Chapter 19, Problem 56P

To determine

The change in the values obtained in heat transfer.

Explanation of Solution

Given:

The diameter of the section $\left(D\right)$ is $10\text{cm}$.

The length of the section $\left(L\right)$ is $12\text{m}$.

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

The temperature of surface $\left(T_s\right)$ is $75°\text{C}$.

The new temperature of surface $\left(T_{s1}\right)$ is $75°\text{C}$.

The emissivity $\left(\epsilon \right)$ is $0.8$.

The time of heat transfer $\left(\Delta T\right)$  is $10\text{h}/\text{day}$.

The average temperature of the surfaces $\left(T_a\right)$ is $0°\text{C}$.

The new average temperature of the surfaces $\left(T_{a1}\right)$ is $-20°\text{C}$.

The efficiency of steam generator $\left(\eta \right)$ is $0.8$.

The Unit cost of energy is $\left(U\right)$ is $\$1.05/\text{therm}$.

The velocity of air is $\left(V\right)$ is $10\text{km}/\text{hr}$.

Calculation:

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

  $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{75°\text{C}+5°\text{C}}{2}\\ =40°\text{C}\end{array}$

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

Obtain the following properties of air corresponding to the temperature of $40°\text{C}$ as follows:

$\begin{array}{c}k=0.02662\text{W}/\text{mK}\\ \mathrm{Pr}=0.7255\\ v=1.702\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\end{array}$

Calculate the Reynolds number $\left(\mathrm{Re}\right)$ using the relation.

  $\begin{array}{c}\mathrm{Re}=\frac{VD}{v}\\ =\frac{\left(10\text{km}/\text{hr}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{hr}}{3600\text{s}}\right)\left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{1.702\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =1.632\times {10}^4\end{array}$

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

    $\begin{array}{c}Nu=0.3+\frac{0.62{\mathrm{Re}}^{0.5}{\mathrm{Pr}}^{1/3}}{{\left[1+{\left(0.4/\mathrm{Pr}\right)}^{2/3}\right]}^{1/4}}{\left[1+{\left(\frac{\mathrm{Re}}{282000}\right)}^{5/8}\right]}^{4/5}\\ =0.3+\frac{0.62{\left(1.632\times {10}^4\right)}^{0.5}{\left(0.7255\right)}^{1/3}}{{\left[1+{\left(0.4/0.7255\right)}^{2/3}\right]}^{1/4}}{\left[1+{\left(\frac{1.632\times {10}^4}{282000}\right)}^{5/8}\right]}^{4/5}\\ =71.19\end{array}$

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

    $\begin{array}{c}h=\frac{k}{D}\left(Nu\right)\\ =\frac{0.02662\text{W}/\text{mK}}{\left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\left(71.19\right)\\ =18.95\text{W}/{\text{m}}^2\text{K}\end{array}$

Calculate the heat loss by convection $\left(q_c\right)$ using the relation.

    $\begin{array}{c}{q}_c=hA\left(T_s-T_{\infty }\right)\\ =h\pi DL\left(T_s-T_{\infty }\right)\\ =18.95\text{W}/{\text{m}}^2\text{K}\times \pi \left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(12\text{m}\right)\left(\left(75°\text{C}+273\right)\text{K}-\left(5°\text{C}+273\right)\text{K}\right)\\ \text{=5001}\text{W}\end{array}$

Calculate the heat loss by radiation $\left(q_r\right)$ using the relation.

    $\begin{array}{c}{q}_c=\epsilon A\sigma \left(T_s^4-T_a^4\right)\\ =\epsilon \pi DL\sigma \left(T_s^4-T_a^4\right)\\ =\left(0.8\right)\pi \left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(12\text{m}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}\right)\left(\begin{array}{l}{\left(75°\text{C}+273\right)}^4{\text{K}}^4-\\ {\left(0°\text{C}+273\right)}^4{\text{K}}^4\end{array}\right)\\ =\text{1558}\text{W}\end{array}$

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

    $\begin{array}{c}Q=q_c+q_r\\ =5001\text{W}+1558\text{W}\\ \text{=6559}\text{W}\end{array}$

Calculate the heat loss by radiation $\left(q_r\right)$ for $-20°\text{C}$ using the relation.

    $\begin{array}{c}{q}_c=\epsilon A\sigma \left(T_s^4-T_{a1}^4\right)\\ =\epsilon \pi DL\sigma \left(T_s^4-T_{a1}^4\right)\\ =\left[\begin{array}{l}\left(0.8\right)\pi \left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(12\text{m}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}\right)\times \\ \left({\left(75°\text{C}+273\right)}^4{\text{K}}^4-{\left(-20°\text{C}+273\right)}^4{\text{K}}^4\right)\end{array}\right]\\ =\text{1807}\text{W}\end{array}$

Calculate the new total heat loss $\left(Q_1\right)$ using the relation.

    $\begin{array}{c}{Q}_1=q_c+q_r\\ =5001\text{W}+1807\text{W}\\ \text{=6807}\text{W}\end{array}$

Calculate %Change in total heat loss $\left(C\right)$ using the relation.

    $\begin{array}{c}C=\frac{Q_1-Q}{Q}\times 100\\ =\frac{6808\text{W}-6559\text{W}}{6559\text{W}}\times 100\\ =\frac{249\text{W}}{6559\text{W}}\times 100\\ =3.80\%\end{array}$

Calculate the heat loss by radiation $\left(q_r\right)$ for $25°\text{C}$ using the relation.

    $\begin{array}{c}{q}_c=\epsilon A\sigma \left(T_s^4-T_{a1}^4\right)\\ =\epsilon \pi DL\sigma \left(T_s^4-T_{s1}^4\right)\\ =\left(0.8\right)\pi \left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(12\text{m}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}\right)\left(\begin{array}{l}{\left(75°\text{C}+273\right)}^4{\text{K}}^4-\\ {\left(25°\text{C}+273\right)}^4{\text{K}}^{\text{4}}\end{array}\right)\\ \text{=1159}\text{W}\end{array}$

Calculate the new total heat loss $\left(Q_1\right)$ using the relation.

    $\begin{array}{c}{Q}_1=q_c+q_r\\ =5001\text{W}+1159\text{W}\\ \text{=6160}\text{W}\end{array}$

Calculate %Change in total heat loss $\left(C\right)$ using the relation.

    $\begin{array}{c}C=\frac{Q-Q_1}{Q}\times 100\\ =\frac{6559\text{W}-6160\text{W}}{6559}\times 100\\ =\frac{399\text{W}}{6559\text{W}}\times 100\\ =6.08\%\end{array}$

Thus, the change in the values obtained in heat transfer is $3.80\%$ for $\left(T_{a1}\right)=-20°\text{C}$ and $6.08\%$ for $\left(T_{s1}\right)=25°\text{C}$ respectively.