# Chapter 12 : Fuel Injection¶

## Example 12.1 Page No : 288¶

In [1]:
#Given:
n  =  6.      #Number of cylinders
bsfc  =  245.      #Brake specific fuel consumption in gm/kWh
bp  =  89.      #Brake power in kW
N  =  2500.      #Engine speed in rpm
s  =  0.84      #Specific gravity of the fuel

#Solution:
m_f  =  bsfc*bp/(1000)      #Fuel consumption in kg/hr
m_f  =  m_f/n      #Fuel consumption per cylinder in kg/hr
m_f  =  (m_f/3600)/(N/(2*60))      #Fuel consumption per cycle in kg
V_f  =  m_f*1000/s      #Volume of fuel consume per cycle in cc

#Results:
print " The quantity of fuel to be injected per cycle per cylinder, V_f   =   %.4f cc"%(V_f)

 The quantity of fuel to be injected per cycle per cylinder, V_f   =   0.0577 cc


## Example 12.2 Page No : 293¶

In [2]:
import math

#Given:
n  =  8.      #Number of cylinders
bp  =  368.      #Brake power in kW
N  =  800.      #Engine speed in rpm
bsfc  =  0.238      #Brake specific fuel consumption in kg/kWh
P1  =  35.
P2  =  60.      #Beginning pressure and maximum pressure in cylinder in bar
P1_i  =  210.
P2_i  =  600.      #Expected pressure and maximum pressure at injector in bar
theta_i  =  12.      #Period of injection in degrees
Cd  =  0.6      #Coefficient of discharge for the injector
s  =  0.85      #Specific gravity of the fuel
P_atm  =  1.013      #Atmospheric pressure in bar

#Solution:
m_f  =  bsfc*bp/(n*60)      #Fuel consumption per cylinder in kg/min
m_f  =  m_f/(N/2)      #Fuel consumption per cycle in kg
t  =  theta_i/360*60/N      #Time for injection in s
m_f  =  m_f/t      #Fuel consumption per cycle in kg/s
deltaP1  =  P1_i-P1      #Pressure difference at beginning in bar
deltaP2  =  P2_i-P2      #Pressure difference at end in bar
deltaP_av  =  (deltaP1+deltaP2)/2      #Average pressure difference in bar
A_f  =  m_f/(Cd*math.sqrt(2*s*1000*deltaP_av*10**5))      #Orifice area of fuel injector in m**2

#Results:
print " The Orifice area of fuel injector, Af   =   %.5f cm**2"%(A_f*10000)

 The Orifice area of fuel injector, Af   =   0.01234 cm**2


## Example 12.3 Page No : 298¶

In [3]:
import math

#Given:
bp  =  15.      #Brake power per cylinder in kW
N  =  2000.      #Engine speed in rpm
bsfc  =  0.272      #Brake specific fuel consumption in kg/kWh
API  =  32.      #American Petroleum Institute specific gravity in degreeAPI
theta_i  =  30.      #Period of injection in degrees
P_i  =  120.      #Fuel injection pressure in bar
P_c  =  30.      #Combustion chamber pressure in bar
Cd  =  0.9      #Coefficient of discharge for the injector
def  specificgravity(API):
return  141.5/(131.5+API)      #Specific gravity(rho) as a function of API

#Solution:
s  =  specificgravity(API)      #Specific gravity of fuel
m_f  =  bsfc*bp*2/(N*60)      #Fuel consumption in kg
t  =  theta_i/360*60/N      #Time for injection in s
m_f  =  m_f/t      #Fuel consumption per cycle in kg/s
A_f  =  m_f/(Cd*math.sqrt(2*s*1000*(P_i-P_c)*10**5))      #Orifice area of fuel injector in m**2
A_f  =  A_f*10**6      #Orifice area of fuel injector in mm**2
d_f  =  math.sqrt(4*A_f/math.pi)      #Diameter of fuel orifice in mm

#Results:
print " The diameter of the fuel orifice, d   =   %.2f mm"%(d_f)

 The diameter of the fuel orifice, d   =   0.56 mm


## Example 12.4 Page No : 303¶

In [4]:
import math

#Given:
s1  =  20.      #Dismath.tance of penetration in cm
t1  =  16.      #Penetration time in millisec
P_i1  =  140.      #Injection pressure in bar
s2  =  s1      #Same dismath.tance of penetration in cm
P_i2  =  220.      #Injection pressure in bar
P_c  =  15.      #Combustion chamber pressure in bar

#Solution:
deltaP1  =  P_i1-P_c      #Pressure difference for 140 bar injection pressure
deltaP2  =  P_i2-P_c      #Pressure difference for 220 bar injection pressure
t2  =  t1*(s2/s1)*math.sqrt(deltaP1/deltaP2)      #Penetration time for 220 bar injection pressure in millisec

#Results:
print " Penetration time for 220 bar injection pressure, t2   =   %.1f milli-seconds"%(t2)
#Answer in the book is wrong

 Penetration time for 220 bar injection pressure, t2   =   12.5 milli-seconds


## Example 12.5 Page No : 308¶

In [5]:
import math

#Given:
V_b  =  7.      #Volume of fuel in the barrel in cc
D_l  =  3.
L_l  =  700.      #Diameter and length of fuel delivery line in mm
V_iv  =  3.      #Volume of fuel in the injection valve in cc
P2  =  200.      #Delivery pressure in bar
P1  =  1.      #Sump pressure in bar
V_d  =  0.15      #Volume to be delivered in cc
C  =  78.8e-6      #Coefficient of compressibility
d  =  8.      #Diameter of the plunger in mm

#Solution:
V_l  =  math.pi/4*D_l**2*L_l*10**-3      #Volume of fuel in delivery line in cc
V1  =  V_b+V_l+V_iv      #Total initial fuel volume in cc
deltaV  =  C*(P2-P1)*V1      #Change in volume due to compression in cc
V_p  =  deltaV+V_d      #Displaced volume by plunger in cc
A_p  =  math.pi/4*d**2*10**-2      #Area of the plunger in cm**2
l  =  V_p/A_p      #Effective stroke of plunger in cm

#Results:
print " The plunger print lacement   =   %.3f cc"%(V_p)
print " The effective stroke of the plunger, l   =   %.2f mm"%(l*10)

 The plunger print lacement   =   0.384 cc
The effective stroke of the plunger, l   =   7.65 mm