Chapter 17, Problem 127RQ

(a)

To determine

The heat transfer rate for the infinitely long fin.

Explanation of Solution

Given:

Thermal conductivity of the fins $\left(k\right)$ is $240\text{W}/\text{m}\cdot \text{K}$.

Heat transfer coefficient $\left(h\right)$ is $\text{250 }\text{W}/{\text{m}}^2\cdot \text{K}$.

Length of the fin $\left(L\right)$ is 50 mm.

Diameter of the fin $\left(D\right)$ is 10 mm.

Calculation:

Determine the perimeter of the circular fin.

  $\begin{array}{c}p=\pi D\\ =\pi \left(0.01\text{ m}\right)\\ =0.03142\text{m}\end{array}$

Determine the cross section area of the circular fin.

  $\begin{array}{c}{A}_c=\frac{\pi D^2}{4}\\ =\frac{\pi {\left(0.01\text{ m}\right)}^2}{4}\\ =7.854\times {10}^{-5}{\text{m}}^2\end{array}$

Determine the value of m.

  $\begin{array}{c}m=\sqrt{\frac{hp}{k{A}_c}}\\ =\sqrt{\frac{\left(250\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.03142\text{ m}\right)}{\left(240\text{W}/\text{m}\cdot °\text{C}\right)\left(7.854\times {10}^{-5}{\text{m}}^2\right)}}\\ =20.41{\text{ m}}^{-1}\end{array}$

Determine the following factor.

  $\begin{array}{c}\sqrt{hpk{A}_c}=\sqrt{\left(250\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.03142\text{m}\right)\left(240\text{W}/\text{m}\cdot °\text{C}\right)\left(7.854\times {10}^{-5}{\text{m}}^2\right)}\\ =0.3848\text{W}/°\text{C}\end{array}$

Determine the heat transfer rate for an infinitely long fin.

  $\begin{array}{c}{\dot{Q}}_{\text{long fin }}=\sqrt{hpk{A}_c}\left(T_b-T_{\infty }\right)\\ =\left(0.3848\text{W}/°\text{C}\right)\left(350°\text{C}-25°\text{C}\right)\\ =125\text{ W}\end{array}$

Thus, the heat transfer rate for the infinitely long fin is $\underset{\_}{125\text{ W}}$.

Write the expression of the temperature variation along the fin.

  $\frac{T(x)-T_{\infty }}{T_b-T_{\infty }}=e^{-mx}$

(b)

To determine

The heat transfer rate for the adiabatic fin tip.

Explanation of Solution

Determine the heat transfer rate for an adiabatic fin tip.

  $\begin{array}{c}{\dot{Q}}_{\text{adiabatic tip}}=\sqrt{hpk{A}_c}\left(T_b-T_{\infty }\right)\tanh mL\\ =\left(0.3848\text{W}/°\text{C}\right)\left(350°\text{C}-25°\text{C}\right)\tanh \left[\left(20.41{\text{ m}}^{-1}\right)\left(0.05\text{ m}\right)\right]\\ =96.3\text{ W}\end{array}$

Thus, the heat transfer rate for the adiabatic fin tip is $\underset{\_}{96.3\text{ W}}$.

Write the expression of the temperature variation along the fin.

  $\frac{T(x)-T_{\infty }}{T_b-T_{\infty }}=\frac{\cosh m(L-x)}{\cosh mL}$

(c)

To determine

The heat transfer rate for the fin with tip temperature of $250°\text{C}$.

Explanation of Solution

Determine the heat transfer rate for fin with tip temperature of $250°\text{C}$.

  $\begin{array}{c}{\dot{Q}}_{\text{specificad temp }}=\sqrt{hpk{A}_c}\left(T_b-T_{\infty }\right)\frac{\cosh mL-\frac{\left(T_L-T_{\infty }\right)}{\left(T_b-T_{\infty }\right)}}{\sinh mL}\\ =\left(0.3848\text{W}/°\text{C}\right)\left(350°\text{C}-25°\text{C}\right)\left(0.725\right)\\ =90.7\text{ W}\end{array}$

Thus, the heat transfer rate for the fin with tip temperature of $250°\text{C}$ is $\underset{\_}{90.7\text{ W}}$.

Write the expression of the temperature variation along the fin.

  $\frac{T(x)-T_{\infty }}{T_b-T_{\infty }}=\frac{\frac{\left(T_L-T_{\infty }\right)}{\left(T_b-T_{\infty }\right)}\sinh mx+\sinh m\left(L-x\right)}{\sinh mL}$

(d)

To determine

The heat transfer rate for fin with convection from the fin tip.

Explanation of Solution

Determine the heat transfer rate for fin with convection from the tip.

  $\begin{array}{c}{\dot{Q}}_{\text{conv tip }}=\sqrt{hpk{A}_c}\left(T_b-T_{\infty }\right)\frac{\sinh mL+\left(\frac{h}{mk}\right)\cosh mL}{\cosh mL+\left(\frac{h}{mk}\right)\sinh mL}\\ =\left(0.3848\text{W}/°\text{C}\right)\left(350°\text{C}-25°\text{C}\right)\left(0.7901\right)\\ =98.8\text{ W}\end{array}$

Thus, the heat transfer rate for fin with convection from the fin tip is $\underset{\_}{98.8\text{ W}}$.

Write the expression of the temperature variation along the fin.

  $\frac{T(x)-T_{\infty }}{T_b-T_{\infty }}=\frac{\cosh m\left(L-x\right)+\left(\frac{h}{mk}\right)\sinh m\left(L-x\right)}{\cosh mL+\left(\frac{h}{mk}\right)\sinh mL}$

Tabulate the values of the temperature variation of the single fin.

$L,\text{ m}$	$T\left(x\right)\text{, }°\text{C}$
	Part (a)	Part (b)	Part (c)	Part (d)
0	350	350	350	350
0.005	318	326	328	325
0.01	290	305	308	304
0.015	264	288	292	285
0.02	241	272	279	270
0.025	220	260	268	256
0.03	201	250	259	246
0.035	184	242	253	237
0.04	169	237	250	231
0.045	155	233	249	227
0.05	142	232	250	224

Plot the temperature variation of the single fin as in Figure (1).