Let's say a car at 60 mph requires roughly 20 hp to maintain its speed (i.e., to overcome rolling resistance and drag).

If this car makes 240 hp and gets about 34 mpg (not too unrealistic), how is it getting 34 mpg if the engine is only at 0.05 efficiency? I know modern automobiles are supposed to have an efficiency limit near 25-30% thanks to Mr. Carnot:

Contour plot of efficiency vs. angular velocity at various load factors. Efficiency increases with load factor with a global maximum at 100% load near 3300 rpm of approximately 0.3.
Chart from here.

I feel like gearing must be part of the answer but I'm having a hard time understanding just how the gearing allows the engine to generate 20 hp while clearly be at a higher efficiency than indicated on the chart for low engine loading.

Maybe I'm not getting exactly what the engine loading is referring to?

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    $\begingroup$ To get you started: 240 hp in this case would be the peak power. That's what you get at one particular engine speed, at wide open throttle. Unless the throttle pedal is all the way down, you're restricting air flow to the engine to reduce the power that it produces and the amount of fuel that it uses. $\endgroup$
    – Dan
    Apr 7 '15 at 22:40

Thermodynamic efficiency vs. fuel economy

When you cite an efficiency of 25-30% for an internal combustion energy, you're talking about the thermodynamic efficiency of the engine. This is, at the theoretical level, based on a temperature differential. It has nothing directly to do with the fuel.

When you cite a fuel economy of 34 miles per gallon, you're now talking about something that depends a great deal on other factors—for example, the energy density of the fuel. How much extractable energy is there in a gallon of gasoline? How about a gallon of antimatter? A gallon of chocolate milk?

Many engines can accept different fuels, or fuel blends, with different energy densities, without a significant change in their thermodynamic efficiency. For example, ethanol is blended with gasoline, but has an energy density about 30% lower than gasoline; a gallon of one is not equal to a gallon of the other.

To operate an engine at a certain thermodynamic efficiency means having a temperature differential; to maintain that temperature differential, you need to add energy at some rate. Getting the same thermodynamic efficiency when your fuel has a lower energy density simply means increasing the fuel delivery rate ($Q$ or $\dot{m}$) so that the energy delivery rate ($q_{in}$) stays the same. This ignores the various nuances of how engines burn different fuels but, generally speaking, there is no direct relationship between the thermodynamic efficiency and the fuel economy.

Significance of engine loading

I feel like gearing must be part of the answer but I'm having a hard time understanding just how the gearing allows the engine to generate 20 hp while clearly be at a higher efficiency than indicated on the chart for low engine loading.

Maybe I'm not getting exactly what the engine loading is referring to?

When the car is not accelerating, engine loading comes from whatever forces are acting against the motion of the engine. Internal friction (pistons, crankshaft, transmission, etc.), external friction (tires on road surface), drag, gravity (when going uphill). "Load" means how much power is required of the engine for the car to have some speed and acceleration.

As you point out, when a vehicle is cruising on the highway, it only needs a small percentage of its total available power output to maintain speed. Unless we're talking really high speeds and/or an exceptionally gutless wonder of a car, cruising on the highway just isn't a high-loading situation. Your confusion seems to come from the fact that vehicles get better fuel economy when cruising at highway speeds than they do while accelerating.

The key thing to realize is that getting better fuel economy doesn't imply that the engine is operating at a higher thermodynamic efficiency because there are many other factors that go into fuel economy. The thermodynamic efficiency of the Carnot cycle is only one of those factors. Another factor is the efficiency of the combustion reaction (which is technically not part of the Carnot cycle). Another is how much power is being used to accelerate the vehicle (useful work) versus how much is being lost to friction, drag and conduction (waste heat, $q_{out}$).

Calculating fuel economy

Consider the following relationship—where does engine loading come in?

$$\text {fuel economy (mpg)} = \dfrac {\text {speed (mph)}} {\text {flow rate (gal/h)}} = \dfrac {v} {Q}$$

In an idealized situation, with no drag on the vehicle, minimal internal and external friction, driving on a horizontal plane, the power required to maintain any speed is effectively zero. This means the load on the engine (when not accelerating) is also effectively zero. The Carnot efficiency is irrelevant at this point, but it would be very low as well. However, the fuel economy would be enormous because you have some $v$ in the numerator with nearly zero $Q$ in the denominator.

The opposite situation is even easier to demonstrate; you can do it at home in your own car. Just floor the accelerator with the transmission in neutral. (Don't actually do this.) Instant high load scenario as you accelerate the hell out of that crankshaft but $v = 0$, so your fuel economy is zero.

Realistic scenarios are more complicated but the long story made short is that the combustion reaction that occurs in the cylinder of the engine is much less efficient during periods of very high acceleration (i.e., near max load). More fuel passes through the engine unburned, or only partly burned, meaning you didn't extract as much energy from the same gallon of fuel. You're still going somewhere and your engine is operating at a higher thermodynamic efficiency due to the load placed on it, but in terms of fuel economy, that benefit is reduced by the cost of lower combustion efficiency. It is even possible, when combustion efficiency is very poor, for that cost to outweigh the benefit of high loading entirely. (This might be expected if the vehicle has been very poorly maintained. In reality, so much depends on the age of the car, the quality of its control units, the gear in which you are accelerating, that it's hard to make an exact prediction for a general scenario.)

The other thing I want to mention is that you have to consider where your power is going. "High engine loading" just means that a lot of the power the engine is capable of producing is being demanded; it doesn't tell you where the power is going. If it's going to fight drag, which increases as the square of velocity, then that's wasted power and wasted fuel. You can deliver it very efficiently but if it's not adding to the speed* of the vehicle, it's not contributing to fuel economy. It only looks efficient when you draw your system boundary around the engine and ignore the purpose of the car.

* Technically, we should consider elevation as well, but fuel economy is typically calculated in terms of horizontal path distance. Gains and losses due to change in elevation are either assumed to cancel out overall or accounted for with some coarse correction factor.


The plot in the question does not apply to the situation being asked. The power capacity of an automotive engine is dynamically changed while driving by both altering the RPM and by throttling the airflow in the intake.

The peak efficiency of a car engine is somewhere midrange, with inefficiencies under both high and low loadings. At 10% loading, the efficiency is still better than half what it is at peak.

But say then we're getting 10% efficiency at 34mpg.. There's no inconsistency there. It just means if you could somehow achieve 100% efficiency, you'd get 340mpg under that load scenario.


Petrol engines have a very poor efficiency at low load... no ifs or buts. The 34mpg efficiency is not quoted at low load.

There is an ideal speed at which maximum mileage occurs... engine efficiency increases as load increases, however, air drag losses increase as the square of the speed.

There are two major reasons for the poor efficiently at low load:

  1. At low load, a significant percentage of the work done is used to overcome engine friction.
  2. At low load, the effective compression ratio is very low, which in turn leads to a very low efficiency.

Power generated by an engine is the product of angular speed by the available torque at that angular speed. The function of the gearbox is merely to match the engine speed to the road speed.

City driving mileage is lower than highway mileage mainly due to continuous braking which dissipates kinetic energy as heat, and idling at traffic lights or stop signs.

  • 1
    $\begingroup$ Any justification/reasons for 1 and 2? Why at low load engine friction is greater? (speed could be the same, so I don't see why engine friction could be greater) Also, what do you mean that at low load the effective compression ratio is low? $\endgroup$
    – ergon
    Mar 1 '17 at 11:01

I can give you an analogy: For a DC electric motor, efficiency is highest at a point LESS THAN its maximum rated power. enter image description here

The loss mechanisms are friction from fast rotation, and magnetic core losses (current going through the iron alloy) as a torque-independent parameter. If it weren't for magnetic losses, the efficiency would be maximum at torque=0. So understand your own loss mechanisms of the motor and see if the same type of curve can apply.


I know it's an old thread, but I can't resist taking a stab at it. Especially when there's so much bad or unrelated info in the other answers.

  1. Thermal efficiency and fuel mileage ARE related - there are other factors to consider though.

    Fuel energy density is a factor. For gasoline engines it's about 44.5 HP-hours (since mpg occurs over some time and torque is instantaneous, I'll use HP, if it bothers you just multiply by 5252 and divide by RPM to translate. Despite all the argument about which measures engine output "better", they are directly related algebraically) and if you use 20 HP and burn 1 gallon of fuel you get 20/44.5=44.9% NET thermal efficiency; if you could do this at 60 MPH you'd get 60/1=60 MPG (MPH/GPH).

  2. One example given above used 10% efficiency at 34 MPG. I'll assume 60 MPH to keep the math simple; 10%x44.5=4.45 HP-hours per gallon. Achieving 34 MPG at 60 MPH requires: 60/34=1.765 gallons per hour of gasoline (MPH/MPG=GPH). If you extract 4.45 HP from 1.765 gallons of gasoline then you use a total of 7.85 HP. You cannot get 340 MPG from a 100% efficient gasoline engine powered car in the real world (even if you allow for the acknowledged impossibility of a 100% efficient engine) because you can't get a car to travel at 60 MPH using less than 8 HP! There are vehicles that get even greater fuel mileage, but there not street worthy cars and the engines certainly run much higher than 10%. Bottom line, the numbers ARE related and you can't just pair up random numbers or you end up with ridiculous claims.
  3. At low highway speeds (again, I like to use 60 MPH to keep the calculations simple), most modern engines operate around 15% to 18% T.E. (and there is some variability) and most modern sedans require somewhere around 15 to 20 BHP at the flywheel. This means MOST mid-size sedans get between 20 and 32 MPG (T.E. x 44.5 x gal. = HP-hours: 15/(.18x44.5)=1.87 gal. and 20/(.15x44.5)=3.00 gal.; 60/1.87=32.0 MPG and 60/3.00= 20.0 MPG).
  4. The reason manufacturers are downsizing engines is that smaller engines tend to operate at greater efficiency AT THE SAME POWER OUTPUT. Cars that get better fuel mileage than those calculated above usually have smaller than 2.0 Liter engines.

Bear in in mind that the chart shown relates to the thermal efficiency of the engine alone and doesn't consider a moving car as a complete system and also that the vertical axis appears to be labeled as a ratio rather than a percentage efficiency so 0.2 on the graph is equivalent to 20%

For a car as a whole, engine load is not independent of speed as drag forces tend to increase with the square of speed.

The main reason that IC engines have lower efficiency at lower loads is that a lot of the energy losses are fairly constant, regardless of the power being produced; these include the power required for things like air induction, pumping fuel, coolant and lubricants and powering electrical systems. So at lower power output these losses represent a much greater proportion of the overall power produced by the engine. For example, say that an engine has a constant requirement of 1 kW for all the ancillaries. If it is producing an output power of 1 kW then these losses are half of the total power being produced, whereas when the engine is developing 50 kW than they are just 2% of the total.

In a larger engine with higher power output these (approximately) constant losses will tend to be larger than in a smaller capacity engine.

Also, the efficiency of the engine doesn't relate directly to mpg. An efficient engine in a heavy car could still have higher fuel consumption than a relatively inefficient engine in a light car. To put it another way, mpg is related to how much energy you need to travel a certain distance but efficiency measures how much of that energy is wasted.

In the diagram 'load' is the torque produced by the engine. For the purposes of testing it will be connected to some adjustable load (e.g., generator and a bank of resistors). The power produced is this torque multiplied by the angular velocity (RPM). Note that all engines have a practical limit on RPM due to the acceleration forces on moving parts (especially valves and connecting rods) so the output power is a function of both the load and the RPM.

To put it another way, 25% load is not the same as 25% of peak power.


Rephrasing your question:
1. How much energy does a car use at 60mph doing 34mpg?
2. What is its efficiency if it requires 20hp to maintain constant speed?

In SI units:
Car at 26.8m/s doing 14.4km/L requires 14.9kW

Empirical data:
* HHV for gasoline approx 47MJ/Kg
* Density of gasoline approx 0.74 Kg/L
=> ~ 35MJ/L

PowerIn = (35MJ/L) * (26.8m/s) / (14km/L)
= (35kJ) * (26.8/s) / (14), NB: MJ=kkJ
= 67kJ/s
= 67kW
Efficiency = 14.9/67
= 22.2%

NB1: The quoted 240hp is the maximum OUTPUT power.

NB2: The % load in the diagram refers to maximum power at that particular speed. It is roughly proportional to volumetric efficiency of the engine, which depends on throttle position, valve timing and manifold resonance characteristics.

  • $\begingroup$ Welcome on the Engineering SE! The text editor has a lot of formatting options, look how beautiful you answer became now. :-) $\endgroup$
    – peterh
    Aug 25 '18 at 6:40
  • $\begingroup$ Sorry! I tried to do a minor formatting edit and ended up destroying most of the previous formatting!<br>Where can I learn more about the editor options? $\endgroup$
    – Juancar70
    Aug 25 '18 at 11:18

That chart is suspect. If you look at the Web page it was taken from, it is clear that the author of that page does not understand the basic terms that he is using, like "load", "power", "torque", etc. The chart may be correct if we interpret "load" as "torque", as it is well known that engine efficiency improves as torque increases and spin decreases (hence, we have "overdrive").


To summarize, engine efficiency is determined, in descending order, by:

  1. Thermodynamics: Second Law of Thermodynamics (Heat Engines). It is impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W . Some amount of heat QC must be exhausted to a cold reservoir. This precludes a perfect heat engine.
  2. Energy content of fuel: The specific energy content of a fuel is the heat energy obtained when a certain quantity is burned (such as a gallon, litre, kilogram). It is sometimes called the heat of combustion. Its theoretical value is determined from the Gibbs free energy which is a thermodynamic potential that measures the "usefulness" or process-initiating work obtainable from a thermodynamic system at a constant temperature and pressure (isothermal, isobaric)(in function of the specific engine).
  3. Specific internal combustion engine efficiency: efficiency of a modern internal combustion engines is determined mainly by air-fuel ratio, compression ratio, valve control systems, engine temperature management, mostly taken care of by closed loop control systems consisted of sensors (air volume,temperature, lambda sensors) connected to microcomputer. These together with the , gear box , determine optimal engine performance at specific throttle position, car load and inclination.(drag coefficient and tires size should be considered as well.)
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    $\begingroup$ Engine efficiency is related to MPG in a complex manner, which is what the OP asked about, and is not addressed here. $\endgroup$
    – Rick
    Aug 4 '15 at 17:19

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