Consider a 1.5-m-high and 2.4-m-wide double-pane window consisting of two 3-mm-thick layers of glass (k = 0.78 W/m⋅K) separated by a 12-mm-wide stagnant airspace (k = 0.026 W/m⋅K). Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at 20°C while the temperature of the outdoors is −5°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1 = 10 W/m2⋅K and h2 = 25 W/m2⋅K, respectively, and disregard any heat transfer by radiation.

kg (glass) = 0.78 W/m⋅K, ka (air) = 0.026 W/m⋅K, h1 = 10 W/m2⋅K, h2 = 25 W/m2⋅K, glass length = Lg = 3 mm = 0.003 m, Lo = 3 mm = 0.003 m, La = 12 mm = 0.012 Height = 1.5 m, width = 2.4 m, room temperature = T = 20°C. Therefore:

Total resistance per unit area is given as:

[tex]R"=\frac{1}{h_1}+\frac{L_g}{k_g}+\frac{L_a}{k_a} +\frac{L_o}{k_g}+\frac{1}{h_2} \\\\R"=\frac{1}{10}+\frac{0.003}{0.78}+\frac{0.012}{0.026} +\frac{0.003}{0.78}+\frac{1}{25} \\\\R"=0.1+0.00385+0.46154+0.00385+0.04\\\\R"=0.60924\ K.m^2/W[/tex]

Area = A = height * width = 1.5 m × 2.4 m = 3.6 m²

The change in temperature = ΔT = 20 °C - (-5 °C) = 25 °C

The rate of heat loss is given as:

[tex]\dot {Q}=A*\frac{\Delta T}{R"}= 3.6*\frac{25}{0.60924}\\ \\\dot {Q}=147.73\ W[/tex]

The inner surface temperature (Ti) is given as:

[tex]T_i=T-\frac{\dot {Q}}{A} *\frac{1}{h_1}\\ \\T_i=20-\frac{147.73}{3.6}*\frac{1}{10}=15.9\ ^oC[/tex]