Chapter 17, Problem 63P

(a)

To determine

The rate of heat transfer through the wall.

Explanation of Solution

Given:

Thickness of the wall is 0.22 m.

Height of the wall is 5 m.

Length of the wall is 8 m.

Temperature at the left surface of the wall is $300°\text{C}$.

Temperature at the right surface of the wall is $100°\text{C}$.

Calculation:

The total thermal resistance of the arrangement is,

  $\begin{array}{c}{R}_{total}=R_A+R_{equ,BC}+R_{equ,DE}+R_F\\ =\frac{L_A}{k_{A}A}+\frac{1}{\frac{L_C}{k_{C}A}+\frac{L_B}{k_{B}A}+\frac{L_C}{k_{C}A}}+\frac{1}{\frac{L_D}{K_{D}A}+\frac{L_E}{k_{E}A}}+\frac{L_F}{k_{F}A}\\ =\frac{0.01\text{ m}}{\left(2\text{ W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}\\ +\frac{1}{\frac{0.05\text{ m}}{\left(20\text{ W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}+\frac{0.05\text{ m}}{\left(\text{8 W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}+\frac{0.05\text{ m}}{\left(20\text{ W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}}\\ +\frac{1}{\frac{0.1\text{ m}}{\left(\text{15 W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}+\frac{0.1\text{ m}}{\left(\text{35 W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}}\\ +\frac{0.06\text{ m}}{\left(2\text{ W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}\\ =0.349°\text{C}/\text{W}\end{array}$

The rate of heat transfer through the wall per $0.12{\text{ m}}^2$ is,

  $\begin{array}{c}\dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{total}}\\ =\frac{\left[300-100\right]°\text{C}}{0.349°\text{C}/\text{W}}\\ =572\text{W}\end{array}$

The total rate of heat transfer through the entire wall is,

  $\begin{array}{c}{\dot{Q}}_{total}=\left(572\text{}\text{W}\right)\left(\frac{5\times 8{\text{ m}}^2}{0.12{\text{ m}}^2}\right)\\ =190.66\text{ kW}\end{array}$

Thus, the rate of heat transfer through the wall is $190.66\text{ kW}$.

(b)

To determine

The temperature at the meeting point of sections B, D and E.

Explanation of Solution

Calculation:

The thermal resistance at left of the meeting point of sections B, D and E is,

  $\begin{array}{c}{R}_{total}=R_A+R_{equ,BC}+R_{equ,DE}+R_F\\ =\frac{L_A}{k_{A}A}+\frac{1}{\frac{L_C}{k_{C}A}+\frac{L_B}{k_{B}A}+\frac{L_C}{k_{C}A}}\\ =\frac{0.01\text{ m}}{\left(2\text{ W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}\\ +\frac{1}{\frac{0.05\text{ m}}{\left(20\text{ W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}+\frac{0.05\text{ m}}{\left(\text{8 W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}+\frac{0.05\text{ m}}{\left(20\text{ W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}}\\ =0.065°\text{C}/\text{W}\end{array}$

The temperature at the meeting point of sections B, D and E is,

  $\begin{array}{l}\dot{Q}=\frac{T_{\infty 1}-T_1}{R_{total}}\\ 572\text{W}=\frac{\left[300-T_1\right]°\text{C}}{0.065°\text{C}/\text{W}}\\ T_1=263°\text{C}\end{array}$

Thus, the temperature at the meeting point of sections B, D and E is $263°\text{C}$.

(c)

To determine

The temperature drop across section F.

Explanation of Solution

Calculation:

Calculate the temperature drop across section F.

  $\begin{array}{c}\Delta T=\dot{Q}{R}_F\\ =\dot{Q}\frac{L_F}{k_{F}A}\\ =\left(572\text{W}\right)\frac{0.06\text{ m}}{\left(2\text{ W/m}\cdot °\text{C}\right)\left(0.12{\text{ m}}^2\right)}\\ =143°\text{C}\end{array}$

Thus, the temperature drop across section F is $143°\text{C}$.