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It's said that ethanol is capable of more power than gasoline in an engine exclusively designed for it. I believe this is because it can handle higher pressure before detonating, meaning higher compression ratio and\or more boost is possible.

My question is, what percentage of a power increase (roughly) would you get in a high compression pure ethanol engine? Specifically I'm asking about sport bike style engines (around 15,000+ rpm naturally aspirated).

Also what would the compression ratio be roughly, and does power just increase in proportion to the compression ratio?

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There's a very handy formula for computing the effective power output of an internal combustion engine:

Effective power =   engine displacement
                  × effective working pressure
                  × engine speed
                ( × unit conversions )
                ( ÷ 2 [ if engine is 4-stroke ] )

The upshot of this formula is that it doesn't matter what fuel is being used - the power output is influenced by an engine's displacement, speed and effective working pressure.

For the purposes of this question, the impact of gasoline vs ethanol will be felt as a change in effective working pressure.

As an example (since I don't have actual data):

If ethanol safely supports a 15:1 CR in a naturally-aspirated application where gasoline can only go up to 10:1 before detonation becomes a concern, one can expect the effective working pressure to improve by up to 50% (of course, the engine's compression ratio will physically have to be altered to take advantage of ethanol).

If all else remains constant, this would translate to an effective power output increase of 50%

  • Well, my way of setting up the calculation may have been flawed, but it looks like the answer was close. ;) – anonymous2 Mar 27 '17 at 17:41
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I found some information from AmericanEnergyIndependence.com:

Because of the compression limitation required to prevent “engine knock”, a typical gasoline engine can only deliver about 25% efficiency — only 25% of the BTU's in a gallon of gasoline are converted to mechanical energy that turns the wheels of the car, the other 75% is lost in waste heat.

An Alcohol Engine can deliver about 40% efficiency — 40% of the BTU's in a gallon of ethanol powering an Alcohol Engine will produce mechanical energy that turns the wheels of the car.

*Note that efficiency refers not to fuel efficiency, but to efficiency per BTU.

Some math:

0.40 / 0.25 = 1.6

In other words, the amount of power you can gain from the same BTU's would be aproximately 1.6 times greater.

Courtesy of Pᴀᴜʟsᴛᴇʀ2: Note that you also have to put in more fuel to get the same number of BTU's. (Pure ethanol has an energy content of 84,530 BTU/gal, whereas standard gasoline has around 120,388 BTU/gal.)

This means that you get approximately the same power per gallon, but because the Ethanol supports a higher compression ratio, you can send more gallons through the engine. The end result is that you get the same amount of power per gallon, but you get a 60% increase in your power per minute.

  • You didn't take into account the energy density of the different fuels. – Pᴀᴜʟsᴛᴇʀ2 Jan 26 '17 at 14:45
  • @Pᴀᴜʟsᴛᴇʀ2, good point, I'm working on something there. – anonymous2 Jan 26 '17 at 14:50
  • So ethanol has only (roughly) 2/3 as much energy, but is 60% more efficient in an internal combustion engine. This means it's somewhere around 107% as efficient overall. But I'm not sure about power. – Guest Jan 26 '17 at 15:17
  • @Guest, you're absolutely right. I'm trying to iron this out, but if I can't, I'll simply delete the answer. It looked like the American Energy Independence thing was on to something, but it appears they might not actually be taking the power into consideration. – anonymous2 Jan 26 '17 at 15:17
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    Actually aren't the power and efficiency improvements the same here, because raising compression increases power for the same amount of fuel. Like turbos being more powerful and fuel efficient I believe (more cylinder pressure). – Guest Jan 26 '17 at 15:33

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