Question

1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inner and outer surface temperatures of the glass are 15°C and -20°C. respectively, on a cold winter day, what is the rate of heat loss through the glass?
2. To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by an air space. If the spacing is 10 mm and the glass surfaces in contact with the air have temperatures of 10° C and - 15°C, what is the rate of heat loss from a 1 m times 2 m window? The thermal conductivity of air is ka = 0.024 W/m*K.

Answer 1

Answer:

1. $\dot Q=19600\ W$

2. $\dot Q=120\ W$

Explanation:

1.

Given:

• height of the window pane, $h=2\ m$

• width of the window pane, $w=1\ m$

• thickness of the pane, $t=5\ mm= 0.005\ m$

• thermal conductivity of the glass pane, $k_g=1.4\ W.m^{-1}.K^{-1}$

• temperature of the inner surface, $T_i=15^{\circ}C$

• temperature of the outer surface, $T_o=-20^{\circ}C$

According to the Fourier's law the rate of heat transfer is given as:

$\dot Q=k_g.A.\frac{dT}{dx}$

here:

A = area through which the heat transfer occurs = $2\times 1=2\ m^2$

dT = temperature difference across the thickness of the surface = $35^{\circ}C$

dx = t = thickness normal to the surface = $0.005\ m$

$\dot Q=1.4\times 2\times \frac{35}{0.005}$

$\dot Q=19600\ W$

2.

• air spacing between two glass panes, $dx=0.01\ m$

• area of each glass pane, $A=2\times 1=2\ m^2$

• thermal conductivity of air, $k_a=0.024\ W.m^{-1}.K^{-1}$

• temperature difference between the surfaces, $dT=25^{\circ}C$

Assuming layered transfer of heat through the air and the air between the glasses is always still:

$\dot Q=k_a.A.\frac{dT}{dx}$

$\dot Q=0.024\times 2\times \frac{25}{0.01}$

$\dot Q=120\ W$