Question 6.7: A spherical vacuum bottle consists of two silvered, concentr...

Equation 6.23

Q_1=-Q_2=\frac{A_1\sigma (T_1^4-T_2^4)}{[1/\epsilon _1(T_1)]+[1/\epsilon _2(T_2)]-1}                 (6.23)

applies for concentric specular spheres. For the small rate of heat leakage expected, it is assumed that the surfaces will be close to 368 K and 294 K. This gives

Q_1=\frac{\pi (0.15)^25.6704\times 10^{-8}(368^4-294^4)}{(1/0.02)+(1/0.02)-1} =0.440 W

If, instead of using the specular formulation, both surfaces are assumed diffuse reflectors with ϵ still 0.02, then Equation 5.23 applies. The denominator of the Q_1 equation becomes

Q_1=\frac{A_1\sigma (T_1^4-T_2^4)}{1/\epsilon _1(T_1)+(A_1/A_2)[1/\epsilon _2(T_2)-1]}                   (5.23)

\frac{1}{\epsilon _1}+ \frac{A_1}{A_2} \left\lgroup\frac{1}{\epsilon _1}-1 \right\rgroup =\frac{1}{0.02} +\left\lgroup\frac{15}{16.3} \right\rgroup^2\left\lgroup\frac{1}{0.02} -1\right\rgroup =91.50

instead of 99 as in the specular case. For diffuse surfaces the heat loss is increased to 0.476 W.