6 Minute Thermal & Fluids ( Problem 25)

[annie]

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.

 

[Art]

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

 

[BORICUAZO]

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.

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

 

[annie]

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.

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

 

[BORICUAZO]

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.

 

[annie]

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

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.

 

[BORICUAZO]

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