Fig, shows a system of two concentric spheres of radii `r_(1)` and` r_(2)` and kept at temperature `T_(1)` and `T_(2) (T_(1) gt T_(2))` respectively. Find the expression for radial rate of flow of heat through the substance.

Explanation

Consider an elementary portion of two spheres between two concentric spherical sheels of radius x and `(x+dx)` fig. Let T and `(T-dT)` be the temperature of the inner and outer part of the elementary portion of the shells. Then rate of flow of heat through the elementary portion is

`(dQ)/(dt) = (-KA[T-(T-dT)])/(dx) = (-KA dT)/(dx)`
Here -ve sign shown that the heat is lost by the spheres.
Surface area of elementary portion of two spheres,
`A = 4 pi x^(2)`
`:. (dQ)/(dt) = -K 4 pi x^(2) (dT)/(dx)`
or `dT = (-1)/(4 pi K) (dQ)/(dt) x^(-2) dx`
Intergrating both the side within the proper limits, we have
`int_(T_1)^(T_2) dt = -(1)/(4 pi K) (dQ)/(dt) int_(r_1)^(r_2) x^(-2) dx`
`[T]_(T_1)^(T_2) = -(1)/(4 pi K) (dQ)/(dt)[-(1)/(x)]_(r_1)^(r_2)`
`T_(2)-T_(1) = (1)/(4 pi K) (dQ)/(dt)[1/(r_2)-1/(r_2)]`
or `T_(1)-T_(2) = (1)/(4 pi K) (dQ)/(dt)(1/(r_1)-1/(r_2))`
`=(1)/(4 pi K) (dQ)/(dt)(((r_2)-(r_1))/(r_(1)r_(2)))`
`(dQ)/(dt) = (4 pi K (T_(1)-T_(2))r_(1)r_(2))/((r_(2)-r_(1))`.