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Mecanismos Fsicos eEquaes de Taxas deTransmisso de Calor

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 Trmica

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 agradientes 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/kgJ ouuU ou

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

+U Energia Trmicau Energia Trmica especfica

Energia trmica transferida por unidadede tempoTaxa detransferncia decalor q W

Energia trmica transferida por unidadede tempo e por unidade de rea

Fluxo de calor 'q' 2/mW

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 configuraeselectrnicas 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 ummeio material. Alis, o transporte radiativo mais eficiente no vcuo.

Modos de Transferncia de Calor


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AplicaesIdentificao 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 sq

Fluxo radiativo til entre as superfcies dos vidros que limitam o espao entre vidros,rad sq

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 uma

placa plana com condutibilidade trmica constante:

Conduo

Forma 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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Radiao

Fluxo de energia que sai devido emisso:

b s E 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 T s 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)

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

surT

sur sur G G T= =


Se = , o fluxo radiativo til a partir da superfcie

devido s trocas de calor por radiao com a envolvente :

( ) ( 4sur4

sSb

''

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

AplicaesArrefecimento de componente electrnica

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 grandeenvolvente,(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 =0.065W

0.158W+0.065W=0.223W

conv s

rad

elec

q CA T T

q

P

=

=

=

(b) Se for conveco forada,

( ) ( )( )2 -4 2h =250W/m K 2.2510 m 60K =3.375W

3.375W+0.065W=3.44W

conv s

elec

q A T T

P

=

=


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Conservao de Energia

Formulaes AlternativasBase 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 temperaturado sistema em estudo.

CONSERVAO DE ENERGIA(Primeira Lei da Termodinmica)


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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

Num instante de tempo:

Notar a representao do sistema atravs de umasuperfcie 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 &&

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 =& ( )TVctdd

Est =&

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 uniformeb) Temperaturas uniformes das regies quente e fria

c) Trocas radiativas entre corpo pequeno eenvolvente grande

d) Perdas da bolacha para o suporte desprezveis

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.

, , , , wrad 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,,w

w sur c w u w l wsur h

d TT 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 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.

Fouriers Lawand the

Heat 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

q

k

x

x

=

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

q

sin

T 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 ofC/m and not C/deg.

( )or ,

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

Cylinder

2r r r r q A q rLq = =

or,

2r r r r q A q rq = =

Sphere24r r r r q A q r q = =

The Heat Equation A differential equation whose solution provides the temperature distribution in a

stationary 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 thecontrol volume (inflow-outflow)

Thermal energygeneration

Change in thermalenergy 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 sinp

T T T T kr k k q c

r r tr r r

+ + + =

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

22

1T Ttx

=

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, requiring

specification 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 x

=

=

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

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 the

temperature 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/m

3).

SCHEMATIC:


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ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal

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

form,

2

2

T 1 T

tx

=

( ) i

0

L

Init ial, 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 the

gradient at x = L decreases with time.

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

=


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Problem 2.28 Surface heat fluxes, heat generation and total rate of radiation

absorption in an irradiated semi-transparent material with aprescribed temperature distribution.

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

Problem: Non-uniform Generation dueto 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 : Non-uniformGeneration (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 usingFouriers 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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Problem : Non-uniformGeneration (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 xA

E E E q 0 q L 1 e .a

= + = + = + & & &


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Specify appropriate form of the heat equation.

Solve for the temperature distribution.

Apply Fouriers Law to determine the heat flux.

Simplest Case: One-Dimensional, Steady-State Conduction with No Thermal Energy Generation

Alternative conduction analysis

Common Geometries:

The Plane Wall: Described in rectangular (x) coordinate. Area

perpendicular to direction of heat transfer is constant (independent ofx).

The Tube Wall: Radial conduction through tube wall.

The Spherical Shell: Radial conduction through shell wall.

Methodology of a Conduction Analysis

Consider a plane wall between two fluids of different temperature:

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 sx

T x T T T L

= +

k


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Heat Flux and Heat Rate:

( ),1 ,2 x s sdT k

q k T T dx L

= =

( ),1 ,2 x s sdT kA

q kA T T dx L

= =

Thermal Resistances and Thermal Circuits:tT

Rq

=

Conduction in a plane wall: ,t condL

RkA

=

Convection: ,1

t convRhA

=

Thermal circuit for plane wall with adjoining fluids:

1 2

1 1tot

LR

h A kA h A= + +

,1 ,2x

tot

T Tq

R

=

Thermal Resistance for Unit Surface Area:

,t condL

Rk

= ,1

t convRh

=

Units: W/KtR 2m K/WtR

Radiation Resistance:

,1

t radr

Rh A

= ,1

t radr

Rh

=

( )( )2 2r s sur s sur h T T T T = + +

Contact Resistance:

,A B

tcx

T TR

q

=

= t ct c

c

RR

A,

Values depend on: Materials A and B, surface finishes, interstitial conditions, andcontact pressure (Tables 3.1 and 3.2)


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Composite Wall with Negligible Contact Resistance:

,1 ,4x

tot

T Tq

R

=

1 4

1 1 1C tot A Btot

A B C

L RL LR

A 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=

Series 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 to x 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:

dx

dTTkxAq

x )()(=

=T

T

x

x dTTkxA

dxq

00

)()(

Tube Wall

Heat Equation:

The Tube Wall

10

d dTkr

r 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

,21 2 2

lnln /

s ss

T T rT r T

r r r

= +


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Heat Flux and Heat Rate:

( )( )

( )( )

( )( )

,1 ,22 1

,1 ,22 1

,1 ,22 1

ln /

22

ln /

22

ln /

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/W

2ln /

Units m K/W2

t cond

t cond

r rR

Lk

r rR

k

=

=

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

Composite Wall withNegligible ContactResistance

( ),1 ,4 ,1 ,4rtot

T Tq UA T T

R

= =

1

Note that

is a constant independent of radius.totUA R

=

But, Uitself is tied to specification of an interface.

( )1

i i tot U A R

=


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Heat Equation

Spherical Shell

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 Rate and 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:

overallr overall

tot

Tq UA T

R

= =

1 ConstanttotUA R=

( )1

Depends oni i tot iU A R A

=

( ) ( )1 2,

1/ 1/

4t condr r

Rk

=


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Critical radius (cylindrical geometry)Isolamento

r1 r

T ,h,1 1

T ,h

r2

h1Lr1p2

1

T

T,1

hLrp2

1

Lk

rr 2p2

/ln( )

Lk1

r2

p2

/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 protectionof turbine blades. Determine maximum blade temperaturewith and without TBC.

Schematic:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constant

properties, (3) Negligible radiation


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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 2

tot,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 2 200 K 3.52 10 W m 500 W m K 1104 K= + =

( ) ( )3 4 2 5 200 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 = + +

Without the TBC,

( )1 1 3 2tot, wo o iInR h L k h 3.20 10 m K W = + + =

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

T 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 are

associated with reliability considerations.


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28

Problem 3.62: Suitability of a composite spherical shell for storing

radioactive 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:

( )Pb1 1

R 1/ 4 35.3 W/m K 0.00150 K/W0.25m 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 3

q=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= + 05 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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29

One-Dimensional, Steady-StateConduction with

Thermal Energy Generation

Implications of Energy Generation

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,

VRI

VEq g

2

==&

&

( )xq exp&


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30

The Plane Wall

Consider one-dimensional, steady-state conductionin a plane wall of constant 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:

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

212 sq L xT x T

k L = +

(3.42)( )2 2

212 sq 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)


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31

Radial SystemsCylindrical (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

+ =

Temperature Distribution Surface Temperature

Overall energy balance:

Or from a surface energy balance:

Solution for Uniform Generation in a Solid Sphere of Constant k

with Convection Cooling:

A summary of temperature distributions is provided in Appendix C

for plane, cylindrical and spherical walls, as well as for solid

cylinders and spheres. Note how boundary conditions are specified

and how they are used to obtain surface temperatures.

32

13

dT q r kr C

dr= +

21

26

Cq rT C

k r= +

0 10 0rdT

C

dr

= = =|

( )2

26

o

o s s

q rT r T C T

k= = +

( )2 2

21

6

o

s

o

q r rT r T

k r

= +

0outg

E E + =

3

o

s

q rT T

h = +

0in out E E =

( )cond o convq r q =3

o

s

q rT T

h = +


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32

Problem 3.91 Thermal conditions in a gas-cooled nuclear reactorwith a tubular thorium fuel rod and a concentric

graphite sheath: (a) Assessment of thermal integrityfor a generation rate of . (b) Evaluation oftemperature distributions in the thorium and graphitefor generation rates in the range .

8 310 W/mq =

8 810 5x10q

Problem 3.91 Thermal conditions in a gas-cooled nuclear reactorwith a tubular thorium fuel rod and a concentric

graphite sheath: (a) Assessment of thermal integrityfor a generation rate of . (b) Evaluation oftemperature distributions in the thorium and graphitefor 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 r

1.

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

2

tot

T Tq

R

=

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 22 1 1

1 2 2

2 1

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

Fin Heat Rate:

( )0| f f c x A sd

q kA h x dAdx

== =


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39

CasoCondio de

fronteira emx =L

Distribuio 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 = ( ) ( ) ( )[ ]

( )LmxLmxmbL

sinh

sinhsinh + ( )

( )LmLm

M bL

sinh

/cosh

(iv) ( ) 0=L xme M

cAk

Phm =2

bc AkPhM =

Fin Performance Parameters Fin Efficiency:

,max

f f

f

f f b

q q

q 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 / 2f A w L t = +

( )/ 2p A t L=( )

( )1

0

21

2f

I 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

=


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40

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

extremidade

isolada

Transmisso de calor

na extremidade

( ) ( ) ( )LLLPhLAhq cctipf =,

P

ALL cc +=

Fin of rectangular cross section with t


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41

Fin Arrays

Representative arrays of(a) rectangular and(b) annular fins.

Total surface area:t f b A NA A= +

Number of fins Area of exposed base (prime surface)

Total heat rate:

,t f f b b b o t b

t o

q N hA hA hAR

= + =

Overall surface efficiency and resistance:

,1b

t o

t o t

Rq hA

= =

( )1 1fo ft

NA

A =

Equivalent Thermal Circuit :

Effect of Surface Contact Resistance:

( )( ),

bt t bo c

t o c

q hAR

= =

( )1

1 1f f

o c

t

NA

A C

=

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

( )( )

,

1t o c

to c

RhA

=


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42

Problem 3.116: Assessment of cooling scheme for gas turbine blade.Determination of whether blade temperatures are less

than the maximum allowable value (1050 C) forprescribed 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-1 0.05 m = 2.39

From Table B.1, . Hence,coshmL=5.51From Table B.1, . Hence,coshmL=5.51

( ) 1200 300 1200 5 51 1037= + =o o oT L C ( ) C/ . C

and, subject to the assumption of an adiabatic tip, the operating conditions are acceptable.

(b) With ( ) ( ) ( )1 / 2

2 4 21 / 2c bM hPkA 250W/m K 0.11m 20W/m K 6 10 m 900 C 517W

= = =

o,

Eq. 3.76 and Table B.1 yield

( )fq M tanh mL 517W 0.983 508W= = =

Hence, b fq q 508W= =

Comments: Radiation losses from the blade surface contribute to reducing the blade

temperatures, but what is the effect of assuming an adiabatic tip condition? Calculate

the tip temperature allowing for convection from the gas.


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43

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

with N=11 fins of prescribed dimensions.

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

with N=11 fins of prescribed dimensions.

cq

85 C,cT =o

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.

Analysis: (a) From the thermal circuit,

c cc

tot t,c t,b t,o

T T T Tq

R R R R

= =+ +

( )2 6 2 2t,c t,cR R / W 2 10 m K / W / 0.02m 0.005 K / W= = =

( )2t,b bR L / k W= ( )W / m K 20.003m /180 0.02m 0.042 K /W= =

From Eqs. (3.103), (3.102), and (3.99) ( )ft,o o f t f bo t t

N A1R , 1 1 , A N A A

h A A= = = +

Af= 2WLf= 2 0.02m 0.015m = 6 10-

m2

Ab = W2 N(tW) = (0.02m)2 11(0.182 10-3 m 0.02m) = 3.6 10-4 m2

At = 6.96 10-3

m2

With mLf= (2h/kt)1/2

Lf= (200 W/m2K/180 W/mK 0.182 10

-3m)

1/2(0.015m) =

1.17, tanh mLf= 0.824 and Eq. (3.87) yields

ff

f

tanh mL 0.8240.704

mL 1.17= = =

o = 0.719,

Rt,o = 2.00 K/W, and

( )

( )c

85 20 Cq 31.8 W

0.005 0.042 2.00 K / W

= =

+ +


44/165

44

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 with

from 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= = =

Transient Conduction:

The Lumped Capacitance Method


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45

Transient Conduction

A heat transfer process for which the temperature varies with time, as wellas 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

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

Why is the assumption never fully realized in practice?

General Lumped Capacitance

Analysis:

Consider a general case,which includes convection,radiation and/or an appliedheat flux at specifiedsurfacesas well as internal energygeneration

( ), , ,, , ,s c s r s h A A A

)t(T)t,r(T r


46/165

46

First Law:

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 &+=

goutin

st EEEtd

TdCV

dt

Ed &&& +==

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

Negligible Radiation ( ), / :T T b a

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

d adt =

Integrating from t=0 to any tand 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

&+= ,

''


47/165

47

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

( ),s cdT

c hA T T dt =

, 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

chA

ThermalResistance, Rt

Lumped ThermalCapacitance, Ct

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

outsto

E Q E dt = ,t

s co

hA dt = ( ) 1 expit

tc

=

(5.8)

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

Assuming radiation exchange with large surroundings,

( )4,s r sur dT

c A T T dt

=

,

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 ofT(t).

1 12 tan tan i

sur sur

TT

T T

+


48/165


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50/165


51/165

51

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

If the lumped capacitance approximation can not be made, consideration mustbe given to spatial, as well as temporal, variations in temperature during thetransient process.

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

Non-dimensionalization of Heat Equation and Initial/Boundary Conditions:

Dimensionless temperature difference: *i i

T T

T T

=

*

xx

LDimensionless coordinate:

The Biot Number: solid

hL

Bi k

( )* * , , f x Fo Bi =

Exact Solution:

( ) ( )* 2 *1

exp cosn n nn

C Fo x

==

( )

4sintan

2 sin 2n

n n n

n n

C Bi

= =+

See Appendix B.3 for first four roots (eigenvalues ) of Eq. (5.39c)1 4,...,

Dimensionless time: *2

tt Fo

L

Fourierthe NumberFo


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52

The One-Term Approximation :( )0.2Fo >

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

( )( )* 21 1exp

o

o

i

T TC Fo

T T

1 1Table 5.1 and as a function ofC Bi

( )Fo Variation of temperature with location (x*) and time :

( )* * *1coso x =

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

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

1.130301.111181.103810.897831.064190.624440.45

1.116351.052791.093140.851581.058040.593240.40

1.102260.989661.082260.801401.051660.559220.35

1.088020.920791.071160.746461.045050.521790.30

1.073650.844731.059840.685591.038190.480090.25

1.059150.759311.048300.616971.031090.432840.20

1.044530.660861.036550.537611.023720.377880.15

1.029800.542281.024580.441681.016090.311050.10

1.026840.514971.022160.419541.014540.295570.091.023870.486001.019730.396031.012970.279130.08

1.020900.455061.017290.370921.011380.261530.07

1.017930.421731.014850.343831.009790.242530.06

1.014950.385371.012400.314261.008190.221760.05

1.011970.345031.009930.281431.006570.198680.04

1.008980.299101.007460.244031.004950.172340.03

1.005990.244461.004980.199501.003310.140950.02

1.003000.173031.002500.141241.001660.099830.01

c11c11c11

EsferaCilindro longoPlaca plana

Bi


53/165

53

Graphical Representation of the One-Term Approximation

The Heisler Charts Plane wall Midplane Temperature:


Change in Thermal Energy Storage:


54/165

54

Radial Systems

Long Rods Heated or Cooled by Convection.

2

/

/o

o

Bi hr k

Fo t r

=

=

( ) ( )

( ) ( )

=

=

== 1

2

Foexp*

,,

nnnon

ii

*

rJcTT

TtrTtr

( )

( ) ( )nno

n

nn

JJ

J

c

21

212

+=

(5.184a)

onn r =

Long rod:

orrr =*

Radial Systems

(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 =

=

Long rod one term approximation (Fo > 0.2):


55/165

55


The Heisler Charts Infinite cylinder Centerline Temperature:




56/165

56

Spherical Systems

Spheres Heated or Cooled by Convection.

2

/

/o

o

Bi hr k

Fo t r

=

=

(5.184a) Sphere:

orrr =*

( ) ( ) ( ) ( )

=

=

==

1

2 *sin*

1Foexp

,,

n

n

nnn

ii

* rr

cTT

TtrTtr

( )( )nnnnn

n

c2sin2 cossin4 =

Bitanco1 = nn

Spherical Systems

(5.184a)


stE Q =

( )o iQ c T T =

(Fo > 0.2)

( )FocTT

TT

i

o*o

211 exp =

=

Sphere one term approximation (Fo > 0.2):

( ) ( ) ( )*

*sinFoexp

*

*sin

1

121

1

11

r

r

r

rc

TT

TT *o

i

* =

=

( )11131

cossin3

1

Q

Q*o

o

=


57/165

57


The Heisler Charts Sphere Center Temperature:




58/165

58

The Semi-Infinite Solid

A solid that is initially of uniform temperature Ti and is assumed to extendto infinity from a surface at which thermal conditions are altered.

Special Cases:

Case 1: Change in Surface Temperature (Ts)

( ) ( )0, ,0s iT t T T x T = =

( ), xerf

2 ts

i s

T x t T

T T

=

( )s is

k T Tq

t

=

Problem formulation

td

Td

x

T

12

2

=

T(x, 0) = Ti

T(, t) = Ti

( )( )

12 22 /

, exp4

erfc2

o

i

o

q t xT x t T

k t

q x x

k t

=

(5.59)

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

( )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 erfc

k k kt

=

+ + (5.60)

Case 3: Convection Heat Transfer ( ),h T


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59

Contact between two semi-infinite bodies

( ) ( )

t

TTk

t

TTk

B

iBsB

A

iAsA

,, =

BpBBApAA

iBBpBBiAApAA

sckck

TckTckT

,,

,,,,

+

+=

Two bodies initially at uniform temperatures,TA and TB, are placed in contact at their freesurfaces

If the contact resistance is neglibible, then thetemperature and the heat flux must be equal atthe contact point

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.

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

=

=


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60

( )

( )

( ) ( )

( )

( )

( ) ( )

( )

( )

bi

o

oa

i

o

o

bi

ai

bai

TT

TtT

TtT

Ty,tT

TT

TtT

TtT

Tx,tT

TT

Ty,tT

TT

Tx,tT

TT

Tx,y,tT

2espessuradeinfinitaplaca




22rrectangulasecodebarra

=

=

( )( )

=

TT

TtxTtxS

i

,, ( )

( )

=

TT

TtxTtxP

i

,, ( )

( )

=

TT

TtrTtrC

i

,,

bespessuradeplanaplacao

aespessuradeplanaplacao

bespessuradeplanaplacao

aespessuradeplanaplacao

barrectangulaodebarrao

Q

Q

Q

Q

Q

Q

Q

Q

Q

Q

22

2222sec

+

=


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61

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

KNOWN: Diameter, density, specific heat and thermal conductivity of Pyrex

spheres in packed bed thermal energy storage system. Convection coefficient and

inlet gas temperature.

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

SCHEMATIC:

Pyrex sphereD = 75 mm, T = 25 Ci o

Gas

T Cg,i o= 300h = 75 W/m -K2

= 2225 kg/m3

k = 1.4 W/m-Kc = 835 J/kg-K


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ASSUMPTIONS: (1) One-dimensional radial conduction in sphere, (2)

Negligible heat transfer to or from a sphere by radiation or conduction due to

contact 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 the

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),

( ) ( )

31

oo1 1 1

Q1

Q3 sin cos

=

With Bi hro/k = 2.01, 1 2.03 and C1 1.48 from Table 5.1. Hence,

( )

( )

3

o

0.1 2.03 0.8370.155

5.3863 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

27 2

0.0375m ln 0.155/1.48t 1,020s

7.54 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 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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63

Problem: 5.82: Use of radiation heat transfer from high intensity lamps

for a prescribed duration (t=30 min) to assessability of firewall to meet safety standards corresponding tomaximum allowable temperatures at the heated (front) andunheated (back) surfaces.

( )210 W/ms

q =

KNOWN: Thickness, initial temperature and thermophysical properties of

concrete firewall. Incident radiant flux and duration of radiant heating.

Maximum allowable surface temperatures at the end of heating.

FIND: If maximum allowable temperatures are exceeded.

qs = 10 W/m24

= 0.25 mx

Concrete, T = 25i o = 2300 kg/m3c = 880 J/kg-Kk = 1.4 W/m-Ks = 1.0

T Cmax o= 325 T Cmaxo= 25

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in wall, (2) Validity of semi-

infinite medium approximation, (3) Negligible convection and radiative exchange

with the surroundings at the irradiated surface, (4) Negligible heat transfer from

the back surface, (5) Constant properties.

ANALYSIS: The thermal response of the wall is described by Eq. (5.59)

( )( )

1/ 2 2o o

i2 q t / q xx x

T x, t T exp erfck 4 t k 2 t

= +

where,7 2

pk / c 6.92 10 m / s = = and for

( )1/ 2

ot 30 min 1800s, 2q t / / k 284.5 K. = = = Hence, at x = 0,

( )T 0,30 min 25 C 284.5 C 309.5 C 325 C= + = <

At ( ) ( )1/ 22 ox 0.25m, x / 4 t 12.54, q x / k 1, 786K, and x / 2 t 3.54. = = = = Hence,

( ) ( ) ( )6T 0.25m, 30min 25 C 284.5 C 3.58 10 1786 C ~ 0 25 C= +


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65

CILINDRO CURTO: 2DPROPRIEDADES CONSTANTES

h CONSTANTE

PARA O CILINDRO INFINITO C(r,t):

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

PARA O CILINDRO CURTO:

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

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


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66

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.

Propriedades da gua:

CLCULO DAS DIMENSES DO CILINDRO:

CLCULO DA TEMPERATURA NO CENTRO DO CILINDRO:


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SOLUO TENTATIVA-ERRO:

Introduction to Convection:Flow and Thermal Considerations


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68

Boundary Layers: Physical Features

Velocity Boundary Layer

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

A region of the flow characterized byshear stresses and velocity gradients.

A region between the surfaceand the free stream whosethickness increases inthe flow direction.

( )0.99

u y

u

=

Why does increase in the flow direction?

Manifested by a surface shearstress that provides a dragforce, .

s

DF

0s y

u

y =

=

s

D s sA

F dA=

How does vary in the flowdirection? Why?

s

2

2

1

=u

C sf

Thermal Boundary Layer

A consequence of heat transferbetween the surface and fluid.

A region of the flow characterizedby 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

Tq k

y=

=

0/f y

s

k T yh

T T

=

If is constant, how do and

h vary in the flow direction?( )sT T sq


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Distinction between Local and

Average Heat Transfer Coefficients Local Heat Flux and Coefficient:

( )sq h T T =

Average Heat Flux and Coefficient for a Uniform Surface Temperature:

( )s sq hA T T =

s

sAq q dA= ( )

ss sA

T T hdA=

1s sA

s

h hdAA

=

For a flat plate in parallel flow:

1 Lo

h hdxL

=

Governing equations

Equao da continuidade ( ) ( ) ( ) 0=

+

+

+

z

w

y

v

x

u

t

Equao de balano da quantidade demovimento

( ) ( )i

j

ij

ij

iji gxx

p

x

uu

t

u

+

+

=

+

ij

k

k

i

j

j

iij

x

u

x

u

x

u

+

=

3

2

Equao de conservao daenergia

( ) ( ) qx

u

x

up

x

Tk

xeu

xe

t j

iij

j

j

jjj

j

+

+

=

+

& Energia interna

2

222222

2

3

2

2

32

+

+

+

+

+

+

+

+

+

+

=

=

+

=

=

z

w

y

v

x

u

y

w

z

v

x

w

z

u

x

v

y

u

z

w

y

v

x

u

x

u

x

u

x

u

x

u

x

u

k

k

j

i

i

j

j

i

j

iij

Dissipaoviscosa de energia


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70

Governing equations

( ) ( ) qx

pu

t

p

x

Tk

xhu

xh

t jj

jjj

j

++

+

+

=

+

& Entalpia especfica

( ) ( ) qx

pu

t

pT

x

Tk

xTu

xcT

tc

jj

jjj

jpp ++

+

+

=

+

& Temperatura

peh +=

Coeficiente de expanso trmica:p

T

=

1

Gs perfeito: = 1/T

Fluido incompressvel:= 0

( ) dpTdTcdh p

1

1 +=

The Boundary Layer Equations

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

negligible body forces.

( ),, pc k

Apply conservation of mass, Newtons 2nd Law of Motion and conservation of energyto 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 v

y x y x

>>

>>


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71

Conservation of Mass:

0u vx 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?

Newtons Second Law of Motion:

2

2

x-direction :

u u dp uu v

x 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 :

0p

y

=

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

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

Conservation of Energy:

22

2p

T T T uc u v k

x y y y

+ = +


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

determining similarity parameters that facilitate application of results obtainedfor 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).

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.

Recast the boundary layer equations by introducing dimensionless forms of theindependent and dependent variables.

* *

* *

* s

s

x yx y

L L

u vu v

V V

T TT

T T

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

+ = +

+ =


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Reynolds NumbeRe the

Pr

r

Prandtl Numberthe

L

p

VL VL

v

c v

k

=

=

For a prescribed geometry,

( )* * *, ,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 Res

f

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?

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

For a prescribed geometry,

( )* * *, ,Re ,PrLT f x y=

( )

( ) **

* *0

* *00

/f y f f s

s s yy

k T y k kT T T Th

T T L T T y L y

=

==

= = = +

The dimensionless local convection coefficient is then

( )*

**

*0

, Re , PrLf y

hL T Nu 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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Boundary Layer Transition

How would you characterize conditions in the laminar region of boundary layerdevelopment? In the turbulent region?

What conditions are associated with transition from laminar to turbulent flow?

Why is the Reynolds number an appropriate parameter for quantifying transition

from laminar to turbulent flow?

Transition criterion for a flat plate in parallel flow:

, critical Reynolds numberRec

x c

u x

location at which transition to turbulence beginscx 5 6

,~ ~

10 Re 3 x 10x c< Rex,c?

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

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?


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The Reynolds Analogy Equivalence of dimensionless momentum and energy equations for

negligible pressure gradient (dp*/dx*~0) and Pr~1:

Advect ion terms Diffusio n

* * 2 ** *

* * *2

1

Re

T T Tu v

x y y

+ =

* * 2 ** *

* * *2

1

Re

u u uu v

x y y

+ =

Hence, for equivalent boundary conditions, the solutions are of the same form:

* *

* *

* *

* *0 0

Re

2

y y

f

u T

u T

y y

C Nu

= =

=

=

=

With Pr = 1, the Reynolds analogy, which relates important parameters of the velocityand thermal boundary layers, is

2fC

St=

or, with the defined asStanton number ,

p

h NuSt

Vc Re Pr =

Modified Reynolds (Chilton-Colburn) Analogy:

An empirical result that extends applicability of the Reynolds analogy:

23Pr 0.6 Pr 60

2f

H

CSt j= <

Figure 8.3f

Noncircular Tubes: Use of hydraulic diameter as characteristic length:

4 ch

AD

P

Since the local convection coefficient varies around the periphery of a tube,approaching zero at its corners, correlations for the fully developed regionare associated with convection coefficients averaged over the peripheryof the tube.

Laminar Flow:

The local Nusselt number is a constant whose value (Table 8.1) depends onthe surface thermal condition and the duct aspect ratio.( )s sT or q

Turbulent Flow:

As a first approximation, the Dittus-Boelter or Gnielinski correlation may be usedwith the hydraulic diameter, irrespective of the surface thermal condition.


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90

Average Nusselt Number for Turbulent Flow in a Circular Tube :

Effects of entry and surface thermal conditions are less pronounced forturbulent flow and can be neglected.

For long tubes :( )/ 60L D >

,D D fd Nu Nu

For short tubes :( )/ 60L D sT TCh or c is an evaporating liquid ( ).cC

Negligible or no change in ( ), , .c c o c iT T T=

Case (c): Ch=Cc.

1 2 1mT T T = =


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119

Heat Exchangers:The Effectiveness NTU Method

General Considerations

Computational Features/Limitations of the LMTD Method:

The LMTD method may be applied to design problems forwhich the fluid flow rates and inlet temperatures, as well asa desired outlet temperature, are prescribed. For a specifiedHX type, the required size (surface area), as well as the other

outlet temperature, are readily determined.

If the LMTD method is used in performance calculations for whichboth outlet temperatures must be determined from knowledge of theinlet temperatures, the solution procedure is iterative.

For both design and performance calculations, the effectiveness-NTUmethod may be used without iteration.


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Definitions

Definitions Heat exchanger effectiveness, :

max

q

q=

0 1

Maximum possible heat rate:

( )max min , ,h i c iq C T T =

min

ifor

h h cC C CC


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122


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123

Design Calculations:

( )min max, / NTU f C C =

Relations Table 11.4 or Figs. 11.14 - 11.19

For all heat exchangers,

with rC

( )1 exp NTU=

( )

or

1n 1NTU =

For Cr= 0, to all HX types.a single relation appliesNTU

Radiao: Consideraes gerais

Estuda-se radiao trmica, cujas origens esto ligadas emisso da matria auma temperatura absoluta T>0

A emisso devida oscilaes e transies electrnicas dos muitos electres queconstituem a matria que, por sua vez, so mantidos pela energia trmica da matria

A emisso corresponde energia transferida da matria (calor) e, portanto, corresponde reduo de energia trmica armazenada na matria.

A radiao tambm pode ser intersectada e absorvida pela matria.

A absoro resulta em transferncia de calor para a matria e, portanto, correspondea um amento de energia trmica armazenada na matria.

Considere um slido temperatura Ts num recintofechado com vcuo, cujas paredes esto temperatura Tsur

Que fenmeno ocorre se Ts > Tsur? Porqu

Que fenmeno ocorre se Ts < Tsur? Porqu


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A emisso de gases ou slidos semi-transparentes ou lquidos umfenmeno volumtrico.A emisso de slidos ou lquidos opacos um fenmeno superficial (com aemisso originria em tomos ou molculas a 1 m da superfcie).

A natureza dual da radiao: Nalguns casos, as manifestaes fsicas da radiao podem ser explicadas

olhando-as como partculas (aka fotes ou quanta).

Noutros casos, a radiao comporta-se como uma onda electromagntica.


c

=

82 998 10 m/soc c= = . x

Em qualquer dos casos, a radiao caracterizada por um comprimento de onda e frequncia que esto relacionados pela velocidade de propagao da radiaono meio em causa, c:

No vcuo:

O espectro electromagntico



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A quantidade de radiao emitida por uma superfcie opaca varia com ocomprimento de onda, podendo falar-se em distribuio espectral em todos oscomprimentos de onda ou de componentes monocromticas/espectrais associadas acomprimentos de onda especficos.


Efeitos direccionais

A radiao emitida por uma superfcie s-lo- em todas asdireces do hemisfrio e segundo uma distribuio direccional

A direco pode ser representada em coordenadas esfricas pelongulo polar ou zenital e pelo ngulo azimutal .



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Intensidade de radiao

A quantidade de radiao emitida por uma superfcie, dA1, e que sepropaga numa direco particular, (,), quantificada em termosde um ngulo slido diferencial associado direco em causa.

2ndAd

r

dAn elemento unitrio de superfcie de uma esfera hipottica na direco (,),




2ndA r d d = sin

2ndAd d d

r = = sin

O ngulo slido tem unidades steradianos (sr).

2 2

0 0 2hemi d d sr

= = /

sin

Intensidade Espectral : a quantidade usada para especificar o fluxo de energia radiante(W/m2) num ngulo slido unitrio numa direco prescrita (W/m2.sr) e num intervalounitrio de comprimentos de onda (W/m2.sr.m).


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( )( )1

e

dqI

dA d d

,, ,

cos

A intensidade espectral, , associada emisso de um elemento de rea unitrio, ,num ngulo slido, (em torno de e ),e num intervalo de comprimento de onda, ,(em torno de ), :

eI, 1dA

d d

eI,

O argumento para definir o fluxo radiativo em termos da rea projectada da superfcieemerge do facto de haver superfcies para as quais, com boa aproximao,

independente da direco: superfcies difusas, e a radiao isotrpica.( )1dA cos

A rea projectada como apareceria

Se observada segundo os ngulos

1dA

,

Quanto vale a rea projectada para ?0 =

Quanto vale a rea projectada para ?2 = /



( ) 1edq

dq I dA d d = , , , cos

A taxa de calor espectral e o fluxo de calor espectral associados emisso a partir deso, respectivamente,

1dA

( ) ( )e edq I d I d d = =, ,, , cos , , cos sin


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Relao da intensidade com poder

emissivo, irradiao e radiosidade

( ) ( )2 2

0 0 e E I d d

= /

, , , cos sin

( )0 E E d

=

( ) ( )eE I = ,

O poder emissivo espectral (W/m2.m) corresponde emisso espectral em todas asdireces possveis:

eE I=

O poder emissivo total (W/m2) corresponde emisso espectral em todas asdireces e comprimentos de onda possveis:

Para superfcies difusas, a emisso isotrpica e:

A intensidade espectral da radiao incidente numa superfcie, definida em termos do ngulo slido unitrio em torno dadireco de incidncia, do intervalo de comprimento de onda,em torno de, , e da rea projectada do receptor,

iI,

d

1dA cos .

Relao da intensidade com poderemissivo, irradiao e radiosidade

( ) ( )2 2

0 0 iG I d d

= /

, , , cos sin

( )0G G d

=

A irradiao espectral vale:( )2W/m m

e a irradiao total :( )2W/m

Quantos e G so expressos se a radiao for difusa?G

A radiosidade de uma superfcie opaca contabilizatoda a radiao que abandona a superfcie em todasas direces e pode incluir as contribuies dareflexo e emisso.


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Radiao de CORPO NEGRO A Cavidade Isotrmica (Hohlraum):parede interior a temperatura uniforme.

(a) Depois de mltiplas reflexes, toda aradiao entrada na cavidade virtualmente absorvida corpo negro.

(b) A emisso a partir da abertura a mxima que se pode atingir, para atemperatura associada cavidade, e difusa (independente da direco) corpo negro.

(c) O campo radiativo no interior da cavidade (efeito acumulado da radiao emitidae da reflectida pela parede da cavidade) tem por efeito assegurar uma irradiaodifusa (correspondente emisso de um corpo negro numa forma igual radiao emergentepela abertura). O campo radiativo no interior da cavidade de corpo negro.Qualquer superfcie no interior da cavidade irradiada de forma difusa: ( )bG E = ,

Esta condio depende da superfcie da cavidade ser acentuadamentereflectora ou absorvedora?

Lei da distribuio de Plank

3 2898 m KT C = = max

( ) ( )( )

15

2 1b b

C E T I T

C T

= =

, ,, ,

exp /

A distribuio espectral do poder emissivo de um corpo negro (determinado teoricamentee confirmado experimentalmente) (Plank):

Primeira constante:8 4 21 3 742 10 W m mC = . x /

Segunda constante: 42 1 439 10 m KC = . x

varia continuamente com e aumenta com T.bE,

A distribuio caracterizada por um mximo para o qual dadopela lei deslocamento de Wien :

max

A quantidade fraccional da emisso total de corpo negro que aparece a baixoscomprimentos de onda aumenta com o aumento de T.


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Lei de Stefan-Boltzmann

( ) ( ) ( )

2 1

1 2 2 1

00 0 4

b o b E d E d F F FT

= = , ,

( ) ( )0

0bE dF f T

T

= =,

O poder emissivo total de um corpo negro obtido integrando a distribuio de Planckem todos os comprimentos de onda possveis.

40b b b E I E d T

= = = ,a lei de Stefan-Boltzmann, em que:

K

a constante se Stefan-Boltzmann, = 5,670 10-8 W/m2.K4

A fraco total da emisso de um corpo negro que est contida num intervalo de comprimentode onda prescrito ou banda :( )1 2

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