Air Standard cycle

The Otto cycle is a set of processes used by spark-ignition internal combustion engines (2-stroke or 4-stroke cycles).

These engines

a) ingest a mixture of fuel and air,

b) compress it,

c) cause it to react, thus effectively adding heat through converting chemical energy into thermal energy,

d) expand the combustion products, and then e) eject the combustion products and replace them with a new charge of fuel and air.

The different processes are shown in Figure,

1. Intake stroke, gasoline vapor, and the air is drawn into engine ( 5-1 ).
2. Compression stroke, p , T increase ( 1-2 ).
3. Combustion (spark), short time, essentially constant volume ( 2-3 ). Model: heat absorbed from a series of reservoirs at temperatures T2 to T3 .
4. Power stroke: expansion ( 3-4 ).
5. Valve exhaust: valve opens, gas escapes.
6. ( 4-1 ) Model: the rejection of heat to series of reservoirs at temperatures T4 to T1.
7. Exhaust stroke, the piston pushes remaining combustion products out of the chamber ( 1-5 ).

The efficiency of an Ideal otto-cycle

The starting point is the general expression for the thermal efficiency of a cycle:

The convention, as previously, is that heat exchange is positive if heat is flowing into the system or engine, so QL is negative. The heat absorbed occurs during combustion when the spark occurs, roughly at constant volume. The heat absorbed can be related to the temperature change from state 2 to state 3.

The heat rejected is given by (for a perfect gas with constant specific heats)

Substituting the expressions for the heat absorbed and rejected in the expression for thermal efficiency yields

We can simplify the above expression using the fact that the processes from 1 to 2 and from 3 to 4 are isentropic:

The quantity (V1/V2=r) is called the compression ratio. In terms of compression ratio, the efficiency of an ideal Otto cycle is:

The ideal Otto cycle efficiency is shown as a function of the compression ratio. As the compression ratio, r, increases, increases, but so does. If T2 is too high, the mixture will ignite without a spark (at the wrong location in the cycle).

#otto cycle
#by Darshan Dodamani

import math
import matplotlib.pyplot as plt

def engine_kinematics(bore,stroke,con_rod,cr,start_crank,end_crank):

 	a=stroke/2
 	R=con_rod/a

 	v_s=math.pi/4*pow(bore,2)*stroke
 	v_c=v_s/(cr-1)

 	sc=math.radians(start_crank)
 	ec=math.radians(end_crank)

 	num_values=50

 	dtheta=(ec-sc)/(num_values-1)

 	v=[]

 	for i in range (0,num_values):
 		theta=sc+i*dtheta
 		term1=0.5*(cr-1)
 		term2=R+1-math.cos(theta)
 		term3=pow(R,2)-pow(math.sin(theta),2)
 		term3=pow(term3,0.5)

 		v.append((1+term1*(term2-term3))*v_c)
 	return v 

bore=0.1
stroke=0.1
con_rod=0.15
cr=12

p1=101325
t1=500
gamma=1.4
t3=2300

v_s=(math.pi/4)*pow(bore,2)*stroke
v_c=v_s/(cr-1)
v1=v_s+v_c

v2=v_c

p2=p1*pow(v1,gamma)/pow(v2,gamma)
rhs=p1*v1/t1
t2=p2*v2/rhs

v_compression=engine_kinematics(bore,stroke,con_rod,cr,180,0)
constant_comp=p1*pow(v1,gamma)

p_compression=[]
print(v_compression)
for v in v_compression:
	p_compression.append(constant_comp/pow(v,gamma))

v3=v2

rhs=p2*v2/t2
p3=rhs*t3/v3

v_expansion=engine_kinematics(bore,stroke,con_rod,cr,0,180)

constant_exp=p3*pow(v3,gamma)

p_expansion=[]
for v in v_expansion:
	p_expansion.append(constant_exp/pow(v,gamma))

v4=v1
p4=p3*pow(v3,gamma)/pow(v4,gamma)
t4=p4*v4/rhs

thermal_effeciency=[(1-1/pow(cr,gamma-1))*100]

print(thermal_effeciency)

plt.plot([v2,v3],[p2,p3])
plt.plot(v_compression,p_compression)
plt.plot(v_expansion,p_expansion)
plt.plot([v4,v1],[p4,p1])
plt.xlabel(\'volume\')
plt.ylabel(\'pressure\')
plt.show()

Conclusion:

It is concluded that the code was written to display the P-V diagram.

The results obtained are

The curve looks pretty smooth when the num_values are more. So, I had mentioned them 50.

The primary task was to display the P-V diagram.

The thermal efficiency is 62.98928245128466