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)
```

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)
```

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)
```

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
```

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)
```