Chapter 20, Problem 90RQ

(a)

To determine

The heat transfer from the outer surface due to convection.

Explanation of Solution

Given:

The power $\left(\dot{Q}\right)$ of the system is $180\text{W}$.

The dimensions of the box are $6\text{in}\times 6\text{in}$.

The ambient temperature $(T_a)$ is $85°\text{F}$.

The outlet temperature $(T_o)$ is $100°\text{F}$.

The flow rate $\left(\epsilon \right)$ of the box is $22\text{cfm}$.

Calculation:

Refer table $\text{A-22}$ E “Properties of air at $1\text{atm}$ pressure ”.

Obtain the following properties of air corresponding to the temperature of $100°\text{F}$.

$\begin{array}{c}k=0.015\text{Btu}/\text{h}\cdot \text{ft}\cdot \text{R}\\ v=0.1809\times {10}^{-5}{\text{ft}}^2/\text{s}\\ \mathrm{Pr}=0.726\\ \rho =0.072\text{lbm}/{\text{ft}}^3\\ {\text{c}}_p=0.2404\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Calculate the volume expansion coefficient $\left(\beta \right)$ using the relation.

    $\begin{array}{c}\beta =\frac{1}{T_m}\\ =\left[\frac{1}{\left(100°\text{F}+\text{460}\right)\text{R}}\right]\\ =0.0017{\text{R}}^{-1}\end{array}$

Calculate the mass flow rate $\left(\dot{\text{m}}\right)$ using the relation.

    $\begin{array}{c}\dot{\text{m}}=\rho \dot{V}\\ =\left(0.072\text{lbm}/{\text{ft}}^3\right)\left(22{\text{ft}}^3/\min \right)\\ =1.602\text{lbm}/\min \end{array}$

Calculate the heat transfer $\left({\dot{Q}}_{forced}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{forced}={\dot{\text{m}}c}_p\left(T_{out}-T_{in}\right)\\ =\left(1.602\text{lbm}/\min \left(\frac{60\text{min}}{1\text{h}}\right)\right)\left(0.2404\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(\begin{array}{l}\left(100\text{°F}+460\right)\text{R}-\\ \left(85\text{°F}+460\right)\text{R}\end{array}\right)\\ =346.6\text{Btu}/\text{h}\end{array}$

Calculate the heat transfer $\left({\dot{Q}}_{conv}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{conv}={\dot{Q}}_{total}-{\dot{Q}}_{forced}\\ =\left(180\times 3.412\right)\text{Btu}/\text{h}-\left(346.6\text{Btu}/\text{h}\right)\\ =268\text{Btu}/\text{h}\end{array}$

Thus, the heat transfer through convection is $268\text{Btu}/\text{h}$.

(b)

To determine

The average temperature of the duct.

Explanation of Solution

Given:

Calculation:

Calculate the characteristic length $\left(L_c\right)$ using the relation.

    $\begin{array}{c}{L}_c=\frac{A_s}{p}\\ =\frac{\left(\left(4\text{ft}\right)\times \left(6\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)\right)}{2\left(\left(4\text{ft}\right)+\left(6\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)\right)}\\ =0.22\text{ft}\end{array}$

Assume surface temperature to be $120°\text{F}$

Calculate the Rayleigh number $\left(\text{Ra}\right)$ using the relation.

  $\begin{array}{c}\text{Ra}=\left(\frac{g\beta \left(T_s-T_a\right)L_c{}^3}{v^2}\mathrm{Pr}\right)\\ =\left[\frac{\left(32.2\text{ft}/{\text{s}}^2\right)\left(0.0017{\text{R}}^{-1}\right)\left(\begin{array}{l}\left(120°\text{F}+460\right)\text{R}-\\ \left(80°\text{F}+460\right)\text{R}\end{array}\right){\left(0.22\text{ft}\right)}^3}{{\left(0.18\times {10}^{-3}{\text{ft}}^2/\text{s}\right)}^2}\left(0.726\right)\right]\\ =5.599\times {10}^5\end{array}$

Calculate the nusselt number $\left(\text{Nu}\right)$ using the relation.

    $\begin{array}{c}\text{Nu}=0.54{\text{Ra}}_L^{1/4}\\ =0.54{\left(5.559\times {10}^5\right)}^{1/4}\\ =14.77\end{array}$

Calculate the heat transfer coefficient $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{L_c}\text{Nu}\\ \text{=}\frac{0.015\text{Btu}/\text{h}\cdot \text{ft}\cdot \text{R}}{0.22\text{ft}}\left(14.77\right)\\ =1.016\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot \text{R}\end{array}$

Calculate the surface area of top $\left(A_{top}\right)$ using the relation.

    $\begin{array}{c}{A}_{top}=\left(\left(4\text{ft}\right)\times \left(6\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)\right)\\ =2{\text{ft}}^2\end{array}$

Calculate the Nusselt number $\left(\text{Nu}\right)$ using the relation.

    $\begin{array}{c}\text{Nu}=0.27{\text{Ra}}_L^{1/4}\\ =0.27{\left(5.559\times {10}^5\right)}^{1/4}\\ =7.386\end{array}$

Calculate the heat transfer coefficient $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{L_c}\text{Nu}\\ \text{=}\frac{0.015\text{Btu}/\text{h}\cdot \text{ft}\cdot \text{R}}{0.22\text{ft}}\left(7.386\right)\\ =0.5082\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot \text{R}\end{array}$

Calculate for vertical side surface:

Calculate the Rayleigh number $\left(\text{Ra}\right)$ using the relation.

  $\begin{array}{c}\text{Ra}=\left(\frac{g\beta \left(T_s-T_a\right)L_c{}^3}{v^2}\mathrm{Pr}\right)\\ =\left[\frac{\left(32.2\text{ft}/{\text{s}}^2\right)\left(0.0017{\text{R}}^{-1}\right)\left(\begin{array}{l}\left(120°\text{F}+460\right)\text{R}-\\ \left(80°\text{F}+460\right)\text{R}\end{array}\right){\left(\left(6\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)\right)}^3}{{\left(0.18\times {10}^{-3}{\text{ft}}^2/\text{s}\right)}^2}\left(0.726\right)\right]\\ =6.379\times {10}^6\end{array}$

Calculate the Nusselt number $\left(\text{Nu}\right)$ using the relation.

    $\begin{array}{c}\text{Nu}={\left[0.825+\frac{0.387{\text{Ra}}^{1/6}}{{\left[1+{\left(\frac{0.492}{\mathrm{Pr}}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{0.387{\left(6.379\times {10}^6\right)}^{1/6}}{{\left[1+{\left(\frac{0.492}{0.726}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ =27.57\end{array}$

Calculate the heat transfer coefficient $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{L_c}\text{Nu}\\ \text{=}\frac{0.015\text{Btu}/\text{h}\cdot \text{ft}\cdot \text{R}}{\left(6\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)}\left(27.57\right)\\ =0.843\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot \text{R}\end{array}$

Calculate the surface Area $\left(A_{side}\right)$ using the relation.

    $\begin{array}{c}{A}_{side}=2\left(\left(4\text{ft}\right)\times \left(6\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)\right)\\ =4{\text{ft}}^2\end{array}$

Calculate the surface temperature $\left(T_s\right)$ using the relation

    $\begin{array}{c}Q={\dot{Q}}_{top}+{\dot{Q}}_{bottom}+{\dot{Q}}_{side}\\ =\left[{\left(hA\right)}_{top}+\left({\left(hA\right)}_{bottom}+{\left(hA\right)}_{side}\right)\right]\left[T_s-T_a\right]\\ 268\text{Btu}/\text{h}=\left[\begin{array}{l}\left(1.016\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot \text{R}\times 2{\text{ft}}^2\right)+\\ \left(0.508\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot \text{R}\times 4{\text{ft}}^2\right)+\\ \left(0.843\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot \text{R}\times 4{\text{ft}}^2\right)\end{array}\right]\left[T_s-\left(80°\text{F}+460\right)\text{R}\right]\\ T_s\text{=}\left(581.7\text{R}-460\right)°\text{F}\\ \text{=121}\text{.7°F}\end{array}$

Thus the surface temperature is $121.7°\text{F}$.