Conversion problem 1

[wchowe]

Hi all,

I have been trying to use a heat exchanger formula and have been unable to get it to work out in both imperial and metric numbers. This leads me to believe I have something wrong. I am trying to solve for flow, using Load, density, Specific Heat,Delta T. When I convert each units and run the equation again in english units I don't get the same answer for flow. I get 1.5 lpm in metric, in english it should come out to 4 gpm,instead I get 2.6 gpm. The coolant is water @20C. Appreciate any help.

Thanks
Bill

 

[wchowe]

Sorry. This should be the correct link to the file example.

Thanks
Bill

 

[IRstuff]

So why are your temperatures so radically different? 4°C = 7.2°F

TTFN
faq731-376

 

[wchowe]

You have to add 32 to the 7.2 to get F! 4C=39.2F

C*9/5 +32

Thanks
Bill

 

[Latexman]

And the flow rate conversion is off. 1.5 Lpm = 0.396 U.S. gpm.

Good luck,
Latexman

 

[Latexman]

It's not temperature! It's temperature difference. IR is right.

Good luck,
Latexman

 

[IRstuff]

DELTA T, not absolute T

You yourself stated that the temperature of the coolant is 20°C, which means that the 4°C is the temperature DIFFERENCE. Temperature difference in Farenheit:

T1*9/5+32 - T2*9/5+32 = (T1-T2)*9/5, where T1-T2=4°C

TTFN
faq731-376

 

[wchowe]

You're right! I was looking at it wrong. 24-20=4C 72.2-68=7.2F. Perfect. But the equation still isn't coming out with 0.396 GPM? Now I am getting 1.42 GPM. I know 1.5lpm=0.396 gpm its not solving. I have to admit, I don't work in this area often.

I am using Q*7.481/(T*60*Cp*p)
T=Delta T
Cp=Specific Heat
p=Density
Q=Load
cu ft of water = 7.481 lbs

Thanks
Bill

 

[IRstuff]

Your spreadsheet say 62.321 lb/ft^3, which is the correct conversion from 998.29kg/m^3

TTFN
faq731-376

 

[wchowe]

I think the conversions are correct, for Q, Cp, p, and Delta T now. I cannot get the same answer in metric and english for flow? 1.5 l/m should come out to 0.396 gpm? Flow is the problem........

Thanks
Bill

 

[imcjoek]

cu ft of water = 7.481 lbs

FWIW, 1ft^3 of water is 7.48 GALLONS. Not Lb(m?).
And 1ft^3 is 62.4 lbm

 

[wchowe]

Yup, agreed. I just wrote too fast, made an error. Density I have right on spread sheet at 62.22 lbs/cu-ft

Thanks
Bill

 

[imcjoek]

Your error is in the metric bit. I get 5.39 L/min and/or 1.42 gal/min.

I can scan my sheet if needed.

 

[wchowe]

I know the flow is not the same, thats the problem I am not resolving. I started with
Q=1500w
Delta T=4 C
Cp=4.182 J/gC
p=998.29 kg/m3
When I use the above in the formula I get 1.5 l/m

The problem occurs when I convert.
If I take the equivalent english units for the above
Q=5118 btu/hr
Delta T=7.2 F
Cp=0.99885 btu/lb F
p=62.321 lb/ft3
insert the new values into the formula, I should come out with 0.396. The problem is I get 1.42 gpm which of course is not equivalent to 1.5 l/m?

I was asking for help, because its not obvious to me where the error is? Using the same formula with either units should give equivalent answers? Correct?
The only thing that I didn't look at was Specific gravity. I believe its not involved becasue waters Sg =1.



Thanks
Bill

 

[imcjoek]

What I'm saying, is that:
Q=1500W
Delta T=4C (or 4K)
Cp=4.182J/gK
P=998.3kg/m^3

Nets you 5.39 L/min.

 

[wchowe]

I am getting 1.5 L/min?

I am using 1500*1000/4*4.182*998.3*60

What am I missing?

Thanks
Bill

 

[IRstuff]

Your last conversion is incorrect. It's supposed to be a multiplication by 60. The natural SI unit for J is W/s. Then you left off the conversion for L to m^3 or kg to gm, i.e., the 1000 should be 1000/1000=1.

TTFN
faq731-376

 

[IRstuff]

I recommend that you get a calculation program that understands units, like Mathcad or SMath. I spent more time trying to figure what you did wrong by hand than it took to type in all the calculations in the Mathcad, which did the conversions seamlessly, and did not require recalculating to get the English conversions.

Oh, see, I got the units wrong, W = J/s, because I did it by hand.

TTFN
faq731-376

 

[imcjoek]

See attached.

I recommend you do unit conversions such that you can visually cancel units, or use a program similar to IRstuff's suggestion.

 

[wchowe]

Thanks for the help. I should have gone back to canceling units, that is basic engineering. Appreciate the help. Have a good day.

Thanks
Bill