Triangular fin performance

  • Thread starter Santorican
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  • #1
Santorican
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Homework Statement


A strait fin fabricated from 2024 AL with k=185w/mK base thickness=3mm, length=15mm. Tb=100C and fluid temp is Tinfinity=20C and h =50 w/m^2K. For the foregoing conditions and a fin of unit width, compare the fin heat rate, efficiency, and volume of a triangular fin


Homework Equations



eta=1/mL*(I1*2*mL)/(Io*2*mL)

mL=(2h/kt)^(1/2)*L



The Attempt at a Solution



For I1 and Io I am trying to find the bessel function solutions they give that e^-xI1 and e^-xIo are equal to something given an x.

I assumed that x=mL

When I calculate mL I get that mL=.201

In the back of the book they say that for x=.2 you get that e^-x*Io and e^-x*I1 are .8268 and .0823 respectively.

Now I have tried solving for Io and I1 using algebra but I cannot for the life of me get the right answers.

The answers to Io and I1 are .205 and 1.042 and the final answer for the efficiency is supposed to be around .98.

So solving for Io first I say that x=e^-x*Io now solving for Io yields Io=e^-x*x

Plugging everything in Io=(e^-.201)*.8268=1.01 which is not the right answer so I assume that I am doing something wrong.

Help?

Thank you
 

Answers and Replies

  • #2
sballard
1
0
Lol, Mech 346?

I googled the same thing and I just figured it out.
If you look at table 3.5 for the triangular fin: nf = I1(2mL)/(mL*I0(2mL))

mL=.201 and 2*mL=.402

I1(.402) roughly= .698/e^-.402=1.04
I0(.402) roughly= .137/e^-.402=.205
 
Last edited:
  • #3
Santorican
11
0
Thank you very much, that helped a lot.
 

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