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 *T*_{a} and
*T*_{b}, 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
*q*_{ab}=*q*_{bc}=*q*_{cd}=*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!