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I have heard that a 120cc engine burns 120cc or 120ml of fuel per minute. Is this true?

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    $\begingroup$ No, it is not true. Where did you hear this? $\endgroup$
    – Mike Sowsun
    Commented Dec 10, 2022 at 3:52
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    $\begingroup$ 120cc is not fuel burn rate. It is the total internal size of your engine's cylinders. When running the engine does not fully flood the cylinders with fuel. Most of what the engine breathes in is air, not fuel. This is true of all internal combustion engines be it for aircraft or motorcycle or car or boat or lawnmower. $\endgroup$
    – slebetman
    Commented Dec 10, 2022 at 18:13
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    $\begingroup$ @slebetman Actually, it is the difference of the spaces in the cylinders between the downstroke and the upstroke position. In other words, it is the space swept by the pistons. $\endgroup$
    – Peter Kämpf
    Commented Dec 10, 2022 at 20:39
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    $\begingroup$ @PeterKämpf also, its the sum of all the cylinders, so a 2 cylinder 120cc engine has 60cc per cylinder, whereas a single-piston 120cc engine has a 120cc space in the one chamber $\endgroup$
    – Criggie
    Commented Dec 10, 2022 at 21:11
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    $\begingroup$ Bore^2 is wrong. It should be (1/2*bore)^2. Bore is the diameter of the cylinder, and the swept area is proportional to the radius^2, just like the area of a circle. $\endgroup$
    – SteveSh
    Commented Dec 11, 2022 at 0:54

5 Answers 5

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This is possible, but still quite extreme fuel consumption.

120ml == 120cc gasoline are ~90g by mass

90g gasoline need about 1320g air in order to fully combust

1320g air are about 1.08 cubic meters at sea level and 15C

Assuming volumetric efficiency of 1.00 (possible with a supercharged engine): 1.08 cubic meters need 9000 cycles of 120cc displacement.

This amounts to 9000RPM for a 2-stroke engine or 18000rpm for a 4-stroke engine.

Rather at the high end, but not unseen in a high-performance engines (e.g. sports motorcycles).

If you allow for a rather rich mixture (e.g. AFR of ~11.0, one will possibly need to run the engine rich at these performance levels) the engine may even not be supercharged and will still be below 10000/20000rpm.

But since we are on the aviation.SE I state that I don't want to fly with an engine pushed this much to the technology limits.

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No, a 120cc engine does not burn 120cc of fuel per minute. It means that the sum of the volumes of space swept by the pistons in each of the cylinders equals 120cc. This is called engine displacement.

The amount of fuel burned will vary with engine power output (i.e. throttle) and likely other factors. An engine with a higher displacement will likely burn more fuel than a smaller one, but if other components or the circumstances are different, this is not necessarily the case.

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    $\begingroup$ It is not the sum of all interiors, but the sum of the space swept by the pistons. Please compare to the Wikipedia article your linked to. $\endgroup$
    – Peter Kämpf
    Commented Dec 10, 2022 at 20:41
  • $\begingroup$ @PeterKämpf I fixed it $\endgroup$
    – Someone
    Commented Jan 12, 2023 at 17:45
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I would say it's unlikely, and is certainly not a rule of thumb. Fuel consumption depends on many variables, including:

  • Engine fuel and cycle type (e.g. 2-stroke petrol/gasoline)
  • Engine speed (RPM) and load
  • Altitude (air density, temperature, pressure)
  • Fuel/air mixture (rich/lean/stoichiometric)

To be clear, “120 cc” refers to the engine displacement, i.e. the combined usable volume of the cylinders (minus the combustion chamber volumes)—basically the amount of air the engine can “breathe” in one cycle. 120 cc would be a pretty small engine, so let assume it's a 2-stroke (1 cycle per revolution) cruising at 2500 RPM at a few thousand feet. Here I'll use the Frink engineering calculator syntax. Air density would be about 0.9 x that at sea level, so the mass airflow could be calculated as follows:

engine_displacement = 120 cc
engine_speed = 2500/min
cycles_per_rev = 1
air_density = 1.204 kg/m^3 * 0.9

engine_airflow = engine_speed / cycles_per_rev * engine_displacement
mass_airflow = engine_airflow * air_density

mass_airflow -> g/s
5.418

For the fuel consumption, let's assume a stoichiometric air-fuel ratio:

fuel_flow = mass_airflow / 14.7
fuel_flow -> g/s
0.36857142857142857143

Converting to volume flow rate:

gasoline_density = 750 kg/m^3
fuel_flow / gasoline_density -> cc/min
29.485714285714285714

So around 30 cc/min for those particular conditions. You could imagine considerably more for a high-power climb at very high engine speeds (potentially even more than 120 cc/min).

Thanks for the interesting question!

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  • $\begingroup$ Nice answer - welcome to Av.SE! $\endgroup$
    – Ralph J ♦
    Commented Dec 12, 2022 at 0:44
  • $\begingroup$ How do you convert mass airflow to G/s $\endgroup$
    – Lelotz
    Commented Dec 13, 2022 at 6:54
  • $\begingroup$ It's not a direct conversion, as it depends on the air-fuel ratio and the density of the fuel. If you know those, it's mass airflow / air-fuel ratio / fuel density. (Note that the air-fuel ratio is a mass ratio, as that reflects the amount of matter in the chemical reaction of combustion, and isn't affected by temperature, pressure, density, etc.) $\endgroup$
    – screwtop
    Commented Dec 15, 2022 at 5:10
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The fuel burn of a very small 2 stroke with a crude carburetor is roughly .6 lbs/hp/hr. You can work it out from that. A 10hp engine running at a cruise power of 70% will use .6 x 7 = 4.2 lbs/hr at cruise power. Which converts to 44.12 cc per minute, or 2.65 L/hr.

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  • $\begingroup$ Do you have a source for these numbers? Is this a lawnmower or outboard? $\endgroup$
    – D Duck
    Commented Dec 10, 2022 at 12:32
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    $\begingroup$ Not at hand. I read it many years ago. It basically goes: 2-strokes in general 5.5 to 6.5, carbureted air cooled 4 strokes around 4.5 +/-, injected 4 strokes around 4+, liquid cooled injected 4 strokes like car engines high 3s, diesels mid to low 3s. Turbo props are similar to 2 strokes around 5 to 6.5. The aircooled Lyc O-290 in my plane is around 4.4 from the performance charts. The turbo compound Wright R3350s were in the mid 3s, almost like diesels. 2 strokes are about the least efficient int comb engines there are. They are used for the power to weight. $\endgroup$
    – John K
    Commented Dec 10, 2022 at 13:26
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    $\begingroup$ @DDuck see here en.wikipedia.org/wiki/Brake-specific_fuel_consumption with pretty good table of assorted engines $\endgroup$
    – fraxinus
    Commented Dec 10, 2022 at 14:25
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No. "120cc" is the swept volume of the cylinder

It represents how much volume of air the engine can breathe per cycle, not how much fuel it might use. It says nothing about fuel consumption.

The cylinders are round and go up and down. They're taking cylinder travel x the surface area of the circle

pi x radius2 x travel.

As a thought exercise, 120cc is a small motorcycle engine, way too small for cars. If it was using 120CC/minute say at 60 MPH/100 KPH, that means 1 litre in 8 minutes/miles = 8 liters per 100km. Or 32 MPG. Awfully low for a 120cc bike. There are large pickup trucks that get better MPG/km/l than that.

However that fuel economy might make sense on an airplane. A very small airplane.

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