Question 295: Thermodynamics & KTG

Question

Two thin metallic spherical shells of radii r
1
and r
2
(r
1
< r
2
) are placed with their centres coinciding. A material of thermal conductivity K is filled in the space between the shells. The inner shell is maintained at temperature $$	heta$$
1
and the outer shell at temperature $$	heta$$
2
($$	heta$$
1
< $$	heta$$
2
). The rate at which heat flows radially through the material is :-

Accepted Answer

Thermal resistance of spherical sheet of thickness dr and radius r is
$$dR = {{dr} \over {K(4\pi {r^2})}}$$
$$R = \int\limits_{{r_1}}^{{r_2}} {{{dr} \over {K(4\pi {r^2})}}} $$
$$R = {1 \over {4\pi K}}\left( {{1 \over {{r_1}}} - {1 \over {{r_2}}}} ight) = {1 \over {4\pi K}}\left( {{{{r_2} - {r_1}} \over {{r_1}{r_2}}}} ight)$$
Thermal current (i) $$ = {{{	heta _2} - {	heta _1}} \over R}$$
$$i = {{4\pi K{r_1}{r_2}} \over {{r_2} - {r_1}}}({	heta _2} - {	heta _1})$$

Answer

$${{4\pi K{r_1}{r_2}({	heta _2} - {	heta _1})} \over {{r_2} - {r_1}}}$$

Answer

$${{\pi {r_1}{r_2}({	heta _2} - {	heta _1})} \over {{r_2} - {r_1}}}$$

Answer

$${{K({	heta _2} - {	heta _1})} \over {{r_2} - {r_1}}}$$

Answer

$${{K({	heta _2} - {	heta _1})({r_2} - {r_1})} \over {4\pi {r_1}{r_2}}}$$

Thermodynamics & KTG

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