Respuesta :
Answer:
2.80321285141
Explanation:
[tex]L_g[/tex] = Thickness of glass = 4.5 mm
[tex]k_g[/tex] = Thermal conductivity of glass = 0.8 W/mK
[tex]R_0[/tex] = Combined thermal resistance = [tex]0.15\times m^2K/W[/tex]
[tex]L_a[/tex] = Thickness of air = 6.6 mm
[tex]k_a[/tex] = Thermal conductivity of air = 0.024 W/mK
The required ratio is the inverse of total thermal resistance
[tex]\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[/tex]
The ratio is 2.80321285141
Answer:
[tex]\frac{\dot Q}{\dot Q'} =2.6668[/tex]
Explanation:
Given:
- area of the each window panes, [tex]A=0.15\ m^2[/tex]
- thickness of each pane, [tex]t_g=4.5\times 10^{-3}\ m[/tex]
- air gap between the two pane of a double pane window, [tex]t_a=6.6\times 10^{-3}\ m[/tex]
- thermal conductivity of glass, [tex]k_g=0.8\ W.m^{-1}.K^{-1}[/tex]
- thermal resistance of the air on the either sides of double pane window, [tex]R_{th}=0.15\ m^2.K.W^{-1}[/tex]
Heat loss through single pane window:
Using Fourier's law of conduction,
[tex]\dot Q=A.dT\div (R_{th}+\frac{t_g}{k} )[/tex]
[tex]\dot Q=0.15\times dT\div (0.15+\frac{4.5\times 10^{-3}}{0.8})[/tex]
[tex]\dot Q=0.9638\ dT\ [W][/tex]
Heat loss through double pane window:
[tex]\dot Q'=dT\times A\div(R_{th}+2\times \frac{t_g}{k}+\frac{t_a}{k_a} )[/tex]
where:
[tex]dT=[/tex] change in temperature
[tex]k_a=[/tex] coefficient of thermal conductivity of air [tex]= 0.026\ W.m^{-1}.K^{-1}[/tex]
[tex]\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})[/tex]
[tex]\dot Q'=0.3614\ dT\ [W][/tex]
Now the ratio:
[tex]\frac{\dot Q}{\dot Q'} =\frac{0.9638(dT)}{0.3614(dT)}[/tex]
[tex]\frac{\dot Q}{\dot Q'} =2.6668[/tex]
