TITLE.PM5

HEAT TRANSFER 845

dharm

\M-therm\Th15-4.pm5

F1–2 is known as ‘configuration factor’ or ‘surface factor’ or ‘view factor’ between the
two radiating surfaces and is a function of geometry only.
Thus, the shape factor may be defined as “The fraction of radiative energy that is diffused
from one surface element and strikes the other surface directly with no intervening reflections.”

Further, Q1–2 = F1–2 A 1 σ T 14 ...(15.94)

Similarly, the rate of radiant energy by A 2 that falls on A 1 , from eqn. (15.88), is given by

QT dA dA

(^212) AA r
4 1212
2
21
− =σ zz
θθ
π
cos cos
The rate of total energy radiated by A 2 is given by
Q 2 = A 2 σ (^) T 24
Hence the fraction of the rate of energy leaving area A 2 and impinging on area A 1 is given by
Q
QA
dA dA
AA r
21
22
1212
2
1
21
− =
zz
cosθθcos
π ...(15.95)
or
Q
Q
(^21) F
2
− = 21 −
F2–1 is the shape factor of A 2 with respect to A 1.
QFAT 21 −−=σ21 2 24 ...(15.96)
From eqns. (15.93) and (15.95), we get
A 1 F1–2 = A 2 F2–1 ...(15.97)
The above result is known as reciprocity theorem. It indicates that the net radiant inter-
change may be evaluated by computing one way configuration factor from either surface to the
other. Thus the net rate of heat transfer between two surfaces A 1 and A 2 is given by
Q 12 = A 1 F1–2 σ (^) ()TT 14 − 24
= A 2 F2–1 σ (^) ()TT 14 − 24 ...(15.98)
It may be noted that eqn. (15.98) is applicable to black surfaces only and must not be used for
surfaces having emissivities very different from unity.
Example 15.26. A body at 1000°C in black surroundings at 500°C has an emissivity of 0.42 at
1000 °C and an emissivity of 0.72 at 500°C. Calculate the rate of heat loss by radiation per m^2.
(i)When the body is assumed to be grey with ε = 0.42.
(ii)When the body is not grey.
Assume that the absorptivity is independent of the surface temperature.
Solution. (i) When the body is grey with ε = 0.42 :
T 1 = 1000 + 273 = 1273 K
T 2 = 500 + 273 = 773 K
ε at 1000°C = 0.42
ε at 500°C = 0.72
σ = 5.67 × 10–8
Heat loss per m^2 by radiation,
q = εσ (^) ()TT 14 − 24
= 0.42 × 5.67 × 10–8 [(1273)^4 – 773)^4 ] = 54893 W
i.e., Heat loss per m^2 by radiation = 54.893 kW. (Ans.)