Chapter9:Carburettors and Fuel Injection in SI Engines

Example 9.1 page no: 279

In [1]:
#given
ma=5                            #Mass flow rate of air per min for a simple jet carburettor in kg/min
mf=0.4                          #Mass flow rate of fuel in kg/min
df=780                          #Density of the fuel in kg/m**3
p1=1.013                        #The initial pressure of air in bar
t1=27                           #The initial temperature of air in degree centigrade
C2=90                           #The air flow velocity in m/s
Cva=0.8                         #The velocity coefficient for the venturi
Cdf=0.6                         #The coefficient of discharge of the main fuel jet 
Cpd=0.75                        #The pressure drop across the fuel metering oriface
Cp=1005                         #The specific heat of gas in J/kgK
g=1.4                           #Adiabatic index
R=287                           #Real gas constant in J/kgK

#Calculations
import math
p2=p1*(1-(C2**2/(Cva**2*2*Cp*(t1+273))))**(g/(g-1))
da1=((p1*10**5)/(R*(t1+273)))
da2=((da1)*(p2/p1)**(1/g))
A2=((ma/60.0)/(da2*C2))*10**4
d2=(4*A2/math.pi)**(1/2.0)
pv=p1-p2
pj=Cpd*pv
Aj=((mf/60.0)/(Cdf*(2*df*pj*10**5)**(1/2.0)))*10**6
dj=(4*Aj/math.pi)**(1/2.0)

#Output
print"The throat diameter of the choke = ",round(d2,3),"cm" 
print"The oriface diameter = ",round(dj,1),"mm" 
The throat diameter of the choke =  3.251 cm
The oriface diameter =  2.2 mm

Example 9.2 page no: 281

In [1]:
#given
Vs=0.002                                 #The swept volume in m**3
nv=75.0                                  #Volumetric efficiency in percent
N=4500.0                                 #The engine rpm
p1=1.013                                 #The initial pressure of air in bar
R=287.0                                  #Real gas constant in J/kgK
t1=15.0                                  #The atmospheric temperature in degree centigrade
Cp=1005.0                                #The specific heat of gas in J/kgK
g=1.4                                    #Adiabatic index
C2=100.0                                 #The air flow velocity at choke in m/s
Cda=0.85                                 #The velocity coefficient for the venturi
Cdf=0.66                                 #The coefficient of discharge of the main fuel jet 
sf=0.75                                  #The specific gravity of fuel
d=0.4                                    #The ratio of the diameter to choke diameter
af=14.0                                  #The air fuel ratio
gf=9.81                                  #The gravitational force constant in m/s**2
Z=0.006                                  #The petrol surface below the choke in m
df=750.0                                 #The density of the fuel in kg/m**3

#Calculations
import math
Va=((nv/100.0)*Vs*N)/(2.0*60.0)
V1=Va/2.0
ma=(p1*10**5*V1)/(R*(t1+273))
p2=p1*(1-(C2**2/(2*Cp*(t1+273))))**(g/(g-1))
da1=((p1*10**5)/(R*(t1+273)))
da2=da1*(p2/p1)**(1/g)
A2=(ma/(da2*C2*Cda))*10**6
D=((A2*4)/(math.pi*0.84))**(1/2.0)
mf=ma/af
pm=(p1-p2-(gf*Z*df/10.0**5))*10**5
Aj=(mf/(Cdf*(2*df*pm)**(1/2.0)))*10**6
dj=(4*Aj/math.pi)**(1/2.0)

#Output
print"The diameter of the choke = ",round(D,2),"mm" 
print"The diameter of the jet in = ",round(dj,2),"mm" 
The diameter of the choke =  22.89 mm
The diameter of the jet in =  1.26 mm

Example 9.3 page no: 283

In [2]:
#given
d=0.08                              #The diameter of the bore in m
L=0.09                              #The length of the stroke in m
N=4000.0                            #The engine rpm
C=84.0                              #The carbon content in the fuel by mass in percent
H=16.0                              #The hydrogen content in the fuel by mass in percent
nv=80.0                             #The volumetric efficiency of the engine in percent
p1=1.0                              #The pressure at ambient condition in bar
t1=25.0                             #The temperature at ambient condition in degree centigrade
p=0.06                              #The depression at venturi throat in bar
ma=0.95                             #The actuat quantity of air supplied
Ra=287.0                            #Real gas constant in J/kgK
Rf=98.0                             #Real gas constant in J/kgK
n=4.0                               #Number of cylinders
Cp=1005.0                           #The specific heat of gas in J/kgK
g=1.4                               #Adiabatic index

#Calculations
import math
V=(math.pi/4.0)*d**2*L*(nv/100.0)*(N/(2.0*60.0))*n
Af=(100/23.0)*((C*(32/12.0))+(H*8))/100.0
mfa=Af*ma
Aaf=mfa
da=(p1*10**5)/(Ra*(t1+273))
dv=(p1*10**5)/(Rf*(t1+273))
ma1=V/((1/da)+(1/(mfa*dv)))
mf1=ma1/mfa
p2=p1-p
C2=(2*Cp*(t1+273)*(1-(p2/p1)**((g-1)/g)))**(1/2.0)
T2=(t1+273)*(p2/p1)**((g-1)/g)
d2=(p2*10**5)/(Ra*T2)
A2=(ma1/(d2*C2))*10**4
d2=(A2*4/math.pi)**(1/2.0)

#Output
print"The fuel flow rate = ",round(mf1,5),"kg/s" 
print"The velocity of air at throat = ",round(C2,1),"m/s" 
print"The throat diameter = ",round(d2*10,2),"mm" 
The fuel flow rate =  0.00379 kg/s
The velocity of air at throat =  102.5 m/s
The throat diameter =  24.75 mm

Example 9.4 page no: 285

In [3]:
#given
d=0.1                                     #The diameter of the bore in m
L=0.12                                    #The length of the stroke in m
N=3000                                    #The engine rpm
d2=0.035                                  #The throat diameter of carburettor venturi in m
nv=80                                     #The volumetric efficiency of the engine in percent
Cda=0.82                                  #The coefficient of discharge of air flow 
p=1.013                                   #The ambient pressure in bar
T=298                                     #The ambient temperature in K
ar=15                                     #The air fuel ratio 
Z=0.005                                   #The top of the jet above the petrol level in the float chamber in m
Cdf=0.7                                   #The coefficient of discharge for fuel flow 
df=750                                    #The specific gravity of the fuel in kg/m**3
R=287                                     #Real gas constant in J/kgK
g=9.81                                    #The gravitational constant in m/s**2
n=4                                       #Number of cylinders 

#Calculations
import math
V=(math.pi/4.0)*d**2*L*(nv/100.0)*(N/(2.0*60.0))*n
da=(p*10**5)/(R*T)
ma=V*da
A2=(math.pi/4.0)*d2**2
P=(ma**2/(Cda**2*A2**2*2*da))/10.0**5
mf=ma/ar
Aj=(mf/(Cdf*(2*df*((P*10**5)-(g*Z*df)))**(1/2.0)))*10**6
dj=(Aj*4/math.pi)**(1/2.0)

#Output 
print"The depression of the throat = ",round(P,3),"bar" 
print"The diameter of the fuel jet of a simple carburettor = ",round(dj,3),"mm" 
The depression of the throat =  0.054 bar
The diameter of the fuel jet of a simple carburettor =  1.953 mm

Example 9.5 page no:286

In [4]:
#given
mf=(6/3600.0)                               #The mass flow rate of fuel in kg/s
df=750                                      #The density of fuel in kg/m**3
Z=0.003                                     #The level in the float chamber below the top of the jet in m
p=1.013                                     #The ambient pressure in bar
T=294                                       #The ambient temperature in K
dj=0.0012                                   #The jet diameter in m
Cdf=0.65                                    #The discharge coefficient of the jet 
Cda=0.8                                     #The discharge coefficient of air 
A=15.3                                      #The air fuel ratio 
g=9.81                                      #The gravitational constant in m/s**2
R=287                                       #Real gas constant in J/kgK
dh=1000                                     #The density of water in kg/m**2

#Calculations
import math
da=(p*10**5)/(R*T)
Ca2=Cda*((2*g*Z*df)/da)**(1/2.0)
Aj=(math.pi/4.0)*dj**2
P=((mf**2/(Cdf**2*Aj**2*2*df))+(g*Z*df))/10.0**5
h=(P*10**5)/(dh*g)
h1=(P*10**5)/g
ma=mf*A
A2=(ma/((Cda*(2*da*(P*10**5))**(1/2.0))))*10**4
d2=((A2*4/math.pi)**(1/2.0))*10

#Output 
print"The critical air velocity = ",round(Ca2,1),"m/s" 
print"The depression at the throat = ",round(h1,1),"mm of H2O" 
print"The effective throat diameter  ",round(d2,2),"mm"
The critical air velocity =  4.9 m/s
The depression at the throat =  351.6 mm of H2O
The effective throat diameter   21.12 mm

Example 9.6 page no: 287

In [22]:
#given
d2=22                            #The venturi throat diameter of a simple carburettor in mm
Cda=0.82                         #The coefficient of air flow 
dj=1.2                           #The fuel orifice diameter in mm
Cdf=0.7                          #The coefficient of fuel flow
Z=0.004                          #The petrol surface below the throat in m
g=9.81                           #The gravitational constant in m/s**2
da=1.2                           #The density of air in kg/m**3
df=750                           #The density of fuel in kg/m**3
P=0.075                          #The pressure drop in bar

#Calculations
A=(Cda/Cdf)*(d2**2/dj**2)*(da/df)**(1/2.0)
A1=(Cda/Cdf)*(d2**2/dj**2)*((da*P)/(df*(P-(g*Z*df)/10.0**5)))**(1/2.0) 
Ca2=(2*g*Z*df/da)**(1/2.0)

#Output
print" (a) The air fuel ratio when the nozzle lip is neglected = ",round(A,2)
print"(b)The air fuel ratio when the nozzle lip is considered = ",round(A1,3)
print"(c) The critical air velocity or minimum velocity required to start the fuel flow = ",round(Ca2,1),"m/s" 
 (a) The air fuel ratio when the nozzle lip is neglected =  15.75
(b)The air fuel ratio when the nozzle lip is considered =  15.78
(c) The critical air velocity or minimum velocity required to start the fuel flow =  7.0 m/s

Example 9.7 page no: 289

In [23]:
#given
h=4000                                   #The altitude of the airplane engine carburettor in m
A=14.7                                   #The air fuel ratio at sea level
ts=22                                    #The temperature at sea level in degree centigrade
R=287                                    #Real gas constant in J/kgK

#Calculations
ta=ts-(0.0065*h)
p=1.013/10.0**0.2083
da=(p*10**5)/(R*(ta+273))
ds=(1.013*10**5)/(R*(ts+273))
Aa=A*(da/ds)**(1/2.0)

#Output
print"The air fuel ratio at altitude = ",round(Aa,2)
The air fuel ratio at altitude =  12.11

Example 9.8 page no: 289

In [17]:
#given
A=14.5                                       #The air fuel ratio
p2=0.825                                     #The pressure at the venturi throat in bar 
p1=1.013                                     #The atmospheric pressure in bar
pd=37.5                                      #The pressure drop to the air cleaner in mm of Hg
ma=260                                       #The mass flow rate of air in kg/h

#Calculations
pa=p1-p2
p21=p1-(pd/750)-pa
pf=pa
pf1=p1-p21
Af=A*(pf/pf1)**(1/2.0)

#Output
print"(a) The throat pressure when the air cleaner is fitted = ",p21,"bar" 
print"(b) The air fuel ratio with the air cleaner fitted = ",round(Af,3)
(a) The throat pressure when the air cleaner is fitted =  0.775 bar
(b) The air fuel ratio with the air cleaner fitted =  12.887

Example 9.9 page no: 291

In [1]:
#given
bp=8                                    #The brake power of the petrol engine in kW
nb=30                                   #The brake thermal efficiency in percent
CV=44000                                #The calorific value of the fuel in kJ/kg
p1=1.013                                #The suction condition of engine pressure in bar
T1=300                                  #The temperature at suction condition in K
Aj=2.5*10**-6                           #The area of jet in m**2
Z=0.008                                 #The nozzle lip in m
g=9.81                                  #The gravitational force constant in m/s**2
A=15                                    #The air fuel ratio
Cda=0.9                                 #The coefficient of air flow
Cdf=0.7                                 #The coefficient of fuel flow
df=750                                  #The density of fuel in kg/m**3
va=0.8                                  #The specific volume of air in m**3/kg

#Calculations
import math
va1=va*T1/273.0
da=1/va
mf=bp/((nb/100.0)*CV)
Cf=mf/(Cdf*df*Aj)
P=((Cf**2*df)/2.0)+(df*g*Z)
Ca=(2*P/da)**(1/2.0)
ma=mf*A
A2=(ma/(Cda*da*Ca))*10**4
d2=(A2*4/math.pi)**(1/2.0)

#Output
print"The venturi throat diameter of the carburator = ",round(d2,2),"cm" 
The venturi throat diameter of the carburator =  2.63 cm