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# Heat Transfer through a Cylinder

A long tube with a uniform heat source is insulated at its outer radius and cooled at its inner radius , and the one-dimensional, radial, steady-state heat transfer is calculated. Use buttons to view a cross section of the tube or plot the temperature as a function of the radius. The heat transfer rate, temperature at position and the outer surface temperature are shown on the cross section diagram. Use sliders to set the coolant temperature, tube thickness , radial position, heat generation, convection coefficient and thermal conductivity.

### DETAILS

For conduction through a cylinder with heat generation, the following assumptions are made:
3. constant thermodynamic properties
4. uniform volumetric heat generation
The heat diffusion equation is solved to determine the radial temperature distribution :
.
The above assumptions reduce this equation to:
.
Separating and integrating yields:
(1) ,
(2) .
Two boundary conditions are needed to determine constants and ; both conditions are evaluated at the outer radius :
(4) ,
(5) .
Evaluating equation (1) at boundary condition (5), evaluating equation (2) at boundary condition (4), and solving for and yields:
,
.
The general solution for the temperature distribution is then:
,
where is temperature, is radius, the subscript refers to the outer surface, is heat generation, and is thermal conductivity.
From Fourier's law, the heat removal rate is:
.
Substituting from and evaluating at the inner radius :
.
Because the tube is insulated at , the rate of heat generated in the tube must equal the rate of removal at . Simplifying yields:
.
The inner surface temperature is calculated using the conservation of energy, since :
,
.
The outer surface temperature is found by evaluating at :
,
where is the coolant temperature, and the subscript refers to the inner surface.
Reference
[1] T. L. Bergman, A. S. Lavine, F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Hoboken, NJ: John Wiley & Sons, 2011.

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