The final publication is available at Springer via:

http://dx.doi.org/10.1007/978-0-85729-926-0

Chapter 13

Otto cycle calculation

Fig. A1.1 T-s and P-v diagrams depicting the Otto cycle

Probably the most well established and modelled compression and expansion process in
the power chamber of an internal combustion engine is the Otto cycle. Of course,
modern engine design is today fully and accurately simulated in computer software, but it
still remains useful to perform ‘long hand’ calculations on idealised gas models to gain
proper understanding of the thermodynamics.
Such calculations are based on idealised gases which are in quasi-equilibrium, expand
and contract adiabatically (dQ=0) or isentropically (dS=0). For such a gas the Gibbs
equation (Appendix A, eqn 1.3) gives:
(5.1)

This means constant enthalpy during the process. Now, for an ideal gas with a constant
specific heat (Cv), and using Boyle’s law the above equation becomes:

(5.2)

Re-arranging produces the following differential equation:

(5.3)

When integrated between states (1) and (2) in fig. A1.1 we get:

244
The final publication is available at Springer via:
http://dx.doi.org/10.1007/978-0-85729-926-0

(5.4)

which can be written in the form:

(5.5)

where k is the specific heat ratio representing the constant pressure value divided by the
constant volume value.
The idealised Otto cycle for a spark ignited gas is shown in fig. 1, as a temperature-
entropy cycle on the left and a pressure-volume cycle on the right. For such a cycle
efficiency can be estimated using:

 (5.6)

But using the definition of specific heat we must have, for heat transfer involving
constant volume processes:
dQin = dmCv(T3-T2) and dQout = dmCv(T4-T1) (5.7)
where dm is the small change in the mass of the gas. Consequently, substituting in
equation (6) we get:

 (5.8)

But for an isentropic processes we have:

and (5.9)

When it is observed that v1=v4 and v3=v2, we see that and the following rather

convenient efficiency formula results:

 

245
The final publication is available at Springer via:
http://dx.doi.org/10.1007/978-0-85729-926-0

where r=v1/v2. So the thermal efficiency in this idealised cycle is dependent only on the
compression ratio r. Hence the reason, that automobile manufacturers seek to achieve
high compression ratios in internal combustion engines.

A typical spark ignition Otto cycle engine could have a compression ratio of the order of
10, and exhibit exhaust temperature and pressure of 200 0C and 200kiloPascals
respectively. If the mechanical work delivered by the engine is known, from
measurement, to be say 1000kJoules per kilogram of fuel, then a comparison of thermal
efficiency of the ‘real’ engine and the Carnot efficiency can be computed as follows:

Thermal efficiency = (60.2%)

where k for the working gas is typically equal to 1.4.

The net work for the complete cycle, given that no work is produced between states 2 and
3 and between states 1 and 4, is:
wnet=w1-2 + w3-4 =cv(T1-T2)+cv(T3-T4)
Therefore 1000=0.717(473-1188+T3-T4) (5.11)
where we have assumed that the specific heat of the combusting vapour is 0.717kJ/
kg.K .
But for the isentropic process 3 4:

(5.12)

Simultaneous solution of equations (5.11) and (5.12) yields T3=3508K and T4=1397K.
Consequently the Carnot efficiency is:

c =

246
The final publication is available at Springer via:
http://dx.doi.org/10.1007/978-0-85729-926-0

In essence the Otto cycle efficiency is considerably less than the Carnot efficiency,
between the same temperature limits, since the heat transfer processes in the Otto cycle
are not reversible.

247