# 6 Minute Thermal & Fluids ( Problem 25)



## annie (Sep 1, 2007)

I am not able to get the fin efficiency as the book. I used the formula rather than the graph and am getting 0.96. I used the formula:

Fin Efficiency = tan h (m(L+t/2))/ m(L+t/2)

The book gets 0.85 and no graph provided.

Did anyone get this problem ? If so help is appreciated.


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## Art (Sep 1, 2007)

try calculating the BTU for each..

fluid side 500 x GPM x Delta T

air side 1.085 x SCFM x Delta T

take the ratio IN/OUT, should be the overall transfer efficiency


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## BORICUAZO (Sep 3, 2007)

annie said:


> I am not able to get the fin efficiency as the book. I used the formula rather than the graph and am getting 0.96. I used the formula:
> Fin Efficiency = tan h (m(L+t/2))/ m(L+t/2)
> 
> The book gets 0.85 and no graph provided.
> ...


Well, here is my way to solve this:

1. Remember to change Lcorr and t from inches to ft.

2. Lcorr = 1.3125 in. = .019375 ft.

3. r = .5 in /2 = .25 in. = .0208 ft.

4. t = .125 in. = .01042 ft.

5. Look for the efficiency in figure 34.5 in the MERM:

6. Lcorr * (h/kt)^1/2 = 0.357 (x coordinate) {h = 3, k = 27}

7. (Lcorr+r)/r = 6.25 (y coordinate)

8. This give me an efficiency of aprox. 0.8


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## annie (Sep 4, 2007)

Well. Thanks. What I did not understand was how you could look for the y coordinate (6.25) on the graph. The values given were for only 2,3,4.

Let me know.

Annie.



IndependencePR said:


> Well, here is my way to solve this:
> 1. Remember to change Lcorr and t from inches to ft.
> 
> 2. Lcorr = 1.3125 in. = .019375 ft.
> ...


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## BORICUAZO (Sep 4, 2007)

annie said:


> Well. Thanks. What I did not understand was how you could look for the y coordinate (6.25) on the graph. The values given were for only 2,3,4.Let me know.
> 
> Annie.


Oh! Sorry. I have to make one correction:

7. (Lcorr+r)/r = 6.25 , corresponds to the line in the graph that we should locate below the "4" line. Lines for values of 2, 3 and 4 are those actually represented in figure 34.5. We should draw a line for the new 6.25 value of (Lcorr+r)/r.

The efficiency is located in the 'y' coordinate.


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## annie (Sep 5, 2007)

Independence,

Thanks.This now makes sense that we have to draw an imaginary line to get the efficiency. Are you getting the same answer with the fin efficiency equation as I am getting 0.96? Sorry I am asking the same question again. I just want to get this concept clear.

Annie



IndependencePR said:


> Oh! Sorry. I have to make one correction:
> 7. (Lcorr+r)/r = 6.25 , corresponds to the line in the graph that we should locate below the "4" line. Lines for values of 2, 3 and 4 are those actually represented in figure 34.5. We should draw a line for the new 6.25 value of (Lcorr+r)/r.
> 
> The efficiency is located in the 'y' coordinate.


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## BORICUAZO (Sep 5, 2007)

annie said:


> Independence,
> Thanks.This now makes sense that we have to draw an imaginary line to get the efficiency. Are you getting the same answer with the fin efficiency equation as I am getting 0.96? Sorry I am asking the same question again. I just want to get this concept clear.
> 
> Annie


My values of (Lcorr+r)/r = 6.25 (graph line) and Lcorr * (h/kt)^1/2 = 0.357 (x coordinate) give me an efficiency of .8 in the y coordinate. The final result for Q = 1,782 Btu/hr. Your value for an efficiency of .96 give me a Q = 2,138 Btu/hr, wich is not the correct answer.

I will select HVAC afternoon, as I did for the last April exam (failed). I think that this type of problem is only for the Thermal and Fluids afternoon session; but, you never know.

Carlos


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