Chapter 21, Problem 137RQ

To determine

The rate of heat transfer by natural convection.

Explanation of Solution

Given:

The temperature of top surface is $\left(T_1\right)$ is $700\text{K}$.

The temperature of base surface is $\left(T_2\right)$ is $950\text{K}$.

The emissivity of base surface ${\epsilon }_2$ is $0.90$.

The temperature of side surface is $\left(T_3\right)$ is $450\text{K}$.

Calculation:

The below figure represents the cubical surface.

Figure-(1)

Consider the top surface is surface 1, the base surface is surface 2 and side surface is surface 3.

The base surface is flat; therefore the value of view factor will be zero.

    $F_{11}=0$

Calculate the view factor from the base or top of side surface.

    $\begin{array}{c}{F}_{11}+F_{12}+F_{13}=1\\ 0+0.20+F_{13}=1\\ F_{13}=0.80\end{array}$

The value of other view factor will be as follows.

    $\begin{array}{c}{F}_{23}=F_{13}\\ =0.80\\ F_{21}=F_{12}\\ =0.20\end{array}$

Use the energy balance for surface 3.

    $\begin{array}{l}\sigma T_3^4=J_3\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(450\text{K}\right)}^4=J_3\\ J_3=2325.05\text{W}/{\text{m}}^{\text{2}}\end{array}$

Use the energy balance for surface 1.

    $\begin{array}{l}\sigma T_1^4=J_1+\frac{1-{\epsilon }_1}{{\epsilon }_1}\left[F_{12}\left(J_1-J_2\right)+F_{13}\left(J_1-J_3\right)\right]\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(700\text{K}\right)}^4=J_1+\frac{1-{\epsilon }_1}{{\epsilon }_1}\left[\begin{array}{l}0.20\left(J_1-J_2\right)\\ +0.80\left(\begin{array}{l}{J}_1-\\ 2325.05\text{W}/{\text{m}}^{\text{2}}\end{array}\right)\end{array}\right]\end{array}$        (I)

Use the energy balance for surface 2.

    $\begin{array}{l}\sigma T_2^4=J_2+\frac{1-{\epsilon }_2}{{\epsilon }_2}\left[F_{21}\left(J_2-J_1\right)+F_{23}\left(J_2-J_3\right)\right]\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(950\text{K}\right)}^4=J_2+\frac{1-0.90}{0.90}\left[\begin{array}{l}0.20\left(J_2-J_1\right)\\ +0.80\left(\begin{array}{l}{J}_2-\\ 2325.05\text{W}/{\text{m}}^{\text{2}}\end{array}\right)\end{array}\right]\end{array}$        (II)

Calculate the heat flux at surface 2.

    $\begin{array}{c}{\dot{Q}}_2=A_2\left[F_{21}\left(J_2-J_1\right)+F_{23}\left(J_2-J_3\right)\right]\\ =\left(9{\text{m}}^2\right)\left[0.20\left(J_2-J_1\right)+0.80\left(J_2-J_3\right)\right]\end{array}$        (III)

Solve equation (I), (II) and (III).

$\begin{array}{l}{\epsilon }_1=0.44\\ J_1=11736\text{W}/{\text{m}}^2\\ J_2=41985\text{W}/{\text{m}}^2\\ J_3=2325\text{W}/{\text{m}}^2\end{array}$

Calculate the rate of heat transfer between the bottom and top surface.

    $\begin{array}{c}{\dot{Q}}_{21}=A_2{F}_{21}\left(J_2-J_1\right)\\ =\left(9{\text{m}}^2\right)\left(0.20\right)\left(41985\text{W}/{\text{m}}^2-11736\text{W}/{\text{m}}^2\right)\\ =54.4\text{kW}\end{array}$

Calculate the rate of heat transfer between the bottom and side surface.

    $\begin{array}{c}{\dot{Q}}_{23}=A_2{F}_{23}\left(J_2-J_1\right)\\ =\left(9{\text{m}}^2\right)\left(0.80\right)\left(41985\text{W}/{\text{m}}^2-2325\text{W}/{\text{m}}^2\right)\\ =285.6\text{kW}\end{array}$

Thus, the emmisivity at the top surface is $0.44$, the heat transfer between the bottom and top surface is $54.4\text{kW}$ and the transfer between the bottom and side surface is $285.6\text{kW}$.