# A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy PS

A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy $\left(k=160\frac{W}{m}\cdot K\right)$. The fin diameter is $D=4$ mm, and the fin is exposed to convective conditions characterized by $h=220\frac{W}{{m}^{2}}\cdot K$. It is reported that the fin efficiency is ${\eta }_{f}=0.65$. Determine the fin length L and the fin effectiveness ${\epsilon }_{f}$ Account for tip convection.

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Deufemiak7
Step 1
Given:
Calculate the parameter m as follows:
$m=\sqrt{\frac{hP}{kA}}$
$=\sqrt{\frac{h×\pi D}{k×\pi \frac{{D}^{2}}{4}}}$
$=\sqrt{\frac{4h}{kD}}$
$=\sqrt{\frac{4×220}{160×0.004}}$
$m=37.08$
Step 2
The expression for the efficiency of the after taking account for convection transfer from the tip of the fin as follows:
${\eta }_{f\in }=\frac{\text{tanh}\left(m{L}_{C}\right)}{m{L}_{C}}$
$0.65=\frac{\mathrm{tan}h\left(37.08{L}_{C}\right)}{37.08{L}_{C}}$
Let $37.08{L}_{C}=x$
$0.65=\frac{\text{tanh}\left(x\right)}{x}$
$0.65=\frac{1}{x}\frac{{e}^{x}-{e}^{x}}{x{e}^{x}+{e}^{-x}}$
$0.65x=\frac{{e}^{2x}-1}{{e}^{2x}+1}$
${e}^{2x}\left(0.65x-1\right)+0.65x+1=0$
By solving the equation, we get
$x=1.34164$
$37.08{L}_{C}=1.34164$
${L}_{C}=0.03618m$
${L}_{C}=36.18mm$
Hence, the length of the fin is 36.18mm
Step 3
Calculate the fin effectiveness
${\epsilon }_{f}=\frac{{q}_{f}}{hA{\theta }_{b}}$
$=\frac{M\text{tanh}\left(m{L}_{C}\right)}{hA{\theta }_{b}}$
$=\frac{\sqrt{hPk}\text{tanh}\left(m{L}_{C}\right\}}{\right\}}\left\{hA\right\}$
$=\frac{2}{\sqrt{\frac{hD}{k}}\text{tanh}\left(m{L}_{C}\right)}$
$=\frac{2}{\sqrt{\frac{220×4×{10}^{-3}}{160}}}\text{tanh}\left(37.08×36.18×{10}^{-3}\right)$

SlabydouluS62
Step 1
Given that
$D=4mm$
$K=160\frac{W}{m-K}$
$h=h=220\frac{W}{{m}^{2}-K}$
${\eta }_{f}=0.65$
We know that
$m=\sqrt{\frac{hP}{KA}}$
For circular fin
$m=\sqrt{\frac{4h}{KD}}$
$m=\sqrt{\frac{4×220}{160×0.004}}$
$m=37.08$
${\eta }_{f}=\frac{\text{tanh}mL}{mL}$
$0.65=\frac{\text{tanh}37.08L}{37.08L}$
By solving above equation we get
$L=36.18mm$
The effectiveness for circular fin given as
$\epsilon =\frac{2\text{tanh}mL}{\sqrt{\frac{hD}{K}}}$
$\epsilon =\frac{2\text{tanh}\left(37.08×0.03618\right)}{\sqrt{\frac{220×0.004}{160}}}$
$\epsilon =23.52$

Don Sumner

Step 1
As per the question:
Thermal conductivity, $k=160\frac{W}{m-K}$
Diameter of the fin, $D=4mm=4×{10}^{-3}m$
Coefficient of convective heat transfer, $h=220\frac{W}{{m}^{2}-K}$
Efficiency of the fin, ${\eta }_{f}=0.65$
Now,
To calculate the length if the fin:
1) ${\eta }_{f}=\frac{\mathrm{tanh}m{L}_{C}}{m{L}_{C}}$
where
${L}_{C}=L+\frac{D}{4}=\left(L+1\right)mm$
Also,
$m=\sqrt{\frac{Ph}{{A}_{C}k}}$
where
$P=\text{Perimeter of the fin}=\pi D$
${A}_{C}=\text{Convection Area}$
${A}_{C}=\frac{\pi {D}^{2}}{4}$
Now,
$m=\sqrt{\frac{\pi Dh}{\frac{\pi {D}^{2}}{4}k}}$
$m=\sqrt{\frac{4h}{Dk}}=\sqrt{\frac{4×220}{4×{10}^{-3}×160}}=37.08{m}^{-3}$
Now, using eqn (1)
$0.65=\frac{\mathrm{tanh}37.08{L}_{C}}{37.08{L}_{C}}$
$\mathrm{tanh}\left(37.08{L}_{C}\right)=24.10{L}_{C}$
Now, by trial-error:
${L}_{C}=0.0362m=36.2mm$
Also, we know that:
${L}_{C}=L+1$
$36.2=L+1$
$L=35.2mm$
Now, the fin effectiveness is given by:
${\epsilon }_{f}=\sqrt{\frac{Pk}{h{A}_{C}}}$
${\epsilon }_{f}=\sqrt{\frac{\pi D×160}{220×\frac{\pi {D}^{2}}{4}}}\mathrm{tanh}m{L}_{C}$
${\epsilon }_{f}\sqrt{\frac{4×160}{220×4×{10}^{-3}}}\mathrm{tanh}\left(37.08×0.0362\right)=23.5$