# Efficiency with a combustion engine

1. Oct 26, 2009

### sciencectn

I've been trying to understand why fuel economy decreases when the engine gets into higher RPMs. That is, if you put the pedal to the metal and quickly accelerate to 40 MPH, your engine operates much less efficiently than if you slowly accelerated to 40 MPH.

Here's where it seems the efficiency loss happens. If you have a certain amount of fuel in a cylinder and ignite it, it puts out a certain amount of force on the piston. Now, if you double the amount of fuel and ignite it, it doesn't necessarily double the force on the piston. Where is the energy being lost? Is the resulting heat from the explosion conducting away, thus resulting in lower gas expansion? Or, is the energy being lost to mechanical friction somewhere?

2. Oct 26, 2009

### Bob S

The simple answer is to study the BSFC (brake specific fuel consumption) map of a stanaard SI (spark ignited) IC (internal combustion) engine. See BSFC map in
http://en.wikipedia.org/wiki/Brake_specific_fuel_consumption
The engine is most efficient at about 30-35% of redline and 80% of maximum torque.

Edit] a gasoline engine's efficiency = 1/(BSFC*0.0122225) = 81.8/BSFC
For BSFC = 205 grams of gasoline per kilowatt-hour, efficiency = 39.9%. This is a very high efficiency. Most SI IC engines are in the 30% range.

Bob S

Last edited: Oct 26, 2009
3. Oct 28, 2009

### Lsos

I'm not so sure that the engine is much less efficient when you accelerate quickly. In fact, I believe that it is MORE efficient. Putting the "pedal to the metal" reduces the pumping losses of your engine, since the throttle is wide open. Throttling the engine is actually quite harmful to efficiency, since the engine has alot of work to do simply to suck in air.

The reason why "pedal to the metal" driving might be "inefficent" is not because of engine inefficencies...but because, well, what good is getting up to 100mph very quickly (and efficiently) if you just have to slam on the brakes a second later so as not to run the red light, and turn all that efficiently gained mechanical energy into heat? Well...the good is that it's much more fun. So in terms of "smiles/ gallon" it is VERY efficient. But, it probably doesn't translate into the best fuel economy...

But then you must look at other efficencies...what good is buying (or making/ designing) a 240hp engine if you're only going to use 10hp? That's grossly inefficent on many levels, and might be considered a travesty in some circles. It is much more efficnet to use as much of that 240hp as possible, and as often as possible.

Of course, this type of behavior might also result in a blown engine after a few years, if it comes to light that it's been worked on and put together with substandard aftermarket parts.

Don't ask me how I know...

4. Oct 28, 2009

### Andrew Mason

A major factor is incomplete combustion. If you floor the accelerator you put too much fuel into the cylinder. For every carbon and every 4 hydrogen molecules that you pump into the combustion chamber you need 2 O_2 molecules. If you don't have enough O2 the fuel is blown out through the exhaust.

AM

5. Oct 28, 2009

### Bob S

One reason the engine gets less efficient at high RPMs is that for a specific power output, higher RPMs means lower torque, and therefore higher intake manifold vacouum. At ~18" of vacuum, the pistons are pulling against a pressure difference during the intake stroke, and are doing work. The best way to improve efficiency is to reduce RPM and increase torque (which reduces intake manifold vacuum). Recall torque = 2 pi RPM/60. In the BSFC plot in
http://en.wikipedia.org/wiki/Brake_specific_fuel_consumption
the engine should be running near the "sweet spot" at ~80% max torque and ~35% redline to get best efficiency.
Bob S

6. Oct 28, 2009

### Lsos

I'm probably thinking of this differently, because for the particular engine I have, higher rpm means higher torque. My torque peak is around 7500 rpm...

Therefore, for the most part, my engine is more efficient at high rpm than at low...

7. Oct 28, 2009

### sophiecentaur

I'm not sure that, with modern engine management and fuel metering / injection, that necessarily follows. It was certainly true when using a carburettor, tho'.

8. Oct 28, 2009

### xxChrisxx

What a pile of bollocks.

Unfortunately for you engine efficiency isn't measured in smiles per gallon. Or how much of max power you are using at any given moment.

Rapid and constant acceleration and deceleration is the most fun but also mose inefficient way to drive. period.

Sitting at WOT is the best way, but only if you are sat at peak BSFC also. If you are revving the t**s off it, you may be at WOT but you are making more bangs than you need to.

Silly post.

Also what on earth car are you driving with the torque peak at 7500 rpm? Unless its some sort of race prepared engine or a bike engine then what on earth is the point of having peak torque that high.

Last edited: Oct 28, 2009
9. Oct 28, 2009

### Andrew Mason

I am sure that modern engine makers have devised clever ways of having a stoichiometric air-fuel ratio. The OP's original question though was:
In that scenario, if the amount of fuel is doubled and it is completely burned, the number of moles of gas produced will roughly double so the pressure will double and the force on the piston will double. So if the force does not double, the fuel could not have been completely burned.

AM

10. Oct 28, 2009

### xxChrisxx

You should get mor ethan 2x the force for double the fuel/air mix burned in the same combustion chamber.

By doing that you've effecively doubled the compression ratio, so you'll have a far higher starting pressure.

11. Oct 28, 2009

### Bob S

This is incorrect reasoning. You need to put the engine on a dynamometer and do a BSFC (brake specific fuel consumption) map.
Bob S

12. Oct 28, 2009

### sciencectn

From Wikipedia: "Combustion generates a great deal of heat, and some of this transfers to the walls of the engine."

This would also explain why your engine overheats when the RPM is in the red zone for too long, because more heat is being lost by conducting into the engine.

So, here's my theory. Say we have a certain amount of fuel in the cylinder. It will produce a large amount of heat upon exploding, and by PV=nRT it's pressure will increase, pushing up on the piston. If we double the fuel (assuming the stoichiometric ratio is ideal), the heat should doubled. However, when the heat difference between the inside and outside of the cylinder is higher, heat will conduct away faster. Thus, your explosion really only generates maybe 1.5 times as much pressure with doubled fuel, because a large amount of heat conducted away into the engine block.

I think power is also generated because the number of moles of product is higher than reactant, but this is only a small part of what generates the push on the piston (and assuming no leaks, it should be perfectly efficient).

13. Oct 28, 2009

### xxChrisxx

Pressure wont increase according to ideal gas law. Compression is isentropic.

The number of moles of product has nothing to do with the 'push' on the piston. It's purely becuase the pressure is higher. You have released gases with higher enthalpy (basically hotter more energetic gasts) meaning higher pressure.

This is the ONLY thing that pushed the piston down.

EDIT: If you are getting very very technical then the chemistry of combustion does play a part, but from what i've read your post isnt referring to anything that deep.

14. Oct 29, 2009

### Lsos

I keep seeing on the web that BSFC closely follows the torque curve, so that's what my logic is going by. Although, as you say, I probably shouldn't make any huge bets on this.

Obviously it's not meant to be the most serious of responses, nevertheless there was nothing untruthful about it.

And the car with the astronomical torque peak: a Honda S2000. Stock engine, but it's not so much the point that the torque peak is that high, but a side effect of having a 2.0 liter engine making 240hp. Anyways, it's served just the way I like it, so I guess ultimately THAT is the point...

15. Oct 29, 2009

### xxChrisxx

I realised the S2000 revved like a maniac, but didn't realise it made peak torque quite that high up the rev range!

I thought you were being a bit flippent so I responded in a slightly harsher tone than I otherwise would have. It appears you werent trying to be flippant so sorry about the tone of the post. As engines are most efficient at high load low revs; though BSFC does track the trend of the torque curve efficieny is not as simple as saying it occures wher ethe torque is highest.

Specific efficiency and acutal efficiency are two different things.

So high up the engine revs in your s2000, you will be using less fuel per watt generated but are unlikely to be requiring all that power. So say 30mph in 2nd vs 30mph in 6th. You may get a higher BSFC reading but you arent requireing the power output.

It's a balance between BSFC and number of combustion events.

Last edited: Oct 29, 2009
16. Oct 29, 2009

### Naty1

You are mixing two issues: fuel use during rapid acceleration and fuel use at increased RPM.

When you accelerate faster you use more POWER...work per unit time..so of course you burn more fuel doing more work to go faster sooner....fuel use versus RPM is also related to power produced but I leave it to engine experts to devine that. One thing for sure: at higher speed and rpm air resistance is greater and that alone uses engine HP and fuel to offset. Also I believe rolling (tire) resistance is also greater at higher speeds.

17. Oct 29, 2009

### Bob S

The thumbnail below shows a BSFC map for a typical spark ignited 2.7L automobile engine. The contours represent lines of constant fuel consumption (grams of gasoline per kWh of energy output). 83 g/kWh is equiv to 100% efficiency. To optimize (minimize) fuel consumption, stay close to the small island near 2200 RPM and 80% of maximum torque.
Running the engine at high RPM and low torque wastes fuel. WOT (wide open throttle) wastes fuel. For city driving, keep RPM below 2000 RPM.
Bob S
Torque is in newton meters. Power (watts) = torque x 2 pi RPM/60
 The BSFC map came from http://sitemaker.umich.edu/mhross/files/fueleff_physicsautossanders.pdf

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Last edited: Oct 29, 2009
18. Oct 30, 2009

### Ranger Mike

Efficiency with a combustion engine

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I've been trying to understand why fuel economy decreases when the engine gets into higher RPMs. That is, if you put the pedal to the metal and quickly accelerate to 40 MPH, your engine operates much less efficiently than if you slowly accelerated to 40 MPH.

Here's where it seems the efficiency loss happens. If you have a certain amount of fuel in a cylinder and ignite it, it puts out a certain amount of force on the piston. Now, if you double the amount of fuel and ignite it, it doesn't necessarily double the force on the piston. Where is the energy being lost? Is the resulting heat from the explosion conducting away, thus resulting in lower gas expansion? Or, is the energy being lost to mechanical friction somewhere?

my opinion

Acceleration and WOT will decrease fuel economy (MPG) , when compared to fuel economy in at steady RPM.
Different acceleration techniques will also impact MPG.
When you double the amount of GASOLINE , all other things being equal..it will NOT produce more power and WILL reduce MPG.

We are dealing with the IC with fixed camshaft profile, some what variable ignition timing and lean burn emission crap.
By adding twice the amount of fuel, the fuel air mixture will be too rich. This means the fuel is not totally burnt on combustion and excess fuel is sucked out on the exhaust stroke. Excess fuel puddles and is not atomized, cools the combustion chamber and fouls the spark plug..all which reduce POWER. So the expected thermal conversion from Liquid fuel energy to useable pressure and Heat byproduct does not occur as efficiently as the fuel air mixture before it was doubled. Your energy loss goes out the tail pipe.
your thinking is good but you forgot one important factor...it is FUEL /AIR mixture and in a normally aspirated engine, the mechanical piston, combustion chamber, camshaft timing etc.. can only fill the cylinder with so much fuel/air...now if you rammed in a double amount of fuel air ( turbo or supercharger) you would get more power for about 15 seconds until the engine grenades...
when you amp up the fuel / air you have to drop the compression ration to prevent detonation...

lastly, why fuel economy decreases when the engine gets into higher RPMs, you ask. As in life everything is compromise and production engines are made to function most efficiently at most commonly used power band. Camshaft profile ignition timing intake runner size, tail pipe length..all are geared for this. When the engine is spun up to higher than " normal" RPM, lets say the normal is 2500 RPM, now you twist it up to 4000 RPM on a 4 cylinder engine , at 2500 RPM , you are lighting each piston 625 times per minute, at 4000 RPM each slug lights 1000 times per minute. to do this you have to use more ( 38% more fuel) fuel / air..and there goes the economy thing..

19. Oct 30, 2009

### austin123

physics is nice topic.In fact, I believe that it is MORE efficient. Putting the "pedal to the metal" reduces the pumping losses of your engine, since the throttle is wide open. Throttling the engine is actually quite harmful to efficiency, since the engine has alot of work to do simply to suck in air.http://www.o2sensorsdirect.com" [Broken]

Last edited by a moderator: May 4, 2017
20. Oct 30, 2009

### Ranger Mike

I believe that it is MORE efficient..
ok
more efficient than what? more efficient when?
finally, efficient, why...?
pumping loss?
this is the tail wagging the dog....what is your definition of pumping loss..compared to what?

21. Oct 30, 2009

### xxChrisxx

"Pedal to the metal" usually implies agressive acceleration and higher speeds basically means the same as "go faster".

If you are using the traditional connotation of pedal to the metal (which is what the OP is using) - you are wrong.
If you just mean that sitting at wide open throttle - then you are correct.

This has been explained earlier, that the engine itsself generates motion most efficiently when its BSFC is lowest. See the map someone posted, this usually occurs at

Antoher way of thinking about it is that the engine works best at high LOAD, low REVS.

Eg. Sitting on the flat ground in 3rd gear, and opening the throttle wide open will cause the car to accelerate quickly and the engine to quickly rev up to read line.

At WOT you have no load on the engine so high revs. Although you have WOT, this is NOT efficient.

Now lets say you come to a steep hill still in 3rd gear, you keep WOT but your car will slows until you are crawling up the hill at a steady speed.

The engine is now under load, so the revs drop to lower in the rev range. This instance of WOT is efficient.

Last edited by a moderator: May 4, 2017