In the Heat Conduction I module the differential equations derived
in examples 1, 2b and 3 were of first order and easily solved by
integration. Let us review them now. In all cases, temperatures are
in degrees Kelvin.
Example 1:
Cartesian geometry
Constant thermal conductivity
One dimensional heat flow in the x-direction
| The temperature distribution T = T(x) was shown
to be governed by the differential equation |
| (1.1) |
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| and the boundary conditions |
| (1.2) |
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| While the constant k is given, the heat flux q
is a constant that must be determined as part of the solution. |
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We obtained the solution
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| (1.3) |
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giving the temperature gradient
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| (1.4) |
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and the heat flux (using equations (1.1) and (1.4))
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| (1.5) |
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units: Watts/m2
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| Example 2b: |
Cylindrical geometry |
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Constant thermal conductivity |
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Radial heat flow in the r-direction |
| The differential equation was |
| (1.6) |
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| where the constant k is given and heat flow
Q is a constant that must be determined as part of the solution.
The boundary conditions were |
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The solution was
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which yields the temperature gradient
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and, using equations (1.6) and (1.9), the constant
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units: Watts/m of length
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Recall that Q is the rate of heat flow through
any cylindrical surface of radius r
per unit length.
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| Example 3: |
Cartesian geometry |
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Thermal conductivity a function of temperature |
| |
One dimensional heat flow in the x-direction |
| The differential equation was |
| (1.11) |
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| where the constant q is to be determined, while
the thermal conductivity k depends on T: |
| (1.12) |
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The boundary conditions were
|
| (1.13) |
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The solution was
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| (1.14) |
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which makes the temperature gradient
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| (1.15) |
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The heat flux may be found by substituting x =
L (and T = TL) in (1.14) and then solving for
q:
|
| (1.16) |
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units: Watts/m2
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