Chapter 17, Problem 129RQ

To determine

The heat transfer rate from a single fin and the increase in the rate of heat transfer.

Explanation of Solution

Given:

Thermal conductivity of the fins $\left(k\right)$ is $230\text{W}/\text{m}\cdot \text{K}$.

Heat transfer coefficient $\left(h\right)$ is $\text{45 }\text{W}/{\text{m}}^2\cdot \text{K}$.

Length of the fin $\left(L\right)$ is 25 mm.

Diameter of the fin $\left(D\right)$ is 4 mm.

Calculation:

Determine the value of mL.

  $\begin{array}{c}mL=\sqrt{\frac{4h}{kD}}L\\ =\sqrt{\frac{4\left(45\text{W}/{\text{m}}^2\cdot °\text{C}\right)}{\left(230\text{W}/\text{m}\cdot °\text{C}\right)\left(0.004\text{ m}\right)}}\left(0.025\text{ m}\right)\\ =0.3497\end{array}$

Determine the area of fin.

  $\begin{array}{c}{A}_{\text{fin }}=\frac{\pi D^4}{96L^2}\left\{{\left[16{\left(\frac{L}{D}\right)}^2+1\right]}^{3/2}-1\right\}\\ =\frac{\pi {\left(0.004\text{ m}\right)}^4}{96{\left(0.025\text{ m}\right)}^2}\left\{{\left[16{\left(\frac{0.025\text{ m}}{0.004\text{ m}}\right)}^2+1\right]}^{3/2}-1\right\}\\ =2.099\times {10}^{-4}{\text{m}}^2\end{array}$

Determine the efficiency of fin.

  $\begin{array}{c}{\eta }_{\text{fin }}=\frac{3}{2mL}\frac{I_1\left(\frac{4}{3}mL\right)}{I_0\left(\frac{4}{3}mL\right)}\\ {\eta }_{\text{fin }}=\frac{3}{2\left(0.3497\right)}\frac{I_1\left(\frac{4}{3}\left(0.3497\right)\right)}{I_0\left(\frac{4}{3}\left(0.3497\right)\right)}\\ {\eta }_{\text{fin }}=0.9738\end{array}$

Determine the heat transfer rate from a single fin.

  $\begin{array}{c}{\dot{Q}}_{\text{fin}}={\eta }_{\text{fin}}h{A}_{\text{fin}}\left(T_b-T_{\infty }\right)\\ =\left(0.9738\right)\left(45\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(2.099\times {10}^{-4}{\text{m}}^2\right)\left(200°\text{C}-25°\text{C}\right)\\ =1.61\text{ W}\end{array}$

Thus, the heat transfer rate from a single fin is $\underset{\_}{1.61\text{ W}}$.

Determine the heat transfer rate for 100 fins.

  $\begin{array}{c}{\dot{Q}}_{\text{fin total }}=\left(100\right)\left(1.61\text{ W}\right)\\ =161\text{ W}\end{array}$

Determine the surface area of the unfinned portion.

  $\begin{array}{c}{A}_{\text{unfin }}=1{\text{ m}}^2-100\left(\frac{\pi }{4}{D}^2\right)\\ =1-100\left(\frac{\pi }{4}{\left(0.004\text{ m}\right)}^2\right)\\ =0.9987{\text{ m}}^2\end{array}$

Determine the heat transfer from the unfinned portion.

  $\begin{array}{c}{\dot{Q}}_{\text{unfin }}=h{A}_{\text{unfin }}\left(T_b-T_{\infty }\right)\\ =\left(0.9987{\text{ m}}^2\right)\left(45\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(200°\text{C}-25°\text{C}\right)\\ =7865\text{W}\end{array}$

Determine the total heat transfer from the surface.

  $\begin{array}{c}{\dot{Q}}_{\text{total }}={\dot{Q}}_{\text{fin total }}+{\dot{Q}}_{\text{unfin }}\\ =161\text{ W}+7865\text{ W}\\ =8026\text{ W}\end{array}$

Determine the heat transfer rate if there was no fin at the surface.

  $\begin{array}{c}{\dot{Q}}_{\text{nofin }}=h{A}_{\text{unfin }}\left(T_b-T_{\infty }\right)\\ =\left(1{\text{ m}}^2\right)\left(45\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(200°\text{C}-25°\text{C}\right)\\ =7875\text{ W}\end{array}$

Determine the increase in heat transfer rate.

  $\begin{array}{c}{\dot{Q}}_{\text{increase }}={\dot{Q}}_{\text{total }}-{\dot{Q}}_{\text{nofin }}\\ =8026\text{ W}-7875\text{ W}\\ =151\text{ W}\end{array}$

Thus, the increase in the rate of heat transfer is $\underset{\_}{151\text{ W}}$.