Chapter 21, Problem 92P

To determine

The net radiation heat transfer rate through the shield under steady conditions.

Explanation of Solution

Given:

The diameter of disks $\left(D\right)$ is $3\text{ft}$.

The distance between two disks $\left(L\right)$ is $2\text{ft}$.

The emissivity of radiation shield $\left({\epsilon }_3\right)$ is $0.15$.

The temperature of disk $1$ is $\left(T_1\right)$ is $1200\text{R}$.

The temperature of disk $2$ is $\left(T_2\right)$ is $700\text{R}$.

The temperature of environment $\left(T_{\infty }\right)$ is $540\text{R}$.

Calculation:

Calculate the radius of the disks $\left(R\right)$ using the relation.

  $\begin{array}{c}R=\frac{D}{2}\\ =\frac{3\text{ft}}{2}\\ =1.5\text{ft}\end{array}$

Calculate the ratio of radius and distance between two disks $\left(r\right)$ using the relation.

    $\begin{array}{c}r=\frac{r}{\left(L/2\right)}\\ =\frac{1.5\text{ft}}{\left(2\text{ft}/2\right)}\\ =\frac{1.5\text{ft}}{1\text{ft}}\\ =1.5\end{array}$

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

Obtain the following values of view factors of different disks corresponding to the ratio of radius to the distance between two disks $1.5$ and its ratio $0.67$ as follows:

$\begin{array}{l}{F}_{32}=0.52\\ F_{13}=0.52\end{array}$

Calculate the view factor from surface $3$ to $4$ $\left(F_{34}\right)$ using the relation.

    $\begin{array}{c}{F}_{34}=1-F_{32}\\ =1-0.52\\ =0.48\end{array}$

Calculate the area of the disk $\left(A_3\right)$ using the relation.

  $\begin{array}{c}{A}_3=\frac{\pi }{4}{D}^2\\ =\frac{\pi }{4}{\left(3\text{ft}\right)}^2\\ =\frac{\pi }{4}\left(9{\text{ft}}^2\right)\\ =7.068{\text{ft}}^2\end{array}$

Calculate the heat transfer between top surface of middle disk and its black surrounding $\left({\dot{Q}}_3\right)$ using the relation.

  $\begin{array}{c}{\dot{Q}}_3=\epsilon A_3\sigma \left[F_{31}\left(T_3^4-T_1^4\right)\right]+\epsilon A_3\sigma \left[F_{32}\left(T_3^4-T_2^4\right)\right]\\ =\epsilon A_3\sigma \left[F_{31}\left(T_3^4-T_1^4\right)+F_{32}\left(T_3^4-T_2^4\right)\right]\\ =1\left(7.068{\text{ft}}^2\right)\left(0.1714\times {10}^{-4}\text{BTU}/\text{h}\cdot {\text{ft}}^2\cdot {\text{R}}^4\right)\left[\begin{array}{l}0.52\left(T_3^4-{\left(1200\text{R}\right)}^4\right)\\ +0.48\left(T_3^4-{\left(540\text{R}\right)}^4\right)\end{array}\right]\\ =\left(1.211\text{BTU}/\text{h}\cdot {\text{R}}^4\right)\left[0.52\left(T_3^4-{\left(1200\text{R}\right)}^4\right)+0.48\left(T_3^4-{\left(540\text{R}\right)}^4\right)\right]\end{array}$

Calculate the heat transfer between bottom surface of middle disk $\left(-{\dot{Q}}_3\right)$ using the relation.

    $\begin{array}{c}-{\dot{Q}}_3=\epsilon A_3\sigma \left[F_{32}\left(T_2^4-T_4^4\right)\right]+\epsilon A_3\sigma \left[F_{34}\left(T_3^4-T_4^4\right)\right]\\ -{\dot{Q}}_3=\left[\begin{array}{l}1\left(7.068{\text{ft}}^2\right)\left(0.1714\times {10}^{-4}\text{BTU}/\text{h}\cdot {\text{ft}}^2\cdot {\text{R}}^4\right)\times \\ \left[0.48\left(T_3^4-{\left(700\text{R}\right)}^4\right)+0.52\left(T_3^4-{\left(540\text{R}\right)}^4\right)\right]\end{array}\right]\\ \left[\begin{array}{l}-\left(1.211\text{BTU}/\text{h}\cdot {\text{R}}^4\right)\times \\ \left[\begin{array}{l}0.52\left(T_3^4-{\left(1200\text{R}\right)}^4\right)\\ +0.48\left(T_3^4-{\left(540\text{R}\right)}^4\right)\end{array}\right]\end{array}\right]=\left(1.211\text{BTU}/\text{h}\cdot {\text{R}}^4\right)\left[\begin{array}{l}0.48\left(T_3^4-{\left(700\text{R}\right)}^4\right)\\ +0.52\left(T_3^4-{\left(540\text{R}\right)}^4\right)\end{array}\right]\\ T_3=894\text{R}\end{array}$

Thus, the net radiation heat transfer rate through the shield under steady conditions is $894\text{R}$.