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Mecanismos FMecanismos Fsicos esicos eEquaEquaes de Taxas dees de Taxas de

Transmisso de CalorTransmisso de Calor


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O que a transferncia/transmisso de calor?

A transferncia/transmisso de calor o trnsito de energia trmicadevido a uma diferena de temperaturas num meio ou entre meios.

O que a energia trmica?A energia trmica est associada translao, rotao, vibrao e aosestados electrnicos dos tomos e molculas que constituem a matria.

Transferncia de Calor e Energia TTransferncia de Calor e Energia Trmicarmica

A energia trmica representa o efeito cumulativo das actividades microscpicase est relacionada com a temperatura da matria.


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UnidadesSmboloSignificado fsicoQuantidade

Transporte de energia trmica devido a

gradientes de temperatura

Transferncia de

Calor

NO confundir ou trocar os significados fsicos de Energia Trmica,

Temperatura e Transferncia de Calor

Energia associada ao comportamentomicroscpico da matria

Energia

Trmica+ J/kgJouuUou

Modo indirecto de determinar aquantidade de energia trmicaarmazenada na matria

Temperatura KC ouT

Quantidade de energia trmica transferidanum intervalo de tempo t > 0

Calor Q J

+

UEnergia TrmicauEnergia Trmica especfica

Energia trmica transferida por unidade

de tempo

Taxa de

transferncia decalor

q W

Energia trmica transferida por unidadede tempo e por unidade de rea

Fluxo de calor 'q' 2/mW


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Conduo: Transferncia de calor num slido ou fluido esttico (gs ou lquido) devida aomovimento aleatrio dos seus tomos, molculas e/ou electres constituintes.

Conveco: Transferncia de calor devida ao efeito combinado do movimentoaleatrio (microscpico) e do movimento macroscpico (adveco)do fluido sobre uma superfcie.

Radiao: Energia que emitida pela matria devido a mudanas das configuraes

electrnicas dos seus tomos ou molculas e que transportada por ondaselectromagnticas (ou por fotes).

A conduo e a conveco exigem a presena de matria e de variaes de temperatura nessemeio material.

Embora a radiao tenha origem na matria, o seu transporte no exige a presena de um

meio material. Alis, o transporte radiativo mais eficiente no vcuo.

Modos de Transferncia de Calor


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AplicaesIdentificaIdentificao de mecanismoso de mecanismos

Problema 1.73(a): Identificao de mecanismos de transferncia de calor para janelas de vidro simples e duplo

Conduo atravs do vidro que tem superfcie interior em contacto com ar exterior na janela de vidro duplo, 2c o n dq

Conveco entre a superfcie interior da janela e o ar interior,1c o n vq

Fluxo radiativo til trocado entre as paredes do quarto e a superfcie interior da janela,1ra dq

Conduo atravs do vidro que tem superfcie interior em contacto com ar interior,1c o n dq

Radiao solar incidente durante o dia: a fraco transmitida pelo vidro duplo menor que a transmitida pelo vidro simples.sq

Conveco entre a superfcie exterior da janela e o ar exterior,2convq

Fluxo radiativo til trocado entre a envolvente e a superfcie exterior da janela,2radq

Conveco no espao entre vidros (janela de vidro duplo),conv sqFluxo radiativo til entre as superfcies dos vidros que limitam o espao entre vidros,rad sq


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2 1x

T TdTq k k

dx L

= =

1 2x

T Tq k

L

=

Taxa de transferncia de calor (W): x xq q A=

Aplicao ao caso de conduo unidimensional, estacionria atravs de umaplaca plana com condutibilidade trmica constante:

ConduoForma geral (vectorial) da Lei de Fourier:

Taxas de Transferncia de Calor

Fluxo de calor (W/m2):

Fluxo de calor2W/m

Condutibilidade trmica

KW/m

Gradiente de temperatura

K/mouC/m


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Conveco

Relao entre conveco e o escoamento sobre uma superfcie e o desenvolvimentodas camadas limite hidrodinmica e trmica:

Lei do arrefecimento de Newton :

( )h sq T T=


h[W/m2

.C] ou [W/m2

.K]: Coeficiente de transferncia de calor por conveco


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2500 - 100000Ebulio ou condensao

50 - 20000Conveco forada - lquidos

25 - 250Conveco forada - gases50 - 1000Conveco natural - lquidos

2 - 25Conveco natural - gases

Gama de valores tpicos do coeficiente deconveco [W m-2 K-1]

Adveco, difuso, conveco

Conveco forada, conveco natural

Calor sensvel e calor latente

Ebulio e condensao


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Radiao

Fluxo de energia que sai devido emisso:4

b sE E T = =

Energia absorvida devida irradiao: absG G=

A transferncia de calor por radiao numa interface gs/slido envolve a emisso deradiao a partir da superfcie e pode tambm envolver a absoro da radiao incidenteda envolvente (irradiao, G ), bem como da conveco (se Ts T)


Gabs[W/m2]: Radiao incidente absorvida

(0 1): Absorsividade da superfcie

G[W/m2]: Irradiao

E [W/m2]: Poder emissivo da superfcie (0 1): Emissividade da superfcieEb [W/m

2]: Poder emissivo de um corpo negro (emissor perfeito) = 5,6710-8 [W m-2 K-4] (constante de Stefan-Boltzmann)


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Irradiao: Caso especial de uma superfcie exposta a umaenvolvente de grandes dimenses com temperatura uniforme,

surT

4sur sur G G T= =


Se = , o fluxo radiativo tila partir da superfcie

devido s trocas de calor por radiao com a envolvente :

( ) 4sur4sSb

''rad TTGTEq ==


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Em alternativa,

Para conveco e radiao combinadas:

( ) ( )conv rad s r s sur q q q h T T h T T = + = + (1.10)


))(( 22 surSsurSr TTTTh ++=

)('' surSrrad TThq =

KmWhr ./2 Coeficiente de transferncia de calor por radiao


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AplicaesArrefecimento de componente electrArrefecimento de componente electrnicanica

Problema 1.31: Dissipao de potncia em chips que operam com uma temperatura superficial de 85Cnum quarto cujas paredes e ar esto a 25C para (a) conveco natural e (b) conveco forada.

Hipteses: (1) Estacionrio,(2) Trocas de radiao entre superfcie pequena e grande

envolvente,(3) Transferncia de calor desprezvel das faces lateraise da superfcie de trs do chip

( ) ( )4 4h s s sur A T T A T T = + elec conv rad P q q= +

( )22 -4 2= 0.015m =2.2510 mA L=

(a) Se for conveco natural,

( ) ( )( )

( ) ( )

5 / 4 5/42 5/4 -4 2

-4 2 -8 2 4 4 4 4

=4.2W/m K 2.2510 m 60K =0.158W

0.60 2.2510 m 5.6710 W/m K 358 -298 K =0.065W0.158W+0.065W=0.223W

conv s

rad

elec

q CA T T

q

P

=

= =

(b) Se for conveco forada,

( ) ( )( )2 -4 2

h =250W/m K 2.2510 m 60K =3.375W3.375W+0.065W=3.44W

conv s

elec

q A T T P

= =


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ConservaConservao de Energiao de Energia


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Formulaes Alternativas

Base temporal:Num instanteouNum intervalo de tempo

Tipo de Sistema:Volume de controloSuperfcie de controlo

Uma ferramenta importante na anlise do fenmeno de transfernciade calor, constituindo geralmente a base para determinar a temperatura

do sistema em estudo.

CONSERVAO DE ENERGIA

(Primeira Lei da Termodinmica)


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Num instante de tempo:Num instante de tempo:

Notar a representao do sistema atravs de umasuperfcie de controlo (linha a tracejado) nasfronteiras.

Fenmenos superficiais

Fenmenos volumtricos

APLICAO A UM VOLUME DE CONTROLO

Taxa de transferncia de energia trmica e/ou mecnica atravs da superfcie de controlo,devido transferncia de calor, escoamento de um fluido ou transferncia de trabalho

Taxa de gerao de energia trmica devido converso de outra forma de energia (e.g.elctrica, nuclear, qumica); converso essa de energia que ocorre no interior do sistema

Taxa de variao de energia armazenada no sistema


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Num instante de tempo:Num instante de tempo:

Notar a representao do sistema atravs de uma

superfcie de controlo (linha a tracejado line) nasfronteiras.

Conservao de energia

APLICAO A UM VOLUME DE CONTROLO

Num intervalo de tempo:

( )bEEEE stoutgin 11.1=+ Cada termo tem unidades [J].

Cada termo tem unidades [J/s] ou [W].


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H um caso especial para o qual no existe massa ou volume contidos na superfcie de controlo

Conservao de Energia (num instante):

Aplica-se em condies estacionrias e transientes

Considere a superfcie de uma parede com transferncia de calor (conduo, conveco e radiao).

0cond conv rad q q q =

( ) ( )4 41 2 2 2 2 0surT T

k T T T T L

=h

Sem massa nem volume, no faz sentido falar em energia armazenada ou em gerao no balano deenergia, mesmo que estes fenmenos ocorram no meio de que a superfcie faz parte.

O BALANO DE ENERGIA SUPERFICIAL

0= outin EE &&


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EXEMPLOS DE APLICAO

Exemplo 1.3: Aplicao resposta trmica de um fio condutor com aquecimento por efeitode Joule (gerao de calor passagem da corrente elctrica).

0=inE& ( ) ( ) ( )[ ]44 surout TTTThLDE += &

2IRE electg =& ( )TVctd

dEst =&

stgoutin EEEE &&&& =+


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EXEMPLOS DE APLICAO

Exemplo 1.43: Processamento trmico de uma bolacha de slica num forno de 2 zonas.

Sabe-se que a bolacha de slica est posicionada no forno com as superfciesinferior e superior expostas, respectivamente, zona quente e zona fria.

Determinar(a) Taxa inicial de aquecimento da bolacha a partir de Twi = 300K,(b) Temperatura em regime estacionrio.

A conveco relevante?

ESQUEMAHipteses:

a) Temperatura da bolacha uniforme

b) Temperaturas uniformes das regies quente e friac) Trocas radiativas entre corpo pequeno e

envolvente grande

d) Perdas da bolacha para o suporte desprezveis


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EXEMPLOS DE APLICAO

Exemplo 1.43: Processamento trmico de uma bolacha de slica num forno de 2 zonas (cont)

ANLISE: No balano de energia bolacha de slica deve contabilizar-se a conveco com o gs ambiente pelassuperfcies inferior (l) e superior (u), as trocas de radiao com as zonas quente e fria e a acumulao de energia.

, , , , w

rad h rad c cv u cv ld T

q q q q cd dt

+ =

Em termos de fluxo (por unidade de rea)

( ) ( ) ( ) ( )4 4 4 4,, ww sur c w u w l wsur hd T

T T T T h T T h T T cd dt

+ =

(a) Como condio inicial temos Tw =Twi = 300K

( )w id T / dt 104 K / s=

3

( ) ( )8 2 4 4 4 8 2 4 4 4 440.65 5.67 10 W / m K 1500 300 K 0.65 5.67 10 W / m K 330 300 K +

( ) ( )2 28W / m K 300 700 K 4 W / m K 300 700 K = ( )w i0.00078 m d T / dt2700kg/m875J/kgK

stoutin EEE &&& =


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EXEMPLOS DE APLICAO

Exemplo 1.43: Processamento trmico de uma bolacha de slica num forno de 2 zonas (cont)

Em regime estacionrio o armazenamento de energia nulo. O balano de energia efectuado com a temperaturada bolacha em regime estacionrio, Tw,ss

( ) ( )4 4 4 4 4 4w,ss w,ss0.65 1500 T K 0.65 330 T K + ( ) ( )2 2w,ss w,ss8W / m K T 700 K 4 W / m K T 700 K 0 =

w,ssT 1251 K=

Para determinar a importncia relativa da conveco, resolver o balano de energia sem conveco. Obtm-se

(dTw/dt)i = 101 K/s e Tw,ss = 1262 K. Logo, a radiao controla a taxa de aquecimento inicial e o regimeestacionrio.


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FourierFouriers Laws Law

and theand theHeat EquationHeat Equation


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A rate equation that allows determination of the conduction heat fluxfrom knowledge of the temperature distribution in a medium.

Fouriers Law

Its most general (vector) form for multidimensional conduction is:

Implications:

Heat transfer is in the direction of decreasing temperature

(basis for minus sign).

Direction of heat transfer is perpendicular to lines of constant

temperature (isotherms).

Heat flux vector may be resolved into orthogonal components.

Fouriers Law serves to define the thermal conductivity of the

medium

Tkq =r

xT

qk xx

=


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Cartesian Coordinates: ( ), ,T x y z

T T Tq k i k j k k

x y z

=

xq yq zq

zq

T T Tq k i k j k k

r r z

=

rq q

Cylindrical Coordinates: ( ), ,T r z

qsinT T Tq k i k j k k r r r

=

rq q

Spherical Coordinates: ( ), ,T r


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In angular coordinates , the temperature gradient is stillbased on temperature change over a length scale and hence hasunits of C/m and not C/deg.

( )or ,

Heat rate for one-dimensional, radial conduction in a cylinder or sphere:

Cylinder2r r r r q A q rLq = =

or,

2r r r r q A q rq = =

Sphere2

4r r r r q A q r q = =


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The Heat Equation A differential equation whose solution provides the temperature distribution in astationary medium.

Based on applying conservation of energy to a differential control volumethrough which energy transfer is exclusively by conduction.

Cartesian Coordinates:

Net transfer of thermal energy into the

control volume (inflow-outflow)

Thermal energy

generation

Change in thermal

energy storage

p

T T T T k k k q c

x x y y z z t

+ + + =


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Spherical Coordinates:

Cylindrical Coordinates:

2

1 1p

T T T T kr k k q c

r r r z z t r

+ + + =

22 2 2 2

1 1 1sin

sin sin p

T T T T kr k k q c

r r tr r r

+ + + =


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One-Dimensional Conduction in a Planar Medium with Constant Propertiesand No Generation

2

2

1T T

tx

=

thermal diffu osivit f the medy iump

k

c


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Boundary and Initial Conditions For transient conduction, heat equation is first order in time, requiringspecification of an initial temperature distribution: ( ) ( )0, ,0tT x t T x= =

Since heat equation is second order in space, two boundary conditionsmust be specified. Some common cases:

Constant Surface Temperature:

( )0, sT t T=

Constant Heat Flux:

0|x sT

k qx

=

=

Applied Flux Insulated Surface

0| 0xT

x =

=

Convection

( )0| 0,xT

k h T T t =

=


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Thermophysical PropertiesThermal Conductivity: A measure of a materials ability to transfer thermalenergy by conduction.

Thermal Diffusivity: A measure of a materials ability to respond to changesin its thermal environment.

Property Tables:Solids: Tables A.1 A.3Gases: Table A.4Liquids: Tables A.5 A.7


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Methodology of a Conduction Analysis Solve appropriate form of heat equation to obtain the temperature

distribution.

Knowing the temperature distribution, apply Fouriers Law to obtain theheat flux at any time, location and direction of interest.

Applications:

Chapter 3: One-Dimensional, Steady-State ConductionChapter 4: Two-Dimensional, Steady-State ConductionChapter 5: Transient Conduction


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Problem 2.46 Thermal response of a plane wall to convection heat transfer.

KNOWN: Plane wall, initially at a uniform temperature, is suddenly exposed to convective heating.

FIND: (a) Differential equation and initial and boundary conditions which may be used to find thetemperature distribution, T(x,t); (b) Sketch T(x,t) for the following conditions: initial (t 0), steady-state (t ), and two intermediate times; (c) Sketch heat fluxes as a function of time at the twosurfaces; (d) Expression for total energy transferred to wall per unit volume (J/m3).

SCHEMATIC:


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ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internalheat generation.

ANALYSIS: (a) For one-dimensional conduction with constant properties, the heat equation has theform,

2

2

T 1 T

tx

=

( ) i

0

L

Initial, t 0: T x,0 T uniform temperature

Boundaries: x=0 T/ x) 0 adiabatic surface

x=L k T/ x) = h T

=

=

( )L,t T surface convection

and theconditions are:

(b) The temperature distributions are shown on the sketch.

Note that the gradient at x = 0 is always zero, since this boundary is adiabatic. Note also that thegradient at x = L decreases with time.


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Dividing both sides by AsL, the energy transferred per unit volume is

c) The heat flux, as a function of time, is shown on the sketch for the surfaces x = 0 and

x = L.

( )txqx ,

( )( )in s 0E hA T T L,t dt

=

d) The total energy transferred to the wall may be expressed asd) The total energy transferred to the wall may be expressed as

in conv s0E q A dt =

( ) 3in0

E hT T L,t dt J/m

V L

=

Problem: NonProblem: Non--uniform Generation dueuniform Generation dueto Radiation Absorptionto Radiation Absorption


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Problem 2.28 Surface heat fluxes, heat generation and total rate of radiationabsorption in an irradiated semi-transparent material with a

prescribed temperature distribution.

KNOWN: Temperature distribution in a semi-transparent medium subjected to radiative flux

to Radiation Absorptionto Radiation Absorption

SCHEMATIC:

FIND: (a) Expressions for the heat flux at the front and rear surfaces, (b) The heat generation rate( )q x ,& and (c) Expression for absorbed radiation per unit surface area.

Problem : NonProblem : Non--uniformuniformGeneration (Cont.)Generation (Cont.)


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Generation (Cont.)( )

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in medium, (3)Constant properties, (4) All laser irradiation is absorbed and can be characterized by an internal

volumetric heat generation term ( )q x .&

ANALYSIS: (a) Knowing the temperature distribution, the surface heat fluxes are found using

Fouriers law,

( ) -axx 2dT A

q k k - a e Bdx

ka

= = +

Front Surface, x=0: ( )xA A

q 0 k + 1 B kBka a

= + = +


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Generation (Cont.)Generation (Cont.)

Alternatively, evaluate gE& by integration over the volume of the medium,

( ) ( )LL L -ax -ax -aL

g 0 0 0

A AE q x dx= Ae dx=- e 1 e .

a a = =

& &

On a unit area basis

( ) ( ) ( )-aLg in out x x AE E E q 0 q L 1 e .a

= + = + = + & & &


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OneOne--Dimensional, SteadyDimensional, Steady--StateStateConduction withoutConduction without

Thermal Energy GenerationThermal Energy Generation


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Specify appropriate form of theSpecify appropriate form of the heat equation.heat equation.

Solve for theSolve for the temperature distributiontemperature distribution..

ApplyApply FourierFouriers Laws Law to determine theto determine the heat flux.heat flux.

Simplest Case:Simplest Case: OneOne--Dimensional, SteadyDimensional, Steady--StateState Conduction withConduction with NoNo Thermal EnergyThermal Energy GenerationGeneration

Alternative conduction analysisAlternative conduction analysis

Common Geometries:Common Geometries:

TheThe Plane Wall:Plane Wall: Described in rectangular (Described in rectangular (xx) coordinate. Area) coordinate. Area

perpendicular to direction of heat transfer is constant (inperpendicular to direction of heat transfer is constant (independent ofdependent ofxx).).

TheThe Tube WallTube Wall: Radial conduction through tube wall.: Radial conduction through tube wall.

TheThe Spherical Shell:Spherical Shell: Radial conduction through shell wall.Radial conduction through shell wall.

Methodology of a Conduction Analysis

The Plane Wall


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Consider a plane wall between two fluids of different temperaturConsider a plane wall between two fluids of different temperature:e:

The Plane Wall

Implications:

0d dT

kdx dx

=

Heat Equation:

( )Heat flux is independent of .xq x

( )Heat rate is independent of .xq x Boundary Conditions: ( ) ( ),1 ,20 ,s sT T T L T = =

Temperature Distribution for Constant :

( ) ( ),1 ,2 ,1s s sT x T T T L= +

k

H t Fl d H t R tHeat Flux and Heat Rate:


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Heat Flux and Heat Rate:Heat Flux and Heat Rate:

( ),1 ,2x s sdT k

q k T T

dx L

= =

( ),1 ,2x s sdT kA

q kA T T dx L

= =

Thermal Resistances and Thermal Circuits:tT

Rq

=

Conduction in a plane wall: ,t cond

LR

kA=

Convection: ,1

t convRhA

=

Thermal circuit for plane wall with adjoining fluids:

1 2

1 1tot

LR

h A kA h A= + +

,1 ,2

x tot

T Tq

R

=


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Thermal Resistance forThermal Resistance for Unit Surface Area:Unit Surface Area:

,t cond

L

R k = ,

1t convR h =

Units: W/KtR 2m K/WtR

Radiation Resistance:

,

1t rad

rR h A= ,

1t rad

rR h =

( )( )2 2r s sur s sur h T T T T = + + Contact Resistance:

, A B

tcx

T TR

q

=

= t ct c

c

RR

,

,

Values depend on: Materials A and B, surface finishes, interstitial conditions, and

contact pressure (Tables 3.1 and 3.2)

Composite WallComposite Wall withwith Negligible Contact Resistance:Negligible Contact Resistance:


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Composite WallC p a withw Negligible Contact Resistance:g g C a a

,1 ,4x

tot

T Tq

R

=

1 4

1 1 1C tot A BtotA B C

L RL LRA h k k k h A

= + + + + =

Overall Heat Transfer Coefficient (U) :

A modified form of Newtons Law of Cooling to encompass multiple resistancesto heat transfer.

x overallq UA T =

1totR

UA=


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SeriesSeries Parallel Composite Wall:Parallel Composite Wall:

Note departure from one-dimensional conditions for .F Gk k

Circuits based on assumption of isothermal surfaces normal tox direction oradiabatic surfaces parallel tox direction provide approximations for .xq


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ALTERNATIVE CONDUCTION ANALYSIS:

STEADY STATE

NO HEAT GENERATION

NO HEAT LOSS FROM THE SIDES

A(x) and k(T)

dxxx qq +=IS TEMPERATURE DISTRIBUTION ONE-DIMENSIONAL?IS IT REASONABLE TO ASSUME ONE-DIMENSIONALTEMPERATURE DISTRIBUTION INx?

FROM THE FOURIERS LAW: dxdTTkxAqx )()(=

=T

T

x

x dTTk

xA

dxq

00

)(

)(

Tube WallTube Wall The Tube Wall


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Heat Equation:Heat Equation:

e ube Wa

1

0

d dT

krr dr dr

=

Is the foregoing conclusion consistent with the energy conservation requirement?

How does vary with ?rq r

What does the form of the heat equation tell us about the variation of with

in the wall?rq

r

Temperature Distribution for Constant :k

( )( )

,1 ,2,2

1 2 2

ln

ln /

s ss

T T rT r T

r r r

= +


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Heat FluxHeat Flux andand Heat Rate:Heat Rate:

( )( )

( )( )

( )( )

,1 ,22 1

,1 ,22 1

,1 ,22 1

ln /

22

ln /

22ln /

r s s

r r s s

r r s s

dT kq k T T dr r r r

kq rq T T

r r

Lkq rLq T T

r r

=

= =

= =

= (3.27)

Conduction Resistance:( )

( )

2 1,

2 1,

ln /

Units K/W2ln /

Units m K/W2

t cond

t cond

r r

R Lk

r rR

k

=

=

Why is it inappropriate to base the thermal resistance on a unit surfacearea?

Composite Wall withComposite Wall with


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Composite Wall withComposite Wall withNegligible ContactNegligible Contact

ResistanceResistance

( ),1 ,4 ,1 ,4rtot

T Tq UA T T

R

= =

1

Note that

is a constant independent of radius.totUA R =

But, U itself is tied to specification of an interface.

( ) 1i i tot U A R =

Spherical Shell


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Heat EquationHeat Equation

p

22

10

d dTr

dr dr r

=

What does the form of the heat equation tell us about the variation ofwith ? Is this result consistent with conservation of energy?rq r

How does vary with ?rq r

Temperature Distribution for Constant :k

( ) ( ) ( )

( )1/

,1 ,1 ,21 2

1

1 /s s s

r rT r T T T

r r

=

Heat flux Heat RateHeat flux Heat Rate andand Thermal Resistance:Thermal Resistance:


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Heat flux, Heat RateHeat flux, Heat Rate andand Thermal Resistance:Thermal Resistance:

( ) ( ) ( ),1 ,22 1 21/ 1/ r s s

dT kq k T T dr r r r= =

( ) ( )( )2 ,1 ,2

1 2

44

1/ 1/ r r s sk

q r q T T r r

= =

Composite Shell:overall

r overalltot

Tq UA T

R

= =

1 ConstanttotUA R =

( ) 1 Depends oni i tot iU A R A=

( ) ( )1 2, 1/ 1/ 4t condr rR

k=

r


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Critical radius (cylindrical geometry)Critical radius (cylindrical geometry)Isolamento

r1 r

T ,h,1 1

T ,h

r2

h1Lr12

1

T

T,1

hLr2

1

Lk

rr 22

/ln( )

Lk1

r2

2

/ln( )r1

(a)

(b)

( )

hLrLk

rr

hLr

TTq revestsemr

21

12

11

1,.,

2

1

2

ln

2

1++

=

( ) ( )

hLrLk

rr

Lk

rr

hLr

TTq revestcomr

2

1

2

ln

2

ln

2

1 2

1

12

11

1,.,

+++

=

2

1

2

11

2

1

rhLrLkrd

Rd tot

=

h

krcrit=0=

rd

Rd tot 02

2

11

2

1322

2

>

+

=

== hkrhkr

tot

rhLrLkrd

Rd

Problem 3 23: Assessment of thermal barrier coating (TBC) for protection


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Problem 3.23: Assessment of thermal barrier coating (TBC) for protectionof turbine blades. Determine maximum blade temperature

with and without TBC.

Schematic:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constantproperties, (3) Negligible radiation

ANALYSIS F i h l h l i i h h TBC i


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ANALYSIS: For a unit area, the total thermal resistance with the TBC is

( ) ( )1 1tot,w o t,c iZr InR h L k R L k h = + + + +

( )3 4 4 4 3 2 3 2tot,wR 10 3.85 10 10 2 10 2 10 m K W 3.69 10 m K W = + + + + =

With a heat flux of

,o ,i 5 2w 3 2tot,w

T T 1300Kq 3.52 10 W m

R 3.69 10 m K W

= = =

the inner and outer surface temperatures of the Inconel are

( )s,i(w) ,i w iT T q h = + ( )5 2 2400 K 3.52 10 W m 500 W m K 1104 K= + =

( ) ( )3 4 2 5 2400 K 2 10 2 10 m K W 3.52 10 W m 1174 K = + + =( ) ( )s,o(w) ,i i wInT T 1 h L k q = + +

i h h C


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Without the TBC,

( )1 1 3 2

tot, wo o iInR h L k h 3.20 10 m K W

= + + =

( )wo ,o ,i tot,woq T T R = = 4.06105W/m2( )wo ,o ,i tot,woq T T R = = 4.06105W/m2

The inner and outer surface temperatures of the Inconel are then

( )s,i(wo) ,i wo iT T q h 1212 K = + =

( ) ( )[ ]s,o(wo) ,i i woInT T 1 h L k q 1293 K = + + =

Use of the TBC facilitates operation of the Inconel below Tmax= 1250 K.

COMMENTS: Since the durability of the TBC decreases with increasingtemperature, which increases with increasing thickness, limits to its thickness areassociated with reliability considerations.

Problem 3.62: Suitability of a composite spherical shell for storing


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y p p gradioactive wastes in oceanic waters.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions,(3) Constant properties at 300K, (4) Negligible contact resistance.

PROPERTIES: Table A-1, Lead: k = 35.3 W/mK, MP = 601K; St.St.: 15.1

W/mK.

ANALYSIS: From the thermal circuit, it follows that

311

tot

T T 4q= q r

R 3 =

&

The thermal resistances are:


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( )Pb1 1

R 1/ 4 35.3 W/m K 0.00150 K/W

0.25m 0.30m

= =

( )St.St.1 1

R 1/ 4 15.1 W/m K 0.000567 K/W0.30m 0.31m

= =

( )2 2 2convR 1/ 4 0.31 m 500 W/m K 0.00166 K/W = =

totR 0.00372 K/W.=

The heat rate is then

( )( )35 3q=5 10 W/m 4 / 3 0.25m 32,725 W =

and the inner surface temperature is

( )1 totT T R q=283K+0.00372K/W 32,725 W= + 405 K < MP = 601K.=

Hence, from the thermal standpoint, the proposal is adequate.

COMMENTS: In fabrication, attention should be given to maintaining a good

thermal contact. A protective outer coating should be applied to prevent long

term corrosion of the stainless steel.


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OneOne--Dimensional, SteadyDimensional, Steady--StateStateConduction withConduction with

Thermal Energy GenerationThermal Energy Generation

Implications of Energy Generation


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p gy

Involves a local (volumetric) source of thermal energy due to conversionfrom another form of energy in a conducting medium.

The source may be uniformly distributed, as in the conversion fromelectrical to thermal energy (Ohmic heating):

or it may be non-uniformly distributed, as in the absorption of radiationpassing through a semi-transparent medium.

Generation affects the temperature distribution in the medium and causesthe heat rate to vary with location, thereby precluding inclusion ofthe medium in a thermal circuit.

For a plane wall,

V

RI

V

Eq

g2

==&

&

( )xq exp&

The Plane Wall


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Consider one-dimensional, steady-state conductionin a plane wall ofconstant k, uniform generation,and asymmetric surface conditions:

Heat Equation:

Is the heat flux independent of x?q

General Solution:

What is the form of the temperature distribution for

0?q=

> 0?q

< 0?q

How does the temperature distribution change with increasing ?q

2

20 0

d dT d T qk q

dx dx dx k

+ = + =

(3.39)2

20 0

d dT d T qk q

dx dx dx k

+ = + =

(3.39)

( ) 2 1 2/ 2T x q k x C x C = + +

Symmetric Surface Conditions or One Surface Insulated:


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What is the temperature gradientat the centerline or the insulatedsurface?

Why does the magnitude of the temperaturegradient increase with increasing x?

Temperature Distribution:

Overall energy balance on the wall

How do we determine the heat rate atx = L?

How do we determine ?sT

( )2 2

21

2 s

q L xT x T

k L

= +

(3.42)( )2 2

21

2 s

q L xT x T

k L

= +

(3.42)

0out gE E + =

( ) 0s s s

s

hA T T q A L

q LT T

h

+ =

= +

(3.46)

( ) 0s s s

s

hA T T q A L

q LT T

h

+ =

= +

(3.46)

Radial SystemsCylindrical (Tube) Wall Spherical Wall (Shell)


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Cylindrical (Tube) Wall Spherical Wall (Shell)

Solid Cylinder (Circular Rod) Solid Sphere

Heat Equations:

Cylindrical

10

d dTkr q

r dr dr

+ =

Spherical

2

2

10

d dTkr q

r dr dr

+ =

Heat Equations:

Cylindrical

10

d dTkr q

r dr dr

+ =

Spherical

2

2

10

d dTkr q

r dr dr

+ =

Solution for Uniform Generation in a Solid Sphere of Constant kwith Convection Cooling:


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Temperature Distribution Surface Temperature

Overall energy balance:

Or from a surface energy balance:

with Convection Cooling:

A summary of temperature distributions is provided in Appendix Cfor plane, cylindrical and spherical walls, as well as for solidcylinders and spheres. Note how boundary conditions are specifiedand how they are used to obtain surface temperatures.

3

2

13

dT q r kr C

dr= +

2

1

26

Cq rT Ck r

= +

0 10 0rdT

Cdr

= = =|

( )

2

26

o

o s s

q r

T r T C T k= = +

( )2 2

21

6

o

s

o

q r rT r T

k r

= +

0out gE E + =

3o

s q rT Th

= +

0in out E E =

( )cond o convq r q =3

o

s

q rT T

h

= +

Problem 3.91 Thermal conditions in a gas-cooled nuclear reactor

with a tubular thorium fuel rod and a concentric

Problem 3.91 Thermal conditions in a gas-cooled nuclear reactor



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graphite sheath: (a) Assessment of thermal integrity

for a generation rate of . (b) Evaluation of

temperature distributions in the thorium and graphite

for generation rates in the range .

8 310 W/mq=

8 810 5x10q


graphite sheath: (a) Assessment of thermal integrity

for a generation rate of . (b) Evaluation of

temperature distributions in the thorium and graphite

for generation rates in the range .

8 310 W/mq=

8 810 5x10q

Schematic:Schematic:

Assumptions: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties, (4) Negligible contact resistance, (5) Negligible radiation, (6) Adiabatic surface at r1.

Properties: Table A.1, Thorium: 2000 ; Table A.2, Graphite: 2300 .mp mpT K T K Properties: Table A.1, Thorium: 2000 ; Table A.2, Graphite: 2300 .mp mpT K T K

Analysis: (a) The outer surface temperature of the fuel, T2, may be determined from the rate equation


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2

tot

T TqR

=

where( )3 2

3

1n / 10.0185 m K/W

2 2tot

g

r rR

k r h = + =

The heat rate may be determined by applying an energy balance to a control surface about the fuelelement,

out gE E=

or, per unit length,out gE E =

Since the interior surface of the element is essentially adiabatic, it follows that

Hence,

With zero heat flux at the inner surface of the fuel element, Eq. C.14 yields

( )2 22 1 17,907 W/mq q r r = =

( )2 17,907 W/m 0.0185 m K/W 600 931totT q R T K K = + = + =

2

2 2 2

2 1 1

1 2 2

2 1

1 1n 931 25 18 938 2.65): 0mL L =

Fin Heat Rate:

( )0| ff c x A sd

q kA h x dA

dx

== =

Condio de Distribuio de temperaturas Taxa de transmisso de


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CasoCondio de

fronteira emx=LDistribuio de temperaturas

/ bTaxa de transmisso de

calor

(i) ( )Lhxd

dk

Lx

=

=

( )[ ] ( )[ ]

( ) ( )Lmkm

hLm

xLmkm

hxLm

sinhcosh

sinhcosh

+

+

( ) ( )

( ) ( )Lmkm

hLm

Lmkm

hLm

M

sinhcosh

coshsinh

+

+

(ii) 0=

=Lxxd

d ( )[ ]( )Lm

xLm

cosh

cosh ( )LmMtanh

(iii) ( ) LL = ( ) ( ) ( )[ ]

( )Lm

xLmxmbL

sinh

sinhsinh + ( )( )Lm

LmM bL

sinh

/cosh

(iv) ( ) 0=L xme M

cAkPhm =2 bc AkPhM=

Fin Performance Parameters Fin Efficiency:


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,max

f ff

f f b

q qq hA

=

How is the efficiency affected by the thermal conductivity of the fin?

Expressions for are provided in Table 3.5 for common geometries.f

( )1/ 2222 / 2fA w L t = +

( )/ 2pA t L=

( )

( )1

0

21

2fI mL

mL I mL =

Fin Effectiveness:

Consider a triangular fin:

,

f

f c b b

q

hA

Fin Resistance:

with , and / f ch k A P

,

1bt f

f f f

Rq hA

=

Correction of fin length to account for heat loss from the tip


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extremidadeisolada

Transmisso de calorna extremidade

( ) ( ) ( )LLLPhLAhq cctipf =,

P

ALL cc +=

Fin of rectangular cross section with t


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0.0 1.0 2.0 3.0 4.0 5.0 0.0

0.2

0.4

0.6

0.8

1.0

1.8 2

4

31.6

1.4

ri

L

ro

t

1i

o

r

rf

0.0 1.0 2.0 3.0 4.0 5.0

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

(a)

(b)

(c)

(d)

(e)

f

t

x

y (x)

Fin Arrays Representative arrays of


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(a) rectangular and(b) annular fins.

Total surface area:t f bA NA A= +

Number of fins Area of exposed base (prime surface)

Total heat rate:

,

bt f f b b b o t b

t o

q N hA hA hAR

= + =

Overall surface efficiency and resistance:

,

1bt o

t o t

R

q hA

= =

( )1 1fo ft

NA

A =

Equivalent Thermal Circuit :


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Effect of Surface Contact Resistance:

( )( ),

bt t bo c

t o c

q hAR

= =

( )1

1 1f fo c

t

NA

A C

=

( )1 , ,1 /f f t c c bC hA R A = +

( )( )

,

1t o c

to c

RhA

=

Problem 3.116: Assessment of cooling scheme for gas turbine blade.Determination of whether blade temperatures are lessthan the maximum allowable value (1050 C) for


83/357

prescribed operating conditions and evaluation of bladecooling rate.

Schematic:

Assumptions: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3)

Adiabatic blade tip, (4) Negligible radiation.

Analysis: Conditions in the blade are determined by Case B of Table 3.4.

(a) With the maximum temperature existing at x=L, Eq. 3.75 yields

( )

b

T L T 1

T T cosh mL

=

( ) ( )1/ 21/ 2 2 4 2

cm hP/kA 250W/m K 0.11m/20W/m K 6 10 m= = = 47.87 m-1( ) ( )

1/ 21/ 2 2 4 2cm hP/kA 250W/m K 0.11m/20W/m K 6 10 m

= = = 47.87 m-1

mL = 47.87 m-10.05 m = 2.39


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Problem 3.132: Determination of maximum allowable power for a 20mm

x 20mm electronic chip whose temperature is not to exceed

when the chip is attached to an air-cooled heat sink

cq

85 C,cT = o

Problem 3.132: Determination of maximum allowable power for a 20mm

x 20mm electronic chip whose temperature is not to exceed

when the chip is attached to an air-cooled heat sink

cq

85 C,cT = o


85/357

with N=11 fins of prescribed dimensions.with N=11 fins of prescribed dimensions.

Schematic:

Assumptions: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal chip, (4)Negligible heat transfer from top surface of chip, (5) Negligible temperature rise for air flow,(6) Uniform convection coefficient associated with air flow through channels and over outersurface of heat sink, (7) Negligible radiation.


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Comments: The heat sink significantly increases the allowable heat dissipation. If it

were not used and heat was simply transferred by convection from the surface of the chip withfrom Part (a) would be replaced by2100 W/m , 2.05 K/Wtoth K R= =

21/hW 25 K/W, yielding 2.60 W.cnv cR q= = =


88/357

Transient Conduction:Transient Conduction:The Lumped CapacitanceThe Lumped Capacitance

MethodMethod

Transient Conduction A heat transfer process for which the temperature varies with time, as well


89/357

as location within a solid.

It is initiated whenever a system experiences a change in operating conditionsand proceeds until a new steady state (thermal equilibrium) is achieved.

It can be induced by changes in: surface convection conditions ( ),,h T

Solution Techniques The Lumped Capacitance Method Exact Solutions The Finite-Difference Method (not to be studied)

surface radiation conditions ( ),,r sur

h T

a surface temperature or heat flux, and/or

internal energy generation.

The Lumped Capacitance Method

B d h i f i ll if di ib i


90/357

Based on the assumption of a spatially uniform temperature distributionthroughout the transient process.

Why is the assumption never fully realized in practice?

General Lumped CapacitanceAnalysis:

Consider a general case,which includes convection,radiation and/or an appliedheat flux at specifiedsurfaces

as well as internal energygeneration

( ), , ,, , ,s c s r s hA A A

)t(T)t,r(T r

First Law:

st EEETdCVEd &&&


91/357

Assuming energy outflow due to convection and radiation and withinflow due to an applied heat flux ,sq

Is this expression applicable in situations for which convection and/orradiation provide for energy inflow?

May h and hrbe assumed to be constant throughout the transient process?

How must such an equation be solved?

gsurr,sr,sc,sh,s

''

h,s E)TT(Ah)TT(hAAqtd

TdCV &+=

goutinst EEE

tdTdCV

dtEd +==

Special Cases (Exact Solutions, )( )0 iT T

N li ibl R di ti ( )/T T b a


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Negligible Radiation ( ), / :T T b a

The non-homogeneous differential equation is transformed into ahomogeneous equation of the form:

da

dt

=

Integrating from t=0 to anytand rearranging,

( ) ( )/

exp 1 expi i

T T b aat at

T T T T

= +

To what does the foregoing equation reduce as steady state is approached?

How else may the steady-state solution be obtained?

CV

Aha

cs

,=

CV

EAqb

ghs

&+= ,

''

Negligible Radiation and Source Terms , 0, 0 :gr sh h E q >> = =

( ),s cdT

c hA T T dt

=

td


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, is c

t

o

c d

hAdt

=

,s c

i i

hAT Texp t

T T c

= =

t

t

=

exp

The thermal time constant is defined as

( )

,

1t

s c

c

hA

ThermalResistance, Rt

Lumped ThermalCapacitance, Ct

The change in thermal energy storage due to the transient process ist

outst

o

E Q E dt =

,

t

s c

o

hA dt = ( ) 1 expit

tc

=

(5.8)

Negligible Convection and Source Terms , 0, 0 :gr sh h E q >> = =


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Assuming radiation exchange with large surroundings,

( )4 4,s r sur dT

c A T T dt

=

,

4 4i

s r T

surTo

tA

c

dT

T Tdt

=

3,

1n 1n4

sur sur i

s r sur sur sur i

T T T T ct

A T T T T T

+ + =

Result necessitates implicit evaluation of T(t).

1 12 tan tan i

sur sur

TT

T T

+

The Biot Number and Validity ofThe Lumped Capacitance Method

The Biot Number: The first of many dimensionless parameters to be


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The Biot Number: The first of many dimensionless parameters to beconsidered.

Definition:chLBi

k

convection or radiation coefficienth

thermal conductivity of t so e dh lik

of the solid ( / or coordinate

associated with maximum spa

char

tial temperature differe

acteristic lengt

e

h

nc )

c sL A

Physical Interpretation:

Criterion for Applicability of Lumped Capacitance Method:

1Bi


96/357


97/357

Problem 5.15: Heating of coated furnace wall during start-up.


98/357

KNOWN: Thickness and properties of furnace wall. Thermal resistance of ceramic coatingon surface of wall exposed to furnace gases. Initialwall temperature.

FIND: (a) Time required for surface of wall to reach a prescribed temperature, (b)Corresponding value of coating surface temperature.

ASSUMPTIONS: (1) Constant properties, (2) Negligible coating thermal capacitance, (3)Negligible radiation.

PROPERTIES: Carbon steel: = 7850 kg/m3, c = 430 J/kgK, k = 60 W/mK.

ANALYSIS: Heat transfer to the wall is determined by the total resistance to heat transferfrom the gas to the surface of the steel, and not simply by the convection resistance.

Hence, with ( )11

1 2 2 2tot f 2

1 1U R R 10 m K/W 20 W/m K.

h 25 W/m K

= = + = + =

2UL 20 W/m K 0.01 mBi 0 0033 1


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Bi 0.0033 1k 60 W/m K

= = =


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Transient Conduction:Transient Conduction:

Spatial Effects and the Role ofSpatial Effects and the Role ofAnalytical SolutionsAnalytical Solutions

Solution to the Heat Equation for a Plane Wall withSymmetrical Convection Conditions

If the lumped capacitance approximation can not be made, consideration must

be given to spatial, as well as temporal, variations in temperature during the


101/357

g p , p , p gtransient process.

For a plane wall with symmetrical convectionconditions and constant properties, the heat

equation and initial/boundary conditions are:2

2

1T T

x t

=

( ),0 iT x T=

0

0x

T

x =

=

( ),

x L

Tk h T L t T

x

=

=

Existence of seven independent variables:

( ), , , , , ,iT T x t T T k h=

How may the functional dependence be simplified?


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The One-Term Approximation :( )0.2Fo>

Variation of midplane temperature (x*= 0) with time :( )Fo

( )( )

( )* 21 1expo

oT T C FoT T


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( )( )1 1po

iT T

1 1Table 5.1 and as a function ofC Bi

( )Fo

Variation of temperature with location (x*

) and time :( )* * *1coso =

Change in thermal energy storage with time:

stE Q =

1 *

1

sin1o oQ Q

=

( )o iQ c T T =

Can the foregoing results be used for a plane wall that is well insulated on oneside and convectively heated or cooled on the other?

Can the foregoing results be used if an isothermal condition isinstantaneously imposed on both surfaces of a plane wall or on one surface ofa wall whose other surface is well insulated?

( )s iT T

1 008980 299101 007460 244031 004950 172340 031.005990.244461.004980.199501.003310.140950.02

1.003000.173031.002500.141241.001660.099830.01

c11c11c11

EsferaCilindro longoPlaca plana

Bi


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-------------------------------------------------

1.130301.111181.103810.897831.064190.624440.45

1.116351.052791.093140.851581.058040.593240.40

1.102260.989661.082260.801401.051660.559220.351.088020.920791.071160.746461.045050.521790.30

1.073650.844731.059840.685591.038190.480090.25

1.059150.759311.048300.616971.031090.432840.20

1.044530.660861.036550.537611.023720.377880.151.029800.542281.024580.441681.016090.311050.10

1.026840.514971.022160.419541.014540.295570.09

1.023870.486001.019730.396031.012970.279130.08

1.020900.455061.017290.370921.011380.261530.071.017930.421731.014850.343831.009790.242530.06

1.014950.385371.012400.314261.008190.221760.05

1.011970.345031.009930.281431.006570.198680.04

1.008980.299101.007460.244031.004950.172340.03

Graphical Representation of the One-Term ApproximationThe Heisler Charts Plane wall

Midplane Temperature:


105/357


106/357

Radial Systems

Long Rods Heated or Cooled by Convection.


107/357

2

/

/o

o

Bi hr k

Fo t r

=

=

( ) ( )( ) ( )

=

=

==

1

2 Foexp*,,

n

nnonii

*rJc

TT

TtrTtr

( )( ) ( )nno

n

nn

JJ

J

c

21

212

+=

(5.184a)

onn r =

Long rod:

orrr =*

Radial Systems

Long rod one term approximation (Fo > 0.2):


108/357

(5.184a)

( ) *o

o

J

Q

Q

1

1121 =


stE Q =

( )o iQ c T T =

(Fo > 0.2)

( ) ( ) ( )*exp* 12111 rJForJcTT

TT o

*oo

i

* =

=

( )FocTT

TT

i

o*o

211 exp=

=

Graphical Representation of the One-Term ApproximationThe Heisler Charts Infinite cylinder

Centerline Temperature:


109/357



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Change in Thermal Energy Storage:


111/357

Spherical Systems

Sphere one term approximation (Fo > 0.2):


112/357

(5.184a)


stE Q =

( )o iQ c T T =

(Fo > 0.2)

( )FocTT

TT

i

o*o

211 exp=

=

( ) ( ) ( )*

*sinFoexp

*

*sin

1

121

1

11

r

r

r

rc

TT

TT *o

i

* =

=

( )11131

cossin3

1

Q

Q *

o

o

=

Graphical Representation of the One-Term ApproximationThe Heisler Charts Sphere

Center Temperature:


113/357



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Change in Thermal Energy Storage:


115/357

( )

( )1

2 22 /

, exp 4o

i

q t x

T x t T k t

q x x

=

Case 2: Uniform Heat Flux( )s oq q =


116/357

erfc2

oq x x

k t

(5.59)

( )

0

0,

x

Tk h T T t

x

=

=

( )

2

2

,

2

2

i

i

T x t T xerfc

T T t

hx h t x h t exp erfck k kt

=

+ + (5.60)

Case 3: Convection Heat Transfer ( ),h T


117/357

Multidimensional Effects Solutions for multidimensional transient conduction can often be expressed

as a product of related one-dimensional solutions for a plane wall, P(x,t),

an infinite cylinder, C(r,t), and/or a semi-infinite solid, S(x,t). See Equations(5.64) to (5.66) and Fig. 5.11.


118/357

Consider superposition of solutions for two-dimensional conduction in ashort cylinder:

( )( ) ( )

( ) ( )

, ,, ,

,

i

Plane Infinitei iWall Cylinder

T r x t T

P x t x C r t T T

T x t T T r,t T x

T T T T

=

=


119/357

( )

( )

( ) ( )

( )( )

( ) ( )( )

( )

bi

o

oa

i

o

o

bi

ai

bai

TTTtT

TtTTy,tT

TTTtT

TtTTx,tT

TT

Ty,tT

TT

Tx,tT

TT

Tx,y,tT

2espessuradeinfinitaplaca




22rrectangulasecodebarra

=

=

bespessuradeplanaplacao

aespessuradeplanaplacao

bespessuradeplanaplacao

aespessuradeplanaplacao

barrectangulaodebarrao

QQ

QQ

Q

Q

Q

Q

Q

Q

22

2222sec

+

=


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( ) ( )

=

TT

TtxTtxS

i

,, ( )

( )

=

TT

TtxTtxP

i

,, ( )

( )

=

TT

TtrTtrC

i

,,


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Problem 5.66: Charging a thermal energy storage system consisting ofa packed bed of Pyrex spheres.


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KNOWN: Diameter, density, specific heat and thermal conductivity of Pyrexspheres in packed bed thermal energy storage system. Convection coefficient andinlet gas temperature.

FIND: Time required for sphere to acquire 90% of maximum possible thermalenergy and the corresponding center and surface temperatures.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction in sphere, (2)Negligible heat transfer to or from a sphere by radiation or conduction due tocontact with adjoining spheres, (3) Constant properties.

ANALYSIS:With Bi h(ro/3)/k = 75 W/m2K (0.0125m)/1.4 W/mK = 0.67,the lumped capacitance method is inappropriate and the approximate (one-term)solution for one-dimensional transient conduction in a sphere is used to obtain thedesired results


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desired results.

To obtain the required time, the specified charging requirement

( )/ 0.9oQ Q = must first be used to obtain the dimensionless center temperature,*.o

From Eq. (5.52),

( ) ( )

3

1oo1 1 1

Q1Q3 sin cos

=

With Bi hro/k = 2.01, 1 2.03 and C11.48 from Table 5.1. Hence,

( )( )

3

o 0.1 2.03 0.837 0.1555.3863 0.896 2.03 0.443 = = =

From Eq. (5.50c), the corresponding time is2o o

211

rt ln

C

=

( )3 7 2k / c 1.4 W / m K / 2225 kg / m 835J / kg K 7.54 10 m / s, = = =

( ) ( )2

0.0375m ln 0.155/1.48t 1 020s


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( )27 2

t 1,020s7.54 10 m /s 2.03

= =

From the definition of *,o the center temperature is

( )o g,i i g,iT T 0.155 T T 300 C 42.7 C 257.3 C= + = =

The surface temperature at the time of interest may be obtained from Eq. (5.50b)

with r 1,

=

( ) ( )o 1

s g,i i g,i1

sin 0.155 0.896T T T T 300 C 275 C 280.9 C

2.03

= + = =

Is use of the one-term approximation appropriate?


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Both requirements are met.

Is the assumption of a semi-infinite solid for a plane wall of finite thicknessappropriate under the foregoing conditions?


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COMMENTS: The foregoing analysis may or may not be conservative, since

heat transfer at the irradiated surface due to convection and net radiationexchange with the environment has been neglected. If the emissivity of thesurface and the temperature of the surroundings are assumed to be = 1 and Tsur= 298K, radiation exchange at Ts= 309.5C would be

( )4 4 2

rad s surq T T 6,080 W / m K, = = which is significant (~ 60% of the prescribed radiation). However, under actualconditions, the wall would likely be exposed to combustion gases and adjoiningwalls at elevated temperatures.

5.89

Um cilindro de cobre, com 100 mm de comprimento e 50 mm de dimetroencontra-se inicialmente temperatura uniforme de 20C.

As duas bases so aquecidas muito rapidamente, a partir de um determinado

instante, ficando temperatura de 500 C, enquanto a superfcie lateral docilindro aquecida por uma corrente de gs a 500 C e com um coeficiente deconveco de 100 W/m2K.


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/

a) Determinar a temperatura do centro do cilindro ao fim de 8 segundos.

b) Atendendo aos parmetros adimensionais que determinam a distribuio detemperaturas nos problemas de difuso transiente do calor, possvel admitirhipteses simplificativas na anlise deste problema?

Apresente uma explicao resumida.

Propriedades do cobre

CILINDRO CURTO: 2DPROPRIEDADES CONSTANTES h CONSTANTE

PARA O CILINDRO INFINITO C(r,t):


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PARA O PLACA PLANA INFINITA P(x,t):

PARA O CILINDRO CURTO:

PARMETROS ADIMENSIONAIS QUE CONTROLAM A CONDUO TRANSIENTE:Fourier e Biot.


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NO CASO DO CILINDRO Bi < 0,1 DESPREZAM-SE GRADIENTESRADIAIS

5.90

Considerando que a carne fica cozida quando atinge uma temperatura de 80C,calcule o tempo necessrio para assar uma pea de carne com 2,25 kg.

Admitir que a pea de carne um cilindro com dimetro igual ao comprimento eque as suas propriedades so equivalentes s de gua lquida.Considere que a carne se encontra inicialmente temperatura de 6C e que atemperatura do forno 175C e o coeficiente de conveco de 15 W/m2K


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temperatura do forno 175C e o coeficiente de conveco de 15 W/m2K.

Propriedades da gua:

CLCULO DAS DIMENSES DO CILINDRO:


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CLCULO DA TEMPERATURA NO CENTRO DO CILINDRO:

SOLUO TENTATIVA-ERRO:


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I t d ti t C tiIntroduction to Convection:


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Introduction to Convection:Introduction to Convection:

Flow and Thermal ConsiderationsFlow and Thermal Considerations

Boundary Layers: Physical Features Velocity Boundary Layer

A consequence of viscous effectsassociated with relative motionbetween a fluid and a surface.


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A region of the flow characterized by

shear stresses and velocity gradients. A region between the surface

and the free stream whosethickness increases in

the flow direction.

( )0.99

u y

u

=

Why does increase in the flow direction?

Manifested by a surface shearstress that provides a drag

force, .

s

DF

0s y

u

y =

=

s

D s s

A

F dA=

How does vary in the flowdirection? Why?

s

2

2

1

=u

C sf

Thermal Boundary Layer

A consequence of heat transfer

between the surface and fluid.

A region of the flow characterizedby temperature gradients and heat


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by temperature gradients and heatfluxes.

A region between the surface andthe free stream whose thicknessincreases in the flow direction.

t

Why does increase in theflow direction?

t

Manifested by a surface heatflux and a convection heattransfer coefficient h .

sq

( )0.99st

s

T T y

T T

=

0s f y

T

q k y =

=

0/f y

s

k T yh

T T

=

If is constant, how do andh vary in the flow direction?

( )sT T sq


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Governing equationsGoverning equations

( ) ( ) qpupTkhuh jj ++

+

+

=

+

& Entalpia especfica

peh +=


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( ) ( ) qx

utx

kx

hux

ht j

jjj

jj

++

+

+

=

+

p p

( ) ( ) qx

pu

t

pT

x

Tk

xTu

xcT

tc

jj

jjj

jpp ++

+

+

=

+

& Temperatura

Coeficiente de expanso trmica:pT

=

1

Gs perfeito: = 1/T

Fluido incompressvel:= 0

( ) dpTdTcdh p

1

1 +=

The Boundary Layer Equations


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Consider concurrent velocity and thermal boundary layer development for steady,two-dimensional, incompressible flow with constant fluid properties andnegligible body forces.

( ),, pc k

Apply conservation of mass, Newtons 2nd Law of Motion and conservation of energy

to a differential control volume and invoke the boundary layer approximations.Velocity Boundary Layer:

Thermal Boundary Layer:

T T

y x

>>

, ,

u v

u u v vy x y x

>>

>>

Conservation of Mass:

0u v

x y

+ =

In the context of flow through a differential control volume, what is the physicalsignificance of the foregoing terms, if each is multiplied by the mass density ofthe fluid?


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Newtons Second Law of Motion:

2

2

x-direction :

u u dp uu v

u dx y

+ = +

What is the physical significance of each term in the foregoing equation?

Why can we express the pressure gradient as dp/dx instead of / ?p x

y-direction :

0py

=

What is the physical significance of each term in the foregoing equation?

Conservation of Energy:

22

2p T T T uc u v k x y y y

+ = +


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p y g g g q

What is the second term on the right-hand side called and under what conditionsmay it be neglected?

Boundary Layer Similarity As applied to the boundary layers, the principle ofsimilitude is based on

determining similarity parameters that facilitate application of results obtained

for a surface experiencing one set of conditions to geometrically similar surfacesexperiencing different conditions. (Recall how introduction of the similarityparametersBi and Fo permitted generalization of results for transient, one-dimensional condition).


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Dependent boundary layer variables of interest are:

and ors q h

For a prescribed geometry, the corresponding independent variables are:

Geometrical: Size (L), Location (x,y)Hydrodynamic: Velocity (V)Fluid Properties:

Hydrodynamic: ,

Thermal : ,pc k

( )

( )

Hence,

, , , , ,

, , , ,s

u f x y L V

f x L V

=

=

( )

( )

and

, , , , , , ,

, , , , , ,

p

p

T f x y L V c k

h f x L V c k

=

=

Key similarity parameters may be inferred by non-dimensionalizing the momentumand energy equations.


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Recast the boundary layer equations by introducing dimensionless forms of the

independent and dependent variables. * *

* *

* s

s

x yx y

L L

u vu v

V V

T TTT T

Neglecting viscous dissipation, the following normalized forms of the x-momentum

and energy equations are obtained: * * * 2 ** ** * * *2

* * 2 ** *

* * *2

1

Re

1

Re Pr

L

L

u u dp uu v

x y dx y

T T Tu v

x y y

+ = +

+ =

Reynolds NumbeRe the

Pr

r

Prandtl Numberthe

L

p

VL VL

v

c v

k

=

=

For a prescribed geometry,

How may the Reynolds and Prandtl numbers be interpreted physically? 0Pr > nnt


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( )* * *, ,ReLu f x y=

*

*

*0 0

s

y y

u V u

y L y

= =

= =

The dimensionless shear stress, or local friction coefficient, is then

*

*

2 *0

2/ 2 Re

sf

L y

uC

V y

=

=

( )*

**

*0

,ReLy

uf x

y=

=

( )*2

,ReRef LL

C f x=

What is the functional dependence of the average friction coefficient,Cf?

For a prescribed geometry,

( )* * *

, ,Re ,PrLT f x y=

( )

( ) *

* *0

* *

/f y f fsk T y k kT T T T

hT T L T T y L y

=

= = = +


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( ) ** 00s s yyT T L T T y L y ==

The dimensionless local convection coefficient is then

( )*

**

*0

, Re , PrLf y

hL TNu f x

k y =

= =

local Nusselt numberNu

What is the functional dependence of the average Nusselt number?

How does the Nusselt number differ from the Biot number?


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What may be said about transition if ReL < Rex,c? If ReL > Rex,c?

Effect of transition on boundary layer thickness and local convection coefficient:


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Why does it increase

significantly with transition to turbulence, despite the increase in the boundary layerthickness?

Why does transition provide a significant increase in the boundary layer thickness?

Why does the convection coefficient decay in the laminar region?

Why does the convection coefficient decay in the turbulent region?


149/357


150/357

Problem 6.28: Determination of heat transfer rate for prescribedturbine blade operating conditions from wind tunnel dataobtained for a geometrically similar but smallerblade. The blade surface area may be assumed to be

directly proportional to its characteristic length .( )sA L

SCHEMATIC:


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ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Surface area A isdirectly proportional to characteristic length L, (4) Negligible radiation, (5) Blade shapes are

geometrically similar.

ANALYSIS: For a prescribed geometry,

( )LhL

Nu f Re , Pr .k

= =

The Reynolds numbers for the blades are

( ) ( )2 2L,1 1 1 1 1 L,2 2 2 2 2Re V L / 15m / s Re V L / 15m / s . = = = =

Hence, with constant properties ( )1 2v v= , L,1 L,2Re Re .= Also, 1 2Pr Pr=Therefore,

2 1Nu Nu=

( ) ( )2 2 2 1 1 1h L / k h L / k=


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( ) ( )2 2 2 1 1 1h L / k h L / k

( )1 1 1

2 12 2 1 s,1

L L q

h hL L A T T= =

The heat rate for the second bladeis then

( )

( )

( )

s,21 2

2 2 2 s,2 12 1 s,1

T TL Aq h A T T q

L A T T

= =

( )

( )( )

s,22 1

s,1

T T 400 35q q 1500 W

T T 300 35

= =

2q 2066 W.=

COMMENTS: (i) The variation infrom Case 1 to Case 2 would cause ReL,2to differ fromReL,1. However, for air and the prescribed temperatures, this non-constant property effect issmall. (ii) If the Reynolds numbers were not equal ( ),1 2Re Re ,L L knowledge of the specific form of

( ),Re PrLf would be needed to determine h2.

Problem 6.35: Use of a local Nusselt number correlation to estimate thesurface temperature of a chip on a circuit board.


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KNOWN: Expression for the local heat transfer coefficient of air at prescribed velocity andtemperature flowing over electronic elements on a circuit board and heat dissipation rate for a 4 4 mmchip located 120mm from the leading edge.

FIND: Surface temperature of the chip surface, Ts.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Power dissipated within chip is lost by convectionacross the upper surface only, (3) Chip surface is isothermal, (4) The average heat transfer coefficientfor the chip surface is equivalent to the local value at x = L.

PROPERTIES: Table A-4, Air (Evaluate properties at the average temperatureof air in the boundarylayer. Assuming Ts= 45C, Tave = (45 + 25)/2 = 35C = 308K. Also, p = 1atm):= 16.69

10-6

m2/s, k = 26.9 10

-3W/mK, Pr = 0.703.


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ANALYSIS: From an energy balance on the chip,

conv gq E 30mW.= =&

Newtons law of cooling for the upper chip surface can be written as

s conv chipT T q / h A= + (2)

where 2chipA .=l

Assuming that the averageheat transfer coefficient ( )h over the chip surface is equivalent to the local

coefficient evaluated at x = L, that is, ( )chip xh h L ,the local coefficient can be evaluated by

applying the prescribed correlation at x = L.0.85

1/ 3xx

h x Vx

Nu 0.04 Prk

= =

0.851/ 3

Lk VL

h 0.04 PrL

=

( )

0.851/ 3 2

L -6 20.0269 W/m K 10 m/s 0.120 m

h 0.04 0.703 107 W/m K.0.120 m 16.69 10 m / s

= =

From Eq. (2), the surface temperature of the chip is

( )2-3 2

sT 25 C 30 10 W/107 W/m K 0.004m 42.5 C.= + =o o

COMMENTS (1) Th ti t d l f T d t l t th i ti i bl


155/357

COMMENTS: (1) The estimated value of Taveused to evaluate the air properties is reasonable.

(2) How else could chiph have been evaluated? Is the assumption of Lh h= reasonable?

External Flow:External Flow:


156/357

The Flat Plate in Parallel FlowThe Flat Plate in Parallel Flow


157/357


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Similarity Solution for Laminar,Constant-Property Flow over an Isothermal Plate

Based on premise that the dimensionless x-velocity component, ,and temperature, , can be represented exclusively interms of a dimensionless similarity parameter

/u u

( ) ( )* /s sT T T T T

( )1/ 2

/y u x


159/357

Similarity permits transformation of the partial differential equations associatedwith the transfer of x-momentum and thermal energy to ordinary differentialequations of the form

3 2

3 22 0

d f d f

fd d + =

( )where / / , u u df d and

2 * *

2 Pr 02+ =d T dT fd d

Similarity Solution for Laminar,Constant-Property Flow over an Isothermal Plate


160/357

Subject to prescribed boundary conditions, numerical solutions to the momentumand energy equations yield the following results for important local boundary layerparameters:

( )1/ 2

- with / 0.99 at 55.0 5

R

. ,

e/

0

x

x

u

u

vx

u

= == =

2

2- with /su d f

u u vx

= =


161/357

2

0 0

s

yy d

= =2 2

0and / 0.332,d f d

= =

, 1/ 2, 2 0.664Re

/ 2 xs x

f xC

u

=

( ) ( )1/ 2* *

0 0- with / / / / x s s

yh q T T k T y k u vx dT d

= =

= = =

* 1/ 3

0and / 0.332 Pr for Pr 0.6,dT d

= = >

1/ 3

r

and

Pt

=

1/ 2 1/ 30.332 Re Prxx xh x

Nuk

= =

How would you characterize relative laminar velocity and thermal boundary layergrowth for a gas? An oil? A liquid metal?

How do the local shear stress and convection coefficient vary with distance fromthe leading edge?

Average Boundary Layer Parameters:

1 x


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, 0

1 xs x sdxx

1/ 2, 1.328 Rexf xC

=

0

1 xx xh

x

h dx=

1/ 2 1/ 30.664 Re Prx xNu =

The effect of variable properties may be considered by evaluating all propertiesat the film temperature.

2s

fT TT +=

Turbulent Flow Local Parameters:

1/ 5

,4 / 5 1/ 3

0.0592 Re

0.0296 Re Prf x x

x x

C

Nu

=

=Empirical

Correlations

How do variations of the local shear stress and convection coefficient withdistance from the leading edge for turbulent flow differ from those for laminar flow?


163/357

( )4 / 5 1/ 30.037 Re 871 PrL LNu =

Average Parameters:

( )101

c

c

x L

L am turbxh h dx h dx

L= +

Substituting expressions for the local coefficients and assuming

5

x ,cRe 5 10 ,= , 1/ 5

0.074 1742

Re Ref L L LC =

( ), ,1/ 5

,

4 / 5 1/ 3

For Re 0 or Re Re ,0.074 Re

0.037 Re Pr

x c c L x c

f L L

L L

L x

C

Nu

=

=

=

Special Cases: Unheated Starting Length (USL)and/or Uniform Heat Flux


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For both uniform surface temperature (UST) and uniform surface heat flux (USF),the effect of the USL on the local Nusselt number may be represented as follows:

( )

0

1/ 30

1 /

Re Pr

x

x ba

m

x x

NuNu

x

Nu C

=

=

=

=4/54/54/54/51/21/21/21/2mm

0.03080.03080.02960.02960.4530.4530.3320.332CC

1/91/91/91/91/31/31/31/3bb9/109/109/109/103/43/43/43/4aa

USFUSFUSTUSTUSFUSFUSTUST

TurbulentTurbulentLaminarLaminar

Sketch the variation of hx versus for two conditions:What effect does an USL have on the local convection coefficient?

( )x 0 and 0. > =

UST:

( )s x sq h T T =

( ) ( )( ) ( ) ( )

2 / 2 12 1 / 2 2

0

lamina

1 /

1 for throughout

= 4 for througho

r flow

turbulent w uflo t

p pp p

L L

LNu Nu LL

p

p

++ +

= =

=

( )= L s sq h A T T


165/357

1

laminar/turbulent flownumerical integration f1

or

c

c

L

x L

L am turbx

h

h h dx h dxL

= +

USF:s

s

x

qT Th

= + s sq q A=

Treatment of Non-Constant Property Effects:

Evaluate properties at the film temperature.

2s

f

T TT

+=

Problem 7.21: Preferred orientation (corresponding to lower heat loss) and thecorresponding heat rate for a surface with adjoining smooth androughened sections.


166/357

SCHEMATIC:

ASSUMPTIONS: (1) Surface B is sufficiently rough to trip the boundary layer when in the upstream position

(Configuration 2); (2) 55 10,Re for flow over A in Configuration 1.x c

PROPERTIES: Table A-4, Air (Tf= 333K, 1 atm): = 19.2 10-6

m2/s, k = 28.7 10

-3

W/mK, Pr = 0.7.

ANALYSIS: Since Configuration (2) results in a turbulent boundary layer over the entiresurface,the lowest heat transfer is associated with Configuration (1).

Find

6L -6 2

u L 20 m/s 1mRe 1.04 10 .

19.2 10 m / s = = =

Hence in Configuration (1), transition will occur just before the rough surface (xc= 0.48m).


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( )L,14 / 56 1/3Nu 0.037 1.04 10 871 0.7 1366

= =

For Configuration (1):L,1

L,1h LNu 6

k.136= =

Hence

( )3 2L,1h 1366 28.7 10 W/m K /1m 39.2 W/m K= = 1568 W=

External Flow:External Flow:Flow over Bluff ObjectsFlow over Bluff Objects


168/357

(Cylinders, Sphere)(Cylinders, Sphere)

The Cylinder in Cross Flow

Conditions depend on special features of boundary layer development, includingonset at a stagnation point and separation, as well as transition to turbulence.


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Stagnation point: Location ofzero velocity and maximum pressure.( )0u=

Followed by boundary layer development under a favorable pressure gradientand hence acceleration of the free stream flow .( )/ 0dp dx< ( )/ 0du dx >

As the rear of the cylinder is approached, the pressure must begin to increase.Hence, there is a minimum in the pressure distribution,p(x), after which boundary

layer development occurs under the influence of an adverse pressure gradient( )/ 0, / 0 .dp dx du dx> D

x

The Local Nusselt Number:

How does the local Nusselt number vary with for ?What conditions are associated with maxima and minima in the variation?

What conditionsare associated with maxima and minima in the variation?

( )


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The Average Nusselt Number( )/ :DNu hD k

Churchill and Bernstein Correlation:

( )

4 / 55 / 81/ 2 1/ 3

1/ 42 / 3

0.62Re Pr Re0.3 1 282,0001 0.4 / Pr

D DDNu

= + + +

Cylinders of Noncircular Cross Section:

1/ 3Re PrmD DNu C=, Table 7.3C m

0.8050.0274104 41050.6180.1934103 41040.4660.68340 41030.3850.9114 40

0.3300.9890.4 4

mCReD

Flow Across Tube Banks

A common geometry fortwo-fluid heat exchangers.

Aligned and Staggered Arrays:


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max

STV VS DT= Aligned:

Staggered: ( ) ( )if 2maxSTV V S D S DD TS DT

=

( ) ( ) ( )if 2max 2

STV V S D S DD TS DD

= or,

Flow Conditions:


175/357

How do convection coefficients vary from row-to-row in an array?

How do flow conditions differ between the two configurations?

Why should an aligned array not be use for ST/SL < 0.7?

Average Nusselt Number for an Isothermal Array:

( )1/ 40.36

2 ,maxRe Pr Pr/ Prm

D D sNu C C =

2

, Table 7.7Table 7.8

C m

C

All properties are evaluated at except for Prs.( )/ 2i oT T+

Aproximado por um tubo102 103

0.400.9010 - 1020.840.0212105 - 2106

0.630.27103 - 2105 (ST/SL 2)

0.600.35 (ST/ S

L)1/5103 - 2105 (S

T/ S

L< 2)

Aproximado por um tuboisolado

102 103

Tubosdesfasados

0.990.980.970.950.920.890.840.760.64Desfasados

0.990.980.970.950.920.900.860.800.70Alinhados

161310754321NL

Fluid Outlet Temperature (To) :

s o

s i T T p

T T DNhexp

T T VN S c

=

T LN N N= x

What may be said about To as ?N

Total Heat Rate:

hA T


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s mq hA T = l

( )sA N DL=

( ) ( )s i s om

s i

s o

T T T T T

T TnT T

=

l

l

Pressure Drop:2

max

2L

Vp N f

=

, Figures 7.13 and 7.14f


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The Sphere Flow over a sphere

Boundary layer development is similar to that for flow over a cylinder,

involving transition and separation.

( ) ( )1/ 41/ 2 2 / 3 0.42 0.4Re 0.06Re Pr / D D D sNu = + +

Figure 7.8DC


179/357

Problem: 7.78 Measurement of combustion gas temperature with a sphericalthermocouple junction.

KNOWN: Velocity and temperature of combustion gases. Diameter and emissivity of thermocouple


180/357

KNOWN: Velocity and temperature of combustion gases. Diameter and emissivity of thermocoupleunction. Combustor temperature.

FIND: (a) Time to achieve 98% of maximum thermocouple temperature rise for negligible radiation, (b)Steady-state thermocouple temperature, (c) Effect of gas velocity and thermocouple emissivity on

measurement error.

SCHEMATIC:

ASSUMPTIONS: (1) Validity of lumped capacitance analysis, (2) Constant properties, (3) Negligibleconduction through lead wires, (4) Radiation exchange between small surface and a large enclosure (parts band c).

PROPERTIES: Thermocouple: 0.1 1.0, k = 100 W/mK, c = 385 J/kgK, = 8920 kg/m

3

; Gases:k = 0.05 W/mK,= 50 10-6m2/s, Pr = 0.69.

ANALYSIS: (a) If the lumped capacitance analysis may be used, it follows from Equation 5.5 that

( )i

s

T TVc D ct ln ln 50

T ThA 6h

= =

.


181/357

Neglecting the viscosity ratio correlation for variable property effects, use of V = 5 m/s with the Whitakercorrelation yields

( ) ( )1/ 2 2 / 3 0.4D D DNu hD k 2 0.4 Re 0.06 Re Pr= = + +

( ) ( )( )( )1/ 2 2 / 3 0.4 20.05W m Kh 2 0.4 100 0.06 100 0.69 328W m K0.001m = + + =

Since Bi = ( )oh r 3 k = 5.5 10-4, the lumped capacitance method may be used.

( ) ( )3

20.001m 8920 kg m 385J kg Kt ln 50 6.83s

6 328W m K= =

(b) Performing an energy balance on the junction, qconv= qrad.

Hence, evaluating radiation exchange from Equation 1.7 and with = 0.5,

( ) ( )4 4s s chA T T A T T =

( ) ( )8 2 4

44 42

0.5 5.67 10 W m K1000 T K T 400 K

328W m K

=

T = 936 K


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Parametric calculations to determine the effects of V and yield the following results:

0 5 10 15 20 25

Velocity, V(m/s)

900

950

1000

Temperature,

T(K)

Emissivity, epsilon = 0.5

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Emissivity

890

910

930

950

970

990

Temperature,

T(K)

Velocity, V = 5 m/s

Since the temperature re

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