Chapter 21, Problem 132RQ

(a)

To determine

The view factor of the surface.

Explanation of Solution

Given:

The diameter $\left(D_a\right)$ of the surface is $20\text{cm}$.

The space $\left(L\right)$ between the surfaces is $10\text{cm}$.

The emissivity $\left({\epsilon }_a\right)$ is $0.6$.

The temperature $\left(T_a\right)$ of the disk is $600°\text{C}$.

The diameter $\left(D_b\right)$ of the surface is $40\text{cm}$.

The emissivity $\left({\epsilon }_b\right)$ is $0.8$.

The temperature $\left(T_b\right)$ of the disk is $200°\text{C}$.

Calculation:

Calculate the constant $\left(A\right)$ using the relation.

    $\begin{array}{c}A=\frac{D_a}{2L}\\ =\frac{\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{2\left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =1\end{array}$

Calculate the constant $\left(B\right)$ using the relation.

    $\begin{array}{l}B=\frac{D_b}{2L}\\ B=\frac{\left(40\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{2\left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ B=2\end{array}$

Calculate the constant $\left(C\right)$ using the relation.

    $\begin{array}{l}C=1+\frac{1+A^2}{B^2}\\ C=1+\frac{1+1^2}{2^2}\\ C=1.5\end{array}$

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

    $\begin{array}{c}{F}_{ab}=0.5{\left(\frac{B}{A}\right)}^2\left\{C-{\left[C^2-4{\left(\frac{A}{B}\right)}^2\right]}^{0.5}\right\}\\ =0.5{\left(\frac{2}{1}\right)}^2\left\{1.5-{\left[{\left(1.5\right)}^2-4{\left(\frac{1}{2}\right)}^2\right]}^{0.5}\right\}\\ =0.764\end{array}$

Thus, the view factor $\left(F_{ab}\right)$ is $0.764$.

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

  $\begin{array}{l}{A}_a{F}_{ab}=A_b{F}_{ba}\\ \left(\frac{\pi D_a{}^2}{4}\right)F_{ab}=\left(\frac{\pi D_b{}^2}{4}\right)F_{ba}\\ D_a{}^2{F}_{ab}=D_b{}^2{F}_{ba}\\ {\left(0.2\text{m}\right)}^2\left(0.764\right)={\left(0.4\text{m}\right)}^2{F}_{ba}\end{array}$

  $\begin{array}{l}{F}_{ba}=\frac{{\left(0.2\text{m}\right)}^2\left(0.764\right)}{{\left(0.4\text{m}\right)}^2}\\ F_{ba}=0.191\end{array}$

Thus, the view factor $\left(F_{ba}\right)$ is $0.191$.

(b)

To determine

The net rate of radiation heat transfer.

Explanation of Solution

Given:

‘

Calculation:

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

    $\begin{array}{c}{A}_a=\pi \frac{{\left[\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right]}^2}{4}\\ =0.031{\text{m}}^2\\ A_b=\pi \frac{{\left[\left(40\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(40\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right]}^2}{4}\\ =0.125{\text{m}}^2\end{array}$

Calculate the net rate of heat transfer $\left({\dot{Q}}_{ab}\right)$ using the relation

  $\begin{array}{c}\dot{Q}=\frac{\sigma \left(T_a{}^4-T_b{}^4\right)}{\left(\frac{1-{\epsilon }_a}{{\epsilon }_a}\right)\frac{1}{A_a}+\frac{1}{A_a{F}_a{}_b}+\left(\frac{1-{\epsilon }_b}{{\epsilon }_b}\right)\frac{1}{A_b}}\\ =\frac{\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\left[{\left(600°\text{C}+273\right)}^4-{\left(200°\text{C}+273\right)}^4{\text{K}}^4\right]}{\left[\frac{1-0.6}{\left(\text{0}\text{.0314}{\text{m}}^2\right)\left(0.6\right)}+\frac{1}{\left(\text{0}\text{.0314}{\text{m}}^2\right)\left(0.764\right)}+\frac{1-0.8}{\left(\text{0}\text{.125}{\text{m}}^2\right)\left(0.8\right)}\right]}\\ =464\text{W}\end{array}$

Thus, the net rate of heat transfer is $464\text{W}$.

(c)

To determine

The net rate of heat exchange between the disks.

Explanation of Solution

Given:

The emissivity $\left({\epsilon }_c\right)$ is $0.7$.

Calculation:

Calculate the temperature $\left(T_c\right)$ using the relation.

    $\begin{array}{l}{\dot{Q}}_{ac}={\dot{Q}}_{cb}\\ \frac{\sigma \left(T_a{}^4-T_c{}^4\right)}{\left(\frac{1-{\epsilon }_a}{{\epsilon }_a}\right)\frac{1}{A_a}+\frac{1}{A_a{F}_{ac}}+\left(\frac{1-{\epsilon }_c}{{\epsilon }_c}\right)\frac{1}{A_c}}=\frac{\sigma \left(T_c{}^4-T_b{}^4\right)}{\left(\frac{1-{\epsilon }_c}{{\epsilon }_c}\right)\frac{1}{A_c}+\frac{1}{A_c{F}_{cb}}+\left(\frac{1-{\epsilon }_b}{{\epsilon }_b}\right)\frac{1}{A_b}}\\ \frac{\left(T_a{}^4-T_c{}^4\right)}{\left(\frac{1-{\epsilon }_a}{{\epsilon }_a}\right)\frac{1}{A_a}+\frac{1}{A_a{F}_{ac}}+0}=\frac{\left(T_c{}^4-T_b{}^4\right)}{0+\frac{1}{A_c{F}_{cb}}+\left(\frac{1-{\epsilon }_b}{{\epsilon }_b}\right)\frac{1}{A_b}}\\ \frac{\left[{\left(600°\text{C}+273\right)}^4-{\left(T_c\right)}^4{\text{K}}^4\right]}{\left(\frac{1-0.6}{0.6}\right)\frac{1}{\left(0.031{\text{m}}^2\right)}+\frac{1}{\left(0.031{\text{m}}^2\right)\left(1\right)}}=\left[\frac{\left[{\left(T_c\right)}^4-{\left(200°\text{C}+273\right)}^4{\text{K}}^4\right]}{\left[\begin{array}{l}\frac{1}{\left(0.125{\text{m}}^2\right)}+\\ \frac{1-0.8}{\left(0.8\right)}\left(\frac{1}{0.125{\text{m}}^2}\right)\end{array}\right]}\right]\\ T_c=605\text{K}\end{array}$

Calculate the net rate of heat transfer $\left(Q_{ab}\right)$ using the relation.

    $\begin{array}{l}{\dot{Q}}_{bc}={\dot{Q}}_{ac}\\ {\dot{Q}}_{ac}=\frac{\sigma \left(T_a{}^4-T_c{}^4\right)}{\left(\frac{1-{\epsilon }_a}{{\epsilon }_a}\right)\frac{1}{A_a}+\frac{1}{A_a{F}_{ac}}+\left(\frac{1-{\epsilon }_c}{{\epsilon }_c}\right)\frac{1}{A_c}}\\ {\dot{Q}}_{ac}=\frac{\left(5.67\times {10}^{-8}{\text{W/m}}^2\cdot {\text{K}}^4\right)\left[{\left(600°\text{C}+273\right)}^4{\text{K}}^4-{\left(605\right)}^4{\text{K}}^4\right]}{\left(\frac{1-0.6}{0.6}\right)\frac{1}{\left(0.031{\text{m}}^2\right)}+\frac{1}{\left(0.031{\text{m}}^2\right)\left(1\right)}+0}\\ {\dot{Q}}_{ac}=477\text{W}\end{array}$

Thus, the heat transfer rate is $477\text{W}$.