Theoretical gas mileage vs actual

99shark473999shark4739 Member Posts: 2
edited April 2015 in Hyundai
I have a 99 hyundai tiburon 2.0L. I got bored and decided to research different equations and measurements for my engine to determine theoretical specs for it. So here's what I have :

My bore is 84mm and stroke is 93.5mm. This means that the volume of air for each cylinder(assuming flat tops and bottoms and everything working out ideally)  is (3.14)(42)(42)(93.5) = 517,892.76 mm^3. Which corresponds to 31.59 in^3 of volume per cylinder, for a total of 126.37 in^3 displacement in the engine. This is because, of course, my engine has 4 cylinders. Now I read the ideal air to fuel ratio  is about 14.6:1, but can be averaged out to around 13:1 for N/A engines, which mine is for the time being. So let's say 13:1. This means that for every 2 crankshaft rotations, my car theoretically pumps 9.03 in^3 of fuel. This is the equivalent of 0.039 gallons. So at an average of about 2,000 Rpm, it would take me less than 30 seconds to consume an entire tank of gas(tank size of 14 gal). Which is most definitely not the case. So either the math is wrong, or I'm missing something big here. Anyone care to elaborate or help me figure that out? 

Comments

  • texasestexases Member Posts: 9,541
    Two big problems:
    1. air/fuel ratio is a mass ratio, not a volume ratio
    2. the engine runs at a vacuum, so less air is admitted than the displacement volume
  • 99shark473999shark4739 Member Posts: 2
    Ah alright. I wasn't aware of the ratio being based on mass. So then volume wise the ratio would be much, much, MUCH higher. That makes more sense. Thank you! 
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