CHAPTER 16 - Supercharging of IC Engines

EXAMPLE 16.1 PAGE 525

In [7]:
from __future__ import division
# Initialisation of Variables
pwu=735#............#Power developed by naturally aspirated engine in kW
afru=12.8#.............#Air fuel ratio for naturally aspirated engine
bsfc=0.350#......#Brake specific fuel consumption in kg/kWh
metau=0.86#...........#Mechanical efficiency of naturally aspirated engine
pi=730#...........#Inlet pressure in mm of Hg absolute
tm=325#...........#Mixture temperature in Kelvin
pr=1.6#.............#Pressure ratio of supercharged engine
etaa=0.7#.............#Adiabatic efficiency of supercharged engine
metas=0.9#..............#Mechanical efficiency of supercharged engine
afrs=12.8#.............#Air fuel ratio for supercharged engine
rhohg=13600#.............#Density of mercury in kg/m**3
R=0.287#...................#Gas constant in kJ/kgK
ga=1.4#................#Degree of freedom for gas
cp=1.005#..................#Specific heat of the fuel
g=9.81#................#Acceleration due to gravity in m/s**2
#calculations
t2=tm*(pr)**((ga-1)/ga)#..............#Ideal temperature for the supercharged engine
t2a=tm+(t2-tm)/etaa#................#Actual temperature for the supercharged engine
wa=cp*(t2a-tm)#.....................#Work of the supercharger
wsup=cp*(t2a-tm)/metas#..............#Work required to drive the supercharger in kJ/kg of air
#When unsupercharged
p1=(pi/1000)*((g*rhohg)/1000)#..............#Inlet pressure in kN/m**2
rhounsup=p1/(R*tm)#
maunsup=(bsfc*pwu*afrs)/3600#...................#Air consumption in kg/s for unsupercharged engine
#When supercharged
rhosup=(pr*p1)/(R*t2a)#
masup=maunsup*(rhosup/rhounsup)#..................#Air consumption in kg/s
Psup=masup*wsup#...............#Power required to run the supercharger in kW
print "The Power required to run the supercharger = %0.2f kW "%Psup
The Power required to run the supercharger = 90.47 kW 

EXAMPLE 16.2 PAGE 526

In [1]:
# Initialisation of Variables
p1=1.0132#..............#Mean pressure at sea level in bar
t1=283#................#Mean temperature at sea level in Kelvin
BP=260#....................#Brake Power output in kW
etaV=0.78#..................#Volumetric efficiency at sea level free air condition
sfc=0.247#............#Specific Fuel consumption in kg/kW.h
afr=17#...................#Air fuel ratio
N=1500#...................#Engine rpm
at=2700#.................#Altitude in mts
p2=0.72#................#Pressure in bar at the given altitude
Psup=0.08#.................#8% power of engine is taken by the supercharger
R=287#...................#Gas constant in J/kgK
t2=32+273#..............#Temperature in Kelvin at the given altitude
#calculations
mf=(sfc*BP)/60#.............#Fuel consumption in kg/min
ma = mf*afr#..................#Air consumption in ig/min
acps = ma/(N/2)#............#Air consumption per stroke in kg
Vs=(acps*R*t1)/(etaV*p1*10**5)#................#Engine capacity in m**3
print "The Engin Capacity = %0.2f m**3 "%Vs
pmb=(BP*6)/(Vs*10*(N/2))#........#Brake Mean Effective Pressure in bar
print "The Brake mean effective pressure  = %0.2f bar"%pmb
gp=BP/(1-Psup)#.................#Gross power produced by supercharged engine in kW
masup=ma*gp/BP#......................#Mass of air required for supercharged engine in kg
matc=masup/(N/2)#..............#Mass of air taken per cycle
pressure=(matc*R*t2)/(etaV*10**5*Vs)#
print "The Increase of pressure required = %0.2f bar"%(pressure-p2)
The Engin Capacity = 0.02 m**3 
The Brake mean effective pressure  = 8.34 bar
The Increase of pressure required = 0.47 bar

EXAMPLE 16.3 PAGE 527

In [2]:
# Initialisation of Variables
ec=3600*10**(-6)#.............#Engine capacity in m**3
pw=13#...............#Power developed in kW per m**3 of free air induced per minute
etaV=0.82#............#Volumetric Efficiency
N=3000#................#Engine rpm
p1=1.0132#...........................#Initial Air pressure in bar
t1=298#........................#Initial Temperature in Kelvin
pr=1.8#.....................#Pressure ratio in rotary compressor
etaC=0.75#.................#Isentropic efficiency of compressor
etaM=0.8#....................#Mechanical efficiency
ga=1.4#.....................#Degree of freedom for the gas
td=4#.......................#The amount by which the temperature is kess than delivery temperature from compressor
R=287#......................#Gas constant in J/kg.K
cp=1.005#.....................#Specific heat capacity
#Calculations
Vs=(ec*N)/2#....................#Swept volume in m**3/min
Vu=Vs*etaV#....................#Unsupercharged volume induced per min
rcdp=pr*p1#........#Rotary compressor delivery pressure
t2=t1*(pr)**((ga-1)/ga)#..............#Ideal temperature for the supercharged engine
t2a=t1+(t2-t1)/etaC#................#Actual temperature for the supercharged engine
ta=t2a-td#............................#Temperature of air at intake to the engine cylinder
V1=(rcdp*Vs*t1)/(p1*ta)#.................#Equivalent volume at 1.0132 bar and 298 K
Vinc=V1-Vs#...........................#Increase in induced Volume of air in m**3/min
ipincai=pw*Vinc#.......................#Increase in IP from air induced in kW
ipinciip=((rcdp-p1)*10**5*Vs)/(60*1000)#...........#Increase in IP due to increased induction pressure kW
ipinctot=ipincai+ipinciip#...............#Total increase in Input Power in kW
bpinc=ipinctot*etaM#....................#Increase in Brake Power of the engine in kW
ma=(rcdp*10**5*Vs)/(60*R*ta)#...................#Mass of air delivered by the compressor kg/s
pc=(ma*cp*(t2a-t1))/etaM#....................#Power required by the compressor
bpincnet=bpinc-pc#..........................#Net Increase in BP
print "The Net increase in Brake Power = %0.2f kW "%bpincnet
The Net increase in Brake Power = 17.60 kW 

EXAMPLE 16.4 PAGE 528

In [3]:
# Initialisation of Variables
p1=1.0132#..............#Mean pressure at sea level in bar
t1=283#................#Mean temperature at sea level in Kelvin
BP=250#....................#Brake Power output in kW
etaV=0.78#..................#Volumetric efficiency at sea level free air condition
sfc=0.245#............#Specific Fuel consumption in kg/kW.h
afr=17#...................#Air fuel ratio
N=1500#...................#Engine rpm
at=2700#.................#Altitude in mts
p2=0.72#................#Pressure in bar at the given altitude
Psup=0.08#.................#8% power of engine is taken by the supercharger
R=287#...................#Gas constant in J/kgK
t2=32+273#..............#Temperature in Kelvin at the given altitude
#calculations
mf=(sfc*BP)/60#.............#Fuel consumption in kg/min
ma = mf*afr#..................#Air consumption in ig/min
acps = ma/(N/2)#............#Air consumption per stroke in kg
Vs=(acps*R*t1)/(etaV*p1*10**5)#................#Engine capacity in m**3
print "The Engine Capacity = %0.2f m**3:"%Vs
pmb=(BP*6)/(Vs*10*(N/2))#........#Brake Mean Effective Pressure in bar
print "The Brake mean effective pressure = %0.2f bar"%pmb
gp=BP/(1-Psup)#.................#Gross power produced by supercharged engine in kW
masup=ma*gp/BP#......................#Mass of air required for supercharged engine in kg
matc=masup/(N/2)#..............#Mass of air taken per cycle
pressure=(matc*R*t2)/(etaV*10**5*Vs)#
print "The Increase of pressure required = %0.2f bar"%(pressure-p2)
The Engine Capacity = 0.02 m**3:
The Brake mean effective pressure = 8.41 bar
The Increase of pressure required = 0.47 bar

EXAMPLE 16.5 PAGE 529

In [4]:
from __future__ import division
from math import pi
# Initialisation of Variables
t1=298#.................#Temperature of the air while entering the compressor in Kelvin
qrej=1210#..............#Amount of heat rejected in cooler in kJ/min
t2=273+65#...............#Temperature of the air leaving the cooler in Kelvin
p2=1.75#.................#Pressure of the air leaving the cooler in bar
n=6#.....................#No of cylinders
d=0.1#...................#Bore of the cylinder in m
l=0.11#...................#Stroke of the cylinder in m
etaV=0.72#................#volumetric efficiency
N=2000#...............#Engine rpm
Tout=150#..................#Torque Output in Nm
etaM=0.8#..................#Mechanical efficiency
R=287#.......................#Gas constant for air in J/kgK
cp=1.005#...................#Specific capacity of air
#calculations
BP=(2*pi*N*Tout)/(60*1000)#...........#Brake power in kW
IP=BP/etaM#..........#Input Power in kW
Vc=(pi/4)*d*d*l#...................#Cylinder Volume in m**3
pmi=(6*IP)/(n*Vc*(N/2)*10)#................#Indicated mean effective pressure
print "The indicated mean effective pressure = %0.2f bar "%(pmi)
Vs=Vc*6*(N/2)#.........................#Engine Swept Volume in m**3/min
Vaa=Vs*etaV#..........................#Aspirated volume of air into engine in m**3/min
maa=(p2*10**5*Vaa)/(R*t2)#..............#Aspirated air mass flow into the engine in kg/min
print "The total aspirated air mass flow into the engine = %0.2f kg/min "%maa
t2a=((((BP/cp)/(qrej/(60*cp)))*t2)-t1)/(((BP/cp)/(qrej/(60*cp)))-1)#
mc=((BP/cp)/(t2a-t1))*60#........................#Air flow into the compressor in kg/min
print "Air flow into the compressor = %0.2f kg/min"%mc
The indicated mean effective pressure = 4.55 bar 
The total aspirated air mass flow into the engine = 6.73 kg/min 
Air flow into the compressor = 16.79 kg/min