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Anonymous Poster

Heat rejection calculation of oil cooled generator

09/16/2008 5:51 PM

I need to work out the heat rejection (in BTU/min or Kw), from an oil cooled aircraft electric generator. If we assume that all of the heat produced by the generator is transfered into the oil then we have these variables at one specific load value:

Oil in temp 128degC.

Oil out temp 151degC.

Oil flow 5.25 imp gals/min.

The oil has a specific gravity of 0.827 and specific heat capacity of 0.521.

Can anybody help to point me in the right direction to work up a formulae. Thanks

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Anonymous Poster
#1

Re: Heat rejection calculation of oil cooled generator

09/16/2008 11:06 PM

The equation is:

Q = m·c·ΔT

Where:

Q = rate of energy transfer (Btu/min)

m = rate of fluid transfer (lbs/min)

c = specific heat capacity of fluid (Btu/lb·°F)

ΔT = difference in temperatures (°F)

So in your case:

m = 5.25 imp gals / min = density of water X volume of water X specific gravity X flowrate = 62.02 [lbs/ft3 @100°F] x 5.25 [imp gals / min] X 0.1605437 [ft3 / imp gal] X 0.827 = 43.23 lbs/min

c = 0.521 Btu/lb·°F

ΔT = 303.8 - 262.4 = 41.4 °F

For a grand total of ...................

Q = 43.23 X 0.521 X 41.4 = 932.5 Btu/min

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Anonymous Poster
#2
In reply to #1

Re: Heat rejection calculation of oil cooled generator

09/16/2008 11:17 PM

Hey ........ wait a minute ................ did I just answer a homework question?

If so you got lucky this time. No more homework answers on this site.

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Anonymous Poster
#3

Re: Heat rejection calculation of oil cooled generator

09/17/2008 6:03 PM

Thanks for the help. No, not a homework question, just something I'm interested in.

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