Chapter 20, Problem 103RQ

To determine

The temperature of the aluminum tube when equilibrium is established.

Explanation of Solution

Given:

The inner diameter of tank $\left(d_i\right)$ is $5\text{cm}$.

The outer diameter of tank $\left(d_o\right)$ is $7\text{cm}$.

The solar radiation absorbed $\left(Q\right)$ is $20\text{W}$.

The temperature of air is $\left(T_{\infty }\right)$ is $30°\text{C}$.

The surface temperature is $\left(T_s\right)$ is $33°\text{C}$.

Calculation:

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

  $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{30°\text{C}+\text{33}°\text{C}}{2}\\ =31.5°\text{C}\end{array}$

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

    $\begin{array}{c}{L}_c=d_o\\ =7\text{cm}\times \frac{1\text{m}}{100\text{cm}}\\ =0.07\text{m}\end{array}$

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

Obtain the following properties of air corresponding to the temperature of $31.5°\text{C}$ as follows:

  $\begin{array}{l}k=0.02599\text{W}/\text{m}\cdot \text{K}\\ v=1.622\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\\ \mathrm{Pr}=0.7278\end{array}$

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

  $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_s-T_{\infty }\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{g\left(\frac{1}{T_f}\right)\left(T_s-T_{\infty }\right)d^3}{v^2}\mathrm{Pr}\\ =\frac{\left[\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1}{\left(31.5°\text{C}+273\right)\text{K}}\right)\left(\begin{array}{l}\left(33°\text{C}+273\right)\text{K}\\ -\left(30°\text{C}+273\right)\text{K}\end{array}\right){\left(7\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^3\right]}{{\left(1.622\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7278\right)\\ =91700\end{array}$

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

    $\begin{array}{c}Nu={\left[0.6+\frac{0.387R{a}^{1/6}}{{\left[1+{\left(\frac{0.559}{\mathrm{Pr}}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.6+\frac{0.387{\left(91700\right)}^{1/6}}{{\left[1+{\left(\frac{0.559}{0.7278}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ =7.626\end{array}$

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

    $\begin{array}{c}h=\frac{k}{d_o}Nu\\ =\frac{0.02599\text{W}/\text{m}\cdot \text{K}}{\left(7\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\left(7.626\right)\\ =2.832\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate rate of heat transfer by convection $\left(Q_c\right)$ using the relation.

  $\begin{array}{c}{\dot{Q}}_c=hA\left(T_s-T_{\infty }\right)\\ =\left(h\right)\left(\pi d_{o}L\right)\left(T_s-T_{\infty }\right)\\ =\left(2.832\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(\pi \left(7\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(1\text{m}\right)\right)\left[\begin{array}{l}\left(T_s°\text{C}+273\right)\text{K}\\ -\left(30°\text{C}+273\right)\text{K}\end{array}\right]\\ \text{=}\left(2.832\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(0.2199{\text{m}}^{\text{2}}\right)\left(T_s-30\right)\text{K}\end{array}$

Calculate rate of heat transfer by radiation $\left(Q_r\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_r=\epsilon A_s\sigma \left(T_s^4-T_{surr}^4\right)\\ =\epsilon \left(\pi d_{o}L\right)\sigma \left(T_s^4-T_{surr}^4\right)\\ =\left[\begin{array}{l}\left(1\right)\left(\pi \left(7\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(1\text{m}\right)\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\\ \times \left[{\left(\left(T_s+273\right)\text{K}\right)}^4-{\left(\left(20+273\right)\text{K}\right)}^4\right]\end{array}\right]\end{array}$

Calculate total rate of heat transfer by radiation $\left(Q\right)$ using the relation.

    $\begin{array}{l}\dot{Q}={\dot{Q}}_c+{\dot{Q}}_r\\ \dot{Q}\text{=}\left[\begin{array}{l}\left(2.832\text{W}/{\text{m}}^2\text{K}\right)\left(0.2199{\text{m}}^{\text{2}}\right)\left(\begin{array}{l}{T}_g-\\ \left(30°\text{C}+273\right)\text{K}\end{array}\right)+\\ \left\{\begin{array}{l}\left(1\right)\left(\pi \left(7\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(1\text{m}\right)\right)\times \\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot \text{K}\right)\times \\ \left({\left(T_g+273\text{K}\right)}^4-{\left(20+273\text{K}\right)}^4\right)\end{array}\right\}\end{array}\right]\\ 20\text{W=}\left[\begin{array}{l}\left(2.832\text{W}/{\text{m}}^2\text{K}\right)\left(0.2199{\text{m}}^{\text{2}}\right)\left(\begin{array}{l}{T}_s-\\ \left(30°\text{C}+273\right)\text{K}\end{array}\right)+\\ \left\{\begin{array}{l}\left(1\right)\left(\pi \left(7\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(1\text{m}\right)\right)\times \\ \left(5.67\times {10}^{-8}{\text{W}/{\text{m}}^2\cdot \text{K}}^4\right)\times \\ \left({\left(T_s+273\text{K}\right)}^4-{\left(20+273\text{K}\right)}^4\right)\end{array}\right\}\end{array}\right]\\ T_s=\left(306.34\text{K}-273\right)°\text{C}\\ T_s=33.34°\text{C}\end{array}$

Assume the aluminum tube temperature to be $45°\text{C}$.

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

  $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{45°\text{C+33}\text{.34}°\text{C}}{2}\\ =39.17°\text{C}\end{array}$

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

Obtain the following properties of air corresponding to the temperature of $39.17°\text{C}$ as follows:

  $\begin{array}{l}k=0.02656\text{W}/\text{m}\cdot \text{K}\\ v=1.694\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\\ \mathrm{Pr}=0.7257\end{array}$

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

    $\begin{array}{c}{L}_c=d\\ =\frac{d_0-d_i}{2}\\ =\frac{\left(7-5\right)\text{cm}}{2}\\ =1\text{cm}\end{array}$

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

  $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_s-T_{\infty }\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{g\left(\frac{1}{T_f}\right)\left(T_s-T_{\infty }\right)d^3}{v^2}\mathrm{Pr}\\ =\frac{\left[\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1}{\left(39.17°\text{C}+273\right)\text{K}}\right)\left(\begin{array}{l}\left(45+273\right)°\text{K}\\ -\left(33.34+273\right)°\text{K}\end{array}\right){\left(1\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^3\right]}{{\left(1.694\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7257\right)\\ =926.5\end{array}$

Calculate geometric factor for concentric cylinders $\left(F_{cyl}\right)$ using the relation.

    $\begin{array}{c}{F}_{cyl}=\frac{{\left[\ln \left(\frac{d_o}{d_i}\right)\right]}^4}{L_c^3{\left(d_i^{-3/5}+d_o^{-3/5}\right)}^5}\\ =\frac{{\left[\ln \left(\frac{7\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{5\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\right)\right]}^4}{{\left(1\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^3{\left({\left(5\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^{-3/5}+{\left(9\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^{-3/5}\right)}^5}\\ =0.08085\end{array}$

Calculate effective thermal conductivity $\left(k_{eff}\right)$ using the relation.

    $\begin{array}{c}{k}_{eff}=0.386k{\left(\frac{\mathrm{Pr}}{0.861+\mathrm{Pr}}\right)}^{1/4}{\left(F_{cyl}Ra\right)}^{1/4}\\ =0.368\left(0.02706\text{W}/\text{m}\cdot \text{K}\right){\left(\frac{0.7238}{0.861+0.7238}\right)}^{1/4}{\left(\left(0.08085\right)\left(926.5\right)\right)}^{1/4}\\ =0.02480\text{W}/\text{m}\cdot \text{K}\end{array}$

Calculate the heat loss from the collector per meter length of the tube $\left(Q\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_c=\frac{2\pi k_{eff}}{\ln \left(\frac{d_o}{d_i}\right)}\left(T_i-T_o\right)\\ =\frac{2\pi k_{eff}}{\ln \left(\frac{d_o}{d_i}\right)}\left(T_i-T_o\right)\\ =\frac{2\pi \left(0.02480\text{W}/\text{m}\cdot \text{K}\right)\left[\left(T_{tube}°\text{C}+273\right)\text{K}-\left(33.34°\text{C}+273\right)\text{K}\right]}{\ln \left(\frac{7\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{5\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\right)}\\ =\frac{2\pi \left(0.02480\text{W}/\text{m}\cdot \text{K}\right)\left(T_{tube}-33.34\text{K}\right)}{\ln \left(\frac{0.07\text{m}}{0.05\text{m}}\right)}\end{array}$

Calculate rate of heat transfer by radiation $\left(Q_r\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_r=\epsilon A_s\sigma \left(T_s^4-T_{surr}^4\right)\\ =\epsilon \left(\pi d_{i}L\right)\sigma \left(T_s^4-T_{surr}^4\right)\\ =\left[\begin{array}{l}\left(1\right)\left(\pi \left(5\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(1\text{m}\right)\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\\ \times \left[{\left(\left(T_{tube}°\text{C}+273\right)\text{K}\right)}^4-{\left(\left(33.34°\text{C}+273\right)\text{K}\right)}^4\right]\end{array}\right]\end{array}$

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

    $\begin{array}{l}\dot{Q}={\dot{Q}}_c+{\dot{Q}}_r\\ 20\text{W}=\left[\begin{array}{l}\frac{2\pi \left(0.02480\text{W}/\text{m}°\text{C}\right)\left(T_{tube}-33.34\text{K}\right)}{\ln \left(\frac{0.07\text{m}}{0.05\text{m}}\right)}\text{+}\\ \left(1\right)\left(0.1517{\text{m}}^{\text{2}}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2{\text{K}}^4\right)\left(\begin{array}{l}{\left(T_{tube}+273\right)}^4{\text{K}}^4+\\ {\left(33.34°\text{C}+273\right)}^4{\text{K}}^4\end{array}\right)\end{array}\right]\\ T_{tube}=\left(319.3\text{K}-273\right)°\text{C}\\ T_{tube}=46.3°\text{C}\end{array}$

Thus, the temperature of the aluminum tube when equilibrium is established is $46.3°\text{C}$.