# Chapter11:Two Stroke Engines¶

## Example 11.1 page no:367¶

In :
#Input data
nsc=75       #The scavenging efficiency of the two stroke engine in percent
ns=20        #The scavenging efficiency is increased by in percent

#Calculations
import math
Rsc=math.log(1/(1-(nsc/100.0)))
nsc1=(nsc/100.0)+((nsc/100.0)*(ns/100.0))
Rsc1=math.log(1/(1-(nsc1)))
Rscr=((Rsc1-Rsc)/Rsc)*100      #Percentage increase in scavenging ratio in persent

#Output
print"The percentage change in the scavenging ratio = ",round(Rscr,1),"percent"

The percentage change in the scavenging ratio =  66.1 percent


## Example 11.2 page no:367¶

In :
#Input data
d=0.12      #The bore diameter of the engine in m
l=0.15      #The stroke length of the engine in m
r=16.0        #The compression ratio
N=2000.0      #The speed of the engine in rpm
mf=(240/60.0) #Actual air flow per min in kg/min
T=300.0       #Air inlet temperature in K
p=1.025     #Exhaust pressure in bar
R=287       #Real gas constant in J/kg

#Calculations
import math
da=(p*10**5)/(R*T)
Vs=((math.pi)*(d**2)*l)/4.0
V=(r/(r-1))*Vs
m=da*V
m1=m*N
Rsc=mf/m1   #Scavenging ratio
nsc=((1-math.exp(-Rsc))*100)
ntr=((nsc/100.0)/Rsc)*100

#Output
print"(a) The scavenging ratio = ",round(Rsc,2)
print"(b) The scavenging efficiency = ",round(nsc,1),"percent "
print"(c) The trapping efficiency = ",round(ntr,1)," percent"

(a) The scavenging ratio =  0.93
(b) The scavenging efficiency =  60.5 percent
(c) The trapping efficiency =  65.1  percent


## Example 11.3 page no: 368¶

In :
#Input data
mf=6.5          #Mass flow rate of fuel in kg/h
N=3000.0          #The speed of the engine in rpm
a=15            #The air fuel ratio
CV=44000        #The calorific value of the fuel in kJ/kg
pm=9            #The mean piston speed in m/s
pmi=4.8         #The mean pressure in bar
nsc=85          #The scavenging efficiency in percent
nm=80           #The mechanical efficiency in percent
R=290.0           #Real gas constant in J/kgK
p=1.03          #The pressure of the mixture in bar
T=288.0           #The temperature of the mixture in K

#Calculations
import math
ma=a*mf
L=((pm*60)/(2*N))*100
mac=mf+ma
mi=(mac)/(nsc/100.0)
da=(p*10**5)/(R*T)
d=(((mi/da)*(4/math.pi)*(1/(L/100.0))*(1/(60*N)))**(1/2.0))*100
ip=(pmi*10**5*(L/100)*((math.pi/4.0)*(d/100)**2)*N)/(60*1000)
bp=ip*(nm/100.0)
nth=(bp/((mf/3600.0)*CV))*100

#Output
print"The diameter of the bore = ",round(d,2),"cm"
print"The length of the stroke = ",L," cm"
print"The brake power = ",round(bp,2)," kW"
print"The brake thermal efficiency =",round(nth,2)," percent "

The diameter of the bore =  8.83 cm
The length of the stroke =  9.0  cm
The brake power =  10.58  kW
The brake thermal efficiency = 13.32  percent


## Example 11.4 page no: 369¶

In :
#Input data
d=0.08            #The diameter of the bore in m
L=0.1             #The length of the stroke in m
r=8.0               #The compression ratio
o=60.0              #The exhaust port open before BDC in degrees
v=60.0              #The exhaust port closes after BDC in degrees
a=15.0              #Air fuel ratio
T=300.0             #The temperature of the mixture entering into the engine in K
p=1.05            #The pressure in the cylinder at the time of clomath.sing
R=290.0             #Real gas constant in J/kgK
ma=150.0           #Mass flow rate of air in kg/h
N=4000.0            #The speed of the engine in rpm

#Calculations
import math
mf=ma/a
mac=ma+mf
r=(L*100)/2.0
Le=(r+(r*math.sin (math.pi/6.0)))/100.0
Vse=(math.pi*d**2*Le)/4.0
V=(r/(r-1))*Vse
V=0.00043                #Value in book after approximation
da=(p*10**5)/(R*T)
m=V*da
mi=m*60*N
Rsc=mac/mi
nsc=(1-(exp(-Rsc)))*100
ntr=nsc/Rsc

#Output
print"The scavenging ratio = ",round(Rsc,2)
print"The scavenging efficiency =",round(nsc,2)," percent "
print"The trapping efficiency = ",round(ntr,2),"percent"

The scavenging ratio =  1.28
The scavenging efficiency = 72.32  percent
The trapping efficiency =  56.3 percent


## Example 11.5 page no: 371¶

In :
#Input data
d=8.25        #The diameter of the bore in cm
L=11.25       #The length of the stroke in cm
r=8.0           #The compression ratio
N=2500.0        #The speed of the engine in rpm
ip=17.0         #Indicated power in kW
a=0.08        #Fuel air ratio
T=345.0         #Inlet temperature mixture in K
p=1.02        #Exhaust pressure in bar
CV=44000.0      #The calorific value of the fuel in kJ/kg
nth=0.29      #Indicated thermal efficiency
M=114.0         #Molar mass of fuel
R=8314.0        #Universal Gas constant in J/kgK

#Calculations
import math
Vs=(math.pi*d**2*L)/4      #Displacement volume in cm**3
V=(r/(r-1))*Vs            #Total cylinder volume in m**3
ps=((29*p*10**5)/(R*T))*(1/(1+a*(29/M)))         #The density of dry air in kg/m**3
nsc=((ip*1000)/((N/60)*V*10**-6*ps*a*CV*1000*nth))*100     #The scavenging efficiency in percent

#Output
print"The scavenging efficiency = ",round(nsc,2)," percent"

The scavenging efficiency =  57.54  percent


## Example 11.6 page no: 372¶

In :
#given
S=15.0        #The speed of the math.piston in m/s
ps=0.35     #The scavenging pressure in bar
pa=1.03     #Atmospheric pressure in bar
r=18.0        #The compression ratio
t=35.0        #The inlet temperature in degree centigrade
Rsc=0.9     #The scavenging ratio
ta=15.0       #The atmospheric temperature in degree centigrade
nc=0.75     #Compressor efficiency
R=287.0       #Real gas constant in J/kgK
Cp=1005.0     #Specific heat of gas in J/kgK

#Calculations
import math
pi=ps+pa    #The scavenging pressure in bar
Ti=(273+ta)+t    #The inlet temperature in K
pr=pa/math.pi    #The ratio of the pressure for calculations
di=(pi*10**5)/(R*Ti)     #The density in kg/m**3
ai=(g*R*Ti)**(1/2.0)            #The sonic velocity in m/s
C=(Rsc)/(2*((r-1)/r)*(ai/S)*(pi/pa)*((2/(g-1))*(((pr)**(2/g))-((pr)**((g+1)/g))))**(1/2.0))
ds=(pa*10**5)/(R*Ti)     #The density in kg/m**3
mep=(ds*Rsc*Cp*Ti*(((pi/pa)**((g-1)/g))-1))/((nc*((r-1)/r))*10**5)      #Mean effective pressure in bar

#Output
print"The flow coefficient = ",round(C,3)
print"The compressor mean effective pressure =  ",round(mep,1),"bar"

The flow coefficient =  0.028
The compressor mean effective pressure =   0.4 bar