Thermodynamic calculation of the P-V cycle of an engine

Let’s take as an example a three liter six cylinder engine.  Then the volume of each cylinder is 0.5 L.  Let us assume a compression ratio of 10:1.  That means that the cylinder volume is 0.05 L or 50 cm3 when the piston is maximally compressed.  So you heat the gas to provide a relatively high pressure.  Let us assume that you can heat the gas in the cylinder to 500 K by burning octane fuel.  Assume that the number of moles in the compressed 0.05 L volume can be calculated from the ideal gas law at 300 K (prior to the heating the pressure should be 1 atm).

However, we must consider that the gas is admitted while the cylinder is in its fully expanded state (state 3) and so the volume is 0.5 L.  Using these initial conditions we calculate the values of pressure and volume for all of the other states of the system.  These can then be plotted on a P-V plot.

The number of moles is calculated for state 3 by assuming that the pressure is 1 atm when the piston is maximally expanded.

n = PV/RT = (1 atm)(0.5 L)/(0.082 L-atm/mole-K)(300 K).

n = 0.02 moles.

The compression ratio dictates P1’ prior to combustion (i.e. at Tcold) is 10 atm. P1’/ P3 = V3/V1.

We can calculate the pressure following combustion from P1/P1’ = Thot/Tcold and it is 16 atm.

We can determine the heat that must be taken up to drive the expansion from the expression

qhot = CvDT = 3/2nRDT = 3/2(0.02 moles)(0.082 L-atm/mole-K)(500 K – 300 K) = 0.5 L-atm  (» 50 J).

The thermodynamic efficiency is

h = 1 - Tcold/Thot

Based on the thermodynamic efficiency of 0.4 we know that the total work is wtotal = nR(Tcold – Thot)ln(V2/V1) = 0.02 L-atm.

We know that V2 = 0.05 L so we can solve for V2

V2 = (0.05 L)exp{0.5 L-atm/(0.02 moles)(0.082 L-atm/mole-K)(200 K)} = (0.05 L)exp{1.52}= (0.05 L)(4.6) = 0.23 L.

Since the total volume is 0.5 L this means that about half of the expansion occurs during the adiabatic step.

Given the relationship V4/V3 = V1/V2 we can calculate

V4 = V1V3/V2 = (0.05)(0.5)/(0.23)= 0.109

P4 = nRTcold/V4 = (0.02)(0.082)(300 K)/0.109 = 4.51 atm.

 

State

P (atm)

V (L)

T (K)

1

16

0.05

500

2

3.5

0.23

500

3

1

0.5

300

4

4.6

0.11

300

 

Using this information along with the definition of an isotherm and an adiabat in P-V space we can draw the following cycle for the engine.


          The description given here was first applied by Carnot and is given in many books as the Carnot cycle.  The point of the P-V plot and the considerations given is that we can design an engine to maximize efficiency.  This must have been of great concern to British steam engine manufacturers since they initially made engines that ran with a Thot of the boiling point of water.  If the ambient temperature is 300 K and Thot is 373K you can calculate the thermodynamic efficiency of the engine and it is not a large number!