(i) Located in large room with red brick wall at a temperature of 27 ℃;

(ii) Enclosed in a 160mm diameter red brick conduit at a temperature of 27℃.

Take ϵ(pipe)=0.79 and ϵ(brick conduit)=0.93

The net radiation heat exchange from pipe surface to room can be expressed as:

$\frac{Q}{L}$=$ε(πD_1 )σ(T_s^4-T_∞^4 )$

=0.79×π×0.08×5.67×$10^(-8)$×$573^4$-$300^4$

=1122.39W/m

The radiation heat exchange between pipe and conduit can be calculated as:

Q=$\frac{A_1 σ(T_s^4-T_∞^4 )}{\frac{1}{ε_1} +(\frac{1}{ε_2} -1) \frac{A_1}{A_2 }}$

When pipe is enclosed within brick conduit:

$\frac{A_1}{A_2}$ =$\frac{D_1}{D_2}$ =$\frac{0.08}{0.16}$=0.5

$\frac{Q}{L}$= $\frac{(π×0.08×5.67×$10^(-8)$×($573^4-300^4$)}{\frac{1}{0.79}+(\frac{1}{0.93}-1)×0.5)}$= …