Two thin metallic spherical shells of radii r1 and r2 (r1 < r2) are placed with their centres coinciding. Material of thermal conductivity K is filled in the space between the shells. The inner shell is maintained at temperature θ1 and the outer shell at temperature θ2(θ1 < θ2). The rate at which heat flows radially through the material is :
(1) \(\frac{4 \pi Kr_1r_2(θ_2 - θ_1)}{r_2 - r_1}\)
(2) \(\frac{\pi r_1r_2(θ_2 - θ_1)}{r_2 - r_1}\)
(3) \(\frac{K(θ_2 - θ_1)}{r_2 - r_1}\)
(4) \(\frac{K(θ_2 - θ_1)(r_2-r_1)}{4 \pi r_1r_2}\)
Correct answer: (1) \(\frac{4 \pi Kr_1r_2(θ_2 - θ_1)}{r_2 - r_1}\)
Explanation:
Thermal resistance of spherical sheet of thickness dr and radius r is
dR = \(\frac{dr}{K(4\pi r^2)}\)
R = \(\int_{r_1}^{r_2}\frac{dr}{K(4\pi r^2)}\)
R = \(\frac{1}{4\pi K}\)\(\Big(\frac{1}{r_1} - \frac{1}{r_2}\Big)\)
= \(\frac{1}{4\pi K}\)\(\Big(\frac{r_2 - r_1}{r_1r_2}\Big)\)
Thermal current (i) = \(\frac{θ_2 - θ_1}{R}\)
i = \(\frac{4 \pi Kr_1r_2(θ_2 - θ_1)}{r_2 - r_1}\)