In: Other
A thermopane window consists of two pieces of glass each 0.1"
thick separated by a layer of dry,
stagnant air .4" thick.
• Find the resistance of a 4 ft × 3 ft thermopane window.
• Suppose the window is made by fastening an aluminum border .1"
thick all around the edge.
Find the total resistance for this configuration.
• As an alternative to the aluminum, we might use a border of .75"
thick white pine. Find the total
resistance for this configuration. (For white pine, "across the
grain", k = .087 Btu/hr ft °F).
• Calculate the heat loss for the two windows and compare with that
for a single pane of .1" glass.
Thermopane | |||
Glass Thickness | 0.1 | 0.00254 | m |
Air | 0.4 | 0.01016 | m |
L | 4 | 1.2192 | m |
B | 3 | 0.9144 | m |
kg | 0.869 | W/mK | |
ka | 0.026 | W/mK | |
R1 | 0.002622 | ||
R2 | 0.350517 | ||
R3 | 0.002622 | ||
Total Resistance | 0.355761 | K/W | |
Aluminium Border | |||
Glass Thickness | 0.1 | 0.00254 | m |
Air | 0.4 | 0.01016 | m |
L | 4 | 1.21412 | m |
B | 3 | 0.90932 | m |
kg | 0.869 | W/mK | |
ka | 0.026 | W/mK | |
kal | 0.022 | ||
R1 | 0.002647 | ||
R2 | 0.350517 | ||
R3 | 0.002647 | ||
R4 | 0.103562 | 0.100914 | |
Total Resistance | 0.456726 | K/W | |
White Pine Border | |||
Glass Thickness | 0.1 | 0.00254 | m |
Air | 0.4 | 0.01016 | m |
L | 4 | 1.20015 | m |
B | 3 | 0.89535 | m |
kg | 0.869 | W/mK | |
ka | 0.026 | W/mK | |
kp | 1.5047 | W/mk | |
R1 | 0.00272 | ||
R2 | 0.350517 | ||
R3 | 0.00272 | ||
R4 | 0.001514 | -0.00121 | |
Total Resistance | 0.358677 | K/W | |
Single Pane | |||
Glass Thickness | 0.1 | 0.00254 | m |
Air | 0.4 | 0.01016 | m |
L | 4 | 1.2192 | m |
B | 3 | 0.9144 | m |
kg | 0.869 | ||
ka | 0.026 | ||
R1 | 0.002622 | ||
R2 | 0.350517 | ||
R3 | |||
Total Resistance | 0.353139 | K/W | |
Temperature Diff | 5 | K | |
Single Pane Heat Loss | 15.78467 | W/m | |
Double Pane Heat Loss | 15.66835 | W/m |