TITLE.PM5

844 ENGINEERING THERMODYNAMICS

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\M-therm\Th15-4.pm5

solid angle at the centre of the other. Let dω 1 be subtended at
dA 1 by dA 2 and dω 2 subtended at dA 2 by dA 1. Then

dω 1 = dA

r

22

2

cosθ , and dω

2 =

dA

r

11

2

cosθ ...(15.86)

The energy leaving dA 1 in the direction given by the
angle per unit solid angle = Ib 1 dA 1 cos θ 1.

where, Ib = Black body intensity, and
dA 1 cos θ 1 = Projection of dA 1 on the line between the
centres.
The rate of radiant energy leaving dA 1 and striking on
dA 2 is given by
dQ 12 − =Ib 1 dA 1 cos θ 1. dω 1

=

IdAdA

r

b 1 1212

2

cosθθcos

...(15.87)

This energy is absorbed by the elementary area dA 2 , since both the surfaces are black. The

quantity of energy radiated by dA 2 and absorbed by dA 1 is given by

dQ

IdAdA

r

b

21

2121

2

− =^2 cosθθcos ...(15.88)

The net rate of transfer of energy between dA 1 and dA 2 is

dQ 12 = dQ1–2 – dQ2–1

=

dA dA

r

12 cos 2 1 cos (^2) IIbb 12
θθ
ej−
But I
E
I
E
b
b
b
b
1
1
2
==^2
ππ
and ...[Eqn. (15.88)]
∴ dQ
dA dA
r
12 =−^12 cos 2 θθ^1 cos^2 ()EEbb 12
π ...(15.89)
or dQ
dA dA
r
12 =−σθθ^12212 TT 12 24
π
cos cos ()
...(15.90)
The rate of total net heat transfer for the total areas A 1 and A 2 is given by
Q 12 = dQ T T
dA dA
(^121) AA r
2
2
(^41212)
2
21
=−σ()zzz cosθθcos ...(15.91)
The rate of radiant energy emitted by A 1 that falls on A 2 , from eqn. (15.87), is given by
QI dA dA
(^12) b AA r
1212
1 2
21
− = zz
cosθθcos
QT dA dA
(^121) AA r
(^41212)
2
21
− =σ zz
θθ
π
cos cos
...(15.92)
The rate of total energy radiated by A 1 is given by,
Q 1 = A 1 σ (^) T 14
Hence the fraction of the rate of energy leaving area A 1 and impinging on area A 2 is given by
Q
QA
dA dA
AA r
12
11
1212
2
1
21
− =
zz
cosθθcos
π ...(15.93)
or Q
Q
(^12) F
1
− = 12 − ...[15.93 (a)]
Fig. 15.52. Radiation heat exchange
between two black surfaces.