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

Gwendolyn Willett

Gwendolyn Willett

Answered question

2021-12-20

A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k=160WmK). The fin diameter is D=4 mm, and the fin is exposed to convective conditions characterized by h=220Wm2K. It is reported that the fin efficiency is ηf=0.65. Determine the fin length L and the fin effectiveness εf Account for tip convection.

Answer & Explanation

Deufemiak7

Deufemiak7

Beginner2021-12-21Added 34 answers

Step 1
Given: D=4mm, k=160WmK, h=220Wm2K, ηf=0.65
Calculate the parameter m as follows:
m=hPkA
=h×πDk×πD24
=4hkD
=4×220160×0.004
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:
ηf=tanh(mLC)mLC
0.65=tanh(37.08LC)37.08LC
Let 37.08LC=x
0.65=tanh(x)x
0.65=1xexexxex+ex
0.65x=e2x1e2x+1
e2x(0.65x1)+0.65x+1=0
By solving the equation, we get
x=1.34164
37.08LC=1.34164
LC=0.03618m
LC=36.18mm
Hence, the length of the fin is 36.18mm
Step 3
Calculate the fin effectiveness
εf=qfhAθb
=Mtanh(mLC)hAθb
=hPktanh(mLC}}{hA}
=2hDktanh(mLC)
=2220×4×103160tanh(37.08×36.18×103)
SlabydouluS62

SlabydouluS62

Skilled2021-12-22Added 52 answers

Step 1
Given that
D=4mm
K=160WmK
h=h=220Wm2K
ηf=0.65
We know that
m=hPKA
For circular fin
m=4hKD
m=4×220160×0.004
m=37.08
ηf=tanhmLmL
0.65=tanh37.08L37.08L
By solving above equation we get
L=36.18mm
The effectiveness for circular fin given as
ε=2tanhmLhDK
ε=2tanh(37.08×0.03618)220×0.004160
ε=23.52
Don Sumner

Don Sumner

Skilled2021-12-27Added 184 answers

Step 1
As per the question:
Thermal conductivity, k=160WmK
Diameter of the fin, D=4mm=4×103m
Coefficient of convective heat transfer, h=220Wm2K
Efficiency of the fin, ηf=0.65
Now,
To calculate the length if the fin:
1) ηf=tanhmLCmLC
where
LC=L+D4=(L+1)mm
Also,
m=PhACk
where
P=Perimeter of the fin=πD
AC=Convection Area
AC=πD24
Now,
m=πDhπD24k
m=4hDk=4×2204×103×160=37.08m3
Now, using eqn (1)
0.65=tanh37.08LC37.08LC
tanh(37.08LC)=24.10LC
Now, by trial-error:
LC=0.0362m=36.2mm
Also, we know that:
LC=L+1
36.2=L+1
L=35.2mm
Now, the fin effectiveness is given by:
εf=PkhAC
εf=πD×160220×πD24tanhmLC
εf4×160220×4×103tanh(37.08×0.0362)=23.5

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