Chapter 21, Problem 134RQ

To determine

The rate of heat transfer.

Explanation of Solution

Given:

The length $\left(H\right)$ of the glass is $2\text{m}$.

The diameter $\left(D\right)$ of the glass is $5\text{m}$.

The length $\left(L\right)$ of the glass is $3\text{cm}$.

The emissivity $\left(\epsilon \right)$ of the glass is $0.9$.

The temperature $\left(T_s\right)$ of the glass is $15°\text{C}$.

The temperature $\left(T_a\right)$ of glass is $5°\text{C}$.

Calculation:

Calculate the bulk mean temperature $\left(T_m\right)$ using the relation.

    $\begin{array}{c}{T}_M=\frac{T_s+T_a}{2}\\ =\frac{15°\text{C}+5°\text{C}}{2}\\ =10°\text{C}\end{array}$

Refer table $\text{A-22}$ “Properties of air at $0.3\text{atm}$ pressure”.

Obtain the following properties of air corresponding to the temperature of $10°\text{C}$.

$\begin{array}{c}k=0.0243\text{W}/\text{m}\cdot \text{K}\\ v=4.75\times {10}^{-5}{\text{m}}^2/\text{s}\\ \mathrm{Pr}=0.733\end{array}$

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

    $\begin{array}{c}\beta =\frac{1}{T_m}\\ =\left[\frac{1}{\left(10°\text{C}+\text{273}\right)\text{K}}\right]\\ =0.00353{\text{K}}^{-1}\end{array}$

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

  $\begin{array}{c}\text{Ra}=\left(\frac{g\beta \left(T_s-T_a\right)L^3}{v^2}\mathrm{Pr}\right)\\ =\left[\frac{9.81\text{m}/{\text{s}}^2\left(0.00353{\text{K}}^{-1}\right)\left(\begin{array}{l}\left(15°\text{C}+273\right)\text{K}-\\ \left(5°\text{C}+273\right)\text{K}\end{array}\right){\left(3\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^3}{{\left(4.75\times {10}^{-5}{\text{m}}^2/\text{s}\right)}^2}\left(0.733\right)\right]\\ =3040\end{array}$

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

  $\begin{array}{c}\text{Nu}=0.197{\text{Ra}}^{1/4}{\left(\frac{H}{L}\right)}^{-}{}^{1/9}\\ =0.197{\left(3040\right)}^{1/4}{\left(\frac{2\text{m}}{3\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\right)}^{-}{}^{1/9}\\ =0.97\end{array}$

Calculate the surface Area $\left(A_s\right)$ using the relation.

    $\begin{array}{c}{A}_s=DL\\ =\left(2\text{m}\right)\left(5\text{m}\right)\\ =10{\text{m}}^2\end{array}$

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

    $\begin{array}{c}{\dot{Q}}_{conv}=k{\text{NuA}}_s\frac{T_1-T_2}{L}\\ =\left[\left(0.02439\text{W/m}\cdot \text{K}\right)\left(0.971\right)\left(10{\text{m}}^2\right)\frac{\left[\left(15°\text{C}+273\right)\text{K}-\left(5°\text{C}+273\right)\right]\text{K}}{\left(3\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\right]\\ =78.9\text{W}\end{array}$

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

  $\begin{array}{c}{\dot{Q}}_{rad}=\frac{A_s\sigma (T_s{}^4-T_{surr}{}^4)}{\frac{1}{{\epsilon }_1}+\frac{1}{{\epsilon }_2}-1}\\ =\left[\frac{\left(10{\text{m}}^2\right)\left(5.67\times {10}^{-8}{\text{W}/\text{m}}^{\text{2}}\cdot {\text{K}}^4\right)\times \left[\begin{array}{l}{\left(15°\text{C}+273\right)}^4{\text{K}}^4-\\ {\left(5°\text{C}+273\right)}^4{\text{K}}^4\end{array}\right]}{\frac{1}{0.9}+\frac{1}{0.9}-1}\right]\\ =421\text{W}\end{array}$

Calculate the total rate of heat rate $\left(\dot{Q}\right)$ using the relation.

      $\begin{array}{c}\dot{Q}={\dot{Q}}_{conv}+{\dot{Q}}_{rad}\\ =\left(78.9+421\right)\text{W}\\ =\text{498}\text{.9}\text{W}\end{array}$

Thus, the net heat rate is $498.9\text{W}$.