# Petrol engine efficiency.

1. Apr 23, 2010

### Spikeywan

When I got my motorbike it had a non-standard rear sprocket with two extra teeth. I decided to swap it for a standard one. I expected to get a better overall mpg, but it didn't change noticeably.

So...

An engine has a red line at 14000 rpm. Peak power is delivered at 12000 rpm. Driving along in top gear at 70 mph, the revs are at 6500 rpm.

Because maxmimum power is delivered at 12k, surely this is the point where the engine is most efficient, and getting the most power from the fuel?

So, when cruising at a constant 70mph would it use less fuel if you changed down a gear or two, to get the engine revs as close to 12k as possible? Or is it better to drive in a high gear, keeping the revs as low as possible?

Why?

2. Apr 23, 2010

### Gordianus

Even though the engine may deliver maximum power at 12000 rpm, best efficiency is roughly obtained when the engine delivers the maximum torque

3. Apr 23, 2010

### rcgldr

The internal drag inside the engine, mostly due to moving air back and forth through the crankcase under the moving pistons is lowest at low rpms. However the engine is more efficient under some load and at higher rpms. As you've discovered the amount fuel rate consumption required to drive at 70mph is about the same within a range of rpms, but running in a much lower gear at 10,000+rpm would result in worse fuel milage. Motorcycles also accelerate to speed so quickly that unless you're in constant stop and go traffic, how fast you accelerate doesn't affect overall fuel milage that much because zero to 60 mph only takes about 4 to 5 seconds under moderate throttle, and zero to 45 mph only 3 to 4 seconds, then you're cruising at a somewhat fuel efficient speed.

4. Apr 23, 2010

### jack action

Like Gordianus said, maximum efficiency is at the maximum torque. Because torque is a measure of the efficiency of the combustion and of the friction losses.

Power is a measure that depends on both torque and rpm. Torque is the result of the energy released by the fuel combustion times the efficiency of the thermodynamic cycle, i.e. how the engine was able to transform that heat into mechanical work. The bigger the efficiency, the lower the fuel needed to create the same torque. The rpm indicates how many of these cycles you do in row, hence the power you're gonna get.

Depending on the engine design (valve event, intake & exhaust design, etc.), an engine often performs better at a certain rpm rather than an other (better efficiency), so it will also have a higher torque at a certain rpm rather than an other.

The friction losses are known to be more or less proportional to the mean piston speed of an engine, which is proportional to the rpm of the engine. This means that an engine @ 5000 rpm has twice the amount of friction losses than @ 2500 rpm. This is the number one reason for reducing the rpm as much as possible when looking for better mileage.

So when you perform a certain task (say cruisin' at 70 mph), you need a certain amount of power (say 20 hp), regardless of the engine's rpm. Whether you're at 5000 rpm or at 10 000 rpm, you're gonna have to adjust your throttle such that your power output is 20 hp. So, before considering anything else, we already know that we need that 20 hp PLUS extra power to compensate friction losses, which are greater at 10 000 rpm than at 5000 rpm. Furthermore, let's say that your engine is more efficient at 5000 rpm. This means that you will need less fuel at that rpm to produce the already lower total power needed. If the opposite was true (better eff at 10 000 rpm), it might be possible that the better efficiency compensate for the friction losses and then you wouldn't see a difference in fuel consumption.

5. Apr 23, 2010

adding to jack actions great explanation, the piston speed is a function of the rpm and stroke. with a shorter stroke the rpm will be higher than one with a longer stroke for those frictional losses, and since torque is also a function of the stroke the longer stroke engine is going to have torque sooner than the short stroke engine.

of cousre this all can be engineered to an extreme where as in F1 the engines tolerances are so tight it requires pre-heating before turning over. oil pressure and viscosity also effect those numbers. idealy you want to have the engine operating in the peak torque band, hence the critical need to have the gear ratios correct and also when changing one component others also need to be looked at. the most common overlooked one is tire dia.

On the race cars I deal with changing the tire size relative to the front/back will cause problems with the braking systems and the transmissions gear selection due to integrated traction systems.

6. Apr 24, 2010

### mender

http://autospeed.com/cms/title_Brake-Specific-Fuel-Consumption/A_110216/article.html

Load makes much more difference in fuel economy than where the torque peak occurs. It would be more accurate to say that the engine should be at about 75% of peak load for best fuel economy at a specific power output.

A quick note: it isn't all about pumping losses like the author is saying; the increase in thermal efficiency caused by the increase in dynamic compression is also very important.

Last edited: Apr 24, 2010
7. Apr 24, 2010

### Spikeywan

Here's a torque graph for you...

8. Apr 24, 2010

### mender

Which is of course at WOT only. That doesn't help very much when discussing what is happening at 25% or even 50% load.

Did you look at the article that I linked? It explains things quite well.

9. Apr 24, 2010

### Spikeywan

Yes. Thanks.

Just processing it all now.

10. Apr 24, 2010

### jack action

Changing the throttle angle is like changing the diameter of the intake port: you're changing the engine design and thus, restricting the flow. So the torque and power curves will also change.

No matter what, the lowest BSFC will always happen at maximum torque, FOR THOSE PARTICULAR CONDITIONS, throughout the rpm range. If BSFC happens at a lower rpm at part load, it is because maximum torque also happens at that same lower rpm.

Torque is proportional to BMEP and the volume of air per revolution that enters the engine:

$$T=BMEP\ V_{eng}$$

BSFC is inversely proportional to BMEP:

$$BSFC=\frac{\left(\rho_{air}/AFR\right)}{BMEP}$$

where $$\rho_{air}$$ is the density of the outside air and $$AFR$$ is the air-fuel ratio.

So when torque is at its maximum, BSFC is at its minimum. Although I must admit that maximum torque usually happens at a rpm slightly different than minimum BSFC, because the volume of air per rev varies also with rpm due to the variation of volumetric efficiency. But they're usually really close.

11. Apr 25, 2010

### mender

So the answer is to select a gear that runs the engine at a high load, and not worry about the rpm that the torque peak at full throttle occurs at.

12. Apr 25, 2010

### Spikeywan

Which is going to be the highest gear you have.

13. Apr 25, 2010

### mender

That your engine will pull without lugging, yes!

14. Apr 25, 2010

### Spikeywan

Hehe. So the obvious answer was correct after all. Thanks guys!

Still, it also explains why the MPG isn't much different between a standard rear sprocket and one with two extra teeth on it.

And for the extra Smiles Per Gallon given by those two extra teeth, they're well worth it.

I've decided that I'm going to fit a +2 rear sprocket again. I just need to wait until I wear the current chain and sprockets out so I can do the change.

15. Apr 25, 2010

### Lsos

And when you throttle the engine, you're increasing the pumping losses by a whole lot, and therefore decreasing the efficiency. It doesn't do you much good to be at the torque peak (point of highest efficiency at WOT), when your throttle is almost closed...

16. Apr 25, 2010

### Spikeywan

If you were cruising at 30 mph through town in top gear, then you come out of town and wanted to accelerate up to say 70, which is better if you want to use as little fuel as possible?

Originally, I thought it would be better to change down lots of gears, and use small throttle openings and all the gears to get up to your new cruising speed, but after reading this thread, it would seem that you're better off leaving it in top gear, and giving it WOT?

17. Apr 25, 2010

### mender

With due consideration for the engine (again, no lugging), yes.

18. Apr 25, 2010

### jack action

Yes, if you want maximum fuel economy. But if you want performance (and not take forever to get to your new speed), you have to downshift such that you get more power from the upper rpm to give you a good acceleration. And performance has a price called fuel consumption.

19. Apr 25, 2010

### rcgldr

Unless you're driving a very low powered vehicle, then maximum miles per amount of fuel used occurs around 55 mph. At any speed below 55mph, you milage is a bit worse. So there's a trade off in how fast to accelerate to 55 mph versus how much time is spent at a lower and less fuel efficient speed.

It's also unlikely that you'd be able to drive at 30 mph in top gear on most sport motorcycles (600cc and up) without lugging the engine. Most motorcycles are relatively overpowered and normally operate well out of peak milage mode at any reasonable speed, so driving technique isn't going to have much effect on milage, unless it's constant stop and go. The highest powered bikes, like the Suzuki Hayabusa and Kawasaki ZX14 get about 35 to 37 mpg (miles per gallon) regardless of driving method. 600cc sport bikes and cruisers will get between 40 to 50 mph, depending on the bike. Some mopeds can get over 50 mpg.

20. Apr 26, 2010