HEAT TRANSFER 841

dharm

\M-therm\Th15-4.pm5

φ

dA 1

dφ

r sinθ

θ r

dθ

rdθ

r sin dθφ

dA = r sin d d 2 2 θθ φ

(b) Illustration for evaluating area dA 2

Fig. 15.51. Radiation from an elementary surface.

The solid angle subtended by dA 2 = dA

r

2

2

∴ The intensity of radiation, I =

dQ

dA dA

r

12

1 22

−

cosθ×

...(15.82)

where dQ1–2 is the rate of radiation heat transfer from dA 1 to dA 2.
It is evident from Fig. 15.51 (b) that,
dA 2 = r dθ (r sin θ dφ)
or dA 2 = r^2 sin θ.dθ.dφ ...(15.83)
From eqns. (15.82) and (15.83), we obtain
dQ1–2 = I dA 1. sin θ. cos θ. dθ. dφ
The total radiation through the hemisphere is given by

Q = I dA 1

φ

φπ

θ

θ π

θθθφ

=

=

=

=

zz 0

2

0

(^2) sin cos dd
= 2π I dA 1
θ
θ π

=
z 0
(^2) sin θ cos θ dθ
= π I dA 1
θ
π
z= 0
(^2) 2 sin θ cos θ dθ
= π I dA 1
θ
π
z= 0
(^2) sin 2θ dθ
or Q = π I dA 1 ...(15.84)
Also Q = E.dA 1
∴ E dA 1 = π I dA 1
or E = πI
i.e., The total emissive power of a diffuse surface is equal to π times its intensity of radiation.