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Gallons Per Minute based on Engine Displacement

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ServoWizard

Electrical
May 21, 2005
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Hello,

I'm trying to calculate the GPH of fuel required to maintain a 6.5:1 Air to Fuel ratio. My engine displaces 400 cubic inches, but being a 4 cycle it actually pumps 200 cubic inches per revolution. The engine will operate at a maximum RPM of 8000.

200 * 8000 = 1,600,000 cubic inches pumped per minute.

1 Gallon = 231 cubic inches.

So here I multiply the single unit by the AFR

231 * 6.5 = 1501.5 (6.5 Units)

1501.5 + 231 = 1732.5 (Total Mixure or 6.5 parts Air + 1 part Fuel)

231 / 1732.5 = .1333 (231 represents 13.3% of Total Mixure)

Now using the same logic with the exception that I reduce the total cubic inches down to 1 unit.

1,600,000 / 6.5 = 246153.8462 (1 Unit = 246153.8462)

6.5 Units + 1 Unit = 1846153.8462 (Total Mixure or 6.5 parts Air + 1 part Fuel)

246153.8462 / 1846153.8462 = .1333 (246153.8462 represents 13.3% of Total Mixure))

246153.8462 / 231 = 1065.601 Gallons Per Minute = WRONG!

I'm obviously in over my head. I would appreciate some help.

Thanks,
Ron
 
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Regards
Pat
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patprimmer,
pontiacjack,

I found an interesting web sight that provides a tune up calculator that will give you a complete baseline fuel system set up. Fuel Injection Enterprises is the name of the company.
Tells you everything you need to know and eliminates the need to become a meteorologist. I have moved on.

Ron
 
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