CL: Conduction Example(s) (cont.)

Another (more complicated) example:

We have now added a second pane of glass (also 5 mm) with an air gap in between (7 mm). The inner and outer surfaces are again kept constant at 12.5 C and -9 C, respectively. What is the rate of heat loss through the window (it is still 1m by 1.5m on a side)? The thermal conductivity of the glass is 1.4 W/mK of air is 0.025 W/mK.

Each region (I, II, and III) may be solved exactly like our previous example, so our solutions for each (stolen from previous example) are:

$\displaystyle{q_{a\to b} = \frac{k_gA}{L_g}(T_a-T_b)}$

$\displaystyle{q_{b\to c} = \frac{k_{air}A}{L_{air}}(T_b-T_c)}$

$\displaystyle{q_{c\to d} = \frac{k_gA}{L_g}(T_c-T_d)}$

Any of these equations would suffice to tell us what the rate of heat flow is since we are at steady state. However, since we only know Ta and Tb, we cannot solve any of them! So...

Rearranging the equations:

$\displaystyle{q_{a\to b}\frac{L_g}{k_gA} = (T_a-T_b)}$

$\displaystyle{q_{b\to c}\frac{L_{air}}{k_{air}A} = (T_b-T_c)}$

$\displaystyle{q_{c\to d} \frac{L_g}{k_gA}= (T_c-T_d)}$

We can now add them (we need to note that qab=qbc=qcd=q):

$\displaystyle{q \left (\frac{L_g}{k_gA} +\frac{L_{air}}{k_{air}A} + \frac{L_g}{k_gA}\right )= (T_a-T_b+T_b-T_c+T_c-T_d)}$

which simplifies to:

$\displaystyle{q \left (\frac{L_g}{k_gA} +\frac{L_{air}}{k_{air}A} + \frac{L_g}{k_gA}\right )= (T_1-T_2)}$

If we now denote (for reasons we will discuss in a minute)

$\displaystyle{R = \left (\frac{L_g}{k_gA} +\frac{L_{air}}{k_{air}A} + \frac{L_g}{k_gA}\right )}$

we can write this solution in a very simple equation:

$\displaystyle{q_{1\to 2} = \frac{(T_1-T_2)}{R}}$

Plugging in our numbers for R:

$\displaystyle{R = \left (\frac{0.005m}{(1.4W/mK)(1.5m^2)} +\frac{0.007m}{(0.025W/mK)(1.5m^2)} + \frac{0.005m}{(1.4W/mK)(1.5m^2)}\right ) = 0.19K/W}$

So our heat flow becomes:

$\displaystyle{q_{1\to 2} = \frac{(12.5C-(-9C))}{0.19K/W} = 112W}$

This is remarkably smaller than the case of a single pane of glass!