Chapter 21, Problem 96P

To determine

The net rate of radiation heat transfer between two surfaces and the horizontal surface and the surroundings.

Explanation of Solution

Given:

The length of edge $\left(w\right)$ is $1.6\text{m}$.

The width of horizontal surface $\left(L_1\right)$ is $0.8\text{m}$.

The width of vertical surface $\left(L_2\right)$ is $1.2\text{m}$.

The emissivity of horizontal surface $\left({\epsilon }_1\right)$ is $0.75$.

The emissivity of vertical surface is $\left({\epsilon }_2\right)$ is $1$.

The emissivity of surrounding $\left({\epsilon }_3\right)$ is $0.85$.

The temperature of horizontal surface $\left(T_1\right)$ is $400\text{K}$.

The temperature of vertical surface $\left(T_2\right)$ is $550\text{K}$.

The temperature of surrounding $\left(T_3\right)$ is $290\text{K}$.

Calculation:

Calculate the ratio of length to width of horizontal surface $\left(r_1\right)$ using the relation.

    $\begin{array}{c}{r}_1=\frac{L_1}{w}\\ =\frac{0.8\text{m}}{1.6\text{m}}\\ =0.5\end{array}$

Calculate the ratio of length to width of vertical surface $\left(r_2\right)$ using the relation.

    $\begin{array}{c}{r}_2=\frac{L_2}{w}\\ =\frac{1.2\text{m}}{1.6\text{m}}\\ =0.75\end{array}$

Refer Table  21-35 “View factor between coaxial parallel disks.”.

Obtain the following values of view factors of different surfaces corresponding to the ratio $r_1=0.5$ and $r_2=0.75$ as follows:

  $F_{12}=0.27$

Calculate the surface area of horizontal surface $\left(A_1\right)$ using the relation.

    $\begin{array}{c}{A}_1=L_1\times w\\ =\left(0.8\text{m}\right)\times \left(1.6\text{m}\right)\\ =1.28{\text{m}}^{\text{2}}\end{array}$

Calculate the surface area of vertical surface $\left(A_1\right)$ using the relation.

    $\begin{array}{c}{A}_2=L_2\times w\\ =\left(1.2\text{m}\right)\times \left(1.6\text{m}\right)\\ =1.92{\text{m}}^{\text{2}}\end{array}$

Calculate the surface area of third connecting surface $\left(A_3\right)$ using the relation.

    $\begin{array}{c}{A}_3=2\frac{L_1\times L_2}{2}+\left(\sqrt{L_1^2+L_2^2}\right)\left(w\right)\\ =2\frac{\left(0.8\text{m}\right)\times \left(1.2\text{m}\right)}{2}+\left(\sqrt{{\left(0.8{\text{m}}^{\text{2}}\right)}^2+{\left(1.2\mathrm{m}\right)}^2}\right)\left(1.6\text{m}\right)\\ =0.96{\text{m}}^{\text{2}}+2.307{\text{m}}^2\\ =3.26{\text{m}}^2\end{array}$

Calculate the view factor from surface $2$ to $1$ $\left(F_{12}\right)$ using the relation.

    $\begin{array}{c}{A}_2{F}_{21}=A_1{F}_{12}\\ \left(1.92{\text{m}}^2\right)F_{21}=\left(1.28{\text{m}}^2\right)\left(0.27\right)\\ F_{12}=0.18\end{array}$

Calculate the view factor from surface $1$ to $3$ $\left(F_{13}\right)$ using the relation.

    $\begin{array}{c}{F}_{11}+F_{12}+F_{13}=1\\ 0+0.27+F_{13}=1\\ F_{13}=1-0.27\\ F_{13}=0.73\end{array}$

Calculate the view factor from surface $2$ to $3$ $\left(F_{23}\right)$ using the relation.

    $\begin{array}{c}{F}_{21}+F_{22}+F_{23}=1\\ 0.18+0+F_{23}=1\\ F_{23}=1-0.18\\ =0.82\end{array}$

Calculate the view factor from surface $3$ to $1$ $\left(F_{31}\right)$ using the relation.

    $\begin{array}{c}{A}_1{F}_{12}=A_3{F}_{31}\\ \left(1.28{\text{m}}^{\text{2}}\right)\left(0.73\right)=\left(3.26{\text{m}}^2\right)F_{31}\\ F_{31}=\frac{0.9344{\text{m}}^2}{3.26{\text{m}}^2}\\ F_{13}=0.29\end{array}$

Calculate the view factor from surface $3$ to $2$ $\left(F_{32}\right)$ using the relation.

    $\begin{array}{c}{A}_2{F}_{23}=A_3{F}_{32}\\ \left(1.92{\text{m}}^{\text{2}}\right)\left(0.82\right)=\left(3.26{\text{m}}^2\right)F_{32}\\ F_{32}=\frac{1.574{\text{m}}^2}{3.26{\text{m}}^2}\\ F_{13}=0.48\end{array}$

Calculate the radiosity of vertical surface $\left(J_2\right)$ using the relation.

    $\begin{array}{c}{J}_2=\sigma T_2^4\\ =\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^4\right){\left(550\text{K}\right)}^4\\ =\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^4\right)\left(9.150\times {10}^{10}{\text{K}}^4\right)\\ =5188\text{W}/{\text{m}}^{\text{2}}\end{array}$

Calculate the radiosity of horizontal surface $\left(J_1\right)$ using the relation.

    $\begin{array}{l}\sigma T_1^4=J_1+\frac{1-{\epsilon }_1}{{\epsilon }_1}\left[F_{12}\left(J_1-J_2\right)+F_{13}\left(J_1-J_3\right)\right]\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(400\text{K}\right)}^4=J_1+\frac{1-0.75}{0.75}\left[\begin{array}{l}0.27\left(J_1-J_2\right)\\ +0.73\left(J_1-J_3\right)\end{array}\right]\\ 1451.52\text{W}/{\text{m}}^{\text{2}}=1.33J_1-0.33\left[0.27\left(5188\text{W}/{\text{m}}^{\text{2}}\right)+0.73J_3\right]\\ J_1=1439\text{W}/{\text{m}}^{\text{2}}+0.181J_3\end{array}$

Calculate the radiosity for third surface $\left(J_3\right)$ using the relation.

  $\begin{array}{l}\sigma T_3^4=J_3+\frac{1-{\epsilon }_3}{{\epsilon }_3}\left[F_{31}\left(J_3-J_1\right)+F_{32}\left(J_3-J_2\right)\right]\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(290\text{K}\right)}^4=1.22J_3-0.176\left[0.77J_1+0.48\left(5188\text{W}/{\text{m}}^{\text{2}}\right)\right]\\ 401.03\text{W}/{\text{m}}^{\text{2}}=1.22J_3-0.135\left(1439\text{W}/{\text{m}}^{\text{2}}+0.181J_3\right)-0.084J_2\\ J_3=811.5\text{W}/{\text{m}}^{\text{2}}\end{array}$

Calculate the radiosity for horizontal surface $\left(J_1\right)$ using the relation.

    $\begin{array}{c}{J}_3=1439\text{W}/{\text{m}}^{\text{2}}+0.181\left(811.5\text{W}/{\text{m}}^{\text{2}}\right)\\ =1587\text{W}/{\text{m}}^{\text{2}}\end{array}$

Calculate the net rate of heat transfer between two surfaces $\left({\dot{Q}}_{12}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{12}=-A_1{F}_{12}\left(J_1-J_2\right)\\ =-\left(1.28{\text{m}}^{\text{2}}\right)\left(0.27\right)\left(1587\text{W}/{\text{m}}^{\text{2}}-5188\text{W}/{\text{m}}^{\text{2}}\right)\\ =-\left(0.3456{\text{m}}^{\text{2}}\right)\left(-3601\text{W}/{\text{m}}^{\text{2}}\right)\\ =1245\text{W}\end{array}$

Calculate the net rate of heat transfer between horizontal surface and surrounding $\left({\dot{Q}}_{13}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{13}=A_1{F}_{13}\left(J_1-J_3\right)\\ =\left(1.28{\text{m}}^{\text{2}}\right)\left(0.73\right)\left(1587\text{W}/{\text{m}}^{\text{2}}-811.5\text{W}/{\text{m}}^{\text{2}}\right)\\ =725\text{W}\end{array}$

Thus, the net rate of radiation heat transfer between two surfaces is $1245\text{W}$ and the net rate of radiation heat transfer between the horizontal surface and the surroundings is $725\text{W}$.