Chapter 21, Problem 138RQ

To determine

The temperature of the side surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces.

Explanation of Solution

Given:

The diameter of cylindrical furnace $\left(D\right)$ is $1.2\text{m}$.

The height of cylindrical furnace $\left(h\right)$ is $1.2\text{m}$.

The emissivity of upper surface $\left({\epsilon }_1\right)$ is $0.7$.

The emissivity of lower surface $\left({\epsilon }_2\right)$ is $0.5$.

The emissivity of side surface $\left({\epsilon }_3\right)$ is $0.4$.

The rate of heat supplied $\left(Q\right)$ is $1400\text{W}$.

The temperature of upper surface $\left(T_1\right)$ is $500\text{K}$.

The temperature of lower surface $\left(T_2\right)$ is $\text{650}\text{K}$.

Calculation:

Calculate the ratio of height of cylindrical furnace $\left(h\right)$ and radius of cylindrical furnace $\left(D\right)$ using the relation.

    $\begin{array}{c}\frac{L}{r}=\frac{L}{\left(\frac{D}{2}\right)}\\ =\frac{1.2\text{m}}{\left(\frac{1.2\text{m}}{2}\right)}\\ =2\end{array}$

Refer Table 21-3 “View factor expressions for some common geometries of finite size (3-D)”.

Calculate the view factor $\left(F_{12}\right)$ using the relation.

    $F_{12}=0.12$

Calculate the view factor $\left(F_{13}\right)$ using the relation.

  $\begin{array}{c}{F}_{11}+F_{12}+F_{13}=1\\ 0+0.17+F_{13}=1\\ F_{13}=1-0.17\\ =0.83\end{array}$

Calculate the view factor $\left(F_{31}\right)$ using the relation.

    $\begin{array}{c}{A}_1{F}_{13}=A_3{F}_{31}\\ \left(\frac{\pi }{4}{D}^2\right)F_{13}=\left(\pi Dh\right)F_{31}\\ \frac{\left(\frac{\pi }{4}{\left(1.2\text{m}\right)}^2\right)\left(0.83\right)}{\left(\pi \left(1.2\text{m}\right)\left(1.2\text{m}\right)\right)}=F_{31}\\ F_{31}=0.21\end{array}$

Refer equation 21-38. Obtain the radiosity using the expression.

    $J_i=\sigma T_i^4$

Calculate the equation 21-38 for surface 1 using the relation.

    $\begin{array}{l}\sigma T_1^4=J_1+\frac{1-{\epsilon }_1}{{\epsilon }_1}\left[F_{12}\left(J_1-J_2\right)+F_{12}\left(J_1-J_3\right)\right]\\ \left(5.67\times {10}^{-8}{\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}}^4\right){\left(500\text{K}\right)}^4=J_1+\frac{1-0.70}{0.70}\left[0.17\left(J_1-J_2\right)+0.83\left(J_1-J_3\right)\right]\end{array}$

Calculate the equation 21-38 for surface 2 using the relation.

    $\begin{array}{c}\sigma T_2^4=J_2+\frac{1-{\epsilon }_2}{{\epsilon }_2}\left[F_{21}\left(J_2-J_1\right)+F_{23}\left(J_2-J_3\right)\right]\\ \left(5.67\times {10}^{-8}{\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}}^4\right){\left(500\text{K}\right)}^4=J_2+\frac{1-0.50}{0.50}\left[0.17\left(J_2-J_1\right)+0.83\left(J_2-J_3\right)\right]\end{array}$

Calculate the equation 21-38 for surface 3 using the relation.

    $\begin{array}{c}\sigma T_3^4=J_3+\frac{1-{\epsilon }_3}{{\epsilon }_3}\left[F_{31}\left(J_2-J_1\right)+F_{23}\left(J_2-J_3\right)\right]\\ \left(5.67\times {10}^{-8}{\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}}^4\right)T_3^4=J_3+\frac{1-0.40}{0.40}\left[0.17\left(J_2-J_1\right)+0.83\left(J_2-J_3\right)\right]\end{array}$

Refer equation 21-50. Obtain the heat transfer on surface 2 using the expression.

    $\begin{array}{l}{Q}_2=A\left[F_{12}\left(J_2-J_1\right)+F_{23}\left(J_2-J_3\right)\right]\\ =\left(\frac{\pi }{4}{\left(1.2\text{m}\right)}^2\right)\left[0.17\left(J_2-J_1\right)+0.83\left(J_2-J_3\right)\right]\end{array}$

Calculate the temperature $\left(T_3\right)$ of side surface, radiosity of surface 1 $\left({\text{J}}_{\text{1}}\right)$ , radiosity of surface 2 $\left({\text{J}}_2\right)$ and radiosity of surface 3 $\left({\text{J}}_3\right)$ solving the above relation.

    $\begin{array}{c}{T}_3=631\text{K}\\ J_1=4974\text{W}/{\text{m}}^{\text{2}}\\ J_2=8883\text{W}/{\text{m}}^{\text{2}}\\ J_3=8193\text{W}/{\text{m}}^{\text{2}}\end{array}$

Calculate the rate of heat transfer between the bottom and the top surface $\left(Q_{21}\right)$ using the relation.

    $\begin{array}{c}{Q}_{12}=A{F}_{12}\left(J_2-J_1\right)\\ =\left(\frac{\pi }{4}{\left(1.2\text{m}\right)}^2\right)\left(0.17\right)\left(8883-4974\right)\text{W}/{\text{m}}^{\text{2}}\\ =751.6\text{W}\end{array}$

Calculate the rate of heat transfer between the bottom and the side surface $\left(Q_{23}\right)$ using the relation.

    $\begin{array}{c}{Q}_{23}=A{F}_{23}\left(J_2-J_3\right)\\ =\left(\frac{\pi }{4}{\left(1.2\text{m}\right)}^2\right)\left(0.83\right)\left(8883-8193\right)\text{W}/{\text{m}}^{\text{2}}\\ =648.0\text{W}\end{array}$

Thus the temperature of the side surface is $631\text{K}$ and the net rates of heat transfer between the top and the bottom surfaces, is $751.6\text{W}$ and between the bottom and side surfaces is $648.0\text{W}$.