Chapter 9 : Spark Ignition Engines

Example 9.1 Page no : 188

In [1]:
import math 
					
#Input data
d = 0.001					#Diameter of the jet in m
vd = 104.					#Venturi depression in cm of water. In textbook, it is given as 10 cm
Cd = 0.65					#Coefficient of discharge 
g = 0.76					#Specific gravity of petrol
w = 1000.					#Weight of water per one cu.m in kg

					
#Calculations
pa = (vd/100)*w					#Venturi depression in kg/m**2
dp = (g*w)					#Density of petrol in kg/m**3
wf = (((3.14*d**2)/4)*Cd*math.sqrt(2*9.81*dp*pa))/10**-3					#Petrol discharge in gm/sec neglecting nozzle lip

					
#Output
print 'The weight of petrol discharged is %3.2f gm/sec'%(wf)

The weight of petrol discharged is 2.01 gm/sec

Example 9.2 Page no : 193

In [2]:
import math 
					
#Input data
d1 = 0.075					#Throat diameter in m
Ca = 0.93					#Coefficient of air flow
d2 = 0.005					#Orifice diameter in m
Cf = 0.68					#Coefficient of fuel discharge
ap = 1.					#Approach factor
dp = 0.15					#Pressure drop in kg/cm**2
da = 1.29					#Density of air in kg/m**3
df = 720					#Density of fuel in kg/m**3

					#Calcultions
w = (((3.14/4)*d1**2)/((3.14/4)*d2**2))*(Ca/Cf)*math.sqrt(da/df)					#The air-fuel ratio neglecting the nozzle lip

					
#Output
print 'The air-fuel ratio neglecting the nozzle lip is %3.1f'%(w)

The air-fuel ratio neglecting the nozzle lip is 13.0

Example 9.3 Page no : 193

In [3]:
import math 
					
#Input data
af = 15.					#Air fuel ratio
dp = 753.					#Density of petrol in kg/m**3
da = 1.28					#Density of air in kg/m**3
C = [0.84,0.7]					#Coefficient of discharge for air and fuel respectively

					
#Calculations
A = 1/(af*(C[1]/C[0])*math.sqrt(dp/da))					#Ratio of areas
d = math.sqrt(A)					#Ratio of diamter of jet to diameter of venturi
x = (1/d)					#Reverse of ratio

					
#Output
print 'The ratio of diameter of jet to diameter of venturi is 1 : %3.1f'%(x)

The ratio of diameter of jet to diameter of venturi is 1 : 17.4

Example 9.4 Page no : 194

In [4]:
import math 
					
#Input data
D = [10.,12.]					#Dimensions of four cylinder in 10 cm* 12 cm
n = 4.					#Four cylinder
N = 2000.					#Speed in r.p.m
d = 0.03					#Venturi throat in m
nv = 70.					#Volumetric efficiency of the engine in percent
Ca = 0.8					#Coefficient of air flow
da = 1.29					#Density of air in kg/m**3

					
#Calculations
Vs = (3.14/4)*(D[0]/100)**2*(D[1]/100)					#Stroke volume of one cylinder in m**3
Va = (Vs*n*(nv/100)*(N/2))					#Volume of air drawn per minute in m**3
w = (Va*da)/60					#Weight of air drawn per sec
dp = ((w/((3.14/4)*d**2*Ca))**2/(2*9.81*da))					#Venturi depression in kg/cm**2

					
#Output
print 'The venturi depression is %3.1f kg/m**2'%(dp)

The venturi depression is 397.7 kg/m**2

Example 9.5 Page no : 198

In [5]:
import math 
					
#Input data
D = [8.25,11.5]					#Dimensions of four cylinder in 8.25 cm* 11.5 cm
n = 4.					#Four cylinder
N = 3000.					#Speed in r.p.m
v = 150.					#Venturi depression in cm of water
nv = 80.					#Volumetric efficiency of the engine in percent
af = 14.					#Air fuel ratio
Ca = 0.84					#Coefficient of air flow
Cf = 0.7					#Coefficient of fuel orifice 
da = 1.29					#Density of air in kg/m**3
df = 700.					#Density of fuel in kg/m**3
dw = 1000.					#Density of water in kg/m**3

					
#Calculations
Va = ((3.14/4)*(D[0]/100)**2*(D[1]/100)*n*(nv/100)*(N/(2*60)))					#Maximum amount of air pasmath.sing through the venturi in m**3
vd = (v/100)*dw					#Venturi depression in kg/m**2
va = (Ca*math.sqrt((2*9.81*vd)/da))					#Velocity of air in m/s
d = math.sqrt((Va/va)*(4/3.14))					#Throat diameter of venturi in m
Af = 1/(af*(va/Va)*math.sqrt(df/da)*(Cf/Ca))					#Area of orifice in m**2
df = math.sqrt((Af*4)/3.14)*1000					#Diameter of orifice in mm

					
#Output
print 'The size of venturi is %i kg/m**2  \
\nThe diameter of fuel orifice is %3.2f mm'%(vd,df)

The size of venturi is 1500 kg/m**2  
The diameter of fuel orifice is 1.35 mm

Example 9.6 Page no : 198

In [6]:
import math 
D = [7.5,10]					#Dimensions of four cylinder in 7.5 cm diameter and 10 cm stroke
n = 6.					#Six cylinder
pC = 84.					#Percentage of carbon in volatile fuel
pH2 = 16.					#Percentage of hydrogen in volatile fuel
dc = (38.5/1000)					#Diameter of the throat of the choke tube in m
N = 3000.					#Speed in r.p.m
nv = 0.8					#Volumetric efficiency in ratio
p = 0.914					#Pressure at the throat of the choke tube in kg/cm**2
T = 15.5+273					#Temperature at the throat of the choke tube in K
Ts = 273.					#Temperature of 0 degree C in K
ps = 1.027					#Atmospheric pressure in kg/cm**2
Ra = 29.27					#Universal gas constant for air in kg.m/kg.K
Rf = 9.9					#Gas constant for fuel in kg.m/kg.K
pO2 = 23.					#Composition by weight of oxygen in air in percent
pN2 = 77.					#Composition by weight of nitogen in air in percent
mO2 = 32.					#Molecular weight of O2
mH2 = 2.					#Molecular weight oh H2
mC = 12.					#Molecular weight of carbon

					
#Calculations
Vm = ((3.14/4)*(D[0]/100)**2*(D[1]/100)*n*(N/2)*nv)					#Volume of mixture supplied per sec in m**3
qa = ((100/pO2)*(((pC/100)*(mO2/mC))+((pH2/100)*(mO2/(2*mH2)))))					#Quantity of air required for complete combustion of fuel in kg
vf = (Rf*Ts)/(ps*10**4)					#Specific volume of volatile fuel in m**3/kg
va = (Ra*Ts)/(ps*10**4)					#Specific volume of air in m**3/kg
wf = (Vm/(qa*va+vf))					#Flow rate of fuel in kg/min
Fc = (wf*60)					#Fuel consumption in kg/hour
da = (p*10**4)/(Ra*T)					#Density of air at the throat of the choke in kg/m**3
Va = ((qa*wf)/((3.14/4)*dc**2*da*60))					#Speed of air at throat in m/s

					
#Output
print 'a) The fuel consumption is %3.1f kg/hr  \
\nb) The speed of the air through the choke is %3.1f m/s'%(Fc,Va)

a) The fuel consumption is 0.0 kg/hr  
b) The speed of the air through the choke is 0.0 m/s

Example 9.7 Page no : 203

In [7]:
import math 
					
#Input data
mf = 7.5					#Consumption of petrol per hour
gf = 0.75					#Specific gravity of fuel
Tf = 25+273					#Temperature of fuel in K
af = 15					#Air fuel ratio
dc = 22					#diameter of choke tube in mm
l = 4					#Top of the jet is 4 mm above the petrol level in the float chamber
Ca = 0.82					#Coefficient of discharge for air 
Cf = 0.7					#Coefficient of discharge for fuel
R = 29.27					#Characteristic gas constant for air in kg.m/kg.K
p = 1.03					#Atmospheric pressure in kg/cm**2

					
#Calculations
da = (p*10**4)/(R*Tf)					#Density of air in kg/m**3 
dp = (gf*1000)					#Density of petrol in kg/cm**3
dpa = ((af*mf*10**6)/(60*60*3.14*Ca*(dc/2)**2))**2/(2*9.81*da)					#Change in pressure in kg/m**2
df = math.sqrt((mf/(60*60*Cf*math.sqrt(2*9.81*dp*(dpa-((l/100)*dp)))))*(4/3.14))*1000					#Diameter of the fuel jet in mm

					
#Output
print 'Diameter of the jet of the carburettor is %3.2f mm'%(df) 

Diameter of the jet of the carburettor is 1.22 mm

Example 9.8 Page no : 207

In [8]:
import math 
					
#Input data
td = 7.5					#Throat diameter in cm
Ca = 0.85					#Coefficient of air flow
fd = 0.5					#Diameter of fuel orifice in cm
Cd = 0.7					#Coefficient of discharge
l = 5					#Nozzle lip in mm
x = 1					#Approach factor
dpa = 0.15					#Pressure drop in kg/cm**2
da = 1.29					#Density of air in kg/m**3
dp = 720					#Density of fuel in kg/m**3

					
#Calculations
afr1 = (((3.14*td**2)/(3.14*fd**2))*(Ca/Cd)*math.sqrt(da/dp))					#Air fuel ratio
afr2 = ((3.14*td**2)/(3.14*fd**2))*(Ca/Cd)*math.sqrt((da*dpa)/(dp*(dpa-((l/100)*(dp/10**6)))))					#Air fuel ratio

					
#Output
print 'The air fuel ratio  \
\na) neglecting nozzle lip is %3.2f  \
\nb) nozzle lip is taken into account is %3.2f'%(afr1,afr2)

The air fuel ratio  
a) neglecting nozzle lip is 11.56  
b) nozzle lip is taken into account is 11.56

Example 9.9 Page no : 209

In [10]:
import math 
					
#Input data
td = 7.5					#Throat diameter in cm
Ca = 0.85					#Coefficient of air flow
fd = 0.5					#Diameter of fuel orifice in cm
Cd = 0.7					#Coefficient of discharge
l = 5.					#Nozzle lip in mm
x = 1.					#Approach factor
dpa = 0.15					#Pressure drop in kg/cm**2
da = 1.29					#Density of air in kg/m**3
dp = 720					#Density of fuel in kg/m**3

					
#Calculations
v = math.sqrt(2*9.81*(l/1000)*(dp/da))					#Velocity of air in m/s

					
#Output
print 'Velocity of air flow is %3.1f m/s'%(v)

Velocity of air flow is 7.4 m/s

Example 9.10 Page no : 212

In [11]:
import math 
					
#Input data
x = 2.8					#Height above the nozzle in mm
va = 58					#Velocity of air in m/s
da = 1.28					#Density of air in kg/m**3
dp = 750					#Density of petrol in kg/m**3
An = 1.8					#Area of cross section of nozzle in mm**2
Cd = 0.6					#Coefficient of discharge of nozzle 
Ca = 0.84					#Coefficient of discharge of air

					
#Calculations
dpa = ((va/Ca)**2*(da/(2*9.81)))					#Change in pressure in kg/m**2
wf = ((An*10**-6)*Cd*math.sqrt(2*9.81*dp*(dpa-((x/1000)*dp))))					#Petrol consumption in kg/sec

					
#Output
print 'Petrol consumption is %3.4f kg/sec'%(wf)

Petrol consumption is 0.0023 kg/sec

Example 9.11 Page no : 214

In [12]:
import math 
					
#Input data
d = 0.155					#Diameter of orifice in mm
Cd = 0.94					#Coefficient of discharge
td = 3.18					#Throat diameter in cm
Ca = 0.84					#Coefficient of discharge
x = 29					#Venturi depression
dw = 0.89					#Minimum depression of water in cm
sa = 1.1					#Specific weight of air in kg/m**3
sg = 0.72					#Specific gravity of petrol
cyd = [7.75,10.75]					#Cylinder dimensions in cm
fc = 10.9					#Fuel consumption in kg/hr
N = 3200					#Speed in r.p.m
n = 4					#Number of cylinders

					
#Calculations
w = (((3.14/4)*td**2)/((3.14/4)*d**2))*(Ca/Cd)*math.sqrt((sa/(sg*1000))*(x/(x-dw)))					#Air fuel ratio
Va = (3.14/4)*(td/100)**2*Ca*math.sqrt(2*9.81*sa*x*6)					#Volume of air drawn/sec
vn = (Va/((3.14/4)*(cyd[0]/100)**2*(cyd[1]/100)*n*(N/(2*60))))*100					#Volumetric efficiency in percent

					
#Output
print 'Air fue ratio is %3.1f  \
\nVolumetric efficiency is %3.1f percent'%(w,vn)

Air fue ratio is 14.9  
Volumetric efficiency is 77.5 percent

Example 9.12 Page no : 219

In [13]:
import math 
					
#Input data
af = 0.066					#Air fuel ratio
p = 0.83					#Pressure at the venturi throat in kg/cm**2
pd = 0.04					#Pressure drop in kg/cm**2
va = 245					#Air flow at sea level in kg per hour

					
#Calculations
dpa = 1.03-p					#Pressure at air cleaner in kg/cm**2
d = (1.03-pd-dpa)					#Throat pressure when the air cleaner is fitted in kg/cm**2
naf = (af*math.sqrt((1.03-d)/dpa))					#New air fuel ratio

					
#Output
print 'a) Throat pressure when the air cleaner is fitted is %3.2f kg/cm**2  \
\nb) New air fuel ratio is %3.4f'%(d,naf)

a) Throat pressure when the air cleaner is fitted is 0.79 kg/cm**2  
b) New air fuel ratio is 0.0723

Example 9.13 Page no : 221

In [14]:
import math 
					
#Input data
x = 3					#Petrol smath.radians(numpy.arcmath.tan(s 3 mm below
Ta = 15.5+273					#Temperature of air in K
pa = 1.027					#Pressure of air in kg/cm**2
R = 29.27					#Characteristic gas constant in kg.m/kg.K 
sg = 0.76					#Specific gravity of fuel
fc = 6.4					#Consumption of fuel in kg/hour
jd = 1.27					#Jet diameter in mm
Cd = 0.6					#Nozzle discharge coefficienct
Ca = 0.8					#Discharge coefficient of air
af = 0.066					#Air fuel ratio

					
#Calculations
df = (sg*1000)					#Density of fuel in kg/m**3
da = (pa*10**4)/(R*Ta)					#Density of air in kg/m**3
va = Ca*math.sqrt((2*9.81*x*df)/(da*1000))					#Critical velocity of air in m/s
dpa = (((fc/(60*60))/((3.14/4)*(jd/1000)**2*Cd))**2/(2*9.81*df))+((x/1000)*df)					#Drop in pressure in kg/m**3
dpaw = (dpa/1000)*100					#Drop in pressure in cm of water
dj = math.sqrt(((fc/(3600*af))/(Ca*math.sqrt(2*9.81*da*dpa)))/(3.14/4))*1000					#Effective throat diameter in mm

					
#Output
print 'Critical air velocity is %3.2f m/sec  \
\nEffective throat diameter of the venturi is %3.1f mm  \
\nThe drop in pressure in the venturi is %3.2f cm of water'%(va,dj,dpaw)

Critical air velocity is 4.85 m/sec  
Effective throat diameter of the venturi is 21.4 mm  
The drop in pressure in the venturi is 36.73 cm of water

Example 9.14 Page no : 226

In [15]:
import math 
					
#Input data
ma = 6.11					#Flow rate of air in kg/min
mf = 0.408					#Flow arte of petrol in kg/min
dp = 768					#Density of petrol in kg/m**3
Ta = 15.5+273					#Temperature of air in K
pa = 1.027					#Pressure of air in kg/cm**2
R = 29.27					#Characteristic gas constant in kg.m/kg.K 
va = 97.5					#Speed of air in m/sec
Cv = 0.84					#Velocity coefficient
g = 1.4					#Ratio of specific heats 
x = 0.8					#pressure at the venturi is 0.8 of the pressure drop at the choke
Cd = 0.66					#Coefficient of discharge

					
#Calculations
rp = (1-((va/Cv)**2/(((2*9.81*g)/(g-1))*R*Ta)))**(g/(g-1))					#Pressure ratio
p2 = (pa*rp)					#Pressure in kg/cm**2
T2 = (Ta*rp**((g-1)/g))					#Temperature in K
da = (p2/(R*T2))*10**4					#Density in kg/m**3
daa = math.sqrt((ma/(60*va*da))/(3.14/4))*1000					#Throat diameter in mm
df = math.sqrt((mf/(60*Cd*math.sqrt(2*9.81*dp*x*(pa-p2)*10**4)))/(3.14/4))*1000					#Orifice diameter in mm

					
#Output
print 'Throat diameter of the choke is %i mm  \
\nThe orifice diameter is %3.2f mm'%(daa,df)

Throat diameter of the choke is 34 mm  
The orifice diameter is 2.05 mm

Example 9.15 Page no : 226

In [16]:
import math 
					
#Input data
ma = 6.8					#Mass flow rate of air in kg/min
mf = 0.45					#Mass flow rate of petrol in kg/min
pa = 1.033					#Pressure of air in kg/cm**2
Ta = 20+273					#Temperature of air in K
va = 97.5					#Velocity of air in m/s
Cv = 0.8					#Velocity coefficient
g = 1.4					#Ratio of specific heats 
R = 29.27					#Characteristic gas constant in kg.m/kg.K 
x = 0.75					#pressure at the venturi is 0.8 of the pressure drop at the choke
Cd = 0.65					#Coefficient of discharge
pw = 800					#Weight of petrol in kg per cu.m

					
#Calculations
rp = (1-((va/Cv)**2/(((2*9.81*g)/(g-1))*R*Ta)))**(g/(g-1))					#Pressure ratio
p2 = (pa*rp)					#Pressure in kg/cm**2
T2 = (Ta*rp**((g-1)/g))					#Temperature in K
da = (p2/(R*T2))*10**4					#Density in kg/m**3
daa = math.sqrt((ma/(60*va*da))/(3.14/4))*100					#Throat diameter in mm
df = math.sqrt((mf/(60*Cd*math.sqrt(2*9.81*pw*x*(pa-p2)*10**4)))/(3.14/4))					#Orifice diameter in mm

					
#Output
print 'Throat diameter of the choke is %3.2f cm  \
\nThe orifice diameter is %3.5f m'%(daa,df)

Throat diameter of the choke is 3.62 cm  
The orifice diameter is 0.00213 m

Example 9.16 Page no : 229

In [17]:
import math 
					
#Input data
n = 6.					#Number of cylinders
d = 100.					#Diameter in mm
L = 100.					#Stroke in mm
N = 1500.					#Speed in r.p.m
ap = 13.5					#Air fuel ratio
Ta = 80.+273					#Temperature of air in K
x = (7./8)					#Ratio of volume drawn
nth = 22.					#Thermal efficiency in percent
p = 76.					#Pressure in cm of mercury
CV = 9000.					#Calorific value of petrol in kcal/kg
l = 1524.					#Altitude in m
dp = 2.54					#Drop in pressure in cm of barometric reading
lx = 274					#Altitude rise in m

					
#Calculations
Vs = (3.14/4)*(d/10)**2*(L/10)					#Swept volume in c.c
Va = (x*Vs)					#Volume of air drawn in per cylinder per stroke in c.c
wa = (Va*10**-6*(N/2)*n)					#Weight of air supplied to the engine per minute in kg
wf = (wa/ap)					#Weight of fuel consumed per minute in kg
q = (wf*CV)					#Heat supplied to the engine per minute in kcal
P = (q*(nth/100)*427)/4500					#Power developed at ground level in H.P
db = (l/lx)*dp					#Drop in barometric reading at an altitude of 1524 m in cm
Pd = (P/p)*(p-db)					#Power developed at 1524 m altitude in H.P

					
#Output
print 'Power developed at the ground level is %i H.P  \
\nPower developed at an altitude of %i m is %i H.P'%(P,l,Pd)

Power developed at the ground level is 43 H.P  
Power developed at an altitude of 1524 m is 35 H.P