Answer:
The relation is such that the product of radial temperature gradient and the square of radius is constant or the radial temperature gradient is inversely proportional to the square of radius.
Explanation:
For this consider following assumptions
- The system is in a steady state condition
- There is no internal energy generation
- There is only 1-D conduction in the radial direction only.
- The system is on constant properties
By using the the Fourier Law in radial direction(spherical coordinates) is given as
[tex]q_r=-kA_r\frac{dT}{dr}[/tex]
Here A_r is the area of the surface, which for the case of sphere is 4πr^2. Thus the equation becomes
[tex]q_r=-4\pi kr^2\frac{dT}{dr}[/tex]
Now in the steady state condition
[tex]\dot{E}_{in}=\dot{E}_{out}[/tex]
This is true due to no internal heat generation and no state change. i.e.
[tex]\dot{E}_{g}=\dot{E}_{st}=0[/tex]
As the energy balance is valid, this indicates
[tex]q_{in}=q_{out}=q_r[/tex]
As the value of r is changing and it is same at two points, this indicate that the function is independent of r such that
[tex]q_r\neq q_r(r)[/tex]
This implies that the function is constant i.e.
[tex]q_r=-4\pi kr^2\frac{dT}{dr}=constant\\[/tex]
As -4πk are already constant, this can be true iff
[tex]r^2\frac{dT}{dr}=constant\\[/tex]
This relation indicates that the product of radial temperature gradient and the square of radius is constant or the radial temperature gradient is inversely proportional to the square of radius.
The distribution is indicated in the attached diagram along with the initial sketch.
