Theoretical Otto cycle

Problem Statement:

The figure below illustrates a theoretical Otto cycle. The compression ratio V1/V2 is 8 and the heat transferred to the working fluid per cycle (we will assume that the air behaves as a diatomic ideal gas) is 25 kJ/mol. At the beginning of the compression (state 1), pressure is 1 atm and temperature is 200C. The heat capacity ratio is γ = 1.4. Calculate:

  1. The pressure and temperature at each state of the cycle.
  2. The work, the heat transferred and the internal energy change during each step of the cycle as well as over the entire cycle per mole of ideal gas.
  3. The cycle efficiency.
  4. The efficiency of a Carnot heat engine operating between the extreme temperatures of the cycle.

Express the results in SI units.

Givens: R = 8.31 J/mol K;

Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!


Solution:

The Otto cycle (see the figure above) consists of four processes:

1-2 Adiabatic compression
2-3 Isochoric heating
3-4 Adiabatic expansion
4-1 Isochoric cooling

The process 0-1 shown in the figure corresponds to the intake stroke: a fuel-air mixture is drawn into the piston.

In this problem we are going to calculate both the pressure and the temperature at the four states of the cycle, as well as the heat transferred, the work and the internal energy change at each step and over the entire cycle. The main given in this problem is the compression ratio: the ratio between the volume of air in the cylinder when the piston is at the bottom of its stroke (maximum volume) and the volume of air when the cylinder is at the top of its stroke (minimum volume).

Using this ratio and the initial pressure and temperature, we can calculate the pressure and temperature for all the other states of the cycle by applying the equation of state of an ideal gas and the adiabatic process equation.

The adiabatic equation for process 1-2 is:

And after substituting the problem givens expressed in SI units we get:

We use the adiabatic process equation expressed as a function of pressure and temperature to find the temperature at state 2 of the cycle:

We isolate T2 then we substitute the temperature T1 expressed in kelvin to get:

We will use the heat transferred to the working fluid to find the temperature of the gas in state 3.
Heat is only transferred to the working fluid during process 2-3, since processes 1-2 and 3-4 are adiabatic (and therefore no heat transfer occurs) and the gas cools down (it releases heat) during process 4-1.

The heat transferred during the isochoric process 2-3 is given by:

Where CV is the molar heat capacity at constant volume for an ideal gas (a diatomic gas in this case).

After substituting and isolating we get:

We will use the fact that process 2-3 is isochoric (volume remains constant) to calculate the pressure in state 3:

As you can see in the figure, this is the maximum pressure in the cycle.

In order to calculate the pressure in state 4 we apply the adiabatic equation to process 3-4. As you can see in the figure above, the volume ratio V4/V3 is equal to the compression ratio V1/V2, so:

Finally, we use the fact that process 4-1 is isochoric to calculate the temperature in state 4:

Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!



Knowing both the temperature and the pressure for each state of the cycle allows us to calculate the work, the heat transferred and the internal energy change during each step of the cycle. In order to do so, we will use the expressions deduced for the processes that constitute the cycle. Click on the following links to see how they are deduced.

1-2 Adiabatic compression

The heat transferred during an adiabatic process is zero by definition. So Q12 = 0.

The work done by the ideal gas during an adiabatic process is equal to the opposite of the internal energy change:

After substituting we get:

As expected, the work done during the adiabatic compression is negative because the volume of the gas drops.

2-3 Isochoric heating

The work done during an isochoric process is zero because the volume is constant. Therefore, applying the first Law of Thermodynamics, we obtain that the heat transferred by the gas is equal to the internal energy change:

That corresponds to the heat transferred by the gas per cycle given in the problem statement.

3-4 Adiabatic expansion

We will use the expressions used for the adiabatic compression 1-2 to calculate the adiabatic expansion:

4-1 Isochoric cooling

Finally, the gas does no work during the isochoric cooling, therefore, the heat transferred is equal to the internal energy change undergone by the gas:

To calculate the total work done by the gas over the cycle, its internal energy change and the heat transferred, we will sum these values for the different processes of the cycle:

And after substituting we obtain:

The internal energy change during a cycle is always zero: it is a state function and as the final and initial states coincides, its change is zero. The first Law of Thermodynamics must apply, so the total heat transferred by the gas has to be equal to the work done.


The cycle efficiency is given by:

Where W is the total work done by the gas and Q1 is the heat absorbed. In the Otto cycle, heat is only absorbed during the isochoric process 2-3. Furthermore, the total work done is the difference between the heat absorbed and the absolute value of the heat given (process 4-1). As a consequence we have:

As you can see, the Otto cycle efficiency does no depends on the amount of working fluid that undergoes the cycle. Finally, after substituting we obtain:

Which of course is less than one because no thermodynamic cycle can have an efficiency of 100%. It would violate the second Law of Thermodynamics if it were the case.


The Otto cycle extreme temperatures in this problem are: T1 = 293 K and T3 = 1876 K. The efficiency of a Carnot heat engine operating between these temperatures would be:

That is greater than the Otto cycle efficiency because the Carnot heat engine has the maximum efficiency that a heat engine can have when operating between two specific temperatures. It is of course less than 100% since no heat engine can transform all the heat it absorbs into work.

The post Theoretical Otto cycle appeared first on YouPhysics