Automotive Thermodynamics

 

Please Note: This problem is way too long to be on an exam. It is intended to show how thermodynamic principles can be applied to a real internal-combustion engine. If something like this were to be on an exam, then it would be just one or two of the steps. Enjoy!

 

The Otto Cycle best describes a gasoline engine. It involves four steps:

 

            1)  Adiabatic Compression (Compression Stroke)

            2)  Isometric Heating (Ignition)

            3)  Adiabatic Expansion (Power Stroke)

            4)  Isometric Cooling (Exhaust and Intake)

 

A Volkswagen Passat has a six-cylinder engine with compression ratio = 10.6, bore = 82.5 mm and stroke = 86.4 mm. Before being compressed, the air-fuel mixture has P = 8.50 x 104 Pa and T = 300 K. QH = 500 J for one cylinder only. Treat the air-fuel mixture as a diatomic ideal gas.

 

The compression ratio is the ratio of the two volumes. The bore is the diameter of the cylinder and the stroke is the distance that the cylinder moves. If bore = 2R and stroke = L, then the total volume is six times the volume of one cylinder:  V = 6 pi R2 L.  This volume will be the difference between the two volumes on the PV-diagram.

 

V1 - V2  =  6  pi  R2 L  =  6  pi  (82.5 x 10 -3 / 2) 2  (86.4 x 10 -3 )  =  2.77 x 10 -3 m3  =  2.77 liters

 

So, this would be called a 2.8-liter engine at a car dealership. (They round up.)

 

V1  =  10.6  V2        and          V1  - V2  =  2.77 liters            Solving simultaneously we get:

 

V1  =  3.06 liters         V2  =  0.29 liters

 

The power stroke is adiabatic so we can use the characteristic equation of an adiabatic process, with gamma = 7/ 5 = 1.4  for a diatomic ideal gas.

 

P1 (V1)1.4  =  P2 (V2) 1.4   

 

Since we know P1, V1 and V2 we can solve this equation:  P2  =  232 x 104  Pa.

 

We can use the Ideal Gas Law to find T2:  T2  =  T1 (P2 V2 / P1 V1)  =  772 K

 

The heat released in the isometric heating is 6 x 500 = 3000 J. Find the value of T3:

 

QV  =  n CV (T3 - T2)         P V  = n R T          n  =  P V / R T

 

QV  =  (P1 V1 / R T1)  CV (T3 - T2)        Solving this we get  T3  =  2156 K

 

Now we can use the Ideal Gas Law to find P3:

 

P3  =  P2 (T3 / T2)  =  648 x 104  Pa.

 

Now for the last point on the PV-diagram.

 

P3 (V3)1.4  =  P4 (V4) 1.4   

 

Since we know P3, V3 and V4 we can solve this equation:  P4  =  23.8 x 104  Pa

 

T4  =  T1 (P4 / P1)  =  840 K

 

Now, let's summarize P, V and T for all four endpoints:

 

                                P (Pa)                               V (m3)                                 T (K)

 

             1              8.50 x 104                         3.06 x 10-3                              300           

                            

             2               232 x 104                         0.29 x 10-3                              772

 

             3               648 x 104                         0.29 x 10-3                             2156

 

             4              23.8 x 104                         3.06 x 10-3                              840

 

Now we can compute the work for one cycle. W(total) = W12 + W23 + W34 + W41

 

Work for an isometric process is zero so W23 = W41 = 0.

 

Work for an adiabatic process using a diatomic gas is W = (5/2) nR (Ti - Tf).

 

So, the total work is W(total) = (5/2) (P1V1/T1) (T3 - T4 + T1 - T2) = 1829.37 J

 

Efficiency:     e  =  W / QH  =  1829.37 / 3000  =  61%

 

                      e (Carnot)  =  (TH - TC) / TH  =  (2156 - 300) / 2156  =  86%

 

                      e (2nd Law)  =  e / e (Carnot)  =  71%

 

The overall efficiency is much lower than 61%, around 35 - 40%  because of mechanical losses.

 

Power Output  =  W f   =  (1829.37) (3600 RPM / 60)  =  109,762 W / 745.7  =  147 hp

 

That was fun. If you have any questions or comments, then please contact me. Thanks.