Calculate the thermal resistance and magnitude of conductive heat flow/flux through a planar wall
Lets look at an example:
The inner and outer surfaces of a glass window are kept at 12.5 C and -9 C, respectively. What is the rate of heat loss through a window that is 5 mm thick and is 1m by 1.5m on a side? The thermal conductivity of the glass is 1.4 W/mK. |
In how many different directions is heat flowing? (i.e., what is the dimensionality of the problem?)
$\displaystyle{\frac{q}{A} = -k\frac{dT}{dx}}$
We can separate these variables and integrate...
$\displaystyle{\int_0^L q dx = \int_{T_1}^{T_2} -kAdT}$
We can reason that q should not change with x (or
else we would have accumulation).
If we also assume that k (and A) are not functions of
T, we get:
$\displaystyle{q\int_0^L dx = -kA\int_{T_1}^{T_2} dT}$
which can be integrated to yield:
$\displaystyle{qL = -kA(T_2-T_1)}$
or
$\displaystyle{q = \frac{kA}{L}(T_1-T_2)}$
Plugging in our numbers...
$\displaystyle{q = \frac{(1.4W/mK)(1m)(1.5m)}{0.005m}(12.5^oC-
(-9^oC))=9.0kW}$
In order to understand why the heat flow within the planar wall must be constant in x at steady state, let's consider some small segment within the wall:
If you consider a fluid flow across this section of our planar wall, it is easy to realize that if v_{1} is not equal to v_{2} then we have more material flowing into (out of) our small region then is flowing out of (into) this region. By definition, if IN does not equal OUT, we have accumulation (unsteady conditions!). |
This idea is very similar to the idea behind the "continuity equation" (conservation of mass) for an incompressible fluid at steady state, which we will see soon:
Also, the General Thermal Energy Balance Equation will prove this to use, at least mathematically. In any case, it is useful to recognize intuitively that in 1-D, the (heat) flow cannot change in the direction that the flow is occurring (at steady state).