#ques1
#theoratical air-fuel ratio for combustion of octane
#combustion equation is
#C8H18 + 12.5O2 + 12.5(3.76) N2 → 8 CO2 + 9H2O + 47.0N2
rm=(12.5+47.0)/1;#air fuel ratio on mole basis
rma=rm*28.97/114.2;#air fuel ratio on mass basis;
print "Theoratical air fuel ratio on mass basis is",round(rma),"kg air/kg fuel"
# ques 2
# molal analysis of the products of combustion
# combustion equation is
# C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 → 8 CO2 + 9H2O + 12.5O2 + 94.0N2
nCO2=8 # number of moles of CO2
nH2O=9 # number of moles of H2O
nO2=12.5 # number of moles of O2
nN2=94.0 # number of moles of N2
moles=nCO2+nH2O+nO2+nN2 # total moles
# molal analysis of products
xCO2=nCO2*100/moles
xH2O=nH2O*100/moles
xO2=nO2*100/moles
xN2=nN2*100/moles
print " CO2 =",round(xCO2,2)
print "\n H2O=",round(xH2O,2)
print "\n O2=",round(xO2,2)
print "\n N2=",round(xN2,2)
print "\n The dew point temp. corresponding to partial pressure of water i.e ",round(xH2O,2),"KPa is 39.7 degree C"
# ques 2
# molal analysis of the products of combustion
# combustion equation is
# C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 → 8 CO2 + 9H2O + 12.5O2 + 94.0N2
pressure=14.7 # in lbf/in^2
nCO2=8 # number of moles of CO2
nH2O=9 # number of moles of H2O
nO2=12.5 # number of moles of O2
nN2=94.0 # number of moles of N2
moles=nCO2+nH2O+nO2+nN2 # total moles
# molal analysis of products
xCO2=nCO2*100/moles
xH2O=nH2O*100/moles
xO2=nO2*100/moles
xN2=nN2*100/moles
print " CO2 =",round(xCO2,2)
print "\n H2O=",round(xH2O,2)
print "\n O2=",round(xO2,2)
print "\n N2=",round(xN2,2)
PressureO2=pressure*xH2O/100 # partial pressure of O2
print "\n The dew point temp. corresponding to partial pressure of water i.e ",round(PressureO2,3),"lbf/in^2 is 104 F degree C"
# ques 3
# to determine the air–fuel ratio on a volumetric basis
# the combustion equation is
# 0.14H2+0.27CO+0.03CH4+0.006O2+0.509N2+0.045CO2+0.259O2+0.259(3.76)N2→0.20H2O+0.345CO2+1.482 N2
mH2=2 # molar mass of H2
mCO=28 # molar mass of CO
mCH4=16 # molar mass of CH4
mO2=32 # molar mass of O2
mN2=28 # molar mass of N2
mCO2=44 # molar mass of CO2
kmolair=0.259 # moles of air in mixture using combustion equation
kmolfuel=0.21 # moles of fuel in mixture using combustion equation
ratioth=kmolair/kmolfuel # theoratical ratio
# For 20% excess air
vratio=ratioth*(1+0.2) # new ratio for excess air on volume basis
print " The A/F ratio on volumetric basis is ",round(vratio,2)
mratio=vratio*28.97 /(0.14*mH2+0.27*mCO+0.03*mCH4+0.006*mO2+0.509*mN2+0.045*mCO2)
print "\n The A/F ratio on mass basis is ",round(mratio,2),"kg air/kg fuel"
# ques 4
# determining the air–fuel ratio and the percent theoretical air
CO2=10.00 # percentage
O2=2.37 # percentage
CO=0.53 # percentage
N2=87.10 # percentage
mCH4=16 # molecular mass of CH4
# using this data the combustion equation is
# a CH4 + b O2 + c N2 → 10.0 CO2 + 0.53 CO + 2.37 O2 + d H2O + 87.1N2
# on balancing we get
# 10.53 CH4 + 23.16 O2 + 87.1N2 →10.0 CO2 + 0.53 CO + 2.37 O2 + 21.06 H2O + 87.1N2
# balanced combustion equation for per Kmole of fuel is:
# CH4 + 2.2O2 + 8.27 N2 → 0.95 CO2 + 0.05 CO + 2H2O + 0.225 O2 + 8.27 N2
molesAir=2.2+8.27 # moles of air
AF= 28.97*molesAir/mCH4 # air fuel ratio on mass basis
print "\n The A/F ratio on mass basis is ",round(AF,2),"kg air/kg fuel"
# Theoritical combustion equation:
# CH4 + 2O2 + 2(3.76)N2 → CO2 + 2H2O + 7.52 N2
molesAirth=2+2*3.76 # moles of air(theoritical)
AFth= 28.97*molesAirth/mCH4 # air fuel ratio on mass basis (theoritical)
print "\n The A/F ratio (theoritical) on mass basis is ",round(AFth,2),"kg air/kg fuel"
print "\n The percent theoretical air is",round(AF*100/AFth,1)
# ques 5
Sulfur=0.6 # percent by mass
Hydrogen=5.7 # percent by mass
Carbon=79.2 # percent by mass
Oxygen=10.0 # percent by mass
Nitrogen=1.5 # percent by mass
Ash=3.0 # percent by mass
mS=32.0 # molecular mass
mH2=2.0 # molecular mass
mC=12.0 # molecular mass
mO2=32.0 # molecular mass
mN2=28.0 # molecular mass
# molal composition per 100 Kg of fuel:
S=Sulfur/mS # molal composition
H=Hydrogen/mH2 # molal composition
C=Carbon/mC # molal composition
O2=Oxygen/mO2 # molal composition
N2=Nitrogen/mN2 # molal composition
#The combustion equation becomes
# 0.02 S + 0.02 O2 →0.02 SO2
# 2.85 H2 + 1.42 O2 →2.85 H2O
# 6.60 C + 6.60 O2 →6.60 CO2
O2req=0.02+1.42+6.60 # kmol O2 required/100 kg fuel
O2present=O2req-O2 # kmol O2 from air/100 kg fuel
AFtheo= (O2present+O2present*3.76)*28.97/100 # AF ration theoritical kg air/kg fuel
print "\n The A/F ratio (theoritical) on mass basis is ",round(AFtheo,2),"kg air/kg fuel"
print "\n The A/F ratio for 30% excess air on mass basis is ",round(AFtheo*1.3,2),"kg air/kg fuel"
#ques6
# determining heat transfer per kilomole of fuel entering combustion chamber
#1-CH4
#2-CO2
#3-H2O
#hf-standard enthalpy of given substance
hf1=-74.873;#kJ
hf2=-393.522;#kJ
hf3=-285.830;#kJ
Qcv=hf2+2*hf3-hf1;#kJ
print "Heat transfer per kilomole of fuel entering combustion chamber is",round(Qcv,3),"kJ"
#the answers in the book is different as they have not printed the decimals in values
#ques7
#calculating enthalpy of water at given pressure and temperature
#1.Assuming steam to be an ideal gas with value of Cp
T1=298.15;#Initial temperature in K
T2=573.15;#final temperature in K
T=(T1+T2)/2;#average temperature in K
Cp=1.79+0.107*T/1000+0.586*(T/1000)**2-.20*(T/1000)**3;#specific heat at constant pressure in kj/kg.K
M=18.015;#mass in kg
dh=M*Cp*(T2-T1);#enthalpy change in kJ/kmol
ho=-241.826;#enthalpy at standard temperature and pressure in kJ/mol
htp1=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol
print " 1. Enthalpy of water at given pressure and temperature using value of Cp =",round(htp1,3),"kJ/kmol"
#2..Assuming steam to be an ideal gas with value for dh
dh=9359;#enthalpy change from table A.9 in kJ/mol
htp2=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol
print " 2. Enthalpy of water at given pressure and temperature assuming value od dh =",round(htp2,3),"kJ/kmol "
#3. Using steam table
dh=M*(2977.5-2547.2);#enthalpy change for gases in kJ/mol
htp3g=dh/1000+ho;
dh=M*(2977.5-104.9);#enthalpy change for liquid in kJ/mol
hl=-285.830;#standard enthalpy for liquid in kJ/kmol
htp31=hl+dh/1000.0;#enthalpy at given temp and pressure in kJ/kmol
print " 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =",round(htp31,3),"kJ/kmol "
print " 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =",round(htp3g,3),"kJ/kmol "
#4.using generalised charts
#htp=ho-(h2*-h2)+(h2*-h1*)+(h1*-h1);
#h2*-h2=Z*R*Tc,
#h2*-h1*=9539 kJ/mol, from part 2
#h1*-h1=0 ,as ideal gas
Z=0.21;#from chart
R=8.3145;#gas constant in SI units
Tc=647.3;#critical temperature in K
htp4=ho+9539/1000-Z*R*Tc/1000;#enthalpy at given temp and pressure in kJ/kmol
print " 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart = ",round(htp4,3),"kJ/kmol"
#the answers in book are different as they have not printed the decimals in values
# ques 8
# Determine the heat transfer
airth=400 # % excess theoritical air
Ti=25 # Entry temp in degree celsius
To=900 # Exit temp in Kelvin
mf=0.25 # specific fuel consumption in kg/s of fuel per megawatt output
# The combustion equation is:
# C8H18(l) + 4(12.5)O2 + 4(12.5)(3.76)N2 → 8CO2 + 9H2O + 37.5O2 + 188.0N2
hfC8H18=-250105 # is the enthalpy of fuel in kJ/kmol
nCO2=8 # moles
hfCO2=-393522 # enthalpy in kJ/kmol
DelhCO2=28030 # enthalpy in kJ/kmol
mC8H18=114.23 # molecular mass
nH2O=9 # moles
hfH2O=-241826 # enthalpy in kJ/kmol
DelhH2O=21937 # enthalpy in kJ/kmol
DelhO2=19241 # enthalpy in kJ/kmol
nO2=37.5 # moles
DelhN2=18225 # enthalpy in kJ/kmol
nN2=188.0 # moles
Eproduct=nCO2*(hfCO2+DelhCO2)+nH2O*(hfH2O+DelhH2O)+nO2*DelhO2+nN2*DelhN2
Wcv=1000*mC8H18/mf # work in kJ/kmol fuel
print " The work is ",Wcv,"kJ/kmol fuel"
Qcv=Eproduct+Wcv-hfC8H18 # using first law
print " The Heat transfer is ",Qcv,"kJ/kmol fuel"
# the answer is slightly different in textbook due to approximation while here the answers are precise
# ques 8
# Determine the heat transfer
airth=400 # % excess theoritical air
Ti=77 # Entry temp in F
To=1100 # Exit temp in F
mf=1 # specific fuel consumption in lb of fuel per horsepower-hour
# The combustion equation is:
# C8H18(l) + 4(12.5)O2 + 4(12.5)(3.76)N2 → 8CO2 + 9H2O + 37.5O2 + 188.0N2
hfC8H18=-107526 # is the enthalpy of fuel in Btu/lb.mol
nCO2=8 # moles
hfCO2=-169184 # enthalpy in Btu/lb.mol
DelhCO2=11391 # enthalpy in Btu/lb.mol
mC8H18=114.23 # molecular mass
nH2O=9 # moles
hfH2O=-103966 # enthalpy in Btu/lb.mol
DelhH2O=8867 # enthalpy in Btu/lb.mol
DelhO2=7784 # enthalpy in Btu/lb.mol
nO2=37.5 # moles
DelhN2=7374 # enthalpy in Btu/lb.mol
nN2=188.0 # moles
Eproduct=nCO2*(hfCO2+DelhCO2)+nH2O*(hfH2O+DelhH2O)+nO2*DelhO2+nN2*DelhN2
Wcv=2544*mC8H18 # work in kJ/kmol fuel
print " The work is ",Wcv,"Btu/lb.mol fuel"
Qcv=Eproduct+Wcv-hfC8H18 # using first law
print " The Heat transfer is ",Qcv,"Btu/lb.mol fuel"
# the answer is slightly different in textbook due to approximation while here the answers are precise
#ques 9
# The combustion equation is
# C2H4 + 3O2 → 2 CO2 + 2H2O(g)
hfC2H4=52467 # enthalpy in KJ/mol
R=8.314 # gas constant
T=298.2 # temperature in Kelvin
hfCO2=-393522 # enthalpy in kJ/kmol
DelhCO2=12906 # enthalpy in kJ/kmol
hfH2O=-241826 # enthalpy in kJ/kmol
DelhH2O=10499 # enthalpy in kJ/kmol
Ureactant=hfC2H4-4*R*T # energy of reactant in KJ
Uproduct=2*(hfCO2+DelhCO2)+2*(hfH2O+DelhH2O) # energy of product in KJ
Q=Uproduct-Ureactant # using first law
print "The Heat transfer is ",Q,"kJ"
# the answer is slightly different in textbook due to approximation while here the answers are precise
#ques 10
# The combustion equation is
# C3H8 + 5O2 → 3 CO2 + 4H2O
hfC3H8g=-103900 # enthalpy of C3H8 in KJ/mol from Table A.10
hfC3H8l=hfC3H8g-44.097*370 # enthalpy of C3H8 liquid as enthalpy of evaporation is given in question
hfH2Ol=285830 # enthalpy of water in KJ/mol
hfH2Og=241826 # enthalpy in KJ/mol
hfCO2=393522 # enthalpy in KJ/mol
nO2=5 # moles
nCO2=3 # moles
nH2O=4 # moles
nC3H8=1 # moles
# part(1) : liquid propane-liquid water
hrP=(nCO2*hfCO2 + nH2O*hfH2Ol -nC3H8*hfC3H8l)/44.097 # higher heating value of liquid propane
print "The higher heating value of liquid propane is",round(hrP,2),"KJ/kg"
# part(2) : Liquid propane–gaseous water
hrP=(nCO2*hfCO2 + nH2O*hfH2Og -nC3H8*hfC3H8l)/44.097 # lower heating value of liquid propane
print "The lower heating value of liquid propane is",round(hrP,2),"KJ/kg"
# part(3) :Gaseous propane–liquid water
hrP=(nCO2*hfCO2 + nH2O*hfH2Ol -nC3H8*hfC3H8g)/44.097 # higher heating value of gaseous propane
print "The higher heating value of gaseous propane is",round(hrP,2),"KJ/kg"
# part(4) :Gaseous propane–Gaseous water
hrP=(nCO2*hfCO2 + nH2O*hfH2Og -nC3H8*hfC3H8g)/44.097 # lower heating value of gaseous propane
print "The lower heating value of gaseous propane is",round(hrP,2),"KJ/kg"
# the answer is slightly different in textbook due to approximation while here the answers are precise
#ques 10
# The combustion equation is
# C3H8(g) + 5O2 → 3 CO2 + 4H2O(g)
hfC3H8=-103900 # enthalpy of C3H8 in KJ/mol from Table A.10
Cp=2.1 # in KJ/Kg.K
T1=500 # temp in Kelvin
T2=298.2 # temp in Kelvin
DelhO2=6086 # enthalpy of O2 in KJ/mol
hfCO2=-393522 # enthalpy in KJ/mol
DelhCO2=8305 # enthalpy of CO2 in KJ/mol
hfH2O=-241826 # enthalpy of H2O in KJ/mol
DelhH2O=6922 # enthalpy in KJ/mol
nO2=5 # moles
nCO2=3 # moles
nH2O=4 # moles
nC3H8=1 # moles
hR500= hfC3H8+Cp*44.097*(T1-T2)+nO2*DelhO2 # in KJ/mol
hP500= nCO2*(hfCO2+DelhCO2)+ nH2O*(DelhH2O+hfH2O) # in KJ/mol
hRP500=hP500-hR500 # in KJ/mol
Hrp500=hRP500/44.097 # KJ/Kg
print " The enthalpy of combustion of gaseous propane at 500 K ",round(Hrp500)
# ques 12
# The question is solved by Trial and Error and using the table , no need for Calculation Thus not solved in Python
# ques 13
# The equation for this chemical reaction is
# C2H4(g) + 3(4) O2 + 3(4)(3.76) N2 → 2 CO2+2H2O(g) + 9O2 + 45.1 N2
T=25+273.15 # temp in degree kelvin
P=0.1 # pressure in Mpa
hfCO2=-393522 # enthalpy in KJ/mol
hfH2O=-241826 # enthalpy of H2O in KJ/mol
HfC2H4=52467 # enthalpy of C2H4 in KJ/mol
hfO2=0 # enthalpy of formation in KJ/mol
SCO2=213.795 # kJ/kmol fuel
SH2O=188.243 # kJ/kmol fuel
SC2H4=219.330 # kJ/kmol fuel
SO2=205.148 # kJ/kmol fuel
DelH=2*hfCO2+2*hfH2O-HfC2H4-3*hfO2 # in KJ/mol
DelS=2*SCO2+2*SH2O-SC2H4-3*SO2 # kJ/kmol fuel
DelG=DelH-T*(DelS) # Gibbs Energy
Wrev=-DelG/28.054 # reversible work in kJ/kg
print "The reversible work is ",round(Wrev,2),"kJ/kg "
# the answer is slightly different in textbook due to approximation while here the answers are precise
# ques
# The question is solved by Trial and Error and using the table , no need for Calculation Thus not solved in Python
#ques15
#calculatng reversible elecromotive force
#1-H2O
#2-H2
#3-O2
#hf-standard enthalpy
#sf-standard entropy
hf1=-285.830;#kJ
hf2=0;#kJ
hf3=0;#kJ
sf1=69.950;#kJ/K
sf2=130.678;#kJ/K
sf3=205.148;#kJ/K
dH=2*hf1-2*hf2-hf3;#change in enthalpy in kJ
dS=2*sf1-2*sf2-sf3;#change in entropy in kJ/K
T=298.15;#temperature in K
dG=dH-T*dS/1000;#change in gibbs free energy in kJ
E=-dG*1000/(96485*4);#emf in V
print" Reversible electromotive Force =",round(E,3)," V"
# the answer is slightly different in textbook due to approximation while here the answers are precise
# ques 16
# to Calculate the combustion efficiency
Tair = 400 # temp in K
Vair = 100 # velocity in m/s
Tfuel = 50 # temp of fuel in ◦C
Tproducts = 1100 # temp of product in K
Vproducts = 150 # velocity of product in m/s
FAactual = 0.0211 # air fuel ratio kg fuel/kg air
# HR + KER= −243737 + 14892nO2
# HP + KEP= −5 068 599 + 120 853nO2
# using these two equations :
no2=(-5068599.0 + 243737.0)/(14892.0-120853) # in kmol O2/kmol fuel
FAideal=114.23/(4.76*no2*28.91) # in kg fuel/kg air
efficomb=FAideal*100/FAactual # Efficiency
print " The combustion efficiency is ",round(efficomb,2),"%"
# the answer is slightly different in textbook due to approximation while here the answers are precise
#ques17
#efficiency of generator and plant
q=325000*(3398.3-856.0);#heat transferred to H2O/kg fuel in kJ/kg
qv=26700.0*33250;#higher heating value in kJ/kg
nst=q/qv*100;#efficiency of steam generator
w=81000.0*3600;#net work done in kJ/kg
nth=w/qv*100.0;#thermal efficiency
print" Efficiency of generator =",round(nst,1),"percent\n"
print" Thermal Efficiency =",round(nth,1)," percent"
# the answer is slightly different in textbook due to approximation while here the answers are precise