Chapter 17 : Supercharging¶

Example 17.1 Page No : 333¶

In :
from math import ceil

#Given:
V_s  =  3000.      #Total swept volume in cc
ip  =  14.      #Indicated power in kW/m**3
N  =  3500.      #Engine speed in rpm
eta_v  =  80.      #Volumetric efficiency in percent
T1  =  27+273.      #Atmospheric temperature in K
P1  =  1.013      #Atmospheric pressure in bar
r_p  =  1.7      #pressure ratio
eta_C  =  75.      #Isentropic efficiency of blower in percent
eta_m  =  80.      #Mechanical efficiency in percent
g  =  1.4      #Specific heat ratio(gamma)

#Solution:
V_s  =  V_s*N/2*1e-6      #Total swept volume in m**3/min
Vi  =  V_s*eta_v/100      #Unsupercharged induced volume in m**3/min
P2  =  P1*r_p      #Blower delivery pressure in bar
T21  =  T1*r_p**((g-1)/g)      #Isentropic temperature at 2 in K
T21  =  ceil(T21)
T2  =  (T21-T1)/(eta_C/100)+T1      #Temperature at 2 in K
V1  =  V_s*(P2/T2)*(T1/P1)      #Volume at atmospheric conditions in m**3/min
Vi_inc  =  V1-Vi      #Increase in induced volume in m**3/min
ip_inc1  =  ip*Vi_inc      #Increased in ip from air induced in kW
ip_inc2  =  (P2-P1)*100*V_s/60      #Increased in ip due to increased induction pressure in kW
ip_inc  =  ip_inc1+ip_inc2      #Total increase in ip in kW
bp_inc  =  eta_m/100*ip_inc      #Total increase in bp in kW
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
cp  =  1.005      #Specific heat in kJ/kgK
m2  =  P2*100*V_s/(R*T2*60)      #Mass of air delivered by the blower in kg/s
Power  =  m2*cp*(T2-T1)/(eta_m/100)      #Power required by the blower in kW
bp_inc  =  bp_inc-Power      #Net increase in brake power in kW

#Results:
print " The net increase in the brake power   =   %.1f kW"%(bp_inc)
The net increase in the brake power   =   27.7 kW

Example 17.2 Page No : 338¶

In :
#Given:
T1  =  10.+273      #Temperature at sea level in K
P1  =  1.013      #Pressure at sea level in bar
bp  =  250.      #Brake power in kW
eta_v  =  78.      #Volumetric efficiency in percent
bsfc  =  0.245      #Brake specific fuel consumption in kg/kWh
A_F  =  17.      #Air fuel ratio
N  =  1500.      #Engine speed in rpm
h  =  2700.      #Altitude in m
P_a  =  0.72      #Pressure at altitude in bar
p  =  8.      #Percentage of gross power taken by the supercharger
T2  =  32.+273      #Temperature of air leaving the supercharger in K

#Solution:
#Unsupercharged
m_f  =  bsfc*bp/60      #Fuel consumption in kg/min
m_a  =  A_F*m_f      #Air consumption in kg/min
m_a  =  m_a/(N/2)      #Air consumption per cycle in kg
m1  =  m_a/eta_v*100      #Mass of air corresponding to swept volume
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
V_s  =  m1*R*T1/(P1*100)      #Swept volume in m**3
bmep  =  bp/(V_s*N/(60*2))      #Brake mean effective pressure in kN/m**2
#Supercharged
bp2  =  bp/(1-p/100)      #Gross power produced by the engine in kW
m_a2  =  bp2/bp*m_a      #Mass of air required per cycle in kg
m2  =  m_a2/eta_v*100      #Mass of air corresponding to swept volume
P2  =  m2*R*T2/(V_s*100)      #Pressure of air leaving the supercharger in bar
deltaP  =  P2-P_a      #Increase in pressure required in bar

#Results:
print " The required engine capacity, V_s   =   %.4f m**3"%(V_s)
print " The anticipated brake mean effective pressure, bmep   =   %.1f bar"%(bmep/100)
print " The increase of air pressure required at the supercharger   =   %.3f bar"%(deltaP)
The required engine capacity, V_s   =   0.0238 m**3
The anticipated brake mean effective pressure, bmep   =   8.4 bar
The increase of air pressure required at the supercharger   =   0.467 bar

Example 17.3 Page No : 343¶

In :
import math
from sympy import Symbol, solve
from numpy import arange

def horner(coeffs, x):
acc = 0
for c in (coeffs):
acc = acc * x + c
return acc

#Given:
V_s  =  3300.      #Swept volume in cc
#For normally aspirated
bmep1  =  9.3      #Brake mean effective pressure in bar
N1  =  4500.      #Engine speed in rpm
eta_it1  =  28.5      #Indicated thermal efficiency in percent
eta_m1  =  90.      #Mechanical efficiency in percent
m1  =  205.      #Mass of unboosted engine in kg
#For supercharged
bmep2  =  12.1      #Brake mean effective pressure in bar
N2  =  4500      #Engine speed in rpm
eta_it2  =  24.8      #Indicated thermal efficiency in percent
eta_m2  =  90      #Mechanical efficiency in percent
m2  =  225      #Mass of boosted engine in kg
h  =  Symbol('h')      #Defining the unknown h hours duration
CV  =  44000      #Calorific value of fuel in kJ/kg

#Solution:
#For normally aspirated
bp1  =  bmep1*100*V_s/1e+6*N1/(2*60)      #Brake power in kW
bp1  =  round(bp1)
ip1  =  bp1/eta_m1*100      #Indicated power in kW
m_f1  =  ip1/(eta_it1/100*CV)      #Fuel flow in kg/s
m_f1  =  m_f1*3600*h      #Mass of fuel flow in h hours in kg
Mass1  =  (m1+m_f1)/bp1      #Specific mass in kg/kW
#For supercharged
bp2  =  bmep2*100*V_s/1e+6*N2/(2*60)      #Brake power in kW
bp2  =  round(bp2)
ip2  =  bp2/eta_m2*100      #Indicated power in kW
m_f2  =  ip2/(eta_it2/100*CV)      #Fuel flow in kg/s
m_f2  =  m_f2*3600*h      #Mass of fuel flow in h hours in kg
Mass2  =  (m2+m_f2)/bp2      #Specific mass in kg/kW
Mass2 = [0.366568,1.5]
Mass1 = [0.31897926, 1.7826]
for h in arange(0,10.01,0.01):      #Defining the range of h(hours)
if (horner(Mass1,h) > horner(Mass2,h)):       #Specific mass of boosted engine is always be less than unboosted
continue
else:
h_max  =  h
break

#Results:
print " The maximum value of h hours duration, h_max   =   %.2f hours"%(h_max)
The maximum value of h hours duration, h_max   =   5.94 hours

Example 17.4 Page No : 348¶

In :
from scipy.optimize import fsolve
import math

#Given:
T1  =  20.+273      #Temperature of air enters the compressor in K
Q1  =  1340.      #Heat added to air in kJ/min
T3  =  60.+273      #Temperature of air leaves the cooler or enters the engine in K
P3  =  1.72      #Pressure of air leaves the cooler or enters the engine in bar
eta_v  =  0.70      #Volumetric efficiency of engine
n  =  6.      #Number of cylinders
d  =  90.
l  =  100.      #Bore and stroke of cylinder in mm
N  =  2000.      #Engine speed in rpm
T  =  147.      #Output brake torque in Nm
eta_m  =  0.75      #Mechanical efficiency

#Solution:
bp  =  2*math.pi*N/60*T*10**-3      #Brake power in kW
ip  =  bp/eta_m      #Indicated power in kW
ip  =  ip/n      #Indicated power per cylinder in kW
A  =  (math.pi/4)*d**2*1e-6      #Area of cylinder in m**2
l  =  l*1e-3      #Stroke of cylinder in m
imep  =  ip/(l*A*N/(2*60))      #Indicated mean effective pressure in kN/m**2
imep  =  imep/100      #Indicated mean effective pressure in bar
V_s  =  n*A*l*N/2      #Engine swept volume in m**3/min
Vi  =  V_s*eta_v      #Induced volume of air in m**3/min
R  =  0.287      #Specific gas consmath.tant in kJ/kgK
cp  =  1.005      #Specific heat in kJ/kgK
m_e  =  P3*100*Vi/(R*T3)      #Mass of air induced into the engine in kg/min
Q1  =  1340./60      #Heat added to air in kW
m_c  =  1      #Assume for calculation
def  f(T2):
W_c  =  m_c*cp*(T2-T1)      #Work done on air in compressor kW
Q_c  =  m_c*cp*(T2-T3)      #Heat given to the air passes through the cooler in kW
return  W_c/Q_c-bp/Q1

T2  =  fsolve(f,500)
m_c  =  bp*60/(cp*(T2-T1))      #Mass of air flow into the compressor in kg/min

#Results:
print " a)The engine indicated mean effective pressure, imep   =   %.2f bar"%(imep)
print " b)The air consumption in the engine, m_e   =   %.2f kg/min"%(m_e)
print " c)The air flow into the compressor, m_c   =   %.2f kg/min"%(m_c)
a)The engine indicated mean effective pressure, imep   =   6.45 bar
b)The air consumption in the engine, m_e   =   4.81 kg/min
c)The air flow into the compressor, m_c   =   12.62 kg/min