Car acceleration, fuel usage and kinetic energy

[VivaLaFisica]

Someone said the following and it just doesn't sit right with me. How do I effectively tell them what's wrong here?

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It applies to physics the same, as can be seen with the Kinetic Energy Formula Ke = 1/2MV^2

http://www.edmunds.com/car-reviews/...mustang-boss-302-vs-2011-ford-mustang-gt.html

As we can see by the track times above, it takes roughly the same time to accelerate from 0 - 30, as it does from 30 - 60 miles per hour.

http://www.frontierpower.com/library/makingsense.htm

From this graph we can see that fuel consumption of an internal combustion engine is quite flat.

So, with this in mind, we can conclude that the same amount of fuel was used to accelerate the car from 0 - 30 as it does 30 - 60.

Since kinetic energy is a product of velocity... Where is this extra energy coming from?

Either the car should take 4 times longer to achieve the velocity, 4 times the fuel, or our cars gain 400% efficiency every time you double the velocity.

“Today’s scientists have substituted mathematics for experiments, and they wander off through equation after equation, and eventually build a structure which has no relation to reality. ”
― Nikola Tesla
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[A.T.]

The car has to deliver more power to the wheels at higher speeds. But the fuel consumption depends on the efficiency of the system, which also varies with speed. So you cannot directly relate fuel consumption to kinetic energy gain.

 

[ViperSRT3g]

You also have to include the effects of the transmission in terms of gears.

 

[russ_watters]

The fuel consumption is not flat, it just looks flat in the graph because of the units used: lbm/hp-hr. The units are chosen to remove the direct impact of rpm difference.

Just picking-off some numbers and eyeballing-it, it appears to be nearly exactly linear, just like the horsepower: doubling with a doubling of rpm.

I recommend you pull the actual numbers off the graph for 1000 and 2000 rpm and compare them.

 

[VivaLaFisica]

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By the graph above, and this one http://hiwaay.net/~bzwilson/prius/pri_bsfc_010.jpg We can see that a car actually uses less fuel at 2000 rpm than it does at 1000 rpm. Since the engine spends the same time through all RPMs of both gears, the energy the motor puts out remains CONSTANT.

Input vs output. All losses are already counted for. Why the 4 times increase in energy in the same time period with the same fuel consumption?
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[cjl]

That fuel usage is for a diesel (your first one at least), first of all, and second of all, it's specific fuel consumption. In other words, fuel usage divided by power. Actual fuel consumption rate is lower at 1000rpm than 2000, but the increase in fuel consumption from 1000 to 2000rpm is less than proportional to the increase in power output.

 

[Lsos]

So, with this in mind, we can conclude that the same amount of fuel was used to accelerate the car from 0 - 30 as it does 30 - 60.

Since kinetic energy is a product of velocity... Where is this extra energy coming from?

Either the car should take 4 times longer to achieve the velocity, 4 times the fuel, or our cars gain 400% efficiency every time you double the velocity.

I used to think that there's some discrepancy there, too, but after looking deeper into it I have found that it is not the case. At these low speeds the results get muddled a lot by other factors, but bear in mind these main ones, just off the top of my head:

-0-30 mph and possibly even 30-60 mph might be limited by traction. It could probably accelerate much faster 0-30 mph than 30-60 mph if traction was greater, but likely the driver is either forced to let off the throttle (therefore burning less gas), or turning the extra energy into tire smoke.

-Depending on gearing and torque curve, the engine could be spinning twice as fast when at 30 mph as when at 60 mph, therefore likely delivering (correct me if I’m wrong, I didn’t go through the math) an average of ~3x as much power between 30-60 mph as 0-30 mph (and burning ~3x as much fuel).