Radiation:

When heat is transferred through radiation, heat transfer rate, Q, is give as

Where F = factor depending upon the view emissivity factors.

σ = Stefan-Boltzmann constant, 0.1713´10^-8 Btu/(h·°F⁴).

If the surfaces are normal to each other then the view factor is 1. The emissivity factor, F_ε, can be calculated as

Where ε_i is an emissivity of body i having surface temperature T_i.

Example 5.18: Two very large walls are at the constant temperature of 800 °F and 1000 °F.

(a) Assuming they are black bodies, how much heat must be removed from the colder wall to maintain a constant temperature?

(b) If these two walls have emissivities of 0.6 and 0.8, respectively, what is the net heat exchange?

Solution: It is important that we use absolute temperatures.

Assuming that heat transfer areas of both walls are one square foot each.

Temperature of hot surface, T₁ = 1000 °F = 1460 °R,

Temperature of cold surface, T₂ = 800 °F = 1260 °R

(a) For black bodies, emissivity values are equal to 1.0

Now the radiation heat transfer can be calculated to be

(b) For a case, where bodies have different values of emissivities, ε₁ = 0.6, and ε₂ = 0.8, F_ε can be found as

And the radiation heat transfer in this case is