Chapter 21, Problem 113P

To determine

The initial rate of radiation heat transfer from the coal bricks to the steaks.

The initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface.

Explanation of Solution

Given:

The diameter of the circular grill $\left(d_i\right)$ is $0.3\text{m}$.

The initial temperature $\left(T_1\right)$ is $950\text{K}$.

The final temperature $\left(T_2\right)$ is $5°\text{C}$.

The distance between the coal bricks and the steaks $\left(L\right)$ is $0.20\text{m}$.

Calculation:

The below figure represent the required diagram.

Figure-(1)

Calculate the ratio of the radius of the circular grill and distance between the coal bricks and the steaks $\left(R_i\right)$ using the relation.

  $\begin{array}{c}{R}_i=\left(R_j\right)\\ =\frac{r_i}{L}\\ =\frac{\left(\frac{d_i}{2}\right)}{L}\\ =\frac{\left(\frac{0.3\text{m}}{2}\right)}{0.20\text{m}}\\ =0.75\end{array}$

Calculate the size factor $\left(S\right)$ using the relation.

  $\begin{array}{c}S=1+\frac{1+R_j^2}{R_i^2}\\ =1+\frac{1+{\left(0.75\right)}^2}{{\left(0.75\right)}^2}\\ =3.7778\end{array}$

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

  $\begin{array}{c}{F}_{12}=\frac{1}{2}\left[S-\left\{S^2-4{\left(\frac{R_j}{R_i}\right)}^{1/2}\right\}\right]\\ =\frac{1}{2}\left[\left(3.7778\right)-{\left\{{\left(3.7778\right)}^2-4{\left(\frac{0.75}{0.75}\right)}^2\right\}}^{1/2}\right]\\ =\frac{1}{2}\left[\left(3.7778\right)-\left(3.2049\right)\right]\\ =0.2864\end{array}$

Calculate the initial rate of radiation heat transfer from the coal bricks to the steaks $\left(Q_{12}\right)$ using the relation.

  $\begin{array}{c}{Q}_{12}=F_{12}{A}_1\sigma \left(T_1^4-T_2^4\right)\\ =F_{12}\left(\frac{\pi d^2}{4}\right)\sigma \left(T_1^4-T_2^4\right)\\ =\left(0.2864\right)\left\{\frac{\pi {\left(0.3\text{m}\right)}^2}{4}\right\}\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left\{{\left(950\text{K}\right)}^4-{\left(5°\text{C}\right)}^4\right\}\\ =\left(0.2864\right)\left\{\frac{\pi {\left(0.3\text{m}\right)}^2}{4}\right\}\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left\{{\left(950\text{K}\right)}^4-{\left(\left(5°\text{C}+273\right)\text{K}\right)}^4\right\}\\ =928.08\text{W}\end{array}$

Thus, the initial rate of radiation heat transfer from the coal bricks to the steaks is $928.08\text{W}$.

Calculate the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface $\left(Q_1\right)$ using the relation.

  $\begin{array}{c}{Q}_1=\left(E_{b1}-E_{b2}\right)\left(\frac{1}{R_{12}}+\frac{1}{R_{13}+R_{23}}\right)\\ =\left(\sigma T_1^4-\sigma T_2^4\right)\left[\frac{1}{\left(\frac{1}{A_1{F}_{12}}\right)}+\frac{1}{\left(\frac{1}{A_2{F}_{13}}\right)+\left(\frac{1}{A_2{F}_{23}}\right)}\right]\\ =\left(\sigma T_1^4-\sigma T_2^4\right)\left[\begin{array}{l}\frac{1}{\left\{\frac{1}{\left(\frac{\pi d_i{}^2}{4}\right)F_{12}}\right\}}+\\ \frac{1}{\left\{\frac{1}{\left(\frac{\pi d_i{}^2}{4}\right)\left(1-F_{12}\right)}\right\}+\left\{\frac{1}{\left(\frac{\pi d_i{}^2}{4}\right)\left(1-F_{12}\right)}\right\}}\end{array}\right]\\ =\left[\begin{array}{c}\left\{\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(950\text{K}\right)}^4-\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(5°\text{C}\right)}^4\right\}\times \\ \left[\begin{array}{l}\frac{1}{\left\{\frac{1}{\left(\frac{\pi {\left(0.3\text{m}\right)}^2}{4}\right)\left(0.2864\right)}\right\}}+\\ \frac{1}{\left\{\frac{1}{\left(\frac{\pi {\left(0.3\text{m}\right)}^2}{4}\right)\left(1-0.2864\right)}\right\}+\left\{\frac{1}{\left(\frac{\pi {\left(0.3\text{m}\right)}^2}{4}\right)\left(1-0.2864\right)}\right\}}\end{array}\right]\end{array}\right]\end{array}$    $\begin{array}{c}{Q}_1=\left\{\begin{array}{l}\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(950\text{K}\right)}^4-\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left\{\left(5°\text{C}+273\right)\text{K}\right\}}^4\end{array}\right\}\left(0.0202{\text{m}}^{\text{2}}+0.0252{\text{m}}^{\text{2}}\right)\\ =\left(46182.504\text{W}/{\text{m}}^{\text{2}}-338.6587\text{W}/{\text{m}}^{\text{2}}\right)\left(0.0454{\text{m}}^{\text{2}}\right)\\ =2081.21\text{W}\end{array}$

Thus, the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface is $2081.21\text{W}$.