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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 |
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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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