Chapter 21, Problem 131RQ

(a)

To determine

The view factor of the surface.

Explanation of Solution

Given:

The diameter $\left(D\right)$ of the surface is $1.5\text{cm}$.

The space $\left(s\right)$ between the surfaces is $3.0\text{cm}$.

Calculation:

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

  $\begin{array}{c}{F}_{12}=1-{\left[1-{\left(\frac{D}{s}\right)}^2\right]}^{0.5}+\left(\frac{D}{s}\right)\left\{{\tan }^{-1}{\left[{\left(\frac{s}{D}\right)}^2-1\right]}^{0.5}\right\}\\ =1-{\left[1-{\left(\frac{1.5\text{cm}}{3\text{cm}}\right)}^2\right]}^{0.5}+\left(\frac{1.5\text{cm}}{3\text{cm}}\right)\left\{{\tan }^{-1}{\left[{\left(\frac{3\text{cm}}{1.5\text{cm}}\right)}^2-1\right]}^{0.5}\right\}\\ =0.658\end{array}$

    $\begin{array}{c}{F}_{ji}=\frac{A_i}{A_j}\left(F_{ij}\right)\\ =\frac{sL}{\pi DL}\left(F_{ij}\right)\\ =\frac{s}{\pi D}\left(F_{ij}\right)\\ =\frac{\left(3\text{cm}\right)}{\pi \left(1.5\text{cm}\right)}\left(0.65\right)\\ =0.41\end{array}$

Thus, the view factor of the surface is $0.41$.

(b)

To determine

The net rate of radiation heat transfer.

Explanation of Solution

Given:

‘

The emissivity $\left({\epsilon }_1\right)$ is $0.8$ at $900°\text{C}$.

The emissivity $\left({\epsilon }_1\right)$ is $0.9$ at $60°\text{C}$.

Calculation:

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

  $\begin{array}{c}\dot{Q}=\frac{\sigma \left(T_i{}^4-T_j{}^4\right)}{\left(\frac{1-{\epsilon }_i}{{\epsilon }_i}\right)\frac{1}{A_i}+\frac{1}{A_1{F}_{ij}}+\left(\frac{1-{\epsilon }_j}{{\epsilon }_j}\right)\frac{1}{A_j}}\\ =\frac{\sigma \left(T_i{}^4-T_j{}^4\right)}{\left(\frac{1-{\epsilon }_i}{{\epsilon }_i}\right)+\frac{1}{F_{ij}}+\left(\frac{1-{\epsilon }_j}{{\epsilon }_j}\right)\frac{A_i}{A_j}}\\ =\frac{\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\left[{\left(1173\right)}^4-{\left(333\right)}^4{\text{K}}^4\right]}{\left[\frac{1-0.8}{\left(0.8\right)}+\frac{1}{\left(0.65\right)}+\frac{1-0.9}{\left(0.9\right)}\right]\frac{\left(3\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}{\pi \left(1.5\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}}\\ =57900\text{W}/{\text{m}}^{\text{2}}\end{array}$

Thus, the net rate of heat transfer is $57900\text{W}/{\text{m}}^{\text{2}}$.

(c)

To determine

The temperature of the tube surface in steady operation.

Explanation of Solution

Given:

The average temperature $\left(T\right)$ is $40°\text{C}$.

The heat transfer coefficient $\left(h\right)$ is $2\text{kW}/{\text{m}}^{\text{2}}\cdot \text{K}$.

Calculation:

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

    $\begin{array}{c}{\dot{q}}_{rad}={\dot{q}}_{conv}\\ h{A}_j\left(T_w-T_j\right)=\frac{\sigma \left(T_i{}^4-T_j{}^4\right)}{\left(\frac{1-{\epsilon }_i}{{\epsilon }_i}\right)+\frac{1}{F_{ij}}+\left(\frac{1-{\epsilon }_j}{{\epsilon }_j}\right)\frac{A_i}{A_j}}\\ \frac{\left(2000\text{W}/{\text{m}}^2\cdot \text{K}\right)\left[\left(T_w\right)-\left(40+273\right)\text{K}\right]}{\frac{\left(3\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}{\pi \left(1.5\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}}=\left[\frac{\begin{array}{l}\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\times \\ \left[{\left(1173\right)}^4-{\left(T_w\right)}^4\text{K}\right]\end{array}}{\begin{array}{l}\left[\frac{1-0.8}{\left(0.8\right)}+\frac{1}{\left(0.65\right)}+\frac{1-0.9}{\left(0.9\right)}\right]\times \\ \frac{\left(3\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}{\pi \left(1.5\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}\end{array}}\right]\\ T_w=\left(331.4-273\right)\text{K}\\ T_w=58.4°\text{C}\end{array}$

Thus, the temperature of the surface is $58.4°\text{C}$.