How to Apply Heat Flow Equations to a Sphere?

AI Thread Summary
To apply heat flow equations to a sphere, start with the derived expression for heat flow along a bar, Q = -KA dT/dx, and adapt it for spherical coordinates. For a spherical heat source, use the equation Q = KA(dT/dr), where A is the surface area, specifically A = 4πr² for a sphere. Integrate the equation while considering the temperature gradient dT/dr at any distance r from the center. Remember that heat flows from hotter to colder regions, and apply boundary conditions such as the outer sphere being held at a constant temperature T_0. This approach will help determine the temperature at the surface of the heat-emitting sphere.
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I have derived expression for the heat flow along a bar with cross-sectional area A, given by 'Q = -KA dT/dx' where K is thermal conductivity constant and T and x refer to temperature and distance measured from the high temperature end of the bar.

I understand this. My problem is when I try to apply it to a sphere:

Say we have a spherical heat source of radius a at the centre of a solid sphere of radius b > a. Take the sphere as having thermal conductivity constant K. The source emits heat equally in all directions at a rate of Q per second. The outside of the outer sphere is help at constant temperature T_0.

How would I determine the temperature at the surface of the heat emitting sphere using the original differential equation?

I'm totally pulling my hair out about this one guys! Any guidance would be greatly appreciated.

Best Regards, James.
 
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Stick to first principles, which is the equation you have started with:
Assume equilibrium conditions.

\dot{Q} = KA\frac{dT}{dr}

where dT/dr is the temperature gradient at any distance r. The area of conduction at any distance r is A = 4\pi r^2.

Separate your variables and integrate accordingly. In case you didn't know, this same procedure can be used to derive your more familiar equation:

\dot{Q} = KA\frac{T_2-T_1}{x_2-x_1}
 
Don't forget that heat flows from hotter to colder! :)
 
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