A thermopane window consists of two glass panes, each 0.50 cm thick, with a 1.0-

Question

A thermopane window consists of two glass panes, each 0.50 cm thick, with a 1.0-cm-thick sealed layer of air in between. (Use 0.8 J/s middot m degree degree C for the thermal conductivity of glass and 0.0234 J/s middot m middot degree C for the thermal conductivity of air to answer the following.) (a) If the inside surface temperature is 21.4 degree C and the outside surface temperature is 0.0 degree C, determine the rate of energy transfer through 1.90 m^2 of the window. (b) Compare your answer to (a) with the rate of energy transfer through 1.90 m^2 of a single 1.0-cm-thick pane of glass. Disregard surface air layers.

Explanation / Answer

(a)

rate of energy transfer is given by

Q/t = KA(T1 - T2)/x

or Q/t = (T1 - T2)/(x/KA, where x/KA is the thermal resistance.

Now, the two window panes and the sealed air layer between are all in series.

So the equivalent thermal resistance is given by,

R = xw/KwA + xw/KwA + xair/KairA

where xw is the thickness of the window panes, Kw is the thermal condcutivity of window panes, xair is the  thickness of air layer and Kair is the thermal condcutivity of air layer.

So, R = (1/A)[2xw/Kw + xair/Kair]

So, rate of energy transfer is,

Q/t = (T1 - T2)/R = (T1 - T2)A/[2xw/Kw + xair/Kair]

or Q/t = (21.40C - 0)(1.9 m2)/[2(0.005m)/(0.8 J/s.m.0C) +(0.01 m)/(0.0234 J/s.m.0C)]

or  Q/t = 92.44 W

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(b)

In this case R = x/KwA , where x = 1 cm, so rate of energy transfer is,

Q/t = (T1 - T2)/(x/KwA) = (21.40C - 0)(1.9 m2)(0.8 J/s.m.0C)/(0.01m)

or Q/t = 3252.8 W is the rate of energy transfer through just the window panes.

This rate is around 35 times more than the panes with sealed air layer.

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