Question

Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m2 to that for a double-pane window with the same area. The glass of a single pane is 4.5 mm thick, and the air space between the two panes of the double-pane window is 6.60 mm thick. The glass has thermal conductivity 0.80 W/m⋅K. The air films on the room and outdoor surfaces of either window have a combined thermal resistance of 0.15 m2⋅K/W. Express your answer using two significant figures.

Answer 1

Answer:

2.80321285141

Explanation:

$L_g$ = Thickness of glass = 4.5 mm

$k_g$ = Thermal conductivity of glass = 0.8 W/mK

$R_0$ = Combined thermal resistance = $0.15\times m^2K/W$

$L_a$ = Thickness of air = 6.6 mm

$k_a$ = Thermal conductivity of air = 0.024 W/mK

The required ratio is the inverse of total thermal resistance

$\dfrac{2(L_g/k_g)+R_0+(L_a/k_a)}{(L_g/k_g)+R_0}\\ =\dfrac{2(4.5\times 10^{-3}/0.8)+0.15+(6.6\times 10^{-3}/0.024)}{(4.5\times 10^{-3}/0.8)+0.15}\\ =2.80321285141$

The ratio is 2.80321285141

Answer 2

Answer:

$\frac{\dot Q}{\dot Q'} =2.6668$

Explanation:

Given:

• area of the each window panes, $A=0.15\ m^2$

• thickness of each pane, $t_g=4.5\times 10^{-3}\ m$

• air gap between the two pane of a double pane window, $t_a=6.6\times 10^{-3}\ m$

• thermal conductivity of glass, $k_g=0.8\ W.m^{-1}.K^{-1}$

• thermal resistance of the air on the either sides of double pane window, $R_{th}=0.15\ m^2.K.W^{-1}$

Heat loss through single pane window:

Using Fourier's law of conduction,

$\dot Q=A.dT\div (R_{th}+\frac{t_g}{k} )$

$\dot Q=0.15\times dT\div (0.15+\frac{4.5\times 10^{-3}}{0.8})$

$\dot Q=0.9638\ dT\ [W]$

Heat loss through double pane window:

$\dot Q'=dT\times A\div(R_{th}+2\times \frac{t_g}{k}+\frac{t_a}{k_a} )$

where:

$dT=$ change in temperature

$k_a=$ coefficient of thermal conductivity of air $= 0.026\ W.m^{-1}.K^{-1}$

$\dot Q'=dT\times 0.15\div (0.15+2\times \frac{4.5\times 10^{-3}}{0.8}+\frac{6.6\times 10^{-3}}{0.026})$

$\dot Q'=0.3614\ dT\ [W]$

Now the ratio:

$\frac{\dot Q}{\dot Q'} =\frac{0.9638(dT)}{0.3614(dT)}$

$\frac{\dot Q}{\dot Q'} =2.6668$