OTTO CYCLE: PHASES, PERFORMANCE, APPLICATIONS, SOLVED EXERCISES - PHYSICAL - 2024

The Otto cycle is a thermodynamic cycle that is made up of two isochoric processes and two adiabatic processes. This cycle occurs on a compressible thermodynamic fluid. It was created by the German engineer Nikolaus Otto in the late 19th century, who perfected the internal combustion engine, the predecessor to the one found in modern automobiles. Later his son Gustav Otto founded the famous BMW company.

The Otto cycle is applied to internal combustion engines that work with a mixture of air and a volatile fuel such as gasoline, gas, or alcohol, and whose combustion is started with an electric spark.

Figure 1. Cars in a Nascar competition. Source: Pixabay.

Phases of the Otto cycle

The steps of the Otto cycle are:

1. Adiabatic compression (without heat exchange with the environment).
2. Absorption of heat energy in isochoric form (without changing the volume).
3. Adiabatic expansion (without heat exchange with the environment).
4. Expulsion of heat energy in isochoric form (without changing the volume).

Figure 2, shown below, shows the different phases of the Otto cycle in a PV (pressure-volume) diagram.

Figure 2. PV diagram of the Otto cycle. Source: self made.

Application

The Otto cycle applies equally to four-stroke and two-stroke internal combustion engines.

-4-stroke engine

This engine consists of one or more pistons in a cylinder, each with one (or two) intake valves and one (or two) exhaust valves.

It is called like this because its operation has precisely four times or well-marked stages that are:

1. The admission.
2. Compression.
3. The explosion.
4. The escape.

These stages or times occur during two turns of the crankshaft, because the piston goes down and up in times 1 and 2, and again goes down and up in times 3 and 4.

Below we describe in detail what happens during these stages.

Step 1

Lowering the piston from the highest point with the intake valves open and the exhaust valves closed, so that the air-fuel mixture is drawn into the piston during its descent.

The intake occurs during step OA of the Otto cycle diagram at atmospheric pressure PA. In this stage the air-fuel mixture has been incorporated, which is the compressible fluid on which the stages AB, BC, CD and DA of the Otto cycle will be applied.

Step 2

Shortly before the piston reaches its lowest point, both valves close. Then it begins to rise in such a way that it compresses the air-fuel mixture. This compression process occurs so fast that it gives up practically no heat to the surroundings. In the Otto cycle it corresponds to the adiabatic process AB.

Step 3

At the highest point of the piston, with the mixture compressed and the valves closed, explosive combustion of the mixture initiated by the spark occurs. This explosion is so fast that the piston has barely descended.

In the Otto cycle it corresponds to the isochoric BC process where heat is injected without appreciable volume change, consequently increasing the pressure of the mixture. Heat is provided by the chemical reaction of combustion of oxygen in the air with fuel.

Step 4

The high pressure mixture expands causing the piston to descend while the valves remain closed. This process occurs so fast that the heat exchange with the outside is negligible.

At this point positive work is done on the piston, which is transmitted by the connecting rod to the crankshaft producing the motive force. In the Otto cycle it corresponds to the adiabatic process CD.

Step 5

During the lower part of the stroke, heat is expelled through the cylinder into the refrigerant, without the volume changing appreciably. In the Otto cycle it corresponds to the DA isochoric process.

Step 6

In the final part of the piston stroke, the burned mixture is expelled by the exhaust valve which remains open, while the intake valve is closed. The escape of burnt gases occurs during step AO in the Otto cycle diagram.

The whole process is repeated with the entry through the intake valve of a new air-fuel mixture.

Figure 3. Four-stroke engine. Source: pixabay

Net work done in the Otto cycle

The Otto cycle works like a heat engine and is run in a clockwise direction.

The work W done by a gas that expands the walls that contain it is calculated by the following formula:

Where Vi is the initial volume and Vf the final volume.

In a thermodynamic cycle the net work corresponds to the area enclosed in the cycle of the P - V diagram.

In the case of the Otto cycle, it corresponds to the mechanical work done from A to B plus the mechanical work done from C to D. Between B and C the work done is zero since there is no change in volume. Similarly between D and A the work is null.

Work done from A to B

Suppose we start from point A, where its volume Va, its pressure Pa and its temperature Ta are known.

From point A to point B an adiabatic compression is carried out. Under quasi-static conditions, adiabatic processes comply with Poisson's law, which states that:

Where γ is adiabatic quotient defined as the quotient between the specific heat at constant pressure and the specific heat at constant volume.

So the work done from A to B would be calculated by the relation:

After taking the integral and using the Poisson's ratio for adiabatic process, we have:

Where r is the compression ratio r = Va / Vb.

Work done from C to D

Similarly the work done from C to D would be calculated by the integral:

Whose result is

Where r = Vd / Vc = Va / Vb is the compression ratio.

The net work will be the sum of the two jobs:

Net heat in Otto cycle

In the processes from A to B and from C to D no heat is exchanged because they are adiabatic processes.

For the process from B to C, no work is done and the heat given by combustion increases the internal energy of the gas and therefore its temperature from Tb to Tc.

Similarly, in the process from D to A there is heat transfer that is also calculated as:

The net heat will be:

performance

The performance or efficiency of a cyclic motor is calculated by finding the quotient between the net work done and the heat supplied to the system for each cycle of operation.

If the previous results are substituted in the previous expression and the assumption is made that the fuel air mixture behaves as an ideal gas, then the theoretical efficiency of the cycle is reached, which only depends on the compression ratio:

Otto cycle solved exercises

-Exercise 1

A 1500 cc four-stroke gasoline engine with a 7.5 compression ratio works in an environment with atmospheric pressure of 100 kPa and 20 degrees Celsius. Determine the net work done per cycle. Assume that the combustion contributes 850 Joules for each gram of air-fuel mixture.

Solution

The net work expression had been previously calculated:

We need to determine the volume and pressure at points B and C of the cycle to determine the net work done.

The volume at point A where the cylinder has been filled with the air-gasoline mixture is the 1500 cc displacement. At point B the volume is Vb = Va / r = 200 cc.

The volume at point C is also 200 cc.

Calculation of pressure at A, B and C

The pressure at point A is atmospheric pressure. The pressure at point B can be calculated using Poisson's ratio for an adiabatic process:

Taking into account that the mixture is predominantly air that can be treated as a diatomic ideal gas, the gamma adiabatic coefficient takes the value 1.4. Then the pressure at point B will be 1837.9 kPa.

The volume of point C is the same as that of point B, that is 200 cc.

The pressure at point C is higher than at point B due to the increase in temperature caused by combustion. To calculate it we need to know how much heat the combustion has contributed.

The heat contributed by combustion is proportional to the amount of mixture that is burned.

Using the ideal gas equation of state:

So the heat contributed by combustion is 1.78 grams x 850 Joules / gram = 1513 Joules. This causes a temperature rise that can be calculated from

Tb can be calculated from the equation of state resulting in 718 K, so for our data, the resulting value of Tc is 1902 K.

The pressure at point C is given by the equation of state applied to that point resulting in 4868.6 kPa.

The net work per cycle is then 838.5 Joules.

-Exercise 2

Determine the efficiency or performance of the motor from Exercise 1. Assuming the motor runs at 3000 rpm, determine the power.

Solution

Dividing the net work by the heat supplied gives an efficiency of 55.4%. This result coincides with that obtained by the direct application of the formula for efficiency as a function of compression ratio.

Power is the work done per unit of time. 3000 rpm equals 50 revolutions per second. But the Otto cycle is completed for every two revolutions of the engine because it is a four-stroke engine, as we explained earlier.

This means that in one second the Otto cycle is repeated 25 times so the work done is 25 x 838.5 Joules in one second.

This corresponds to 20.9 kilowatts of power equivalent to 28 horsepower.

References

1. Thermodynamic cycles. Recovered from: fis.puc.cl
2. Martín, T. and Serrano, A. Otto cycle. Recovered from: 2.montes.upm.es.
3. Sevilla University. Wiki of the Department of Applied Physics Otto cycle case study. Recovered from: laplace.us.es.
4. Wikipedia. Otto cycle. Recovered from: es.wikipedia.com
5. Wikipedia. Otto engine. Recovered from: es.wikipedia.com