I have heard that a 120cc engine burns 120cc or 120ml of fuel per minute. Is this true?
This is possible, but still quite extreme fuel consumption.
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.
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.
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!
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.
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.