Ppt Aero 6sem HT

8/9/2019 Ppt Aero 6sem HT

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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

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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! =?


=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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=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


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?


31/232

= ?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


32/232

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!

33


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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

−= +

34


35/232

[ ]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


36/232

(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


37/232

( )'$ 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

<>

=

37


38/232

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

38


39/232

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

39


40/232

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$"@$

−×= × ×

=

40


41/232

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

41


42/232

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


43/232

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

43


44/232

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−= ×

44


45/232

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

45


46/232

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


47/232

re

+

T Te """" )3

T T

ρ ρ

− × × × ∞

∞

− =− L&rom HMT data 'ook +a*e /o!?;

9e kno()

Characteristics len*th c>

- A

=

47


48/232

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'


49/232

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


50/232

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



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 =


51/232

&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


hAt

C >

+

T Te

T T ρ ρ

− × × × ∞

∞

− =−PPP!@1A


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

51


52/232

immerse' in a flui' at 1 °C 9ith heat transfer co"efficient ; 3Bm ! /2 Estimate the timere


53/232

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


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


54/232

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)


- =

c

+"++$ =

- +"++2 m=9e kno()

Biot num'er cih-

B5

=!)"$$: +"++2

#$"@!×=

B i 1!?< × 17 .= 7!1


hAt

C >

+

T Te

T T ρ ρ

− × × × ∞

∞

− =−PPP!@1A



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=


55/232

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=


56/232

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

56


57/232

t =


58/232

+

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

− − × × × −⇒ = ×

× × ×=

58


59/232

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=


60/232

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


61/232

Case ?ii@

Temperature @TxA at a distance of 7!716I m from mid planeL,efer HMT data 'ook +a*e /o!


62/232

: 2)"! )+ m / s"α −= ×

Solution&or Sphere)


- =

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!


63/232

+

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⇒ =

63


64/232

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!


65/232

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

65


66/232

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*!

66


67/232

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!

67


68/232

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!

68


69/232

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


70/232

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 (


71/232

! #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(

71


72/232

&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

72


73/232

'+"#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

73


74/232

"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(

74


75/232

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 = × !=

75


76/232

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


77/232

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!


$

! #

#

! +")*"@: )+

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

77


78/232

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

=

78


79/232

$

# #

#

+"@)$"@$ )+

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

−

−

−

−

⇒ × = ×

= ×= ×

×=

=

79


80/232

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(


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

80


81/232


82/232


83/232

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


84/232

;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 (


85/232

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()

85


86/232

,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!


87/232

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


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"

−×= ×

= < ××


"ocal /usselt num'er / x 7!==6 @,eA7!I @+rA7!===

87


88/232

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


$

#

#

2"# 21e

2)"+@ )+1e 2 :+:@")* # )+

Since 1e # )+ ,flow is laminar"

−×= ×

= ××


/ 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

88


89/232


90/232

/ 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


91/232

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(


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

91


92/232

,e - value is 6!7; × 17 I ) so C value is 7!76


93/232

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 @


94/232

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


95/232

2

h +"+$): "

+"!: )+

Heat transfer coefficient h = ** %/m 5

−×= ×

Case ?ii@

Tu'e diameter -


96/232

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


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(

96


97/232

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!

97


98/232

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 / – =!


99/232

+

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


100/232

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 / =!


101/232

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

+=

= °°

×=

×

101


102/232

"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


103/232

#vera*e temperature T m ?7 °C elocity 7!


104/232

elocity =I mDs %nner diameter - i ? cm 7!7?m

Outer diameter - o < cm 7!7


105/232

/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(

105


106/232

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 ===


107/232

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


108/232

'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

⇒

⇒

××

×

108


109/232

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


110/232

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

$

Date post:	01-Jun-2018
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