Recap of the Heat Conduction I Module

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)
 
and the boundary conditions
(1.2)
 
While the constant k is given, the heat flux q is a constant that must be determined as part of the solution.
We obtained the solution
(1.3)
 
giving the temperature gradient
(1.4)
 
and the heat flux (using equations (1.1) and (1.4))
(1.5)
units: Watts/m2

Example 2b: Cylindrical geometry
  Constant thermal conductivity
  Radial heat flow in the r-direction


The differential equation was
(1.6)
 
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
(1.7)
 
The solution was
(1.8)
 
which yields the temperature gradient
(1.9)
 
and, using equations (1.6) and (1.9), the constant
(1.10)
units: Watts/m of length
Recall that Q is the rate of heat flow through any cylindrical surface of radius r per unit length.

Example 3: Cartesian geometry
  Thermal conductivity a function of temperature
  One dimensional heat flow in the x-direction

 
The differential equation was
(1.11)
 
where the constant q is to be determined, while the thermal conductivity k depends on T:
(1.12)
The boundary conditions were
(1.13)
The solution was
(1.14)
which makes the temperature gradient
(1.15)
The heat flux may be found by substituting x = L (and T = TL) in (1.14) and then solving for q:
(1.16)
units: Watts/m2


Copyright 1998-2001 Rensselaer Polytechnic Institute. All Rights Reserved.