Radiation Heattransfer

RADIATION HEATTRANSFER

Radiation heat transfer is the dominant heat transfer mechanism in everyday life:solarradiation is the source of energy sustaining life and every object at finitesurfacetemperature radiates electromagnetic waves. In engineering applications,radiationbecomes important in high temperature processes such as combustion. But radiationalsocan be important in low temperature processes such as natural convection. In thischapterwewilldevelopfinitevolumemethodtosolveradiationheattransferbytakingenergybalance due to radiationcontribution.

11.1RadiationTransferEquation

Detail derivation of radiative transfer equation can be found in standardtextbook

[Brewster, 1996]andwillnotberepeatedhere.Theintensityofaradiationbeamtravellingalong s-directioninamediumisgivenby

,

The' first term in the RIIS indicates the loss of intensity due to absorption andscattering

ofradiationbeam.Theextinctioncoefficient,'isthesumofabsorptioncoefficient(K)

and scattering coefficient(cr),

=K+cr(11.2)

Thesecond term istheradiationintensitygainduetoemissionandin-scattering,'

sKib+Jt<I>drl'(11.3)41t411

r "

The beam travelling alongthe,,s direction at a location z is shown in Fig.11.1.

S'

s

z

z

Figure 11.1 Radiation beam in s-direction at a location z

c

-J·· :(..''./(.) ;

'i -·· ····;, fr

Radiation intensity is a function of direction and its location. Therefore, inI-dimensionalradiation, i1=I(s,i'). Note that intensity does not depend on time because radiationbeamtravels with the speed of light. Integrating the radiative transfer equation over acontrolvolumesurroundingaspatiallocationatzandacontrolanglesurroundingradiationbeamalong a direction s, weget G

J.J'\acliI1·dvd.01 = [_._J(-I1+S' vd.Cl'

(11.4)

a,lt.rd/ J.

t,.Q Av dj(t:701. v

.j,J ; -

A...O;(i_

Noting that the beam direction shoul· be resolved into a coordinate system, whichis

moreconvenientinanalysis,wecanexpresstherateofchangeinbeamintensityi1along,.c

..'()"

S-dlfeCtlOnas

/;·I\1 /--c 'f_

dI\.(1r_.·i- j ,cl··p( 1.5) ----(..,J,,,,1.

ds)I'(. "} [

)

L,v

Invoking the divergence theorem, therefore, the LHS of Eqn.(11.4) can be writtenas

(11.6)

where N is the total number of control volume surfaces. In I-dimensional radiation,there are two surfaces; one at north side and the oth¢r at the south side of control volumeasshown inFig.11.2Theunitvectoralongthebeamdirectionis,intermsofusualsphericalcoordinatesystems,

r-,,,.

s1)=isin8cos8+jsin8sin<I>+kcos8

"-t- ,+-v " c ,-1 c,_l.,.·J-· 12.-cC,-,..,(-,·,.\

(11.7)

-+I

s

Northsurface

Southsurface

Figure 11.2 1-D controlvolume

ThustheLHSbecomes

(11.8)

t.d

Theradiationintensityisassumeduniformoverthegivensolidangleandthegiven

surface area. For 1-dimensional radiation surface area must remain constant. Theunit vector along the beam direction at the north and south control volume surface isidentical

and is given by Eqn.(11. 7). The solid angle, dfJ.1,is'

Now consider theintegral,

,1..

dfJ.1sin8d8d<I>(11.9)1'

<1>+e'4>+e+

s1 •kdfJ.1 =cos8sin8d8d<I>=d<I> cos8sin8d8=

w.1<1>-e-<I>-e-(11.10)

-(<!>+ -<I>-)(cos2 e+ -cos2 e-)7--n(cos2 e+ -cos2 e-)=Di Tl

Insphericalcoordinatesystem,<I> variesfromOto2nand8variesfromOton.In1-dimensional radiation, radiation intensity does not depend on <p. Note that

e::::;7t/

DiOfor

Di ::::;Ofor

2(11.11)

7t

-2

Integrating the RHS of radiative transfer equation over a control volume andcontrolangle, wehave

(11.12)

Inthisexpression,

t1.0.1 =-2n(cos8+ -case-)0

t'lv=APt'lz

frrJ trf'I·

(11.13)..,.

.:. i •

&'

}:}-ctf.J

)

f)

Nowcollectingallterms,integratedradiativetransferequation,becomes

(11.14)

Fewmorestepsarestillrequiredtoexpressthisequationintofinalfinitevolume equation.

-··I3

' I

Evaluation oflntensity at the control volumesurface

Radiative intensity at the north and south control volume surface are notexplicitlydefined and thus require interpolation using intensity at the center of controlvolumes.The simplest method is to assume the intensity of the down stream intensity calledstepscheme (Chai,et.al.). This is similar to the upwind scheme used in convectionhea(transferprocess.

z

North surfaceSouthSurface

Figure 11.3 Beam intensity at a control volumeboundaries

/-)I1=I1I =IifD1 O

"'I =II =IifD1 O

(11.15)

This approximation implies that radiation intensity remains uniform in thecontrolvolume including the boundary and becomes less accurate as the control volumesize

mcreasesl .,,,,

,p) -yv,j,',,J v 16g ,

Linearization of SourceTerm,;,-l=f'T

Mev.,..,-1-,,Iv./,,..,.

pv- _, I;,

For gray medium, the source term;ean be written as------_

A 11S1 - ()11Icrp J11.<I>11.d01·)_

41t\<11t

In the modified sourceterm

4

(11.17)

Thusthefinitevolumeequationbecomes(leftasanexercise)

a1I1=a1I1+a1I1 +b1

where

a= max(O,-AnD)

a =max(o,A.o)/t'r

a=max(O,AnD)+max(O,-A.D)+Apitn1 b =sAp.i:\n1,/,1i'

(11.19)

Theradiationintensitycanbesolvedbyusingthefinitevolumeequationforalldirectionsandforallcontrolvolumesexcepttheintensityattheboundaries.Theseboundary intensities are found by considering boundaryconditions.

BoundaryConditions

Consider the top boundary at z = Zmax· The radiation beams arriving at this surfacefromthe interior are

L

\\·max(o,o)(11.20)

Where !@:is the radiation intensity in the control volume next to the boundary control

s

volume.Thetotalradiationenergyleavingthetopsurfaceisthenemissivepowerfrom

thesurfaceplusthereflectedincident radiation beam.Foradiffusegraysurface,radiationintensityleavingthetopsurfaceisrelatedbythefollowingequation,

ETA.crT;+(1-ET)1<kA.max(o,o)=A.1tl',,(11.21)

i=l ,5

--,;-,...•..,.

--ti'.-

J-.f--.·.

p·

Notethatthedirectioncosineisnegativeforthebeamsleavingthetopsurface.Asimilarexpression can be obtained for the lower boundary surfac intensity. The directioncosineof the beam leaving the bottom surface ispositive.

EBAncrT;+(1-EB)in max(o,-o)=And(11.22)

i=l N(

Othertypeofappropriateboundaryconditionshouldbeusedtodeterminetheradiative •/(,

boundaryconditions.Theseboundary conditionswillbediscussedlater..7 ;:-

RadiativeEquilibrium

,

The emission contribution in the source term, lbP,is

_crT:

bP-

(11.23)

"f17I r;i°,

7t

wherecr is the Stefan-Holtzmanconsa&nt.

lov,,,._.,-·,.(

p·).,,v,;:·I'-.·· S'5

Ifthetemperatureofthemediumissolelydeterminedbytheradiationprocess,radiationequilibriumconditionisrequired.Thenthemissionisequaltotheirradiation,i.e.,

I_I J11 dQ1'\(11.24)

bPP,i

7t47!•·.·

...'),,.

Numerical Procedures

I •I Il

Numericalprocedureneededtosolveradiativetransferequationissimilartoothertransport processes. Radiation beam intensities are assumed at each control volumes andateachbeamdirections.Boundaryconditionsarespecifiedatbothendsofspatialcoordinate.Dependingonthebeamdirection,finitevolumeequationsaresolvedbysweeping either upward (increasing z) or downward (decreasing z) z-directions.

Calculationisrepeateduntilconvergenceisobtainedforallbeamdirectionsineverycontrolvolumes.

Example 11.1

AgasmediumiscontainedbetweentwolargeplanesurfacesasshowninFig.11.4.Thegasisgrayandisotropic.Thetopandbottomsurfacesaremaintainedatdifferenttemperatures. Heat transfer rate from the hot surface to the cold surfaces is influenced bythe radiation properties of the gas and the surface of the plates.

gas H

Figure 11.4 Radiation between two parallelsurfaces

SinceI-dimensionalradiation,radiationintensityinalldirectionsis functionofitslocation. Figure 11.5 shows radiation beam passing through a control volume withthespecifieddirections.

ThereareL1numberofdivisionsinthe8-direction andthusL1numberofradiation beams. SinceI-dimensional inz-direction, radiationbeamisindependent on<!>-direction andanyincrement of.1q>canbeused.Inthepresent case21tisused.Thedistance

between the plate is 2 m, and the temperature of the top plate is 900 K and bottom plate is

Figure 11.5 Beam directions and controlangles

1000K.Theemissivityofbothplatesis1.0anddiffusesurfaceisassumed.Nnumberof uniform control volumes and Ll number of uniform control angles are used.The

probability function<1>11

is 1.0 for gray medium. Absorption coefficient (1e) andscattering

coefficient (cr) are both 0.1 m. Error tolerance for convergence is set at 10-6.

MtlTLAB

AForifatl program based on the finite volume equation is developed andnamed

--RABID.FOR. Numerical results with N=20, Ll=20, 1e=O. l and cr=0.1 are shown intable

11.1intermsoftemperature,dimensionlesstemperatureslipandnormalizedheattransfer

rate. The temperature of the gas is determined by Eq.(11.23) with the aide ofEq.(11.24). The dimensionless temperature slip is definedby

L\T = T4-T,i

T4-T4

BT

The normalized heat transfer rate is defined by

(11.25)

q..=

q

-

qmax

(11.26)

where q is the actual radiation energy flowing along the z-direction at the givencontrolvolume and Qmaxis the maximum possible radiation energy flow when the mediumis non-participating (no absorption indno scattering).Thus

qp=LJ1s'

LJi

LI'LI

·kdn1=

cosesineded<1>=

l=I 1',.0}l=I Ml1(11.27)

and

q*mustbeconstantsince1-dimensiona)5'-f-ec,/11- s{/{1••

(11.28)

Numerical results were obtained by the step scheme with 12 iterations.Resultsshow that q• at the boundary is slightly higher than the values in the interiorcontrolvolumes.Theexactsolutionofthisproblemgivesq*=0.7458[HeasletandWarming].Dimensionlesstemperatureslipisingood agreementwiththeexactsolution.Toimprovethe accuracy further by using the step approximation for interface radiationintensity,finergridpointsareneeded.

ExponentialScheme

Inthestepscheme,radiationintensityat theinterfaceisequaltointensityatthecenter of control volume. Exponential scheme is to account the effects of absorption andscattering while the beam travels from the center of control volume to the boundary surface.UsingtheexactsolutionofEq.(11.1),itcanbeshownthatforabeamtravellingupward (D 0 ), interface intensitiesare r

t=lexp(-f3.Pds)+s.P {t-exp(-P,.Pds)}

f3m,P

r;r:expH.,ds'.)+:·'-exp(-.,ds:))

m,S

(11.29)

where ds.• aredistancesgivenby

ds' =0.5LizP

ncos91

ds' =0.5Liz.

scose'

(11.30)

Thusstepschemeisequaltozerotraveldistanceandrequiresverysmallcontrolvolumesize.

Figure 11.6ExponentialScheme

Introducing these interface intensities, finite volume equation for the upwardtravellingbeams become (left as anexercise)

p pssp(11.31)

The coefficients and source termsare

0m,S

.6t:f

Thefinitevolumeequationforthedownwardtravellingbeams(D'.':;'.;0)is

The coefficients and source termsare

(11.33)

In the aboveexpression

ds'=0.5AzN

O

andds1 =O.SAzP

(11.35)

Jcos8i

sicos8i

Numerical results for the example problem with exponential scheme arealsoshown in Table 11.1. Better agreement with the exact solution is obtained bytheexponential scheme. However, number of iteration is increased to 64. To economizethecomputertime,itisnecessarytousethestepschemeatthebeginningofcalculationandthenswitchestotheexponentialschemelaterincalculation.Forexample,17iterationswere required to have the same results when the step scheme was used first(error

tolerance of10-3

and then switched to the exponential scheme (error tolerance of1o-6).

Table 11.1 Numerical Results for Example11.1

(a)step scheme, (b) exponential scheme, (c) mixedscheme

0.7353286

0.7355548

0.7355027

0.7351256

ITER=17cc)

QSTAR=

0.7442943

0.7443101

0.7443072

0.7442797

963.9070

956.1823

948.5531

939.9907

0.5857453

0.5076134

0.4304170

0.3433715

Table 11.2 Dimensionless heat flux, q•

f{ ::. (VD

OpticalthicknessK0 / Heat flux,q• / OpticalthicknessK0 / Heat flux,q•
0.1 / 0.9157 / 0.8 / 0.6046
0.2 / 0.8491 IP. f'f ?1 / 1.0 / 0.5532,jJ--iJ
0.3 / 0.7934 / 1.5 / 0.4572
0.4 / 0.7458 u,1f0 / 2.0 / 0.3900 o,3tNo
0.5 / 0.7040 / 2.5 / 0.3401
0.6 / 0.6672 6'• 6t1? / 3.0 / 0.3016 O I > 0/JJ

xn=H, ForK

1.0,q•=-

4/3--

01.42089+K

/·::._!i··(S­

Figure below shows dimensionless temperature distribution and heat flux ingraygas contained between infinite black parallel plates [Heaslet and Warming].(a)Temperature distribution: (b) energyflux.

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C..l...

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c:

·;;;

Optical thickness,

KO

c:•2

Cl.>

c

0

a-.o1.0

)(

:2.8

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E' :·...

"c:'...6

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"....'.

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0

•;;;

c:

."§'.2

0

.2.4.6.81.0

Relative optical depth. KlKo

(al

0.51.01.52.0

Optical thickness,Ko

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2.53.0

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t.,'•,

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