# Chapter 15 : Fuel Injection¶

## Example 15.1 Page no : 274¶

In [1]:
import math

#Input data
n = 6.						#Number of cylinders
p = 720.					#Horse power in h.p
N = 180.					#Speed in r.p.m
f = 250.					#Fuel rate in gm per horse power hour

#Calculations
w = (((f/1000)*p)/((N/2)*60*n))*1000					#Weight of fuel per cycle in gm/cycle

#Outptut
print 'The quantity of fuel to be injected per cylinder is %3.2f gm/cycle'%(w)

The quantity of fuel to be injected per cylinder is 5.56 gm/cycle


## Example 15.2 Page no : 278¶

In [2]:
import math

#Input data
n = 4					#Number of cylinders
fc = 0.215					#Brake specific fuel consumption in kg/B.H.P hour
BHP = 400					#Brake horse power in B.H.P
N = 250					#Speed in r.p.m
sg = 0.9					#Specific gravity

#Calculations
Fc = (fc*BHP)					#Fuel consumption per hour in kg/hr
Fcy = (Fc/n)					#Fuel consumption per cylinder in kg/hr
Fcyc = ((Fcy/(60*(N/2)))/(sg*1000))*10**6					#Fuel consumption per cycle in kg. In textbook it is given wrong as 0.0287 instead of 3.185

#Output
print 'The quantity of fuel to be injected per cycle per cylinder is %3.3f c.c'%(Fcyc)

The quantity of fuel to be injected per cycle per cylinder is 3.185 c.c


## Example 15.3 Page no : 279¶

In [3]:
import math

#Input data
n = 4.					#Number of cylinders
p = 450.					#Brake Horse power in B.H.P
N = 200.					#Speed in r.p.m
f = 0.2					#Fuel rate in kg per horse power hour
g = 0.9					#Specific gravity of fuel

#Output
Fc = (p*f)					#Fuel consumption per hour in kg/hr
Fcy = (Fc/n)					#Fuel consumption per cylinder in kg/hr
Fcyc = (Fcy/(60*(N/2)))					#Fuel consumption per cycle in kg
q = (Fcyc/(g*1000))*10**6					#Quantity of fuel injected per cylinder per cycle in c.c

#Output
print 'The quantity of fuel to be injected per cycle per cylinder is %3.3f c.c'%(q)

The quantity of fuel to be injected per cycle per cylinder is 4.167 c.c


## Example 15.4 Page no : 284¶

In [4]:
import math

#Input data
#Data from problem 1
n = 6.					#Number of cylinders
p = 720.					#Horse power in h.p
N = 180.					#Speed in r.p.m
f = 250.					#Fuel rate in gm per horse power hour

Vo = 20.					#Volume of oil in the suction chamber in c.c
dp = 80.					#Discharge pressure in kg/cm**2
voi = 6.					#Volume of oil in the injector in c.c
g = 0.9					#Specific gravity of oil
b = 78.8*10**-6					#Coefficient of compressibility in cm**2/kg when pressure is taken as atmospheric

#Calculations
w = (((f/1000)*p)/((N/2)*60*n))*1000					#Weight of fuel per cycle in gm/cycle
Va = (w/g)					#Volume of air per cycle in c.c
V1 = (Vo+Va)					#Initial volume in c.c
dV12 = (b*V1*dp)					#Change in volume in c.c
#Assuming in accordance with average practice that s = 2d, nv = 0.94 and full load in this pump type x = 0.5
d = ((voi+dV12)/((3.14/4)*2*0.94*0.5))**(1./3)					#Diameter in cm
l = (2*d)					#Stroke in cm

#Output
print 'The diameter of the pump is %3.2f cm  \
\nThe total stroke is %3.2f cm'%(d,l)

The diameter of the pump is 2.03 cm
The total stroke is 4.06 cm


## Example 15.5 Page no : 287¶

In [5]:
import math

#Input data
p = 110.					#Oil pressure in kg/cm**2
pc = 25.					#Pressure in the combustion chamber in kg/cm**2
q = 0.805					#Velocity coefficient. In textbook it is given wrong as 9.805
d = 0.906					#Specific gravity

#Calculations
v = (37.1*q*math.sqrt((p-pc)/d))					#Velocity in m/s

#Output
print 'The velocity of injection is %3.0f m/s'%(v)

The velocity of injection is 289 m/s


## Example 15.6 Page no : 290¶

In [6]:
import math

#Input data
Vf = 6.2					#Volume of fuel in c.c
l = 65					#Length of fuel line in cm
di = 2.5					#Inner diameter in mm
V = 2.75					#Volume of fuel in the injector valve in c.c
Vd = 0.15					#Volume of fuel to be delivered in c.c. In textbook it is given wrong as 0.047
p = 140					#Pressure in kg/cm**2
pp = 1					#Pump pressure in kg/cm**2
patm = 1.03					#Atmospheric pressure in kg/cm**2
b = 78.8*10**-6					#Coefficient of compressibility in cm**2/kg when pressure is taken as atmospheric

#Calculations
V1 = (Vf+(3.14/4)*(di/10)**2*l+V)					#Initial volume in c.c
dV = ((b*V1*(p-pp)/patm))					#Change in volume in c.c
d = (dV+Vd)					#Total print lacement of the plunger in c.c

#Output
print 'The total print lacement of the plunger is %3.3f c.c'%(d)

The total print lacement of the plunger is 0.279 c.c


## Example 15.7 Page no : 292¶

In [7]:
import math

#Input data
Vf = 6.75					#Volume of fuel in c.c
l = 65					#Length of fuel line in cm
di = 2.5					#Inner diameter in mm
V = 2.45					#Volume of fuel in the injector valve in c.c
Vd = 0.15					#Volume of fuel to be delivered in c.c.
p = 150					#Pressure in kg/cm**2
pp = 1					#Pump pressure in kg/cm**2
patm = 1.03					#Atmospheric pressure in kg/cm**2
b = 78.8*10**-6					#Coefficient of compressibility in cm**2/kg when pressure is taken as atmospheric

#Calculations
V1 = (Vf+(3.14/4)*(di/10)**2*l+V)					#Initial volume in c.c
dV = ((b*V1*(p-pp)/patm))					#Change in volume in c.c
d = (dV+Vd)					#Total print lacement of the plunger in c.c

#Output
print 'The total print lacement of the plunger is %3.3f c.c'%(d)

The total print lacement of the plunger is 0.291 c.c


## Example 15.8 Page no : 295¶

In [8]:
import math

#Input data
Vf = 6.75					#Volume of fuel in c.c
l = 65					#Length of fuel line in cm
di = 2.5					#Inner diameter in mm
V = 2.45					#Volume of fuel in the injector valve in c.c
Vd = 0.15					#Volume of fuel to be delivered in c.c.
p = 150					#Pressure in kg/cm**2
pp = 1					#Pump pressure in kg/cm**2
patm = 1.03					#Atmospheric pressure in kg/cm**2
b = 78.8*10**-6					#Coefficient of compressibility in cm**2/kg when pressure is taken as atmospheric
dp = 0.75					#Diameter of the plunger in cm

#Calculations
V1 = (Vf+(3.14/4)*(di/10)**2*l+V)					#Initial volume in c.c
dV = ((b*V1*(p-pp)/patm))					#Change in volume in c.c
d = (dV+Vd)					#Total print lacement of the plunger in c.c
s = ((4/3.14)*(d/dp**2))*10					#Stroke in mm

#Output
print 'The effective plunger stroke is %3.1f mm'%(s)

The effective plunger stroke is 6.6 mm


## Example 15.9 Page no : 298¶

In [10]:
import math

#Input data
n = 6.					#Number of cylinders
p = 300.					#Horse power in H.P
N = 1200.					#Speed in r.p.m
f = 0.2					#Fuel rate in kg per B.H.P hour
ip = 200.					#Injection pressure in kg/cm**2
cp = 40.					#Pressure in the combustion chamber in kg/cm**2
pic = 33.					#Period of injection of the crank angle in degrees
g = 0.83					#Specific gravity of fuel. In textbook, it is given wrong as 0.89
Cd = 0.9					#Coefficient of discharge

#Output
Fc = (p*f)					#Fuel consumption per hour in kg/hr
Fcy = (Fc/n)					#Fuel consumption per cylinder in kg/hr
Fcyc = (Fcy/(60*(N/2)))					#Fuel consumption per cycle in kg
q = (Fcyc/(g*1000))*10**6					#Quantity of fuel injected per cylinder per cycle in c.c
I = ((pic/360.)*(1/N)*60)					#Injection period in sec
df = (g/1000)					#Density of fuel in kg/m**3
v = math.sqrt(2*981*((ip-cp)/df))					#Velocity of fuel through orifice in m/s
A = (q/(Cd*v*I))					#Area of orifice in cm**2
d = math.sqrt(A/(3.14/4))*10					#Diameter in mm

#Output
print 'The diameter of the math.single orifice injector is %3.2f mm'%(d)

The diameter of the math.single orifice injector is 0.73 mm


## Example 15.10 Page no : 302¶

In [11]:
import math

#Input data
n = 6					#Number of cylinders
d = 11.5					#Bore in cm
l = 14					#Stroke in cm
af = 16					#Air fuel ratio
pa = 1.03					#Pressure of air intake in kg/cm**2
Ta = 24+273					#Temperature of air intake in K
nv = 76.5					#Volumetric efficiency in percent
R = 29.27					#Characteristic gas constant in kg.m/kg.K
N = 1500					#Speed in r.p.m
ip = 125					#Injection pressure in kg/cm**2
cp = 40					#Compression pressure in kg/cm**2
q = 18.5					#Fuel injection occupies 18.5 degrees of crenk travel
fsw = 760					#Fuel specific weight in kg/m**2
dc = 0.94					#Orifice discharge coefficient

#Calculations
Vs = ((3.14/4)*d**2*l)					#Stroke volume in c.c
Va = (Vs*(nv/100))					#Volume of air supplied in c.c
wa = ((pa*10**4*Va*10**-6)/(R*Ta))					#Weight of air supplied per cylinder per cycle in kg
wf = (wa/af)					#Weight of fuel injected per cylinder per cycle in kg
I = ((60*q)/(N*360))					#Injection time per cycle in sec
F = (wf/I)					#Fuel injected per cylinder per sec in kg/sec
Af = (F/(dc*math.sqrt(2*9.81*fsw*(ip-cp)*10**4)))					#Area of orifice in sq.m
df = math.sqrt(Af/(3.14/4))*1000					#Diameter of orifice in mm

#Output
print 'Maximum amount of fuel injected per cylinder per sec is %3.2f kg/sec  \
\nDiameter of orifice is %3.3f mm'%(F,df)

Maximum amount of fuel injected per cylinder per sec is 0.04 kg/sec
Diameter of orifice is 0.694 mm
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