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 ???

my 4yr old could tell you that it depends on how much the throtle is open, but assuming full throttle, is it about 2mL, 5mL, how much???

Answer 1

The engine control unit uses a formula and a large number of lookup tables to determine the pulse width for given operating conditions. The equation will be a series of many factors multiplied by each other. Many of these factors will come from lookup tables.

Pulse width = (Base pulse width) x (Factor A) x (Factor B)

In order to calculate the pulse width, the ECU first looks up the base pulse width in a lookup table. Base pulse width is a function of engine speed (RPM) and load (which can be calculated from manifold absolute pressure).
So, since we know that base pulse width is a function of load and RPM, and that pulse width = (base pulse width) x (factor X)
Factor X is many variables i.e temp, oxygen density,etc.

The pulse width is typically measured in milliseonds, combined with the delivery rate of your injectors, measured in cc/min.
If at idle, your V-8 5.0L engine base pulse width is 2 ms, and your injectors are of the garden variety 300cc/min you get 0.1cc/sec. If you idle for a minute, you use up 48cc or 1.6 oz. If you idle for 15min. you burn a more than a quart of gas. Think Limos.

Answer 2

You're obviously a very smart man, and I'm probably going to sound like a real rube to you, because I'm working off of memory from about 20 years ago. BUT:

The burn ratio for optimum efficiency is 1 part gasoline to 14 parts air, BUT to get a spark to fire it, the ratio has to be 1 part gasoline to 8 parts air. In other words, the fuel charge has to be way richer than optimum to get it to ignite, thus the hydrocarbon emissions.

Computers have come a long way in reducing emissions, but the basic dilemna remains.

Answer 3

It relies upon on the dimensions of the engine, the dimensions of the automobile, no rely if there's a hill, and so on. and so on. yet you may still get a minimum of an concept. Air weighs a million.3 grams per liter. a typical gasoline combustion is C7H16 + 11O2 -> 7CO2 + 8 H2O. The molecular weight of heptane is one hundred; the completed molecular weight of the oxygen is 352, and the completed weight of the air containing that's 1760. If the stroke quantity is V liters (engine at finished capacity), then the gasoline weight would be a million.3(one hundred/1760)V grams.

Answer 4

This is in a sense a trick question. First, it varies drastically by the displacement of the engine. Second, gasoline (vapors) mixed w/ air, only enter the combustion chamber every other stroke. Unless you are talking about a two stroke engine (weedeater, dirtbike, etc.) Also there are other variables to consider, like temperature of the engine and ambient air temperature. Therefore your question is technically impossible to answer.

Answer 5

the actual answer is none,,only a mixture,,which quickly turns in to a vapor enters the engine,,combustion chamber,,which in turn is ignited by the spark that occurs to the plug,,for that cylinder,,at that moment,,if actual wet gas went to the engine it would flood,,spark plugs wont fire when wet,,even with gas,,although some wet gas does not get vaporized as it enters the chamber,the ratio,,is too small to actually measure,,i own a repair shop,,and have been asked this question a dozen times,,there is a margin of error,,in all calculations for this ,,and all measurements for fuel and air ratio,,for all engines,,i hope this help,s.

Answer 6

Depends on the size of the motor and the size of the carburetor and also how much you stomp down on the gas pedal

Answer 7

Depends on how much you have your "pedal to the metal". Light foot, not much, heavy foot, lots. Pops