Question 7.6: Thermal Resistance Consider heat transfer through an insulat...
Thermal Resistance
Consider heat transfer through an insulated wall as shown in Figure 7.24. The wall is made of a layer of brick with thermal conductivity k_{1} and two layers of foam with thermal conductivity k_{2} for insulation. The left surface of the wall is at temperature T_{1} and exposed to air with heat transfer coefficient h_{1}. The right surface of the wall is at temperature T_{2} and exposed to air with heat transfer coefficient h_{2}. Assume that k_{1}=0.5 \mathrm{~W} /\left(\mathrm{m} \cdot{ }^{\circ} \mathrm{C}\right), k_{2}=0.17 \mathrm{~W} /\left(\mathrm{m} \cdot{ }^{\circ} \mathrm{C}\right), h_{1}=h_{2}=10 \mathrm{~W} /\left(\mathrm{m}^{2 .}{ }^{\circ} \mathrm{C}\right), T_{1}= 38^{\circ} \mathrm{C}, and T_{2}=20^{\circ} \mathrm{C}. The thickness of the brick layer is 0.1 \mathrm{~m}, the thickness of each foam layer is 0.03 \mathrm{~m}, and the cross-sectional area of the wall is 16 \mathrm{~m}^{2}.
a. Determine the heat flow rate through the wall.
b. Determine the temperature distribution through the wall.
a. The heat transfer through the insulated wall can be represented using a thermal circuit with five thermal resistances connected in series as shown in Figure 7.25. Two modes of heat transfer, conduction and convection, are involved. The corresponding thermal resistances are
\begin{aligned} & R_{1}=\frac{1}{h_{1} A}=\frac{1}{10 \times 16}=6.25 \times 10^{-3\circ} \mathrm{C} \cdot \mathrm{s} / \mathrm{J} \\ & R_{2}=R_{4}=\frac{L_{2}}{k_{2} A}=\frac{0.03}{0.17 \times 16}=1.10 \times 10^{-2 \circ} \mathrm{C} \cdot \mathrm{s} / \mathrm{J} \\ & R_{3}=\frac{L_{1}}{k_{1} A}=\frac{0.1}{0.5 \times 16}=1.25 \times 10^{-2\circ} \mathrm{C} \cdot \mathrm{s} / \mathrm{J} \\ & R_{5}=\frac{1}{h_{2} A}=\frac{1}{10 \times 16}=6.25 \times 10^{-3\circ} \mathrm{C} \cdot \mathrm{s} / \mathrm{J} \end{aligned}
The total thermal resistance is
R_{\mathrm{eq}}=\sum\limits_{i=1}^{5} R_{i}=4.70 \times 10^{-2 \circ} \mathrm{C} \cdot \mathrm{s} / \mathrm{J} .
Thus, the heat flow rate through the insulated wall is
q_{\mathrm{h}}=\frac{\Delta T}{R_{\mathrm{eq}}}=\frac{38-20}{4.70 \times 10^{-2}}=382.98 \mathrm{~W}.
b. Note that the heat flow rate stays the same through the insulated wall. Thus, from left to right, the heat flow rate through each layer is
\begin{array}{ll} \text { Air: } & q_{\mathrm{h}}=R_{1}\left(T_{1}-T_{3}\right) \\ \text { Foam: } & q_{\mathrm{h}}=R_{2}\left(T_{3}-T_{4}\right) \\ \text { Brick: } & q_{\mathrm{h}}=R_{3}\left(T_{4}-T_{5}\right) \\ \text { Foam: } & q_{\mathrm{h}}=R_{4}\left(T_{5}-T_{6}\right) \\ \text { Air: } & q_{\mathrm{h}}=R_{5}\left(T_{6}-T_{2}\right) \end{array}
With the given temperatures T_{1} and T_{2}, we have T_{3}=35.61^{\circ} \mathrm{C}, T_{4}=31.40^{\circ} \mathrm{C}, T_{5}=26.61^{\circ} \mathrm{C}, and T_{6}=22.40^{\circ} \mathrm{C}. Figure 7.26 shows the temperature distribution through the wall. Note that the temperatures shown in Figure 7.26 are the values when the heat transfer process reaches steady state.
