A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k=160W/mK). The fin diameter is 4mm, and the fin is exposed to convective conditions characterized by h=220W/m^2K. It is reported that the fin efficiency is eta_f=0.65. Determine the fin length L and the fin effectiveness epsilon_f.

A Pin Fin Of Uniform Cross Sectional Area Is Fabricated Of An Aluminum Alloy

This question aims to find the length of the pin fin of a uniform fabricated aluminum alloy and its effectiveness in accounting for tip convection.

The question is based on the concepts of convection heat transfer. Convection heat transfer is the movement of heat from one medium to another due to fluid motion. We can calculate the heat transfer using the thermal conductivity of the metal, its efficiency, and heat transfer coefficient.

Expert Answer

The information is given in the problem to find the length $L$ of the fin; its effectiveness $\varepsilon_f$ is given as follows:

\[ \text{Thermal Conductivity, $k$}\ =\ 160\ W/mK \]

\[ \text{Diameter, $D$}\ =\ 4 mm \]

\[ \text{Fin Efficiency , $\eta_f$}\ =\ 0.65 \]

\[ \text{Heat Transfer Coefficient, $h$}\ =\ 220\ W/m^2K \]

a) To find the length $L$ of the fin, we will use the efficiency formula given as:

\[ \eta_f = \dfrac{ \tanh mL_c} {m L_c} \]

$m$ is the effective mass of the fin. We can find the value for $m$ by using this formula:

\[ m = \sqrt{ \dfrac{4 h} {D k}} \]

Substituting the values, we get:

\[ m = \sqrt{ \dfrac{4 \times 220} {4 \times 10^{-3} \times 160}} \]

By solving, we get:

\[ m = 37.08\ m^ {-3} \]

Putting this value of effective mass $m$ in the formula for efficiency, we get:

\[ 0.65 = \dfrac{ \tanh (37.08 \times L_c)} {37.08\ L_c} \]

Solving for $L_c$, we get:

\[ L_c = 36.2\ mm \]

$L_c$ is the convection length of the fin. To find the length $L$ of the fin, we can use the following formula:

\[ L = L_c\ -\ \dfrac {D} {4} \]

\[ L = 36.2\ -\ \dfrac {4} {4} \]

\[ L = 35.2\ mm \]

b) The formula gives the fin effectiveness $\varepsilon_f$:

\[ \varepsilon_f = \dfrac{ \tanh (m L_c)} {\sqrt {\dfrac {D h} {4 k}}} \]

Putting the value in the above equation, we get:

\[ \varepsilon_f = \dfrac {\tanh (37.08 \times 0.0362)}{\sqrt{ \dfrac{0.004 \times 220} {4 \times 160}}} \]

By solving this equation we get the value of effectiveness of the fin $\varepsilon_f$:

\[ \varepsilon_f = 23.52 \]

Numerical Result

The length $L$ of the fin is calculated to be:

\[ L = 35.2\ mm \]

The effectiveness of the fin $\varepsilon_f$ is calculated to be:

\[ \varepsilon_f = 23.52 \]

Example

The diameter of an aluminum alloy is $3mm$ and its convection length $L_c=25.6mm$. Find the length $L$.

\[ \text{Diameter, $D$}\ =\ 3\ mm \]

\[ \text{Convection Length, $L_c$}\ =\ 25.6\ mm \]

Using the formula for finding length $L$, we get:

\[ L\ =\ L_c\ -\ \dfrac {D} {4} \]

\[ L\ =\ 25.6\ -\ \dfrac {3} {4} \]

\[ L\ =\ 24.85\ mm \]

The length $L$ is calculated to be $24.85mm$.

Previous Question < > Next Question