What is the radiative heat flux between two walls that are
maintained at 117 and 63 degrees Fahrenheit?
Assume that the walls have emissivities of 0.75 and 0.38. How does this radiation compare to situation
when there is radiation protective shield (emissivity 0.046) placed between
these two walls?
Solution:
For perfect black bodies, radiative flux is given by 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.
If these bodies are not perfect 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.
Without any protective radiation shield: For emissivity values of 0.75 and 0.38, F12
is calculated as 0.337. Then plugging
in the values of T1 and T2 as 577 and 523 degrees Rankin,
Q12 is found to be 20.77 BTUs per hour.
With protective shield: For given emissivity values, F13 and F32
are calculated as 0.0453 and 0.0428. At
steady state the amount of heat transferred through walls 1 and 3 is equal to
that transferred between walls 3 and 2.
Thus, temperature of the protective radiation shield, T3 is
calculated to be 552.34 degrees Rankin.
And the associated value of radiative flux is 1.355 BTUs per hour.