For conduction through a cylinder with heat generation, the following assumptions are made:

1. steady-state conduction

2. one-dimensional radial conduction

3. constant thermodynamic properties

4. uniform volumetric heat generation

5. outer surface is adiabatic

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.

[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.