Question 14.1: The exterior concrete wall of a residential building has a s......
The exterior concrete wall of a residential building has a surface area A of 620 m² and thickness \delta of 360 mm. The wall has a thermal insulation on its outer surface. The thermal conductivity of concrete k is 2 W/mK and of insulation k_{i} is 0.04 W/mK. The indoor and outdoor air temperatures are t_{i} = 18°C and t_{o} = −12°C, respectively. The heat-transfer coefficients are h_{i} = 15 W/m²K (on the inner surface) and h_{o} = 25 W/m²K (on the outer surface). Determine (1) the rate of heat loss through the wall without insulation and (2) the thickness of the insulation required to reduce the rate of heat loss by 60%.
1. Uninsulated concrete wall: thickness δ = 0.36 m, thermal conductivity k = 2 W/m K, h_{i} = 15 W/m²K, h_{o} = 25 W/m²K
a. Overall heat-transfer coefficient for heat transfer from the warm indoor air to cold outdoor air across the wall
U=(1/h_{i}+\mathbf{d}/k+1/h_{o})^{-1}=(1/15+0.36/2+1/25)^{-1}=3.488\ \mathrm{W/m^{2}K}b. Total thermal resistance of the uninsulated wall
R=1/h_{i}+\mathbf{d}/k+1/h_{\mathrm{o}}\,{or}\,R=1/U=1/3.488=0.287\,\mathrm{m^{2}K/W}c. Rate of heat loss through the uninsulated wall
Q=A(t_{\mathrm{i}}-t_{\mathrm{o}})/R=U A~(t_{\mathrm{i}}-t_{\mathrm{o}})=3.488\,^{*}\,620\,^{*}\left[18-(-12)\right]=64,877\,\mathrm{W}2. Insulated concrete wall (insulation with thermal conductivity k_{i} = 0.04 W/m K)
a. Rate of heat loss through the insulated wall is reduced by 60%. Thus,
Q_{\mathrm{ins}}=(1-0.6)\ Q=0.4\,^{*}{\, 64,877=25,951\ W}b. Required thickness of the insulation is calculated as follows:
i. Total thermal resistance of the insulated wall
ii. Thermal resistance of the insulation
(\mathbf{d}/k)_{\mathrm{ins}}\,=\,R_{\mathrm{ins}}-R\,=\,1.395\,\,-\,\,0.287\,=\,1.108\,\,\mathrm{m^{2}K/W}iii. Thickness of the insulation
\mathbf{d}_{\mathrm{ins}}=(R_{\mathrm{ins}}-R)^{*}\,k_{\mathrm{ins}}=1.108\,^{*}\,0.04=0.0443\,\mathrm{m}=44.3\,\mathrm{mm}\,