Calculate the heat transfer between two large walls through
radiation. It is assumed that colder
wall is maintained at 800 degrees Fahrenheit, while hot wall is at 1000 degrees
Fahrenheit. (a). Assume that the walls
are perfect black bodies. (b). Assume that the walls have emissivities of 0.6
and 0.8.
Solution:
For perfect black bodies, this equation reduces to sigma
times A1 times (T14 minus T24). Please note that sigma is Stefan-Boltzmann
constant, and is equal to 0.173 times 10-8 BTU per hour per square
foot per degree Rankine4. It
is important that we use absolute values of the temperature in calculations.
If the two planes are not black bodies and have different
emissivities, the net exchange of energy is corrected by a factor, Fe that is
given by reciprocal of (1 over e1 plus 1
over e2 minus
1).
Assume A1 is 1 square foot.
Case (a): Plugging
in the values of T1 and T2 as 1460 and 1260 degrees
Rankin, Q is found to be 3,466 BTUs per hour.
Case (b): For
emissivity values of 0.6 and 0.8, Fe is calculated as 0.552. Thus Q is 0.552 times 3,466 equals to 1808 BTUs per hour.