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)
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)
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)
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)
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)
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)
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)
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)
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)
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)