10.10 A fuel gas on a dry basis contains 80% methane, 6% ethane, 4% propane, 2%
ID: 702153 • Letter: 1
Question
10.10 A fuel gas on a dry basis contains 80% methane, 6% ethane, 4% propane, 2% oxygen and 896 nitrogen. The fuel saturated with water vapour at 300 K and 101.3 kPa and measuring 100 m3 is burned with 50% excess air supplied at 305 K and 101.3 kPa with a relative saturation of 60%. The vapour pressure of water is given by the Antoine equation: n PS 16.26205 3799.887-46.854 where P is in kPa and temperature is in K. Calculate the following: (a) The analysis of the gas leaving the combustion chamber (b) The volume of the gas (in m3) leaving the furnace at 550 K and 101.3 kPa assuming complete combustion (c) The dew point of the flue gas. (a) CO2 = 6.61%, 02 = 6.45%, N2 71.88%,H20-15.06% (b) 2784.93 m3 (c) 327.56 KExplanation / Answer
a) Applying Ideal Law, PV=nRT on fuel gas
=>101.3*10^-2*100*1000=n*.082*300
=> n=4117 moles
Using given Antoine equation to get Saturated Pressure of water at feed condition
=>ln(P)=16.26-3799/(300-46.85)
=>Psat=3.17 KPa
Total pressure = 101.3 Kpa
=> Mole fraction of water in fuel gas = 3.17/101.3=.031
=>Moles of water in fuel gas = 0.031*4117=127.6 Moles
Dry moles remaining of fuel gas = 4117-127.6=3989.4
=> Moles of Methane in Fuel gas = 80/100*3989.4=3191.52moles
=>Moles of Ethane in Fuel gas=6/100*3989.4=239.36 moles
=>Moles of Propane in Fuel gas=4/100*3989.4=159.57 moles
=>Moles of oxygen in Fuel gas=2/100*3989.4=79.78 moles
=>Moles of Nitrogen in Fuel gas=8/100*3989.4=319.15 moles
Combustion reactions:
Methane -> CH4+2*O2=>CO2+2H2O
Ethane -> C2H6+3.5*O2=>2CO2+3H2O
Propane -> C3H8+5*O2=>3CO2+4H2O
Because, for fully combustion, Methane requires two moles O2, Ethane requires 3.5 moles O2, Propane requires 5 moles of O2
Therefore, for full combustion, total O2 moles required = 2*3191.52+3.5*239.36+5*159 = 8018.65 moles
% excess air = 150%
=> Total moles of Oxygen in air = 8018.65*150/100=12027.95 moles
Air consists of 21% O2 and 79% N2
=>Total Moles of N2 in air = 12027.95/.21*.79=45248 moles
At 305 K, saturated Pressure of water = 4.68 Kpa
Relative saturation =60%
=> Vapor pressure of water = 60/100*4.68=2.8 Kpa
=>Mole fraction of water in air = 2.8/101.3=0.027
Let moles of water in air = n
=> n/(n+45248+12027.95)=.027
n,moles of water in air=1589.3 moles
Total Moles CO2 produce in Methane combustion = 1*3191.52 = 3191.52 moles
Total moles H2O Produce in Methane combustion = 2*3191.52=6383.04 moles
Total Moles CO2 produce in Ethane combustion = 2*239.3 = 478.6 moles
Total moles H2O Produce in Ethane combustion = 3*239.3=717.9 moles
Total Moles CO2 produce in Propane combustion = 3*159.5 = 478.5 moles
Total moles H2O Produce in Propane combustion = 4*159.5=638 moles
Flue gas consists of CO2, O2, N2, H2O
Total moles of component i in flue gas =Initial amount+Generate in reaction-consumed in reaction
Total Moles of CO2 in flue gas =3191.5+478.6+478.5=4148.6
Total Moles of O2 in flue gas = 79.78+12027.95-8018.65=4089.08
Total Moles of N2 in flue gas = 319.15+45248=45567.15
Total Moles of H2O in flue gas = 127.6+1589.3+6383.04+717.9+638=9455.84
Total Moles in flue gas = 9455.84+45567.5+4089.08+4148.6=63261 moles
=>% CO2 in flue gas = 4148.6/63261=6.61%
=>% O2 in flue gas = 4089.08/63261=6.45%
=>% N2 in flue gas = 45567.5/63261=71.8%
=>% H2O in flue gas = 9455.8/63261=15.06%
b) Applying PV=nRT
=>101.3*10^-2*V=63261*.082*550
=>V=2784.93 m3
c)
Mole fraction of water in water in flue gas = 15.06/100
=>Vapor Pressure of water in flue gas = 15.06/100*101.3=15.25 Kpa
at Dew point, Vapor pressure of water=saturation Pressure of water
=>ln(15.25)=16.26-3799.88/(T-46.85)
=>T=327.56 K
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