Otto Cycle Efficiency Calculator
Work out the air standard efficiency of a petrol engine cycle from the compression ratio and the heat added.
What it does
Every petrol engine is a heat engine that burns fuel, pushes a piston down and gets a turn of the crankshaft out of it. The Otto cycle is the textbook version of that machine, named after Nikolaus Otto and used in the physics teaching of engines to this day. It is not a real engine, but it captures the four steps that every spark ignition engine goes through.
The thermodynamics text behind this page sets out the cycle as two isochores and two adiabats. The mixture is drawn in and compressed adiabatically, the spark is struck and heat is added at constant volume, the hot gas expands adiabatically and does its work, and then heat is rejected at constant volume as the piston pushes the burnt gas out.
The efficiency comes out surprisingly clean. Because the two adiabatic legs cancel out of the heat balance, the thermal efficiency depends only on the compression ratio and the specific heat ratio, written as one minus one over the ratio raised to the power of gamma minus one. With air at 1.4 and a ratio of 8 that gives 56.47 %, and the UC Davis course notes behind this page show the same algebra from the temperature relations along the adiabats.
This page runs the full cycle rather than only the efficiency. Give it a compression ratio, a specific heat ratio, the temperature at the start of compression and the heat added, and it returns the efficiency, the three other temperatures in the cycle, the net work, the mean effective pressure and the Carnot limit at the peak temperature.
The limits are clear. The air standard cycle assumes air is the working fluid, that combustion is a heat transfer from outside, and that every process is internally reversible. A real engine has heat losses, friction, incomplete combustion and a finite burning time, so the real efficiency sits well below the number this page gives.
How to use
- Choose metric or imperial for the temperatures, the heat added and the answer.
- Enter the compression ratio, which is the largest cylinder volume divided by the smallest.
- Enter the specific heat ratio. Use 1.4 for air, or about 1.3 for the burnt gases of a real fuel mixture.
- Enter the temperature at the start of the compression stroke, in kelvin or in degrees Fahrenheit.
- Enter the heat added, which is the energy from the fuel for each kilogram of air in the cylinder.
- Read the efficiency first, then the four temperatures so you can see how far the cycle swings, and the mean effective pressure if you want to know what the engine is working against.
Formula
- eta
- Thermal efficiency of the cycle, the share of the heat added that leaves as work.
- r
- Compression ratio, the largest volume divided by the smallest.
- gamma
- Specific heat ratio of the working fluid, 1.4 for air.
- cv
- Specific heat at constant volume of the working fluid, in kJ per kg per kelvin.
- Q
- Heat added during the constant volume part of the cycle, in kJ per kg.
The specific heat at constant volume follows from the gas constant and the specific heat ratio, since cv equals R divided by gamma minus one, and the gas constant for air is 0.287 kJ per kg per kelvin. The net work is the heat added minus the heat rejected, and the mean effective pressure is that work divided by the swept volume, which is the largest volume times one minus one over the compression ratio. The Carnot limit quoted here is one minus the starting temperature over the peak temperature, the best any engine could do with that peak temperature.
Worked example
Example: a compression ratio of 8 with 800 kJ per kg of heat added
Defaults: metric units, a compression ratio of 8, a specific heat ratio of 1.4, a starting temperature of 300 K and 800 kJ per kg of heat added.
- The compression ratio raised to the power of gamma minus one is 8 raised to 0.4, which is 2.2974.
- Efficiency: 1 - 1 / 2.2974 = 0.5647, so 56.47 %.
- The temperature after compression is 300 x 2.2974 = 689.2 K.
- The specific heat at constant volume is 0.287 / 0.4 = 0.7175 kJ per kg per kelvin, so the heat added lifts the temperature by 800 / 0.7175 = 1115.0 K.
- The peak temperature is 689.2 + 1115.0 = 1804.2 K, and the expansion brings it down to 1804.2 / 2.2974 = 785.3 K.
- The net work is the heat added times the efficiency, 800 x 0.56473 = 451.78 kJ per kg, which over a swept volume of 0.7435 m3 per kg is a mean effective pressure of 607.6 kPa.
- The tool shows 56.47 %, 689.2 K, 1,804.2 K, 785.3 K, 451.78 kJ per kg, 607.6 kPa and a Carnot limit of 83.37 %.
Reference table
| Compression ratio | Efficiency | Peak temperature | Net work | Mean effective pressure |
|---|---|---|---|---|
| 6 | 51.16 % | 1,729.3 K | 409.31 kJ per kg | 578.0 kPa |
| 8 | 56.47 % | 1,804.2 K | 451.78 kJ per kg | 607.6 kPa |
| 10 | 60.19 % | 1,868.5 K | 481.51 kJ per kg | 629.6 kPa |
| 12 | 62.99 % | 1,925.6 K | 503.91 kJ per kg | 646.9 kPa |
Limits
- The air standard cycle assumes air as the working fluid, so it ignores the change in the gas properties caused by combustion and the dissociation of the burnt gases at high temperature.
- It treats combustion as a heat transfer from an external source, so the real timing of the burn and the pressure during it are not represented.
- Every process is taken as internally reversible, which leaves out friction, heat loss through the cylinder walls and the pumping of the piston.
- The efficiency is an upper bound for the cycle, not a prediction of what a real engine will deliver.
Common mistakes
- Expecting the ideal efficiency from a real engine. A modern petrol engine runs well below 56 % because of the losses the model leaves out.
- Forgetting that the efficiency depends on the specific heat ratio as well as the compression ratio, so 1.4 for air and 1.3 for burnt gases give different numbers.
- Mixing the Otto cycle with the Diesel cycle. The Diesel cycle adds heat at constant pressure, and its efficiency is a different formula.
- Using the temperature at the start of compression in degrees Fahrenheit without converting, which puts the whole cycle out.
- Forgetting that a higher compression ratio raises the temperature after compression, which is the reason high octane fuel exists and the reason knocking limits the ratio.
Try it yourself
Quick checks. Answers sit tucked under each question.
How does the efficiency change if the compression ratio goes from 8 to 10?
It rises from 56.47 % to 60.19 %, and the net work rises from 451.78 to 481.51 kJ per kg for the same heat added. The efficiency follows one minus one over r to the power of gamma minus one, so the gain gets smaller each time the ratio goes up.
What happens if the same engine only takes 600 kJ per kg of heat?
The efficiency stays at 56.47 % because it depends only on the compression ratio and the specific heat ratio, but the peak temperature falls from 1,804.2 K to 1,525.5 K and the net work falls from 451.78 to 338.83 kJ per kg. Less fuel means less work, not a different efficiency.
What is the Carnot limit for the default cycle?
83.37 %, which is one minus the starting temperature over the peak temperature, 1 - 300 / 1804.2. The ideal Otto cycle at 56.47 % leaves a lot of room below that limit, which is where the real losses have to fit.
Questions
What is the Otto cycle?
The idealised petrol engine cycle: adiabatic compression, heat added at constant volume, adiabatic expansion and heat rejected at constant volume. The text behind this page describes it as two isochores and two adiabats.
Why does a higher compression ratio raise the efficiency?
Because the efficiency is one minus one over the compression ratio raised to the power of gamma minus one. A bigger ratio means more of the heat added comes out as work. The limit is knock, because the higher temperature after compression can set the mixture off before the spark arrives.
What is the specific heat ratio?
The ratio of the specific heat at constant pressure to the specific heat at constant volume. It is 1.4 for air, and closer to 1.3 for the burnt gases of a real mixture, which is one of the reasons the ideal number is generous.
What is the mean effective pressure?
The net work of the cycle divided by the swept volume, and it is a way of comparing engines of different sizes. A higher figure means more work for the same displacement.
How does the Otto cycle differ from the Diesel cycle?
In the Otto cycle the fuel is burned at constant volume after the air has been compressed, while in the Diesel cycle the fuel is injected at constant pressure during the expansion. The Otto cycle is more efficient at equal compression ratios, and the Diesel cycle is better suited to heavy loads.
What is the air standard assumption?
That air is the working fluid, that combustion is treated as heat transfer from an external source, and that every process is internally reversible. It makes the algebra clean and the answer an upper bound rather than a forecast.
Sources
- Physics LibreTexts, 11.4: the Otto cycle, with the efficiency relation
- Physics LibreTexts, University of California Davis physics 9B: engines and thermal efficiency
- MIT OpenCourseWare 16.050: thermal energy, part 2 on power and propulsion cycles