What quantity of gasoline is shot into a cylinder for each stroke?

Question

I already know PV=nRT, and how to calculate the gas ratios, stoichiometry of the moles of liquid --> gas, the increase in air pressure, and the (pie)(r)^2 Area of the piston. How much gasoline is squirted into each cylinder/stroke ???

assume a 2.0L engine/4cylinders=0.5L/cylinde... from a chem standpoint, 0.5L/22.42L/mol = .0223mol, and since oxygen is 21% of atmosphere, then 0.0223(.21)=4.68x10^-3 mol oxygen/cylinder. multiply this by mol fraction of 1/16 octane/oxygen yields 2.93x10^-4mol octane, which is aprx. 0.3mL per stroke at full bore. at 1000rpm, this is 300mL per minute, and 18L per hour (4.7gallons/hr). anyone dispute or have a better calculation, or a suggestion?

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now, I just need to factor in enthalpy (heat produced by combustion), and the increase in pressure from stoichiometry and PV=nRT as function of heat increase, and i will have the Force of the stroke. (over the Area of the piston, of course). I did this in the past, and I recall about 110pounds per stroke under full throttle ... it's been years, though since I calculated that. Remember, the engine is at 1000-6000rpm.

Using your formula of 1.3(100/1760)=.07g=.07mL per stroke, which is almost 0.1mL per stroke. A flaw in my calculation, though, is that i forgot to multiply by 4cylinders; that would be 4.7gallons(4cyl)=18.8gallons/hr. your version arrives at (.07mL)(1000rpm)(60min)(4cly)=16.8L/hr=4.42gallons/hr. Your version is more logical. My objective is to estimate the quantity of fuel injected.

Answer 1

The ratio of air mass to gasoline mass is 14.7 to 1. You could google "14.7 gasoline air" and get lots of citations on that.

For the volume of air, divide the engine's displacement (e.g 2.0 liters) by X number of cylinders. For the mass of air, divide its average molecular weight, 29.2 g/mole by molar volume at STP (22.4 l/mole) = 1.3 g air/liter

If you need a liquid volume, use a density of 0.74 g/cc.

So per liter of cylinder displacement, there will be 0.088 g or 0.1195 cc of gasoline used in every two rotations of the engines (for a 4-stroke).

The slightly tricky part is to estimate the efficiency of filling the cylinder at full throttle - about 85-90%. i.e. when the cylinder starts its compression stroke there is LESS than atmospheric pressure inside. Has to be - if there was no pressure gradient, they would be no flow.

The trickier part is how to figure volumetric flow rates under partial load - when the throttle is mostly closed and much less than atmospheric pressure is being compressed. (At least the same air-fuel ratio still applies.

Answer 2

It depends on the size of the engine, the size of the car, whether there is a hill, etc. etc. But one can get at least an idea. Air weighs 1.3 grams per liter. A typical gasoline combustion is C7H16 + 11O2 -> 7CO2 + 8 H2O. The molecular weight of heptane is 100; the total molecular weight of the oxygen is 352, and the total weight of the air containing it is 1760. If the stroke volume is V liters (engine at full power), then the fuel weight would be 1.3(100/1760)V grams.

Answer 3

It depends on how much power you want to get from that stroke. More if you are accelerating or pulling a load or going uphill than if you are idling or coasting. Use a different, more readily available set of details to get an estimate. 20 miles per gallon, 60 miles per hour, 2400 RPM, 6-cylinder 4-stroke cycle engine. That's about .00002 gallons per revolution of the engine.

Answer 4

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Answer 5

You are obfuscating with the amount of information in the question. Take miles per gallon at a given speed. Take rpms to maintain that speed. From that you can get a gallons/minute solution. Divide gallons per minute by number of cylinders for your gallons/cylinder answer.