
Q A W m
−⇒ =
+ + + +
−⇒ =
+ + + +
−⇒ =
+ + + +
= −
The ne*ative si*n indicates that the heat flo(s from the outside into the cold room!
!2 A 9all is constructe' of several layers2 The first layer consists of masonry +ric =cm2 thic of thermal con'uctivity =266 3Bm/> the secon' layer consists of ! cm thic mortar of thermal con'uctivity =26 3Bm/> the thir' layer consists of ; cm thic limestone of thermal con'uctivity =2 ; 3Bm/ an' the outer layer consists of 12 cm thic plaster of thermal con'uctivity =26 3Bm/2 The heat transfer coefficient on the interioran' e)terior of the 9all are 26 3Bm / an' 11 3Bm / respectively2 nterior roomtemperature is °C an' outsi'e air temperature is " °C2
Calculate
a@ .verall heat transfer coefficient+@ .verall thermal resistancec@ The rate of heat transfer'@ The temperature at the Function +et9een the mortar an' the limestone2
iven Data
Thickness of masonry " 1 67cm 7!67 mThermal conductivity 1 7!
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overall T Q
R
∆= L&rom e$uation @1=A @orA LHMT -ata 'ook pa*e /o! =?
9here∆T T a – T '
=1 6 ?
1 6 = ?
=1 6 ?
1 6 = ?
1 1
1 1
6:I 6
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1 6 = = ? ? I I1 6
1 6 = ?
1
1a
1
1
1
1 6
1
6:I.T 1 E ,
1D
6:ID
1D
6:I=?!I<
1DI!<
6;;!;
a b
a b
a
a
a a
a
T T T T T T T T T T T T Q
R R R R R R
T T Q R
h A h A
T Q A
h
T
T K
T T Q
R
− − − − −−⇒ = = = = = =
−⇒ = =
−⇒ =
−⇒ =
⇒ =−
⇒ =
Q
6 1
11 1
1
6;;!; ,
T LQ
L k A K A
−= =
Q
6
1
1
6
6
6 =
6
= 66
6 6
6
=
6
6
=
=
6;;!;D
6;;!;=?!I<
7!677!T Q A
L K
T
T
T Q
R
T LQ
L K A K A
T Q A
L K
T
T
−⇒ =
−⇒ =
⇒ =−
⇒
−= =
−⇒ =
−⇒ =
⇒ =
Q
Temperature 'et(een Mortar and limestone @T = is 6K
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52 A steam to li
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=
?71!I; 17
Q
W
−−
⇒ = ×
×Heat transfer Q = ' 2#"2 )+L.ve si*n indicates that the heat flo(s from) outside to inside(e kno(Heat transfer # @T a – T ' A L&rom e$uation /o! @1?A
a
=
6
E#@T A
6I!6 17
6I!6 @ ?7A
bT
K
⇒ −− ×= × −
&verall heat transfer co' efficient ( = 2# %/m
Temperature drop @T = – T ?A across the scale is *iven 'y[ ]= ?
=
TE T
6I!6 177!771I
=K!;
s ale
T Q T
R
T
T !
∆= ∆ −
∆× =
⇒ ∆ = °
2 A surface 9all is ma'e up of ! layers one of fire +ric > one of insulating +ric an' oneof re' +ric 2 The inner an' outer surface temperatures are (== °C an' != °Crespectively2 The respective co"efficient of thermal con'uctivity of the layers are 12 >=215 an' =2( 3Bm/ an' the thic ness of =cm> ; cm an' 11 cm2 Assuming close +on'ingof the layers at the interfaces2 %in' the heat loss per s
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=1 6
1 6 =
1 ?
=1 6
1 6 =
1 1
T E1 1
a b
a b
L L L R
h A K A K A K A h A
T L L L
h A K A K A K A h A
= + + + +
−⇒+ + + +
LConvective heat transfer co.efficient h a) h ' are not *iven!So ne*lect that terms
1 ?
=1 6
1 6 =
1 ?
=1 6
1 6 =
6
T E
TD
11K= =7=
7!6 7!7; 7!111!6 7!1? 7!:
D 1711!6I?< D
T L L L
K A K A K A
T Q A
L L L K K K
Q A W m
−⇒
+ +
−=+ +
−=+ +
=
?ii@ nterface temperatures ?T an' T ! @
9e kno( that) interface temperatures relation6 = = ?1 ? 1 6
1 6 =
1 6
1
11
1
1 6
1
1
!!!!!!@ A
@ A
,
T T T T T T T T Q A
R R R R
T T A Q R
L K A
T T Q
L K A
− −− −⇒ = = = =
−⇒ =
=
−⇒ =
where
1 6
1
1
6
6
6 =
6
T D# E
11K=1711!6I?< 7!6
1!6
177?!?IK
T L K
T
T K
T T Q
R
−
−=
=
−=
Similarly,
(here
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66
6
6 =
6
6
6 =
6
6
=
=
,
T D# E
177?!?IK1711!6I?<
7!7;7!1?
?6
L K A
T T Q L K A
T L K
T
T K
=
−⇒ =
−⇒
−=
=
62 A furnace 9all ma'e up of :2 cm of fire plate an' =26 cm of mil' steel plate2 nsi'esurface e)pose' to hot gas at 6 = °C an' outsi'e air temperature : °C2 The convectiveheat transfer co"efficient for inner si'e is 6= 3Bm /2 The convective heat transfer co"efficient for outer si'e is ;3Bm /2 Calculate the heat lost per s
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=1 6
1 6 =
a
=1 6
1 6 =
1 1
T E1 1
a b
b
a b
L L L R
h A K A K A K A h A
T L L L
h A K A K A K A h A
= + + + +
−⇒+ + + +
LThe term " = is not *iven so ne*lect that term
a
=1 6
1 6 =
a
1 6
1 6
6
T E
1 1
T E 1 1
:6= =77D
1 7!7KI 7!77
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:2 A mil' steel tan of 9all thic ness 1=mm contains 9ater at (= °C2 Calculate the rate of heat loss per m of tan surface area 9hen the atmospheric temperature is 1 °C2 The
thermal con'uctivity of mil' steel is = 3Bm/ an' the heat transfer co"efficient forinsi'e an' outsi'e the tan is ;== an' 11 3Bm / respectively2 Calculate also thetemperature of the outsi'e surface of the tan 2
iven Data
Thickness of (all " 1 17mm 7!71 m%nside temperature of (ater T a 7° J 6K= =
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1 61 6
1
1 a
1
!!!!!!@ A
1@ A ,
=
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= 7!K< 9Dm Ta 6
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6 66
6 6
66a
6 6
6
!!!!!@6A
7!1, 16!I 7!I
7!71< CD9
a b
a b
a a
a
R R R
R R
L K A
R
×=+
= =× ×=
66'
6 6
6
7!1,
1;!I 7!I
7!717; CD9
b b
b
L K A
R
= =× ×
=
6
6
7!71< 7!717;@6A
7!71< 7!717;
7!77
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1 6
1
6
6
6 =
6
=
=
@BA
6:7!IK 16K!
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6
1 6
6
# E 1m7!1 7!7?7!K 7!?;
66!
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T Q
R∆=
9here∆T T 1 – T ?
=1 6
1 6 =
1 ?
=1 6
1 6 =
1 1
T E
1 1
a b
a b
L L L R
h A K A K A K A h A
T L L L
h A K A K A K A h A
= + + + +
−⇒
+ + + +
/e*lectin* unkno(n terms @h a and h ' A
1 ?
=1 6
1 6 =
6
6
T E
:6= ?6= # E 1m
7!76 7!1I7 7!61
I77 E
7!17;<
?
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112 A thic 9alle' tu+e of stainless steel J/ ::2; KBhr m °CL mm D an' = mm.D is covere' 9ith a mm layer of as+estos J/ =2;; KBhr m °CL2 f the insi'e 9all
temperature of the pipe is maintaine' at = °C an' the outsi'e of the insulator at 5 °C2Calculate the heat loss per meter length of the pipe2
iven Data
%nner diameter of steel d 1 6I mm%nner radius r 1 16!I mm ⇒ 7!716I mOuter diameter - 6 I7 mmOuter radius r 6 6I mm ⇒ 7!76I m,adius r = r 6 J 6I mm I7 mm ⇒ 7!7I m
Thermal conductivity of stainless steel 1 KK!;I k8Dhr m°C KK!;I=
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a
=6
1 6
1 6
a
=6
1 6
1 6
T E
16
T D" E
16
b
b
T Q
r r In In
r r L K K
T
r r In In
r r K K
π
π
−⇒
+
−⇒
+
II7 . ?ID E
7!76I 7!7I
1 7!716I 7!76I6 61!
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9here∆T T a – T '
=6 ?
a 1 1 1 6 6 = = ?
a
=6 ?
a 1 1 1 6 6 = = ?
1 1 1 1 1 1
6 h
T E
1 1 1 1 1 1
6 h
b
b
b
r r r R In In In
L r K r K r K r h r
T Q
r r r In In In
L r K r K r K r h r
π
π
= + + + + −
⇒ + + + +
LThe terms = and r ? are not *iven) so ne*lect that terms
a
=6
a 1 1 1 6 6 =
T E
1 1 1 1 1
6 h
I;: . =7= E
1 1 1 7!7=;1 1 7!7
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Heat loss throu*h hollo( sphere is *iven 'yoverall T
Q R∆
= L&rom e$uation /o!@1 A or HMTdata 'ook +a*e /o!=? > =I
9here∆T T a – T '
6 61 1 6 6 6 =a 1 =
a
6 61 1 6 6 6 =a 1 =
1 1 1 1 1 1 1 1 1
? h
T E
1 1 1 1 1 1 1 1 1
? h
b
b
b
R K r r K r r r h r
T Q
K r r K r r r h r
π
π
= + − + − + −
⇒ + − + − +
ha) h ' not *iven so ne*lect that terms!
1 1 6 6 6 =
1 1 1 1 1 1 1?
KK= .=6=
1 1 1 1 1 1 1?
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= ?6
=1 6
a 1 1 6 = ?
1 1 1
6 h b
r r r In In In
r r r R
L r K K K h r π
= + + +
1 ?
= ?6
=1 6
a 1 1 6 = ?
1 ?
T E
1 1 16 h
T E
16
b
T Q
r r r In In In
r r r
L r K K K h r
T r
In
L
π
π
−⇒
+ + + +
−⇒
Heat transfer coefficients h ,h are not .iven"a b So ne.lect that terms"
= ?6
=1 6
1 6 =
r r In In
r r r
K K K
+ +
I6= . 6:= E
7!7?II 7!1=II 7!1KII1 7!7?7 7!7?II 7!1=II
6 ?K 7!I 7!6I
D ??;!; 9Dm
Q In In In
L
Q L
π
⇒ + +
⇒ =
Heat transfer D" ??;!; 9Dm!
152 A hollo9 sphere has insi'e surface temperature of !== °C an' then outsi'e surfacetemperature of != °C2 f / 1; 3Bm/2 Calculate ?i@ heat lost +y con'uction for insi'e'iameter of cm an' outsi'e 'iameter of 1 cm ?ii@ heat lost +y con'uction> if e
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Solution
?i@ #eat lost ?G@
Heat flo( overall T
Q R
∆= L&rom HMT data 'ook +a*e
/o!=? > =I9here
∆T T a – T ' @orA T1 – T 6
6 61 1 6a 1 6
1 6
6 61 1 6a 1 6
1 1 1 1 1 1
? h
T E
1 1 1 1 1 1 ? h
b
b
R K r r r h r
T Q
K r r r h r
π
π
= + − + −
⇒
+ − + LThe terms h a) h ' not *iven so ne*lect that terms !
1 6
1 1 6
1 1 1 1?
IK= .=7= E
1 1 1 1? 1; 7!76I 7!7KI
E 66:7!66 9
T T Q
K r r π
π
−⇒ =
−
⇒ −
⇒
?ii@ #eat loss ? f the area is e
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( )
1 61
1 61
6 61 1 6
1
6 6
1
" ,E
C#
6
IK= . =7=
7!7I1; 6 @7!76I 7!7KI A
=;1K!7=
T T Q
L KA
T T Q
L
K r r
Q
Q W
π
π
− =
−=
+
=
× +=
Q for plain wall
Derive an e)pression of Critical Ra'ius of nsulation %or A Cylin'er2
Consider a cylinder havin* thermal conductivity ! "et r 1 and r 7 inner and outer radiiof insulation!
Heat transfer 71
r
6
iT T Q
Inr
KLπ
∞−=
L&rom e$uation /o!@=A
Considerin* h 'e the outside heat transfer co.efficient!
π π
π π
∞
∞
−∴ +
=−
⇒ = +
i
+
)
+
+ +
i
+
)
+
T TQ =r
Inr )
2 5- A h
Here A 2 r -
T TQ
r In
r )
2 5- 2 r -h
To find the critical radius of insulation) differentiate (ith respect to r 7 and e$uate itto 0ero!
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i 2+ +
+ +
) +
i
2+ +
+ c
) )+ T T 3
2 5-r 2 h-r dQdr r ) )
In2 5- r 2 h-r
since T T 3 +
) ) +
2 5-r 2 h-r
5 r r
h
π π
π π
π π
∞
∞
− − − ⇒ = +
− ≠
⇒ − =
⇒ = =
1 2 A 9ire of 6 mm 'iameter 9ith mm thic insulation ?/ =211 3Bm/@2 f theconvective heat transfer co"efficient +et9een the insulating surface an' air is 3Bm 8>fin' the critical thic ness of insulation2 An' also fin' the percentage of change in theheat transfer rate if the critical ra'ius is use'2
iven Data
d1 < mmr 1 = mm 7!77= mr 6 r 1 J 6 = J 6 I mm 7!77I m
7!11 9Dm h ' 6I 9Dm 6
Solution
1! Critical radius c5
r h
= L&rom e$uation /o!@61A
c
c
+"))r !"! )+ m
2#
r !"! )+ m
−
−
= = ×
= ×
Critical thickness r c – r 1
!"! )+ +"++)"! )+ m
−−
= × −= ×
'cCritical thic4ness t = )"! )+ or3 )"! mm×
6! Heat transfer throu*h an insulated (ire is *iven 'y
a b)
2
)
) b 2
T TQ
r In
r ) )2 - 5 h r π
−= +
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[ ]a b
a b
6rom H7T data boo4 8a.e 9o" #
2 - T T 3 =+"++#
In)+"++
+")) 2# +"++#
2 - T T 3Q) =
)2"$!
π
π
− + ×
−
Heat flo( throu*h an insulated (ire (hen critical radius is used is *iven 'y
[ ]a b2 2 cc
)
) b c
T TQ r r
r In
r ) )2 - 5 h r π
−= → +
a b
a b2
2 - T T 3 =
!"! )+In +"++ )
+")) 2# !"! )+2 - T T 3
Q =)2"#:2
π
π
−
−
− × + × ×
−
∴ +ercenta*e of increase in heat flo( 'y usin*
Critical radius 2 ))
Q Q)++
Q− ×
) ) )++)2"#: )2"$!
))2"$!
+"##;
− ×=
=
nternal #eat eneration 0 %ormulae use'%or plane 9all
12 Surface temperature w
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(hereT∞ . &luid temperature)
$ . Heat *eneration) 9Dm=
" – Thickness) mh . Heat transfer co.efficient) 9Dm 6
– Thermal conductivity) 9Dm !%or Cylin'er
12 #eat generationQ
<>
=
2 Ma)imum temperature2
ma w
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( )'$ 2
2
1esistivity -en.th1esistance of wire 1 =
Area
:+ )+ )+ ) =)+!
1 +"+@@
π −
−
×
× × ××
= Ω
9e kno( that %6,
@677A6 × @7!7 AQ = @$+ %
Heat *enerated 2Q @$+
< > d -!π = = ×
( )2
$
@$+<
)+ )!
< #$+ )+ % / m
π −=
× ×
= ×
Su'stitutin* $ value in E$uation @#A
2
ma c
c
#$+ )+$ )"# )+ 3T T *
! )@ T @@"# 5
−× × ×= = +×
=
1:2 A sphere of 1== mm 'iameter> having thermal con'uctivity of =21; 3Bm/2 The outersurface temperature is ; °C an' = 3Bm of energy is release' 'ue to heat source2Calculate12 #eat generate'
2 Temperature at the centre of the sphere2
iven
-iameter of sphere d 177 mmr I7 mm 7!7I7 m
Thermal conductivity 7!1; 9Dm Surface temperature T ( ; °C J 6K= 6;1 Ener*y released 6I7 9Dm 6
Solution
Heat *eneratedQ
<>
=
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π
π π
π π
⇒ =
⇒ =
⇒ =
⇒ =
× × ×⇒ = ×
⇒
2
2
2
2
Q / A< / A Here Q/A = 2#+ %/m
>
Q / A< / A Here Q/A = 2#+ %/m>
2#+< / A
! / r
< 2#+! r ! / r
2#+ ! +"+#+3<
! / +"#+3
< = )#,+++ %/m
Temperature at the centre of the sphere2
c w
2
c
thetemperature of the ro' is measure' at t9o points = cm apart are foun' to +e 1 = °Can' 1== °C2 The convective heat transfer co"efficient +et9een the ro' an' thesurroun'ing air is != 3Bm /2 Calculate the thermal con'uctivity of the ro' material2
iven Data
#tmospheric Temperature T ∞ 6I °C J 6K= 6 ; -istance x 67 cm 7!67 mBase temperature T ' 1I7 °C J 6K= ?6= %ntermediate temperature T 177 °C J 6K= =K= Heat transfer co.efficient h =7 9Dm 6 !
Solution
Since the rod is lon*) it is treated as lon* fin! So) temperature distri'ution
b
T Te
T T−∞
∞
− =−mx
L&rom HMT data 'ook @C+ A
+a*e /o!?1
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m +"2+3
'm +"2+3
')
: ' 2@* e
!2 ' 2@*
+"$ = e In +"$3= 'm +"2+3 ' +"#) = 'm +"2+3
m = 2"## m
− ×
×
⇒ =
⇒⇒ ×⇒ ×
9e kno( that)
06rom H7T data boo4
C853 8a.e 9o"!)
h82"## = """"""""""""" A3
5A
hP m
KA=
h – heat transfer co.efficient =7 9Dm 6 + – +erimeter πd π × 7!7I7
+")#:P m=2 A Area d
!π − =
2 = +"+#+3!π
2 A )"@$ )+ m−= ×
'
+ +")#:A3 2"##
5 )"@$ )++ +")#:
$"#+ =5 )"@$ )+
5 = $@": %/m5
−×
⇒ = × ××
⇒ × ×
1(2 An aluminium alloy fin of : mm thic an' = mm long protru'es from a 9all> 9hichis maintaine' at 1 = °C2 The am+ient air temperature is °C2 The heat transfercoefficient an' con'uctivity of the fin material are 15= 3Bm / an' 3Bm/ respectively2 Determine
12 Temperature at the en' of the fin22 Temperature at the mi''le of the fin2
!2 Total heat 'issipate' +y the fin2
iven
Thickness t Kmm 7!77K m"en*th " I7 mm 7!7I7 m
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Base temperature T ' 167 °C J 6K= = = #m'ient temperature T ∞ 66 ° J 6K= 6 I Heat transfer co.efficient h 1?7 9Dm
6
Thermal conductivity II 9Dm !
Solution
"en*th of the fin is I7 mm! So) this is short fin type pro'lem! #ssume end isinsulated!
9e kno(Temperature distri'ution LShort fin) end insulated
b
T T cos h m 0- '""""""" A3
T T cos h m-3
∞
∞
− =−
L&rom HMT data 'ook +a*e /o!?1
?i@ Temperature at the en' of the fin> *ut ) 8
b
b
T ' T cos h m 0-'-A3
T T cos h m-3
T ' T ) """ )3
T T cos h m-3
where
h8 m =
5A 8 = 8erimeter = 2 - Appro 3 = 2 +"+#+
8 = +") m
∞
∞
∞
∞
⇒ =−
⇒ =−
××
# – #rea "en*th × thickness 7!7I7 × 7!77K ! 2 A "# )+ m−= ×
h8 m =
5A⇒
!)!+ +")
## "# )+
m 2$"@$
−×= × ×
=
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b
b
-
T ' T ))3
T T cos h 2$"@ +"+#+3
T ' T ) T T 2"+#
T ' 2@# )
@ ' 2@# 2"+# T ' 2@# = !:"*
T = !2"* 5
Temperature at the end of the fin T !2"* 5
∞
∞
∞
∞
=
⇒ =− ×
⇒ =−
⇒ =
⇒
⇒
=
?ii@ Temperature of the mi''le of the fin>
+ut x "D6 in E$uation @#A
[ ]
b
b
T ' T cos hm 0-'-/2A3
T T cos h m-3
+"+#+cos h 2$"@ +"+#+ '
T ' T 2 T T cos h 2$"@ +"+#+3
T' 2@# )"2 !
@ ' 2@# 2"+!@T ' 2@#
+"$+2#@ '2@#
T #!"+! 5
∞
∞
∞
∞
⇒ =−
⇒ =− ×
⇒ =
⇒ ==
Temperature at the middle of the fin
- / 2T #!"+! 5= =
?iii@ Total heat 'issipate'
L&rom HMT data 'ook +a*e /o!?1
)/2b
'! )/ 2
Q = h85A3 T T 3 tan h m-3
0)!+ +") ## "# )+ @ 2@#3 tan h 2$"@ +"+#+3
Q = !!"! %
∞⇒ −⇒ × × × × × −
× ×
=2 Ten thin +rass fins ?/ 1== 3Bm/@> =2: mm thic are place' a)ially on a 1 m longan' 6= mm 'iameter engine cylin'er 9hich is surroun'e' +y ! °C2 The fins aree)ten'e' 12 cm from the cylin'er surface an' the heat transfer co"efficient +et9een
cylin'er an' atmospheric air is 1 3Bm /2 Calculate the rate of heat transfer an' thetemperature at the en' of fins 9hen the cylin'er surface is at 16= °C2
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iven
/um'er of fins 17Thermal conductivity 177 9Dm Thickness of the fin t 7!KI mm 7!KI × 17 .= m"en*th of en*ine cylinder 1m-iameter of the cylinder d
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Heat transfer from unfinned surface due to convection is
2
b
2
Q h A T
= h d- ' )+ t -3 T T 3
0 Area of unfinned surface = Area of cylinder ' Area offin
= )# 0 +"+$+ ) 0)+ +":# )+ )"# )+ 0!
π
π
∞
− −
= ∆× × × −
× × × − × × × ×
Q
' ++
2Q :#"* % """""""""" C3=
So) Total heat transfer 1 J 6 I;1 J =KI!;
Total heat transfer Q @#$"* %=
9e kno( that)
Temperature distri'ution Lshort fin) end insulated
b
T T cos h m 0-'T T cos h m-3
∞
∞
− =−L&rom HMT data 'ook +a*e /o!?1
Temperature at the end of fin) so put x "
12 Aluminium fins 12 cm 9i'e an' 1= mm thic are place' on a 2 cm 'iameter tu+eto 'issipate the heat2 The tu+e surface temperature is 1:= °C am+ient temperature is
=°C2 Calculate the heat loss per fin2 Ta e h 1!= 3Bm C an' / == 3Bm C foraluminium2
iven
9ide of the fin ' 1!I cm 1!I × 17 .6 mThickness t 17 mm 17 × 17 .= m-iameter of the tu'e d 6!I cm 6!I × 17 .6 m
Surface temperature T ' 1K7°C J 6K= ??= #m'ient temperature T ∞ 67 °C J 6K= 6 = Heat transfer co.efficient h 1=7 9Dm 6 °CThermal conductivity 677 9Dm °C
Solution
#ssume fin end is insulated) so this is short fin end insulated type pro'lem!Heat transfer Lshort fin) end insulated
@h+ #A1D6 @T ' . T ∞A tan h @m"A PP!!@1A L&rom HMT data 'ook +a*e /o!?19here
# – #rea Breadth × thickness
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2
! 2
2
)"# )+ )+ )+
A )"# )+ m
8 8erimeter 2 b t3 = 20 )"# )+ 3 )+ )+ 3
8 = +"+# m
h8 m =
5A
) + +"+# =
2++ )"#
− −
−
− −
= × × ×= ×
− = +× + ×
×× '!
')
)+
m = )!": m
×
! )/ 2
'2
)3 Q = 0) + +"+# 2++ )"# )+
!! '2@ 3 tan h )!": )"# )+ 3 Q )!" %
−⇒ × × × ×
× × × ×=
2 A straight rectangular fin has a length of ! mm> thic ness of 125 mm2 The thermalcon'uctivity is 3Bm °C2 The fin is e)pose' to a convection environment at = °C an' h
== 3Bm °C2 Calculate the heat loss for a +ase temperature of 1 = °C2
iven
"en*th " =I mm 7!7=I mThickness t 1!? mm 7!771? mThermal conductivity II 9Dm °C&luid temperature T ∞ 67 °C J 6K= 6 = Base temperature T ' 1I7 °C J 6K= ?6= Heat transfer co.efficient h I77 9Dm 6 !
Solution
"en*th of the fin is =I mm) so this is short fin type pro'lem! #ssume end is insulated!
Heat transferred LShort fin) end insulated
@h+ #A1D6
@T ' . T ∞A tan h @m"A PP!@1AL&rom HMT data 'ook +a*e /o!?1
9here+ – +erimeter 6 × "en*th @#pproximatelyA
6 × 7!7=I8 = +"+: m
# – #rea "en*th × thickness 7!7=I × 7!771?
# 2 A !"@ )+ m−= ×
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h8m
5A=
'#
')
#++ +"+: =
## !"@ )+
m = )) "@ m
×× ×
Su'stitutin* h) p) ) #) T ' ) T∞) m) " values in e$uation @1A'# )/ 2)3 Q = 0#++ +"+: ## !"+ )+
!2 ' 2@ 3 tan h )) "@ +"+ #3
Q = @"* %
⇒ × × × ×× × ×
!2 A heating unit ma'e in the form of a cylin'er is 6 cm 'iameter an' 12 m long2 t isprovi'e' 9ith = longitu'inal fins ! mm thic 9hich protru'e = mm from the surfaceof the cylin'er2 The temperature at the +ase of the fin is ;= °C2 The am+ienttemperatures is °C2 The film heat transfer co"efficient from the cylin'er an' fins tothe surroun'ing air is 1= 3Bm /2 Calculate the rate of heat transfer from the finne'9all to the surroun'ing2 Ta e / (= 3Bm/2
iven
-iameter of the cylinder d < cm 7!7< m"en*th of the cylinder 1!6 m
/um'er of fins 67Thickness of fin @tA = mm 7!77= m"en*th of fin " I7 mm 7!7I7 mBase temperature T ' ;7 °C J 6K= =I= #m'ient temperature T ∞ 6I °C J 6K= 6 ; &ilm heat transfer co.efficient h 17 9Dm 6 Thermal conductivity 7 9Dm !
Solution
"en*th of the fins is I7 mm! #ssume end is insulated! So this is short fin end insulatedtype pro'lem!
9e kno(Heat transferred Lshort fin) end insulated
@h+ #A1D6 @T ' . T ∞A tan h @m"A PP!!@1A L&rom HMT data 'ook +a*e /o! ?1
9here+ – +erimeter 6 × "en*th of the cylinder
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6 × 1!6 8 2"! m=
# – #rea "en*th of the cylinder × thickness of fin
1!6 × 7!77= 2 A "$ )+ m−= ×
h8m
5A=
'
')
)+ 2"! =
@+ "$ )+
m = *"$ m
×× ×
)/ 2)3 Q = 0)+ 2"! @+ "$ )+ # ' 2@*3 tan h *"$ +"+#+3
Q $2")$ %
−⇒ × × × ×
× × ×=
Heat transferred per fin
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re
+
T Te """" )3
T T
ρ ρ
− × × × ∞
∞
− =− L&rom HMT data 'ook +a*e /o!?;
9e kno()
Characteristics len*th c>
- A
=
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c
ht
C -
+)2+
t@++ +"+) 2:++
T'T)3 e
T T
#2 ' 2* e
:: ' 2*')2+
In +"!*@3 = t@++ +"+) 2:++
t = )!!"*$ s
ρ ρ
− × × × ∞
∞ − × × ×
⇒ =−
⇒ =
⇒ ×× ×
⇒
Time re$uired for the cu'e to reach 6I7 °C is 1??!;< s!
2 A copper plate mm thic is heate' up to 5== °C an'
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Biot num'er cih-
B5
=
)++ +"++)*$
×= B i !2"#@ )+ +")−× <
Biot num'er value is less than 7!1! So this is lumped heat analysis type pro'lem!
&or lumped parameter system)
hAt
C >
+
T Te
T T ρ ρ
− × × × ∞
∞
− =−PPP!@1A
L&rom HMT data 'ook +a*e /o!?;9e kno()
Characteristics len*th " c >
A
c
ht
C -
+
)++t
$+ +"++) **++
T'T)3 e
T T
2 ' + e
$: ' +
t = @2"! s
ρ ρ
− × × × ∞
∞− × × ×
⇒ =−
⇒ =
⇒
Time re$uired for the plate to reach I7 °C is 6!?= s!62 A 1 cm 'iameter long +ar initially at a uniform temperature of 5= ° C is place' in a
me'ium at 6 = °C 9ith a convective co"efficient of 3Bm /2 Determine the timere
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Characteristic "en*th c1
-2
=
c
+"+$ =2
- +"+ m=9e kno()
Biot num'er cih-
B5
=22 +"+
2+×=
B i 7!7== 7!1
Biot num'er value is less than 7!1! So this is lumped heat analysis type pro'lem!&or lumped parameter system)
hAt
C >
+
T Te
T T ρ ρ
− × × × ∞
∞
− =−PPP!@1A
L&rom HMT data 'ook +a*e /o!?;9e kno()
Characteristics len*th " c >
A
c
ht
C -
+
22 t)+#+ +"+ #*+
T'T)3 e
T T
#2* ' @2 e
) ' @2
t = $+"* s
ρ ρ − × × × ∞
∞− ×
× ×
⇒ =−
⇒ =
⇒
Time re$uired for the cu'e to reach 6II °C is =
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&inal temperature T ∞ 177 °C J 6K= =K= %ntermediate temperature T 1I7 °C J 6K= ?6= Heat transfer co.efficient h 17 9Dm
6
To fin'
Time re$uired for the 'all to reach 1I7 °CL&rom HMT data 'ook +a*e /o!1
Solution
-ensity of steel is K;== k*Dm =
:* 4. / m ρ =&or sphere)
Characteristic "en*th c 1- =
c
+"+2# =
- *" )+ m−= ×9e kno()
Biot num'er cih-
B5
=
)+ *" )+
#
−× ×=
B i 6!=; × 17 .= 7!1
Biot num'er value is less than 7!1! So this is lumped heat analysis type pro'lem!&or lumped parameter system)
hAt
C >
+
T Te
T T ρ ρ
− × × × ∞
∞
− =−PPP!@1A
L&rom HMT data 'ook +a*e /o!?;9e kno()
Characteristics len*th " c > A
c
ht
C -
+
)+t
!$+ *" )+ :*
T'T)3 e
T T
!2 ' : e
:2 ' :!2 ' : )+
In t:2 ' : !$+ *" )+ :*
t = #*!+"#! s
ρ ρ
−
− × × × ∞
∞− × × × ×
−
⇒ =−
⇒ =
−⇒ = ×× × ×⇒
Time re$uired for the 'all to reach 1I7 °C is I;?7!I? s!;2 An aluminium sphere mass 2 g an' initially at a temperature of (= o is su''enly
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immerse' in a flui' at 1 °C 9ith heat transfer co"efficient ; 3Bm ! /2 Estimate the timere
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B i K!?1 × 17 .= 7!1
Biot num'er value is less than 7!1! So this is lumped heat analysis type pro'lem!
&or lumped parameter system)
hAt
C >
+
T Te
T T ρ ρ
− × × × ∞
∞
− =−PPP!@1A
L&rom HMT data 'ook +a*e /o!?;9e kno()
Characteristics len*th " c > A
c
ht
C -
+
#*t
@++ +"+2$2 2:++
T'T)3 e
T T
$* ' 2** e
#$ ' 2**$* ' 2** #*
In t#$ ' 2** @++ +"+2$2 2:++
t = ) ##" $ s
ρ ρ
− × × × ∞
∞− × × ×
⇒ =−
⇒ =
− ⇒ = × × ×
⇒ Time re$uired to cool the aluminium to I °C is 1=II!< s!
(2 Alloy steel +all of mm 'iameter heate' to ;== °C is
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Heat transfer co.efficient h 1I7 k8Dhr m 6
2
2
)#+ )+++
$++ s m 5!)"$$ % /m 5
×=
=Solution
Case ?i@ Temperature of +all after 1= sec2
&or sphere)
Characteristic "en*th c1
- =
c
+"++$ =
- +"++2 m=9e kno()
Biot num'er cih-
B5
=!)"$$: +"++2
#$"@!×=
B i 1!?< × 17 .= 7!1
Biot num'er value is less than 7!1! So this is lumped heat analysis type pro'lem!&or lumped parameter system)
hAt
C >
+
T Te
T T ρ ρ
− × × × ∞
∞
− =−PPP!@1A
L&rom HMT data 'ook +a*e /o!?;9e kno()
Characteristics len*th " c >
A
c
h tC -
+
!)"$$:)+
!#+ +"++2 :*$+
T'T)3 e """""""""" 23
T T
T ' : e)+: ' :
T = )+ 2"@# 5
ρ ρ − × × × ∞
∞− × × ×
⇒ =−
⇒ =
⇒
Case ?ii@ Time for +all to cool to 5== °C
∴T ?77 °C J 6K=
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c
ht
C -
+!)"$$:
t!#+ +"++2 :*$+
T'T23 e """"""" 23
T T
$: ' : e)+: ' :
$: ' : !)"$$: In t
)+: ' : !#+ +"++2 :*$+
t = )! "*!@ s
ρ ρ
− × × × ∞
∞ − × × ×
⇒ =−
⇒ =
− ⇒ = × × × ⇒
!=2 A large 9all cm thic has uniform temperature != °C initially an' the 9alltemperature is su''enly raise' an' maintaine' at 5== °C2 %in'
12 The temperature at a 'epth of =2; cm from the surface of the 9all after 1= s22 nstantaneous heat flo9 rate through that surface per m per hour2
Ta e α =2==; m Bhr> / 6 3Bm °C2
iven
Thickness " 6 cm 7!76 m%nitial temperature T i =7°C J 6K= =7= Surface temperature T 7 ?77 °C J 6K=
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Case ?i@
&or semi infinite solid!+
i +
T Terf
T T 2 at− = −
L&rom HMT data 'ook +a*e /o! I7+
i +
T T erf D3 """"""" )3
T T−
⇒ =−9here)
D2 at
=
+ut x 7!77; m) t 17 s) α 6!66 × 17 .< m 6Ds!
'$
+"++* D =
2 2"22 )+ )+
D = +"*!*
⇒× ×
Q 7!;?;) correspondin* erf @QA is 7!KK7<
erf D3 = +"::+$⇒
L,efer HMT data 'ook +a*e /o!I6+
i +
T 'T)3 +"::+$
T T
T ' $: +"::+$
+ ' $:T ' $:
+"::+$' :+
T = *:"*# 5
⇒ =−
⇒ =
⇒ =
⇒
Case ?ii@
%nstantaneous heat flo(
[ ]2
! t+ i5 T T< e
a t
α
π
− −=
L&rom HMT data 'ook +a*e /o!I7
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t =
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+
i +
T Terf
T T 2 at− = −
L&rom HMT data 'ook +a*e /o! I7+
i +
T T erf D3 where,
T T−
⇒ =−
D2 at
=
$2 )* erf D3
)+2 )*−
⇒ =−
⇒ 7!?=6 erf @QA
⇒ erf @QA 7!?=6
erf @QA 7!?=6) correspondin* Q is 7!?1
⇒ D +"!)=
9e kno(
D2 at
=
'#
22
2 #
+"+!# +"!) =
2 )"$$ )+ t
+"+!#3 +"!)3
23 )"$$ )+ t
t = )*)"!2 s
−
⇒× ×
⇒ = × × ×⇒
Time re$uired to reach =I7 °C is 1;1!?6 s!
2 nstantaneous heat flo9
[ ]2
! t+ i5 T T< ea t
α
π
− −=
L&rom HMT data 'ook +a*e /o!I7t =7 minutes @5ivenAt 1;77 s
2
#+"+!#3
! )"$$ )+ )*++
'#
2
!*"# )* )+2 3 < e
)"$$ )+ )*++
< )+@:2#"! % / m
π
− − × × × −⇒ = ×
× × ×=
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L/e*ative si*n sho(s that heat lost from the in*ot !
!2 Total heat energy
+ i
#
t< 250T T
:2++2 !*"# )* )+2 3
)"$$ )+
τ πα
π −
= −
= × − × × ×LTime is *iven) 6 hr K677 s
$ 2< *+ "# )+ / mτ = − ×L/e*ative si*n sho(s that heat lost from the in*ot! 2 A large steel plate cm thic is initially at a uniform temperature of 5== °C2 t issu''enly e)pose' on +oth si'es to a surroun'ing at 6= °C 9ith convective heat transferco"efficient of ; 3Bm /2 Calculate the centre line temperature an' the temperatureinsi'e the plate 12 cm from theme' plane after ! minutes2
Ta e / for steel 5 2 3Bm/> α for steel =2=5! m Bhr2
iven
Thickness " I cm 7!7I m%nitial temperature T i ?77 °C J 6K=
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Biot num'er cih-
B5
=
2*# +"+2#!2"#×=
iB +")$:#⇒ =
7!1 B i 177) So this is infinite solid type pro'lem!nfinite Soli's
Case ?i@
LTo calculate centre line temperature @orA Mid plane temperature for infinite plate)refer HMT data 'ook +a*e /o!I Heisler chart !
2c
'#
2
c
tD a is 6ourier number =
-
)")@ )+ )*+ =
+"+2#3
D a is 6ourier number = "!2
h-Curve
5
α →
× ×
→
=
c
2*# +"+2#+")$:
!2"#
h-Curve +")$:
5
×= =
= =
Q axis value is =!?6) curve value is 7!1
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Case ?ii@
Temperature @TxA at a distance of 7!716I m from mid planeL,efer HMT data 'ook +a*e /o!
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: 2)"! )+ m / s"α −= ×
Solution&or Sphere)
Characteristic "en*th c1
- =
c
+"+# =
- +"+)$ m=9e kno()
Biot num'erc
i
h-
B 5=$ +"+)$
+"$×=
iB +")$⇒ =7!1 B i 177) So this is infinite solid type pro'lem!
nfinite Soli's
LTo calculate centre line temperature for sphere) refer HMT data 'ook +a*e /o!
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+
i
+
+
T T +"*$
T T
T 2:* +"*$2@ 2:*
T 2@+"@ 5
∞
∞
−⇒ =−
−⇒ =−⇒ =
Center line temperature T 7 6 7! !
!52 A long steel cylin'er 1 cm 'iameter an' initially at = °C is place' into furnace at; = °C 9ith h 15= 3Bm /2 Calculate the time re
1! Time @tA re$uired for the axis temperature to reach ;77 °C!6! Correspondin* temperature @T tA at a radius of I!? cm!
Solution
&or Cylinder)
Characteristic "en*th c1 +"+$
-2 2
= =
c- +"+ m=9e kno()
Biot num'er B i ch-5
)!+ +"+
2)×=
i B +"2⇒ =
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7!1 B i 177) So this is infinite solid type pro'lem!nfinite Soli's
Case ?i@
+
A is temperature or3 T *++ CCentre line temperature
= °
To ;77 °C J 6K= 17K= Time @tA R
L,efer HMT data 'ook +a*e /o!
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r +"+#!Curve +"@
1 +"+$
h1D a is =5
)!+ +"+$ = +"!
2)
= = =
× =
Curve value is 7! ) Q axis value is 7!?) correspondin* 2 axis value is 7!;?!
r
+
T T E a is = +"*!
T T∞
∞
−⇒ =−
r
+
r
r
T T +"*!T T
T )+@ +"*!
)+: )+@
T )+:$"2 5
∞
∞
−⇒ =−
−⇒ =−⇒ =
1! Time re$uired for the axis temperature to reach ;77 °C is 6 ?I! s!6! Temperature @Tr A at a radius of I!? cm is 17K
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U$ T 0
%ntroduction
&ree convection in atmosphere free convection on a vertical flat plate
Empirical relation in free convection
&orced convection
"aminar and tur'ulent convective heat transfer analysis in flo(s
'et(een parallel plates) over a flat plate and in a circular pipe!
Empirical relations)
#pplication of numerical techni$ues in pro'lem solvin*!
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C.$VECT .$
*ART 0 A
12 3hat is 'imensional analysis4
-imensional analysis is a mathematical method (hich makes use of the study of thedimensions for solvin* several en*ineerin* pro'lems! This method can 'e applied to all typesof fluid resistances) heat flo( pro'lems in fluid mechanics and thermodynamics!
2 State -uc ingham π theorem2
Buckin*ham π theorem states as &ollo(s: 3%f there are n varia'les in a dimensionallyhomo*eneous e$uation and if these contain m fundamental dimensions) then the varia'les arearran*ed into @n – mA dimensionless terms! These dimensionless terms are called π terms!
!2 3hat are all the a'vantages of 'imensional analysis4
1! %t expresses the functional relationship 'et(een the varia'les in dimensional terms!6! %t ena'les *ettin* up a theoretical solution in a simplified dimensionless form!=! The results of one series of tests can 'e applied to a lar*e num'er of other similar
pro'lems (ith the help of dimensional analysis!
52 3hat are all the limitations of 'imensional analysis4
1! The complete information is not provided 'y dimensional analysis! %t only indicatesthat there is some relationship 'et(een the parameters!
6! /o information is *iven a'out the internal mechanism of physical phenomenon!=! -imensional analysis does not *ive any clue re*ardin* the selection of varia'les!
2 Define Reynol's num+er ?Re@2
%t is defined as the ratio of inertia force to viscous force!Inertia force
1e>iscous force
=
62 Define pran'tl num+er ?*r@2
%t is the ratio of the momentum diffusivity of the thermal diffusivity!
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7omentum diffusivity8r
Thermal diffusivity=
:2 Define $usselt num+er ?$u@2
%t is defined as the ratio of the heat flo( 'y convection process under an unittemperature *radient to the heat flo( rate 'y conduction under an unit temperature *radientthrou*h a stationary thickness @"A of metre!
/usselt num'er @/uA convcond
Q"
Q
;2 Define rash of num+er ? r@2
%t is defined as the ratio of product of inertia force and 'uoyancy force to the s$uare of viscous force!
2
Inertia force Buyoyancy forceFr
>iscous force3×=
(2 Define Stanton num+er ?St@2
%t is the ratio of nusselt num'er to the product of ,eynolds num'er and prandtlnum'er!
9uSt
1e 8r =
×
1=2 3hat is meant +y $e9tonion an' non 0 $e9tonion flui's4
The fluids (hich o'ey the /e(tonFs "a( of viscosity are called /e(tonion fluids andthose (hich do not o'ey are called non – ne(tonion fluids!
112 3hat is meant +y laminar flo9 an' tur+ulent flo94
8aminar flo9 "aminar flo( is sometimes called stream line flo(! %n this type of flo() thefluid moves in layers and each fluid particle follo(s a smooth continuous path! The fluid particles in each layer remain in an orderly se$uence (ithout mixin* (ith each other!
Tur+ulent flo9 %n addition to the laminar type of flo() a distinct irre*ular flo( is fre$uencyo'served in nature! This type of flo( is called tur'ulent flo(! The path of any individual
particle is 0i* – 0a* and irre*ular! &i*! sho(s the instantaneous velocity in laminar andtur'ulent flo(!
1 2 3hat is hy'ro'ynamic +oun'ary layer4
%n hydrodynamic 'oundary layer) velocity of the fluid is less than N of free streamvelocity!
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1!2 3hat is thermal +oun'ary layer4
%n thermal 'oundary layer) temperature of the fluid is less than N of free streamvelocity!
152 Define convection2
Convection is a process of heat transfer that (ill occur 'et(een a solid surface and afluid medium (hen they are at different temperatures!
1 2 State $e9ton7s la9 of convection2
Heat transfer from the movin* fluid to solid surface is *iven 'y the e$uation h # @T ( – T ∞A
This e$uation is referred to as /e(tonFs la( of coolin*!9hereh – "ocal heat transfer coefficient in 9Dm 6 !# – Surface area in m 6
T ( – Surface @orA 9all temperature in T∞ . Temperature of fluid in !
162 3hat is meant +y free or natural convection4
%f the fluid motion is produced due to chan*e in density resultin* from temperature*radients) the mode of heat transfer is said to 'e free or natural convection!
1:2 3hat is force' convection4
%f the fluid motion is artificially created 'y means of an external force like a 'lo(er or fan) that type of heat transfer is kno(n as forced convection!
1;2 Accor'ing to $e9ton7s la9 of cooling the amount of heat transfer from a soli' surface of area A at temperature T 9 to a flui' at a temperature T ∞ is given +y 2
#ns : h # @T ( – T ∞A
1(2 3hat is the form of e
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The thickness of the 'oundary layer has 'een defined as the distance from the surfaceat (hich the local velocity or temperature reaches N of the external velocity or
temperature!
*ART 0 -
12 Air at = °C> at a pressure of 1 +ar is flo9ing over a flat plate at a velocity of ! mBs2 if
the plate maintaine' at 6= °C> calculate the heat transfer per unit 9i'th of the plate2Assuming the length of the plate along the flo9 of air is m2
iven &luid temperature T ∞ 67 °C) +ressure p 1 'ar) elocity = mDs)
+late surface temperature T (
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! #1e #" :: )+ # )+= × < ×,eynolds num'er value is less than I × 17 I ) so this is laminar flo(!
&or flat plate) "aminar flo()"ocal /usselt /um'er /u x 7!==6 @,eA7!I @+rA7!===
! +"# +"
s
9u +" 2 #" :: )+ 3 +"$@@3
9u ):#"2:
%e 4now that,
h --ocal 9usselt 9umber 9u
5
= × ×=
×=
sh 2):#"2:2$"#$ )+
−×
⇒ =×
"ocal heat transfer coefficient h x 6!=6K 9Dm 6 9e kno()
#vera*e heat transfer coefficient h 6 × hxh 2 2" 2:= ×
h ?! calculate the follo9ing at ) !== mm2
12 #y'ro'ynamic +oun'ary layer thic ness>2 Thermal +oun'ary layer thic ness>
!2 8ocal friction coefficient>52 Average friction coefficient>
2 8ocal heat transfer coefficient62 Average heat transfer coefficient>:2 #eat transfer2
iven &luid temperature T ∞ 67 °C elocity = mDs 9ide 9 1 m
Surface temperature T ( ;7 °C-istance x =77 mm 7!= m
Solution 9e kno(
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&ilm temperature wf T T
T2
∞+=
f
'$ 2
'
*+ 2+2
T #+ C
8roperties of air at #+ C
Gensity = )"+@ 4./m
5inematic viscosity v = ):"@# )+ m / s8randtl number 8r =+"$@*
Thermal conductivity 5 = 2*"2$ )+ % /m5
ρ
+=
= °°
×
×
9e kno()
,eynolds num'er ,e (-v
$
! #
+"):"@# )+
1e #"+) )+ # )+
−×= ×
= × < ×
Since ,e I × 17 I ) flo( is laminar
&or &lat plate) laminar flo()
12 #y'ro'ynamic +oun'ary layer thic ness
+"#h
! +"#
h
# 1e3
= # +" #"+) )+ 3
$": )+ m
δ
δ
−
−
−
= × ×× × ×
= ×2 Thermal +oun'ary layer thic ness
( )
+"TD h
+"TD
TD
8r3
$": )+ +"$@*3
:"# )+ m
δ δ
δ
δ
−
− −
−
=⇒ = ×
= ×
!2 8ocal %riction coefficient
+"#f
! +"#
'f
C +"$$! 1e3
= +"$$! #"+) )+ 3
C = 2"@$ )+
−
−
=×
×
52 Average friction coefficient
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'+"#f-
! +"#
'
f-
C )" 2* 1e3
= )" 2* #"+) )+ 3
= #"@ )+
C #"@ )+
−
−
=×
×= ×
2 8ocal heat transfer coefficient ?h ) @
"ocal /usselt /um'er /u x 7!==6 @,eA7!I @+rA7!===
! +"+" 2 #"+) )+ 3 +"$@*3
9u $#"@
= ×
=9e kno(
"ocal /usselt /um'er
[ ]2
2
h -9u
5h +"
$#"@ = - = +" m2 "2$ )+
h $"2+ %/m 5
-ocal heat transfer coefficient h $"2+ % / m 5
−
×=
×= ×⇒ =
=
Q
62 Average heat transfer coefficient ?h@
2
h 2 h
2 $"2+
h )2"!) % / m 5
= ×= ×=
:2 #eat transfer
9e kno( that)wQ h A T T 3
= )2"!) ) +" 3 *+'2+3
Q = 2 " * %atts
∞= −× ×
!2 Air at != °C flo9s over a flat plate at a velocity of mBs2 The plate is m long an' 12m 9i'e2 Calculate the follo9ing
12 -oun'ary layer thic ness at the trailing e'ge of the plate>2 Total 'rag force>
!2 Total mass flo9 rate through the +oun'ary layer +et9een ) 5= cm an' ) ;cm2
iven &luid temperature T ∞ =7°C elocity 6 mDs
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"en*th " 6 m 9ide 9 9 1!I m
To fin'12 -oun'ary layer thic ness
2 Total 'rag force2!2 Total mass flo9 rate through the +oun'ary layer +et9een ) 5= cm an' ) ;
cm2Solution +roperties of air at =7 °C
$ 2
)")$# 4./m
v )$ )+ m / s8r +":+)5 2$":# )+ % /m5
ρ −
== ×=
= × −9e kno()
,eynolds num'er (-
1ev
=
$
# #
#
2 2)$ )+
1e 2"# )+ # )+
Since 1e # )+ ,flow is laminar
−×= ×
= × < ××
&or flat plate) laminar flo() Lfrom HMT data 'ook) +a*e /o!
Hydrodynamic 'oundary layer thickness
+"#h
# +"#
h
# 1e3
= # 2 2"# )+ 3+"+2 m
δ
δ
−
−
= × ×× × ×
=
Thermal 'oundary layer thickness)
+"t h 8r3δ δ
−×'+"
TD
=+"+2 +":+)3+"+22# mδ
×=
9e kno()#vera*e friction coefficient)
+"#f-
# +"#
'f-
C )" 2* 1e3
= )" 2* 2"# )+ 3
C 2"$# )+
−
−
=× ×
= ×
9e kno(
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f- 2
'2
' 2
'
tC
(2
t 2"$# )+
)")$# 232
Avera.e shear stress t = $") )+ 9/ m
Gra. force = Area Avera.e shear stress
= 2 )"# $") )+Gra. force = +"+)* 9
Gra. force on two sides of the plate= +"+)*
ρ =
⇒ × = ×
⇒ ××
× × ×
×2= +"+ $ 9
Total mass flo( rate 'et(een x ?7 cm and x ;I cm!
[ ]h h#
m ( *# !+*
ρ δ δ ∆ = = − =
Hydrodynamic 'oundary layer thickness
+"#h +"#
+"#
# 1e3
( = # +"*#
v
δ −=−
= × ×× × ×
+"#
$
HD +"*#
'+"#h =+"!+
+"#
+"#
$
HD +"!+
'
2 +"*## +"*#
)$ )++"+) + m
= # 1e3
(# +"!+v
2 +"!+# +"!+
)$ )+
*"@ )+ m
#)3 m= )")$# 2 +"+) + *"@ )+
*m = #"@: )+ 5. /
δ
δ
δ
−
=
−
−
−
−=
−
× = × × × =
× ×
× = × × ÷ × = × × ÷×
= ×
⇒ ∆ × × − × ∆ × s,
52 Air at != °C> %lo9s over a flat plate at a velocity of 5 mBs2 The plate measures = × !=
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cm an' is maintaine' at a uniform temperature of (= °C2 Compare the heat loss fromthe plate 9hen the air flo9s
?a@ *arallel to = cm>?+@ *arallel to != cmAlso calculate the percentage of heat loss2
iven &luid temperature T ∞ =7°C elocity ? mDs +late dimensions I7 cm × =7 cm
2+"#+ +" + m= × Surface temperature T ( 7°C
Solution &ilm temperature wf T TT 2∞+=
f
@+ +
2T $+ C
+=
= °
+roperties of air at
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2
h -9(
5
h +"#+@#" #2*"@$ )+
-ocal heat transfer coefficient h #"#2 %/m 5
−
=
×= ×=
9e kno(
#vera*e heat transfer coefficient h 2 h= ×
2
) w
)
h 2 #"#2
h ))"+! %/m 5
Heat transfer Q h A T T 3))"+! +"# +" 3 @+ +3
Q @@" $ %
∞
⇒ = ×=
= −= × × × −
=
Case ?ii@ 9hen the flo( is parallel to =7 cm side!
,eynolds num'er ,e (-v
$
! #
#
! +")*"@: )+
1e = $" )+ # )+Since 1e # )+ , flow is laminar
−×= ×
× < ××
&or flat plate) laminar flo()"ocal /usselt /um'er
+"# +"
! +"# +"
9( +" 2 1e3 +"$@$3
+" 2 $" 2 )+ 3 +"$@$39( :!"++*
== ×
=
9e kno( that) / xh -5
2
h +" +:!"++*2*"@$ )+
h :")!) %/m 5
−×= ×
⇒ =
2-ocal heat transfer coefficient h :")!) %/m 5=#vera*e heat transfer coefficient h 6 ×hx
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2
2 w
w
2
h 2 :")!
h )!"2* %/m 5
%e 4nowHeat transfer Q h A T T 3
h - % T T 3
)!"2* +" +"# $ + 3Q )2*"#%
∞
∞
= ×=
= × × −= × × −= × × × −
=
Case ?iii@2 )
)
Q Q; heat loss = )++Q
)2*"#'@@" $ = )++
@@" $; heat loss = 2@" ;
− ×
×
2 Air at 5= °C is flo9s over a flat plate of =2( m at a velocity of ! mBs2 Calculate thefollo9ing
12 .verall 'rag coefficient2 Average shear stress>
!2 Compare the average shear stress 9ith local shear stress ?shear stress at thetrailing e'ge@
iven %lui' temperature T ∞ 5= °C 8ength 8 =2( m Velocity U ! mBs2
Solution
+roperties of air at ?7 °C:
'$ 2
'
)")2* 5./m
= )$"@$ )+ m / s8r +"$@@
5 2$"#$ )+ %/m5
ρ
ν
=×
== ×
9e kno()
,eynolds num'er(-
1ev
=
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$
# #
#
+"@)$"@$ )+
1e )"#@ )+ # )+Since 1e # )+ , flow is laminar
−×= ×
= × < ××
&or plate) laminar flo()
-ra* coefficient @orA #vera*e skin friction coefficient
+"#f-
# +"#
f-
C )" 2* 1e3
)" 2* )"#@ )+ 3
C " )+
−
−
−
= ×= × ×
= ×9e kno(
#vera*e friction coefficient f- 2C
(2
τ ρ
=
2
f-
' 2
2
(C
2" )+ )")2* 3
=2
Avera.e shear stress = +"+)$ 9/m
ρ τ
τ
= ×
× × ×
9e kno()
"ocal skin friction coefficient
+"#f
# +"#
f
C +"$$! 1e3
+"$$! )"#@ )+ 3
C )"$$ )+
−
−
−
= ×= × ×
= ×(e kno(
"ocal skin friction coefficient f 2C (2
τ ρ =
2
2
2
2
2
)"$$ )+)")2* 3
2*"! )+ 9/m
-ocal shear stress *"! )+ 9/ m
-ocal shear stress *"! )+ 9/ m
Avera.e shear stress +"+)$ 9 / m+"#2
τ
τ
τ
τ
τ
−
−
−
−
⇒ × = ×
= ×= ×
×=
=
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62 Air at (= °C flo9s over a flat plate at a velocity of 6 mBs2 The plate is 1m long an' =2m 9i'e2 The pressure of the air is 6 $B 2 f the plate is maintaine' at a temperature of
:= °C> estimate the rate of heat remove' form the plate2
iven &luid temperature T ∞ 6 7°C elocity < mDs! "en*th " 1 m
9ide 9 7!I m +ressure of air + < k/Dm 6
2$ )+ 9/ m= ×+late surface temperature T ( K7°C
To fin' Heat removed from the plate
Solution
9e kno(
&ilm temperature wf T T
T2
∞+=
f
:+ 2@+2
T )*+ C
+=
= °
+roperties of air at 1;7 °C @#t atmospheric pressureA
'$ 2
'
+":@@ 5./m
= 2"!@ )+ m / s8r +"$*)
5 :"*+ )+ %/m5
ρ
ν
=×
== ×
$ote +ressure other than atmospheric pressure is *iven) so kinematic viscosity (ill vary (ith pressure! +r) ) C p are same for all pressures!
inematic viscosityatm
atm.iven
88ν ν = ×
[ ]
$2
# 2$
) bar 2"!@ )+
$ )+ 9 / m Atmospheric pressure = ) bar
)+ 9 /m2"!@ )+
$ )+ 9/ m
ν −
−
⇒ = × ×
= × ××
Q
# 2
'! 2
) bar = ) )+ 9 / m
5inematic viscosity v = #")!# )+ m / s"
× ×
Q
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2
2
) w
w
)
h :*"* % / m 5
Avera.e heat transfer coefficienth=:*"* %/m 5Head transfer Q h A T T 3
h - % T T 3
= :*"* +"* ) ++ ' !+3Q )$ @+"! %
∞
∞
=
= × × += × × × +
× × ×=
Case ?ii@ Entire plate is tur'ulent flo(:
"ocal nusselt num'er /ux 7!76 < × @,eA7!; × @+rA7!===
/ x 7!76 < × @1!6
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;2 Air at = °C flo9s over a flat plate at 6= °C 9ith a free stream velocity of 6 mBs2Determine the value of the average convective heat transfer coefficient upto a length of
1 m in the flo9 'irection2iven &luid temperature T ∞ 67 °C
+late temperature T (
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2
2
h )):#"2:
2$"#$ )+
-ocal nusselt numberJ 9( !"$# %/m 5 Avera.e heat transfer coefficientJ h = 2 h
2 !"$#
h @" ) %/m 5
−×= ×
=×
= ×=
(2 Air at °C at the atmospheric pressure is flo9ing over a flat plate at ! mBs2 f theplate is 1 m 9i'e an' the temperature T 9 : °C2 Calculate the follo9ing at a location of 1m from lea'ing e'ge2
i2 #y'ro'ynamic +oun'ary layer thic ness>
ii2 8ocal friction coefficient>iii2 Thermal +oun'ary layer thic ness>iv2 8ocal heat transfer coefficient
iven &luid temperature T ∞ 6I °C elocity = mDs 9ide 9 1 m
+late surface temperature T ( KI°C-istance 1 m
To fin'
1! Hydrodynamic 'oundary layer thickness!6! "ocal friction coefficient=! Thermal 'oundary layer thickness?! "ocal heat transfer coefficient
Solution 9e kno(
wf
f
$ 2
'
T T6ilm temperature T
2
:# 2# 2 5 = #+ C2
T #+ C
8roperties of air at #+ CGensity = )"+@
5inematic viscosity )+ m / s
8randtl number 8r = +"$@*
Thermal conductivity 5 = 2*"2$ ) %/m
ρ
ν
∞
−
+=
+= = °
= °°
=17.95×
× 5
9e kno()
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,eynolds num'er ,e(-
0 = - )mv
Q
#$
) )"$: )+):"@# )+ −
×= = ××# #
#
1e )"$: )+ # )+
Since 1e # )+ ,flow is laminar
= × < ××
&or flat plate) laminar flo()
1! Hydrodynamic 'oundary layer thickness)
+"#h
# +"#
h
# 1e3 = # ) )"$: )+ 3
+"+)22 m
δ
δ
−−
= × ×× × ×
=
6! "ocal friction coefficient
'+"#f
# +"#
f
C +"$!! 1e3
= +"$!! )"$: )+ 3
C )"$2 )+
−
−
=×
= ×
=! Thermal 'oundary layer thickness)
+"TD h
+"
TD
8r3
+"+)22 +"$@*3+"+) :#
δ δ
δ
−
−
= ×= ×=
?! "ocal heat transfer coefficient @h xA:
9e kno(
"ocal nusselt num'er / x 7!==6 @,eA7!I @+rA7!===
7!==6 @1!
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1=2 Atmospheric air at !== / 9ith a velocity of 2 mBs flo9s over a flat plate of length 8
m an' 9i'th 3 1m maintaine' at uniform temperature of 5== /2 Calculate thelocal heat transfer coefficient at 1 m length an' the average heat transfer coefficientfrom 8 = to 8 m2 Also fin' the heat transfer>
iven &luid temperature T ∞ =77 elocity 6!I mDs
Total "en*th " 6 m9idth 9 1 m
Surface temperature T ( ?77
To fin'
1! "ocal heat transfer coefficient at " 1 m6! #vera*e heat transfer coefficient at " 6 m=! Heat transfer
Solution
Case ?i@ "ocal heat transfer coefficient at " 1m
wf
f
T T6ilm temperature T
2
!++ ++ #+ 52T :: C
∞+=
+= == °
'$ 2
'
8roperties of air at :: C *+ C
= ) 5./m
= 2)"+@ )+ m / s8r = +"$@2
5 = +"!: )+ %/m5
%e 4now(-
1eynolds number 1e =v
ρ
ν
° ≈ °
×
×
$
#
#
2"# )
2)"+@ )+1e ))*# @"!# # )+
Since 1e # )+ ,flow is laminar"
−×= ×
= < ××
&or flat plate) laminar flo()
"ocal /usselt num'er / x 7!==6 @,eA7!I @+rA7!===
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7!==6 @11;I= !IA7!I @7!< 6A7!===
/ x 171!1;
9e kno()"ocal nusselt num'er
h -9(
5=
171!1; h )
+"!: )+ −××
hx =!7;=6 9Dm 6
⇒ "ocal heat transfer coefficient h x =!7; 9Dm 6
Case ?ii@ #vera*e heat transfer coefficient at " 6m
,eynolds num'er ,e (-v
$
#
#
2"# 21e
2)"+@ )+1e 2 :+:@")* # )+
Since 1e # )+ ,flow is laminar"
−×= ×
= ××
&or flat plate) laminar flo()
/ x 7!==6 @,eA7!I @+rA7!===
7!==6 @6=K7K !1;A7!I @7!< 6A7!=== / x 1?=
h -%e 4now that, 9(
5h 2
)! =+"!: )+ −
=
×⇒ ×
"ocal heat transfer coefficient h x 6!1K 9Dm 6 9e kno( that)
#vera*e heat transfer coefficient h 6 × hxh 6 × 6!1Kh ?!=I 9Dm 6
#vera*e heat transfer coefficient h ?!=I 9Dm 6
Case ?iii@ Heat transfer h # @T ( . T ∞A ?!=I × 6 × 1 @?77 – =77A
[ ]- = 2mK %= )mQ ;K7 9!
112 %or a particular engine> the un'ersi'e of the cran case can +e i'ealiIe' as a flat plat
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/ x 1I6?!<
9e kno( that) h -9(5=
2
h +"*)#2!"$ 0 - = +"*m
2*"2$ )+h # "*# %/m 5
−×= ×
=
Q
"ocal heat transfer coefficient h x I=!;I 9Dm 6 &or tur'ulent flo() flat plate
#vera*e heat transfer coefficient h 1!6? h x
h 1!6? × I=!;Ih %ormula use' for %lo9 over cylin'ers an' spheres
1! &ilm temperature %f T T
T2
∞+= 9here T ∞ . &luid temperature °C
T ( – +late surface temperature °C
6! ,eynolds num'er(G
9(v
= 9here – elocity) mDs
- . -iameter) m ν . inematic viscosity) m 6Ds
=! /usselt num'er / C @,eAm
@+rA7!===
?! /usselt num'er / hG5
I! Heat transfer h × # × @T( . T ∞A9here # G-π
%or sphere
/usselt num'er / 7!=K @,eA 7!<
Heat transfer h # @T ( . T ∞A
90
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9here # ? πr 6
1 2 Air at 1 °C> != mBh flo9s over a cylin'er of 5== mm 'iameter an' 1 == mm height9ith surface temperature of 5 °C2 Calculate the heat loss2
iven &luid temperature T ∞ 1I °C elocity =7 mDh
+ )+ m
$++ s( *" m/s
×=
=
-iameter - ?77 mm 7!? m"en*th " 1I77 mm 1!I m+late surface temperature T ( ?I °C
To fin' Heat loss!Solution 9e kno(
&ilm temperature wf T T
T2
∞+=!# )#
2+=
f T + C= °
+roperties of air at =7 °C : L&rom HMT data 'ook) +a*e /o!66-ensity ρ 1!1
$
*" +"!)$ )+ −
×= ×#
G1e 2"+* )+= ×
9e kno(
/usselt /um'er /u C @,eA m @+rA7!===
L&rom HMT data 'ook) +a*e /o!17I
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,e - value is 6!7; × 17 I ) so C value is 7!76
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f
'$ 2
'
) + +
2
T *+ C8roperties of air at *+ C
= ) 5./m
= 2)"+@ )+ m / s8r = +"$@2
5 = +"!: )+ %/m5%e 4now
(G1eynolds number 1e =
ρ
ν
ν
+=
= °°
×
×
+"2 +"+:+$$ "*2
2)"+@ )+1e $$ "*2
−×= =×
=9e kno(
/usselt /um'er /u 7!=K @,eA 7!<
7!=K @
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152 Air at 5= °C flo9s over a tu+e 9ith a velocity of != mBs2 The tu+e surface temperatureis 1 = °C2 Calculate the heat transfer for the follo9ing cases2
12 Tu+e coul' +e s
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2
h +"+$): "
+"!: )+
Heat transfer coefficient h = ** %/m 5
−×= ×
Case ?ii@
Tu'e diameter -
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1 2 n a surface con'enser> 9ater flo9s through staggere' tu+es 9hile the air is passe'
in cross flo9 over the tu+es2 The temperature an' velocity of air are != °C an' ; mBsrespectively2 The longitu'inal an' transverse pitches are mm an' = mmrespectively2 The tu+e outsi'e 'iameter is 1; mm an' tu+e surface temperature is (= °C2Calculate the heat transfer coefficient2
iven &luid temperature T ∞ =7°C elocity ; mDs
"on*itudinal pitch) S p 66mm 7!766 mTransverse pitch) S n 67mm 7!767 m-iameter - 1;mm 7!71; mTu'e surface temperature T ( 7°C
Solutionw
f
f
T T6ilm temperature T
2@+ +
2
T $+ C
∞−=
+=
= °
'$ 2
'
8roperties of air at $+ C
= )"+$+ 5./m
= )*"@: )+ m / s8r +"$@$
5 = 2*"@$ )+ % / m5
ρ
ν
°
×=
×
9e kno(
Maximum velocity max n
n
S(
S G× −
ma
ma
+"+2+ ( *
+"+2+ +"+)*
( = *+ m/s
⇒ = × −
9e kno(
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ma
$
!
n
n
p
p
( G1eynolds 9umber 1e =
*+ +"+)* )*"@: )+
1e :"# )+S +"+2+
)"))G +"+)*S
)"))GS +"+22
)"22G +"+)*
S )"22G
ν
−
×
×=×
= ×
= =
=
= =
=
pnSS
)"))" )"22,G G
= = correspondin* C) n values are 7!I1; and 7!II< respectively!
L&rom HMT data 'ook) +a*e /o!11?C 7!I1;n 7!II<
9e kno()
/usselt /um'er /u 1!1= @+rA 7!===LC @,eA n
L&rom HMT data 'ook) +a*e /o!11?
+" ! +"##$ 9u = )") +"$@$3 0+"#)* :"# )+ 3 9u = 2$$"⇒ × × × ×
9e kno(
/usselt /um'erhG
9u5
=
'h +"+)* 2$$" =
2*"@$ )+×⇒ ×
Heat transfer coefficient h ?6;!< 9Dm 6 !%ormulae use' for flo9 through Cylin'ers ? nternal flo9@
1! Bulk mean temperature
mi mom
T TT
2+=
Tmi %nlet temperature °C)9here
Tmo Outlet temperature °C!
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6! ,eynolds /um'er(G
1eν
=
%f ,eynolds num'er value is less than 6=77) flo( is laminar! %f ,eynolds num'er values is*reater than 6=77) flo( is tur'ulent!
=! "aminar &lo(: /usselt /um'er / – =!
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+
i
%here G &uter diameter
G ' Inner diameter
−
K! Heat transfer
h # @T ( – T mA (here # π × - × " @orA m C p @Tmo – T miA
9here T ( – Tu'e (all temperature °C) Tm – Mean temperature °C! Tmi – %nlet temperature °C Tmo – Outlet temperature °C!
;! Mass flo( ratem . ρ × # × *Ds
9here ρ . -ensity) *Dm =
# – #rea) 2 2G , m!π
– elocity) mDs
162 3hen =26 /g of 9ater per minute is passe' through a tu+e of cm 'iameter> it isfoun' to +e heate' from = °C to 6= °C2 The heating is achieve' +y con'ensing steam onthe surface of the tu+e an' su+se
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m
'$ 2
'
8
2
2+ $+
2
T !+ C8roperties of water at !+ C
= @@# 5./m
= +"$#: )+ m / s
8r = !" !+
5 = $2* )+ %/m5C !"):* 5 /5.5 = !):* /5.5
7ass flow rate m = A (
m ( = A
+"+) =
@@# +"+23!
>elocit
ρ
ν
ρ
ρ
π
+=
= °°
×
×=
⇒
×
y ( = +"+ ) m/s"et us first determine the type of flo(
(G1e
ν =
$+"+ ) +"+21e +"$#: )+ 1e @! "$Since 1e 2 ++, flow is laminar
−×⇒ = ×=
&or laminar flo()
/usselt num'er / =!
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w m
%e 4now that Q = h A= h G - T T 3
= )$:)"2 =))!"@ +"+2 - @+'!+3- = !"$2m
π
π
∆Τ× × × × −
× × × ×
1:2 3ater at = °C enters = mm 'iameter an' 5 m long tu+e 9ith a velocity of =2; mBs2The tu+e 9all is maintaine' at a constant temperature of (= °C2 Determine the heattransfer coefficient an' the total amount of heat transferre' if e)ist 9ater temperatureis := °C2
iven
%nner temperature of (ater T mi I7 °C-iameter - I7mm 7!7I m"en*th " ? m
elocity 7!; mDsTotal (all temperature T ( 7°CExit temperature of (ater T mo K7°C
To fin'
1! Heat transfer coefficient @hA6! Heat transfer @ A
Solution
Bulk mean temperaturemi mo
m
T TT 2
+=
m
'$ 2
'
#+ :+2
T $+ C
8roperties of water at $+ C
= @*# 5./m
= +"!:* )+ m / s8r "+2+
5 = $#)" )+ %/m5
ρ
ν
+=
= °°
×=
×
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"et us first determine the type of flo(:
'$
(G1e
+"* +"+# =
+"!:* )+
ν =
××
!1e *" $ )+Since 1e L 2 ++, flow is turbulent
= ×
!
- !*+
G +"+#-
*+ L $+
G1e = *" $ )+ )+,+++8r "+2+ +"$ 8r )$+
= =
=
× >= ⇒
-G
ratio is *reater than
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#vera*e temperature T m ?7 °C elocity 7!
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elocity =I mDs %nner diameter - i ? cm 7!7?m
Outer diameter - o < cm 7!7
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/u 7!76= @,eA 7!; @+rAn
This is heatin* process so) n 7!?+"* +"!
e
2
9u = +"+2 ! :#+3 +":+)3 9u = )+2"@
hG%e 4now 9u =
5h +"+2
)+2"@2$":# )+
h = ) :": %/m 5"
−
⇒ × ×
×= ×⇒
=2 Air at != °C> 6 mBs flo9s over a rectangular section of siIe !== × ;== mm2 Calculatethe heat lea age per meter length per unit temperature 'ifference2
iven #ir temperature T m =7°C
elocity < mDs #rea # =77 × ;77 mm 6
# 7!6? m 6
To fin'1! Heat leaka*e per metre len*th per unit temperature difference!
Solution
'! 2
'
8roperties of air at + C
= )")$# 5./m
= )$ )+ m / s
8r = +":+)
5 = 2$":# )+ % / m5
ρ
ν
°
×
×
E$uivalent diameter for =77 × ;77 mm 6 cross section is *iven 'y
e
e
!A ! +" +"*3G
8 2 +" +"*3%here 8 ' 8erimeter = 2 - %3
G +"! $ m
× ×= = +
⇒ =
9e kno(
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e
$
!
(G1eynolds 9umber 1e =
$ +"! $ )$ )+
1e = )$" )+
ν
−×= ××
Since ,e 6=77) flo( is tur'ulent!
&or tur'ulent flo( *eneral e$uation is @,e 17777A /u 7!76= @,eA 7!; @+rAn
#ssumin* the pipe (all temperature to 'e hi*her than a temperature! So heatin* process ⇒ n 7!?
! +"* +"! 9u = +"+2 )$" )+ 3 +":+)39u 2@!"@$⇒ ×
=9e kno(
e
'
hG9usselt 9umber 9u =
5h +"! $
2@!"@$ =2$":# )+
×⇒ ×Heat transfer coefficient ⇒ h 1;!7 9Dm 6 Heat leaka*e per unit per len*th per unit temperature difference
h + [ ])*"+@ 2 +" +"*× ×
= !K 9
12 Air at !!!/> 12 +ar pressure> flo9 through 1 cm 'iameter tu+e2 The surfacetemperature of the tu+e is maintaine' at 5== °/ an' mass flo9 rate is : gBhr2Calculate the heat transfer rate for 12 m length of the tu+e2
iven #ir temperature T m ===
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Solution
Since the pressure is not much a'ove atmospheric) physical properties of air may 'e taken atatmospheric condition
'$ 2
'
8roperties of air at $+ C
= )"+$+ 5./m
= )*"@: )+ m / s8r = +"$@$
5 = 2*"@$ )+ %/m5(G
1eynolds number 1e =
ρ
ν
ν
°
×
×
9e kno(
7ass flow rate m p ( ∆2
2
'$
+"+2+ = )"+$+ G (!
+"+2+ = )"+$+ +")23 (!
( = )"$$* m/s
(G)3 1e =
)"$$* +")2
)*"@: )+1e = )+##)"
π
π
ν
× × ×
⇒ × × ×
⇒
⇒
×= ×
Since ,e 6=77) so flo( is tur'ulent&or tur'ulent flo() *eneral e$uation is @,e 17777A
+"* +"!9u +"+2 1e3 +"$@$3= × × /u =6!
'
2
w m
w m
hG%e 4now 9u =5
h +")2 2"@ =
2*"@$ )+ h = :"@! %/m 5
Heat transfer rate Q = h A T T 3
h G -3 T T 3
:"@! +")2 )"#3 )2: $+3Q ++"*2 %
π π
×⇒
×⇒
−= × × × × −= × × × × −=
2 = /gBhr of air are coole' from 1== °C to != °C +y flo9ing through a !2 cm inner
107
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'iameter pipe coil +ent in to a heli) of =26 m 'iameter2 Calculate the value of air si'eheat transfer coefficient if the properties of air at 6 °C are
/ =2= (; 3Bm/ µ =2==! /gBhr 0 m*r =2:ρ 12=55 /gBm !
iven Mass flo( rate in 67I k*Dhr 2+#
5./ s in = +"+#$ 5./s$++
=
%nlet temperature of air T mi 177 °COutlet temperature of air T mo =7°C-iameter - =!I cm 7!7=I m
Mean temperature mi momT T
T $# C2+= = °
To fin' Heat transfer coefficient @hA
Solution
,eynolds /um'er ,e (G
ν inematic viscosity
µ ν ρ
=
: 2
+"++ 5. / s m$++
)"+!! 5./m
v :"@* )+ m / s7ass flow rate in = A ( ρ
−
−
= ×
2
+"+#$ )"+!! G (!π
= × × ×2+"+#$ )"+!! +"+ #3 (
!π = × × ×
':
$
( = ##": m/s(G
)3 1e =
##": +"+ #=
:"@* )+
1e = 2"!! )+
ν
⇒
⇒
××
×
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Since ,e 6=77) flo( is tur'ulent&or tur'ulent flo() *eneral e$uation is @,e 17777A
+"* +"
$ +"* +"
9u +"+2 1e3 8r3
This is coolin. process, so n = +"
9u = +"+2 2"!! )+ 3 +":39u 2$$)":
= × ×
⇒ × × ×=
9e kno( that)hG
9u5
=h +"+ #
2$$)":
+"+2@*
×=
Heat transfer coefficient h 66!2 n a long annulus ?!21 cm D an' cm .D@ the air is heate' +y maintaining thetemperature of the outer surface of inner tu+e at = °C2 The air enters at 16 °C an'leaves at ! °C2 ts flo9 rate is != mBs2 Estimate the heat transfer coefficient +et9een airan' the inner tu+e2
iven %nner diameter - i =!16I cm 7!7=16I m Outer diameter - o I cm 7!7I m
Tu'e (all temperature T ( I7 °C%nner temperature of air T mi 1
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Hydraulic or e$uivalent diameter
[ ]( ) ( )
2 2i
ho i
o i o i
o i
o i
! G G!A !G 8 G G
G G G GG G 3
G G
π
π
× − = = +
+ − −= += − 7!7I – 7!7=16I
- h 7!71;KI mh
$