# Chapter 7 - Performance of IC Engines¶

## Example 1 - pg 7.19¶

In [2]:
#pg 7.19
#calculate the brake torque and Power
#Input data
N=1500.;#Engine speed in rpm
L=900.;#Length of brake arm in mm
g=9.81;#Gravitational force in N/m**2
pi=3.14;#Mathematical constant

#Calculations
T=((p*g)*(L/1000.));#Braking torque in Nm
P=((T/1000)*((2*3.14*N)/60));#Power available at the brakes of the engine in kW

#Output
print '(a) Brake torque is (Nm) = ',round(T,1)
print '(b)Power available at the brakes of the engine is (kW) = ',round(P,2)
print 'The answers given in textbook are wrong. Please verify using a calculator'

(a) Brake torque is (Nm) =  971.2
(b)Power available at the brakes of the engine is (kW) =  152.48
The answers given in textbook are wrong. Please verify using a calculator


## Example 2 - pg 7.19¶

In [4]:
#pg 7.19
#calculate the power available
#Input data
N=700.;#Engine speed in rpm
D=0.6;#Diameter of brake drum in m
d=0.05;#Diameter of rope in m
g=9.81;#Gravitational constant in N/m**2
pi=3.14;#Mathematical constant

#Calculations
P=(((W-S)*g*pi*(D+d))/1000)*(N/60);#Power in kW

#Output
print 'The power available at the brakes is (kW) = ',round(P,3)

The power available at the brakes is (kW) =  7.125


## Example 3 - pg 7.20¶

In [5]:
#pg 7.20
#calculate the brake thermal efficiency
#Input data
W=950.;#Load on hydraulic dynamometer in N
C=7500.;#Dynamometer constant
f=10.5;#Fuel used per hour in kg
h=50000.;#Calorific value of fuel in kJ/kg
N=400.;#Engine speed in rpm

#Calculations
P=(W*N)/C;#Power available at the brakes in kW
H=P*60;#Heat equivalent of power at brakes in kJ/min
Hf=(f*h)/60;#Heat supplied by fuel per minute in kJ/min
n=(H/Hf)*100;#Brake thermal efficiency in percentage

#Output
print ' Brake thermal efficiency of the engine is (percent) = ',round(n,2)

 Brake thermal efficiency of the engine is (percent) =  34.74


## Example 4 - pg 7.21¶

In [6]:
#pg 7.21
#calculate the specific fuel consumption and Brake mean effective pressure
#Input data
import math
n1=50.5;#Air standard efficiency in percentage
n2=50.;#Brake thermal efficiency in percentage
N=3000.;#Engine speed in rpm
H=10500.;#Heating value of fuel in kcal/kg
T=7.2;#Torque developed in kgf*m
B=6.3;#Bore diameter in cm
S=0.09;#stroke in m

#Calculations
nbt=(n1/100)*(n2/100.);#Brake thermal efficiency in percentage
B1=(2*(22./7)*N*T)/4500.;#Brake horse power in kW
B2=B1/4;#Brake horse power per cylinder in kW
Bsf=(4500*60)/(H*427.*nbt);#Brake specific fuel consumption in kg/BHP hr
bmep=(B2*4500)/(S*(math.pi*B**2. /4.)*(N/2.));#Brake mean effective pressure in kgf/cm**2

#Output
print '(a)Specific fuel consumption is (kg/BHP hr) = ',round(Bsf,3)
print '(b)Brake mean effective pressure is (kgf/cm^2) = ',round(bmep,3)

(a)Specific fuel consumption is (kg/BHP hr) =  0.238
(b)Brake mean effective pressure is (kgf/cm^2) =  8.066


## Example 5 - pg 7.22¶

In [8]:
#pg 7.22
#calculate the Mechanical efficiency
#Input data
W=30.;#The net dynamometer load in kg
N=2400.;#Speed in rpm
FHP=6.5;#Engine power in hp

#Calculations
BHP=(2*3.14*R*N*W)/4500;#Brake horse power in kW
IHP=BHP+FHP;#Indicated horse power in kW
nm=(BHP/IHP)*100;#Mechanical efficiency in percentage

#Output
print 'Mechanical efficiency of the engine is (percent) = ',round(nm,2)

Mechanical efficiency of the engine is (percent) =  88.54


## Example 6 - pg 7.22¶

In [9]:
#pg 7.22
#calculate the indicated, brake and Friction horse powers
#Input data
import math
d=25.;#Diameter of cylinder in cm
l=0.4;#Stroke of piston in m
N=200.;#Speed in rpm
m=10.;#Misfires per minute
M=6.2;#Mean effective pressure in kgf/cm**2
nm=0.8;#Mechanical efficiency in percent

#Calculations
np=(N/2)-m;#Number of power strokes per minute
A=(math.pi*d**2)/4;#Area of the cylinder
I=(M*l*A*np)/4500.;#Indicated horse power in kW
B=I*nm;#Brake horse power in kW
F=I-B;#Friction horse power in kW

#Output
print '(a)The indicated horse power is (kW) = ',round(I,2)
print '(b)The brake horse power is (kW) = ',round(B,2)
print '(c)Friction horse power is (kW) = ',round(F,2)

(a)The indicated horse power is (kW) =  24.35
(b)The brake horse power is (kW) =  19.48
(c)Friction horse power is (kW) =  4.87


## Example 7 - pg 7.23¶

In [10]:
#pg 7.23
#calculate the average piston speed
#Input data
import math
I=5.;#Indicated power developed by single cylinder of 2 stroke petrol engine
M=6.5;#Mean effective pressure in bar
d=0.1;#Diameter of piston in m

#Calculations
A=(math.pi*d**2)/4;#Area of the cylinder
LN=(I*1000*60.)/(M*10**5*A);#Product of length of stroke and engine speed
S=2*LN;#Average piston speed in m/s

#Output
print 'The average piston speed is (m/s) = ',round(S,2)

The average piston speed is (m/s) =  117.53


## Example 8 - pg 7.24¶

In [11]:
#pg 7.24
#calculate the diameter and stroke length
#Input data
P=60.;#Power developed by oil engine in kW
M=6.5;#Mean effective pressure in kgf/cm**2
N=85.;#Number of explosions per minute
r=1.75;#Ratio of stroke to bore diameter
nm=0.8;#Mechanical efficiency

#Calculations
I=P/nm;#Indicated horse power
d=((I*100*4*4500.)/(M*r*3.14*N))**(1./3);#Bore diameter in cm
l=r*d;#Stroke length in cm

#Output
print '(a)Diameter of the bore is (cm) = ',round(d,2)
print '(b)Stroke length of the piston is (cm) = ',round(l,3)

(a)Diameter of the bore is (cm) =  35.43
(b)Stroke length of the piston is (cm) =  61.999


## Example 9 - pg 7.24¶

In [13]:
#pg 7.24
#calculate the bore diameter and stroke length
#Input data
I=45.;#Power developed by two cylinder internal combustion engine operating on two stroke principle
N=1100.;#Speed in rpm
M=6.;#Mean effective pressure in kgf/cm**2
r=1.3;#Ratio of stroke to the bore
nc=2.;#Number of cylinders

#Calculations
d=((I*4500*4)/(M*(r/100)*3.14*N*nc))**(1./3);#Diameter of the bore in cm
l=1.3*d;#Stroke length in cm

#Output
print '(a)The bore diameter of the cylinder is (cm) = ',round(d,2)
print '(b)Stroke length of the piston is (cm) = ',round(l,2)

(a)The bore diameter of the cylinder is (cm) =  11.46
(b)Stroke length of the piston is (cm) =  14.89


## Example 10 - pg 7.25¶

In [14]:
#pg 7.25
#calculate the volumetric efficiency
#Input data
d=6.;#Diameter of the bore in cm
l=9.;#Length of the stroke in cm
m=0.00025;#Mass of charge admitted in each suction stroke
R=29.27;#Gas constant Kgfm/kg K
p=1.;#Normal pressure in kgf/cm**2
T=273.;#Temperature in K

#Calculations
V=(m*R*T)*10**6/(p*10**4);#Volume of charge admitted in each cycle in m**3
Vs=(3.14*d**2*l)/4;#Swept volume of the cylinder
nv=(V/Vs)*100;#Volumetric efficiency in percentage

#Output
print 'The volumetric efficiency is (percent) = ',round(nv,1)

The volumetric efficiency is (percent) =  78.5


## Example 11 - pg 7.26¶

In [16]:
#pg 7.26
#calculate the volumetric efficiency of the engine
#Input data
import math
d=0.12;#Diameter of the bore in m
l=0.13;#Length of stroke in m
N=2500.;#Speed of the engine in rpm
d1=0.06;#Diameter of the orifice in m
Cd=0.70;#Discharge coefficient of orifice
hw=33.;#Heat causing air flow through orifice in cm of water
p=760.;#Barometric reading in mm of Hg
T1=298.;#Ambient temperature in degree K
p1=1.013;#Pressure of air at the end of suction in bar
T2=22.;#Temperature of air at the end of suction in degree C
R=0.287;#Universal gas constant
n=6.;#Number of cylinders in the engine
n1=1250.;#Number of strokes per minute for a four stroke engine operating at 2500 rpm

#Calculations
V=(math.pi*d**2*l)/4;#Swept volume of piston in m**3
Ao=(math.pi*d1**2)/4;#Area of the orifice in m**2
rho=p1*10**5/((R*T1)*1000);#Density of air at 1.013 bar and 22 degrees C
Va=840.*Cd*Ao*(hw/rho)**(1./2);#Volume of air passing through the orifice in m**3/min
V1=8.734/n;#Actual volume of air per cylinder in m**3/min
As=V1/n1;#Air supplied per cycle per cylinder in m**3
nv=(As/V)*100;#Volumetric efficiency of the engine in percentage

#Output
print 'The volumetric efficiency of the engine is (percent) = ',round(nv,2)

The volumetric efficiency of the engine is (percent) =  79.21


## Example 12 - pg 7.27¶

In [18]:
#pg 7.27
#calculate the air standard efficiency and Indicated power, thermal efficiency
#Input data
import math
d=0.15;#Diameter of the piston in m
l=0.19;#Length of the stroke in m
V=0.00091;#Clearance volume in m**3
N=250.;#Speed of the engine in rpm
M=6.5;#Indicated mean effective pressure in bar
c=6.3;#Gas consumption in m**3/hr
H=16000.;#Calorific value of the has in kJ/m**3
r1=1.4;#Polytropic index

#Calculations
Vs=(math.pi*d**2*l)/4;#Swept volume in m**3
Vt=Vs+V;#Total cylinder volume in m**3
r=Vt/V;#Compression ratio
na=(1-(1/r**(r1-1)))*100;#Air standard efficiency in percent
A=(math.pi*d**2)/4;#Area of the bore in m
I=(M*10**5*l*A*N)/(1000*60);#Indicated power in kW
Hs=(c*H)/(60*60);#Heat supplied per second
nt=(I/Hs)*100;#Indicated thermal efficiency in percent

#Output
print '(a)The air standard efficiency is (percent) = ',round(na,1)
print '(b)Indicated power is (kW) = ',round(I,3)
print '(c)Indicated thermal efficiency is (percent) = ',round(nt,1)

(a)The air standard efficiency is (percent) =  46.1
(b)Indicated power is (kW) =  9.093
(c)Indicated thermal efficiency is (percent) =  32.5


## Example 13 - pg 7.28¶

In [19]:
#pg 7.28
#calculate the diameter in all cases
#Input data
import math
ma=6.;#Air supplied per minute by a single jet carburetor in kg/min
mf=0.44;#Mass flow rate of petrol in kg/min
s=0.74;#Specific gravity of petrol in kg/m**3
p1=1.;#Initial pressure of air in bar
T1=300.;#Initial temperature of air in K
Ci=1.35;#Isentropic coefficient of air
V=90.;#Speed of air in the venturi in m/s
Vc=0.85;#Velocity coefficient of the venturi in m/s
Cf=0.66;#Coefficient of discharge for the jet
Cp=1005.;#Coefficient of pressure in J/kg K
n=1.35;#Isentropic coefficient of air
R=0.281;#Real gas constant in Nm/kg K
rhof=740.;#Density of fuel in mm of Hg

#Calculations
p2=(1-((V/Vc)**(2)/(2*T1*Cp)))**((n)/(n-1));#Pressure at the venturi in bar
V1=((R*T1)/(p1*10**5))*1000;#Initial volume in m**3/kg
V2=V1*((p1/p2)**(0.741));#Final volume in m**3/kg
A2=((ma*V2)/(V*60.))*10**4;#Throat area of venturi in cm**2
d=((A2*4.)/math.pi)**(0.5);#Diameter of venturi in cm
deltaPa=1-p2;#Pressure drop causing air flow in bar
deltaPf=0.8*deltaPa;#Pressure drop causing fuel flow in bar
Af=(mf/60.)*(10**4)/((Cf)*(2*rhof*deltaPf*10**5)**(1./2));#Area through which fuel flows in cm**2
df=((Af*(4/math.pi))**(1./2))*10.;#Diameter of fuel jet in mm

print '(a)The diameter of the venturi of the venturi if the air speed is 90 m/s is (cm) = ',round(d,2)
print '(b)The diameter of the jet if the pressure drop at the jet is 0.8 times the pressure drop at the venturi is (mm) = ',round(df,3)

(a)The diameter of the venturi of the venturi if the air speed is 90 m/s is (cm) =  3.55
(b)The diameter of the jet if the pressure drop at the jet is 0.8 times the pressure drop at the venturi is (mm) =  2.218


## Example 14 - pg 7.30¶

In [21]:
#pg 7.30
#calculate the weight of fuel
#Input data
r=14.;#The compression ratio of a diesel engine
Vc=1.;#Clearance volume in m**3
c=0.08;#Fuel supply cut off point
nr=0.55;#Relative efficiency
H=10000.;#Calorific value of fuel in kcal/kg
r1=1.4;#Ratio of specific heat of air
Vs=13.;#Stroke volume in m**3

#Calculations
rho=Vc+(c*Vs);#Cut off ratio
na=1-(1*(rho**r1-1)/((r**(r1-1)*r1)*(rho-1)));#Air standard efficiency of diesel cycle in percent
In=(na*nr);#Indicated thermal efficiency in percent
H1=(4500*60)/(In*427.);#Heat in fuel supplied/1HP hr
W=H1/10**4;#Weight of fuel required/1HP hr

#Output
print 'The weight of fuel required per 1HP hr is (kg) = ',round(W,4)

The weight of fuel required per 1HP hr is (kg) =  0.1947


## Example 15 - pg 7.31¶

In [24]:
#pg 7.31
#calculate the quantity of fuel
#Input data
P=120;#Power developed by a six cykinder four stroke diesel engine
N=2400;#Speed in rpm
f=0.2;#Brake specific fuel consumption in kg/kWh
s=0.85;#Specific gravity of fuel

#Calculations
F=f*P;#Fuel consumed per hour in kg
F1=F/6;#Fuel consumed per cylinder in kg/h
n=(N*60.)/2;#Number of cycles per hour
F2=(F1/n)*10**3;#Fuel consumption per cycle in gm
V=F2/s;#Volume of fuel to be injected per cycle in cc

#Output
print 'The quantity of fuel to be injected per cycle per cylinder is (cc) = ',round(V,4)

The quantity of fuel to be injected per cycle per cylinder is (cc) =  0.0654


## Example 16 - pg 7.32¶

In [25]:
#pg 7.32
#calculate the diameter of the orifice
#Input data
P=20.;#Power developed by a four stroke diesel engine per cylinder in kW
N=2000.;#Operating speed of the diesel engine in rpm
s=0.25;#Specific fuel consumption in kh/kW
p1=180.;#Pressure of fuel injected in bar
d=25.;#Distance travelled by crank in degrees
p2=38.;#Pressure in the combustion chamber in bar
Cd=0.85;#Coefficient of velocity
A=30.;#API in degrees

#Calculations
T=d/(360.*(N/60));#Duration of fuel injection in s
SG=(141.5/(131.5+A))*10**3;#Specific gravity of fuel
V=Cd*(2*(p1-p2)*10**5/SG)**(1./2);#Velocity of fuel injection in m/s
Vf=(s/60.)*P/((N/2)*SG);#Volume of fuel injected per cycle in m**3/cycle
Na=Vf/(V*T);#Nozzle orifice area in m**2
d=(((4*Na)/3.14)**(1./2))*10**3;#Diameter of the orifice of the fuel injector in mm

#Output
print 'The diameter of the orifice is (mm) = ',round(d,4)

The diameter of the orifice is (mm) =  0.6165


## Example 17 - pg 7.33¶

In [26]:
#pg 7.33
#calculate the total orifice area
#Input data
P=200.;#Power developed by a six cylinder diesel engine in kW
N=2000.;#Operating speed of the engine in rpm
bs=0.2;#The brake specific fuel consumption in kg/kWh
p1=35.;#The pressure of air in the cylinder at the beginning of injection in bar
p2=55.;#Maximum cylinder pressure in bar
p3=180.;#Initial injection pressure in bar
p4=520.;#Maximum pressure at the injector in bar
Cd=0.75;#Coefficient of discharge
S=850.;#Specific gravity of fuel
p5=1.;#Atmospheric pressure in bar
a=16.;#The crank angle over which injection takes place in degrees

#Calculations
Po=P/6.;#Power output per cylinder in kW
F=(Po*bs)/60.;#Fuel consumed per cylinder in kg/min
Fi=F/(N/2.);#Fuel injected per cycle in kg
T=a/(360.*(N/60));#Duration of injection in s
deltaP1=p3-p1;#Pressure difference at the beginning of injection in bar
deltaP2=p4-p2;#Pressure difference at the end of injection in bar
avP=(deltaP1+deltaP2)/2;#Average pressure difference in bar
V=Cd*(2.*(avP*10**5)/S)**(1./2);#Velocity of injection of fuel jet in m/s
Vo=Fi/S;#Volume of fuel injected per cycle in m**3/cycle
A=(Vo/(V*T));#Area of fuel orifices in m**2

#Output
print 'The total orifice area required per injector if the injection takes place over 16 degree crank angle is (m^2) = ',round(A,11)

The total orifice area required per injector if the injection takes place over 16 degree crank angle is (m^2) =  4.8796e-07


## Example 18 - pg 7.34¶

In [27]:
#pg 7.34
#calculate the indicated mean effective pressure and indicated power
#Input data
A=450.;#Area of indicator diagram in mm^2
l=60.;#Length of indicator diagram in mm
s=1.1;#Spring number in bar/mm
d=0.1;#Diameter of piston in m
L=0.13;#Length of stroke in m
N=400.;#Operating speed of the engine in rpm

#Calculations
Av=A/l;#Average height of indicator diagram in mm
pm=Av*s;#Mean effective pressure in bar
np=N/2.;#Number of power strokes per minute for a four stroke diesel engine
Ar=(3.14*d**2)/4;#Area of the piston in m^2
I=(pm*10**5*L*Ar*np)/(1000*60);#Indicated power in kW

#Output
print '(a)The indicated mean effective pressure is (bar) = ',pm
print '(b)Indicated power is (kW) = ',round(I,2)

(a)The indicated mean effective pressure is (bar) =  8.25
(b)Indicated power is (kW) =  2.81


## Example 19 - pg 7.35¶

In [28]:
#pg 7.35
#calculate the brake, Indicated horse power and Thermal efficiency
#Input data
d=25.;#Diameter of the bore in cm
l=0.4;#Stroke length in m
N=300.;#Operating speed of the engine in rpm
n=120.;#Number of explosions per minute
pm=6.7;#Mean effective pressure in kgf/cm**2
R=0.75;#Radius of brake drum in m
f=0.22;#Fuel supplied per minute in m**3
C=4500.;#Calorific value of fuel in kcal/m**3

#Calculations
BHP=(2*3.14*R*N*Tnet)/4500;#Brake horse power in kW
A=(3.14*d**2)/4;#Area of the cylinder in cm**2
IHP=(pm*l*A*n)/4500;#Indicated horse power in kW
H=f*C;#Heat supplied by fuel per minute in kcal
nt1=((IHP*C)/(990*427))*100;#Thermal efficiency on IHP basis in percent
nt2=((BHP*C)/(990*427))*100;#Thermal efficiency on BHP basis in percent

#Output
print '(a)The brake horse power is (kW) = ',round(BHP,2)
print '(b)Indicated horse power is (kW) = ',round(IHP,3)
print '(c)Thermal efficiency on IHP basis is (percent) = ',round(nt1,2)
print '(d)Thermal efficiency on BHP basis is (percent) = ',round(nt2,2)

(a)The brake horse power is (kW) =  28.26
(b)Indicated horse power is (kW) =  35.063
(c)Thermal efficiency on IHP basis is (percent) =  37.33
(d)Thermal efficiency on BHP basis is (percent) =  30.08


## Example 20 - pg 7.36¶

In [29]:
#pg 7.36
#calculate the brake, Indicated horse power and Thermal efficiency
#Input data
D=0.6;#Brake wheel diameter of a constant speed compression ignition engine operating on four stroke cycle in m
t=0.01;#Thickness of brake band in m
N=500.;#Operating speed of the engine in rpm
W=20.;#Load on brake band in kgf
l=6.25;#Length of indicator diagram in cm
A=4.35;#Area of indicator diagram in cm**2
Sn=11.;#Spring number in kgf/cm**2/cm
d=10.;#Diameter of the bore in cm
L=0.13;#Length of the stroke in m
F=0.23;#Specific fuel consumption in kg/BHP hr
CV=10000.;#Heating value of fuel in kcal/kg

#Calculations
BHP=(3.14*(D+t)*N*(W-S))/4500;#Brake horse power in kW
MEP=(A*Sn)/l;#Mean effective pressure in kgf/cm**2
Ar=(3.14*d**2)/4;#Area of the cylinder in cm**2
np=N/2;#Number of explosions per minute
IHP=(MEP*L*Ar*np)/4500;#Indicated horse power in kW
nm=(BHP/IHP)*100;#Mechanical efficiency in percentage
Wf=F*BHP;#Fuel consumption per hr in kg/hr
nt=((IHP*4500*60)/(Wf*CV*427))*100;#Indicated thermal efficiency in percentage
nb=((BHP*4500*60)/(Wf*CV*427))*100;#Brake thermal efficiency in kW

#Output
print '(a)The brake horse power is (kW) = ',round(BHP,2)
print '(b)Indicated horse power is (kW) = ',round(IHP,3)
print '(c)Mechanical efficiency is (percent) = ',round(nm,1)
print '(d)Indicated thermal efficiency is (percent) = ',round(nt,0)
print '(e)Brake thermal efficiency is (percent) = ',round(nb,1)

(a)The brake horse power is (kW) =  3.62
(b)Indicated horse power is (kW) =  4.341
(c)Mechanical efficiency is (percent) =  83.4
(d)Indicated thermal efficiency is (percent) =  33.0
(e)Brake thermal efficiency is (percent) =  27.5


## Example 21 - pg 7.38¶

In [31]:
#pg 7.38
#calculate the indicated thermal efficiency
#Input data
N=1200.;#Operating speed of a four cylinder engine in rpm
BHP=25.3;#The brake horse power when all 4 cylinders are operating in kW
T=10.5;#The average torque when one cylinder was cut out in mkgf
CV=10000.;#Calorific value of the fuel used in kcal/kg
f=0.25;#The amount of petrol used in engine per BHP hour
J=427.;#

#Calculations
BHP1=(2*3.14*N*T)/4500.;#BHP for 3 cylinders when 1 cylinder is cut out in kW
IHP=BHP-BHP1;#IHP of one cylinder in kW
IHPt=IHP*4.;#Total IHP of the engine with 4 cylinders
Wf=(f*BHP)/60.;#Fuel used per minute in kg
ni=((IHPt*4500.)/(Wf*CV*J))*100;#Indicated thermal efficiency in percent
nm=(BHP/IHPt)*100;#Mechanical efficiency in percent
nb=(IHPt*nm)/100;#Brake thermal efficiency in percent

#Output
print 'The indicated thermal efficiency is (percent) = ',round(ni,1)
print 'Mechanical efficiency is (percent) = ',round(nm,1)
print 'Brake thermal efficiency is (percent) = ',round(nb,1)

The indicated thermal efficiency is (percent) =  30.9
Mechanical efficiency is (percent) =  82.0
Brake thermal efficiency is (percent) =  25.3


## Example 22 - pg 7.39¶

In [32]:
#pg 7.39
#calculate the IHP of the engine and Mechanical efficiency
#Input data
B=32.;#Brake horse power in kW with all cylinders working
B1=21.6;#BHP with number 1 cylinder cut out in kW
B2=22.3;#BHP with number 2 cylinder cut out in kW
B3=22.5;#BHP with number 3 cylinder cut out in kW
B4=23.;#BHP with number 4 cylinder cut out in kW

#Calculations
I1=B-B1;#Indicated horse power of number 1 cylinder in kW
I2=B-B2;#IHP of number 2 cylinder in kW
I3=B-B3;#IHP of number 3 cylinder in kW
I4=B-B4;#IHP of number 4 cylinder in kW
I=I1+I2+I3+I4;#Total IHP of the engine in kW
nm=(B/I)*100;#Mechanical efficiency in percent

#Output
print '(a)The IHP of the engine is (kW) = ',I
print '(b)Mechanical efficiency is (percent) = ',round(nm,1)

(a)The IHP of the engine is (kW) =  38.6
(b)Mechanical efficiency is (percent) =  82.9


## Example 23 - pg 7.40¶

In [33]:
#pg 7.40
#calculate the Compression ratio, indicated thermal efficiency, brake specific fuel consumption and bore diameter
#Input data
r=15.;#The air fuel ratio by weight
CV=45000.;#Calorific value of fuel in kJ/kg
nm=85.;#Mechanical efficiency of 4 stroke 4 cylinder engine in percent
na=53.;#Air standard efficiency of the engine in percent
nr=65.;#Relative efficiency of the engine in percent
nv=80.;#Volumetric efficiency of the engine in percent
r1=1.3;#Stroke to bore ratio
p1=1.;#Suction pressure in bar
T=303.;#Suction temperature in K
S=3000.;#The operating speed of the engine in rpm
P=75.;#Power at brakes in kW
r2=1.4;#Ratio of specific heats for air
R1=0.287;#Characteristic gas constant for air fuel mixture in kJ/kg K

#Calculations
R=(1/(1-(na/100)))**(1/(r2-1));#Compression ratio of the engine
nti=((na/100)*(nr/100))*100;#The indicated thermal efficiency in percent
Pi=P/(nm/100);#Indicated power in kW
F=Pi/((nti*CV)/100);#Fuel per second injected in kg/sec
B=F/P;#Brake specific fuel consumption in kg/kWsec
A=1+r;#Mass of fuel mixture entering the engine foe every one kg of fuel in kg
m=A*F;#Mass of air fuel mixture per second in kg
V=(m*R1*T)/(p1*10**5/1000);#Volume of air fuel mixture supplied to the engine per sec
Vs=V/(nv/100);#Swept volume per second in m**3/sec
d=((Vs*2*60*4)/(S*3.14*r1*4))**(1./3)*1000;#Diameter of the bore in mm
L=r1*d;#Stroke length in mm

#Output
print '(a)Compression ratio = ',round(R,1)
print '(b)Indicated thermal efficiency is (percent) = ',nti
print '(c)Brake specific fuel consumption is (kg/kW sec) = ',round(B,7)
print '(d)Bore diameter of the engine is (mm) = ',round(d,2)
print '(e)Stroke length of the engine is (mm) = ',round(L,1)

(a)Compression ratio =  6.6
(b)Indicated thermal efficiency is (percent) =  34.45
(c)Brake specific fuel consumption is (kg/kW sec) =  7.59e-05
(d)Bore diameter of the engine is (mm) =  98.99
(e)Stroke length of the engine is (mm) =  128.7


## Example 24 - pg 7.42¶

In [34]:
#pg 7.42
#calculate the power and efficiency in all cases
#Input data
d=0.3;#Diameter of the bore in m
L=0.45;#Stroke length in m
N=220.;#Operating speed of the engine in rpm
T=3600.;#Duration of trial in sec
F=7.;#Fuel consumption in kg per minute
CV=45000.;#Calorific value of fuel in kJ/kg
A=320.;#Area of indicator diagram in mm**2
l=60.;#Length of indicator diagram in mm
S=1.1;#Spring index in bar/mm
W=130.;#Net load on brakes in kg
D=1.65;#Diameter of brake drum in m
W1=500.;#Total weight of jacket cooling water in kg
t=40.;#Temperature rise of jacket cooling water in degrees celsius
t1=300.;#Temperature of exhaust gases in degrees celsius
ma=300.;#Air consumption in kg
sg=1.004;#Specific heat of exhaust gas in kJ/kgK
sw=4.185;#Specific heat of water in kJ/kgK
t2=25.;#Room temperature in degrees celsius
g=9.81;#gravity

#Calculations
P=(W*g*3.14*D*N)/(1000*60);#Power available at brakes in kW
pm=(A*S)/l;#Mean effective pressure in bar
I=(pm*10**5*L*((3.14*d**2)/4)*N)/(1000.*2*60);#Indicated power developed in kW
nm=(P/I)*100;#Mechanical efficiency in percent
nt=(P/((F/T)*CV))*100;#Brake thermal efficiency in percent
ni=(I/((F/T)*CV))*100;#Indicated thermal efficiency in percent
Hs=F*CV;#Heat supplied on one hour basis
Hp=P*T;#Heat equivalent of brake power in kJ
Hf=(I-P)*3600;#Heat lost in friction in kJ
Hc=W1*t*sw;#Heat carried away by cooling water in kJ
He=(ma+F)*(t1-t2)*sg;#Heat carried away by exhaust gas in kJ
Hu=Hs-(Hp+Hf+Hc+He);#Heat unaccounted in kJ
nb=(He/Hs)*100;#Heat equivalent of power at brakes in percent
nf=(Hf/Hs)*100;#Heat lost in friction in percent
nw=(Hc/Hs)*100;#Heat removed by jacket water in percent
ne=(He/Hs)*100;#Heat carried away by exhaust gases in percent
nu=(Hu/Hs)*100;#Heat unaccounted in percent

#Output
print '(a)Power available at brakes is (kW) = ',round(P,2)
print '(b)Indicated power developed is (kW) = ',round(I,2)
print '(c)Mechanical efficiency is (percent) = ',nm
print '(d)Brake Thermal efficiency is (percent) = ',round(nt,2)
print '(e)Indicated thermal efficiency is (percent) = ',round(ni,2)
print 'Heat balance :'
print 'Heat supplied by fuel (kJ/hr) = ',Hs
print 'Heat equivalent of power of brakes (percent) = ',round(nb,1)
print 'Heat equivalent of loss in friction (percent) = ',round(nf,1)
print 'Heat equivalent of removed through jacket (percent) = ',round(nw,1)
print 'Heat equivalent of carried away by gases (percent) = ',round(ne,2)
print 'Heat equivalent of unaccounted (percent) = ',round(nu,1)

(a)Power available at brakes is (kW) =  24.23
(b)Indicated power developed is (kW) =  34.19
(c)Mechanical efficiency is (percent) =  70.85
(d)Brake Thermal efficiency is (percent) =  27.69
(e)Indicated thermal efficiency is (percent) =  39.08
Heat balance :
Heat supplied by fuel (kJ/hr) =  315000.0
Heat equivalent of power of brakes (percent) =  26.9
Heat equivalent of loss in friction (percent) =  11.4
Heat equivalent of removed through jacket (percent) =  26.6
Heat equivalent of carried away by gases (percent) =  26.91
Heat equivalent of unaccounted (percent) =  7.4


## Example 25 - pg 7.46¶

In [35]:
#pg 7.46
#calculate the Indicated, brake horse power
#Input data
d=25.;#The bore diameter of a single cylinder 4 stroke engine in cm
l=0.38;#Stroke length in m
t=3600.;#Duration of test in sec
r=19710.;#Total number of revolutions
F=6.25;#Fuel oil used in kg
A=5.7;#Area of indicator diagram in cm**2
L=7.6;#Length of indicator diagram in cm
S=8.35;#Spring number in kgf/cm**3
P=63.5;#Net load on brake drum in kg
R=1.2;#Radius of brake drum in m
Ww=5.7;#Rate of coolant flow in kg/min
deltaT=44.;#Temperature rise of coolant in degrees celsius
T1=15.5;#Atmospheric temperature in degrees celsius
As=30.;#Air supplied per kg of fuel
CV=10600.;#Calorific value of fuel in kcal/kg
Te=390.;#Exhaust gas temperature in degrees celsius
sm=0.25;#Mean specific heat of exhaust gas

#Calculations
Hs=(F*CV)/60.;#Heat supplied by fuel per minute in kcal
pm=(A*S)/L;#Mean effective pressure in kgf/cm**2
I=(pm*l*(3.14*d**2)*r)/(4*60.*2*4500);#Indicated horse power in kW
B=(P*R*2*3.14*r)/(4500*60);#Brake horse power in kW
Hei=(I*4500)/427.;#Heat equivalent of IHP/min in kcal
Heb=(B*4500)/427.;#Heat equivalent of BHP/min in kcal
Hf=Hei-Heb;#Heat in friction per minute in kcal
Hc=Ww*deltaT;#Heat carried away by coolant in kcal
We=(F+(As*F))/60.;#Weight of exhaust gases per minute
He=We*(Te-T1)*sm;#Heat carried away by exhaust gases in kcal

#Output
print '(a)Indicated horse power is (kcal) = ',round(I,2)
print '(b)Brake horse power developed is (kcal) = ',round(B,2)
print '(c)Heat equivalent of friction is (kcal) = ',round(Hf,1)

(a)Indicated horse power is (kcal) =  42.62
(b)Brake horse power developed is (kcal) =  34.93
(c)Heat equivalent of friction is (kcal) =  81.0


## Example 26 - pg 7.48¶

In [36]:
#pg 7.48
#calculate the percentage of heat carried away
#Input
F=10.;#Quantity of fuel supplied during the trial of a diesel engine in kg/hr
CV=42500.;#Calorific value of fuel in kJ/kg
r=20.;#Air fuel ratio
T=20.;#Ambient temperature in degrees celsius
mw=585.;#Water circulated through the gas calorimeter in litres/hr
T1=35.;#Temperature rise of water through the calorimeter in degrees celsius
T2=95.;#Temperature of gases at exit from the calorimeter in degrees celsius
se=1.05;#Specific heat of exhaust gases in kJ/kgK
sw=4.186;#Specific heat of water in kJ/kgK

#Calculations
M=(F/60.)*(r+1);#Mass of exhaust gases formed per minute
H=((mw/60.)*sw*T1)+(M*se*(T2-T));#Heat carried away by the exhaust gases per minute in kJ/min
Hs=(F/60.)*CV;#Heat supplied by fuel per minute in kJ/min
nh=(H/Hs)*100;#Percentage of heat carried away by the exhaust gas

#Output
print 'Percentage of heat carried away by exhaust gas is (percent) = ',round(nh,2)

Percentage of heat carried away by exhaust gas is (percent) =  24.06


## Example 27 - pg 7.49¶

In [37]:
#pg 7.49
#calculate the Percentage of heat carried away
#Input data
F=11.;#Fuel used per hour observed during the trial of a single cylinder four stroke diesel engine in kg
mc=85.;#Carbon present in the fuel in percent
mh=14.;#Hydrogen present in the fuel in percent
mn=1.;#Non combustibles present in the fuel in percent
CV=50000.;#Calorific value of fuel in kJ/kg
Vc=8.5;#Percentage of carbon dioxide present in exhaust gas by Volumetric analysis
Vo=10.;#Oxygen present in exhaust gases in percent
Vn=81.5;#Nitrogen present in exhaust gases in percent
Te=400.;#Temperature of exhaust gases in degrees celsius
se=1.05;#Specific heat of exhaust gas in kJ/kg
Pp=0.030;#Partial pressure of steam in the exhaust in bar
Ta=20.;#Ambient temperature in degrees celsius
hs=2545.6;#Enthalpy of saturated steam in kJ/kg
Tsa=24.1;#Saturation temperature from graph in degrees celcius
Cp=2.1;#Specific heat in kJ/kg K
hst=3335.;#Enthalpy of super heated steam in kJ/kg
F1=9.
#Calculations
Ma=(Vn*mc)/(33.*Vc);#Mass of air supplied per kg of fuel in kg
Me=Ma+1;#Mass of exhaust gases formed per kg of fuel in kg
me=(Me*F)/60.;#Mass of exhaust gases formed per minute in kg
ms=F1*(mh/100.);#Mass of steam formed per kg of fuel in kg
ms1=(ms*F)/60.;#Mass of steam formed per minute in kg
mde=me-ms1;#Mass of dry exhaust gases formed per minute in kg
H=mde*se*(Te-Ta);#Heat carried away by the dry exhaust gases per minute in kJ/min
Es=hs+(Cp*(Te-Tsa));#Enthalpy of superheated steam in kJ/kg
He=ms1*hst;#Heat carried away by steam in the exhaust gases in kJ/min
Hl=H+He;#Total heat lost through dry exhaust gases and steam in kJ/min
Hf=(F/60.)*CV;#Heat supplied by fuel per minute in kJ/min
nh=(Hl/Hf)*100.;#Percentage of heat carried away by exhaust gases

#Output
print 'Percentage of heat carried away by exhaust gases is (percent) = ',round(nh,1)

Percentage of heat carried away by exhaust gases is (percent) =  27.9


## Example 28 - pg 7.51¶

In [38]:
#pg 7.51
#calculate the net increase in brake power
#Input data
C=0.0033;#The capacity of a four stroke engine of compression ignition type
I=13.;#Average indicated power developed in kW/m**3
N=3500.;#Operating speed of the engine
nv=80.;#Volumetric efficiency in percentage
p1=1.013;#Initial pressure in bar
T1=298.;#Initial temperature in K
r=1.75;#Pressure ratio of the engine
ni=75.;#The isentropic efficiency in percentage
nm=80.;#mechanical efficiency in percentage
r1=1.4;#Polytropic index

#Calculations
Vs=(N/2.)*C;#Swept volume in m**3/min
Vi=Vs*(nv/100);#Unsupercharged engine inducted volume in m**3/min
Pb=p1*r;#Blower delivery pressure in bar
T2s=((r)**((r1-1)/r1))*T1;#Final temperature in K
T2=((T2s-T1)/(ni/100.))+T1;#Blower delivery temperature in K
Ve=((Pb*Vs)*T1)/(T2*p1);#Equivalent volume at 1.013 bar and 298K in m**3/min
Vin=Ve-Vi;#Increase in inducted volume of air in m**3/min
Pin=Vin*I;#Increase in indicated power due to extra air inducted in kW
Pinp=((Pb-p1)*Vs*100.)/60.;#Increase in indicated power due to increase in induction pressure in kW
Pt=Pin+Pinp;#Total increase in indicated power in kW
nb=Pt*(nm/100.);#Total increase in brake power efficiency in kW
ma=(Pb*Vs*100.)/(60*0.287*T2);#Mass of air delivered by the blower in kg/s
Wb=ma*1.005*(T2-T1);#Work input to air by blower in kW
Pb1=Wb/(nv/100.);#Power required to drive the blower in kW
Pb2=nb-Pb1;#Net increase in brake power in kW

#Output
print 'The net increase in brake power is (kW) = ',round(Pb2,2)

The net increase in brake power is (kW) =  29.15