Given:
outer radius, R' = 10 m
inner diameter, d = 2 m
inner radius, R = [tex]\frac{d}{2}[/tex] = 1 m
surface temperature, T' = [tex]20^{\circ}C[/tex]
Thermal conductivity of soil, K = 0.52 W/mK
Solution:
To calculate the thermal temperature of conductor, T, we know amount of heat, Q is given by :
Q = [tex]\frac{T - T'}{\frac{R' - R}{4\pi KRR'}}[/tex]
500 = [tex](T - 20)\frac{4\pi \times0.52\times 1\times 10}{10 - 1}[/tex]
T = 68.86 +20 = [tex]88.865^{\circ}C[/tex]
Therefore, outside surface temperature is [tex]88.865^{\circ}C[/tex]
