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Something that has bothered me for a long time is why a 600 hp engine uses more fuel per kilometer than a 80 hp engine. Let pretend I have two equal cars (same shape same weight etc) except for the engine, one is 600 hp and other is 100 hp. If I manage the throttle to accelerate both with the same acceleration, my knowledge says that will require the same energy to get to a certain speed, so why will the 600 hp will use more fuel/energy?

Also if I am right is it possible that in the future cars will be so efficient that a big engine will only consume more if pushed to the limit?

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Unfortunately, I see a premise here that is difficult to prove true or false. Mainly because it is almost impossible to use a 600hp engine in a way similar to an 80hp engine and compare them directly.

Any car that has a 600hp engine will have other components to support it. Compared with an 80hp car, that might be a larger, more robust transmission, larger exhaust, stronger engine mounts, bigger wheels, etc. All of these make it difficult to determine how much of the efficiency difference is due to the behavior of the engine and how much is due to moving around more mass.

It is difficult to imagine a case where you have the benefit of having the power of a large engine usefully available to you without also having the penalty of increased mass.

In the realm of physics, there is no reason that a large engine in isolation couldn't be as efficient (in a power delivered per fuel consumed sense) as a smaller engine. But real engines are not developed to maximize that value at the expense of everything else, and that value doesn't immediately translate to vehicle fuel efficiency ($mpg$ or $l/100km$). Real engines are built to have reasonable fuel consumption while also holding development costs, production costs, maintenance costs, and total vehicle production costs low as well. A fuel efficient vehicle is created by not focusing only on the engine, but on many other aspects of the car as well.

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It's true that a 600 hp engine does not necessarily have to be generating that amount of power in order to get from point A to point B. But even though you manage the throttles so that both cars accelerate at the same rate, the larger engine has greater internal cylinder displacement, and therefore consumes more air and fuel per engine revolution than the smaller engine.

With every revolution of the crankshaft, the larger engine will draw in and expel more air and fuel than the smaller engine. The result is wasted energy. The first point in this link explains this more clearly: http://www.driverside.com/auto-library/top_10_factors_contributing_to_fuel_economy-317

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The answer is that a big engine is less efficient at low power than a small one, because it has to lug around a lot of extra stuff (bigger cylinders, more cylinders, bigger and more bearings and so on), all of which involve frictional and other losses, which the smaller engine does not have.

This is really an engineering question rather than a physics one however: I think a physicist's answer would be that a big engine could be as efficient at low power as a small one: for instance make a big engine by strapping two small ones together, and at low power just do not run half of the cylinders. That's fine in theory but an engineer will point out that you still pay the frictional costs (I think some modern cars more-or-less do do this though).

Edit. Actually, I think that if you consider a big engine running at low power but not doing any clever tricks (see below), then it necessarily has to run with a lower temperature differential, and thus has lower best-case thermodynamic efficiency than a smaller engine running much hotter. So I was wrong: there is a physics reason, although engineering can get around some of it.

'Clever tricks' mean things like not running some of the cylinders at all, or firing only every nth stroke, or so on. Strapping two smaller engines together as I suggested above is such a trick, of course.

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To generate more power, an engine needs to be able to burn more fuel; laws of physics, chemistry and thermodynamics dictate this requires a larger displacement (bigger volume in which you burn the fuel).

Larger volume = larger area over which you generate friction, more more importantly, more air being pulled through per stroke. Some engines with a lot of cylinders have actually been engineered to "shut off" a certain number of cylinders when running in low power mode. This is done by closing the valves for the entire stroke. Rather than sucking in air, compressing it, and letting it go through the exhaust (which takes energy, even if you didn't burn any fuel), these cylinders keep the air inside - it loses a bit of energy due to friction, but much less than is wasted by moving air.

See for example variable displacement technology and this article on the Mercedes Benz engine which claims up to 30% improved efficiency because of this technology. That demonstrates that the air flow is indeed a major component of the efficiency of small displacement engines.

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It’s all about engine's efficiency. According to wikipedia:

gasoline engine's efficiency = 1/(BSFC × 0.0122225)

(Also Actual efficiency can be lower or higher than the engine’s average due to varying operating conditions.)

To calculate BSFC (Brake specific fuel consumption) use the formula: BSFC=r/P (where: r is the fuel consumption rate in grams per second (g/s) and P is the power produced in watts.)

And after merging 2 formula we have:

r (fuel consumption)=BSFC×P=P/( gasoline engine's efficiency× 0.0122225)

So for those two car ( car A with 600hp power and car B with 80hp power) if we consider: r(A)=r(B) --> engine's efficiency(A)=7.5×engine's efficiency(B)

So for each engine with more power you should have more engine's efficiency in order to fuel consumption be at fixed level.

Now real example:

Car A: Audi 2.5 litre TDI (1990) power=88kW efficiency=42.5 --> fuel consumption= 198 (g/kW·h)

Car B: Volkswagen 3.3 V8 TDI (2000) power=165kW efficiency=41.1 --> fuel consumption=205

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  • $\begingroup$ Your answer unnecessarily complicates the answer and does not answer the question as well. $\endgroup$ – Yashas Feb 11 '17 at 6:21

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