Express the total mass of the coal when the ash is substituted.
mtotal=100−mash (I)
Here, mass of ash is mash.
Express the mass fraction of carbon.
mfC=mCmtotal (II)
Here, mass of carbon is mC.
Express the mass fraction of hydrogen.
mfH2=mH2mtotal (III)
Here, mass of hydrogen is mH2.
Express the mass fraction of oxygen.
mfO2=mO2mtotal (IV)
Here, mass of oxygen is mO2.
Express the mass fraction of nitrogen.
mfN2=mN2mtotal (V)
Here, mass of nitrogen is mN2.
Express the mass fraction of sulphur.
mfS=mSmtotal (VI)
Here, mass of sulphur is mS.
Express the number of moles of carbon.
NC=mfCMC (VII)
Here, molar mass of carbon is MC.
Express the number of moles of hydrogen.
NH2=mfH2MH2 (VIII)
Here, molar mass of hydrogen is MH2.
Express the number of moles of oxygen.
NO2=mfO2MO2 (IX)
Here, molar mass of oxygen is MO2.
Express the number of moles of nitrogen.
NN2=mfN2MN2 (X)
Here, molar mass of nitrogen is MN2.
Express the number of moles of sulphur.
NS=mfSMS (XI)
Here, molar mass of sulphur is MS.
Express the total number of moles.
Nm=NC+NH2+NO2+NN2+NS (XII)
Express the mole fraction of carbon.
yC=NCNm (XIII)
Express the mole fraction of hydrogen.
yH2=NH2Nm (XIV)
Express the mole fraction of oxygen.
yO2=NO2Nm (XV)
Express the mole fraction of nitrogen.
yN2=NN2Nm (XVI)
Express the mole fraction of sulphur.
yS=NSNm (XVII)
Express the heat transfer from the combustion chamber.
q=hC=HP−HR=∑NPh¯f,Po−∑NRh¯f,Ro=(Nh¯fo)CO2+(Nh¯fo)H2O+(Nh¯fo)SO2 (XVIII)
Here, number of moles of products is NP, number of moles of reactants is NR, enthalpy of formation is h¯f, enthalpy of combustion is hC, enthalpy of products and reactants is HP and HR respectively.
Express apparent molecular weight of the coal.
Mm=mmNm=NCMC+NH2MH2+NO2MO2+NN2MN2+NSMSNC+NH2+NO2+NN2+NS (XIX)
Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is NC,NH2,NO2,NN2 and NS respectively.
Express higher value of coal.
HHV=−hCmm=−hCNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS (XX)
Express lower value of coal.
LHV=−hCmm=−hCNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS (XXI)
Conclusion:
Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.
MC=12 kg/kmolMH2=2 kg/kmolMO2=32 kg/kmolMS=32 kg/kmol
Mair=29 kg/kmolMN2=28 kg/kmol
Here, molar mass of air is Mair.
Substitute 5 for mash in Equation (I).
mtotal=100−5=95 kg
Substitute 61.40 kg for mC and 95 kg for mtotal in Equation (II).
mfC=61.40 kg95 kg=0.6463=64.63 kg
Substitute 5.79 kg for mH2 and 95 kg for mtotal in Equation (III).
mfH2=5.79 kg95 kg=0.06095=6.095 kg
Substitute 25.31 kg for mO2 and 95 kg for mtotal in Equation (IV).
mfO2=25.31 kg95 kg=0.2664=26.64 kg
Substitute 1.09 kg for mN2 and 95 kg for mtotal in Equation (V).
mfN2=1.09 kg95 kg=0.01147=1.147 kg
Substitute 1.41 kg for mS and 95 kg for mtotal in Equation (VI).
mfS=1.41 kg95 kg=0.01484=1.484 kg
Substitute 64.63 kg for mC and 12 kg/kmol for MC in Equation (VII).
NC=64.63 kg12 kg/kmol=5.386 kmol
Substitute 6.095 kg for mH2 and 2 kg/kmol for MH2 in Equation (VIII).
NH2=6.095 kg2 kg/kmol=3.048 kmol
Substitute 26.64 kg for mO2 and 32 kg/kmol for MO2 in Equation (IX).
NO2=26.64 kg32 kg/kmol=0.8325 kmol
Substitute 1.147 kg for mN2 and 28 kg/kmol for MN2 in Equation (X).
NN2=1.147 kg28 kg/kmol=0.04096 kmol
Substitute 1.484 kg for mS and 32 kg/kmol for MS in Equation (XI).
NS=1.484 kg32 kg/kmol=0.04638 kmol
Substitute 5.386 kmol for NC, 3.048 kmol for NH2, 0.8325 kmol for NO2, 0.04096 kmol for NN2 and 0.04638 kmol for NS in Equation (XII).
Nm=5.386 kmol+3.048 kmol+0.8325 kmol+0.04096 kmol+0.04638 kmol=9.354 kmol
Substitute 5.386 kmol for NC and 9.354 kmol for Nm in Equation (XIII).
yC=5.386 kmol9.354 kmol=0.5758
Substitute 3.048 kmol for NH2 and 9.354 kmol for Nm in Equation (XIV).
yH2=3.048 kmol9.354 kmol=0.3258
Substitute 0.8325 kmol for NO2 and 9.354 kmol for Nm in Equation (XV).
yO2=0.8325 kmol9.354 kmol=0.0890
Substitute 0.04096 kmol for NN2 and 9.354 kmol for Nm in Equation (XVI).
yN2=0.04096 kmol9.354 kmol=0.00438
Substitute 0.04638 kmol for NS and 9.354 kmol for Nm in Equation (XVII).
yS=0.04638 kmol9.354 kmol=0.00496
Express the combustion equation.
[0.5758C+0.3258H2+0.0890O2+0.00438N2+0.00496S+ath(O2+3.76N2)→0.5758CO2+0.3258H2O+0.00496SO2+kN2] (XXII)
Perform the species balancing:
Oxygen balance:
0.0890+ath=0.5758+0.5(0.3258)+0.00496ath=0.6547
Nitrogen balance:
k=0.00438+3.76(0.6547)=2.466
Substitute 0.6547 for ath and 2.466 for k in Equation (XXII).
[0.5758C+0.3258H2+0.0890O2+0.00438N2+0.00496S+0.6547(O2+3.76N2)→{0.5758CO2+0.3258H2O+0.00496SO2+2.466N2}] (XXIII)
Refer Equation (XIII) and write the number of moles of carbon dioxide, water and sulfur oxide.
NCO2=0.5758 kmolNH2O=0.3258 kmolNSO2=0.00496 kmol
Refer Table A-26, “enthalpy of formation, Gibbs function of formation and entropy at 25°C,1 atm”, and write the enthalpy of formation of carbon dioxide, water and ethane.
h¯f,CO2o=−393,520 kJ/kmolh¯f,H2Oo(l)=−285,830 kJ/kmolh¯f,SO2o=−297,100 kJ/kmol
Substitute 0.5758 kmol for NCO2, −393,520 kJ/kmol for h¯f,CO2o, 0.3258 kmol for NH2O, −285,830 kJ/kmol for h¯f,H2Oo, 0.00496 kmol for NSO2 and −297,100 kJ/kmol for h¯f,SO2o in Equation (XVIII).
hC=[(0.5758 kmol)(−393,520 kJ/kmol)+(0.3258 kmol)(−285,830 kJ/kmol)+(0.00496 kmol)(−297,100 kJ/kmol)]=−321,200 kJ/kmol
Refer Equation (XXIII), and write the number of moles.
NC=0.5758 kmolNH2=0.3258 kmolNO2=0.0890 kmolNN2=0.00438 kmol
NS=0.00496 kmol
Substitute 0.5758 kmol for NC, 0.3258 kmol for NH2, 0.0890 kmol for NO2, 0.00438 kmol for NN2, 0.00496 kmol for NS, 12 kg/kmol for MC, 2 kg/kmol for MH2, 32 kg/kmol for MO2, 32 kg/kmol for MS and 28 kg/kmol for MN2 in Equation (XIX).
Mm=[(0.5758 kmol)(12 kg/kmol)+(0.3258 kmol)(2 kg/kmol)+(0.0890 kmol)(32 kg/kmol)+(0.00438 kmol)(28 kg/kmol)+(0.00496 kmol)(32 kg/kmol)]0.5758 kmol+0.3258 kmol+0.0890 kmol+0.00438 kmol+0.00496 kmol=10.69 kg1 kmol=10.69 kg/kmol
Substitute −321,200 kJ/kmol for hC, 0.5758 kmol for NC, 0.3258 kmol for NH2, 0.0890 kmol for NO2, 0.00438 kmol for NN2, 0.00496 kmol for NS, 12 kg/kmol for MC, 2 kg/kmol for MH2, 32 kg/kmol for MO2, 32 kg/kmol for MS and 28 kg/kmol for MN2 in Equation (XX).
HHV=−(−321,200 kJ/kmol)[(0.5758 kmol)(12 kg/kmol)+(0.3258 kmol)(2 kg/kmol)+(0.0890 kmol)(32 kg/kmol)+(0.00438 kmol)(28 kg/kmol)+(0.00496 kmol)(32 kg/kmol)]=321,200 kJ/kmol10.69 kg/kmol=30,000 kJ/kg
Hence, the higher heating value of a coal is 30,000 kJ/kg.
For the LHV (lower heating value), the water in the products is taken to be vapor.
Refer Table A-26, “enthalpy of formation, Gibbs function of formation and entropy at 25°C,1 atm”, and write the enthalpy of gaseous water.
h¯f,H2Oo(g)=−241,820 kJ/kmol
Substitute 0.5758 kmol for NCO2, −393,520 kJ/kmol for h¯f,CO2o, 0.3258 kmol for NH2O, −241,820 kJ/kmol for h¯f,H2Oo, 0.00496 kmol for NSO2 and −297,100 kJ/kmol for h¯f,SO2o in Equation (XVIII).
hC=[(0.5758 kmol)(−393,520 kJ/kmol)+(0.3258 kmol)(−241,820 kJ/kmol)+(0.00496 kmol)(−297,100 kJ/kmol)]=−306,850 kJ/kmol
Substitute −306,850 kJ/kmol for hC, 0.5758 kmol for NC, 0.3258 kmol for NH2, 0.0890 kmol for NO2, 0.00438 kmol for NN2, 0.00496 kmol for NS, 12 kg/kmol for MC, 2 kg/kmol for MH2, 32 kg/kmol for MO2, 32 kg/kmol for MS and 28 kg/kmol for MN2 in Equation (XXI).
LHV=−(−306,850 kJ/kmol)[(0.5758 kmol)(12 kg/kmol)+(0.3258 kmol)(2 kg/kmol)+(0.0890 kmol)(32 kg/kmol)+(0.00438 kmol)(28 kg/kmol)+(0.00496 kmol)(32 kg/kmol)]=306,850 kJ/kmol10.69 kg/kmol=28,700 kJ/kg
Hence, the lower heating value of a coal is 28,700 kJ/kg.