Chapter 20, Problem 34P

To determine

The equilibrium temperature of the absorber plate.

Explanation of Solution

Given:

The length of the plate $\left(l\right)$ is $1.2\text{m}$.

The width of plate  is $\left(w\right)$ is $\text{0}\text{.8}\text{m}$.

The temperature of air $\left(T_{\infty }\right)$ is $25°\text{C}$.

The temperature of sky $\left(T_{sky}\right)$ is $10°\text{C}$.

The surface temperature $\left(T_s\right)$ is $115°\text{C}$.

The solar radiation $\left(q\right)$ is $600\text{W}/{\text{m}}^{\text{2}}$.

The emissivity $\left(\epsilon \right)$ is $0.98$.

The absorptivity $\left(\alpha \right)$ is $0.98$.

Calculation:

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

  $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{115°\text{C}+25°\text{C}}{2}\\ =70°\text{C}\end{array}$

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

Obtain the following properties of air corresponding to the temperature of $70°\text{C}$ as follows:

$\begin{array}{c}k=0.02881\text{W}/\text{m}\cdot \text{K}\\ v=1.995\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\\ \mathrm{Pr}=0.7177\end{array}$

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

    $\begin{array}{c}{L}_c=\frac{A}{p}\\ =\frac{lw}{2\left(l+w\right)}\\ =\frac{\left(1.2\text{m}\right)\left(0.8\text{m}\right)}{2\left[\left(1.2\text{m}\right)+\left(0.8\text{m}\right)\right]}\\ =0.24\text{m}\end{array}$

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

  $\begin{array}{c}\text{Ra=}\frac{g\beta \left(T_s-T_{\infty }\right)L_c^2}{v^2}\mathrm{Pr}\\ =\frac{g\left(\frac{1}{T_f}\right)\left(T_s-T_{\infty }\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81{\text{m}/\text{s}}^2\right)\left(\frac{1}{\left(70°\text{C}+273\right)\text{K}}\right)\left(\begin{array}{l}\left(115°\text{C}+273\right)\text{K}-\\ \left(25°\text{C}+273\right)\text{K}\end{array}\right){\left(0.24\text{m}\right)}^3}{{\left(1.995\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7177\right)\\ =6.414\times {10}^7\end{array}$

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

    $\begin{array}{c}Nu=0.15R{a}^{1/3}\\ =0.15{\left(6.414\times {10}^7\right)}^{1/3}\\ =60.04\end{array}$

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

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02881\text{W}/\text{m}\cdot \text{K}}{\left(0.24\text{m}\right)}\left(60.04\right)\\ =7.208\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the rate of heat loss $\left(Q\right)$ using the relation.

    $\begin{array}{c}Q=\alpha qA\\ =\left(0.87\right)\left(600\text{W}/{\text{m}}^{\text{2}}\right)\left(0.8\text{m}\right)\left(1.2\text{m}\right)\\ =501.1\text{W}\end{array}$

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

    $\begin{array}{c}Q=hA\left(T_s-T_{\infty }\right)+\epsilon A\sigma \left[{\left(\left(T_s+273\right)\text{K}\right)}^4-{\left(\left(T_{sky}+273\right)\text{K}\right)}^4\right]\\ 501.1\text{W}=\left[\begin{array}{l}7.208\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\left(\left(0.8\text{m}\right)\left(1.2\text{m}\right)\right)\left(T_s-\left(25°\text{C}+273\right)\text{K}\right)+\\ \left(0.09\right)\left(\left(0.8\text{m}\right)\left(1.2\text{m}\right)\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2{\text{K}}^4\right)\left[\begin{array}{l}{\left(\left(T_s+273\right)\text{K}\right)}^4-\\ {\left(\left(10+273\right)\text{K}\right)}^4\end{array}\right]\end{array}\right]\\ T_s=\left(362.7\text{K}-273\right)°\text{C}\\ =89.7°\text{C}\end{array}$

This is not nearby assumed temperature of $115°\text{C}$ so repeat the calculation using $T_s=95°\text{C}$.

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

  $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{95°\text{C}+25°\text{C}}{2}\\ =60°\text{C}\end{array}$

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

Obtain the following properties of air corresponding to the temperature of $70°\text{C}$ as follows:

$\begin{array}{c}k=0.02808\text{W}/\text{m}\cdot \text{K}\\ v=1.896\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\\ \mathrm{Pr}=0.7202\end{array}$

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

    $\begin{array}{c}{L}_c=\frac{A}{p}\\ =\frac{lw}{2\left(l+w\right)}\\ =\frac{\left(1.2\text{m}\right)\left(0.8\text{m}\right)}{2\left[\left(1.2\text{m}\right)+\left(0.8\text{m}\right)\right]}\\ =0.24\text{m}\end{array}$

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

  $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_s-T_{\infty }\right)L_c^2}{v^2}\mathrm{Pr}\\ =\frac{g\left(\frac{1}{T_f}\right)\left(T_s-T_{\infty }\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81{\text{m}/\text{s}}^2\right)\left(\frac{1}{\left(60°\text{C}+273\right)\text{K}}\right)\left(\begin{array}{l}\left(95°\text{C}+273\right)\text{K}-\\ \left(25°\text{C}+273\right)\text{K}\end{array}\right){\left(0.24\text{m}\right)}^3}{{\left(1.995\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7177\right)\\ =5.711\times {10}^7\end{array}$

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

    $\begin{array}{c}Nu=0.15R{a}^{1/3}\\ =0.15{\left(5.711\times {10}^7\right)}^{1/3}\\ =57.77\end{array}$

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

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02808\text{W}/\text{m}\cdot \text{K}}{\left(0.24\text{m}\right)}\left(57.77\right)\\ =6.759\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the rate of heat loss $\left(Q\right)$ using the relation.

    $\begin{array}{c}Q=\alpha qA\\ =\left(0.87\right)\left(600\text{W}/{\text{m}}^{\text{2}}\right)\left(0.8\text{m}\right)\left(1.2\text{m}\right)\\ =501.1\text{W}\end{array}$

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

    $\begin{array}{c}Q=hA\left(T_s-T_{\infty }\right)+\epsilon A\sigma \left[{\left(\left(T_s+273\right)\text{K}\right)}^4-{\left(\left(T_{sky}+273\right)\text{K}\right)}^4\right]\\ 501.1\text{W}=\left[\begin{array}{l}6.759\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\left(\left(0.8\text{m}\right)\left(1.2\text{m}\right)\right)\left(T_s-\left(25°\text{C}+273\right)\text{K}\right)\\ +\left(0.09\right)\left(\left(0.8\text{m}\right)\left(1.2\text{m}\right)\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2{\text{K}}^4\right)\left[\begin{array}{l}{\left(\left(T_s+273\right)\text{K}\right)}^4-\\ {\left(\left(10°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\end{array}\right]\\ T_s=\left(366.5\text{K}-273\right)°\text{C}\\ =93.5°\text{C}\end{array}$

Repeat the whole calculation with $\alpha =0.28$ and $\epsilon =0.07$.

Calculate the rate of heat loss $\left(Q\right)$ using the relation.

    $\begin{array}{c}Q=\alpha qA\\ =\left(0.28\right)\left(600\text{W}/{\text{m}}^{\text{2}}\right)\left(0.8\text{m}\right)\left(1.2\text{m}\right)\\ =161.3\text{W}\end{array}$

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

    $\begin{array}{c}Q=hA\left(T_s-T_{\infty }\right)+\epsilon A\sigma \left[{\left(\left(T_s+273\right)\text{K}\right)}^4-{\left(\left(T_{sky}+273\right)\text{K}\right)}^4\right]\\ 161.3\text{W}=\left[\begin{array}{l}6.629\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\left(\left(0.8\text{m}\right)\left(1.2\text{m}\right)\right)\left(T_s-\left(25°\text{C}+273\right)\text{K}\right)+\\ \left(0.07\right)\left(\left(0.8\text{m}\right)\left(1.2\text{m}\right)\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2{\text{K}}^4\right)\times \\ \left[{\left(\left(T_s+273\right)\text{K}\right)}^4-{\left(\left(10+273\right)\text{K}\right)}^4\right]\end{array}\right]\\ T_s=\left(320.8\text{K}-273\right)°\text{C}\\ =47.8°\text{C}\end{array}$

Thus, the equilibrium temperature of the absorber plate is $47.8°\text{C}$.