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Efficiency of a modern internal combustion engine

  1. Feb 29, 2008 #1
    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":


    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?
    Last edited: Feb 29, 2008
  2. jcsd
  3. Feb 29, 2008 #2
    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%
  4. Feb 29, 2008 #3
    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?
  5. Feb 29, 2008 #4
    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.
    Last edited: Feb 29, 2008
  6. Feb 29, 2008 #5


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    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.
    Last edited: Feb 29, 2008
  7. Feb 29, 2008 #6
    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.

    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.
  8. Feb 29, 2008 #7


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    Typical internal combustion engines lose around 60% of their total energy developed to heat through the exhaust and radiator.
  9. Feb 29, 2008 #8
    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.
  10. Feb 29, 2008 #9


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    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.

    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).
  11. Feb 29, 2008 #10
    In fact you've even changed the numbers! Before it was 15-20% efficiency (and therefore 80-85% heat), now it's 60% heat. :smile:
  12. Feb 29, 2008 #11


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    Read it and weep:


  13. Feb 29, 2008 #12
    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).

    Why would they limit speed? This makes no marketing sense. Engine won't explode so easily at such a small power output.
    Last edited: Feb 29, 2008
  14. Feb 29, 2008 #13
    Can't go at maximum speed though for legal reasons, I better take measurements as I accelerate at full throttle at 100 km/h.
  15. Feb 29, 2008 #14
    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%.
  16. Feb 29, 2008 #15


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    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.

  17. Feb 29, 2008 #16
    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.
    Last edited: Feb 29, 2008
  18. Feb 29, 2008 #17
    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


    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.
    Last edited: Feb 29, 2008
  19. Feb 29, 2008 #18
    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.
    Last edited: Feb 29, 2008
  20. Feb 29, 2008 #19
  21. Mar 1, 2008 #20


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    Oh really? What is your load then?
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