Answer:
A)Q = 1208.33 W/m²
B)K = 0.138 W/m.K
Explanation:
We are given;
inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K
outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K
Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K
Thickness, L = 4mm = 0.004m
convection heat transfer coefficient ; hi = 25 W/(m².K)
A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;
Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]
Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;
Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]
Q = 583.33 + 625
Q = 1208.33 W/m²
B) The formula for thermal conductivity is;
K = (QL)/(AΔT)
Where;
K is the thermal conductivity in W/m.K
Q is the amount of heat transferred through the material
L is the distance between the two isothermal planes
A is the area of the surface in square meters
ΔT is the difference in temperature in Kelvin
ΔT = 298K - 263K = 35K
Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;
k = (Q/A) x (L/ΔT)
K = 1208.33 x (0.004/35)
K = 0.138 W/m.K
