Question

A stream of methane at 120.90 kg/hr is burned in a boiling tank. The boiler is fed 37.30 % air, in excess. Solve for the water composition of the flue gas (wet basis), in mass percentage.

Air can be treated to be 78 % N₂, 21 % O₂ and 1% Argon, where all percentages are in mole percents. The average molecular weight for air of this composition is 28.97 kg/kgmol

Answer 1

Theory: this is a simple conbusion and mole balance equation. We will use the mole balnce techniques to determine the final answers;

Solution:

A) writing the atomic blance equation for reaction:

CH4+ 2O2 --> CO2 + 2H2O

37.3% excess air

For 1 moles of CH4 we need = 2 moles of O2

actual moles of CH4= mass/molar mass = 120.9/16= 7.55625 kmol

so moles of O2 required = 7.55625*2= 15.1125 kmol

But we are giving 37.3% extra oxygen so O2 given = 15.1125*(1.373) = 20.74946 kmoles

Please note that air 21% O2 and has N2 (78% by mol) and 1% Ar also N2 and Ar do not react in system and leaves unreacted

so mole of N2 into the sytem = moles of O2*78/21 = 20.74946*78/21 = 77.06943 kmoles of N2

so mole of Ar into the sytem = moles of O2*1/21 = 20.74946*1/21 = 0.98807 kmoles of Ar

now moles of Co2 formed = moles of CH4 reacted = 7.55625 kmol

mass of CO2 = moles *molar mass = 7.55625*44 =332.475 Kg

moles of O2 in flue gas = O2 inlet - O2 reacted = 20.74946-15.1125=5.636963 kmoles

mass of O2 = mles of O2*molar mass = 5.636963*32 = 180.3828 Kg

mole of N2 in flue gas = 77.06943 kmoles

mass of N2 in flue gas = moles*molar mass = 77.06943*28 = 2157.944 kg

moles of Ar in flue gas = 0.98807 kmol

mass of Ar = moles*molar mass = 0.98807*39.95 = 39.47338 kg

moles of water formed = 2*moles of CH4 reacted = 2*7.55625 = 15.1125 kmol

mass of water formed = mole* molar mass = 15.1125*18 = 272.025 Kg

total mass of flue gas including water = mass of CO2+O2+N2+Ar +H2O

=332.475+180.3828+2157.944+39.47338+272.025= 2982.3 Kg

Flue gas composition:

mass% of CO2 = mass of CO2/total mass *100 = 332.475/2982.3*100 = 11.15%

mass% of O2= mass of O2/total mass *100 = 180.3828/2982.3*100 = 6.05%

mass% of N2 = mass of N2/total mass *100 = 2157.944/2982.3*100 = 72.36%

mass% of Ar= mass of Ar/total mass *100 = 39.47338/2982.3*100 = 1.32%

mass% of H2O= mass of H2O/total mass *100 = 272.025/2982.3*100 = 9.12%

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