1.

Solar radiation of 1000 W/m2 is incident on a grey opaque surface with emissivity of 0.4 and emissive power of 400 W/m2 . The ratiosity of the surface will be : (a) 940 W/m3 (b) 850 W/m2 (c) 760 W/m2 (d) 670 W/m2

Answer»

(c) 760 W/m2 

Here G = 1000 W/m2 

Eb = 400 W/m2 (Assume)

Radiosity, J = E + pG

p = Reflectivity

as surface is opaque.

So, t = 0 (Transmitivity)

So, p = 1 - \(\alpha\) = 1 - \(\varepsilon\) = 1 - 0.4 = 0.6

(by Krischoff’s law)

So, J = \(\varepsilon\)Eb + pG

= 0.4 x 400 + 0.6 x 100

= 760 W/m2

Note : In this question emissive power = 400 W/m2 is considered for black surface.

Otherwise, for opaque surface emissive power = 400 W/m2 , surface turned out to be black.



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