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1-D Steady Heat Conduction

Copyright Brian G. Higgins (2009)

Introduction

The analytical solution of steady 1-D conduction problems is relatively straightforward if the problem islinear. A general solution to the differential equation can be determined by direct integration. However, ifthe conduction problem involves composite layers, the algebra required to satisfy the boundary condi-tions can become overwhelming. In this tutorial we illustrate how Mathematica can be used to handle allthe algebraic manipulations. The problem we will consider is steady 1-D heat conduction in an insulatedpipe (see Section 2.4, Whitaker,1983) . The inner radius of the pipe is R0 , the outer radius is R1 . Thepipe is filled with a fluid at temperature T A . For safety reasons the outer surface of the shell is coveredwith an insulating material. A schematic of the geometry is shown below:

R 1

R 2 R 3

Formulation

The steady state heat conduction is defined by

(1) 2 T = 0

For our geometry it makes sense to use cylindrical coordinates. The Laplacian 2 T in cylindrical coordi-nates is given by

(2) 2 T =1

r

!

! rr

! T

! r+

1

r 2

! 2 T

! q 2+

! 2 T

! z 2

where ( r, q , z ) are the radial, azimuthal and axial coordinate directions. The gradient operator incylindrical coordinates is

(3) T = e r! T

! r+ e q

1

r

! T

! q+ e z

! T

! z

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where He r , e q , e z L are the unit vectors tangent to the orthogonal coordinates.

If we take the heat flux to be in the radial direction only, then the heat conduction problem is 1-D andgiven by:

(4)

1

r

rr

T 1

r= 0, R 0 < r < R1

1

r

rr

T 2

r= 0, R 1 < r < R2

The boundary conditions are

(5)

BC1:k 1 T 1

rHR0 L = h 0 @T 1 HR0 L - T A D

BC2:T 1 HR1 L = T 2 HR1 L

BC3:k 1

T 1 r

HR1 L = k 2

T 2 r

HR1 L

BC4 : - k 2 T 2

rHR2 L = h 2 @T 2 HR2 L - T B D

We can integrate the ODEs in each region to get the general solution

(6)

T 1 Hr L = C1 Log @r D + C2

T 2 Hr L = C3 Log @r D + C4

Thus we have 4 constants and 4 BCs. In the next section we show how to use Mathematica to find theconstants.

Mathematica Assisted Solution

The following code clears all key variables

8T = ., h = ., k = ., T = ., r = ., A = ., B = . R2 ;We then define lists to hold all the equations and variables and use Solve to determine the constantsC i :

2 SteadyHeatCond_1D.nb

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eqns = Flatten @8BC1, BC2, BC3, BC4

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(7)q i = - k i T i

r

is continuous across the discontinuity in material properties. We define the following functions for theheat flux:

q1 @r_ D : = H- k 1 D@T1 @s D, s D . soln . s r L ; R 0 r R1q1 @r_ D : = H- k 2 D@T2 @s D, s D . soln . s r L ; R 1 r R2Here is a plot of the heat flux q i :

Plot @Evaluate @q1 @r D .soln D, 8r, R 0 , R 2