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I'm trying to calculate the fuel consumption for l550e engine. In its datasheet you can see this graph:

enter image description here

At 6500 RPM, engine power is around 48hp and the SFC is 345g/HPh. Ok, then the fuel consumption rate is:

fuel consumption rate: 345g/HPh * 48HP = 16560g/h

Am I right? We can assume the density of gasoline is 0.755g/ml based on here. Thus this engine at this RPM consumes 22Lit/h. right? Isn't it too much high?

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1 Answer 1

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Your math is right.

For comparison, the Rotax 582 which is also a two-stroke engine, outputs 48 kW (take-off performance) and uses 26.5 L/h. The stated specific fuel consumption is 425 g/kWh (317 g/HPh), which is close enough to the Limbach number.

Small two-stroke gasoline engines are generally less efficient than four-stroke ones.

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  • $\begingroup$ Yes 2 strokes have an SFC roughly 25% higher than 4 strokes, more or less comparable to turboprops. $\endgroup$
    – John K
    Commented Feb 9, 2022 at 21:08

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