# Chapter15:Air Capacity and SuperCharging¶

## Example 15.1 page no: 474¶

In [2]:
#Input data
Vs=0.0028       #Swept volume in m**3
N=3000          #Speed of the engine in rpm
ip=12.5         #The average indicated power developed in kW/m**3
nv=85           #Volumetric efficiency in percent
p1=1.013        #The atmospheric pressure in bar
T1=288          #The atmospheric temperature in K
ni=74           #Isentropic efficiency in percent
pr=1.6          #The pressure ratio
nm=78           #All mechanical efficiencies in percent
R=287           #Real gas constant in J/kgK
Cp=1.005        #The specific heat of gas in kJ/kgK

#Calculations
Vs1=(Vs*(N/2.0))         #Volume swept by the piston per minute in m**3/min
Vi=(nv/100.0)*Vs1        #Unsupercharged induced volume in m**3/min
p2=pr*p1               #Blower delivery pressure in bar
T21=T1*(p2/p1)**((g-1)/g)         #Temperature after isentropic compression in K
T2=T1+((T21-T1)/((ni/100.0)))       #Blower delivery temperature in K
Ve=(Vs1*p2*T1)/(T2*p1)            #Equivalent volume at 1.013 bar and 15 degree centigrade in m**3/min
nv1=(Ve/Vs1)*100                  #Volumetric efficiency of supercharged engine in percent
Vii=Ve-Vi                         #Increase in induced volume in m**3/min
ipa=ip*Vii                        #Increase in ip from air induced in kW
ipi=((p2-p1)*10**5*Vs1)/(60*1000)       #Increase in ip due to increased induction pressure in kW
ipt=ipa+ipi                             #Total increase in ip in kW
bp=ipt*(nm/100.0)                              #Increase in engine bp in kW
ma=(p2*(Vs1/60.0)*10**5)/(R*T2)                #Mass of air delivered per second by blower in kg/s
P=ma*Cp*(T2-T1)                              #Power input to blower in kW
Pd=P/(nm/100.0)                                #Power required to drive the blower in kW
bpn=bp-Pd                                    #Net increase in bp in kW
bpu=ip*Vi*(80/100.0)                           #The bp of unsupercharged engine in kW
bpp=(bpn/(bpu))*100                          #Percentage increase in bp in percent

#Output
print"The volumetric efficiency of supercharged engine = ",round(nv1,0),"percent"
print"The increase in brake power by supercharging = ",round(bpn,1)," kW "
print"The percentage increase in brake power = ",round(bpp,1)," percent "

The volumetric efficiency of supercharged engine =  134.0 percent
The increase in brake power by supercharging =  15.1  kW
The percentage increase in brake power =  42.3  percent


## Example 15.2 page no: 477¶

In [1]:
#Input data
p=1.013       #The pressure at the sea level in bar
T=283         #The temperature at the sea level in K
bp=275.0        #Brake power in kW
N=1800.0        #The speed of the engine in rpm
a=20          #Air fuel ratio
R=287         #The real gas constant in J/kgK
bsfc=0.24     #Brake specific fuel consumption in kg/kWh
nv=80         #Volumetric efficiency in percent
p2=0.75       #The atmospheric pressure at altitude in bar
P=9           #The power consumed by supercharger of the total power produced by the engine in percent
T2=303        #The temperature of air leaving the supercharger in K

#Calculations
mf=(bsfc*bp)/60.0
ma1=mf*a
ma=(2/N)*ma1
dai=(p*10**5)/(R*T)
Vd=(ma/(dai*(nv/100.0)))
pmb=(bp*2*60*1000)/(Vd*N*10**5)
GP=bp/(1-0.09)
ma2=(ma1/bp)*GP
ma1=(ma2*2)/N
p21=((R*T2*ma1)/((nv/100.0)*Vd))/10.0**5
pi=p21-p2

#Output
print"(a) The engine capacity Vd = ",round(Vd,3),"m**3"
print"The bmep of the unsupercharged engine = ",round(pmb,3),"bar"
print"(b) Increase in air pressure required in the supercharged = ",round(pi,3),"bar"

(a) The engine capacity Vd =  0.024 m**3
The bmep of the unsupercharged engine =  7.483 bar
(b) Increase in air pressure required in the supercharged =  0.442 bar


## Example 15.3 page no: 479¶

In [2]:
#Input data
Vs=0.003          #Swept volume in m**3
bmep=9            #Brake mean effective pressure in bar
N=4000            #The speed of the engine in rpm
ni=30.0           #Indicated thermal efficiency in percent
nm=90             #Mechanical efficiency in percent
bmep1=12          #The brake mean effective pressure of other engine in bar
N1=4000           #The speed of other engine in rpm
ni1=25            #The indicated thermal efficiency of other engine in percent
nm1=91            #The mechanical efficiency of other engine in percent
m=200             #The mass of naturally aspired engine in kg
m1=220            #The mass of supercharged engine in kg
CV=44000          #The calorific value of the fuel in kJ/kg

#Calculations
bp=(bmep*10**5*Vs*N)/(2.0*60.0*1000)
ip=bp/(nm/100.0)
mf=(ip)/((ni/100.0)*CV)
bp1=(bmep1*10**5*Vs*N1)/(2.0*60.0*1000)
ip1=bp1/(nm1/100.0)
mf1=ip1/((ni1/100.0)*CV)
mf2=mf*3600
mf3=mf1*3600
x=((200/90.0)-(220/120.0))/((43.2/120.0)-(27.27/90.0))

#Output
print"The maximum hours required for supply of sufficient fuel = ",round(x,3),"hr"

The maximum hours required for supply of sufficient fuel =  6.823 hr


## Example 15.4 Page no 480¶

In [4]:
#Input data
d=0.1             #The diameter of the bore in m
L=0.12            #The length of the stroke in m
N=3000            #The speed of the engine in rpm
n=4               #Number of cylinders
R=287             #Real gas constant in J/kgK
t=120             #Output Torque in Nm
nm=85             #The mechanical efficiency of the engine in percent
T1=288            #The inlet temperature of air into compressor in K
p1=1              #The inlet pressure of air into compressor in bar
Q=1200            #Heat rejected rate in kJ/min
T=328             #The outlet temperature of air in K
p=1.7             #The outlet pressure of air in bar
nv=90             #Volumetric efficiency in percent
Cp=1.005          #Specific heat of gas in kJ/kg

#Calculations
import math
bp=(2*math.pi*N*t)/(60.0*1000.0)         #The brake power in kW
ip=bp/(nm/100.0)                       #The indicated power in kW
pmi=((ip*2*60*1000*4)/(L*(math.pi*d**2)*N*n))/10.0**5         #The mean effective pressure in bar
Vs=(math.pi/4.0)*d**2*L               #Swept volume in m**3
Vs1=Vs*(N/2.0)*n                      #Volume swept by the piston per min
V1=(nv/100.0)*Vs1                     #Rate of volume flow of air into the engine in m**3/min
me=((p*10**5*V1)/(R*T))*60          #Rate of mass flow of air into the engine in kg/h
E=Q/60.0                              #Energy balance in the after cooling in kJ/s
T2=((bp/E)*T-T1)/((bp/E)-1)         #The outlet temperature of air in K
mc=((bp)/(Cp*(T2-T1)))*3600         #Mass flow rate in kg/h
maf=mc-me                           #Rate of air flow available to the consumer in kg/h

#Output
print"(a) The imep of the supercharged engine = ",round(pmi,3),"bar"
print"(b) The rate of air consumed by the engine = ",round(me,1),"kg/h"
print"(c) The rate of air flow available to the consumer = ",round(maf,1),"kg/h"

(a) The imep of the supercharged engine =  4.706 bar
(b) The rate of air consumed by the engine =  551.5 kg/h
(c) The rate of air flow available to the consumer =  1033.5 kg/h


## Example 15.5 page no: 482¶

In [6]:
#Input data
Vs=0.0045       #Swept volume in m**3
N=4000.0          #The speed of the engine in rpm
nv=150.0          #Overall volumetric efficiency in percent
ni=90.0           #Isentropic efficiency of the compressor in percent
nm=85.0           #Mechanical efficiency in percent
T=330.0           #The temperature of compressed air after cooler in K
p2=1.8          #The pressure of the compressed air in bar
T1=290.0          #The ambient temperature of air in K
p1=1.0            #The pressure of the ambient condition in bar
R=287.0           #The real gas constant in J/kgK
Cp=1.005        #The specific heat of gas in kJ/kgK

#Calculations
T21=T1*(p2/p1)**((g-1)/g)
T2=T1+((T21-T1)/(ni/100.0))
Vs1=Vs*(N/(2*60))                  # m**3/s
Va=(nv/100)*Vs1
d=(p1*10**5)/(R*T1)                # kg/m**3
ma=d*Va                            # kg/s
Q=ma*Cp*(T2-T)                     # kJ/s
P=ma*Cp*(T2-T1)                    # kW
Pa=P/(nm/100.0)

#Output
print "(a) The rate of heat rejected from the engine after cooler = ",round(Q,2),"kJ/s"
print"(b) The power absorbed by the supercharger from the engine = ",round(Pa,1),"kW"

(a) The rate of heat rejected from the engine after cooler =  5.14 kJ/s
(b) The power absorbed by the supercharger from the engine =  18.8 kW


## Example 15.6 page no: 483¶

In [22]:
#Input data
p1=0.98         #The inlet pressure of air in bar
T1=290.0          #The inlet temperature of air in K
p2=1.8          #The pressure of air delivered to the engine in bar
a=20.0            #The air fuel ratio
T3=850.0          #The temperature of the exhaust gases leaving the engine in K
p3=1.6          #The pressure of the exhaust gases leaving the engine in bar
p4=1.03         #The turbine exhaust pressure in bar
nc=80.0           #The isentropic efficiency of compressor in percent
nt=85.0           #The isentropic efficiency of turbine in percent
Cpa=1.005       #The specific heat of air in kJ/kgK
Cpg=1.15        #The specific heat of gas in kJ/kgK
g=1.33          #isentropic index

#Calculations
T21=T1*(p2/p1)**(0.286)    #value taken in book (g-1/g)=0.286
T2=T1+((T21-T1)/(nc/100.0))
T22=T2-273
T41=T3*(p4/p3)**((g-1)/g)
T4=T3-((nt/100.0)*(T3-T41))
T44=T4-273
mf=1.0                              # kg/s
ma=mf*a                           # kg/s
Wc=ma*Cpa*(T2-T1)                 # kW
mg=ma+mf                          #Mass flow rate of gas in kg/s
Wt=mg*Cpg*(T3-T4)
Pt=(Wc/Wt)*100

#Output
print"(a) The temperature of the air leaving the compressor = ",round(T22,0),"degree centigrade"
print"(b) The temperature of gases leaving the turbine = ",round(T44,0),"degree centigrade"
print"(c) The mechanical power used to run the turbocharger = ",round(Pt,1),"percent"

(a) The temperature of the air leaving the compressor =  86.0 degree centigrade
(b) The temperature of gases leaving the turbine =  502.0 degree centigrade
(c) The mechanical power used to run the turbocharger =  76.6 percent


## Example 15.7 page no: 485¶

In [34]:
#Input data
a=14.0          #Air fuel ratio
T1=288          #The ambient temperature of air in K
T2=(288-23)     #The evaporation of fuel cause 23 degree C drop in mixture temperature in K
p=1.3           #Pressure ratio
nc=75           #The isentropic efficiency of the compressor in percent
Cpm=1.05        #The specific heat of the mixture in kJ/kgK
Cpa=1.005           #The specific heat of air in kJ/kgK
h=1.4           #Isentropic index
ma=1            #Mass flow rate of air in kg/s

#Calculations
T31=T2*p**((g-1)/g)
T3=T2+((T31-T2)/(nc/100.0))
mm=1+(1/a)
Wc1=mm*Cpm*(T3-T2)
T21=T1*p**((h-1)/h)
T4=T1+((T21-T1)/(nc/100.0))
T4_=317        #approx value taken in book of T4=317
Wc2=ma*Cpa*(T4_-T1)
T5=T4-23
Ps=((Wc2-round(Wc1,0))*100)/Wc2

#Output
print"(a) The power required by the compressor  before the supercharger = ",round(Wc1,0),"kW/kg of air per second"
print"(b) The power required by the compressor after the supercharger = ",round(Wc2,1),"kW/kg of air per second"
print"Percentage of turbine power used to run the compressor = ",round(Ps,3),"percent"

(a) The power required by the compressor  before the supercharger =  27.0 kW/kg of air per second
(b) The power required by the compressor after the supercharger =  29.1 kW/kg of air per second
Percentage of turbine power used to run the compressor =  7.36 percent