Chapter 11, Problem 28P

To determine

Calculate the reduction in heat transfer rate.

Answer to Problem 28P

The reduction in heat transfer rate is $97\%$.

Explanation of Solution

Given data:

Area of the concrete wall, $A=1000{\text{ft}}^{\text{2}}$.

Thickness of the concrete wall, $L_1=6\text{in}$.

Thickness of the insulation batt, $L_2=2\text{in}$.

Inside temperature of concrete wall, $T_1=20\text{C}$.

Outside temperature of concrete wall, $T_2=0\text{C}$.

From Table 11.3 in the textbook, the thermal conductivity of the concrete wall is, $k_1=1.4\frac{\text{W}}{\text{m}\cdot \text{K}}$.

The thermal conductivity of the insulation batt is, $k_2=0.03\frac{\text{Btu}}{\text{h}\cdot \text{ft}\cdot \text{°F}}$.

Formula used:

The formula for the heat transfer is,

$q=kA\frac{T_1-T_2}{L}$ (1)

Here,

$k$ is the thermal conductivity,

$A$ is the area of cross section,

$L$ is the thickness of the material,

$T_1$ is the inside temperature of the material, and

$T_2$ is the outside temperature of the material,

Calculation:

Convert the unit of area of cross section into meter.

$\begin{array}{c}A=1000{\text{ft}}^{\text{2}}\left[\therefore 1\text{ft}=0.3048\text{m}\right]\\ =\left(1000\right){\left(0.3048\text{ m}\right)}^2\\ =\left(1000\right)\left(0.0929\right){\text{m}}^{\text{2}}\\ A=92.9{\text{m}}^{\text{2}}\end{array}$

Convert the unit of thickness into meter for $L_1$.

$\begin{array}{c}{L}_1=6\text{in}\left[\therefore 1\text{in}=0.0254\text{m}\right]\\ =\left(6\text{in}\right)\left(\frac{0.0254\text{m}}{1\text{in}}\right)\\ L_1=0.1524\text{m}\end{array}$

Convert the unit of thickness into meter for $L_2$,

$\begin{array}{c}{L}_2=2\text{in}\\ =\left(2\text{in}\right)\left(\frac{0.0254\text{m}}{1\text{in}}\right)\\ L_2=0.0508\text{m}\end{array}$

Substitute $1.4\frac{\text{W}}{\text{m}\cdot \text{K}}$ for $k_1$, $92.9{\text{m}}^{\text{2}}$ for $A$, $20\text{C}$ for $T_1$, $0\text{C}$ for $T_2$, and $0.1524\text{m}$ for $L$ in equation (1) to find heat transfer rate of concrete wall without insulation $q_1$.

$\begin{array}{l}{q}_1=\left(1.4\frac{\text{W}}{\text{m}\cdot \text{K}}\right)\left(92.9{\text{m}}^{\text{2}}\right)\left(\frac{\left(20\text{C}-0\text{C}\right)}{0.1524\text{m}}\right)\\ q_1=\left(1.4\frac{\text{W}}{\text{m}\cdot \text{K}}\right)\left(92.9{\text{m}}^{\text{2}}\right)\left(\frac{20\text{C}}{0.1524\text{m}}\right)\end{array}$ (2)

From the truth the temperature difference in $\text{C}$ is equals to the temperature difference in K.

Substitute $20\text{K}$ for $20\text{C}$ in equation (2),

$\begin{array}{l}{q}_1=\left(1.4\frac{\text{W}}{\text{m}\cdot \text{K}}\right)\left(92.9{\text{m}}^{\text{2}}\right)\left(\frac{20\text{K}}{0.1524\text{m}}\right)\\ q_1=17070\text{W}\end{array}$

Change the unit of thermal conductivity for insulation batt,

$\begin{array}{c}{k}_2=0.03\frac{\text{Btu}}{\text{h}\cdot \text{ft}\cdot \text{°F}}\left[\therefore 1\text{W}=3.4123\frac{\text{Btu}}{\text{h}}\right]\\ =\left(0.03\frac{\text{Btu}}{\text{h}\cdot \text{ft}\cdot \text{°F}}\right)\left(\frac{1\text{W}}{3.4123\frac{\text{Btu}}{\text{h}}}\right)\\ k_2=8.792\times {10}^{-3}\frac{\text{W}}{\text{ft}\cdot \text{°F}}\end{array}$

For the temperature difference $\left(\text{1}\text{°F}=\text{1}\text{°R}\right)$. Therefore,

$\begin{array}{c}{k}_2=8.792\times {10}^{-3}\frac{\text{W}}{\text{ft}\cdot \text{°R}}\left[\therefore 1\text{K}=\frac{5}{9}\text{°R}\right]\\ =\left(8.792\times {10}^{-3}\frac{\text{W}}{\text{ft}\cdot \text{°R}}\right)\left(\frac{\frac{5}{9}\text{°R}}{1\text{K}}\right)\left[\therefore 1\text{m}=3.28\text{ft}\right]\\ =\left(4.885\times {10}^{-3}\frac{\text{W}}{\text{ft}\cdot \text{K}}\right)\left(\frac{3.28\text{ft}}{1\text{m}}\right)\\ k_2=0.016\frac{\text{W}}{\text{m}\cdot \text{K}}\end{array}$

Rearrange equation (1), we get

$q=\frac{\left(T_1-T_2\right)A}{\left(\frac{L}{k}\right)}$ (3)

Here,

$\frac{L}{k}$ is the thermal insulation.

From equation (3), the expression for the heat transfer rate of the concrete wall with insulation batt is,

$\begin{array}{c}{q}_2=\frac{\left(T_1-T_2\right)A}{\left(\frac{L_1}{k_1}\right)}+\frac{\left(T_1-T_2\right)A}{\left(\frac{L_2}{k_2}\right)}\\ q_2=\left(T_1-T_2\right)A\left[\frac{1}{\frac{L_1}{k_1}+\frac{L_2}{k_2}}\right]\end{array}$ (4)

Substitute $1.4\frac{\text{W}}{\text{m}\cdot \text{K}}$ for $k_1$, $0.016\frac{\text{W}}{\text{m}\cdot \text{K}}$ for $k_2$, $92.9{\text{m}}^{\text{2}}$ for $A$, $20\text{C}$ for $T_1$, $0\text{C}$ for $T_2$, $0.1524\text{m}$ for $L_1$, and $0.0508\text{m}$ for $L_2$ in equation (4) to find $q_2$.

$\begin{array}{c}{q}_2=\left(20\text{C}-0\text{C}\right)\left(92.9{\text{m}}^{\text{2}}\right)\left[\frac{1}{\frac{0.1524\text{m}}{1.4\frac{\text{W}}{\text{m}\cdot \text{K}}}+\frac{0.0508\text{m}}{0.016\frac{\text{W}}{\text{m}\cdot \text{K}}}}\right]\\ =\left(20\text{K}\right)\left(92.9{\text{m}}^{\text{2}}\right)\left[\frac{1}{\frac{0.1524\text{m}}{1.4\frac{\text{W}}{\text{m}\cdot \text{K}}}+\frac{0.0508\text{m}}{0.016\frac{\text{W}}{\text{m}\cdot \text{K}}}}\right]\\ =\left(20\text{K}\right)\left(92.9{\text{m}}^{\text{2}}\right)\left(0.3045\frac{\text{W}}{{\text{m}}^{\text{2}}\cdot \text{K}}\right)\\ q_2=566\text{W}\end{array}$

Hence the reduction in heat transfer is,

$\%\text{reduction}\text{in}\text{q}=\frac{q_1-q_2}{q_1}\times 100$ (5)

Substitute $17070\text{W}$ for $q_1$ and $566\text{W}$ for $q_2$ in equation (5) to find % reduction in q,

$\begin{array}{c}\%\text{reduction}\text{in}\text{q}=\frac{\left(17070\text{W}\right)-\left(566\text{W}\right)}{\left(17070\text{W}\right)}\times 100\\ =0.97\times 100\\ =97\%\end{array}$

Therefore, the reduction in heat transfer rate is $97\%$.

Conclusion:

Hence, the reduction in heat transfer rate is $97\%$.