Chapter 17, Problem 130RQ

To determine

The increase in heat transfer from the tube per meter of its length.

Explanation of Solution

Given:

Thermal conductivity of the fins $\left(k\right)$ is $180\text{W}/\text{m}\cdot \text{K}$.

Heat transfer coefficient $\left(h\right)$ is $\text{60 }\text{W}/{\text{m}}^2\cdot \text{K}$.

Space between the fins $\left(s\right)$ is 3 mm.

Outer diameter of the tube $\left(D_1\right)$ is 3 cm.

Outer diameter of the fin $\left(D_2\right)$ is 6 cm.

Number of fins $\left(n\right)$ is 200.

Calculation:

Determine the surface area in case of no fins.

  $\begin{array}{c}{A}_{\text{no fin }}=\pi D_{1}L\\ =\pi \left(0.03\text{ m}\right)\left(1\text{ m}\right)\\ =0.0942{\text{ m}}^2\end{array}$

Determine the heat transfer in case of no fins.

  $\begin{array}{c}{\dot{Q}}_{\text{no fin }}=h{A}_{\text{no fin }}\left(T_b-T_{\infty }\right)\\ =\left(60\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.0942{\text{ m}}^2\right)\left(120°\text{C}-25°\text{C}\right)\\ =537\text{ W}\end{array}$

Determine the length.

  $\begin{array}{c}L=\frac{\left(D_2-D_1\right)}{2}\\ =\frac{\left(0.06\text{ m}-0.03\text{ m}\right)}{2}\\ =0.015\text{ m}\end{array}$

Determine the following factor.

  $\begin{array}{c}\frac{r_2+\left(\frac{t}{2}\right)}{r_1}=\frac{0.03\text{ m}+\left(\frac{0.002\text{ m}}{2}\right)}{0.015\text{ m}}\\ =2.07\end{array}$

Determine the following factor.

  $\begin{array}{c}{L}_c^{3/2}{\left(\frac{h}{k{A}_p}\right)}^{1/2}=\left(L+\frac{t}{2}\right)\sqrt{\frac{h}{kt}}\\ =\left(0.015\text{ m}+\frac{0.002\text{ m}}{2}\right)\sqrt{\frac{60\text{W}/{\text{m}}^2\cdot °\text{C}}{\left(180\text{W}/\text{m}\cdot °\text{C}\right)\left(0.002\text{ m}\right)}}\\ =0.207\text{ m}\end{array}$

Obtain the efficiency of these circular fins from the efficiency curve of Figure 17-44 as 0.96.

Determine the area of fin.

  $\begin{array}{c}{A}_{\text{fin}}=2\pi \left(r_2^2-r_1^2\right)+2\pi r_{2}t\\ =2\pi \left({\left(0.03\text{ m}\right)}^2-{\left(0.015\text{ m}\right)}^2\right)+2\pi \left(0.03\text{ m}\right)\left(0.002\text{ m}\right)\\ =0.004624{\text{ m}}^2\end{array}$

Determine the heat transfer from a single fin.

  $\begin{array}{c}{\dot{Q}}_{\text{fin }}={\eta }_{\text{fin }}{\dot{Q}}_{\text{fin, max }}\\ ={\eta }_{\text{fin }}h{A}_{\text{fin }}\left(T_b-T_{\infty }\right)\\ =0.96\left(60\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.004624{\text{ m}}^2\right)\left(120°\text{C}-25°\text{C}\right)\\ =25.3\text{ W}\end{array}$

Determine the surface area of a single unfinned portion of the tube.

  $\begin{array}{c}{A}_{\text{unfin }}=\pi D_{1}s\\ =\pi \left(0.03\text{ m}\right)\left(0.003\text{ m}\right)\\ =0.000283{\text{ m}}^2\end{array}$

Determine the heat transfer from a single unfinned portion of the tube.

  $\begin{array}{c}{\dot{Q}}_{\text{unfin }}=h{A}_{\text{unfin }}\left(T_b-T_{\infty }\right)\\ =\left(60\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.000283{\text{ m}}^2\right)\left(120°\text{C}-25°\text{C}\right)\\ =1.6\text{ W}\end{array}$

Determine the total heat transfer from the finned tube.

  $\begin{array}{c}{\dot{Q}}_{\text{total fin }}=n\left({\dot{Q}}_{\text{fin }}+{\dot{Q}}_{\text{unfin }}\right)\\ =200\left(25.3\text{ W}+1.6\text{ W}\right)\\ =5380\text{ W}\end{array}$

Determine the increase in heat transfer from the tube.

  $\begin{array}{c}{\dot{Q}}_{\text{increase }}={\dot{Q}}_{\text{total fin }}-{\dot{Q}}_{\text{no fin }}\\ =5380\text{ W}-537\text{ W}\\ =4843\text{ W}\end{array}$

Thus, the increase in heat transfer from the tube is $\underset{\_}{4843\text{ W}}$.