A mixture of fuel containing 0.550 mol CH4/mol, o.330 mol caHavmol, and o.120 mo

Question

A mixture of fuel containing 0.550 mol CH4/mol, o.330 mol caHavmol, and o.120 mol csH12/mol at 52.0"c kPa (gauge) enters at of where reacts with Determine the volumetric flow rate of air to the in SCFM (standard cubic feet per minute) by following the steps below. Assume that the gases behave ideally. What is the molar flow rate of fuel entering the furnace? Number mol fuel min What is the molar flow rate of CH4 entering the fumace? Number mol min What is the molar flow rate of C3H8 entering the furnace? Number mol C H min (Scroll down for more questions.)

Explanation / Answer

Flow rate, V= 62 L/min, P= 123Kpa(g)=123+101.3 Kpa= 224.3 Kpa= 224.3/101.3 atm=2.214 atm

T= 52+273= 325K, R=0.0821 L.atm/mole.K , n= no of moles of gas mixture= PV/RT= 2.214*62/(0.0821*325) =5.144 moles/min

Hence molar flow rate of CH4= 5.144*0.550 =2.83

Molar flow rate of C3H8= 5.144*0.330= 1.70

Molar flow rate of C5H12= 5.144*0.120= 0.61

The reactions of combustion are

CH4+2O2------àCO2 + 2H2O, moles of oxygen required per mole of CH4=2, moles of O2 required for 2.83 moles of CH4= 2.83*2= 5.76 moles

C3H8+ 5O2 ----à3CO2+4H2O, moles of O2 required= 5*1.70= 8.5 moles

C5H12+ 8O2-----à5CO2+ 6H2O, moles of O2 required = 8*0.61= 4.88 moles

Total moles of oxygen required= 5.76+8.5+4.88=19.14 moles of oxygen. Air contains 21% O2 and 79% N2, moles of air required= 19.14/0.21 =91.14 moles

1 mole of gas any gas at STP occupies 22.4 L

91.14 moles of air occupies 22.4*91.14 L/min=429 L/min at STP. 429 L/min = 429*0.0353 ft3/min=15.14 ft3/min

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