On 1/14/2024 1:56 PM, Adam Deem via KRnet wrote:
I don't know about a fuel flow chart, but the math to calculate it is
pretty simple. The 2180 is 75 hp takeoff power at 3600RPM.
2180cc=.077cu ft. 3600RPM=1800 intakes/min for 138.6 cu ft/min or
8316 cu ft/hr. 8316x.08 lbs/cu ft air=665.28 lbs air/hr. Correction
for normally aspirated volumetric efficiency 665.28x.95=632 lb/hr.
632lbs air+60 lbs gasoline/hr = 100% power at 10.5:1 F/A ratio.
60lbs/6 lbs/gal = 10.0gph. Best power ratio would be 12:1. You want
your fuel system to deliver enough fuel in any scenario on takeoff
(SL@10.5:1 standard day) with 150% safety margin to avoid starving the
engine under worst case scenario (winter day with very low DA)
For that motor you'd want to run a fuel flow system test and verify it
can deliver at least 15gph. For comparison, my electric pump fed
system (Facet 3psi cube pumps) delivers a verified 19gph through the
system. I hope this is helpful.
Adam Deem
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A bit less cumbersome math / rule of thumb that is quite accurate for
the type of engines we use is 1/2 pound of fuel per horsepower per
hour. The 0-200 at 100 hp / 2 = 50 pounds of fuel / 6 ppg = 8.33 gph at
full throttle at takeoff if its actually producing 100 hp. A VW
producing 75 hp on takeoff would be 75/2=37.5 / 6ppg = 6.25 gph.
Multiply that number by 1.5 for minimum acceptable fuel flow. This
holds true for any % of power hp fuel flow.
The 0-200 for example at 65% power at cruise is 65hp / 2 =32.5 pounds of
fuel / 6 =5.42 gph at an acceptable fuel / air mixture (moderate lean
mixture of 14:1)
A 75 hp VW at 65% power cruise = 48.75 hp / 2 = 24.37 pounds of fuel / 6
= 4 gph fuel burn. I've never flown a VW engine but I'm guessing that
is pretty close.
Larry Flesner
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