A hollow sphere of inner radius R and outer r | vedclass

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(C) The thermal resistance $R_{th}$ of a spherical shell of inner radius $r_1$,outer radius $r_2$,and thermal conductivity $K$ is given by $R_{th} = \

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(C) The thermal resistance $R_{th}$ of a spherical shell of inner radius $r_1$,outer radius $r_2$,and thermal conductivity $K$ is given by $R_{th} = \frac{1}{4\pi K} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.For the inner sphere $(R_1 = R, R_2 = 2R)$: $R_{th1} = \frac{1}{4\pi K} \left( \frac{1}{R} - \frac{1}{2R} \right) = \frac{1}{4\pi K} \left( \frac{1}{2R} \right) = \frac{1}{8\pi KR}$.For the outer sphere $(R_2 = 2R, R_3 = 3R)$: $R_{th2} = \frac{1}{4\pi K} \left( \frac{1}{2R} - \frac{1}{3R} \right) = \frac{1}{4\pi K} \left( \frac{1}{6R} \right) = \frac{1}{24\pi KR}$.In steady state,the heat current $H$ is the same through both spheres: $H = \frac{T_{int} - 0}{R_{th1}} = \frac{100 - T_{int}}{R_{th2}}$.Substituting the resistances: $\frac{T_{int}}{1/(8\pi KR)} = \frac{100 - T_{int}}{1/(24\pi KR)}$.$8\pi KR \cdot T_{int} = 24\pi KR \cdot (100 - T_{int})$.$T_{int} = 3(100 - T_{int}) \implies T_{int} = 300 - 3T_{int}$.$4T_{int} = 300 \implies T_{int} = 75^oC$.

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