# Chapter 19 : Two Stroke Engines¶

## Example 19.3 Page no : 334¶

In :
import math

#Input data
d = 11.25					#Bore in cm
l = 15.					#Stroke in cm
r = 7.					#Compression ratio
N = 1800.					#Speed in r.p.m
a = 4.5					#Air supply in kg/min
Ta = 72.+273					#Temperature of air in K
af = 14.3					#Air fuel ratio
ep = 1.					#Exhaust pressure in kg/cm**2
R = 29.27					#Characteristic gas constant in kg.m/kg.degree C

#Calculations
Vc = ((r/(r-1))*(3.14/4)*(d/100)**2*(l/100))					#Swept volume in m**3
Wa = (Vc*N*ep*10**4)/(R*Ta)					#Ideal air capacity in kg/min
sr = (a/Wa)					#Scavenging ratio
sn = (1-math.exp(-sr))					#Scavenging efficiency
nt = (sn/sr)					#Trapping efficiency

#Output
print 'a) Ideal air capacity is %3.2f kg/min  \
\nb) Scavenging ratio is %3.2f  \
\nc) Scavenging efficiency is %3.3f  \
\nd) Trapping efficiency is %3.2f'%(Wa,sr,sn,nt)

a) Ideal air capacity is 3.10 kg/min
b) Scavenging ratio is 1.45
c) Scavenging efficiency is 0.766
d) Trapping efficiency is 0.53