Theoretical Diesel cycle

Problem Statement:

The figure below illustrates a theoretical Diesel cycle. The compression ratio V1/V2 is 18 and the heat transferred to the working fluid per cycle (we will assume that air behaves as a diatomic ideal gas) is 50 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 exchanged 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 this Diesel cycle.

Express the results in SI units.

Givens: R = 8.31 J/mol K;

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Solution:

The theoretical Diesel cycle is an idealized version of a diesel car engine. It is assumed that the engine’s working fluid is an ideal gas (air) and that all processes it undergoes are reversible.

The Diesel cycle (see the figure above) consists of four processes:
1-2 Adiabatic compression (isentropic)
2-3 Isobaric expansion
3-4 Adiabatic expansion
4-1 Isochoric cooling

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 exchanged, 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 along with the initial pressure and temperature, we can calculate the pressure and temperature for each state of the cycle by applying both the equation of state for an ideal gas and the adiabatic process equation.

The adiabatic equation applied to process 1-2 gives:

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

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

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

Pressure is the same in states 3 and 2, because this process is isobaric, so we have:

We will use the heat transferred to the working fluid per mole over the cycle to find the gas temperature in state 3. Heat transfers occur only during processes 2-3 and 4-1 of the Diesel cycle. The temperature drops in the last one, so heat can only be transferred to the working fluid during process 2-3, that is isobaric, so we have:

Where Cp is the molar heat capacity at constant pressure of an ideal gas (a diatomic gas in this case).

After substituting we get:

First, we will calculate the relationship between the volume of the gas in states 4 and 3, and later we will calculate the pressure and temperature in state 4:

We can apply the adiabatic equation to process 3-4 as well as the volume relationship we have just found to get:

Finally, the temperature in state 4 is given by:

To get an idea of whether the results we have calculated are correct, we can draw the isotherm (for an ideal gas they are hyperbolae) that pass through each one of the four states of the Diesel cycle:

The higher the isotherm curve, the hotter its temperature. So the hottest temperature in the Diesel cycle is achieved at state 3. This agrees with the results we have obtained.


To calculate the work, the heat transferred and the internal energy change at each step of the cycle 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 minus the internal energy change:

After substituting we get:

As expected, the work done by the gas during the adiabatic compression is negative because its volume decreases.

2-3 Isobaric expansion

The work done, the heat transferred and the internal energy change of an ideal gas that undergoes a constant pressure process are given by:

And after substituting the givens, we have:

This heat is obviously equal to the heat absorbed by the gas over the cycle.

3-4 Adiabatic expansion

We will use the same expressions used for the adiabatic compression 1-2:

The internal energy change of the ideal gas is negative because its temperature drops, and the work it does is positive because its volume increases (the gas expands).

4-1 Isochoric cooling

During this process, the work done, the heat transferred and the internal energy change are respectively given by:

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 states of the cycle:

And after substituting we obtain:

The internal energy change during a cycle is always zero: the internal energy 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 net work output.

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The cycle efficiency is given by:

Where W is the net work output and Q1 is the heat absorbed (positive) during the cycle. In the Diesel cycle, heat is only absorbed during the isobaric expansion 2-3. Furthermore, the net work output is the difference between the heat absorbed and the absolute value of the heat discharged. Therefore we have:

As you can see, the Diesel cycle efficiency does no depends on the amount of working fluid that undergoes the cycle.

Finally, after substituting both the heat capacity ratio and the temperatures previously calculated we get:

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 Diesel cycle extreme temperatures in this problem are (as you can see in the upper figure): T1 = 293 K and T3 = 2650 K. The efficiency of a Carnot heat engine operating between these temperatures would be:

That is greater than the Diesel 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.

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