Answer :
Answer:
1. [tex]\dot Q=19600\ W[/tex]
2. [tex]\dot Q=120\ W[/tex]
Explanation:
1.
Given:
- height of the window pane, [tex]h=2\ m[/tex]
- width of the window pane, [tex]w=1\ m[/tex]
- thickness of the pane, [tex]t=5\ mm= 0.005\ m[/tex]
- thermal conductivity of the glass pane, [tex]k_g=1.4\ W.m^{-1}.K^{-1}[/tex]
- temperature of the inner surface, [tex]T_i=15^{\circ}C[/tex]
- temperature of the outer surface, [tex]T_o=-20^{\circ}C[/tex]
According to the Fourier's law the rate of heat transfer is given as:
[tex]\dot Q=k_g.A.\frac{dT}{dx}[/tex]
here:
A = area through which the heat transfer occurs = [tex]2\times 1=2\ m^2[/tex]
dT = temperature difference across the thickness of the surface = [tex]35^{\circ}C[/tex]
dx = t = thickness normal to the surface = [tex]0.005\ m[/tex]
[tex]\dot Q=1.4\times 2\times \frac{35}{0.005}[/tex]
[tex]\dot Q=19600\ W[/tex]
2.
- air spacing between two glass panes, [tex]dx=0.01\ m[/tex]
- area of each glass pane, [tex]A=2\times 1=2\ m^2[/tex]
- thermal conductivity of air, [tex]k_a=0.024\ W.m^{-1}.K^{-1}[/tex]
- temperature difference between the surfaces, [tex]dT=25^{\circ}C[/tex]
Assuming layered transfer of heat through the air and the air between the glasses is always still:
[tex]\dot Q=k_a.A.\frac{dT}{dx}[/tex]
[tex]\dot Q=0.024\times 2\times \frac{25}{0.01}[/tex]
[tex]\dot Q=120\ W[/tex]