# Chapter 18 : Super Charging¶

## Example 18.1 Page no : 327¶

In :
import math

#Input data
n = 6.					#Number of cylinders
d = 9.					#Bore in cm
l = 10.					#Stroke in cm
N = 2500.					#Speed in r.p.m
Ta = 25.+273					#Temperature of air entering the compressor in K
q = 16800.					#Heat rate in kcal/hour
T = 60.+273					#Temperature of air leaving the cooler in K
p = 1.6					#Pressure of air leaving the cooler in kg/cm**2
t = 14.5					#Engine torque in kg.m
nv = 75.					#Volumetric efficiency in percent
nm = 74.					#Mechanical efficiency in percent
R = 29.27					#Characteristic gas constant in kg.m/kg.K
Cp = 0.24					#Specific heat at constant pressure n kcal/kg.K

#Calculations
BHP = (2*3.14*N*t)/4500					#Brake horse power in B.H.P
IHP = (BHP/(nm/100))					#Indicated horse power in I.H.P
pm = ((IHP*4500)/((l/100)*(3.14/4)*d**2*(N/2)*n))					#Mean effective pressure in kg/cm**2
Vs = (n*(3.14/4)*(d/100)**2*(l/100)*(N/2))					#Swept volume in m**3/min
Va = (Vs*(nv/100))					#Aspirated Volume of air into engine in m**3/min
ma = (p*10**4*Va)/(R*T)					#Aspirated mass flow into the engine in kg/min
mcdT = ((BHP*4500/427)/Cp)					#Product of mass flow rate and change in temperature
msdT = ((q/60)/Cp)					#Product of mass flow rate and change in temperature
x = (mcdT/msdT)					#Ratio
T2 = ((Ta-(x*T)))/(1-x)					#Temperature in K
mc = (msdT/(T2-T))					#Air flow in kg/min

#Output
print 'a) the mean effective pressure is %3.2f kg/cm**2  \
\nb) the air consumption is %3.3f kg/min  \
\nc) the air flow into the compressor is %3.2f kg/min'%(pm,ma,mc)

a) the mean effective pressure is 6.45 kg/cm**2
b) the air consumption is 5.871 kg/min
c) the air flow into the compressor is 30.14 kg/min


## Example 18.2 Page no : 328¶

In :
import math

#Input data
IMEP = 10.					#Indicated mean effective pressure in kg/cm**2
x = 20.					#Mixture strength 20% richer math.tan chemically correct
pIMEP = 0.41					#Pumping Indicated mean effective pressure in kg/cm**2
p1 = 0.97					#Charge pressure at the beginning of compression in kg/cm**2
T1 = 100.+273					#Charge temperature at the beginning of compression in K
pm = 0.91					#Mean pressure during the conduction stroke in kg/cm**2
bn = 70.					#Blower adiabatic efficiency in percent
T2 = 50.					#Temperature of the charge after delivery by the blower in degree C
dp = 0.07					#Pressure drop in kg/cm**2
pi = 1.47					#Charge pressure in the cylinder during the induction stroke in kg/cm**2
Ta = 15.+273					#Atomspheric temperature in K
pa = 1.03					#Atmospheric pressure in kg/cm**2
g = 1.4					#Ratio of specific heats

#Calculations
T2x = ((((pi/pa)**((g-1)/g)-1)/(bn/100))+1)*Ta+T2					#Temperature in K
rIMEP = ((pi/pa)*(T1/T2x))					#Ratio of I.M.E.P
gIMEP = (rIMEP*IMEP)					#Gross I.M.E.P in kg/cm**2
nsIMEP = (gIMEP+(pi-pa))					#Net I.M.E.P supercharged in kg/cm**2
nuIMEP = (IMEP-pIMEP)					#Net I.M.E.P unsupercharged in kg/cm**2
iIMEP = (nsIMEP-nuIMEP)					#Increase in I.M.E.P in kg/cm**2
pei = (iIMEP*100)/nuIMEP					#Percentage increase

#Output
print 'Percentage increase in the net I.M.E.P due to supercharging is %3.1f percent'%(pei)

Percentage increase in the net I.M.E.P due to supercharging is 49.9 percent


## Example 18.3 Page no : 331¶

In :
import math

#Input data
l = 4.5					#Capacity in litres
P = 20.					#Power in H.P per m**3 of free air induced per minute
N = 1700.					#Speed in r.p.m
nv = 75.					#Volumetric efficiency in percent
Ta = 27.+273					#Atomspheric temperature in K
pa = 1.03					#Atmospheric pressure in kg/cm**2
pr = 1.75					#Pressure ratio
ie = 70.					#Isentropic efficiency in percent
nm = 75.					#Mechanical efficiency in percent
g = 1.4					#Ratio of specific heats
nb = 80.					#Efficiency of blower in percent
R = 29.27					#Characteristic gas constant in kg.m/kg.K
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
Vs = (l/1000*(N/2))					#Swept volume in m**3/min
uVs = ((nm/100)*Vs)					#Unsupercharged swept volume in m**3/min
dp = (pr*pa)					#Blower delivery pressure in kg/cm**2
Tc = (Ta*pr**((g-1)/g))					#Temperature after isentropic compression in K
dT = (Ta+(Tc-Ta)/(ie/100))					#Blow delivery temperature in K
Va = (Vs*(dp*Ta)/(pa*dT))					#Equivalent volume at free air condition in m**3/min
iiv = (Va-uVs)					#Increase in the induced volume in m**3/min
iIHP = (P*iiv)					#ncrease in I.H.P
iBHP = (iIHP*(nm/100))					#Increase in B.H.P
ma = (dp*10**4*Vs)/(R*dT)					#Mass of air delivered by blower in kg/min
HP = (ma*Cp*(dT-Ta)*J)/(4500*(80./100))					#H.P required for blower
nibhp = (iBHP-HP)					#Net increse in engine b.h.p
pei = (nibhp/(P*uVs))*100					#Percentage increase

#Output
print 'Percentage net increase in B.H.P is %3.1f percent'%(pei)

Percentage net increase in B.H.P is 42.1 percent