Q. 28

A room has a window fitted with a single 1.0 m × 2.0 m glass of thickness 2 mm.

(a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and that outside is 10°C.

(b) The glass is now replaced by two glass panes, each having a thickness of 1 mm and separated by a distance of 1 mm.

Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0 J s–1m–1 °C–1and that of air = 0.025 J s–1m–1 °C–1.

Answer :

The diagram is shown –


w7.PNG


(a) Given


Thickness, l = 2 mm = 0.0002 m


Temperature inside the room = 32°C


outside = 10°C


Dimensions of wall = 1.0 m × 2.0 m


Rate of flow of heat


=


Where,


∆T = is change in temperature between the two sides of the window.


A= Area of cross section of the window


K = thermal conductivity of the window


L = length of the window


=


= 8000J/s


(b). Resistance of glass-


The equivalent circuit for the two glass panes and air becomes


w8.PNG


Here resistance of glass Rg =


And of air Rair =


Since, these are connected in series, equivalent resistance becomes


Reqv =


Thermal conductivity of window glass = 1.0 J s–1m–1 °C–1


And of air, = 0.025 J s–1m–1 °C–1.


Substituting values


Reqv = ( + )


= ( + )


= 0.021


Now, rate of heat flow



=


= 380.95


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