Chapter 21, Problem 94P

To determine

The rates of radiation heat transfer between the cylinders and between the hot cylinder and the surroundings.

Explanation of Solution

Given:

The diameter of cylinders $\left(D\right)$ is $20\text{cm}$.

The distance between cylinders $\left(s\right)$ is $30\text{cm}$.

The temperature of cylinder $1$ $\left(T_1\right)$ is $425\text{K}$.

The temperature of cylinder $2$ $\left(T_2\right)$ is $275\text{K}$.

The temperature of surrounding back body $\left(T_3\right)$ is $300\text{K}$.

The length of the cylinder $\left(L\right)$ is $1\text{m}$.

The emissivity $\left(\epsilon \right)$ is $1$.

Calculation:

Calculate the length $1$ $\left(L_1\right)$ using the relation.

  $\begin{array}{c}{L}_1^2=s^2+L^2\\ L_1=\sqrt{s^2+L^2}\\ =\sqrt{{\left(30\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2+{\left(1\text{m}\right)}^2}\\ =1.04\text{m}\end{array}$

Calculate the length $2$ $\left(L_2\right)$ using the relation.

    $\begin{array}{c}{L}_1^2=s^2+L^2\\ L_1=\sqrt{s^2+L^2}\\ =\sqrt{{\left(30\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2+{\left(1\text{m}\right)}^2}\\ =1.04\text{m}\end{array}$

Calculate the surface of string $\left(P\right)$ using the relation.

    $\begin{array}{c}P=\frac{\pi }{2}D\\ =\frac{\pi }{2}\left(20\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =0.314\text{m}\end{array}$

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

    $\begin{array}{c}{F}_{12}=\frac{{\displaystyle \sum \left(\text{Crossed strings}\right)-}{\displaystyle \sum \left(\text{Uncrossed strings}\right)}}{2P}\\ =\frac{\left(L_1+L_2\right)-\left(s+s\right)}{2P}\\ =\frac{\left(1.04\text{m}+\text{1}\text{.04}\text{m}\right)-2\left(30\mathrm{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{2\left(0.314\text{m}\right)}\\ =2.35\end{array}$

Calculate the view factor between hot cylinder and surrounding $\left(F_{13}\right)$ using the relation.

    $\begin{array}{c}{F}_{13}=1-F_{12}\\ =1-2.35\\ =-1.35\end{array}$

Calculate the area of the cylinder $\left(A\right)$ using the relation.

    $\begin{array}{c}A=\frac{\pi }{2}DL\\ =\frac{\pi }{2}\left(20\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1\text{m}\right)\\ =\frac{\pi }{2}0.2{\text{m}}^2\\ =\text{0}\text{.314}{\text{m}}^2\end{array}$

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

    $\begin{array}{c}{\dot{Q}}_{12}=A{F}_{12}\sigma \left(T_1^4-T_2^4\right)\\ =\left(0.314{\text{m}}^{\text{2}}\right)\left(2.355\right)\left(5.67\times {10}^{-8}{\text{W}/{\text{m}}^2\cdot \text{K}}^4\right)\left[{\left(425\text{K}\right)}^4-{\left(275\text{K}\right)}^4\right]\\ =\left(0.74{\text{m}}^2\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\left(2.7\times {10}^{10}\text{K}\right)\\ =1132.86\text{W}\end{array}$

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

    $\begin{array}{c}{A}_1=\pi DL\\ =\pi \left(20\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1\text{m}\right)\\ =\pi \left(20{\text{cm}}^2\times \frac{{10}^{-2}{\text{m}}^2}{1{\text{cm}}^2}\right)\\ =0.628{\text{m}}^{\text{2}}\end{array}$

Calculate the rate of heat transfer between cylinder surface and surroundings.

  $\begin{array}{c}{\dot{Q}}_{13}=A_1{F}_{13}\sigma \left(T_1^4-T_3^4\right)\\ =\left(0.628{\text{m}}^2\right)\left(-1.35\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^4\right)\left[{\left(425\text{K}\right)}^4-{\left(275\text{K}\right)}^4\right]\\ =\left(-4.8\times {10}^{-8}\text{W}/{\text{K}}^4\right)\left(2.452\times {10}^{10}{\text{K}}^4\right)\\ =-1178.68\text{W}\end{array}$

Thus, the radiation heat transfer between the cylinders is $1132.86\text{W}$ and between the hot cylinder and the surroundings is $-1178.68\text{W}$.