# A thermopane window consists of 2 glass panes,each 0.5cm thick,with 1cm thick sealed layer of air in between. Inside 23°C outside 0°C. Determine the rate of energy transfer through 1m² of the window?

Jul 9, 2018

see below

#### Explanation:

in a first approch you can consider only conduction with the fourier law:
$\frac{Q}{t} = \frac{C}{s} \times S \times \Delta T$
for more layers of isolant you have:
$\frac{Q}{t} = \frac{S \times \Delta T}{{s}_{1} / {C}_{1} + {s}_{2} / {C}_{2} + {s}_{3} / {C}_{3}}$.

Where S is the area of the window =$1 {m}^{2}$
$\Delta T$ the total differnce of temperature = 23°C=23 K
s are the thickness of glasses and of air and C the thermal conductivity of glass $0 , 5 \frac{W}{m K}$ and of air $0 , 02 \frac{W}{m K}$
$\frac{Q}{t} = \frac{S \times \Delta T}{{s}_{1} / {C}_{1} + {s}_{2} / {C}_{2} + {s}_{3} / {C}_{3}} = \frac{1 {m}^{2} \times 23 K}{\frac{0.005 m}{\frac{0.5 W}{m K}} + \frac{0.01 m}{0.02 \frac{W}{m K}} + \frac{0.005 m}{\frac{0.5 W}{m K}}} = \frac{1 {m}^{2} \times 23 K}{\left(0.01 {m}^{2} \times \frac{K}{W}\right) + \left(0.5 {m}^{2} \times \frac{K}{W}\right) + \left(0.01 {m}^{2} \times \frac{K}{W}\right)} = \frac{1 {m}^{2} \times 23 K}{\left(0.52 {m}^{2} \times \frac{K}{W}\right)} = 45 W$

It seems that only the layer of air is useful to stop the energy transfer, but there are three strong approximations:
1) if the layer is bigger than 1 mm you have convention beyond conduction in air so you could suppose that air layer is only few mm.
2) in the outside of the window and in the inside of the window there are two layer of air that oppose to energy transfer
3) you have trasfer of heat also throught wood or the material with which the window frame is made.
Finally You can consider about 20-40 W of rate of energy trasfer