# CHAPTER 17 - Testing and performance of I.C. engines¶

## EXAMPLE 17.1 PAGE 570¶

In [34]:
from __future__ import division
from math import pi
# Initialisation of Variables
Pmi=6#.....................#Mean effective pressure in bar
N=1000#....................#Engine rpm
d=0.11#.....................#Diameter of piston in m
l=0.14#.....................#Stroke length in m
n=1#........................#No of cylinders
k=1#........................#k=1 for two stroke engine
#Calculations
V=l*(pi/4)*d*d#.............#Volume of the cylinder in m**3
IP=(n*Pmi*V*k*10*N)/6#.........#Indicated Power developed in kW
print "Indicated power developed = %0.2f kW"%IP

Indicated power developed = 13.30 kW


## EXAMPLE 17.2 PAGE 571¶

In [35]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=4#........................#No of cylinders
P=14.7#....................#Power developed in kW
N=1000#....................#Engine speed in rpm
Pmi=5.5#....................#Mean effective pressure in bar
lbyd=1.5#...................#Ratio of stroke to bore
k=0.5#.......................#For four stroke engine
#Calculations
d=((P*6)/(n*Pmi*N*k*10*(pi/4)*lbyd))**(1/3)#......................#Calculation of bore in m
l=lbyd*d#................................#Calculation of stroke in m
print "The bore of the engine = %0.2f mm"%(d*1000)
print "The stroke of the engine = %0.2f mm"%(l*1000)

The bore of the engine = 87.96 mm
The stroke of the engine = 131.94 mm


## EXAMPLE 17.3 PAGE 572¶

In [36]:
from __future__ import division
from math import pi
# Initialisation of Variables
Db=0.6#.....................#Diameter of the brake wheel in m
d=0.026#......................#Diameter of the rope in m
N=450#......................#Engine speed in rpm
#Calculations
BP=((W-S)*pi*(Db+d)*N)/(60*1000)#...............#Brake Power in KW
print "Brake Power = %0.2f KW"%(BP)

Brake Power = 2.51 KW


## EXAMPLE 17.4 PAGE 572¶

In [37]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=4#........................#No of cylinders
k=0.5#......................#For four stroke engine
Tb=160#.....................#Max brake torque in Nm
N=3000#......................#Engine rpm
Pm=9.6#....................#Brake mean effective pressure in bar
#Calculations
D=((2*pi*N*Tb*6)/(60*1000*Pm*(pi/4)*N*k*10))**(1/3)#.....................#Bore of engine in m
L=D#...................#Given that the stroke is equal to bore
Disp=(pi/4)*D*D*L#....................................#Displacement in m**3
print "Since it is given that the stroke is equal to bore, their value is = %0.2f mm"%(D*1000)
print "The engine displacement = %0.2f m**3"%(Disp)

Since it is given that the stroke is equal to bore, their value is = 138.67 mm
The engine displacement = 0.00 m**3


## EXAMPLE 17.5 PAGE 572¶

In [38]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=6#.....................#No of cylinders
Pmb=6#....................#Brake mean effective pressure in bar
N=1000#..................#Engine rpm
k=0.5#.......................#For four stroke engine
Wce=820#.................#Work during compression and expansion in kW
Wie=50#...................#Work during intake and exhaust in kW
f=150#......................#Rubbing friction in engiine in kW
WnetT=40#...................#Net work done by turbine in kW
#Calculations
BP=Wce-(Wie+f+WnetT)#.....................#Net work available or brake power in kW
D=((BP*6)/(n*Pmb*(pi/4)*N*k*10))**(1/3)#......................#Bore of engine in m
L=D#.........................................#Given that bore is equal to stroke
print "Since the stroke and bore are equal, their value is = %0.2f mm"%(D*1000)

Since the stroke and bore are equal, their value is = 290.90 mm


## EXAMPLE 17.6 PAGE 573¶

In [39]:
from __future__ import division
from math import pi
# Initialisation of Variables
Cm=50150#............................#Heating value of methane in kJ/kg
Co=44880#............................#Heating value of octane in kJ/kg
#Calculations
#Since Energy supplied is proportional to mass of fuel supplied time calorific value of the fuel supplied
ratioP=Cm/Co#.........................#Ratio of the power input of the engine with methane fuel to that with octane
print "Ratio of the power input of the engine with methane fuel to that with octane : %0.2f"%ratioP

Ratio of the power input of the engine with methane fuel to that with octane : 1.12


## EXAMPLE 17.7 PAGE 573¶

In [40]:
from __future__ import division
from math import pi
# Initialisation of Variables
N=2000#...........................#Engine rpm
k=0.5#..............................#Four stroke engine
Disp=0.025#........................#Engine displacement in m**3
Pmb=6#..............................#Brake mean effective pressure in bar
mf=0.018#............................#Fuel consumption in kg/s
Cf=42000#............................#Calorific value of fuel in kJ/kg
#Calcuations
BP=(Pmb*Disp*N*k*10)/(6)#................#Brake power in kW
etaBT=BP/(mf*Cf)#.................#Brake thermal efficiency
print "The Brake power = %0.2f kW"%(BP)
print "Brake thermal efficiency = %0.2f %%"%(etaBT*100)

The Brake power = 250.00 kW
Brake thermal efficiency = 33.07 %


## EXAMPLE 17.8 PAGE 574¶

In [41]:
from __future__ import division
from math import pi
# Initialisation of Variables
T=175#.......................#Torque due to brake load in Nm
N=500#.........................#Engine speed in rpm
#calcuations
BP=(2*pi*N*T)/(60*1000)#.......................#Brake power developed by engine in kW
print "Brake power developed by engine = %0.2f kW"%(BP)

Brake power developed by engine = 9.16 kW


## EXAMPLE 17.9 PAGE 575¶

In [42]:
from __future__ import division
from math import pi
# Initialisation of Variables
D=0.3#.............................#Bore of engine cylinder in m
L=0.45#............................#Stroke of engine cylinder in m
N=300#.............................#Engine rpm
Pmi=6#.............................#Indicated mean effective pressure in bar
Db=1.8#............................#Diameter of brake drum in m
d=0.02#............................#Brake rope diameter
k=0.5#.............................#Four stroke engine
n=1#...............................#No of cylinders
#Calculations
IP=(n*Pmi*L*(pi/4)*D*D*N*k*10)/6#......................#Indicated power in kW
BP=(Nbl*pi*(Db+d)*N)/60#...............................#Brake power in kW
etam=BP/IP#.............................................#Mechanical efficiency
print "Indicated power = %0.2f kW"%(IP)
print "Brake power = %0.2f kW"%(BP)
print "Mechanical efficiency = %0.2f %%"%(etam*100)

Indicated power = 47.71 kW
Brake power = 42.88 kW
Mechanical efficiency = 89.88 %


## EXAMPLE 17.10 PAGE 576¶

In [43]:
from __future__ import division
from math import pi
# Initialisation of Variables
Db=0.7#.............................#Diameter of brake pulley in m
d=0.025#............................#Diameter of the rope in m
W=50#...............................#Load on the tight side of the rope in kg
N=900#..............................#Engine rpm
mf=4#...............................#Rate of fuel consumption in kg/h
C=44000#............................#Calorific value of fuel in kJ/kg
g=9.81#.............................#Acceleration due to gravity in m/s**2
#Calculations
BP=(((W*g)-S)*pi*(Db+d)*N)/(60*1000)#........................#Brake power in kW
bsfc=mf/BP#...................................................#Brake specific fuel consumption in kJ/hW-h
print "Brake specific fuel consumption = %0.2f kJ/hW-h"%bsfc
etathB=(BP*3600)/(mf*C)#.............................................#Brake thermal efficiency
print "Brake thermal efficiency : %0.2f %%"%(etathB*100)

Brake specific fuel consumption = 0.27 kJ/hW-h
Brake thermal efficiency : 30.78 %


## EXAMPLE 17.11 PAGE 576¶

In [44]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=4#.......................#No of cylinders
k=0.5#.....................#Four stroke engine
r=8#.......................#Compression ratio
d=0.1#.....................#Engine bore in m
l=0.1#.....................#Engine stroke in m
etaV=0.75#.................#Volumetric efficiency
N=4800#....................#Engine rpm
afr=15#....................#Air fuel ratio
C=42000000#................#Calorific value of fuel
rhoa=1.12#.................#Atmospheric density in kg/m**3
Pmi=10#....................#Mean effective pressure in bar
etamech=0.8#...............#Mechanical efficiency
#Calculations
IP=(n*Pmi*l*(pi/4)*d*d*N*k*10)/6#.................#Indicated power in kW
Ac=n*(pi/4)*d*d*l*(N/2)*(etaV/60)#.....................#Air consumption in m**3/s
ma=Ac*rhoa#........................................#Mass flow of air in kg/s
mf=ma/afr#.........................................#Mass flow of fuel in kg/s
etath=(IP*1000)/(mf*C)#...................................#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etath*100)
BP=IP*etamech#.....................................#Brake Power in kW
print "Brake power = %0.2f kW"%BP

Indicated thermal efficiency = 42.52 %
Brake power = 100.53 kW


## EXAMPLE 7.12 PAGE 577¶

In [45]:
from __future__ import division
from math import pi
# Initialisation of Variables
N=1800#...................#Engine rpm
l=0.11#...................#Engine stroke in m
d=0.085#..................#Engine bore in m
ma=0.56#..................#Air flow rate in kg/min
BP=6#.....................#Brake power developed in kW
afr=20#...................#Air fuel ratio
C=42550#..................#Calorific value of fuel in kJ/kg
rhof=1.18#................#Density of fuel in kg/m**3
#calculations
V=(pi/4)*d*d*l*(N/2)#.....................#Volume displacemt in m**3/min
Ma=V*rhof#.................................#Mass of air in kg/min
etaV=ma/Ma#................................#Volumetric efficiency
fc=ma/afr#.................................#Fuel concumption
bsfc=(fc*60)/BP#...........................#Brake specific fuel consumption in kg/kWh
print "The volumetric efficiency = %0.2f %%"%(etaV*100)
print "Brake specific fuel consumption = %0.2f kg/kWh"%(bsfc)

The volumetric efficiency = 84.48 %
Brake specific fuel consumption = 0.28 kg/kWh


## EXAMPLE 17.13 PAGE 578¶

In [46]:
from __future__ import division
from math import pi
# Initialisation of Variables
pmicover=6.5#....................#Mean effective pressure on cover side in bar
pmicrank=7#......................#Mean effective pressure on crank side in bar
D=0.2#...........................#Engine bore in m
l=0.35#..........................#Engine stroke in m
drod=0.02#.........................#Diameter of piston rod in m
Db=1.2#............................#Brake wheel diameter in m
d=0.02#............................#Brake rope diameter in m
k=0.5#.............................#Four stroke engine
N=420#.......................#Engine rpm
#calculations
Acover=(pi/4)*D*D#.......................#Area of cylinder on the cover side in m**2
Acrank=(pi/4)*((D**2)-(drod**2))#..........#Effective area of cylinder on the crank end side in m**2
IPcover=(pmicover*l*Acover*N*k*10)/6#................#Indicated power on the cover end side in kW
IPcrank=(pmicrank*l*Acrank*N*k*10)/6#................#Indicated power on the crank end side in kW
IPtotal=IPcover+IPcrank#....................#TOtal
BP=((W-S)*pi*(Db+d)*N)/(60*1000)#...................#Brake power in kW
etamech=BP/IPtotal#..................................#Mechanical efficiency
print "The mechanical efficiency = %0.2f %%"%(etamech*100)

The mechanical efficiency = 63.59 %


## EXAMPLE 17.14 PAGE 579¶

In [47]:
from __future__ import division
from math import pi
# Initialisation of Variables
BP=14.7#........................#Brake power in kW
p1=0.9#.........................#Suction pressure in bar
etamech=0.8#....................#Mechanical efficiency
r=5#............................#Compression ratio
p3=24#..........................#maximum explosion pressure in bar
N=1000#.........................#Engine rpm
rld=1.5#........................#Ratio of length and stroke
ic=1.35#........................#Index of compression curve
ie=1.3#.........................#Index of expansion curve
k=0.5#..........................#Four stroke engine
#calculations
p2=(r**ic)*p1#......................#intermediate pressure = %0.2f bar) during compression
p4=p3/(r**ie)#......................#Intermediate pressure = %0.2f bar) during expansion
pm=((((p3-r*p4)/(ie-1))-((p2-p1*r)/(ic-1)))*(10**5))/(r-1)#...........#Mean effective pressure in N/m**2
pmb=pm/100000#........................................#Mean effective pressure in bar
IP=BP/etamech#........................................#Indicated power in kW
D=((IP*6*4)/(pmb*rld*(pi)*N*k*10))**(1/3)#............#Engine bore in m
L=rld*D#..............................................#Engine stroke in m
print "Engine bore = %0.2f mm"%(D*1000)
print "Engine stroke = %0.2f mm"%(L*1000)

Engine bore = 152.98 mm
Engine stroke = 229.47 mm


## EXAMPLE 17.15 PAGE 580¶

In [48]:
from __future__ import division
from math import pi
#Initialisation of Variables
IP=30#........................#Indicated power in kW
BP=26#........................#Brake power in kW
N=1000#.......................#Engine rpm
fpbph=0.35#...................#Fuel per brake power hour in kg/B.P.h
C=43900#......................#Calorific value of fuel used in kJ/kg
#Calculations
mf=BP*fpbph#.............#Fuel consumption per hour in kg/h
etaIth=IP/((mf/3600)*C)#.................#Indicated thermal efficiency
etaBth=BP/((mf/3600)*C)#.................#Indicated brake efficiency
etamech=BP/IP#...........................#Mechanical efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etaIth*100)
print "Indicated brake efficiency = %0.2f %%"%(etaBth*100)
print "Mechanical efficiency = %0.2f %%"%(etamech*100)

Indicated thermal efficiency = 27.03 %
Indicated brake efficiency = 23.43 %
Mechanical efficiency = 86.67 %


## EXAMPLE 17.16 PAGE 581¶

In [49]:
from __future__ import division
from math import pi
#Initialisation of Variables
Db=0.75#.....................#Diameter of brake pulley in m
d=0.05#......................#Rope diameter in m
cf=4.2#......................#Consumption of fuel in kg/h
N=1000#......................#Engine rpm
C=43900#.....................#Calorific value of fuel in kJ/kg
#Calculations
BP=((W-S)*pi*(Db+d)*N)/(60*1000)#...............#Brake power in kW
bsfc=cf/BP#......................................#Brake specific fuel consumption  in kg/kWh
etabth=BP/((cf/3600)*C)#.........................#Brake thermal efficiency
print "Brake specific fuel consumption  = %0.2f kg/kWh"%(bsfc)
print "Brake thermal efficiency = %0.2f %%"%(etabth*100)

Brake specific fuel consumption  = 0.29 kg/kWh
Brake thermal efficiency = 28.62 %


## EXAMPLE 17.17 PAGE 581¶

In [50]:
from __future__ import division
from math import pi
#Initialisation of Variables
n=6#.............................#No of cylinders
D=0.09#..........................#Bore of cylinder in m
L=0.1#...........................#Stroke length in m
r=7#.............................#Compression ratio
etarel=0.55#.....................#Relative efficiency
isfc=0.3#........................#Indicated specific fuel consumption in kg/kWh
imep=8.6#........................#Indicated mean effective pressure in bar
N=2500#..........................#Engine speed
ga=1.4#..........................#Degree of freedom for air
k=0.5#...........................#Four stroke engine
#calculations
etastan=1-1/(r**(ga-1))#...................#Air standard efficiency
etath=etarel*etastan#.....................#Indicated thermal efficiency
C=3600/(etath*isfc)#.......................#Calorific value of fuel in kJ/kg
IP=(n*imep*L*D*D*(pi/4)*N*k*10)/6#................#Indicated power in kW
fc=IP*isfc#.............................#Fuel consumption in kg/h
print "The calorific value of the fuel used = %0.2f kJ/kg"%(C)
print "Fuel consumption = %0.2f (kg/h"%(fc)

The calorific value of the fuel used = 40341.03 kJ/kg
Fuel consumption = 20.52 (kg/h


## EXAMPLE 17.18 PAGE 582¶

In [51]:
from __future__ import division
from math import pi
# Initialisation of Variables
BP=30#........................#Brake power in kW
pmi=8#.........................#Mean effective pressure in bar
etamech=0.8#....................#Mechanical efficiency
n=4#............................#No of cylinders
N=2500#.........................#Engine rpm
rld=1.5#........................#Ratio of length and stroke
etabth=0.28#......................#Brake thermal efficiency
k=1#..........................#Two stroke engine
C=43900#.........................#Calorific value of fuel in kJ/kg
#calculations
IP=BP/etamech#........................................#Indicated power in kW
D=((IP*6*4)/(pmi*n*rld*(pi)*N*k*10))**(1/3)#............#Engine bore in m
L=rld*D#..............................................#Engine stroke in m
print "Engine bore = %0.2f mm"%(D*1000)
print "Engine stroke = %0.2f mm"%(L*1000)
mf=BP/(etabth*C)#..............................#Fuel consumption in kg/s
print "Fuel consumption = %0.2f kg/h"%(mf*3600)

Engine bore = 62.04 mm
Engine stroke = 93.05 mm
Fuel consumption = 8.79 kg/h


## EXAMPLE 17.19 PAGE 583¶

In [52]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=6#......................#No of cylinders
pdpc=700*10**(-6)#.................#Piston displacement per cylinder in m**3
P=78#............................#Power developed in kW
N=3200#.............................#Engine rpm
mf=27#.............................#Fuel consumption in kg/h
C=44000#...........................#Calorific value of fuel in kJ/kg
afr=12#..............................#Air fuel ratio
p1=0.9#..........................#Intake air pressure
pa=p1#
t1=305#...............................#Intake air temperature
ta=t1#
R=0.287#.....................#Gas constant in kJ/kgK
#Calculations
ma=afr*mf#............................#maaa of air in kg/h
Va=(ma*R*t1)/(p1*100)#.............#Volume of air intake in m**3/h
Vs=pdpc*n*(N/2)*60#.....................#Swept volume in m**3/h
etaV=Va/Vs#.............................#Volumetric efficiency
print "The volumetric efficiency of the engine = %0.2f %%"%(etaV*100)
etabt=P/(mf*(C/3600))#...................#Brake thermal efficiency
print "The brake thermal efficiency of the engine = %0.2f %%"%(etabt*100)
Tb=(P*60)/(2*pi*N)#..........................#Brake torque in kNm
print "Brake torque = %0.2f kNm"%(Tb)

The volumetric efficiency of the engine = 78.16 %
The brake thermal efficiency of the engine = 23.64 %
Brake torque = 0.23 kNm


## EXAMPLE 17.20 PAGE 583¶

In [53]:
from __future__ import division
from math import pi,floor
# Initialisation of Variables
n=6#.......................#No of cylinders
Vs=1.75*10**(-3)#..............#Stroke volume in m**3
IP=26.3#.....................#Indicated power in kW
Ne=504#.......................#Expected Engine rpm
Pmi=6#........................#Mean effective pressure in bar
k=0.5#.........................#Four stroke engine
#Calculations
Na=floor((IP*6)/(n*Pmi*Vs*k*10))#.......................#Actual Engine rpm
af=(Na*n)/2#.......................#Actual no of fires in min
ef=(Ne*n)/2#.......................#Expected no of fires in min
Nm=ef-af#........................#No of misfires/min
nm=Nm/n#....................#Average no of times each cylinder misfires in one min
print "Average no of times each cylinder misfires in one min : %0.2f"%nm

Average no of times each cylinder misfires in one min : 2.00


## EXAMPLE 17.21 PAGE 584¶

In [1]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=4#.......................#No of cylinders
D=0.075#.......................#Engine bore in m
L=0.09#........................#Engine length in m
err=39/8#.......................#Engine to rear axle ratio
Dw=0.65#.........................#Wheel diameter in m
pc=0.227#.......................#Petrol consumption in kg
pmi=5.625#.........................#Mean effective pressure in bar
C=43470#..............................#Calorific value of petrol in kJ/kg
k=0.5#.............................#Four stroke engine
sc=48#............................#Speed of the car in km/h
d=3.2#.............................#Distance covered by car in km
#Calculations
sc1=sc*(1000/60)#...................#Speed of the car in m/min
Nt=sc1/(pi*Dw)#......................#Revolutions made by tire per min
Ne=Nt*err#............................#Speed of engine shaft
IP=(n*pmi*L*(pi/4)*D*D*Ne*k*10)/6#........#Indicated power in kW
print "Indicated power = %0.2f kW"%(IP)
sc2=sc/60#.......................#Speed of the car in km/min
t=d/sc2#..........................#Time for covering 3.2 km in min
fc=pc/(t*60)#.....................#Fuel consumed per second in kg
etait=IP/(fc*C)#...............#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etait*100)

Indicated power = 14.24 kW
Indicated thermal efficiency = 34.63 %


## EXAMPLE 17.22 PAGE 584¶

In [2]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=1#.................#No of cylinders
D=0.25#................#Engine bore in m
L=0.4#.................#Engine stroke in m
pmg=7#.................#Gross mean effective pressure in bar
pmp=0.5#...............#Pumping mean effective pressure in bar
N=250#..................#Engine rpm
Db=1.5#................#Effective diameter of the brake in m
Nl=1080#..............#Net load on the brake in N
fh=10#.................#Fuel used per hour in kg
C=44300#...............#Calorific value of fuel in kJ/kg
k=0.5#.................#Four stroke engine
#Calculations
mf=fh/3600#.........................#Fuel used per second in kg
pm=pmg-pmp#.......................#Net pressure
IP=(n*pm*L*(pi/4)*D*D*N*k*10)/6#..........#/Indicated power in kW
print "Indicated power = %0.2f kW"%(IP)
BP=((Nl)*pi*Db*N)/(60*1000)#...............#Brake power in kW
print "Brake power = %0.2f kW"%(BP)
etamech=BP/IP#...........................#Mechanical efficiency
print "Mechanical efficiency = %0.2f %%"%(etamech*100)
etath=IP/(mf*C)#.........................#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etath*100)

Indicated power = 26.59 kW
Brake power = 21.21 kW
Mechanical efficiency = 79.75 %
Indicated thermal efficiency = 21.61 %


## EXAMPLE 17.23 PAGE 585¶

In [3]:
from __future__ import division
from math import pi
# Initialisation of Variables
etabth=0.3#....................#Brake thermal efficiency
afrw=20#........................#Air fuel ratio by weight
C=41800#.........................#Calorific value of fuel used in kJ/kg
R=287#........................#Gas constant in J/kg
#Calculations
Wp=etabth*C#...................#Work produced per kg of fuel in kJ
p1=1.0132#
t=273+15#............#STP conditions in bar and Kelvin
V=(afrw*t*R)/(p1*10**5)#.......#Volume of air used in m**3
pmb=(Wp*1000)/(V*10**5)#........#Brake mean effective pressure in bar
print "Brake mean effective pressure = %0.2f bar"%(pmb)

Brake mean effective pressure = 553.38 bar


## EXAMPLE 17.24 PAGE 585¶

In [4]:
from __future__ import division
from math import pi
# Initialisation of Variables
v1=0.216#.....................#Gas consumption in m**3/min
pw=75#........................#Pressure of gas in mm of water
t1=290#......................#Temperature of gas in K
ac=2.84#....................#Air consumption in kg/min
br=745#......................#Barometer reading in m of Hg
D=0.25#.....................#Engine bore in m
L=0.475#......................#Engine stroke in m
N=240#........................#Engine rpm
R=287#......................#Gas constant for air in J/kgK
#Calculations
p1=br+(pw/13.6)#...................#Pressure of gas in mm of mercury
p2=760
t2=273#.....................#NTP conditions in mm of Hg and Kelvin
v2=(p1*v1*t2)/(t1*p2)#...............#Volume of gas used at NTP in m**3
gs=v2/(N/2)#.........................#Gas used per stroke in m**3
v=(ac*R*t2)/(1.0132*10**5)#...........#Volume occupied by air at NTP in m**3/min
aps=v/(N/2)#...........................#Air used per stroke
Va=gs+aps#.....................#Actual volume of mixture in m**3 drawn per stroke at NTP
Vs=(pi/4)*D*D*L#...............#Swept volume in mm**3
etaV=(Va/Vs)#...................#Volumetric efficiency
print "Volumetric efficiency = %0.2f %%"%(etaV*100)

Volumetric efficiency = 85.67 %


## EXAMPLE 17.25 PAGE 586¶

In [5]:
from __future__ import division
from math import pi
# Initialisation of Variables
t=1#...................#Duration of trial in hrs
Rev=14000#.............#Revolutions
nmc=500#...............#Number of missed cycles
mep=7.5#................#Mean effective pressure in bar
gc=20000#...............#Gas consumption in litres
lcv=21#.................#LCV of gas at supply condition in kJ/litre
D=0.25#.................#Engine bore in m
L=0.4#.................#Engine stroke in m
r=6.5#..................#Compression ratio
n=1#......................#No of cylinders
Cb=4#......................#Effective brake Circumference
k=0.5#....................#Four stroke engine
ga=1.4#......................#Degree of freedom
#Calculations
N=Rev/60#..............#Engine rpm
Vg=gc/3600#.............#Fuel consumption in litres/s
Na=((Rev/2)-nmc)/60#................#Working cycles per min
IP=(n*mep*L*(pi/4)*D*D*Na*10)/6#............#indicated power in kW
print "Indicated power = %0.2f kW"%(IP)
BP=((bl)*Cb*N)/(60*1000)#...............#Brake power in kW
print "Brake power = %0.2f kW"%(BP)
etamech=BP/IP#...........................#Mechanical efficiency
print "Mechanical efficiency = %0.2f %%"%(etamech*100)
etath=IP/(Vg*lcv)#.........................#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etath*100)
etast=1-(1/r**(ga-1))#............#Air standard efficiency
etarel=etath/etast#............#Relative efficiency
print "Relative efficiency = %0.2f %%"%(etarel*100)

Indicated power = 26.59 kW
Brake power = 22.87 kW
Mechanical efficiency = 86.00 %
Indicated thermal efficiency = 22.79 %
Relative efficiency = 43.24 %


## EXAMPLE 17.26 PAGE 587¶

In [6]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=1.3#...................#Index of compression
pa=1.4#
pb=3.6#
posa=(1/4)#..........#Point a - the position 1/4 of the stroke
posb=(3/4)#..........#Point b - the position 3/4 of the stroke
ga=1.4#...............#Degree of freedom for gas
etarel=0.4#...................#Relative efficiency
C=18800#....................#Calorific value of fuel in kJ/m**3
#Calculations
r=1+((((pb/pa)**(1/n))-1)/(posb-(((pb/pa)**(1/n))*(posa))))#.........#Compression ratio
print "The compression ratio : %0.2f"%r
etast=1-(1/r**(ga-1))#............#Air standard efficiency
etath=etarel*etast#............#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)
v=1/(etath*C)#...............#Gas consumption per IP sec
print "Gas consumption = %0.2f m**3/IP hour"%(v*3600)

The compression ratio : 5.58
Thermal efficiency = 19.89 %
Gas consumption = 0.96 m**3/IP hour


## EXAMPLE 17.27 PAGE 587¶

In [7]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=6#.....................#No of cylinders
r=5#..................#Compression ratio
Vc=0.000115#................#Clearance volume of each cylinder in m**3
fc=10.5#.....................#Fuel consumed in kg/h
C=41800#......................#Calorific value of fuel in  kJ/kg
N=2500#.......................#Engine speed in rpm
er=0.65#.......................#Efficiency ratio
ga=1.4#........................#Degree of freedom
#calculations
etast=1-(1/r**(ga-1))#...............................#Air standard efficiency
etath=etast*er#.................................#Thermal efficiency
IP=etath*(fc/3600)*C#..........................#Indicated power in kW
Wnet=(IP*(10**3)*60)/(n*(N/2))#..............#Net work froom one cycle per cylinder in N-m
Vs=(r-1)*Vc#......................#Swept volume in m**3
pm=Wnet/(Vs*10**5)#...................#Mean effective pressure developed
print "Mean effective pressure developed = %0.2f bar"%(pm)

Mean effective pressure developed = 6.54 bar


## EXAMPLE 17.28 PAGE 588¶

In [8]:
from __future__ import division
from math import pi
# Initialisation of Variables
D=0.2#.................#Engine bore in m
L=0.25#...............#Engine stroke in m
n=2#......................#No of cylinders
r=13#......................#Compression ratio
fc=14#..................#Fuel consumption in kg/h
N=300#....................#Engine rpm
etarel=0.65#..............#Relative efficiency
etamech=0.76#.............#Mechanical efficiency
co=0.05#.....................#Cut off of the stroke
C=41800#.....................#Calorific value of fuel in kJ/kg
k=1#........................#Two stroke engine
ga=1.4#.......................#Degree of freedom
#calculations
rho=1+(co*(r-1))#
etast=1-((1/(r**(ga-1)))*(1/ga)*((rho**ga)-1)*(1/(rho-1)))#............#Air standard efficiency
etath=etarel*etast#........................#Thermal efficiency
IP=etath*(fc/3600)*C#........................#Indicated power in kW
BP=etamech*IP#................................#Brake power in kW
pmi=(6*IP)/(n*N*L*(pi/4)*D*D*k*10)#............#mean effective pressure in bar
print "Mean effective pressure = %0.2f bar"%(pmi)
pmb=pmi*etamech#...........................#Brake mean effective pressure in bar
print "Brake mean effective pressure = %0.2f bar"%(pmb)

Mean effective pressure = 8.11 bar
Brake mean effective pressure = 6.16 bar


## EXAMPLE 17.29 PAGE 589¶

In [9]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=4#.................#No of cylinders
C=45200#..................#calorific value of fuel in kJ/kg
etamech=0.82#...............#Mechanical efficiency
etarel=0.7#.................#Relative efficiency
etast=0.52#...............#Air standard efficiency
etav=0.78#...............#Volumetric efficiency
sbr=1.25#...................#Stroke bore ratio
N=2400#...................#Engine rpm
p=1#.......................#Suction pressure in bar
t=298#....................#Suction temperature in bar
BP=72#...................#Brake power in kW
ga=1.4#......................#Degree of freedom
afr=16#.................#Air fuel ratio
R=287#.......................#Gas constant in J/kg
#calculations
r=(1/(1-etast))**(1/(ga-1))#............#Compression ratio
print "The compression ratio : %0.2f"%r
etath=etast*etarel#......................#Indicated thermal efficiency
print "Indicated thermal efficiency : %0.2f %%"%(etath*100)
IP=BP/etamech#....................#Indicated power in kW
mf=IP/(etath*C)#......................#Fuel consumption in kg/s
bsfc=mf/BP#......................#Brake specific fuel consumption in kg/kWs
print "Brake specific fuel consumption = %0.2f kg/kWs"%(bsfc*3600)
mafm=afr+1#......................#Mass of air fuel mixture in kg/kg of fuel
mafm1=mafm*mf#....................#Mass of air fuel mixture when mf amount of fuel is supplied to engine per second
v=(mafm1*R*t)/(p*10**5)#.......................#/Volume of air fuel mixture supplied to the engine in m**3
Vs=v/etav#..............................#Swept volume in m**3
D=((Vs)/((pi/4)*sbr*n*(N/(2*60))))**(1/3)#............#Engine bore in m
print "Engine bore = %0.2f mm"%(D*1000)
print "Engine stroke = %0.2f mm"%(D*1000*sbr)

The compression ratio : 6.26
Indicated thermal efficiency : 36.40 %
Brake specific fuel consumption = 0.27 kg/kWs
Engine bore = 108.20 mm
Engine stroke = 135.25 mm


## EXAMPLE 17.30 PAGE 589¶

In [10]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=1#.......................#No of cylinders
D=0.18#...................#Engine bore in m
L=0.34#....................#Engine stroke in m
N=400#......................#Engine rpm
mepw=6.4#.................#Mean effective pressure of working loop in bar
mepp=0.36#..................#Mean effective pressure of pumping loop in bar
mepd=0.64#.................#Mean effective pressure (dead cycle) iin bar
fs=46#................#Firing strokes per min
#calculations
pminet=mepw-mepp#..........#Net indicated mean effective pressure in bar
IPnet=(n*pminet*(pi/4)*L*D*D*fs*4*10)/6#.............#Net indicated power output in kW
ppdc=(n*pminet*L*(pi/4)*D*D*10*dc)/6#.............#Pumping power of dead cycles in kW
FP=IPnet-ppdc#...........................#Frictional power in kW
IP=(n*pminet*L*(pi/4)*D*D*(N/2)*10)/6#...............#Indicated power in kW
BP=IP-FP#..................#Brake power in kW
print "Brake power = %0.2f kW"%(BP)
etamech=BP/IP#.................#Mechanical efficiency
print "Mechanical efficiency = %0.2f %%"%(etamech*100)

Brake power = 14.81 kW
Mechanical efficiency = 85.00 %


## EXAMPLE 17.31 PAGE 590¶

In [11]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=1#..............#No of cylinders
B=0.32#...............#Engine bore in m
L=0.42#..............#Engine stroke in m
N=200#................#Engine rpm
Nk=90#..................#No of explosions per min
v1=11.68#............#Gas used in m**3/h
pg=170#................#Pressure of gas in mm of water
br=755#................#Barometer reading in mm of Hg
pmi=6.2#.................#Mean effective pressure in bar
C=21600#.......................#Calorific value of gas in kJ/kg
bl=2040#......................#Net load on brake in N
Db=1.2#......................#Brake drum diameter in m
t1=298#.....................#Ambient temperature in Kelvin
#Calculations
IP=(n*pmi*L*(pi/4)*B*B*Nk*10)/6#..........................#Indicated power in kW
BP=(bl*pi*Db*N)/(60*1000)#.........................#Brake power in kW
etamech=(BP/IP)#...................#Mechanical efficiency
print "Mechanical efficiency = %0.2f %%"%(etamech*100)
p1=br+(pg/13.6)#.................#In mm of Hg
p2=760#
t2=273#...................#NTP conditions in mm of Hg and Kelvin
v2=(p1*v1*t2)/(p2*t1)#
etabth=BP/((v2/3600)*C)#..............#Brake thermal efficiency
print "Brake thermal efficiency = %0.2f %%"%(etabth*100)

Mechanical efficiency = 81.61 %
Brake thermal efficiency = 39.54 %


## EXAMPLE 17.32 PAGE 591¶

In [12]:
from __future__ import division
from math import pi,sqrt
# Initialisation of Variables
n=1#...................#No of cylinders
d=0.032#................#Diameter of circular orifice in m
Cd=0.62#.............#Co efficient of discharge
hw=150#.................#Pressure across orfice in mm of water
t=20+273#..............#Temperature of air in the room in Kelvin
p=1.0132#.................#Ambient pressure in bar
pd=0.00178#............#Piston displacement in m**3
R=287#....................#Gas constant in J/kg
r=6.5#..................#Compression ratio
fc=0.135#................#Fuel consumption in kg/min
C=43900#.................#Calorific value of fuel in kJ/kg
BP=28#................#Brake power in kW
N=2500#...................#Engine rpm
k=0.5#....................#Four stroke engine
g=9.81#.......................#Acceleration due to gravity in m/s**2
rhow=1000#....................#Density of water in kg/m**3
ga=1.4#........................#Degree of freedom
#calculations
mbyv=(p*10**5)/(R*t)#
pw=(hw/rhow)*rhow#....................#Pressure across orifice in kg/m**2
H=pw/mbyv#........................#Head of air column causing the flow in m
ma=Cd*(pi/4)*d*d*sqrt(2*g*H)#................#Air flow through orifice in m**3/s
maps=(ma*60)/(N/2)#........................#Air consumption per stroke
etav=maps/pd#.................#Volumetric efficiency
print "Volumetric efficiency = %0.2f %%"%(etav*100)
ac=ma*60*mbyv#...............#Mass of air drawn into cylinder per min in kg
afr=ac/fc#...................#Air fuel ratio
print "Air fuel ratio : %0.2f"%afr
pmb=(6*BP)/(n*pd*N*k*10)#...................#Mean effective pressure in bar
print "Mean effective pressure = %0.2f bar"%(pmb)
etast=1-(1/(r**(ga-1)))#...............#Air standard efficiency
etabth=BP/((fc/60)*C)#...............#Brake thermal efficiency
etarel=etabth/etast#.................#Relative efficiency
print "Relative efficiency = %0.2f %%"%(etarel*100)

Volumetric efficiency = 66.45 %
Air fuel ratio : 13.20
Mean effective pressure = 7.55 bar
Relative efficiency = 53.79 %


## EXAMPLE 17.33 PAGE 591¶

In [13]:
from __future__ import division
from math import pi
# Initialisation of Variables
N=400#.................#Engine rpm
n=1#....................#no of cylinders
W=370#.................#Load on the brake in N
Db=1.2#.................#Diameter of the brake drum
mf=2.8#.................#Fuel consumption in kg/h
C=41800#..................#Calorific value of fuel in kJ/kg
D=0.16#...................#Engine bore in m
L=0.2#....................#Engine stroke in m
k=0.5#.....................#Four stroke engine
Sc=1#....................#Spring constant in bar/mm
l=40#....................#Length of diagram in mm
aic=300#.................#Area of indicator card in mm**2
#Calculations
pmi=aic*(Sc/l)#..................#Mean effective pressure in bar
IP=(n*pmi*L*(pi/4)*D*D*k*N*10)/6#..............#Indicated power in kW
print "Indicated power = %0.2f kW"%(pmi)
BP=((W-S)*pi*Db*N)/(60*1000)#............#Brake power in kW
print "Brake power = %0.2f kW"%(BP)
pmb=(BP*6)/(n*L*D*D*(pi/4)*k*N*10)#...........#Brake mean effective pressure in bar
print "Brake mean effective pressure = %0.2f bar"%(pmb)
bsfc=mf/BP#.................#Brake specific fuel consumption in kg/BP h
print "Brake specific fuel consumption = %0.2f kg/BP h"%bsfc
etabth=BP/((mf/3600)*C)#..................#Brake thermal efficiency
print "Brake thermal efficiency = %0.2f %%"%(etabth*100)
etaith=IP/((mf/3600)*C)#....................#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etaith*100)

Indicated power = 7.50 kW
Brake power = 8.04 kW
Brake mean effective pressure = 6.00 bar
Brake specific fuel consumption = 0.35 kg/BP h
Brake thermal efficiency = 24.74 %
Indicated thermal efficiency = 30.92 %


## EXAMPLE 17.34 PAGE 592¶

In [14]:
from __future__ import division
from math import pi
# Initialisation of Variables
R=287#................#Gas constant in J/kg K
n=4#...................#No of cylinders
D=0.0825#..............#Engine bore in m
L=0.13#................#Engine stroke in m
BP=28#..................#Brake power in kW
N=1500#.................#Engine rpm
afrth=14.8#...............#theoretical air fuel ratio
C=45980#..................#Calorific value of fuel in kJ/kg
etamech=0.9#.............#Mechanical efficiency
ap=70#..................#Percentage of Volume of air in he cylinder
fr=20#..................#Percentage richness of the fuel
p1=1.0132#.................#Ambient pressure in bar
pc=762#...................#Pressure in the cylinder in mm of Hg
tc=273+15.5#...............#Temperature in the cylinder in Kelvin
k=0.5#..................#Four stroke engine
#Calculations
Vs=(pi/4)*D*D*L#.......................#Swept volume in m**3
va=(ap/100)*Vs#.....................#Volume of air drawn in m**3
p=(pc/760)*p1#
m=(p*(10**5)*va)/(R*tc)#...................#Mass of air per stroke per cylinder
tmau=m*(N/2)*n#...................#Theoretical mass of air used per minute in kg
tmfu=tmau/afrth#..................#Theoretical mass of fyel used per min in kg
mf=(tmfu/60)*((100+fr)/100)#...............#Mass of fuel burnt per second in kg
IP=BP/etamech#.........................#Indicated power in kW
etaith=IP/(mf*C)#.....................#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etaith*100)
pmb=(BP*6)/(n*L*D*D*(pi/4)*N*10*k)#...............#Mean effective pressure in bar
print "Mean effective pressure = %0.2f bar"%(pmb)

Indicated thermal efficiency = 27.96 %
Mean effective pressure = 8.06 bar


## EXAMPLE 17.35 PAGE 593¶

In [15]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=1#..................#No of cylinders
D=0.2#............#Engine bore in m
L=0.4#..............#Engine stroke in m
Nt=9400#...............#Total no of revolutions
Ne=4200#...............#Total no of explosions
t=40#...................#Duration of testing in min
Nk=Ne/t#...............#No of explosions
Db=1.6#.................#Diameter of brake wheel in m
d=0.02#................#Diameter of rope in m
gu=8.5#..................#Gas used in m**3/sec
C=15900#...............#Calorific value of fuel in kJ/kg
Vg=(gu/(t*60))#.................#Volume of gas used in m**3/sec
aic=550#.....................#Area of indicator diagram mm**2
l=72#.......................#Length of indicator diagram in mm
s=0.8#.....................#Spring number in bar/mm
#calculations
pmi=(aic*s)/l#................#Mean effective pressure in bar
IP=(n*pmi*L*D*D*(pi/4)*Nk*10)/6#............#Indicated power in kW
print "Indicated power = %0.2f kW"%(IP)
BP=(bl*pi*(Db+d)*(Nt/t))/(60*1000)#...............#Brake power in kW
print "Brake power = %0.2f kW"%(BP)
etaith=IP/(Vg*C)#...............#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etaith*100)
etabth=BP/(Vg*C)#...............#Brake thermal efficiency
print "Brake thermal efficiency = %0.2f %%"%(etabth*100)

Indicated power = 13.44 kW
Brake power = 10.76 kW
Indicated thermal efficiency = 23.87 %
Brake thermal efficiency = 19.11 %


## EXAMPLE 17.36 PAGE 594¶

In [16]:
from __future__ import division
from math import pi,sqrt
# Initialisation of Variables
n=6#....................#No of cylinders
D=0.125#................#Engine bore in m
L=0.125#...............#Engine stroke in m
N=2400#.................#Engine rpm
W=490#...............#Load on the dynamometer in N
CD=16100#...............#Dynamometer constant
d0=0.055#...................#Air orifice diameter in m
Cd=0.66#...................#Co efficient of discharge
hw=310#.................#Head causing flow through prifice in mm of water
br=760#................#Barometer reading in mm of Hg
t=298#..................#Ambient temperature in Kelvin
fc=22.1#..................#Fuel consumption per hour in kg
C=45100#..................#Calorific value of fuel used in kJ/kg
perc=85#...................#Percentage of carbon in the fuel
perh=15#...................#Percentage of hydrogen in the fuel
p1=1.013#....................#Pressure of air at the end of suction stroke in bar
t1=298#......................#Temperature of air the the end of suction stroke in Kelvin
k=0.5#.......................#Four stroke engine
R=287#.......................#Gas constant in J/kgK
#calculations
BP=W*(N/CD)#................#Brake power in kW
pmb=(BP*6)/(L*D*D*k*10*N*n*(pi/4))#................#Brake mean effective pressure in bar
print "Brake mean effective pressure = %0.2f bar"%(pmb)
bsfc=fc/BP#.......................#Brake specific fuel consumption in kg/kWh
print "Brake specific fuel consumption = %0.2f kg/kWh"%(bsfc)
etathb=BP/((fc/3600)*C)#......................#Brake thermal efficiency
print "Brake thermal efficiency = %0.2f %%"%(etathb*100)
Vst=(pi/4)*D*D*L#..............#Stroke volume in m**3
Val=840*(pi/4)*d0*d0*Cd*sqrt((hw/10)/((p1*10**5)/(R*t1)))#............#Volume of air passing through orifice of air box per min
Vac=Val/n#.........................#Actual volume of air per cylinder in m**3/min
asps=Vac/(N/2)#.......................#Air supplied per stroke per cylinder in m**3
etav=asps/Vst#....................#Volumetric efficiency
print "Volumetric efficiency = %0.2f %%"%(etav*100)
Qa=(100/23)*(((perc/100)*(8/3))+((perh/100)*(8/1)))#.....................#Quantity of air required per kg of fuel combustion
aqas=(Val*((p1*10**5)/(R*t1))*60)/fc#....................#Actual quantity of air supplied per kg of fuel
pe=(aqas-Qa)/Qa#....................#Fraction of excess air supplied to engine
print "Percentage of excess air supplied : %0.2f"%(pe*100)

Brake mean effective pressure = 3.97 bar
Brake specific fuel consumption = 0.30 kg/kWh
Brake thermal efficiency = 26.38 %
Volumetric efficiency = 61.01 %
Percentage of excess air supplied : 43.76


## EXAMPLE 17.37 PAGE 595¶

In [17]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=1#......................#No of cylinders
D=0.3#....................#Engine bore in m
L=0.45#....................#Engine stroke in m
mf=8.8#...................#Fuel consumption in kg/h
C=41800#...................#Calorific value of fuel in kJ/kg
N=200#....................#Engine rpm
pmi=5.8#....................#Mean effective pressure in bar
Db=1.22#...................#Diameter of brake drum in m
k=0.5#........................#four stroke engine
mw=650#......................#Mass of cooling water in kg
cpw=4.18#....................#Specific heat capacity of water
delt=22#......................#Temperature rise
#Calculations
IP=(n*L*D*D*k*10*pmi*N*(pi/4))/6#...............#Indicated power in kW
BP=(bl*pi*Db*N)/(60*1000)#..................#Brake power in kW
etamech=BP/IP#............#Mechanical efficiency
print "Mechanical efficiency = %0.2f %%"%(etamech*100)
etathb=BP/((mf/3600)*C)#...................#Brake thermal efficiency
print "Brake thermal efficiency = %0.2f %%"%(etathb*100)
#Heat supplied
hip=IP*3600#...........#Heat equivalent of IP in kJ/h
hcw=mw*cpw*delt#..........#Heat carried away by cooling water
hf=mf*C#................#heat supplied by fuel
hex=hf-hip-hcw#..........#Heat carried by exhaust gasses
pf=100#
pip=(hip/hf)*100#
pcw=(hcw/hf)*100#
pex=(hex/hf)*100
print "HEAT BALANCE TABLE : "
print "_______________________________________________________________________"
print "Item                                      kJ                  Percent"
print "_______________________________________________________________________"
print "Heat supplied by fuel                     %d             %0.2f"%(hf,pf)
print "Heat absorbed in IP                       %d             %0.2f"%(hip,pip)
print "Heat taken away by cooling water          %d             %0.2f"%(hcw,pcw)
print "Heat carried away by exhaust gases        %d             %0.2f"%(hex,pex)

Mechanical efficiency = 77.28 %
Brake thermal efficiency = 23.26 %
HEAT BALANCE TABLE :
_______________________________________________________________________
Item                                      kJ                  Percent
_______________________________________________________________________
Heat supplied by fuel                     367840             100.00
Heat absorbed in IP                       110694             30.09
Heat taken away by cooling water          59774             16.25
Heat carried away by exhaust gases        197371             53.66


## EXAMPLE 17.38 PAGE 596¶

In [18]:
from __future__ import division
from math import pi
# Initialisation of Variables
r=15#................#Compression ratio
n=1#...................#No of cylinders
mf=10.2#..................#Fuel consumption in kg/h
C=43890#.................#Calorific value of fuel in kJ/kg
ma=3.8#.................#Consumption of air in kg/min
N=1900#...................#Engine rpm
T=186#....................#Torque on brake drum in Nm
mw=15.5#.................#Mass of cooling water used in kg/min
delt=36#..................#temperature rise
tg=410#..................#Exhaust gas temperature in Celsius
tr=20#...................#Room temperature in Celsius
cp=1.17#.................#Specific heat capacity for exhaust gases kJ/kgK
cpw=4.18#..................#Specific heat capacity for water in kJ/kgK
#calculations
BP=(2*pi*N*T)/(60*1000)#................#Brake power in kW
print "Brake power = %0.2f kW"%BP
bsfc=mf/BP#.............................#Brake specific fuel consumption in kg/kWh
print "Brake specific fuel consumption = %0.2f kg/kWh:"%bsfc
etabth=BP/((mf/3600)*C)#....................#Brake thermal efficiency
print "Brake thermal efficiency = %0.2f %%"%(etabth*100)
#Heat supplied
mg=(mf/60)+ma#....................#Mass of exhaust gases in kg/min
hbp=BP*60#...........#Heat equivalent of BP in kJ/min
hcw=mw*cpw*delt#..........#Heat carried away by cooling water
hf=(mf/60)*C#................#heat supplied by fuel
hex=mg*cp*(tg-tr)#..........#Heat carried by exhaust gasses
ha=round(hf)-round(hbp+hex+hcw)#............#Unaccounted heat
pf=100#
pbp=(hbp/hf)*100#
pcw=(hcw/hf)*100#
pex=(hex/hf)*100#
pa=(ha/hf)*100#
print "HEAT BALANCE TABLE:"
print "_______________________________________________________________________"
print "Item                                      kJ                  Percent"
print "_______________________________________________________________________"
print "Heat supplied by fuel                     %d             %0.2f"%(hf,pf)
print "Heat absorbed in BP                       %d             %0.2f"%(hbp,pbp)
print "Heat taken away by cooling water          %d             %0.2f"%(hcw,pcw)
print "Heat carried away by exhaust gases        %d             %0.2f"%(hex,pex)
print "Unaccounted heat                          %d             %0.2f"%(ha,pa)

Brake power = 37.01 kW
Brake specific fuel consumption = 0.28 kg/kWh:
Brake thermal efficiency = 29.76 %
HEAT BALANCE TABLE:
_______________________________________________________________________
Item                                      kJ                  Percent
_______________________________________________________________________
Heat supplied by fuel                     7461             100.00
Heat absorbed in BP                       2220             29.76
Heat taken away by cooling water          2332             31.26
Heat carried away by exhaust gases        1811             24.28
Unaccounted heat                          1097             14.70


## EXAMPLE 17.39 PAGE 597¶

In [19]:
from __future__ import division
from math import pi
# Initialisation of Variables
Cpw=4.18#..............#Specific heat of water in kJ/kgK
n=1#................#No of cylinders
N=350#.......#Engine rpm
pmi=3.1#..........#Mean effective pressure in bar
mf=1.52#............#Fuel consumption in kg
mw=162#..............#Mass of cooling water
tw1=30#...............#Water inlet temperature in C
tw2=55#................#Water outlet temperature in C
ma=32#..................#Mass of air used per kg of fuel in kg
tr=25#.................#Room temperature in C
tg=305#.................#Exhaust temperature in C
D=0.2#.................#Engine bore in m
L=0.28#.................#Engine stroke in m
Db=1#......................#Brake drum diameter in m
ms=1.4#......................#Mass of steam formed per kg of fuel exhaust in kg
C=43900#...................#Calorirfic value of fuel in kJ/kg
Cps=2.09#..................#Specific heat of steamm in exhaust in kJ/kgK
Cpg=1.0#...................#Specific heat of dry exhaust gases in kJ/kgK
k=1#....................#Two stroke engiine
t=20#.....................#Duration of testing in min
#Calculations
IP=(n*pmi*N*D*D*L*k*10*(pi/4))/6#...................#Indicated power in kW
print "Indicated power = %0.2f kW"%IP
BP=(bl*pi*Db*N)/(60*1000)#......................#Brake power in kW
print "Brake power = %0.2f kW"%BP
#Heat supplied
hf=mf*C#................#heat supplied by fuel
hip=IP*60*t#...........#Heat equivalent of BP in kJ/min
hcw=mw*Cpw*(tw2-tw1)#..........#Heat carried away by cooling water
mg=mf+(ma*mf)#....................#Mass of exhaust gases in kg/min
mst=mf*ms#..................#Mass of steam formed
hg=(mg-mst)*Cpg*(tg-tr)#..........#Heat carried by exhaust gasses
hst=mst*(417.5+2257.9+(Cps*(305-99.6)))#....................#Heat carried by exhaust steam, the obtained values are from steam table and hence are constants at NTP
ha=round(hf)-round(hip+hg+hst+hcw)#............#Unaccounted heat
pf=100#
pip=(hip/hf)*100#
pcw=(hcw/hf)*100#
pg=(hg/hf)*100#
pa=(ha/hf)*100#
pst=(hst/hf)*100#
print "HEAT BALANCE TABLE"
print "_______________________________________________________________________"
print "Item                                          kJ               Percent"
print "_______________________________________________________________________"
print "Heat supplied by fuel                       %d             %f"%(hf,pf)
print "Heat absorbed in IP                         %d             %f"%(hip,pip)
print "Heat taken away by cooling water            %d             %f"%(hcw,pcw)
print "Heat carried away by dry exhaust gases      %d             %f"%(hg,pg)
print "Heat carried away by steam in exhaust gases %d             %f"%(hst,pst)
print "Unaccounted heat                            %d             %f"%(ha,pa)

Indicated power = 15.91 kW
Brake power = 11.73 kW
HEAT BALANCE TABLE
_______________________________________________________________________
Item                                          kJ               Percent
_______________________________________________________________________
Heat supplied by fuel                       66728             100.000000
Heat absorbed in IP                         19088             28.606158
Heat taken away by cooling water            16929             25.370159
Heat carried away by dry exhaust gases      13448             20.154897
Heat carried away by steam in exhaust gases 6606             9.901049
Unaccounted heat                            10655             15.967810


## EXAMPLE 17.40 PAGE 598¶

In [20]:
from __future__ import division
from math import pi,sqrt
# Initialisation of Variables
n=6#................#No of cylinders
D=0.1#...............#Engine bore in m
L=0.14#...............#Engine stroke in m
N=2500#...............#Engine rpm
k=0.5#..................#Four stroke
br=76#..................#Barometer reading in cm of Hg
d0=3.3/100#..............#Orifice diameter in m
Cd=0.62#.................#Co efficient of discharge of orifice
pd=14#...................#Pressure drop across orifice in cm of Hg
tr=25#...............#Room temperature in C
mf=0.32#................#Fuel consumption in kg/min
rhohg=13600#.................#Density of Hg in kg/m**3
R=0.287#...................#gas constant in kJ/kgK
g=9.81#.................#Acceleration due to gravity in m/s**2
CD=17000#....................#dynamometer constant
#Calculations
Vs=(pi/4)*D*D*L*(N/2)*(n/60)#..............#Swept volume in m**3
rhoa=br1/(R*(tr+273))#...............#Density of air
pd1=(pd/100)*rhohg*g#......................#Conversion of pd into N/m**2
ha=pd1/(rhoa*g)#.......................#Head of air causing flow in m
Va=Cd*(pi/4)*d0*d0*sqrt(2*g*ha)#............#Volume of air passing through orifice of air box per min
etav=Va/Vs#....................#Volumetric efficiency
print "Volumetric efficiency = %0.2f %%"%(etav*100)
BP=bl*(N/CD)#................#Brake power in kW
pmb=(BP*6)/(L*D*D*k*10*N*n*(pi/4))#................#Brake mean effective pressure in bar
print "Brake mean effective pressure = %0.2f bar"%(pmb)
T=(BP*60*1000)/(2*pi*N)#....................#Engine torque in N-m
print "Engine torque = %0.2f N-m"%T
bsfc=(mf*60)/BP#.......................#Brake specific fuel consumption in kg/kWh
print "Brake specific fuel consumption = %0.2f kg/kWh"%bsfc

Volumetric efficiency = 68.49 %
Brake mean effective pressure = 5.14 bar
Engine torque = 269.63 N-m
Brake specific fuel consumption = 0.27 kg/kWh


## EXAMPLE 17.41 PAGE 598¶

In [21]:
from __future__ import division
from math import pi
# Initialisation of Variables
Cpw=4.18#..............#Specific heat of water in kJ/kgK
n=1#................#No of cylinders
N=350#.......#Engine rpm
pmi=2.74#..........#Mean effective pressure in bar
mf=4.22#............#Fuel consumption in kg
mw=495#..............#Mass of cooling water
tw1=13#...............#Water inlet temperature in C
tw2=38#................#Water outlet temperature in C
ma=135#..................#Mass of air used in kg/h
tr=20#.................#Room temperature in C
tg=370#.................#Exhaust temperature in C
D=0.2#.................#Engine bore in m
L=0.28#.................#Engine stroke in m
Db=1#......................#Brake drum diameter in m
C=44670#...................#Calorirfic value of fuel in kJ/kg
Cps=2.093#..................#Specific heat of steamm in exhaust in kJ/kgK
Cpg=1.005#...................#Specific heat of dry exhaust gases in kJ/kgK
k=1#....................#Two stroke engiine
t=60#.....................#Duration of testing in min
perh=15#.................#Percentage of H2 in the fuel
#Calculations
IP=(n*pmi*N*D*D*L*k*10*(pi/4))/6#...................#Indicated power in kW
print "Indicated power = %0.2f kW"%IP
BP=(bl*pi*Db*N)/(60*1000)#......................#Brake power in kW
etaith=(IP)/((mf/3600)*C)#.................#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etaith*100)
#Heat supplied
hf=(mf/t)*C#................#heat supplied by fuel
hbp=BP*t#...........#Heat equivalent of BP in kJ/min
hcw=(mw/60)*Cpw*(tw2-tw1)#..........#Heat carried away by cooling water
mg=(mf+ma)/t#....................#Mass of exhaust gases in kg/min
mst=9*(perh/100)*(mf/60)#..................#Mass of steam formed
mdg=mg-mst#..............................#Mass of dry exhaust gases per min
hg=(mdg)*Cpg*(tg-tr)#..........#Heat carried by exhaust gasses
hst=mst*(417.5+2257.9+(Cps*(305-99.6)))#....................#Heat carried by exhaust steam, the obtained values are from steam table and hence are constants at NTP
ha=round(hf)-round(hbp+hg+hst+hcw)#............#Unaccounted heat
pf=100#
pbp=(hbp/hf)*100#
pcw=(hcw/hf)*100#
pg=(hg/hf)*100#
pa=(ha/hf)*100#
pst=(hst/hf)*100#
print "HEAT BALANCE TABLE"
print "_______________________________________________________________________"
print "Item                                          kJ               Percent"
print "_______________________________________________________________________"
print "Heat supplied by fuel                       %0.2f             %0.2f"%(hf,pf)
print "Heat equivalent of BP                       %0.2f             %0.2f"%(hbp,pbp)
print "Heat taken away by cooling water            %0.2f             %0.2f"%(hcw,pcw)
print "Heat carried away by dry exhaust gases      %0.2f             %0.2f"%(hg,pg)
print "Heat carried away by steam in exhaust gases %0.2f             %0.2f"%(hst,pst)
print "Unaccounted heat                            %0.2f             %0.2f"%(ha,pa)

Indicated power = 14.06 kW
Indicated thermal efficiency = 26.85 %
HEAT BALANCE TABLE
_______________________________________________________________________
Item                                          kJ               Percent
_______________________________________________________________________
Heat supplied by fuel                       3141.79             100.00
Heat equivalent of BP                       659.73             21.00
Heat taken away by cooling water            862.12             27.44
Heat carried away by dry exhaust gases      782.78             24.92
Heat carried away by steam in exhaust gases 294.85             9.38
Unaccounted heat                            543.00             17.28


## EXAMPLE 17.42 PAGE 599¶

In [22]:
from __future__ import division
from math import pi
# Initialisation of Variables
Cpw=4.18#..............#Specific heat of water in kJ/kgK
n=1#................#No of cylinders
N=350#.......#Engine rpm
pmi=2.8#..........#Mean effective pressure in bar
mf=4.3#............#Fuel consumption in kg
mw=500#..............#Mass of cooling water
tw1=25#...............#Water inlet temperature in C
tw2=50#................#Water outlet temperature in C
ma=33#..................#Mass of air used per kg of fuel in kg
tr=25#.................#Room temperature in C
tg=400#.................#Exhaust temperature in C
D=0.22#.................#Engine bore in m
L=0.28#.................#Engine stroke in m
Db=1#......................#Brake drum diameter in m
C=43900#...................#Calorirfic value of fuel in kJ/kg
Cps=2.09#..................#Specific heat of steamm in exhaust in kJ/kgK
Cpg=1.0#...................#Specific heat of dry exhaust gases in kJ/kgK
k=1#....................#Two stroke engiine
perh=15#...................#Percentage of hydrogen
#Calculations
IP=(n*pmi*N*D*D*L*k*10*(pi/4))/6#...................#Indicated power in kW
print "Indicated power = %0.2f kW"%IP
BP=(bl*pi*Db*N)/(60*1000)#......................#Brake power in kW
print "Brake power = %0.2f kW"%BP
#Heat supplied
hf=(mf/60)*C#................#heat supplied by fuel
hip=IP*60#...........#Heat equivalent of BP in kJ/min
hcw=(mw/60)*Cpw*(tw2-tw1)#..........#Heat carried away by cooling water
mg=(mf+(mf*ma))/60#....................#Mass of exhaust gases in kg/min
mst=9*(perh/100)*(mf/60)#..................#Mass of steam formed
mdg=mg-mst#..............................#Mass of dry exhaust gases per min
hg=(mdg)*Cpg*(tg-tr)#..........#Heat carried by exhaust gasses
hst=mst*(417.5+2257.9+(Cps*(400-99.6)))#....................#Heat carried by exhaust steam, the obtained values are from steam tables at NTP
mg=mf+(ma*mf)#....................#Mass of exhaust gases in kg/min
ha=round(hf)-round(hip+hg+hst+hcw)#............#Unaccounted heat
pf=100#
pip=(hip/hf)*100#
pcw=(hcw/hf)*100#
pg=(hg/hf)*100#
pa=(ha/hf)*100#
pst=(hst/hf)*100#
print "HEAT BALANCE TABLE : "
print "_______________________________________________________________________"
print "Item                                          kJ               Percent"
print "_______________________________________________________________________"
print "Heat supplied by fuel                       %d             %0.2f"%(hf,pf)
print "Heat absorbed in IP                         %d             %0.2f"%(hip,pip)
print "Heat taken away by cooling water            %d             %0.2f"%(hcw,pcw)
print "Heat carried away by dry exhaust gases      %d             %0.2f"%(hg,pg)
print "Heat carried away by steam in exhaust gases %d             %0.2f"%(hst,pst)
print "Unaccounted heat                            %d             %0.2f"%(ha,pa)

Indicated power = 17.38 kW
Brake power = 10.81 kW
HEAT BALANCE TABLE :
_______________________________________________________________________
Item                                          kJ               Percent
_______________________________________________________________________
Heat supplied by fuel                       3146             100.00
Heat absorbed in IP                         1043             33.15
Heat taken away by cooling water            870             27.68
Heat carried away by dry exhaust gases      877             27.89
Heat carried away by steam in exhaust gases 319             10.16
Unaccounted heat                            35             1.11


## EXAMPLE 17.43 PAGE 600¶

In [23]:
from __future__ import division
from math import pi
# Initialisation of Variables
I=210#.....................#Output of generator in A
V=200#.....................#Generator voltage in V
etag=0.82#.................#Generator efficiency
mf=11.2#....................#Fuel used in kg/h
C=42600#.......................#Calorific value of fuel in kJ/kg
afr=18#....................#Air fuel ratio
mc=580#.....................#Mass of water through calorimeter in kg/h
delt=36#....................#Temperature raise in C
tg=98#........................#Temperature of exhaust in C
ta=20#.......................#Ambient temperature in C
phcw=0.32#.....................#Heat lost to cooling jacket is 32% of heat supplied
cpe=1.05#...................#Specific heat of exhause gases in kJ/kgK
cpw=4.18#...................#Specific heat of water in kJ/kgK
#Calculations
pow=V*I#......................#Total power generated in W
BP=(pow/1000)/etag#..................#Brake power in kW
hf=(mf/60)*C#...................#Heat supplied to the engine
hbp=BP*60#.........................#Heat equivalent of BP
mg=(mf/60)*(afr+1)#...............#Mass of exhaust gases formed per min in kg
hg=((mc/60)*cpw*(delt))+(mg*cpe*(tg-ta))#..........#Heat carried by exhaust gases per min
hcw=phcw*hf#...................#Heat lost to cooling jacket
ha=hf-(hcw+hg+hbp)#...................#Unaccounted heat
pf=100#pbp=(hbp/hf)*100#pcw=(hcw/hf)*100#pg=(hg/hf)*100#pa=(ha/hf)*100
print "HEAT BALANCE TABLE: "
print "_______________________________________________________________________"
print "Item                                      kJ                  Percent"
print "_______________________________________________________________________"
print "Heat supplied by fuel                     %d             %0.2f"%(hf,pf)
print "Heat absorbed in BP                       %d             %0.2f"%(hbp,pbp)
print "Heat taken away by cooling water          %d             %0.2f"%(hcw,pcw)
print "Heat carried away by exhaust gases        %d             %0.2f"%(hg,pg)
print "Unaccounted heat                          %d             %0.2f"%(ha,pa)

HEAT BALANCE TABLE:
_______________________________________________________________________
Item                                      kJ                  Percent
_______________________________________________________________________
Heat supplied by fuel                     7951             100.00
Heat absorbed in BP                       3073             21.00
Heat taken away by cooling water          2544             27.68
Heat carried away by exhaust gases        1745             27.89
Unaccounted heat                          589             1.11


## EXAMPLE 17.44 PAGE 601¶

In [24]:
from __future__ import division
from math import pi
# Initialisation of Variables
D=0.34#..............#Engine bore in m
k=0.5#...............#Four stroke engine
n=1#..................#No of cylinders
L=0.44#................#Engine stroke in m
Ne=400#................#Engine rpm
aic=465#..............#Area of indicator diagram in mm**2
l=60#..................#Length of diagram in mm
s=0.6#...............#Spring constant in bar/mm
CD=7460#................#Dynamometer constant
mf=10.6#................#Fuel used in kg/h
Ca=49500#.................#Calorific value of fuel in kJ/kg
mw=25#...................#Cooling water circulated in kg/min
cpw=4.18#..................#Specific heat capacity of water in kJ/kgC
delt=25#..................#Rise in temperature of water
#Mass analysis of fuel
C=84#..................#Percentage of carbon
H=15#..................#Percentage of hydrogen
In=1#..................#Percentage of incombustible
#Volume analysis of exhaust gases
CO2=9#..................#Percentage of caron dioxide
O=10#..................#Percentage of oxygen
N=81#..................#Percentage of nitrogen
tg=400#................#Temperature of exhaust gases in C
cpg=1.05#..............#Specific heat of exhaust gases in kJ/kgC
tr=25#.................#Temperature of room in C
ppst=0.03#..............#Partial pressure of steam in exhaust gases in bar
cpst=2.1#.................#Specific heat of superheated steam in kJ/kgC
#Calculations
pmi=(aic*s)/l#................#Mean effective pressure in bar
IP=(n*pmi*L*D*D*k*10*Ne*(pi/4))/6#...............#Indicated power in kW
BP=(W*Ne)/CD#.......................#Brake power in kW
FP=IP-BP#...........................#Frictional power in kW
hf=(mf/60)*Ca#...................#Heat supplied in kJ per min
hbp=BP*60#.....................#Heat equivalent of Brake power in kW
hfp=FP*60#......................#heat equivalent of frictional power in kW
hcw=mw*cpw*delt#..................#Heat carried away by cooling water
ma1=(N*C)/(33*(CO2))#...........#Mass of air supplied per kg of fuel
mg1=ma1+1#.......................#Mass of exhaust gases per kg of fuel
mg=mg1*(mf/60)#.................#Mass of exhaust gas formed per min
mst1=9*(H/100)#.................#Mass of steam formed per kg of fuel
mst=mst1*(mf/60)#..................#Mass of steam formed per min
mdg=mg-mst#.......................#Mass of dry exhaust gas
Hex=mdg*cpg*(tg-tr)#...............#Heat carried by exhaust gases
hst=(2545.5+(cpst*(tg-24.1)))*mst#................#Heat carried by steam in exhaust gases in kJ/kg.....The values are from steam tables corresponding to the partial pressure 0.03 and temperature 400 Celsius
ha=hf-(hbp+hfp+hcw+Hex+hst)#.....................#Unaccounted heat
pf=100#
pbp=(hbp/hf)*100#
pfp=(hfp/hf)*100#
pcw=(hcw/hf)*100#
pex=(Hex/hf)*100#
pa=(ha/hf)*100#
pst=(hst/hf)*100#
print "HEAT BALANCE TABLE : "
print "_______________________________________________________________________"
print "Item                                          kJ               Percent"
print "_______________________________________________________________________"
print "Heat supplied by fuel                       %d             %0.2f"%(hf,pf)
print "Heat absorbed in BP                         %d             %0.2f"%(hbp,pbp)
print "Heat lost by FP                             %d              %0.2f"%(hfp,pfp)
print "Heat taken away by cooling water            %d             %0.2f"%(hcw,pcw)
print "Heat carried away by dry exhaust gases      %d             %0.2f"%(Hex,pex)
print "Heat carried away by steam in exhaust gases %d              %0.2f"%(hst,pst)
print "Unaccounted heat                            %d               %0.2f"%(ha,pa)

HEAT BALANCE TABLE :
_______________________________________________________________________
Item                                          kJ               Percent
_______________________________________________________________________
Heat supplied by fuel                       8745             100.00
Heat absorbed in BP                         3056             34.95
Heat lost by FP                             658              7.53
Heat taken away by cooling water            2612             29.87
Heat carried away by dry exhaust gases      1569             17.94
Heat carried away by steam in exhaust gases 795              9.10
Unaccounted heat                            52               0.60


## EXAMPLE 17.45 PAGE 602¶

In [25]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=4#........................#No of cylinders
ga=1.4#...................#Degree of freedom
D=0.075#..................#Engine bore in m
L=0.1#...................#Engine stroke in m
mf=6#.......................#Fuel consumption in kg/h
C=83600#..................#Calorific value of fuel used
Vc=0.0001#.................#Clearence volume in m**3
BP=15.6#.................#Brake power wilh all cylinder working in kW
BP1=11.1#...................#Brake power wilh cylinder no 1 cutout in kW
BP2=11.03#...................#Brake power wilh cylinder no 2 cutout in kW
BP3=10.88#...................#Brake power wilh cylinder no 3 cutout in kW
BP4=10.66#...................#Brake power wilh cylinder no 4 cutout in kW
#Calculations
IP1=BP-BP1#...........................#Indicated power produced in cylinder 1 in kW
IP2=BP-BP2#...........................#Indicated power produced in cylinder 2 in kW
IP3=BP-BP3#...........................#Indicated power produced in cylinder 3 in kW
IP4=BP-BP4#...........................#Indicated power produced in cylinder 4 in kW
IP=IP1+IP2+IP3+IP4#.............................#Total Indicated power produced in kW
etamech=BP/IP#..............................#Mechanical efficiency
print "Mechanical efficiency = %0.2f %%"%(etamech*100)
etaith=IP/((mf/3600)*C)#.....................#Indicated thermal efficiency
print "Indicated thermal efficiency = %0.2f %%"%(etaith*100)
Vs=(pi/4)*D*D*L#..........................#Stroke volume in m**3
r=(Vs+Vc)/Vc#................................#Compression ratio
etast=1-(1/(r**(ga-1)))#...........................#Air standard efficiency
print "Air standard efficiency = %0.2f %%"%(etast*100)

Mechanical efficiency = 83.29 %
Indicated thermal efficiency = 13.44 %
Air standard efficiency = 49.13 %


## EXAMPLE 17.46 PAGE 603¶

In [26]:
from __future__ import division
from math import pi
# Initialisation of Variables
n=4#..........................#No of cylinders
D=0.06#......................#Engine bore in m
L=0.09#.......................#Engine stroke in m
N=2800#.......................#Engine rpm
Ta=0.37#.......................#Length of torque arm in m
spgr=0.74#.....................#Specific graviy of fuel
fc=8.986#......................#Fuel consumption in ltrs/h
mf=fc*spgr#.....................#Fuel consumed in kg/h
C=44100#......................#Calorific value of fuel in kJ/kg
BP1=110#...................#Brake load with cylinder no 1 cutout in N
BP2=107#...................#Brake load with cylinder no 2 cutout in N
BP3=104#...................#Brake load with cylinder no 3 cutout in N
BP4=110#...................#Brake load with cylinder no 4 cutout in N
k=0.5#.....................#Four stroke engine
#Calculations
T=BPnl*Ta#.......................#Engine torque in N-m
print "Engine torque = %0.2f Nm:"%T
BP=(2*pi*N*T)/(60*1000)#..........................#Brake power in kW
pmb=(BP*6)/(n*D*D*L*N*10*(pi/4)*k)#...................#Brake mean effective pressure in bar
print "Brake mean effective pressure = %0.2f bar"% pmb
etabth=BP/((mf/3600)*C)#...........................#Brake thermal efficiency
print "Brake thermal efficiency = %0.2f %%"%(etabth*100)
sfc=mf/BP#.......................#Specific fuel consumption in kg/kWh
print "Specific fuel consumption = %0.2f kg/kWh"%sfc
IP1=BPnl-BP1#...........................#Indicated power produced in cylinder 1 in kW
IP2=BPnl-BP2#...........................#Indicated power produced in cylinder 2 in kW
IP3=BPnl-BP3#...........................#Indicated power produced in cylinder 3 in kW
IP4=BPnl-BP4#...........................#Indicated power produced in cylinder 4 in kW
IP=IP1+IP2+IP3+IP4#.............................#Total Indicated power produced in kW
etamech=BPnl/IP#..............................#Mechanical efficiency
print "Mechanical efficiency = %0.2f %%"%(etamech*100)
pmi=pmb/etamech#............................#Indicated mean effective pressure in bar
print "Indicated mean effective pressure = %0.2f bar "%pmi

Engine torque = 59.20 Nm:
Brake mean effective pressure = 7.31 bar
Brake thermal efficiency = 21.31 %
Specific fuel consumption = 0.38 kg/kWh
Mechanical efficiency = 76.56 %
Indicated mean effective pressure = 9.55 bar