16. radiation transfer

7/28/2019 16. Radiation Transfer

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1

Determination of Heat Exchange Between

Black SurfacesElectrical Network Analogy for Blackbody

Heat Exchange

Rewrite the Stefan-Boltzmann result for radiation

exchange between two black surfaces

( )212-112-1 bb

EEFSQ =&

into a form: ( )

2-11

212-1

FS

EEQ bb

1

=&

16. RADIATION HEAT EXCHANGE BETWEEN

SURFACES


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2

As for the heat conduction, analogy with Ohm's law:

21Q& = current

bE = potential

211

1

FS= space resistance, related to a good view

how the objects see each other

( )

2-11

212-1

FS

EEQ bb1=

&

1Q&

2Q&

422 TEb =411 TEb = 21Q&

Radiation network: A circuit representing the exchangeequation is shown below:


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Radiation network for an enclosure of black surfaces

Each surface is at temperatures T1, T2 and T3 or

potentialsEb1, Eb2 and Eb3.

Each surface exchanges

radiation with other surfaces

that it "sees"

iQ& = net heat leaving surface i

= sum of heat exchanged

with all surfaces

An enclosure of3 black surfaces.

1

32

211

1

=FS

R

322

1

=FS

R

311

1

=FS

R

1

2

3


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( ) = == =

3

1

44

111

3

1 1111)(

j jjj bjbjTTFSEEFSQ &

==

3

1 11

11

1j j

bjb

FS

EEQ&

Apply Stefan-Boltzmann result and sum

3-11

31

2-11

211

FS

EE

FS

EEQ bbbb

11

+

=&

For steady state: 021 =++ 3QQQ &&&

1

2

3

Figure (a)

1

2

3

211

1

=FS

R

3221

= FS

R

311

1

= FSR

1

2

3


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Apply Stefan-Boltzmann result for n surfaces

( ) ==

n

jbjbijiii EEFSQ

1

&

==

n

j jii

bjbii

FS

EEQ

1 1&

For steady state: 0.......21 =++++++ ni QQQQQ &&&&& 3

n

4

32

1

6.10Fig.


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Radiation Exchange Between Gray Surfaces

A graybody absorbs and reflects radiation

Objectives:

(1) Determine the net heat transfer from each surface

(2) Determine the heat transfer between any two surfaces

in a multi-surface enclosure

Types of problems: Two:

(1) Specified surface temperature: Determine heat rate

(2) Specified surface heat rate or insulated surface:

Determine surface temperature


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Assumptions:

Opaque surface, 0

Gray surfaces, a = , r = 1 - a = 1 -

Nonparticipating medium

Uniform surface temperature

Uniform surface emissivity

Diffuse emission, irradiation and reflection

Procedure:

(1) Define node, surface heat fluxand surface resistance

(2) Introduce the concept ofradiation networkand

electrical circuit analogy(3) Apply the network method to radiation exchange in

(i) Two-surface enclosure

(ii) Three- and multi-surface enclosures


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For the ith surface of an enclosure:

iE = emissive power W/m2

iG = irradiation W/m2

iJ = radiosity W/m2

iq& = net flux leaving surface [W/m2]

(difference between what hits

the surface and what leaves it)

At the fictitious surface:

Incoming irradiation = iG

Outgoing radiosity = iJ

G

G

GG E

J

10.2Fig.

rG

tG

iJ iG

surfacethi

fictitioussurface

iq


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From the definition of iJ :

(a)iiii GrEJ +=

(a) becomes iibiii GEJ )1( +=

Solve fori

Gi

biii

i 1

EJG

=

Conservation of energy at the fictitious surface:

iii GJq=

&

and express radiation heat rate SqQ ii .&& =

Use: biii EEiii ar == 11 and

S

JEQ

i

i

ibii

=1

&

substitute


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ii

ii

S

R

=1

iQ& = current

iR = resistance

)( ibi JE = potential drop

iR is known as surface resistance

The drop ( ibi JE ) takes place at surface To determine iQ

& from (a), must know the radiosity

iJ

iJ

surfacethi

fictitioussurface

biE

Qi

ii

iSiR

= 1or i

ibi

i R

JE

Q

=&

(a)


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11

Two-Surface Enclosures

Enclosure is formed by

two gray surfaces

Uniform emissivity

Uniform surface

temperature

2J

222 ,, TEb

1J

111

,, TEbQ1

Q2

12Qfictitioussurface

Examples:

(1) Two infinitely large parallel plates(2) Two infinitely long concentric cylinders

(3) A convex body enclosed by a surface


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Given: Temperature, emissivity and area of each surface

Objective: Determine the net heat exchange 21Q& between

the two surfaces

Conservation of energy at each surface

21 QQQ 2-1&&& ==

1Q& = energy added at the back of surfaces 1

2Q& = energy added at the back of surfaces 2

Sign convention: Heat entering enclosure is positive

Heat leaving enclosure is negative


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1R and 2R are surface resistances

11

11

S

1R

=

22

22

S

1R

=

we have three equations for four unknowns

.,,, 2121 QQJJ&& Need a 4th equation.

Consider radiation exchange between the fictitioussurfaces:

21Q& leaving1 and reaching21J= fraction of

- fraction of 2J leaving 2 and reaching 2

2222 RJEQ b )( =&

Apply (a) to each surface

1111 RJEQ b )( =&


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2122121121 JFSJFSQ =&

Compare equation for two black surfaces:

21-2212-112-1 bb EFSEFSQ =&

For a gray surface radiosity iJ replaces blackbody

emissive power biE

The fictitious surfaces act as if they were black surfaces

with potentials 1J and 2J instead of 1bE and 2bE

Use the reciprocal rule

)( 212-112-1 JJFSQ =&


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2-1

21

2-11

21

12 1 R

JJ

FS

JJ

Q

=

=&

21R = the space or view resistance defined as

2-112-1

FSR 1=

.We have the 4th equation for determining 21Q&

And can complement the network:

2121

212-1

RRR

EEQ bb

++

=&

1111 RJEQ b )( =&

2222 )( RJEQ b =&

2-12112 ) RJ(JQ =&

21Q&

2Q&

1Q&


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This result applies to any two-surface enclosure, based

on:

(1) Gray surfaces, a =

(2) Opaque surfaces, t = 0

(3) Diffuse emission, absorption, irradiation and

reflection(4) Uniform emissivity and temperature for each surface

(5) Nonparticipating medium

Use Stefan-Boltzmann law and definitions of resistances

21 SFSS

TTQQQ

1 2

2

2-11

1

42

41

212-1 )(11)(1

)(

++

=== &&& (A)


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A gray surface adds a surface resistance

The configuration of the two surfaces is reflected in theview factor 12F

Equation (A) will be applied to three special cases:

Radiation network

1J1bE 2J 2bE1-211

FS 2221

S

1

11

S1

1-2QQ

1 2Q

21 SFSS

TTQQQ

1 2

2

2-11

1

4241212-1 )(11)(1

)(

++

=== &&& (A)


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111

)(

21

44

21 +

=

21 TTSQ&

1S

(2) Infinitely long concentric cylinders or concentric

spheres: 2S surrounds , 121 =

221

44

/)1)((/1)(

+=

21

2112-1

SSTTSQ&

)( 4

2

4

1112-1

TTSQ = &

(1) Infinitely large parallel plates: 121 = SSS == 21

(3) Small convex surfaceS1 surrounded by very large

surface 2S 0/ 21 SS and 121 =F


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Radiation Shields

Radiation shield: Used to

reduce radiation heat transfer

Example: Place a radiation

shield between two largeparallel plates

There are 6 resistances: 2 space and 4 surface

1J 31J 32J 2J

1 31 32 2

21 shield3


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Ohms analogy gives the net heat transfer rate 21Q& :

( ) ( ) ( ) ( )

22

2

2-332-33

2-3

1-33

1-3

3-1111

1

42412-1 111111

SFSSSFSS

TTQ

++

+

++

=&

3bE 3-2J3-1J 2J1J1bE 2bE

11

11

S

3-13

3-11

S

3-23

3-21

S

22

21

S

1-31

1

FS 3-23

1

FS


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21

For concentric cylinders and spheres 12331 == FF

( )( ) ( )( )111111 2-31-3312211

42

411

2-1 +++=

SSSSTTSQ&

NOTE:

Additional parallel shields reduces the heat loss further

Each added shield adds three resistances

2111

)(

2-31-321

42

41

2-1 +++

=

1TTS

Q&

For large parallel plates: SSSS === 321 and

12331 == F


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22

If all surfaces have the same emissivity:

Nshields reduce the heat rate

by a factor of (N+ 1)

One shield reduces the heat transfer rate

by a factor of 2

1 3 2

T1 T3 T2

How we can determine the shield temperature?

Lets assume identical emissivities 1 = 2 = 3 =

1

1

1

TT

1

1

1

TTq

23

42

43

31

42

41

+

=

+

=& ( 424143 TT

2

1T =


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Three-Surface Enclosures - gray

Given:

Temperature, emissivity

and area of each surface

Objective:

Determine the heat transfer

rate supplied at the back ofeach surface:

3,21 , QQQ&&&

T1, S1, 1

T2,S

2,2

T3,

S3,3


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Assumptions:

(1) Gray surfaces, a =

(2) Opaque surfaces, 0=t

(3) Diffuse emission, absorption, irradiation and

reflection

(4) Uniform emissivity and temperature for each surface

(5) Nonparticipating medium

i

ibii

R

JEQ

=&

For each surface, we can write an equation

1

111

R

JEQ b

=&

2

222

R

JEQ b

=&

3

333

R

JEQ b

=&

T1, S

1,

1

T2,S

2,2

T3,

S3,3


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We have three unknown radiosities J1, J2, J3.

Solutions to these equations give their values.

Conservation of energy at each node:

2

3

34

FS

1

JJ

FS

1

JJ

S

1

JT

+

=

22

2

122

12

22

2

22

1

3

34

FS

1JJ

FS

1JJ

S

1JT

+=

11

1

211

21

11

1

11

3

233

23

133

13

33

3

33

+

=

FS

1

JJ

FS

1

JJ

S

1

JT4


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Lets proceed from the definition of radiosity:

( ) 114

111 G1TJ +=

13331222111111 FSJFSJFSJSG ++=

31132112111111 FSJFSJFSJSG ++=

Using reciprocity rule

( )( )44444 344444 21

1G

31321211114

111 FJFJFJ1TJ +++=

After substituting for G1 :

T1, S1, 1

T2,S

2,2

T3,

S3,3


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Similarly for surfaces 2 and 3:

( )( )44444 344444 212

2221212222G

332

4

FJFJFJ

1

T

J +++=

( )( )44444 344444 21

3

3231313333

G

3324

FJFJFJ1TJ +++=

( )( )44444 344444 21

1G

31321211114

111 FJFJFJ1TJ +++=

Adding the equation for surface 1:

we have a complete set of equations for radiosities J1, J2, J3:

T1, S1, 1

T2,S

2,2

T3,

S3,3


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( ) += = 3

1jj1j1

4111 FJ1TJ

( ) +==

3

1j j2j2

4

222

FJ1TJ

( ) +==

3

1jj3j3

4333 FJ1TJ

T1, S1, 1

T2,S

2,2

T3,

S3,3

Now we can return to the equations for conservation of

energy in the individual nodes

C ti f t h d


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Conservation of energy at each node:

1

3

3

4

FS

1

JJ

FS

1

JJ

S

1

JTQ

+

=

=

11

1

211

21

11

1111&

2

3

34

FS

1

JJ

FS

1

JJ

S

1

JTQ

+

=

=

22

2

122

12

22

2

222

&

3

233

23

133

13

33

3

333

+

=

=FS

1

JJ

FS

1

JJ

S

1

JT

Q

4

&


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Three-Surface Enclosures with an Insulated

Surface - Refractor

The insulated surface is called a

re-radiating surface or refractor

213, QandQ,QT 21&&&Find

Objective:

2J

3J

333 ,, bET

222 ,, bET

111 ,, bT

1J

1

2

3

insulation

10.17aFig.

1

2

3

insulation


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( )iii

ii

i

i4

ii GJS

S

1

JTQ =

=&

Lets first remind the gray body surface:

iJ

surfacethi

fictitioussurface

biE

Qi

iii

SiR = 1radiation that

leaves the surfaceradiation that

irradiates the surface

energy deficit must be compensated

for by heating the surface iQ&

In case of a refractor, i.e. insulated surface: 0QQ 3i == &&

ii4

ibi GJTE === ( ) ii4

iii G1TJ +=also


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2121 QQQ == &&&

1

2

3

T1, S1, 1

T2, S2, 2

TR, SR, R

can be written as a radiative heat transfer rate between

two potentials - Eb1 andEb2 . From nodeJ1 the heat flux

follows two paths (a) and (b).

22

2ekv

11

1

b2b121

S

1R

S

1

EEQQ

++

== &&baekv R

1

R

1

R

1+=

1Q&


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22

21

R22R1121111

1

b2b121

S

1

FS1

FS1FS

1

S

1

EEQQ

+

++

+

==

&&

211a

FS

1R

=

RR11b

FS

1

FS

1R

+=22


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To determine the refractor surface temperature TR, the

radiosity JR must be determined.

RR4

RbR GJTE ===

From the energy balance in theR node it follows:

R22

2R

R11

1RR

FS

1

JJ

FS

1

JJ0Q

+

==&


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Radiosities J1 andJ2 are determined from nodes 1 and 2:

=

11

1

14

11

S

1

JTQ&

11

11

411

1

SQTJ = &

=

22

2

2

4

22

S

1JTQ&

22

22422 1

SQTJ = &

In case of identical surfaces 1 and 2 and their view factors:

+

==

R22

2R

R11

1R

R

FS

1

JJ

FS

1

JJ

0Q& 4

R

21

R T2

JJ

J =

+

=

4

RR

JT =

16. radiation transfer

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