Chapter 21, Problem 133RQ

(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 $40\text{cm}$.

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

The temperature $\left(T_a\right)$ of the disk is $800°\text{C}$.

The diameter $\left(D_b\right)$ of the surface is $60\text{cm}$.

The emissivity $\left({\epsilon }_b\right)$ is $0.4$.

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

Calculation:

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

    $\begin{array}{l}A=\frac{D}{L}\\ =\frac{\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{\left(40\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =0.5\end{array}$

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

    $\begin{array}{l}B=\frac{D}{L}\\ B=\frac{\left(60\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{\left(40\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ B=1.5\end{array}$

Calculate the view factor $\left(F\right)$ using the relation.

    $\begin{array}{c}{F}_{ab}=\frac{1}{2A}\left\{{\left[{\left(B+A\right)}^2+4\right]}^{0.5}-{\left[{\left(B-A\right)}^2+4\right]}^{0.5}\right\}\\ =\frac{1}{2A}\left\{{\left[{\left(1.5+0.5\right)}^2+4\right]}^{0.5}-{\left[{\left(1.5-0.5\right)}^2+4\right]}^{0.5}\right\}\\ =0.592\end{array}$

Thus, the view factor is $0.592$.

(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.

    $\begin{array}{c}{A}_a=\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =0.04{\text{m}}^2\\ A_b=\left(60\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(60\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =0.36{\text{m}}^2\end{array}$

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

  $\begin{array}{c}\dot{Q}=\frac{\sigma \left(T_a{}^4-T_b{}^4\right)}{\left(\frac{1-{\epsilon }_a}{{\epsilon }_a}\right)\frac{1}{A_a}+\frac{1}{A_a{F}_a{}_b}+\left(\frac{1-{\epsilon }_b}{{\epsilon }_b}\right)\frac{1}{A_b}}\\ =\frac{\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\left[{\left(800°\text{C}+273\right)}^4-{\left(200°\text{C}+273\right)}^4{\text{K}}^4\right]}{\left[\frac{1-0.8}{\left(\text{0}\text{.04}{\text{m}}^2\right)\left(0.8\right)}+\frac{1}{\left(\text{0}\text{.04}{\text{m}}^2\right)\left(0.592\right)}+\frac{1-0.4}{\left(\text{0}\text{.36}{\text{m}}^2\right)\left(0.4\right)}\right]}\\ =1374\text{W}\end{array}$

Thus, the net rate of heat transfer is $1374\text{W}$.

(c)

To determine

The net rate of heat exchange between the disks.

Explanation of Solution

Given:

The emissivity $\left({\epsilon }_c\right)$ is $0.1$.

The diameter $\left(D\right)$ is $2.0\text{m}$

The length $\left(L\right)$ is $20\text{cm}$.

Calculation:

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

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

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

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

Calculate the view factors $\left(F\right)$ using the relation.

    $\begin{array}{c}{F}_{ab}=\frac{1}{2A}\left\{{\left[{\left(C+A\right)}^2+4\right]}^{0.5}-{\left[{\left(C-A\right)}^2+4\right]}^{0.5}\right\}\\ =\frac{1}{2\times 1}\left\{{\left[{\left(10+1\right)}^2+4\right]}^{0.5}-{\left[{\left(10-1\right)}^2+4\right]}^{0.5}\right\}\\ =0.98\end{array}$

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

    $\begin{array}{l}B=\frac{D}{L}\\ B=\frac{\left(60\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ B=3\end{array}$

Calculate the view factors $\left(F\right)$ using the relation.

    $\begin{array}{c}{F}_{bc}=\frac{1}{2A}\left\{{\left[{\left(C+B\right)}^2+4\right]}^{0.5}-{\left[{\left(C-B\right)}^2+4\right]}^{0.5}\right\}\\ =\frac{1}{2\times 1}\left\{{\left[{\left(10+3\right)}^2+4\right]}^{0.5}-{\left[{\left(10-3\right)}^2+4\right]}^{0.5}\right\}\\ =0.59\end{array}$

Calculate the view factor $\left(F\right)$ using the relation.

    $\begin{array}{c}{A}_b{F}_{bc}=A_c{F}_{cb}\\ \left(0.36\right)\left(0.979\right)=\left(4\right)F_{cb}\\ F_{cb}=0.08\end{array}$

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

    $\begin{array}{c}{\dot{Q}}_{ac}={\dot{Q}}_{cb}\\ \frac{\sigma \left(T_a{}^4-T_c{}^4\right)}{\left(\frac{1-{\epsilon }_a}{{\epsilon }_a}\right)\frac{1}{A_a}+\frac{1}{A_a{F}_{ac}}+\left(\frac{1-{\epsilon }_c}{{\epsilon }_c}\right)\frac{1}{A_c}}=\frac{\sigma \left(T_c{}^4-T_b{}^4\right)}{\left(\frac{1-{\epsilon }_c}{{\epsilon }_c}\right)\frac{1}{A_c}+\frac{1}{A_c{F}_{cb}}+\left(\frac{1-{\epsilon }_b}{{\epsilon }_b}\right)\frac{1}{A_b}}\\ \frac{\left[\begin{array}{l}{\left(800°\text{C}+273\right)}^4-\\ {\left(T_c\right)}^4\end{array}\right]}{\left[\begin{array}{l}\left(\frac{1-0.8}{0.8}\right)\frac{1}{\left(0.04{\text{m}}^2\right)}+\\ \frac{1}{\left(0.04{\text{m}}^2\right)\left(0.981\right)}+\\ \left(\frac{1-0.1}{0.1}\right)\frac{1}{4{\text{m}}^2}\end{array}\right]}=\left[\frac{\left[{\left(T_c\right)}^4-{\left(200°\text{C}+273\right)}^4\text{K}\right]}{\left[\begin{array}{l}\frac{1}{\left(4{\text{m}}^2\right)}\frac{1-0.1}{\left(0.1\right)}+\\ \left(\frac{1}{\left(4{\text{m}}^2\right)\left(0.0881\right)}\right)+\left(\frac{1-0.4}{0.4}\right)\frac{1}{0.36{\text{m}}^2}\end{array}\right]}\right]\\ T_c=\left(\left(754\text{K}-273\right)\right)°\text{C}\\ T_c=481°\text{C}\end{array}$

Thus, the temperature is $481°\text{C}$.