Efficiency of a modern internal combustion engine

[Ulysees]

This is an attempt to calculate the efficiency of a modern gasoline engine (Volkswagen's 1.6 litre FSI).

The result seems too good to be true. Please let me know if there is any error.

At a constant 100 km/h on a flat motorway, the consumption of my car reads 5.7 l/100 km. The engine revs at exactly 3500 rpm. From the power curve below, the power output at this rpm is 54 kW = 54000 joules per second

The consumption of 5.7 l/100 km at 100 km/h translates into 5.7 litres per hour, or 5.7 / 3600 = 0.00158 litres per second.

Therefore the energy per litre is 54000 joules / 0.00158 litres = 34.2 megajoules per litre

According to wikipedia, "gasoline contains about 34.6 megajoules per litre":

http://en.wikipedia.org/wiki/Gasoline

Therefore the efficiency of my engine is:

efficiency = output energy / input energy = 34.2 / 34.6 = 98.8%

But someone said internal combustion engines have a typical efficiency of 15-20%. Was that about old technology in fact?

Or is the 1.6 FSI so incredibly efficient?

 

[Ulysees]

In fact we should remove a significant figure from the result, because the consumption measurement only had two significant figures.

So the efficiency is 99%, with an accuracy of 0.05/5.7 = +/-0.9%

 

[trambolin]

I would like to give a hint by asking, then how come we need to cool our engine so badly if the energy efficiency is that good? In other words, where is this heat coming from, if 99% percent is spent on the torque generation?

 

[Ulysees]

That's logical. But it does not locate the error in the maths above, if one exists. Up to 2% of energy goes to heat according to this calculation, maybe that's a lot of heat.

 

[Mech_Engineer]

Your method for calculating the efficiency is flawed because you are calculating it based on a FULL THROTTLE dyno graph of the engine. When you are going at a constant speed, unless you are going the drag-limited top speed of the car, you will not be using full-throttle, and therefore your engine is developing far less power than you estimated. At 100km/hr (62 mi/hr) I would estimate your power requirement to be around 30-40 horsepower.

I would suggest estimating your car's aerodynamic drag, and using that to calculate the power it is using to maintain constant speed. That should give you an efficiency of around 30% for your vehicle.

 

[Ulysees]

I would suggest estimating your car's aerodynamic drag, and using that to calculate the power it is using to maintain constant speed. That should give you an efficiency of around 30% for your vehicle.

Wait a minute, this is about engine efficiency, not the whole car's efficiency. We're only looking at the speed because the computer that displays consumption is looking at speed to do its calculation.

Your method for calculating the efficiency is flawed because you are calculating it based on a FULL THROTTLE dyno graph of the engine. When you are going at a constant speed, unless you are going the drag-limited top speed of the car, you will not be using full-throttle

So the graph is not actual power, it's power when accelerating as fast as possible, ie if the rpm is rising as fast as possible? I better step on it then and see what happens . :)

Or reach the maximum speed of the car and take readings there.

 

[Mech_Engineer]

That's logical. But it does not locate the error in the maths above, if one exists. Up to 2% of energy goes to heat according to this calculation, maybe that's a lot of heat.

Typical internal combustion engines lose around 60% of their total energy developed to heat through the exhaust and radiator.

 

[Ulysees]

Typical internal combustion engines lose around 60% of their total energy developed to heat through the exhaust and radiator.

My friend Mech_Engineer, substantiate your statements please. This is not like bible reading where the pastor speaks and the others accept, we can't accept figures without some substantiation.

 

[Mech_Engineer]

Wait a minute, this is about engine efficiency, not the whole car's efficiency. We're only looking at the speed because the computer that displays consumption is looking at speed to do its calculation.

That's what I'm talking about. Because your engine is putting power through your transmission and differential to the ground, the only efficiency you can calculate is the "system" effciciency, which would be total output to the ground, divided by the total chemical energy input. Unless you have precise efficiency ratings of all of the other components in your system (transmission, differential, etc.) you cannot measure the effciency of just the engine unless it's bolted on a dyno by itself.

So the graph is not actual power, it's power when accelerating as fast as possible, ie if the rpm is rising as fast as possible? I better step on it then and see what happens . :)

Or reach the maximum speed of the car and take readings there.

Stepping on it won't help you, because then you have to try and precisely measure your acceleration AND the drag force on your car, AND the total friction in your system to estimate where all of the power is going.

...and getting to your vehicle's top speed won't help either, because I suspect it will have a computer-controlled speed limiter in place which will prevent you from getting to the car's drag-limited top speed (if it can even get there, some cars run out of revs before they hit the drag limited top speed).

 

[Ulysees]

In fact you've even changed the numbers! Before it was 15-20% efficiency (and therefore 80-85% heat), now it's 60% heat.

 

[Mech_Engineer]

My friend Mech_Engineer, substantiate your statements please. This is not like bible reading where the pastor speaks and the others accept, we can't accept figures without some substantiation.

Read it and weep:

http://en.wikipedia.org/wiki/Internal_combustion_engine#Engine_Efficiency

Most internal combustion engines waste about 36% of the energy in gasoline as heat lost to the cooling system and another 38% through the exhaust. The rest, about 6%, is lost to friction.

 

[Ulysees]

That's what I'm talking about. Because your engine is putting power through your transmission and differential to the ground, the only efficiency you can calculate is the "system" effciciency, which would be total output to the ground, divided by the total chemical energy input.

That's totally wrong, look at the math in the first post: consumption is read on a digital dial, it's derived by the engine control unit as it squirts fuel and divides squirt duration by the speed, while power is read from a graph, it's the power output on the axis of engine (when at full throttle according to your correction).

...and getting to your vehicle's top speed won't help either, because I suspect it will have a computer-controlled speed limiter in place which will prevent you from getting to the car's drag-limited top speed

Why would they limit speed? This makes no marketing sense. Engine won't explode so easily at such a small power output.

 

[Ulysees]

Can't go at maximum speed though for legal reasons, I better take measurements as I accelerate at full throttle at 100 km/h.

 

[TVP45]

My friend Mech_Engineer, substantiate your statements please. This is not like bible reading where the pastor speaks and the others accept, we can't accept figures without some substantiation.

Ulysees,
Look at the SAE literature for data on IC efficiency. Current numbers, depending on exactly how you calculate the efficiency, range from about 10% to over 30%.

 

[FredGarvin]

The fact of the matter is is that you have no idea what the load on your engine is. Your choice of power output is not correct. That is why it was suggested that you try to look at aerodynamic drag and work backwards from there. Since you have no idea of what the load is, there is no way you can estimate efficiency.

http://www.auto-ware.com/combust_bytes/eng_sci.htm

 

[Ulysees]

This is getting silly, of course you know what your load is. It's on the graph! Both the torque and power are there, it's the requirement for full-throttle that has not been met yet.

To go from New York to Washington, don't go via Australia.

Just step on that pedal, and repeat measurements is the answer.

 

[trambolin]

No need to get nervous. Let me start from the beginning. Because there are a couple of possibilities that can cause this contradiction.

Let's say you are going at a constant speed. So the net force on the car is zero. That means the engine is producing power just enough to equal the air resistance, road friction and internal losses. If that requires 54 kW, you can not go faster because that is the max torque that you can get out of the engine, right? But we know that your car accelerates beautifully after 100 km/h. But we know that this must happen at around 240-260 km/h . So something is wrong here. And that is the part where we say

The consumption of 5.7 l/100 km at 100 km/h translates into 5.7 litres per hour, or 5.7 / 3600 = 0.00158 litres per second.

I have some problem interpreting the data that is you are doing CONSTANT 100 km/h and you are still at 3500 rpm. From that I understand that you are in second or third gear. Because that particular car when operated at the optimum rpm is capable of crazy stuff. (Nice choice by the way!) Then we have the error. Because this clearly say you are not at the 54kW point. Please check the animation in

http://science.howstuffworks.com/fpte5.htm

It would rather be enough instead of trying to explain in words here.

Other possibility is that since the consumption dial on your car does averaging of some time interval, you are making peaks of 3500 rpm so that the average is not affected much.

As a side remark I would also like to state that these curves are recorded form the axle and not from the road data i.e. it is connected to a electrical motor and from the braking current the power is recorded.

Hope this clarifies but if not please elaborate more about the problem, your example is really fruitful about these concepts which sometimes easily overlooked.

 

[Ulysees]

Thanks the effort, it's actually been resolved already: it's because the curve shows full-throttle torque and power (ie when you keep your foot all the way on the accelerator pedal and therefore most likely accelerate, not when you're going at a constant speed well below max speed).

For the record, that car does indeed show 3500 rpm at 100 km/h on 5th gear. That is the optimal speed for low consumption (the reading of consumption matches manufacturer data too, I found later). To truly be at that 54 kW point, you would have to be keeping your foot on the pedal all the way the moment the speed crosses 100 km/h. Efficiency would not be the maximum possible under these stressful conditions, it would only give some idea. But you have no way to measure the maximum efficiency without opening up the engine output part and connecting something to the axle.

 

[Ulysees]

Please go to my other topic below too, if you have any detailed knowledge of cars and control engineering:

https://www.physicsforums.com/showthread.php?p=1581945#post1581945

 

[FredGarvin]

This is getting silly, of course you know what your load is. It's on the graph! Both the torque and power are there, it's the requirement for full-throttle that has not been met yet.

To go from New York to Washington, don't go via Australia.

Just step on that pedal, and repeat measurements is the answer.

Oh really? What is your load then?

 

[russ_watters]

In fact you've even changed the numbers! Before it was 15-20% efficiency (and therefore 80-85% heat), now it's 60% heat.

Reread his posts a little closer. The 60% was loss through the radiator and tailpipe only. That's not the only source of loss in a car. A significant amount of heat is also lost in the engine block itself and in the power transmission system. That's where the other 20-25% is lost.

 

[russ_watters]

This is getting silly, of course you know what your load is. It's on the graph! Both the torque and power are there, it's the requirement for full-throttle that has not been met yet.

To go from New York to Washington, don't go via Australia.

Just step on that pedal, and repeat measurements is the answer.

If you have a dyno available to you, that'll be easy...

 

[Ulysees]

Oh really? What is your load then?

154 Nm with the foot on the pedal all the way. But I don't want to use load, because there's a simpler way:

- Read power (energy per second at axle)
- Read consumption (fuel per second)
- Divide the two.

 

[Ulysees]

If you have a dyno available to you, that'll be easy...

Not sure what a dyno is, you mean this?

http://www.collingdaleperformance.com/home.html

 

[OmCheeto]

This isn't a homework question is it?
From my calculations, a 2004 GTI should only require 9.9 kilowatts of powers to maintain a speed of 100km/hr. This is purely from aerodynamic drag.

So I would say your chart is representing pure energy conversion of the fuel.
The fuel energy to propulsion efficiency appears to be about 18%
If you factor in rolling resistance, the efficiency would go up of course, but I'm guessing not by much.


references:
http://www.carfolio.com/specifications/models/car/?car=122001
http://en.wikipedia.org/wiki/Drag_(physics )

 

[Nesha]

@Ulysees
Your attempt to calculate the efficiency of engine is not correct in many ways. Efficiency is calculate trough this formula:
E=Pe/Hd*Gh
Pe-efective power (measured on dyno table)
Hd~44000kJ/kg
Gh-Mass flow rate of fuel to engine
With your data I calculated E>1, and that is not possible (first low of termodynamic). Specific consumption of 4 stroke engine is around 250-320[g/kWh], and that is data that you don't have. On picture is Otto cycle and if you know a little termodynamic you will understand that your calculation is not correct.
http://img85.imageshack.us/img85/6602/masine02un4.th.gif

 

[Ulysees]

Thanks the effort, it's actually been resolved already, it's because the curve shows full-throttle power, not the power when running at constant speed, so we need to repeat with the pedal all the way down. The formula you give is what has been effectively used, if you care to read carefully, save that litres are used instead of kilograms. This formula is correct but it is not the definition of efficiency (defined as an output energy divided by an input energy). Your formula is the correct way to calculate efficiency from the fuel's known energy per kg instead per lt.

 

[Nesha]

Pe is output energy (measured on dyno table), and Hd*Gh is input energy (kJ/kg *kg/s=kW).You can find density for gasoline so you can find mass of fuel.

 

[Mech_Engineer]

The fact is, you're still not getting exactly what is wrong with your approach for determining your total vehicle efficiency.

1) You cannot floor the accelerator and take a reading instantaneously at 100km/hr UNLESS you also have a nice sensitive accelerometer in your car. This is because while accelerating your load is a combination of acceleration force, air drag force, and other frictional losses. The load (and therefore fuel consumption) will be far different in a car that accelerating at full speed through 100km/hr, and one that is cruising at a constant speed of 100 km/hr.

2) The Dyno graph is not a graph of the load, it is simply a manufacturer produced graph that show's your engine's power output as a function of RPM while at 100% throttle. It doesn't tell you anything about the load on the engine, or what the efficiency losses will be in anything that might be bolted to it after installation in the vehicle (transmission, differentials, etc.)

To CORRECTLY estimate your vehicle's efficiency the best route in my opinion will be to drive at a constant speed (100km/hr) and somehow describe the forces it is fighting. This means get an estimate of your vehicle's drag coefficient at 100km/hr, and analytically calculate drag force, rolling friction (NOT static friction), and estimate losses in the transmission and other mechanical components. Unfortunately this will only give you a gross estimate, which if you're careful may come within the range you should be expecting.

The only way to definitively measue your vehicle's efficiency would be a wind tunnel with a Dyno in it.

 

[Ulysees]

1) You cannot floor the accelerator and take a reading instantaneously at 100km/hr UNLESS you also have a nice sensitive accelerometer in your car.

Sorry mate, but you just don't get it: from the instantaneous consumption which is shown on the panel (this is an exact value if consumption is constant as in the original post and an approximate value if consumption varies), we can eliminate considerations like air drag, acceleration etc.

The only way to definitively measure your vehicle's efficiency would be a wind tunnel with a Dyno in it.

Except we were not after the vehicle's efficiency, we were after the engine's efficiency. And that's where the manufacturer helps, by giving us the maximum power curve of the engine and only the engine, so the manufacturer helps us eliminate considerations of air drag, transmission friction, road quality, road slope and so on by giving us direct engine characteristics, so we can focus on the engine only and calculate its efficiency from just:

- rpm
- instantaneous consumption
- speed (which translates the consumption reading from litres per metre to litres per second)
- ensuring all are read under full-throttle

I measured yesterday a consumption increase from 6 to 16 lt/100km as you accelerate from 80 to 110 km/h, and the moment you cross 100 km/h, it is 14.7 lt/100km (+/-0.4, after repeating several times). This is 3 times more consumption than in the original post. I won't bother to calculate efficiency again because it's obvious the result will be 3 times less. Around 30% efficiency. End of story.

 

[Ulysees]

You'll be laughing in the end when you realize how simple it really is.

 

[Ulysees]

I think the complications you get drawn to arise because you imagine the car on a dyno and that raw dyno measurements are shown on the power/torque curves. But such curves would be car-specific, not engine-specific.

Manufacturer gives power/torque from the axle, the engine output. Where that torque goes, how much goes to fight friction, how much goes into acceleration, makes little or no difference in instantaneous consumption per second, or efficiency.

 

[Mech_Engineer]

Based on your previous posts, you are even farther from understanding the dynamics of the problem that I thought, and it's pretty obvious that you have no interest in learning why you are wrong.

 

[Ulysees]

Would it help if I drew a system diagram for you?

 

[Mech_Engineer]

I think it would be a lot more useful for you to first define exactly what efficiency you are looking for, and how you plan to calculate it.