# Chapter 10:Intoduction to Internal Combustion engines¶

## example 10.1;pg no: 387¶

In :
#cal of indicated power,,mechanical efficiency
#intiation of all variables
# Chapter 9
import math
print"Example 10.1, Page:387  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 1")
k=20.*10**6;#spring constant in N/m^2
N=2000.;#engine rpm
print("from stroke to bore ratio i.e L/D=1.2 and cylinder diameter=bore,i.e D=12 cm")
D=12.*10**-2;#cylinder diameter in m
print("stroke(L)=1.2*D in m")
L=1.2*D
print("Area of indicator diagram(A)=30*10^-4 m^2")
A=30.*10**-4;
print("length of indicator diagram(l)=(1/2)*L in m")
l=(1./2.)*L
print("mean effective pressure(mep)=A*k/l in N/m^2")
mep=A*k/l
print("cross-section area of piston(Ap)=%pi*D^2/4 in m^2")
Ap=math.pi*D**2./4.
print("for one cylinder indicated power(IP)=mep*Ap*L*N/(2*60) in W")
IP=mep*Ap*L*N/(2.*60.)
IP=4.*IP
print("for four cylinder total indicated power(IP)=4*IP in W"),round(IP,2)
print("frictional power loss(FP)=0.10*IP in W")
FP=0.10*IP
BP=IP-FP
print("brake power available(BP)=indicated power-frictional power=IP-FP in W"),round(BP,2)
n=BP/IP
print("therefore,mechanical efficiency of engine(n)=brake power/indicated power=BP/IP="),round(BP/IP,2)
print("in percentage"),round(n*100,2)
print("so indicated power=90477.8 W")
print("and mechanical efficiency=90%")

Example 10.1, Page:387

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 1
from stroke to bore ratio i.e L/D=1.2 and cylinder diameter=bore,i.e D=12 cm
stroke(L)=1.2*D in m
Area of indicator diagram(A)=30*10^-4 m^2
length of indicator diagram(l)=(1/2)*L in m
mean effective pressure(mep)=A*k/l in N/m^2
cross-section area of piston(Ap)=%pi*D^2/4 in m^2
for one cylinder indicated power(IP)=mep*Ap*L*N/(2*60) in W
for four cylinder total indicated power(IP)=4*IP in W 90477.87
frictional power loss(FP)=0.10*IP in W
brake power available(BP)=indicated power-frictional power=IP-FP in W 81430.08
therefore,mechanical efficiency of engine(n)=brake power/indicated power=BP/IP= 0.9
in percentage 90.0
so indicated power=90477.8 W
and mechanical efficiency=90%


## example 10.2;pg no: 388¶

In :
#cal of power required to drive
#intiation of all variables
# Chapter 10
import math
print"Example 10.2, Page:388  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 2")
A=40*10**-4;#area of indicator diagram in m^2
l=8*10**-2;#length of indicator diagram in m
D=15*10**-2;#bore of cylinder in m
L=20*10**-2;#stroke in m
k=1.5*10**8;#spring constant in pa/m
N=100;#pump motor rpm
print("reciprocating pump is work absorbing machine having its mechanism similar to the piston-cylinder arrangement in IC engine.")
print("mean effective pressure(mep)=A*k/l in pa")
mep=A*k/l
print("indicator power(IP)=Ap*L*mep*N*1*2/60 in W")
Ap=math.pi*D**2/4
IP=Ap*L*mep*N*2/60
print("(it is double acting so let us assume total power to be double of that in single acting)"),round(IP,2)
print("so power required to drive=88.36 KW")

Example 10.2, Page:388

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 2
reciprocating pump is work absorbing machine having its mechanism similar to the piston-cylinder arrangement in IC engine.
mean effective pressure(mep)=A*k/l in pa
indicator power(IP)=Ap*L*mep*N*1*2/60 in W
(it is double acting so let us assume total power to be double of that in single acting) 88357.29
so power required to drive=88.36 KW


## example 10.3;pg no: 388¶

In :
#cal of indicated power,frictional power loss,mechanical efficiency
#intiation of all variables
# Chapter 10
print"Example 10.3, Page:388  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 3")
n=0.9;#mechanical efficiency of engine
BP=38;#brake power in KW
IP=BP/n
print("indicated power(IP)=brake power/mechanical efficiency in KW="),round(IP,2)
FP=IP-BP
print("frictional power loss(FP)=IP-BP in KW"),round(FP,2)
print("brake power at quater load(BPq)=0.25*BP in KW")
BPq=0.25*BP
IP=BPq+FP;
n1=BPq/IP
print("mechanical efficiency(n1)=BPq/IP"),round(n1,2)
print("in percentage"),round(n1*100,2)
print("so indicated power=42.22 KW")
print("frictional power loss=4.22 KW")
print("mechanical efficiency=69.24%")

Example 10.3, Page:388

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 3
indicated power(IP)=brake power/mechanical efficiency in KW= 42.22
frictional power loss(FP)=IP-BP in KW 4.22
brake power at quater load(BPq)=0.25*BP in KW
mechanical efficiency(n1)=BPq/IP 0.69
in percentage 69.23
so indicated power=42.22 KW
frictional power loss=4.22 KW
mechanical efficiency=69.24%


## example 10.4;pg no: 389¶

In :
#cal of brake power,fuel consumption
#intiation of all variables
# Chapter 10
import math
print"Example 10.4, Page:389  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 4")
m=0.25;#specific fuel consumption in kg/KWh
Pb_mep=1.5*1000;#brake mean effective pressure of each cylinder in Kpa
N=100;#engine rpm
D=85*10**-2;#bore of cylinder in m
L=220*10**-2;#stroke in m
C=43*10**3;#calorific value of diesel in KJ/kg
A=math.pi*D**2/4;
BP=Pb_mep*L*A*N/60
print("brake power of engine(BP) in MW="),round(BP,2)
print("so brake power is 3.121 MW")
print("The fuel consumption in kg/hr(m)=m*BP in kg/hr")
m=m*BP
print("In order to find out brake thermal efficiency the heat input from fuel per second is required.")
print("heat from fuel(Q)in KJ/s")
print("Q=m*C/3600")
Q=m*C/3600
print("energy to brake power=3120.97 KW")
n=BP/Q
print("brake thermal efficiency(n)="),round(n,2)
print("in percentage"),round(n*100,2)
print("so fuel consumption=780.24 kg/hr,brake thermal efficiency=33.49%")
print("NOTE=>In book,it is given that brake power in MW is 31.21 but in actual it is 3120.97 KW or 3.121 MW which is corrected above.")

Example 10.4, Page:389

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 4
brake power of engine(BP) in MW= 3120.98
so brake power is 3.121 MW
The fuel consumption in kg/hr(m)=m*BP in kg/hr
In order to find out brake thermal efficiency the heat input from fuel per second is required.
heat from fuel(Q)in KJ/s
Q=m*C/3600
energy to brake power=3120.97 KW
brake thermal efficiency(n)= 0.33
in percentage 33.49
so fuel consumption=780.24 kg/hr,brake thermal efficiency=33.49%
NOTE=>In book,it is given that brake power in MW is 31.21 but in actual it is 3120.97 KW or 3.121 MW which is corrected above.


## example 10.5;pg no: 389¶

In :
#cal of brake thermal efficiency,brake power,volumetric efficiency
#intiation of all variables
# Chapter 10
import math
print"Example 10.5, Page:389  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 5")
Pb_mep=6*10**5;#brake mean effective pressure in pa
N=600;#engine rpm
m=0.25;#specific fuel consumption in kg/KWh
D=20*10**-2;#bore of cylinder in m
L=30*10**-2;#stroke in m
k=26;#air to fuel ratio
C=43*10**3;#calorific value in KJ/kg
R=0.287;#gas constant in KJ/kg K
T=(27+273);#ambient temperature in K
P=1*10**2;#ambient pressure in Kpa
n=3600/(m*C)
print("brake thermal efficiency(n)=3600/(m*C)"),round(n,2)
print("in percentage"),round(n*100,2)
print("brake power(BP)in KW")
A=math.pi*D**2/4;
BP=4*Pb_mep*L*A*N/60000
print("BP="),round(BP,2)
print("brake specific fuel consumption,m=mf/BP")
print("so mf=m*BP in kg/hr")
mf=m*BP
print("air consumption(ma)from given air fuel ratio=k*mf in kg/hr")
ma=k*mf
print("ma in kg/min")
ma=ma/60
print("using perfect gas equation,")
print("P*Va=ma*R*T")
print("sa Va=ma*R*T/P in m^3/min")
Va=ma*R*T/P
print("swept volume(Vs)=%pi*D^2*L/4 in m^3")
Vs=math.pi*D**2*L/4
n_vol=Va/(Vs*(N/2)*4)
print("volumetric efficiency,n_vol=Va/(Vs*(N/2)*no. of cylinder)"),round(n_vol,2)
print("in percentage"),round(n_vol*100,2)
print("NOTE=>In this question,while calculating swept volume in book,values of D=0.30 m and L=0.4 m is taken which is wrong.Hence above solution is solve taking right values given in book which is D=0.20 m and L=0.3 m,so the volumetric efficiency vary accordingly.")

Example 10.5, Page:389

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 5
brake thermal efficiency(n)=3600/(m*C) 0.33
in percentage 33.49
brake power(BP)in KW
BP= 226.19
brake specific fuel consumption,m=mf/BP
so mf=m*BP in kg/hr
air consumption(ma)from given air fuel ratio=k*mf in kg/hr
ma in kg/min
using perfect gas equation,
P*Va=ma*R*T
sa Va=ma*R*T/P in m^3/min
swept volume(Vs)=%pi*D^2*L/4 in m^3
volumetric efficiency,n_vol=Va/(Vs*(N/2)*no. of cylinder) 1.87
in percentage 186.55
NOTE=>In this question,while calculating swept volume in book,values of D=0.30 m and L=0.4 m is taken which is wrong.Hence above solution is solve taking right values given in book which is D=0.20 m and L=0.3 m,so the volumetric efficiency vary accordingly.


## example 10.6;pg no: 390¶

In :
#cal of brake power
#intiation of all variables
# Chapter 9
import math
print"Example 10.6, Page:390  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 6")
N=3000;#engine rpm
m=5;#fuel consumption in litre/hr
r=19;#air-fuel ratio
sg=0.7;#specific gravity of fuel
V=500;#piston speed in m/min
P_imep=6*10**5;#indicated mean effective pressure in pa
P=1.013*10**5;#ambient pressure in pa
T=(15+273);#ambient temperature in K
n_vol=0.7;#volumetric efficiency
n_mech=0.8;#mechanical efficiency
R=0.287;#gas constant for gas in KJ/kg K
print("let the bore diameter be (D) m")
print("piston speed(V)=2*L*N")
print("so L=V/(2*N) in m")
L=V/(2*N)
L=0.0833;#approx.
print("volumetric efficiency,n_vol=air sucked/(swept volume * no. of cylinder)")
print("so air sucked/D^2=n_vol*(%pi*L/4)*N*2 in m^3/min")
n_vol*(math.pi*L/4)*N*2
print("so air sucked =274.78*D^2 m^3/min")
print("air requirement(ma),kg/min=A/F ratio*fuel consumption per min")
print("so ma=r*m in kg/min")
ma=r*m*sg/60
print("using perfect gas equation,P*Va=ma*R*T")
print("so Va=ma*R*T/P in m^3/min")
Va=ma*R*T*1000/P
print("ideally,air sucked=Va")
print("so 274.78*D^2=0.906")
print("D=sqrt(0.906/274.78) in m")
D=math.sqrt(0.906/274.78)
print("indicated power(IP)=P_imep*L*A*N*no.of cylinders in KW")
IP=P_imep*L*(math.pi*D**2/4)*(N/60)*2/1000
print("brake power=indicated power*mechanical efficiency")
BP=IP*n_mech
print("BP=IP*n_mech in KW"),round(BP,2)
print("so brake power=10.34 KW")

Example 10.6, Page:390

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 6
let the bore diameter be (D) m
piston speed(V)=2*L*N
so L=V/(2*N) in m
volumetric efficiency,n_vol=air sucked/(swept volume * no. of cylinder)
so air sucked/D^2=n_vol*(%pi*L/4)*N*2 in m^3/min
so air sucked =274.78*D^2 m^3/min
air requirement(ma),kg/min=A/F ratio*fuel consumption per min
so ma=r*m in kg/min
using perfect gas equation,P*Va=ma*R*T
so Va=ma*R*T/P in m^3/min
ideally,air sucked=Va
so 274.78*D^2=0.906
D=sqrt(0.906/274.78) in m
indicated power(IP)=P_imep*L*A*N*no.of cylinders in KW
brake power=indicated power*mechanical efficiency
BP=IP*n_mech in KW 10.35
so brake power=10.34 KW


## example 10.7;pg no: 391¶

In :
#cal of power and efficiency
#intiation of all variables
# Chapter 10
import math
print"Example 10.7, Page:391  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 7")
N=3000;#speed of rotation in rpm
D=20*10**-2;#bore in m
L=30*10**-2;#stroke in m
m=0.15;#fuel supplying rate in kg/min
C=43;#calorific value of fuel in MJ/kg
FP=5;#friction power in KW
g=9.81;#acceleration due to gravity in m/s^2
print("After switching off fuel supply the capacity of motor required to run engine will be the friction power required at this speed of engine.")
print("friction power(FP)=5 KW")
BP=2*math.pi*N*(M*g*r)*10**-3/60
print("brake power(BP) in KW="),round(BP,2)
IP=BP+FP
print("indicated power(IP) in KW="),round(IP,2)
n_mech=BP/IP
print("mechanical efficiency(n_mech)="),round(n_mech,2)
print("in percentage"),round(n_mech*100,2)
bsfc=m*60/BP
print("brake specific fuel consumption(bsfc)=specific fuel consumption/brake power in kg/KW hr="),round(bsfc,2)
n_bte=3600/(bsfc*C*1000)
print("brake thermal efficiency(n_bte)="),round(n_bte,2)
print("in percentage"),round(n_bte*100,2)
print("also,mechanical efficiency(n_mech)=brake thermal efficiency/indicated thermal efficiency")
n_ite=n_bte/n_mech
print("indicated thermal efficiency(n_ite)="),round(n_ite,2)
print("in percentage"),round(n_ite*100,2)
print("indicated power(IP)=P_imep*L*A*N")
P_imep=IP/(L*(math.pi*D**2/4)*N/60)
print("so P_imep in Kpa="),round(P_imep,2)
print("Also,mechanical efficiency=P_bmep/P_imep")
n_mech=0.8604;#mechanical efficiency
P_bmep=P_imep*n_mech
print("so P_bmep in Kpa="),round(P_bmep,2)
print("brake power=30.82 KW")
print("indicated power=35.82 KW")
print("mechanical efficiency=86.04%")
print("brake thermal efficiency=28.67%")
print("indicated thermal efficiency=33.32%")
print("brake mean effective pressure=65.39 Kpa")
print("indicated mean effective pressure=76.01 Kpa")

Example 10.7, Page:391

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 7
After switching off fuel supply the capacity of motor required to run engine will be the friction power required at this speed of engine.
friction power(FP)=5 KW
brake power(BP) in KW= 30.82
indicated power(IP) in KW= 35.82
mechanical efficiency(n_mech)= 0.86
in percentage 86.04
brake specific fuel consumption(bsfc)=specific fuel consumption/brake power in kg/KW hr= 0.29
brake thermal efficiency(n_bte)= 0.29
in percentage 28.67
also,mechanical efficiency(n_mech)=brake thermal efficiency/indicated thermal efficiency
indicated thermal efficiency(n_ite)= 0.33
in percentage 33.32
indicated power(IP)=P_imep*L*A*N
so P_imep in Kpa= 76.01
Also,mechanical efficiency=P_bmep/P_imep
so P_bmep in Kpa= 65.4
brake power=30.82 KW
indicated power=35.82 KW
mechanical efficiency=86.04%
brake thermal efficiency=28.67%
indicated thermal efficiency=33.32%
brake mean effective pressure=65.39 Kpa
indicated mean effective pressure=76.01 Kpa


## example 10.8;pg no: 392¶

In :
#cal of brake thermal efficiency,volumetric effeciency
#intiation of all variables
# Chapter 10
import math
print"Example 10.8, Page:392  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 8")
N=300.;#engine rpm
BP=250.;#brake power in KW
D=30.*10**-2;#bore in m
L=25.*10**-2;#stroke in m
m=1.;#fuel consumption in kg/min
r=10.;#airfuel ratio
P_imep=0.8;#indicated mean effective pressure in pa
C=43.*10**3;#calorific value of fuel in KJ/kg
P=1.013*10**5;#ambient pressure in K
R=0.287;#gas constant in KJ/kg K
T=(27.+273.);#ambient temperature in K
IP=P_imep*L*(math.pi*D**2/4)*N*4*10**3/60
print("indicated power(IP) in KW="),round(IP,2)
print("mechanical efficiency(n_mech)=brake power/indicated power")
n_mech=BP/IP
print("so n_mech="),round(n_mech,2)
print("in percentage "),round(n_mech*100,2)
print("brake specific fuel consumption(bsfc)=m*60/BP in kg/KW hr")
bsfc=m*60./BP
n_bte=3600./(bsfc*C)
print("brake thermal efficiency(n_bte)="),round(n_bte,2)
print("in percentage"),round(n_bte*100,2)
print("swept volume(Vs)=%pi*D^2*L/4 in m^3")
Vs=math.pi*D**2*L/4
print("mass of air corresponding to above swept volume,using perfect gas equation")
print("P*Vs=ma*R*T")
print("so ma=(P*Vs)/(R*T) in kg")
ma=(P*Vs)/(R*T*1000)
ma=0.02;#approx.
print("volumetric effeciency(n_vol)=mass of air taken per minute/mass corresponding to swept volume per minute")
print("so mass of air taken per minute in kg/min ")
1*10
print("mass corresponding to swept volume per minute in kg/min")
ma*4*N/2
print("so volumetric efficiency "),round(10./12.,4)
print("in percentage"),round((10./12.)*100.,4)
print("so indicated power =282.74 KW,mechanical efficiency=88.42%")
print("brake thermal efficiency=34.88%,volumetric efficiency=83.33%")

Example 10.8, Page:392

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 8
indicated power(IP) in KW= 282.74
mechanical efficiency(n_mech)=brake power/indicated power
so n_mech= 0.88
in percentage  88.42
brake specific fuel consumption(bsfc)=m*60/BP in kg/KW hr
brake thermal efficiency(n_bte)= 0.35
in percentage 34.88
swept volume(Vs)=%pi*D^2*L/4 in m^3
mass of air corresponding to above swept volume,using perfect gas equation
P*Vs=ma*R*T
so ma=(P*Vs)/(R*T) in kg
volumetric effeciency(n_vol)=mass of air taken per minute/mass corresponding to swept volume per minute
so mass of air taken per minute in kg/min
mass corresponding to swept volume per minute in kg/min
so volumetric efficiency  0.8333
in percentage 83.3333
so indicated power =282.74 KW,mechanical efficiency=88.42%
brake thermal efficiency=34.88%,volumetric efficiency=83.33%


## example 10.9;pg no: 393¶

In :
#cal of indicated power,brake power,mechanical efficiency
#intiation of all variables
# Chapter 10
import math
print"Example 10.9, Page:393  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 9")
h=10.;#height of indicator diagram in mm
k=25.;#indicator constant in KN/m^2  per mm
N=300.;#engine rpm
Vs=1.5*10**-2;#swept volume in m^3
M=60.;#effective brake load upon dynamometer in kg
r=50.*10**-2;#effective brake drum radius in m
m=0.12;#fuel consumption in kg/min
C=42.*10**3;#calorific value in KJ/kg
mw=6.;#circulating water rate in kg/min
T1=35.;#cooling water entering temperature in degree celcius
T2=70.;#cooling water leaving temperature in degree celcius
Eg=30.;#exhaust gases leaving energy in KJ/s
Cw=4.18;#specific heat of water in KJ/kg K
g=9.81;#accelaration due to gravity in m/s^2
print("indicated mean effective pressure(P_imeb)=h*k in Kpa")
P_imeb=h*k
IP=P_imeb*Vs*N/(2*60)
print("indicated power(IP)=P_imeb*L*A*N/2 in KW") ,round(IP,3)
BP=2*math.pi*N*(M*g*r*10**-3)/(2*60)
print("brake power(BP)=2*%pi*N*T in KW"),round(BP,2)
n_mech=BP/IP
print("mechanical efficiency(n_mech)="),round(n_mech,2)
print("in percentage"),round(n_mech*100,2)
print("so indicated power=9.375 KW")
print("brake power=4.62 KW")
print("mechanical efficiency=49.28%")
print("energy liberated from fuel(Ef)=C*m/60 in KJ/s")
Ef=C*m/60
print("energy available as brake power(BP)=4.62 KW")
print("energy to coolant(Ec)=(mw/M)*Cw*(T2-T1) in KW")
Ec=(mw/M)*Cw*(T2-T1)
print("energy carried by exhaust gases(Eg)=30 KJ/s")
Ef-BP-Ec-Eg
print("unaccounted energy loss in KW="),round(Ef-BP-Ec-Eg,2)
print("NOTE=>overall energy balance sheet is attached as jpg file with this code.")

Example 10.9, Page:393

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 9
indicated mean effective pressure(P_imeb)=h*k in Kpa
indicated power(IP)=P_imeb*L*A*N/2 in KW 9.375
brake power(BP)=2*%pi*N*T in KW 4.62
mechanical efficiency(n_mech)= 0.49
in percentage 49.31
so indicated power=9.375 KW
brake power=4.62 KW
mechanical efficiency=49.28%
energy liberated from fuel(Ef)=C*m/60 in KJ/s
energy available as brake power(BP)=4.62 KW
energy to coolant(Ec)=(mw/M)*Cw*(T2-T1) in KW
energy carried by exhaust gases(Eg)=30 KJ/s
unaccounted energy loss in KW= 34.75
NOTE=>overall energy balance sheet is attached as jpg file with this code.


## example 10.10;pg no: 394¶

In :
#cal of brake power,brake specific fuel consumption,indicated  thermal efficiency
#intiation of all variables
# Chapter 10
import math
print"Example 10.10, Page:394  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 10")
m=4.;#mass of fuel consumed in kg
N=1500.;#engine rpm
mw=15.;#water circulation rate in kg/min
T1=27.;#cooling water inlet temperature in degree celcius
T2=50.;#cooling water outlet temperature in degree celcius
ma=150.;#mass of air consumed in kg
T_exhaust=400.;#exhaust temperature in degree celcius
T_atm=27.;#atmospheric temperature in degree celcius
Cg=1.25;#mean specific heat of exhaust gases in KJ/kg K
n_mech=0.9;#mechanical efficiency
T=300.*10**-3;#brake torque in N
C=42.*10**3;#calorific value in KJ/kg
Cw=4.18;#specific heat of water in KJ/kg K
BP=2.*math.pi*N*T/60
print("brake power(BP)=2*%pi*N*T in KW"),round(BP,2)
print("so brake power=47.124 KW")
bsfc=m*60/(mw*BP)
print("brake specific fuel consumption(bsfc) in kg/KW hr="),round(bsfc,2)
IP=BP/n_mech
print("indicated power(IP) in Kw="),round(IP,2)
n_ite=IP*mw*60/(m*C)
print("indicated  thermal efficiency(n_ite)="),round(n_ite,2)
print("in percentage"),round(n_ite*100,2)
print("so indicated thermal efficiency=28.05%")
Qf=(m/mw)*C
print("heat available from fuel(Qf)=(m/mw)*C in KJ/min"),round(Qf,2)
BP=BP*60
print("energy consumed as brake power(BP) in KJ/min="),round(BP,2)
Qw=mw*Cw*(T2-T1)
print("energy carried by cooling water(Qw) in KJ/min="),round(Qw,2)
Qg=(ma+m)*Cg*(T_exhaust-T_atm)/mw
print("energy carried by exhaust gases(Qg) in KJ/min="),round(Qg,2)
Qf-(BP+Qw+Qg)
print("unaccounted energy loss in KJ/min"),round(Qf-(BP+Qw+Qg),2)
print("NOTE=>Heat balance sheet on per minute basis is attached as jpg file with this code.")

Example 10.10, Page:394

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 10
brake power(BP)=2*%pi*N*T in KW 47.12
so brake power=47.124 KW
brake specific fuel consumption(bsfc) in kg/KW hr= 0.34
indicated power(IP) in Kw= 52.36
indicated  thermal efficiency(n_ite)= 0.28
in percentage 28.05
so indicated thermal efficiency=28.05%
heat available from fuel(Qf)=(m/mw)*C in KJ/min 11200.0
energy consumed as brake power(BP) in KJ/min= 2827.43
energy carried by cooling water(Qw) in KJ/min= 1442.1
energy carried by exhaust gases(Qg) in KJ/min= 4786.83
unaccounted energy loss in KJ/min 2143.63
NOTE=>Heat balance sheet on per minute basis is attached as jpg file with this code.


## example 10.11;pg no: 395¶

In :
#cal of indicated power and mechanical efficiency
#intiation of all variables
# Chapter 10
import math
print"Example 10.11, Page:395  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 11")
BP=50.;#brake power output at full load in KW
BP1=40.1;#brake power output of 1st cylinder in KW
BP2=39.5;#brake power output of 2nd cylinder in KW
BP3=39.1;#brake power output of 3rd cylinder in KW
BP4=39.6;#brake power output of 4th cylinder in KW
BP5=39.8;#brake power output of 5th cylinder in KW
BP6=40.;#brake power output of 6th cylinder in KW
print("indicated power of 1st cylinder=BP-BP1 in KW")
BP-BP1
print("indicated power of 2nd cylinder=BP-BP2 in KW")
BP-BP2
print("indicated power of 3rd cylinder=BP-BP3 in KW")
BP-BP3
print("indicated power of 4th cylinder=BP-BP4 in KW")
BP-BP4
print("indicated power of 5th cylinder=BP-BP5 in KW")
BP-BP5
print("indicated power of 6th cylinder=BP-BP6 in KW")
BP-BP6
IP=9.9+10.5+10.9+10.4+10.2+10
print(" total indicated power(IP)in KW="),round(IP,2)
n_mech=BP/IP
print("mechanical efficiency(n_mech)="),round(n_mech,2)
print("in percentage"),round(n_mech*100,2)
print("so indicated power=61.9 KW")
print("mechanical efficiency=80.77%")

Example 10.11, Page:395

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 11
indicated power of 1st cylinder=BP-BP1 in KW
indicated power of 2nd cylinder=BP-BP2 in KW
indicated power of 3rd cylinder=BP-BP3 in KW
indicated power of 4th cylinder=BP-BP4 in KW
indicated power of 5th cylinder=BP-BP5 in KW
indicated power of 6th cylinder=BP-BP6 in KW
total indicated power(IP)in KW= 61.9
mechanical efficiency(n_mech)= 0.81
in percentage 80.78
so indicated power=61.9 KW
mechanical efficiency=80.77%


## example 10.12;pg no: 396¶

In :
#cal of brake power,indicated power,heat balance sheet
#intiation of all variables
# Chapter 9
import math
print"Example 10.12, Page:396  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 12")
m=0.189;#fuel consumption rate in kg/min
C=43.*10**3;#fuel calorific value in KJ/kg
k=12.;#air to fuel ratio
T_exhaust=600.;#exhaust gas temperature in degree celcius
mw=18.;#cooling water flow rate in kg/min
T1=27.;#cooling water entering temperature in degree celcius
T2=50.;#cooling water leaving temperature in degree celcius
T_atm=27.;#atmospheric air temperature
Cg=1.02;#specific heat of exhaust gas in KJ/kg K
Cw=4.18;#specific heat of water in KJ/kg K
BP=2.*math.pi*N*F*r*10**-3/60.
print("brake power output of engine(BP) in KW="),round(BP,2)
BP1=2.*math.pi*N*F1*r*10**-3/60.
print("brake power when cylinder 1 is cut(BP1) in KW="),round(BP1,2)
IP1=BP-BP1
print("so indicated power of first cylinder(IP1) in KW="),round(IP1,2)
BP2=2.*math.pi*N*F2*r*10**-3/60.
print("brake power when cylinder 2 is cut(BP2) in KW="),round(BP2,2)
IP2=BP-BP2
print("so indicated power of second cylinder(IP2) in KW="),round(IP2,2)
BP3=2.*math.pi*N*F3*r*10**-3/60.
print("brake power when cylinder 3 is cut(BP3) in KW="),round(BP3,2)
IP3=BP-BP3
print("so indicated power of third cylinder(IP3) in KW="),round(IP3,2)
BP4=2.*math.pi*N*F4*r*10**-3/60.
print("brake power when cylinder 4 is cut(BP4) in KW="),round(BP4,2)
IP4=BP-BP4
print("so indicated power of fourth cylinder(IP4) in KW="),round(IP4,2)
IP=IP1+IP2+IP3+IP4
print("now total indicated power(IP) in KW"),round(IP,2)
n_mech=BP/IP
print("engine mechanical efficiency(n_mech)="),round(n_mech,2)
print("in percentage"),round(n_mech*100,2)
print("so BP=19.63 KW,IP=23 KW,n_mech=83.35%")
Qf=m*C
print("heat liberated by fuel(Qf)=m*C in KJ/min"),round(Qf,2)
Qg=(k+1)*m*Cg*(T_exhaust-T_atm)
print("heat carried by exhaust gases(Qg) in KJ/min="),round(Qg,2)
Qw=mw*Cw*(T2-T1)
print("heat carried by cooling water(Qw) in KJ/min="),round(Qw,2)
BP=19.63*60
print("energy to brake power(BP) in KJ/min="),round(19.63*60,2)
Qf-(Qg+Qw+BP)
print("unaccounted losses in KJ/min"),round(Qf-(Qg+Qw+BP),2)
print("NOTE=>Heat balance sheet on per minute basis is attached as jpg file with this code. ")

Example 10.12, Page:396

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 12
brake power output of engine(BP) in KW= 19.63
brake power when cylinder 1 is cut(BP1) in KW= 13.74
so indicated power of first cylinder(IP1) in KW= 5.89
brake power when cylinder 2 is cut(BP2) in KW= 14.14
so indicated power of second cylinder(IP2) in KW= 5.5
brake power when cylinder 3 is cut(BP3) in KW= 14.29
so indicated power of third cylinder(IP3) in KW= 5.34
brake power when cylinder 4 is cut(BP4) in KW= 13.35
so indicated power of fourth cylinder(IP4) in KW= 6.28
now total indicated power(IP) in KW 23.01
engine mechanical efficiency(n_mech)= 0.85
in percentage 85.32
so BP=19.63 KW,IP=23 KW,n_mech=83.35%
heat liberated by fuel(Qf)=m*C in KJ/min 8127.0
heat carried by exhaust gases(Qg) in KJ/min= 1436.02
heat carried by cooling water(Qw) in KJ/min= 1730.52
energy to brake power(BP) in KJ/min= 1177.8
unaccounted losses in KJ/min 3782.66
NOTE=>Heat balance sheet on per minute basis is attached as jpg file with this code.


## example 10.13;pg no: 397¶

In :
#cal of thermal efficiency and heat balance sheet
#intiation of all variables
# Chapter 10
import math
print"Example 10.13, Page:397  \n \n"
print("Engineering Thermodynamics by Onkar Singh Chapter 10 Example 13")
D=20.*10**-2;#cylinder diameter in m
L=28.*10**-2;#stroke in m
m=4.22;#mass of fuel used in kg
C=44670.;#calorific value of fuel in KJ/kg
N=21000./60.;#engine rpm
mep=2.74*10**5;#mean effective pressure in pa
F=600.;#net brake load applied to a drum of 100 cm diameter in N
mw=495.;#total mass of cooling water in kg
T1=13.;#cooling water inlet temperature in degree celcius
T2=38.;#cooling water outlet temperature in degree celcius
ma=135.;#mass of air used in kg
T_air=20.;#temperature of air in test room in degree celcius
T_exhaust=370.;#temperature of exhaust gases in degree celcius
Cp_gases=1.005;#specific heat of gases in KJ/kg K
Cp_steam=2.093;#specific heat of steam at atmospheric pressure in KJ/kg K
Cpw=4.18;#specific heat of water in KJ/kg K
print("brake power(BP)=2*%pi*N*T in KW")
BP=2*math.pi*N*F*r/60000
print("indicated power(IP)=(mep*L*A*N)/60000 in KW")
IP=(mep*L*(math.pi*D**2/4)*N)/60000
Q=m*C/3600
print("or Q in KJ/min")
Q=Q*60
n_th=IP/Q
print("thermal efficiency(n_th)= "),round(n_th,2)
print("in percentage"),round(n_th*100,2)
BP=BP*60
print("B> heat equivalent of brake power(BP) in KJ/min= "),round(10.996*60,2)
print("C> heat loss to cooling water(Qw)=mw*Cpw*(T2-T1) in KJ/min")
Qw=mw*Cpw*(T2-T1)/60
print("heat carried by exhaust gases=heat carried by steam in exhaust gases+heat carried by fuel gases(dry gases) in exhaust gases")
print("mass of exhaust gases(mg)=mass of air+mass of fuel in kg/min")
print("mg=(ma+m)/60")
mg=(ma+m)/60
print("mass of steam in exhaust gases in kg/min")
9*(0.15*m/60)
print("mass of dry exhaust gases in kg/min")
mg-0.095
0.095*(Cpw*(100-T_air)+2256.9+Cp_steam*(T_exhaust-100))
print("D> heat carried by steam in exhaust in KJ/min"),round(0.095*(Cpw*(100-T_air)+2256.9+Cp_steam*(T_exhaust-100)),2)
Qg=2.225*Cp_gases*(T_exhaust-T_air)
print("E> heat carried by fuel gases(dry gases) in exhaust gases(Qg) in KJ/min"),round(Qg,2)
print("F> unaccounted loss=A-B-C-D-E in KJ/min"),round(3141.79-659.76-862.13-299.86-782.64,2)
print("NOTE># on per minute basis is attached as jpg file with this code.")

Example 10.13, Page:397

Engineering Thermodynamics by Onkar Singh Chapter 10 Example 13
brake power(BP)=2*%pi*N*T in KW
indicated power(IP)=(mep*L*A*N)/60000 in KW
A> heat added(Q)=m*C/3600 in KJ/s 52.36
or Q in KJ/min
thermal efficiency(n_th)=  0.27
in percentage 26.85
B> heat equivalent of brake power(BP) in KJ/min=  659.76
C> heat loss to cooling water(Qw)=mw*Cpw*(T2-T1) in KJ/min
heat carried by exhaust gases=heat carried by steam in exhaust gases+heat carried by fuel gases(dry gases) in exhaust gases
mass of exhaust gases(mg)=mass of air+mass of fuel in kg/min
mg=(ma+m)/60
mass of steam in exhaust gases in kg/min
mass of dry exhaust gases in kg/min
D> heat carried by steam in exhaust in KJ/min 299.86
E> heat carried by fuel gases(dry gases) in exhaust gases(Qg) in KJ/min 782.64
F> unaccounted loss=A-B-C-D-E in KJ/min 537.4
NOTE># on per minute basis is attached as jpg file with this code.