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Butane (C4H10) is burned with excess air. The exhaust gas from the combustion is analyzed and...

Butane (C4H10) is burned with excess air. The exhaust gas from the combustion is analyzed and has the following composition: 4.5 mole% O2, 73.8 mole% N2, 11.824% H2O, 0.236 mole% C4H10, 8.48 mole% CO2, 1.16 mole% CO. Determine: 1. (8 points) The fractional conversion of C4H10 2. (8 points) The ratio of exhaust produced to C4H10 fed 3. (9 points) The percent excess air fed to the process. Draw a flow chart.

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Answer #1

The reaction is C4H10+6.5O2----> 4CO2+ 5H2O

Basis : 100 moles of exhaust gases

Exhaust gases contains 73.8 moles and comes from air supplied for combustion. Nitrogen at the inlet= 73.8 moles

79 moles of nitrogen is there in 100 moles of air

73.8 moles of nitrogen is there in 100*73.8/79=93.42 moles of air supplied

air contains 21% oxygen, hence oxygen actually supplied = 93.42*0.21=19.62 moles

moles un reacted = 4.5 moles

Oxygen reacted = 19.62-4.5 moles = 15.12 moles

air supplied = (15.12/0.21)= 72

C4H10 to be reacted = 15.12/6.5 ( 1 mole of C4H10 requires 6.5 moles of oxygen)=2.32

C4H10 remaining =0.236

C4H10 reacted= 2.32-0.236=2.084 moles

Fractional conversion =100*2.084/2.32=89.8%

C4H10 fed /Exhaust= 2.32/100 =0.0232

Excess air : (Excess air/ Stoichiometric air)*100= 100*(93.42-72)/72=29.75%

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