# Chatper 11 : Four Stroke Spark Ignition Engine¶

## Example 11.1 Page no : 239¶

In [1]:
import math

#Input data
d = 0.0625					#Diameter in m
L = 0.09					#Stroke in m
nv = 0.75					#Volumetric efficiency
p = 1.03					#Pressure at N.T.P in kg/cm**2
T = 273					#Temperature at N.T.P in K
R = 29.27					#Characteristic gas constant in kg.m/kg.degree C

#Calculations
Vs = ((3.14/4)*d**2*L)					#Swept volume in cu.m
V = (nv*Vs)					#Volume of charge at N.T.P in cu.m
w = (p*10**4*V)/(R*T)					#Weight of the charge in kg/cycle

#Output
print 'The weight of the charge is %3.6f kg/cycle'%(w)

The weight of the charge is 0.000267 kg/cycle


## Example 11.2 Page no : 242¶

In [3]:
import math

#Input data
n = 9.					#Number of cylinder
d = 0.145					#Bore in m
l = 0.19					#Stroke in m
r = 5.9					#Compression ratio
bhp = 460.					#Brake horse power in B.H.P
N = 2000.					#Speed in r.p.m
x = 20.					#Percentage rich in mixture
CV = 11200.					#Calorific value in kcal/kg
pC = 85.3					#Percentage of carbon
pH2 = 14.7					#Percentage of Hydrogen
nv = 70.					#Volumetric efficiency in percent
T = 15.+273					#Temperature in K
nm = 90.					#Mechanical efficiency in percent
wO2 = 23.3					#Percentage of oxygen by weight in air
da = 1.29					#Density of air in kg/m**3
mC = 12.					#Molecular weight of carbon
mO2 = 32.					#Molecular weight of O2
mH2 = 2.					#Molecular weight of H2

#Calculations
hihp = ((bhp/(nm/100))*(4500./427))					#Heat equivalent in kcal
Vs = ((3.14/4)*d**2*l*(N/2)*n)					#Swept volume in c.m per min
cw = (Vs/da)					#Charge weight of air per minute in kg
ma = (100/wO2)*((pC/100)*(mO2/mC)+(pH2/100)*(mO2/(2*mH2)))					#Wt. of air required per kg of fuel in kg
mf = (cw/ma)					#Minimum fuel inkg
ith = (hihp/(mf*(100+x)/100*CV))*100					#Indicated thermal efficiciency in percent'

#Output
print 'Indicated thermal efficiency of the engine is %3.1f percent'%(ith)

Indicated thermal efficiency of the engine is 27.1 percent


## Example 11.3 Page no : 244¶

In [4]:
import math

#Input data
n = 8.					#Number of cylinders
d = 8.57					#Bore in cm
l = 8.25					#Stroke in cm
r = 7.					#Compression ratio
N = 4000.					#Speed in r.p.m
la = 53.35					#Length of the arm in cm
t = 10.					#Test duration in min
br = 40.8					#Beam reading in kg
gas = 0.455					#gasoline in kg. In textbook, it is given wrong as 4.55
CV = 11400.					#Calorific value in kcal/kg
Ta = 21.+273					#Temperature of air in K
pa = 1.027					#Pressure of air in kg/cm**2
wa = 5.44					#Rate of air in kg/min
J = 427.					#Mechanical equivalent of heat in kg.m/kcal
R = 29.27					#Characteristic gas constant in kg.m/kg.K

#Calculations
bhp = (2*3.14*N*br*la)/(4500*100)					#Brake horse power in B.H.P
pb = (bhp*4500)/((n/2)*(l/100)*(3.14/4)*d**2*N)					#Brake mean effective pressure in kg/cm**2
bsfc = (gas*60)/bhp					#Brake specific fuel consumption in kg/b.h.p.hr
bsac = ((wa*60)/bhp)					#Brake specific fuel consumption in kg/b.h.p.hr
nb = ((bhp*4500)/(J*gas*CV))*100					#Brake thermal efficiency in percent
Vd = ((3.14/4)*d**2*l)					#Piston print lacement in c.c/cycle
Pd = (Vd/10**6)*(N/2)*n					#Piston print lacement in m**3/min
Va = ((wa*R*Ta)/(pa*10**4))					#Volume of air used in m**3/min
nv = (Va/Pd)*100					#Volumetric efficiency in percent
af = (wa/gas)					#Air fel ratio

#Output
print 'a) the B.H.P delivered s %3.0f h.p  \
\nb) the b.m.e.p is %3.1f kg/cm**2  \
\nc) the b.s.f.c is %3.3f kg/b.h.p.hr  \
\nd) the brake specific air consumption is %3.3f kg/b.h.p.hr  \
\ne) the brake thermal efficiency is %3.1f percent  \
\nf) the volumetric efficiency is %3.0f percent  \
\ng) the air fuel ratio is %3.2f'%(bhp,pb,bsfc,bsac,nb,nv,af)

a) the B.H.P delivered s 122 h.p
b) the b.m.e.p is 7.2 kg/cm**2
c) the b.s.f.c is 0.225 kg/b.h.p.hr
d) the brake specific air consumption is 2.686 kg/b.h.p.hr
e) the brake thermal efficiency is 24.7 percent
f) the volumetric efficiency is  60 percent
g) the air fuel ratio is 11.96


## Example 11.4 Page no : 246¶

In [5]:
import math

#Input data
n = 4.					#Number of cylinders
N = 2000.					#Speed in r.p.m
m = 13.15					#Mass of fuel in kg/hour
Vd = 655.5					#Displacement volume in c.c
da = 1.2					#Density of air in kg/m**3
d = 12.7					#Manometer depression in cm
#Qa = 0.231*math.sqrt(ha); Qa is the flow rate in cu.m/min and ha is the pressure difference in metres

#Calculations
Qa = (0.231*math.sqrt(((d*1000)/da)/100))					#Flow rate in cu.m/min
Wa = (Qa*da)					#Weight of air in kg/min
Va = (Qa*(2/N)*(1/n))*10**6					#Volume of air drawn in per cycle per cylinder in c.c
nv = (Va/Vd)*100					#Volumetric efficiency in percent
af = (Wa/(m/60))					#Air fuel ratio

#Output
print 'a) the weight of air drawn is %3.3f kg/min  \
\nb) volumetric efficiency taking air into account is %3.1f percent  \
\nc) the air-fuel ratio is %i'%(Wa,nv,af)

a) the weight of air drawn is 2.852 kg/min
b) volumetric efficiency taking air into account is 90.6 percent
c) the air-fuel ratio is 13


## Example 11.5 Page no : 248¶

In [6]:
import math

#Input data
d = 10.					#Diameter in cm
l = 15.					#Stroke in cm
r = 6.					#Compression ratio
ihp = 20.					#Indicated horse power in h.p
N = 1000.					#Speed in r.p.m
n = 4.					#Number of cylinders
nt = 30.					#Thermal efficiency in percent
CV = 10000.					#Calorific value in kca/kg
g = 1.4					#Ratio of specific heats

#Output
Vs = ((3.14/4)*d**2*l)					#Swept volume in c.c
Vc = (Vs/(r-1))					#Clearance volume in c.c
na = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent
pm = ((ihp*4500)/((l/100)*(3.14/4)*(d/100)**2*(N/2)*n))					#Pressure in kg/cm**2
pc = (ihp*4500*60)/(427*(nt/100)*CV)					#Petrol consumption in kg/hr

#Output
print 'Clearance volume is %3.1f c.c  \
\nThe air standard efficiency is %3.1f percent  \
\nPetrol consumption is %3.2f kg/hr'%(Vc,na,pc)

Clearance volume is 235.5 c.c
The air standard efficiency is 51.2 percent
Petrol consumption is 4.22 kg/hr


## Example 11.6 Page no : 253¶

In [7]:
import math

#Input data
n = 6.					#Number of cylinders
P = 62.					#Power in HP
N = 3000.					#Speed in r.p.m
nv = 85.					#Volumetric efficiency in percent
nt = 25.					#Thermal efficiency in percent
CV = 10500.					#Calorific value in kcal/kg
af = 15.					#Air fuel ratio
T = 273.					#standard atmosphere temperature in K
p = 1.03					#standard atmosphere pressure in kg/cm**2
R = 29.27					#Characteristic gas constant in kg.m/kg.K
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
q = (P*4500)/(J*(nt/100))					#Heat supplied in kcal/min
F = (q/CV)					#Fuel supplied per minute in kg
Fc = (F/N)*(2/n)					#Fuel supplied per cycle per cylinder in kg
wt = (af*Fc)					#Weight of air supplied per cycle in kg
d = ((((wt)*R*T)/(p*10**4*(3.14/4)*(nv/100)))**(1./3))*100					#Diameter in cm

#Output
print 'Cylinder bore  =  stroke  =  %3.2f cm'%(d)

Cylinder bore  =  stroke  =  7.84 cm