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One-dimensional heat conduction in a spherical coordinate system can be solved by introducing a new dependent variable. Let us consider a sphere with radius of ro and a uniform initial temperature of <math>{T_i}</math>. It is exposed to a fluid with a temperature of <math>{T_\infty }</math> (<math>{T_\infty } < {T_i}</math>) and the convective heat transfer coefficient between the fluid and finite slab is <math>h</math>. Assuming that there is no internal heat generation and constant thermophysical properties, the governing equation is
	
<center><math>\frac{1}{r}\frac{{{\partial ^2}(rT)}}{{\partial {r^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}},{\rm{   }}0 < x < {r_o},{\rm{ }}t > 0    \qquad \qquad(1)</math></center>

subject to the following boundary and initial conditions
 	
<center><math>\frac{{\partial T}}{{\partial r}} = 0,{\rm{   }}r = 0{\rm{ (axisymmetric)}}    \qquad \qquad(2)</math></center>

	<center><math> - k\frac{{\partial T}}{{\partial r}} = h(T - {T_\infty }),{\rm{   }}r = {r_o}    \qquad \qquad(3)</math></center>

	<center><math>T = {T_i},{\rm{   }}0 < r < {r_o},{\rm{ }}t = 0    \qquad \qquad(4)</math></center>

By using the following dimensionless variables 
<math>\theta  = \frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}},{\rm{ }}R = \frac{r}{{{r_o}}},{\rm{ Fo}} = \frac{{\alpha t}}{{r_o^2}},{\rm{ Bi}} = \frac{{h{r_o}}}{k} </math> 

eqs. (1) – (4) can be nondimensionalized as
	
<center><math>\frac{1}{R}\frac{{{\partial ^2}(R\theta )}}{{\partial {R^2}}} = \frac{{\partial \theta }}{{\partial {\rm{Fo}}}},{\rm{   }}0 < R < 1,{\rm{ Fo}} > 0    \qquad \qquad(5)</math></center>

	<center><math>\frac{{\partial \theta }}{{\partial R}} = 0,{\rm{   }}R = 0    \qquad \qquad(6)</math></center>

	<center><math> - \frac{{\partial \theta }}{{\partial R}} = {\rm{Bi}}\theta ,{\rm{   R}} = 1    \qquad \qquad(7)</math></center>

	<center><math>\theta  = 1,{\rm{   }}0 < R < 1,{\rm{ Fo}} = 0    \qquad \qquad(8)</math></center>

Defining a new dependent variable 

	<center><math>U = R\theta     \qquad \qquad(9)</math></center>

eqs. (5) – (8) become

	<center><math>\frac{{{\partial ^2}U}}{{\partial {R^2}}} = \frac{{\partial U}}{{\partial {\rm{Fo}}}},{\rm{   }}0 < R < 1,{\rm{ Fo}} > 0    \qquad \qquad(10)</math></center>
	
	<center><math>\frac{{\partial \theta }}{{\partial R}} = 0,{\rm{   }}R = 0    \qquad \qquad(11)</math></center>

	<center><math> - \frac{{\partial U}}{{\partial R}} = ({\rm{Bi - 1)}}\theta ,{\rm{   R}} = 1    \qquad \qquad(12)</math></center>

	<center><math>U = R,{\rm{   }}0 < R < 1,{\rm{ Fo}} = 0    \qquad \qquad(13)</math></center>

It can be seen that eqs. (10) – (13) are identical to the case of heat conduction in a finite slab with a non-uniform initial temperature. This problem can be readily solved by using the method of separation of variables (see Problem 3.28). After the solution is obtained, one can change the dependent variable back to <math>\theta </math> and the result is

	<center><math>\theta  = \frac{1}{R}\sum\limits_{n = 1}^\infty  {\frac{{4[\sin ({\lambda _n}) - {\lambda _n}\cos ({\lambda _n})]}}{{{\lambda _n}[2{\lambda _n} - \sin (2{\lambda _n})]}}} \sin ({\lambda _n}R){e^{ - \lambda _n^2{\rm{Fo}}}}    \qquad \qquad(14)</math></center>

where the eigenvalue is the positive root of the following equation

 	<center><math>1 - {\lambda _n}\cot {\lambda _n} = {\rm{Bi}}    \qquad \qquad(15)</math></center>

==References==
Faghri, A., Zhang, Y., and Howell, J. R., 2010, ''Advanced  Heat and Mass Transfer'', Global Digital Press, Columbia, MO.

==Further Reading==

==External Links==

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