Question

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.

Answer 1

The answer is below

Step-by-step explanation:

Given:

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:

`R`

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:

`\dot {Q}=A*(\Delta T)/(R`

The inner surface temperature (Ti) is given as:

`T_i=T-\frac{\dot {Q}}{A} *(1)/(h_1)\\ \\T_i=20-(147.73)/(3.6)*(1)/(10)=15.9\ ^oC`