Chapter 11 : Carburetion¶

Example 11.1 Page No : 243¶

In :
from scipy.optimize import fsolve
import math

#Given:
m_a  =  5.      #Mass of air in kg/min
P1  =  1.013      #Pressure of air in bar
T1  =  27.+273      #Temperature of air in K
C1  =  0
C2  =  90.      #Flow velocity at opening and throat in m/s
Cv  =  0.8      #Velocity coefficient
cp  =  1.005      #Specific heat at consmath.tant pressure in kJ/kgK
g  =  1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 11.32
#Defining, y as a function of P2
#This function is the difference of C2 actual and C2 given
def  f(P2):
C2_act  =  0.8*math.sqrt(2*cp*1000*T1*(1-(P2/P1)**((g-1)/g)))
return  C2_act-C2

#The function f is solve for zero to get the value of P2
P2  =  fsolve(f,1)
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
rho1  =  P1*100/(R*T1)      #Mass density at opening in kg/m**3
rho2  =  rho1*(P2/P1)**(1/g)      #Mass density at throat in kg/m**3
A2  =  m_a/(60*rho2*C2)      #Cross section area at throat in m**2
d2  =  math.sqrt(4*A2/math.pi)      #Diameter of cross section in m

#Results:
print " The throat diameter of the choke, d2   =   %.2f cm"%(d2*100)
The throat diameter of the choke, d2   =   3.25 cm

Example 11.2 Page No : 248¶

In :
from scipy.optimize import fsolve
import math

#Given:
m_a  =  6.
m_f  =  .45      #Mass of air and fuel in kg/min
rho_f  =  740.      #Density of fuel in kg/m**3
P1  =  1.013      #Pressure of air in bar
T1  =  27.+273      #Temperature of air in K
C2  =  92.      #Flow velocity at throat in m/s
Cv  =  0.8      #Velocity coefficient
Cd_f  =  0.60      #Coefficient of discharge of fuel
cp  =  1.005      #Specific heat at consmath.tant pressure in kJ/kgK
g  =  1.4      #Specific heat ratio(gamma)

#Solution:
#Defining, y as a function of P2
#This function is the difference of C2 actual and C2 given
def  f(P2):
C2_act  =  Cv*math.sqrt(2*cp*10**3*T1*(1-(P2/P1)**((g-1)/g)))
return  C2_act-C2

#The function f is solve for zero to get the value of P2
P2  =  fsolve(f,1)
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
rho1  =  P1*100/(R*T1)      #Mass density at opening in kg/m**3
rho2  =  rho1*(P2/P1)**(1/g)      #Mass density at throat in kg/m**3
A2  =  m_a/(60*rho2*C2)      #Cross section area at throat in m**2
d2  =  math.sqrt(4*A2/math.pi)      #Diameter of cross section in m
deltaP_v  =  P1-P2      #Pressure drop at venturi in bar
deltaP_f  =  0.75*deltaP_v      #Given, Pressure drop at fuel metering orifice in bar
A_f  =  m_f/(60*Cd_f*math.sqrt(2*rho_f*deltaP_f*10**5))      #Area of cross section of fuel nozzle in m**2
d_f  =  math.sqrt(4*A_f/math.pi)      #Diameter of cross section of fuel nozzle in m

#Results:
print " The throat diameter of the choke, d2   =   %.3f cm"%(d2*100)
print " The orifice diameter, d_f   =   %.2f mm"%(d_f*1000)
The throat diameter of the choke, d2   =   3.526 cm
The orifice diameter, d_f   =   2.34 mm

Example 11.3 Page No : 253¶

In :
import math

#Given:
d  =  10.
l  =  12.      #Bore and stroke in cm
n  =  4.      #Number of cylinders
N  =  2000.      #Speed of engine in rpm
d2  =  3.      #Diameter of throat in cm
eta_vol  =  70.      #Volumetric efficiency
rho_a  =  1.2      #Density of air in kg/m**3
Cd_a  =  0.8      #Coefficient of discharge for air

#Solution:
V_s  =  (math.pi/4)*d**2*l*n*10**-6      #Swept volume of engine in m**3
V_act  =  eta_vol*V_s*N/(2*100*60)      #Actual volume sucked in m**3/s
m_a  =  V_act*rho_a      #Mass of air sucked in kg/s
deltaP_v  =  (m_a*4/(Cd_a*math.pi*d2**2*10**-4))**2/(2*rho_a)      #Pressure drop at venturi in pascal

#Results:
print " The suction at the throat   =   %.4f bar"%(deltaP_v/10**5)
#Answer in the book is wrong
The suction at the throat   =   0.0363 bar

Example 11.4 Page No : 258¶

In :
#Given:
m_f  =  7.5      #Mass of fuel in kg/hr
s  =  0.75      #Specific gravity of the fuel
T1  =  25.+273      #Temperature of air in K
A_F  =  15.      #Air fuel ratio
d  =  22.      #Diameter of choke tube in mm
z  =  4.      #Elevation of the jet in mm
Cd_a  =  0.82
Cd_f  =  0.7      #Coefficient of discharge for air and fuel
P1  =  1.013      #Pressure of air in bar
g  =  9.81      #Accelaration due to gravity in m/s**2

#Solution:
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
rho_a  =  P1*100/(R*T1)      #Mass density of air in kg/m**3
rho_f  =  s*1000      #Mass density of fuel in kg/m**3
m_a  =  A_F*m_f/3600      #Mass of air in kg/s
deltaP_v  =  (m_a*4/(Cd_a*math.pi*d**2*10**-6))**2/(2*rho_a)      #Pressure drop at venturi in pascal
A_f  =  m_f/(3600*Cd_f*math.sqrt(2*rho_f*(deltaP_v-z*10**-3*g*rho_f)))      #Area of cross section of fuel nozzle in m**2
d_f  =  math.sqrt(4*A_f/math.pi)      #Diameter of cross section of fuel nozzle in m

#Results:
print " The diameter of the fuel jet of a simple carburettor, d_f   =   %.3f mm"%(d_f*1000)
#Answer in the book is wrong
The diameter of the fuel jet of a simple carburettor, d_f   =   1.228 mm

Example 11.5 Page No : 263¶

In :
from scipy.optimize import fsolve
import math
from sympy import Symbol,solve

#Given:
V_s  =  1489.      #Capacity of the engine in cc
N  =  4200.      #Speed of the engine in rpm
eta_v  =  70.      #Volumetric efficiency
A_F  =  13.      #Air fuel ratio
C2  =  90.      #Flow velocity at throat in m/s
Cd_a  =  0.85
Cd_f  =  0.66      #Coefficient of discharge for air and fuel
s  =  0.74      #Specific gravity of the fuel
z  =  6.      #Elevation of the jet in mm
P1  =  1.013      #Pressure of air in bar
T1  =  27.+273      #Temperature of air in K
g  =  1.4      #Specific heat ratio(gamma)
cp  =  1.005      #Specific heat at consmath.tant pressure in kJ/kgK

#Solution:
V_s  =  V_s*10**-6      #Swept volume in m**3
V_act  =  eta_v*V_s*N/(2*100*60)      #Actual volume sucked in m**3/s
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
m_a  =  P1*10**5*V_act/(R*10**3*T1)      #Mass of air sucked in kg/s
#Defining, y as a function of P2
#This function is the difference of C2 actual and C2 given
def  f(P2):
C2_act  =  math.sqrt(2*cp*10**3*T1*(1-(P2/P1)**((g-1)/g)))
return  C2_act-C2

#The function f is solve for zero to get the value of P2
P2  =  fsolve(f,1)
V2  =  V_act*(P1/P2)**(1/g)      #Volume at throat in m**3/s
A2  =  V2/(C2*Cd_a)      #Cross section area at throat in m**2
d2  =  Symbol('d2')      #Defining the diameter of choke as unknown in m
d_f  =  d2/2.5      #Given
d2  =  solve(math.pi/4*(d2**2-d_f**2)-A2)      #Diameter of choke in m
m_f  =  m_a/A_F      #Mass of fuel sucked in kg/s
A_f  =  m_f/(Cd_f*math.sqrt(2*s*1000*(P1*10**5-P2*10**5-z*10**-3*9.81*s*1000)))      #Area of cross section of fuel nozzle in m**2
d_f  =  math.sqrt(4*A_f/math.pi)      #Diameter of cross section of fuel nozzle in m

#Results:
print " The diameter of the fuel jet of a simple carburettor, D_jet   =   %.2f mm"%(d_f*1000)
The diameter of the fuel jet of a simple carburettor, D_jet   =   1.56 mm

Example 11.6 Page No : 268¶

In :
from scipy.optimize import fsolve
from sympy import Symbol, solve
import math

#Given:
V_s  =  1710.      #Capacity of the engine in cc
N  =  5400.      #Speed of the engine in rpm
eta_vol  =  70.      #Volumetric efficiency
n  =  2.      #Number of carburettor
A_F  =  13.      #Air fuel ratio
C2  =  107.      #Flow velocity at throat in m/s
Cd_a  =  0.85
Cd_f  =  0.66      #Coefficient of discharge for air and fuel
s  =  0.75      #Specific gravity of the fuel
z  =  6.      #Elevation of the jet in mm
P1  =  1.013      #Pressure of air in bar
T1  =  27.+273      #Temperature of air in K
g  =  1.4      #Specific heat ratio(gamma)
cp  =  1.005      #Specific heat at consmath.tant pressure in kJ/kgK

#Solution:
V_s  =  V_s*10**-6      #Swept volume in m**3
V_act  =  eta_vol*V_s*N/(2*100*60)      #Actual volume sucked in m**3/s
V_act  =  V_act/n      #Actual volume of air sucked through each carburettor in m**3/s
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
m_a  =  P1*10**5*V_act/(R*10**3*T1)      #Mass of air sucked in kg/s
#Defining, y as a function of P2
#This function is the difference of C2 actual and C2 given
def  f(P2):
C2_act  =  math.sqrt(2*cp*10**3*T1*(1-(P2/P1)**((g-1)/g)))
return  C2_act-C2

#The function f is solve for zero to get the value of P2
P2  =  fsolve(f,1)
V2  =  V_act*(P1/P2)**(1/g)      #Volume at throat in m**3/s
A2  =  V2/(C2*Cd_a)      #Cross section area at throat in m**2
d2  =  Symbol('d2')      #Defining the diameter of choke as unknown in m
d_f  =  d2/2.5      #Given
d2  =  solve(math.pi/4*(d2**2-d_f**2)-A2)      #Diameter of choke in m
m_f  =  m_a/A_F      #Mass of fuel sucked in kg/s
A_f  =  m_f/(Cd_f*math.sqrt(2*s*1000*(P1*10**5-P2*10**5-z*10**-3*9.81*s*1000)))      #Area of cross section of fuel nozzle in m**2
d_f  =  math.sqrt(4*A_f/math.pi)      #Diameter of cross section of fuel nozzle in m

#Results:
print " The diameter of the choke tube, D   =   %.2f cm"%(d2*100)
print " The diameter of the fuel jet of a simple carburettor, D_f   =   %.2f mm"%(d_f*1000)
The diameter of the choke tube, D   =   -2.17 cm
The diameter of the fuel jet of a simple carburettor, D_f   =   1.23 mm

Example 11.7 Page No : 273¶

In :
from scipy.optimize import fsolve
import math

#Given:
ha  =  5000.      #Altitude in m
A_F  =  14.      #Air fuel ratio at sea level
P1  =  1.013      #Pressure of air in bar at sea level
T1  =  27.+273      #Temperature of air in K at sea level
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
def  f1(h):
return  ts-0.0065*h      #Temperature(t in degreeC) as a function of altitude(h in m)
def  f2(P):
return  19200*math.log10(1.013/P)      #Altitude(h in m) as a function of pressure(P in bar)

#Solution:
ts  =  T1-273      #Sea level temperature in degreeC
T2  =  f1(ha)      #Temperature at altitude(ha   =   5000 m) in degreeC
T2  =  T2+273      #in K
#Defining, y as a function of P
#This function is the difference of function(f2) and ha given
def  f(P):
return  f2(P)-ha

#The function f is solve for zero to get the value of P2
P2  =  fsolve(f,1)
rho_a  =  P2/(R*T2)      #Density of air at altitude in kg/m**3
rho_s  =  P1/(R*T1)      #Density of air at sea level in kg/m**3
A_F_a  =  A_F*math.sqrt(rho_a/rho_s)      #Air fuel ratio at altitude

#Results:
print " The air fuel ratio supplied at 5000 m altitude by a carburettor   =   %.2f"%(A_F_a)
The air fuel ratio supplied at 5000 m altitude by a carburettor   =   10.99

Example 11.8 Page No : 278¶

In :
import math

#Given:
d2  =  20.
d_f  =  1.25      #Diameter of throat and fuel nozzle in mm
Cd_a  =  0.85
Cd_f  =  0.66      #Coefficient of discharge for air and fuel
z  =  5.      #Elevation of the jet in mm
rho_a  =  1.2
rho_f  =  750.      #Mass density of air and fuel in kg/m**3
deltaP_a  =  0.07      #Pressure drop of air in bar
g  =  9.806      #Accelaration due to gravity in m/s**2

#Solution:
#(a)Nozzle lip is neglected
A_f  =  (math.pi/4)*d_f**2;A2  =  (math.pi/4)*d2**2      #Area of cross section of fuel nozzle and venturi in mm**2
m_a1  =  Cd_a*A2*math.sqrt(2*rho_a*deltaP_a);m_f1  =  Cd_f*A_f*math.sqrt(2*rho_f*deltaP_a)      #Air flow and fuel flow
A_F1  =  m_a1/m_f1      #Air fuel ratio
#(b)Nozzle lip is taken into account
m_a2  =  m_a1      #Air flow remain same
m_f2  =  Cd_f*A_f*math.sqrt(2*rho_f*(deltaP_a-z*10**-3*g*rho_f*10**-5))      #Fuel flow
A_F2  =  m_a2/m_f2      #Air fuel ratio
#(c)Minimum velocity required to start the fuel flow when nozzle lip is provided
#To just start the fuel flow pressure difference in venturi must be equivalent to the nozzle lip pressure
deltaP_a  =  z*10**-3*g*rho_f      #Pressure difference in N/m**2
C2  =  math.sqrt(2*deltaP_a/rho_a)      #Minimum velocity of air at throat in m/s

#Results:
print " The air fuel ratio when the nozzle lip is neglected   =   %.1f"%(A_F1)
print " The air fuel ratio when the nozzle lip is taken into account   =   %.3f"%(A_F2)
print " The minimum velocity required to start the fuel flow when lip is provided   =   %.2f m/s"%(C2)
The air fuel ratio when the nozzle lip is neglected   =   13.2
The air fuel ratio when the nozzle lip is taken into account   =   13.223
The minimum velocity required to start the fuel flow when lip is provided   =   7.83 m/s

Example 11.9 Page No : 283¶

In :
import math
from sympy import Symbol, solve

#Given:
A_F  =  14.      #Air fuel ratio at sea level
P2  =  0.834      #Pressure at venturi throat without an air cleaner in bar
P1  =  1.013      #Pressure of air in bar at sea level
deltaP_ac  =  30.      #Pressure drop to air cleaner in mm of mercury
m_a  =  250.      #Air flow in kg/hr

#Solution:
#No air cleaner
deltaP_a1  =  P1-P2      #Pressure difference at venturi throat in bar
#When air cleaner is fitted
deltaP_ac  =  deltaP_ac/750      #Pressure drop to air cleaner in bar
p  =  Symbol('p')      #Defining pressure at venturi throat with an air cleaner as unknown in bar
deltaP_a2  =  P1-deltaP_ac-p      #Pressure difference at venturi throat in bar
#For same air flow and consmath.tant coefficients pressure difference in above two cases is same
p  =  solve(deltaP_a2-deltaP_a1)      #Pressure at venturi throat with an air cleaner in bar
deltaP_f  =  P1-p      #Pressure difference at venturi throat when air cleaner is fitted in bar
A_F_f  =  A_F*math.sqrt(deltaP_a1/deltaP_f)      #Air fuel ratio when air cleaner is fitted

#Results:
print " a)The throat pressure when the air cleaner is fitted, P   =   %.3f bar"%(p)
print " b)Air fuel ratio with air cleaner is fitted   =   %.2f"%(A_F_f)
a)The throat pressure when the air cleaner is fitted, P   =   0.794 bar
b)Air fuel ratio with air cleaner is fitted   =   12.66