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Fan Horsepower equation derivationHelpful Member!(3) 

wilg (Mechanical) (OP)
23 Dec 03 13:28
  The equation for Fan Horsepower is
BHP = (cfm x static friction x specific gravity)/(6356 x motor efficiency)....How is the 6356 derived ?
KHilgefort (Mechanical)
24 Dec 03 9:33
I usually see this figure in terms of eff = (volume flowrate in cfm) X (pressure in inches of water) / 6356 / (brake horsepower).

R.A. Wallis in "Axial Flow Fans" similarly puts the equation as Shaft h.p. = (5.2) X (in. water) X (cu. ft/min) / (33,000 X efficiency).

If you convert the horsepower units to ft-lb/min (33,013 per hp) and convert the pressure to lb/ft^2 (5.202 per inch H2O), you get 33013/5.202 = 6346.  This at least gets you to a dimensionless value for the efficiency.
lilliput1 (Mechanical)
24 Dec 03 13:35
The efficiency to get bhp is fan efficiency, not motor efficiency. A good apporoximation for fan efficiency is 0.65

For pumps
bhp = (gpm x ft wg TDH)/(3960 x eff)

Here alse 0.65 is a good approximation for pump efficiency.

On projects we HVAC engineers have to give electrical loads to the Elect Engineer on the project. We use the above formula + guick estimate from experience of pressure drops to come up with motor hp. We want to be safe at the initial stage so we always pick mhp conservatively. Later on as plans are developed, actual pressure drops are calculated & fans/pumps are selected on actual fan/pump curves.
Helpful Member!  quark (Mechanical)
24 Dec 03 23:56
There are two mistakes in your equation. First one, as suggested by lilliput, it is fan efficiency you have to use to get bhp. Secondly you can omit specific gravity as this is inbuilt in the equation.

1HP = 33000 ft-lbf/min = 33000 (cuft/min)x(lbf/sq.ft)

and 1 inch. wc = 5.192 lbf/sq.ft (this is where sg is included)

therefore, 1 HP = 33000 (cu.ft/min)x inches wc/5.192

which gives, 1HP = 6356 cfm x inches wc

So fan BHP = cfmx dp across fan/(6356xeff. blower)

PS: I HP is the amount of power required to lift a weight of 76 kgs to a height of 1 meter in 1 second. I don't know exact definition in IP units but you can convert to get the above said value.


Helpful Member!(2)  fredt (Mechanical)
25 Dec 03 6:39
You guys should go metric then you get rid of all these odd constants.  Fan power in kW is simply:

Flow cu.m/s X pressure kPa (kilopascals) / Fan efficiency.

It should be noted that this ignores compressibility effects as does the English unit equations that have been referenced.

KHilgefort:  I have been looking for a copy of R.A.Wallis "Axial Flow Fans" which is now out of print. Let me know if you are prepared to sell yours.
IRstuff (Aerospace)
25 Dec 03 12:32
Have you tried to do a used book search?


quark (Mechanical)
27 Dec 03 1:21
FredT's comment is not totally correct. We do require some conversion, but a simple one.

We generally speak of Pascals when we deal with fans and we do require a conversion factor to convert Pa to kPa or from Watts to kW. Still, if one insists about measuring fan pressure in kPa we will end up with gauges as big as the fans themselves .

Question: Is Tower of Babel a possibility?


fredt (Mechanical)
27 Dec 03 3:55

Explain to me how the units effect gauge size.
quark (Mechanical)
28 Dec 03 1:23
I am sorry for the false statement.


wilg (Mechanical) (OP)
29 Dec 03 12:50
Excellent answers to a difficult question, many
engineers know the constants but can't derive them.
I'm surprised at the level of intelligence that
corresponds on this engineering chat line.
kmpillai (Mechanical)
30 Dec 03 4:44
Dear fredt,
Please see the following link, to get the R.A.Wallis book


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