Chapter 20, Problem 22P

(a)

To determine

The plate midpoint temperature for a highly polished surface.

Explanation of Solution

Given:

The length of thin vertical plate $\left(l\right)$ is $0.5\text{m}$.

The heat flux $\left(Q\right)$ is $1000\text{W}/{\text{m}}^{\text{2}}$.

The temperature of air $\left(T_{\infty }\right)$ is $5°\text{C}$.

The film temperature $\left(T_f\right)$ is $30°\text{C}$.

The emissivity is $\left(\epsilon \right)$ is $0.02$.

Calculation:

Refer Table A-22 “Properties of gases at 1 atm pressure”.

Obtain the following properties of air corresponding to the temperature of $30°\text{C}$ as follows:

$\begin{array}{c}k=0.02588\text{W}/\text{m}\cdot \text{K}\\ v=1.608\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\\ \mathrm{Pr}=0.7282\end{array}$

Calculate the midpoint temperature $\left(T_{L/2}\right)$ using the relation.

    $\begin{array}{c}{T}_{L/2}=2T_f-T_{\infty }\\ =2\left(30°\text{C}\right)-5°\text{C}\\ =\text{55}°\text{C}\end{array}$

Calculate Characteristic length $\left(L_c\right)$ using the relation.

    $\begin{array}{c}{L}_c=l\\ =0.5\text{m}\end{array}$

Calculate the Rayleigh number $\left(\text{Ra}\right)$ using the relation.

  $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_{L/2}-T_{\infty }\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{g\left(\frac{1}{T_f}\right)\left(T_{L/2}-T_{\infty }\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81{\text{m}/\text{s}}^2\right)\left(\frac{1}{\left(30°\text{C}+273\right)\text{K}}\right)\left(\begin{array}{l}\left(55°\text{C}+273\right)\text{K}-\\ \left(5°\text{C}+273\right)\text{K}\end{array}\right){\left(0.5\text{m}\right)}^3}{{\left(1.608\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7282\right)\\ =5.699\times {10}^8\end{array}$

Calculate the Nusselt number $\left(Nu\right)$ for vertical plate using the relation.

    $\begin{array}{c}Nu={\left[0.825+\frac{0.387R{a}^{1/6}}{{\left[1+{\left(\frac{0.492}{\mathrm{Pr}}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{0.387{\left(5.699\times {10}^8\right)}^{1/6}}{{\left[1+{\left(\frac{0.492}{0.7282}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ =103.7\end{array}$

Calculate the heat transfer coefficient $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02588\text{W}/\text{m}\cdot \text{K}}{\left(0.5\text{m}\right)}\left(103.7\right)\\ =5.368\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the plate midpoint temperature $\left(T_{L/2}\right)$ using the relation.

  $\begin{array}{c}Q=h\left(T_{L/2}-T_{\infty }\right)+\epsilon \sigma \left(T_{L/2}^4-T_{surr}^4\right)\\ 1000\text{W}/{\text{m}}^{\text{2}}=\left[\begin{array}{l}\left(5.368\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(\begin{array}{l}\left(T_{L/2}+273\right)\text{K}-\\ \left(5°\text{C}+273\right)\text{K}\end{array}\right)+\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}\right)\left(0.02\right)\left(\begin{array}{l}{\left(T_{L/2}+273\right)}^4{\text{K}}^4-\\ {\left(5°\text{C}+273\right)}^4{\text{K}}^4\end{array}\right)\end{array}\right]\\ T_f=\left(420.2\text{K}-273\right)°\text{C}\\ =147.2°\text{C}\end{array}$

Thus, the plate midpoint temperature for a highly polished surface is $147.2°\text{C}$.

(b)

To determine

The plate midpoint temperature for a black oxidized surface.

Explanation of Solution

Given:

The length of thin vertical plate $\left(l\right)$ is $0.5\text{m}$.

The heat flux $\left(Q\right)$ is $1000\text{W}/{\text{m}}^{\text{2}}$.

The temperature of air $\left(T_{\infty }\right)$ is $5°\text{C}$.

The film temperature $\left(T_f\right)$ is $30°\text{C}$.

The emissivity is $\left(\epsilon \right)$ is $0.78$.

Calculation:

Refer Table A-22 “Properties of gases at 1 atm pressure”.

Obtain the following properties of air corresponding to the temperature of $30°\text{C}$ as follows:

$\begin{array}{c}k=0.02588\text{W}/\text{m}\cdot \text{K}\\ v=1.608\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\\ \mathrm{Pr}=0.7282\end{array}$

Calculate the midpoint temperature $\left(T_{L/2}\right)$ using the relation.

    $\begin{array}{c}{T}_{L/2}=2T_f-T_{\infty }\\ =2\left(30°\text{C}\right)-5°\text{C}\\ =\text{55}°\text{C}\end{array}$

Calculate Characteristic length $\left(L_c\right)$ using the relation.

    $\begin{array}{c}{L}_c=l\\ =0.5\text{m}\end{array}$

Calculate the Rayleigh number $\left(\text{Ra}\right)$ using the relation.

  $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_{L/2}-T_{\infty }\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{g\left(\frac{1}{T_f}\right)\left(T_{L/2}-T_{\infty }\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81{\text{m}/\text{s}}^2\right)\left(\frac{1}{\left(30°\text{C}+273\right)\text{K}}\right)\left(\begin{array}{l}\left(55°\text{C}+273\right)\text{K}-\\ \left(5°\text{C}+273\right)\text{K}\end{array}\right){\left(0.5\text{m}\right)}^3}{{\left(1.608\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7282\right)\\ =5.699\times {10}^8\end{array}$

Calculate the Nusselt number $\left(Nu\right)$ for vertical plate using the relation.

    $\begin{array}{c}Nu={\left[0.825+\frac{0.387R{a}^{1/6}}{{\left[1+{\left(\frac{0.492}{\mathrm{Pr}}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{0.387{\left(5.699\times {10}^8\right)}^{1/6}}{{\left[1+{\left(\frac{0.492}{0.7282}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ =103.7\end{array}$

Calculate the heat transfer coefficient $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02588\text{W}/\text{m}\cdot \text{K}}{\left(0.5\text{m}\right)}\left(103.7\right)\\ =5.368\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the plate midpoint temperature $\left(T_{L/2}\right)$ using the relation.

  $\begin{array}{c}Q=h\left(T_{L/2}-T_{\infty }\right)+\epsilon \sigma \left(T_{L/2}^4-T_{surr}^4\right)\\ 1000\text{W}/{\text{m}}^{\text{2}}=\left[\begin{array}{l}\left(5.368\text{W}/{\text{m}}^{\text{2}}\cdot \text{°K}\right)\left(\begin{array}{l}\left(T_{L/2}+273\right)\text{K}-\\ \left(5°\text{C}+273\right)\text{K}\end{array}\right)+\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}\right)\left(0.78\right)\left(\begin{array}{l}{\left(T_{L/2}+273\right)}^4{\text{K}}^4-\\ {\left(5°\text{C}+273\right)}^4{\text{K}}^4\end{array}\right)\end{array}\right]\\ T_f=\left(361.5\text{K}-273\right)°\text{C}\\ =88.5°\text{C}\end{array}$

Thus, the plate midpoint temperature for a highly polished surface is $88.5°\text{C}$.