MEL 725 Pow er-Pl ant Steam Generator s (3-0 -0) Dr. Prabal Talukdar A ss i s t ant Pr o f essor Department o f Mechanical Eng ineeri ng IIT Delhi Radiation Exchange between Surfaces
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MEL 725
Power-Plant Steam Generators (3-0-0)
Dr. Prabal Talukdar
Assistant Professor Department of Mechanical Engineering
IIT Delhi
Radiation Exchange between
Surfaces
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View/Configuration Factor
• Also called shape factors.
• Radiation exchange between two or more
surfaces depends strongly on the surface
geometries and orientations, as well as ontheir radiative properties and temperature.
• To compute radiation exchange between
any two surfaces, we must first introducethe concept of a view factor.
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Definition• The view factor Fij is defined as the fraction of the
radiation leaving a surface i that is intercepted by j.
dA jcosθ j
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• Rate at which radiation leaves dAi and isintercepted by dA j may be expressed as
dqi→ j=Ii cosθi dAi dω j-i
i = intensity of radiation leaving surface idω j-i = soild angle subtended by dA j when viewed from dAi
= cosθ j dA j /R2
dqi→ j= Ii(cosθi cosθ j/R2 )dAidA j
• Assuming surface i emits and reflects diffusely
dqi→ j=Ji(cosθi cosθ j/πR2 )dAidA j
Derivation
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• The total rate at which radiation leaves surface i
and is intercepted by j may then be obtained byintegrating over the two surfaces. That is
where it is assumed that the radiosity Ji is
uniform over the surface Ai.
Fij = radiation that leaves Ai and is intercepted by A j/total
radiation leaving Ai
= qi→ j / Ai
Ji
jiA A
2
ji
i ji dAdAR
coscos
Jq i j∫ ∫ π
θθ
=→
Derivation (cont’d)
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Final Expression
ji
A A
2 ji
i
ij dAdAR
coscos
A1F
i j∫ ∫ π
θθ=
ji
A A
2 ji
j
ji dAdAR coscos
A1F
i j
∫ ∫ π θθ=
Similarly, the view factor F ji
defined as the fraction of the radiation
that leaves A j and intercepted by Ai can be expressed as
Assn: diffuse emitters and reflectors and have uniform radiosity
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View Factor Relations• Reciprocity Relation:
AiFij = A jF ji
• Summation Rule:
• What is Fii ?• Value of Fii for a convex or flat surface?
Fii = 0
1F N
1 j
ij =∑=
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View Factor Relations (cont’d)• Radiation exchange in an enclosure of N surfaces: N2
view factors required
• Summation rule can be applied to get N equations which
gives N view factors• Application of Reciprocity relation for N(N-1)/2 timesgives N(N-1)/2 view factors
• So we need essentially N2-N-N(N-1)/2
⎥
⎥⎥⎥
⎦
⎤
⎢
⎢⎢⎢
⎣
⎡
−−
−−−−
−−−−
NN2 N1 N
N22221
N11211
FFF
FFF
FFF
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Example• For a 3 surface enclosure, we need
32-3-3(3-1)/2=3
F12 =F11=
F21=
F22= 1
2
10
A1/A2
1-A1/A2
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View Factor Relation• Additive nature of view factor
∑
∑∑
∑
=
=
=
=
=
=
=
n
1k
k
n
1k
kik
)i( j
n
1k
kik )i( j j
n
1k
ik ) j(i
A
FA
F
FAFA
FF
i
j
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Problem
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Black Body Radiation Exchange
ijii ji F)JA(q =→
bj ji ji j
biiji ji
EFAq
EFAq
=
=
→
→
For a black surface Ji = Ebi
Similarly,
Net radiative exchange between the two surfaces:
)TT(FAEFAEFAq q q 4
j
4
iiji bj ji j biijii j jiij −σ=−=−= →→
Rate at which radiation leaves surface i and intercepted by j
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Problem
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Chart
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• To use the last equation, surface radiosity Ji must known. To
determine this quantity, it is necessary to consider exchange
between surfaces of the enclosure
• Total rate at which radiation reaches surface i from all surfaces,
• or from Reciprocity relation,
• Cancelling the area Ai and substituting in
qi = Ai(Ji – Gi)
Radiation Exchange between
Surfaces• To use the last equation, surface radiosity Ji must known. To
determine this quantity, it is necessary to consider exchange
between surfaces of the enclosure
• Total rate at which radiation reaches surface i from all surfaces,
• or from Reciprocity relation,
∑=
= N
1 j
j j jiii JAFGA
∑=
= N
1 j
jiijii JAFGA
∑=
−= N
1 j
jijiii )JFJ(Aq
∑∑ == −=
N
1 j jiji
N
1 jijii )JFJF(Aq
Summation rule
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Radiation Exchange between
Surfaces (cont’d)
∑∑==
=−= N
1 j
ij j
N
1 j
iijii q )JJ(FAq
iii
i bii
A/)1(
JEq
εε−
−=
∑= −−=εε− − N
1 j1
iji
ji
iii
i bi
)FA(JJ
A/)1(JE
Preferable
when temperature is known
Preferable when qi is known
∑= −−= N
1 j1
iji
jii
)FA(JJq
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Set of Algebraic Equation
N N NNi Ni11 N
i NiNiii11i
1 N N1ii1111
CJaJaJa
CJaJaJa
CJaJaJa
=+−−−++−−−+
−−−−−−−−−−−−−−−−−−−−−
−−−−−−−−−−−−−−−−−−−−−
=+−−−++−−−+
−−−−−−−−−−−−−−−−−−−−−
−−−−−−−−−−−−−−−−−−−−−
=+−−−++−−−+
[ ][ ] [ ]CJA =In Matrix form,
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Problem
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Problem (cont’d)
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Problem (cont’d)
222
22 b2
A/)1(
JEq
εε−
−=
∑=
−
−=
εε−− N
1 j1
iji
ji
iii
i bi
)FA(
JJ
A/)1(
JE
1232
321
212
12
222
22 b
)FA(JJ
)FA(JJ
A/)1(JE −− −+−=εε− −
For the hypothetical surface, J3 = Eb3
So, only unknowns are J2
and J1
24
22 b
2433 b3
m/W7348TE
m/W459TEJ
=σ=
=σ==F21= ?
F23 = ?
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Problem (cont’d)121212 FAFA =
Y/L = 10/1 = 10
X/L = 1/1 = 1
F12 = 0.39
F21 = (A1/A2)F12
= (1x10/15) x 0.39
= 0.26
F13 = ?
F13 = 1-F12= 1 – 0.39
= 0.61
F31 = (A1/A3) x F13
=(10/(2(10.1)0.61
= 0.305F23 = ? F23 = 0.41
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Problem (cont’d)7536J67.1J26.0 21 −=−
510002J39.0J10
)FA(
JJ
)FA(
JJ
A/)1(
JE
21
1131
31
1121
21
111
11 b
−=+−⇒
−+
−=
εε−−
−−
J2 = 12528 W/m2
kW7.77A/)1(
JEq 222
22 b2 −=
εε−−=
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The Two-Surface Enclosure
22
2
12111
1
42
41
2112
A
1
FA
1
A
1
)TT(q q q
εε−
++ε
ε−−σ
=−==
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Large (Infinite Parallel Plates)
A1,T1,ε1
A2,T2,ε2
q12 = ?
111
)TT(q
21
4
2
4
112
−ε
+ε
−σ=
Small convex object in a Large cavity q12