Efficiency of Triangular Fin for Heat Transfer

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In summary, the conversation discusses the comparison of fin heat rate, efficiency, and volume for a triangular fin made of 2024 AL with k=185w/mK, base thickness=3mm and length=15mm. The temperature of the fin is Tb=100C and the fluid temperature is Tinfinity=20C with h=50 w/m^2K. The equations used to solve for the efficiency are eta=1/mL*(I1*2*mL)/(Io*2*mL) and mL=(2h/kt)^(1/2)*L. The final answer for efficiency is supposed to be around .98 and the solution is found by solving for Io and I1 using algebra. The answers for Io
  • #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
 
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  • #2
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
Thank you very much, that helped a lot.
 

1. What is the purpose of a triangular fin?

A triangular fin is a type of hydrodynamic structure that is used to improve the performance of an object moving through a fluid, such as a boat or an aircraft. The main purpose of a triangular fin is to provide stability and control, as well as to reduce drag and increase maneuverability.

2. How does the shape of a triangular fin affect its performance?

The shape of a triangular fin plays a crucial role in its performance. The triangular shape allows for efficient flow of fluid around the fin, reducing drag and increasing lift. The angle of the triangular fin also determines the amount of lift and drag produced, which can be adjusted to optimize performance for different conditions.

3. What factors affect the performance of a triangular fin?

The performance of a triangular fin is affected by several factors, including its size, shape, angle, and placement on the object. The properties of the fluid, such as density and viscosity, can also impact the performance of the fin. Additionally, the speed and direction of the object's movement through the fluid can affect the performance of the fin.

4. How can the performance of a triangular fin be improved?

The performance of a triangular fin can be improved by making adjustments to its size, shape, and angle. By optimizing these parameters, the fin can produce more lift and reduce drag, resulting in improved stability and control. Additionally, the placement of the fin on the object can also be adjusted to improve its performance in specific conditions.

5. What are some real-world applications of triangular fin performance?

Triangular fin performance has many real-world applications, including in the fields of aerospace, marine, and automotive engineering. Triangular fins are commonly used on airplanes, boats, and cars to improve their stability, control, and maneuverability. They are also used in sports equipment, such as surfboards and sailboards, to enhance performance in the water.

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