Chapter 20, Problem 30P

(a)

To determine

The average temperature of the hot surface of the board when it is vertical.

Explanation of Solution

Given:

The length of the board $\left(l\right)$ is $20\text{cm}$.

The width of the board $\left(w\right)$ is $15\text{cm}$.

The temperature of the room $\left(T_{\infty }\right)$ is $20°\text{C}$.

The heat dissipated by the board $\left(\dot{Q}\right)$ is $8\text{W}$.

The emissivity of the surface of the board $\left(\epsilon \right)$ is $0.8$.

The film temperature $\left(T_f\right)$ is $32.5°\text{C}$.

Calculation:

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

    $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ T_s=2T_f-T_{\infty }\\ =2\left(32.5°\text{C}\right)-\left(20°\text{C}\right)\\ =45°\text{C}\end{array}$

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

Obtain the following values of different properties corresponding film temperature $32.5°\text{C}$ to the as follows:

  $\begin{array}{l}k=0.02607\text{W}/\text{m}\cdot \text{K}\\ \mathrm{Pr}=0.7275\\ v=1.631\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\end{array}$

Calculate the volume expansion coefficient $\left(\beta \right)$ using the relation.

    $\begin{array}{c}\beta =\frac{1}{T}\\ =\frac{1}{\left(32.5°\text{C}+\text{273}\right)\text{K}}\\ =\frac{1}{305.5\text{K}}\\ =0.003273{\text{K}}^{-1}\end{array}$

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

    $\begin{array}{c}{L}_c=l\\ =20\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\\ =0.2\text{m}\end{array}$

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

    $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_s-T_{\infty }\right)L_c{}^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.003273{\text{K}}^{-1}\right)\left(\begin{array}{l}45°\text{C}\\ -20°\text{C}\end{array}\right){\left(0.2\text{m}\right)}^3}{{\left(1.631\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7275\right)\\ =\frac{\left(0.03210\text{m}/{\text{s}}^2\cdot \text{K}\right)\left[\begin{array}{r}\left(45°\text{C}+273\right)\text{K}\\ -\left(20°\text{C}+273\right)\text{K}\end{array}\right]\left(8\times {10}^{-3}{\text{m}}^{\text{3}}\right)}{{\left(1.631\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7275\right)\\ =1.756\times {10}^6\end{array}$

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

    $\begin{array}{c}\text{Nu}={\left[0.825+\frac{0.389{\left(\text{Ra}\right)}^{1/6}}{{\left[1+{\left(0.492/\text{Pr}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{0.389{\left(1.756\times {10}^7\right)}^{1/6}}{{\left[1+{\left(0.492/0.7275\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{6.271}{1.19}\right]}^2\\ =37.14\end{array}$

Calculate the area of heat transfer $\left(A\right)$ using the relation.

    $\begin{array}{c}A=l\times w\\ =20\text{cm}\times 15\text{cm}\\ =\text{300}{\text{cm}}^2\times \frac{{10}^{-4}{\text{m}}^{\text{2}}}{1{\text{cm}}^2}\\ =0.03{\text{m}}^{\text{2}}\end{array}$

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

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02607\text{W}/\text{m}\cdot \text{K}}{0.2\text{m}}\left(37.14\right)\\ =4.841\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

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

    $\begin{array}{c}\dot{Q}=hA\left(T_{sa}-T_{\infty }\right)+\epsilon A\sigma \left(T_{sa}^4-T_{surr}^4\right)\\ \left(8\text{W}\right)=\left[\begin{array}{l}\left(4.841\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.03{\text{m}}^2\right)\left(\begin{array}{l}{T}_{sa}\\ -20°\text{C}\end{array}\right)\\ +\left(0.8\right)\left(0.03{\text{m}}^{\text{2}}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left[\begin{array}{l}{\left(\left(T_{sa}+273\right)\text{K}\right)}^4\\ -{\left(\left(20°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\end{array}\right]\\ =\left(0.145\text{W}/\text{K}\right)\left(\begin{array}{l}{T}_{sa}\\ -20°\text{C}\end{array}\right)+\left(0.1360\times {10}^{-8}\text{W}/{\text{K}}^{\text{4}}\right)\left[\begin{array}{l}{\left(\left(T_{sa}+273\right)\text{K}\right)}^4\\ -{\left(\left(20°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\\ T_{sa}=\left(319.6\text{K}-273\right)°\text{C}\\ T_{sa}=46.6°\text{C}\end{array}$

By use of trial and error approach the surface temperature obtained is $46.6\text{°C}$, which is nearest to $45°\text{C}$.

Thus, the average temperature of the hot surface of the board is $\underset{\_}{46.6°\text{C}}$ when it is vertical.

(b)

To determine

The average temperature of the hot surface of the board when it is horizontal and hot surface facing up.

Explanation of Solution

Calculation:

Calculate the characteristic length $\left(L_{c2}\right)$ using the relation.

    $\begin{array}{c}{L}_{c2}=\frac{A}{p}\\ =\frac{0.03{\text{m}}^{\text{2}}}{2\left(l+w\right)}\\ =\frac{0.0\text{3}{\text{m}}^{\text{2}}}{2\left(\left(20\text{cm}\times \frac{{10}^{-2}\text{cm}}{1\text{m}}\right)+\left(\text{15}\text{cm}\times \frac{{10}^{-2}\text{cm}}{1\text{m}}\right)\right)}\\ =0.0429\text{m}\end{array}$

Calculate the Rayleigh number $\left({\text{Ra}}_2\right)$ using the relation.

    $\begin{array}{c}{\text{Ra}}_2=\frac{g\beta \left(T_s-T_{\infty }\right)L_{c2}{}^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.003273{\text{K}}^{-1}\right)\left(\begin{array}{l}45°\text{C}\\ -20°\text{C}\end{array}\right){\left(0.0429\text{m}\right)}^3}{{\left(1.631\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7275\right)\\ =\frac{\left(0.03210\text{m}/{\text{s}}^2\cdot \text{K}\right)\left[\begin{array}{r}\left(45°\text{C}+273\right)\text{K}\\ -\left(20°\text{C}+273\right)\text{K}\end{array}\right]\left(7.895\times {10}^{-5}{\text{m}}^{\text{3}}\right)}{{\left(1.631\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7275\right)\\ =1.728\times {10}^5\end{array}$

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

    $\begin{array}{c}{\text{Nu}}_2=0.54{\left({\text{Ra}}_2\right)}^{1/4}\\ =0.54{\left(1.728\times {10}^5\right)}^{1/4}\\ =11.01\end{array}$

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

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02607\text{W}/\text{m}\cdot \text{K}}{0.0429\text{m}}\left(11.01\right)\\ =6.69\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

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

    $\begin{array}{c}\dot{Q}=hA\left(T_{s2}-T_{\infty }\right)+\epsilon A\sigma \left(T_{s2}^4-T_{surr}^4\right)\\ \left(8\text{W}\right)=\left[\begin{array}{l}\left(6.69\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.03{\text{m}}^2\right)\left(\begin{array}{l}{T}_{s2}\\ -20°\text{C}\end{array}\right)\\ +\left(0.8\right)\left(0.03{\text{m}}^{\text{2}}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left[\begin{array}{l}{\left(\left(T_{s2}+273\right)\text{K}\right)}^4\\ -{\left(\left(20°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\end{array}\right]\\ =\left(0.2007\text{W}/\text{K}\right)\left(\begin{array}{l}{T}_{s2}\\ -25°\text{C}\end{array}\right)+\left(0.13608\times {10}^{-8}\text{W}/{\text{K}}^{\text{4}}\right)\left[\begin{array}{l}{\left(\left(T_{s2}+273\right)\text{K}\right)}^4\\ -{\left(\left(20°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\\ T_{s2}=\left(315.6\text{K}-273\right)°\text{C}\\ T_{s2}=42.6°\text{C}\end{array}$

Thus, the average temperature of the hot surface of the board is $\underset{\_}{42.6°\text{C}}$ when it is horizontal and hot surface facing up.

(c)

To determine

The average temperature of the hot surface of the board when it is horizontal and hot surface facing down.

Explanation of Solution

Calculation:

Calculate the characteristic length $\left(L_{c3}\right)$ using the relation.

    $\begin{array}{c}{L}_{c3}=\frac{A}{p}\\ =\frac{0.03{\text{m}}^{\text{2}}}{2\left(l+w\right)}\\ =\frac{0.0\text{3}{\text{m}}^{\text{2}}}{2\left(20\text{cm}\times \frac{{10}^{-2}\text{cm}}{1\text{m}}+\text{15}\text{cm}\times \frac{{10}^{-2}\text{cm}}{1\text{m}}\right)}\\ =0.0429\text{m}\end{array}$

Calculate the Rayleigh number $\left({\text{Ra}}_2\right)$ using the relation.

    $\begin{array}{c}{\text{Ra}}_2=\frac{g\beta \left(T_s-T_{\infty }\right)L_{c2}{}^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.003273{\text{K}}^{-1}\right)\left(\begin{array}{l}45°\text{C}\\ -20°\text{C}\end{array}\right){\left(0.0429\text{m}\right)}^3}{{\left(1.631\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7275\right)\\ =\frac{\left(0.03210\text{m}/{\text{s}}^2\cdot \text{K}\right)\left[\begin{array}{r}\left(45°\text{C}+273\right)\text{K}\\ -\left(20°\text{C}+273\right)\text{K}\end{array}\right]\left(7.895\times {10}^{-5}{\text{m}}^{\text{3}}\right)}{{\left(1.631\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7275\right)\\ =1.728\times {10}^5\end{array}$

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

    $\begin{array}{c}{\text{Nu}}_3=0.54{\left(\text{Ra}\right)}^{1/4}\\ =0.54{\left(1.728\times {10}^5\right)}^{1/4}\\ =5.505\end{array}$

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

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02607\text{W}/\text{m}\cdot \text{K}}{0.0429\text{m}}\left(5.505\right)\\ =3.345\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

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

    $\begin{array}{c}\dot{Q}=hA\left(T_{s3}-T_{\infty }\right)+\epsilon A\sigma \left(T_{s3}^4-T_{surr}^4\right)\\ \left(8\text{W}\right)=\left[\begin{array}{l}\left(3.345\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.03{\text{m}}^2\right)\left(\begin{array}{l}{T}_{s3}\\ -20°\text{C}\end{array}\right)\\ +\left(0.8\right)\left(0.03{\text{m}}^{\text{2}}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left[\begin{array}{l}{\left(\left(T_{s3}+273\right)\text{K}\right)}^4\\ -{\left(\left(20°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\end{array}\right]\\ =\left(0.2007\text{W}/\text{K}\right)\left(\begin{array}{l}{T}_{s3}\\ -25°\text{C}\end{array}\right)+\left(0.13608\times {10}^{-8}\text{W}/{\text{K}}^{\text{4}}\right)\left[\begin{array}{l}{\left(\left(T_{s3}+273\right)\text{K}\right)}^4\\ -{\left(\left(20°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\\ T_{s3}=\left(323.7\text{K}-273\right)°\text{C}\\ T_{s3}=50.7°\text{C}\end{array}$

Thus, the average temperature of the hot surface of the board is $\underset{\_}{50.7°\text{C}}$ when it is horizontal and hot surface facing down.