Chapter 17, Problem 124RQ

(a)

To determine

The rate of heat transfer across the warehouse wall.

Explanation of Solution

Given:

Width of the refrigerator wall $\left(w\right)$ is 5 m.

Height of the refrigerator wall $\left(L\right)$ is 10 m.

Thickness of the aluminum $\left(t_1\right)$ is 1 cm.

Thickness of the fiberglass $\left(t_2\right)$ is 8 cm.

Thickness of the gypsum board $\left(t_3\right)$ is 3 cm.

Thermal conductivity of aluminum $\left(k_1\right)$ is $200\text{W}/\text{m}\cdot \text{K}$.

Thermal conductivity of aluminum $\left(k_2\right)$ is $0.038\text{W}/\text{m}\cdot \text{K}$.

Thermal conductivity of aluminum $\left(k_3\right)$ is $0.48\text{W}/\text{m}\cdot \text{K}$.

Calculation:

Determine the overall heat transfer coefficient.

  $\begin{array}{c}{U}_1=\frac{1}{\frac{1}{h_i}+\frac{L_{Al}}{k_{Al}}+\frac{L_{fg}}{k_{fg}}+\frac{L_{gy}}{k_{gy}}+\frac{1}{h_o}}\\ =\frac{1}{\left\{\begin{array}{l}\frac{1}{40\text{W}/{\text{m}}^2\cdot °\text{C}}+\frac{0.01\text{ m}}{200\text{W}/{\text{m}}^2\cdot °\text{C}}+\frac{0.08\text{ m}}{0.038\text{W}/{\text{m}}^2\cdot °\text{C}}\\ +\frac{0.03\text{ m}}{0.48\text{W}/{\text{m}}^2\cdot °\text{C}}+\frac{1}{40\text{W}/{\text{m}}^2\cdot °\text{C}}\end{array}\right\}}\\ =0.451\text{W}/{\text{m}}^2\cdot °\text{C}\end{array}$

Determine the rate of heat transfer through the warehouse.

  $\begin{array}{c}{\dot{Q}}_1=U_{1}A\left(T_o-T_i\right)\\ =\left(0.451\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(5\text{ m}\times 10\text{ m}\right)\left[20°\text{C}-\left(-10°\text{C}\right)\right]\\ =676\text{ W}\end{array}$

Thus, the rate of heat transfer through the warehouse is $\underset{\_}{676\text{ W}}$.

(b)

To determine

The rate of heat transfer for the bolted wall.

Explanation of Solution

Determine the rate of heat transfer with the consideration of metal bolts.

  $\begin{array}{c}{\dot{Q}}_1=U_1{A}_1\left(T_o-T_i\right)\\ =\left(0.451\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left[10\text{ m}\times 5\text{ m}-400\times 0.25\pi {\left(0.02\text{ m}\right)}^2\right]\left(20°\text{C}-\left(-10°\text{C}\right)\right)\\ =674.8\text{ W}\end{array}$

Determine the overall heat transfer coefficient.

  $\begin{array}{c}{U}_2=\frac{1}{\frac{1}{h_i}+\frac{L_{bolts}}{k_{bolts}}+\frac{1}{h_o}}\\ =\frac{1}{\frac{1}{40\text{W}/{\text{m}}^2\cdot °\text{C}}+\frac{0.12\text{ m}}{43\text{W}/{\text{m}}^2\cdot °\text{C}}+\frac{1}{40\text{W}/{\text{m}}^2\cdot °\text{C}}}\\ =18.94\text{W}/{\text{m}}^2\cdot °\text{C}\end{array}$

Determine the rate of heat transfer with the consideration of metal bolts.

  $\begin{array}{c}{\dot{Q}}_2=U_2{A}_2\left(T_o-T_i\right)\\ =\left(18.94\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left[400\times 0.25\pi {\left(0.02\text{ m}\right)}^2\right]\left[20°\text{C}-\left(-10°\text{C}\right)\right]\\ =71.4\text{ W}\end{array}$

Determine the total rate of heat transfer.

  $\begin{array}{c}\dot{Q}={\dot{Q}}_1+{\dot{Q}}_2\\ =674.8\text{ W}+71.4\text{ W}\\ =746\text{ W}\end{array}$

Thus, the rate of heat transfer for the bolted wall is $\underset{\_}{746\text{ W}}$.

(c)

To determine

The percent change in the rate of heat transfer across the wall due to metal bolts.

Explanation of Solution

Determine the percent change in the rate of heat transfer across the wall due to metal bolts.

  $\begin{array}{c}\%\text{ change }=\frac{746\text{ W}-676\text{ W}}{676\text{ W}}\\ =0.103\times 100\%\\ =10.3\%\end{array}$

Thus, the percent change in the rate of heat transfer across the wall due to metal bolts is $\underset{\_}{10.3\%}$.