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Fuel usage to accelerate to 20 mphby zanes
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#1
Feb913, 04:36 PM

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In an effort to approach the EPA on the subject of eliminating unnecessary stop signs in an effort to reduce fuel consumption nationally, I am trying to determine approximately how much fuel is required to accelerate from a dead stop to 20 mph. With all the variables (vehicle weight, acceleration rate, etc.) , I would simply like to take a small car (3000 lbs) that would normally be in the 30mpg range for city driving at a "normal" acceleration rate. With such a number in hand, calculating the minimum fuel savings given the traffic counts at any given stop sign could be easily computed. I strongly suspect that there are millions of unnecessary stop signs in the U.S., and that eliminating them would save many millions of gallons of wasted fuel. Any help would be greatly appreciated. Zanes



#2
Feb1013, 10:15 PM

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P: 22,239

Welcome to PF.
Do you know Newton's equations of motion? The definition (mathematical) of work? You'll need to make an assumption about how fast the acceleration is, but after that, it is a pretty straightforward calculation. Why don't you give it a try and we'll see if/where you get hung up. 


#3
Feb1113, 12:15 PM

P: 3

I think this site is out of my depth. In math I didn't get much beyond the Pythagorean Theorem, and finding the hypotenuse doesn't seem to be germaine to the problem. Looks like Plan B, empirical data, is in my future. If I accelerate my car to 20mph, then decelerate to 0mph 100 times, then refill my tank, then drive the same route at 20mph, filling my tank a second time, I can get my answer (and wear out my brakes.) Thank you for responding, though. Sincerely, Zane Suedmeyer



#4
Feb1113, 12:50 PM

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P: 11,589

Fuel usage to accelerate to 20 mph
I don't think this experiment will give reasonable data with just 100 stops.
The kinetic energy of 1000kg, moving at 10m/s (a bit more than 20mph), is E=1/2mv^2 = 50kJ. Cars are quite inefficient, so we can assume at least 100kJ required energy in the fuel. Petrol has an energy density of about 35MJ/l, therefore 100kJ correspond to 1/350l or about 3ml. With 1.4€ to 1.7€ per liter (roughly the price in Germany), this costs 0.4 to 0.5 (euro)cents. All calculations are easy without a calculator. SI units are so nice ;). 


#5
Feb1113, 06:55 PM

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P: 22,239

This site can help anyone who is willing to learn, regardless of their current knowledge level. 


#6
Feb1213, 12:22 AM

P: 186

You might want to consider converting mfb's excellent answer in to CO2 emissions. You would have a politically more attractive argument if you use CO2 rather than cents.



#7
Feb1213, 03:26 AM

P: 31

Only the very largest of ships diesel engines achieve a thermal efficiency of 50% the conversion factor for a car engine is more like 25% at reduced throttle openings.



#8
Feb1213, 07:52 AM

P: 276

I venture that 70% of the driving population drive in such a way as to negate the advantages of new fuel mileage technologies.
They will not change driving habits to get the most out of the technology. 


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