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842 ENGINEERING THERMODYNAMICS

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15.5.9.2. Lambert’s Cosine Law
The law states that the total emissive power Eθ from a radiating plane surface in any direction
is directly proportional to the cosine of the angle of emission. The angle of emission θ is the angle
subtended by the normal to the radiating surface and the direction vector of emission of the receiv-
ing surface. If En be the total emissive power of the radiating surface in the direction of its normal,
then
Eθ = En cos θ ...(15.85)
The above equation is true only for diffuse radiation surface. The radiation emanating from a
point on a surface is termed diffused if the intensity, I is constant. This law is also known as Lambert’s
law of diffuse radiation.
Example 15.23. The effective temperature of a body having an area of 0.12 m^2 is 527°C.
Calculate the following :
(i)The total rate of energy emission.
(ii)The intensity of normal radiation, and
(iii)The wavelength of maximum monochromatic emissive power.
Solution. Given : A = 0.12 m^2 ; T = 527 + 273 = 800 K
(i) The total rate of energy emission, Eb :
Eb = σ AT^4 W (watts) ...[Eqn. 15.64 (a)]

= 5.67 × 10–8 × 0.12 × (800)^4 = 5.67 × 0.12 ×

800

100

F^4

HG

I

KJ = 2786.9 W. (Ans.)

(ii)The intensity of normal radiation, Ibn :

Ibn = Eb

π

, where Eb is in W/m^2 K^4

= σ

ππ

T^4

4

567 800

=^100

×F

HG

I

KJ

.

= 7392.5 W/m^2 .sr. (Ans.)

(iii)The wavelength of maximum monochromatic emissive power, λmax :

From Wien’s displacement law,

λmax.T = 2898 μm K ...[Eqn. 15.80]

or λmax =
2898 2898
T 800
= = 3.622 μm. (Ans.)
Example 15.24. Assuming the sun to be a black body emitting radiation with maximum
intensity at λ = 0.49 μm, calculate the following :
(i)The surface temperature of the sun, and
(ii)The heat flux at surface of the sun.
Solution. Given : λmax = 0.49 μm
(i) The surface temperature of the sun, T :
According to Wien’s displacement law,
λmax. T = 2898 μmK

∴ T =

2898 2898

λmax 048.

= = 5914 K. (Ans.)

(ii)The heat flux at the surface of the sun, (E)sun :

(E)sun = σT^4 = 5.67 × 10–8 T^4 = 5.67

T

100

F^4

HG

I

KJ