Chapter 14, Problem 59P

(a)

To determine

The rate of heat flow through the wall.

Answer to Problem 59P

The rate of heat flow through the wall is $320\text{W}$.

Explanation of Solution

Write the expression for the thermal resistance foe the material.

$R_{\text{Wood}}=\frac{d_W}{K_{W}A}$ . (I)

Here, $R_{\text{wood}}$ is the thermal resistance of the wood, $d_W$ is the thickness of the wood, $K_W$ thermal conductivity of the wood, $A$ is the area of cross section.

Write the expression for the thermal resistance of the insulation.

$R_{\text{insulation}}=\frac{d_i}{K_{i}A}$ . (II)

Here, $R_i$ is the thermal resistance of the insulation, $d_i$ is the thickness of the insulation, $K_W$ thermal conductivity of the insulation, $A$ is the area of cross section.

Write the expression for the rate of heat of flow.

$\wp =\frac{\Delta T}{R_W+R_i}$                                                                                                            (III)

Here, $\wp$ is the rate of heat flow, $\Delta T$ is the change in temperature, $R_W$ is the thermal resistance of the wood, $R_i$ is the thermal resistance of the insulation.

Substitute equation I and II in equation III,

$\begin{array}{c}\wp =\frac{A\Delta T}{\frac{d_W}{k_W}+\frac{d_i}{k_i}}\\ =\frac{lb\left(T_2-T_1\right)}{\frac{d_W}{k_W}+\frac{d_i}{k_i}}\end{array}$ (IV)

Conclusion:

Substitute $2.74\text{m}$ for $l$, $3.66\text{m}$ for $b$, $23.0°\text{C}$ for $T_1$, $-5.00°\text{C}$ for $T_2$, $1.00\text{cm}$ for $d_W$, $3.00\text{cm}$ for $d_{}$, $0.13\text{W}/\text{m}\cdot \text{K}$ for $k_W$ and $0.038\text{W}/\text{m}\cdot \text{K}$ for $k_i$ in the above equation IV to find $\wp$.

$\begin{array}{c}\wp =\frac{\left(2.74\text{m}\right)\left(3.66\text{m}\right)\left[23.0°\text{C}-\left(-5.00°\text{C}\right)\right]}{\frac{1.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}}{0.13\text{W}/\left(\text{m}\cdot \text{K}\right)}+\frac{3.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}}{0.038\text{W}/\left(\text{m}\cdot \text{K}\right)}}\\ =320\text{W}\end{array}$

Therefore, The rate of heat flow through the wall is $320\text{W}$.

(b)

To determine

The amount of heat flows out of the wall.

Answer to Problem 59P

The amount of heat flows out of the wall is $18\text{kW}$.

Explanation of Solution

Write the rate of heat flow.

${\wp }_g=\frac{\Delta T}{R_g}$ (V)

Here, ${\wp }_g$ is the rate of heat flow, $\Delta T$ is the change in temperature, $R_g$ is the thermal resistance of the glass.

Write the expression for the thermal resistance of the glass.

$R_{\text{glass}}=\frac{d_g}{K_{g}A}$ . (VI)

Here, $R_{\text{glass}}$ is the thermal resistance of the glass $d_{\text{glass}}$ is the thickness of the glass,  $K_g$ thermal conductivity of the glass, $A$ is the area of cross section.

Substitute equation VI in equation V,

$\begin{array}{c}{\wp }_g=\frac{A\Delta T}{\frac{d_g}{K_g}}\\ =\frac{K_{g}A\Delta T}{d_g}\end{array}$

Write the expression for the total rate of heat flow.

${\wp }_{\text{total}}=\frac{K_{g}A\Delta T}{d_g}+\frac{\wp }{2}$

Here, ${\wp }_{\text{total}}$ is the total rate of heat flow,

Conclusion:

 Substitute $0.63\text{W}/\text{m}\cdot \text{K}$ for $K_g$, $2.74\text{m}$ for $l$, $3.66\text{m}$ for $b$, $23.0°\text{C}$ for $T_1$, $-5.00°\text{C}$ for $T_2$ and $0.500\text{cm}$ for $d_g$ In the above equation to find ${\wp }_{\text{total}}$.

$\begin{array}{c}{\wp }_{\text{total}}=\frac{0.63\text{W}/\text{m}\cdot \text{K}\left(1/2\right)\left(2.74\text{m}\right)\left(3.66\text{m}\right)\left(23.0°\text{C}-\left(-5.00°\text{C}\right)\right)}{0.500\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}}\\ =18\text{kW}\end{array}$

Therefore, the amount of heat flows out of the wall is $18\text{kW}$.