Chapter 17, Problem 62P

(a)

To determine

The rate of heat transfer through the wall made of solid bricks.

Explanation of Solution

Given:

Thickness of the wall is 10 in.

Height of the wall is 10 ft.

Length of the wall is 30 ft.

Thermal conductivity of the solid bricks is $0.40\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}$.

Cross section of the solid brick is $7\text{ in}\times \text{7 in}$.

Thermal conductivity of the bricks with air holes is $0.015\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}$.

Cross section of the brick used with air holes is $\text{1}\text{.5 in}\times 1.5\text{ in}$.

Length of the bricks is 9 in.

Thickness of plaster is 0.5 in.

Thermal conductivity of the plaster is $0.010\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}$.

Indoor temperature is $80°\text{F}$.

Outdoor temperature is $30°\text{F}$.

The inside heat transfer coefficient is $1.5\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}$.

The outside heat transfer coefficient is $4\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}$.

Calculation:

The total thermal resistance of the arrangement is,

  $\begin{array}{c}{R}_{total}=R_{conv,1}+2R_{plaster}+R_{mid}+R_{conv,2}\\ =\frac{1}{h_{1}A}+\frac{2L_1}{k_{plaster}A}+\frac{1}{\frac{L_2}{k_{plaster}A}+\frac{L_2}{k_{plaster}A}+\frac{L_2}{k_{brick}A}}+\frac{1}{h_{2}A}\\ =\left\{\begin{array}{c}\frac{1}{\left(1.5\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right){\left(\frac{7.5}{12}\text{ ft}\right)}^2}\\ +\frac{2\left(0.5/12\text{ ft}\right)}{\left(0.010\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right){\left(\frac{7.5}{12}\text{ ft}\right)}^2}+\frac{1}{\left(4\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right){\left(\frac{7.5}{12}\text{ ft}\right)}^2}\\ +\frac{1}{\left[\begin{array}{l}\frac{9/12\text{ft}}{\left(0.010\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right)\left(\frac{7.5}{12}\text{ft}\times \frac{0.5}{12\text{ft}}\right)}+\frac{9/12\text{ft}}{\left(0.010\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right)\left(\frac{7}{12}\text{ft}\times \frac{0.5}{12\text{ft}}\right)}\\ +\frac{9/12\text{ft}}{\left(0.40\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right){\left(\frac{7}{12}\text{ft}\right)}^2}\end{array}\right]}\end{array}\right\}\\ =9.7937\text{h}\cdot °\text{F}/\text{Btu}\end{array}$

The rate of heat transfer through the wall per $0.3906{\text{ ft}}^2$ is,

  $\begin{array}{c}\dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{total}}\\ =\frac{\left[80-30\right]°\text{F}}{9.7937\text{h}\cdot °\text{F}/\text{Btu}}\\ =5.105\text{Btu}/\text{h}\end{array}$

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

  $\begin{array}{c}{\dot{Q}}_{total}=\left(5.105\text{Btu}/\text{h}\right)\left(\frac{30\times 10{\text{ ft}}^2}{0.3906{\text{ ft}}^2}\right)\\ =3921\text{Btu}/\text{h}\end{array}$

Thus, the rate of heat transfer through the wall made of solid bricks is $3921\text{Btu}/\text{h}$.

(b)

To determine

The rate of heat transfer through the wall made of bricks with air holes.

Explanation of Solution

Calculation:

The total thermal resistance of the arrangement is,

  $\begin{array}{c}{R}_{total}=R_{conv,1}+2R_{plaster}+R_{mid}+R_{conv,2}\\ =\frac{1}{h_{1}A}+\frac{2L_1}{k_{plaster}A}+\frac{1}{\frac{L_2}{k_{plaster}A}+\frac{L_2}{k_{plaster}A}+\frac{L_2}{k_{air}A}+\frac{L_2}{k_{brick}A}}+\frac{1}{h_{2}A}\\ =\left\{\begin{array}{c}\frac{1}{\left(1.5\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right){\left(\frac{7.5}{12}\text{ ft}\right)}^2}\\ +\frac{2\left(0.5/12\text{ ft}\right)}{\left(0.010\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right){\left(\frac{7.5}{12}\text{ ft}\right)}^2}+\frac{1}{\left(4\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right){\left(\frac{7.5}{12}\text{ ft}\right)}^2}\\ +\frac{1}{\left[\begin{array}{l}\frac{9/12\text{ft}}{\left(0.010\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right)\left(\frac{7.5}{12}\text{ft}\times \frac{0.5}{12\text{ft}}\right)}+\frac{9/12\text{ft}}{\left(0.010\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right)\left(\frac{7}{12}\text{ft}\times \frac{0.5}{12\text{ft}}\right)}\\ +\frac{9/12\text{ft}}{\left(0.40\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right)\left[{\left(\frac{7}{12}\text{ft}\right)}^2-9{\left(\frac{1.5}{12}\text{ft}\right)}^2\right]}+\frac{9/12\text{ft}}{\left(0.15\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right)\left[9{\left(\frac{1.5}{12}\text{ft}\right)}^2\right]}\end{array}\right]}\end{array}\right\}\\ =13.098\text{h}\cdot °\text{F}/\text{Btu}\end{array}$

The rate of heat transfer through the wall per $0.3906{\text{ ft}}^2$ is,

  $\begin{array}{c}\dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{total}}\\ =\frac{\left[80-30\right]°\text{F}}{\text{13}\text{.098h}\cdot °\text{F}/\text{Btu}}\\ =3.817\text{Btu}/\text{h}\end{array}$

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

  $\begin{array}{c}{\dot{Q}}_{total}=\left(3.817\text{Btu}/\text{h}\right)\left(\frac{30\times 10{\text{ ft}}^2}{0.3906{\text{ ft}}^2}\right)\\ =2932\text{Btu}/\text{h}\end{array}$

Thus, the rate of heat transfer through the wall made of bricks with air holes is $2932\text{Btu}/\text{h}$.