The Pv and Ts diagrams of an ideal Otto cycle are shown on the left.
The working fluid contained in the cylinder of the car engine forms a closed system.
(1) Determine the power generated by this fourstrokefourcylinder sparkignition (SI) engine
In an Otto cycle, process 12 is an isentropic compression process. According to
the data given, temperature and pressure at state 1 are
T_{1} = 15^{o}C
P_{1} = 100 kPa
Also, the Ts diagram shows the temperature of the air reaches its maximum
value at state 3. That is,
T_{3} = 1800^{o}C
In the four processes of the Otto cycle, only processes 12 and
34 have work interaction. The energy balance in these two
processes are,
w_{12} = u_{2}  u_{1}
w_{34} = u_{4} 
u_{3}
The net work output is the sum of the work at process 12 and process
34. w_{net }= w_{12} +
w_{34} =
(u_{1}  u_{2}) +
(u_{3}  u_{4})
= c_{v}(T_{1}  T_{2}) + c_{v}(T_{3} 
T_{4})
The coldairstandard assumption states that the working fluid is air
and modeled as an ideal gas, and all the processes are reversible. Hence,
the relation between state 1 and 2 is
T_{2} can be determined from the above relation.
T_{2} = 2.14T_{1} = 2.14(15 + 273) = 616.3 K = 343.3^{o}C
Since v_{1 } = v_{4} and v_{2} = v_{3},
and process 34 is an isentropic process,
T_{3} is given as 1800^{o}C. Substitute T_{1},
T_{2}, and
T_{3} to the above equation gives T_{4}.
T_{4} = (288/616.3)(1800
+ 273)
= 968.7 K = 695.7^{o}C
Substitute all the data to the expression of the net work gives,
w_{net }= c_{v12}(T_{1}  T_{2}) +
c_{v34}(T_{3} 
T_{4})
=
0.718(288  616.3) + 0.718(2073  968.7)
= 557.2 kJ/kg
