A mixture of fuel containing 0.530 mol CH4/mol, 0.320 mol C3H8/mol, and 0.150 mo

Question

A mixture of fuel containing 0.530 mol CH4/mol, 0.320 mol C3H8/mol, and 0.150 mol C5H12/mol at 54.0°C and 123 kPa (gauge) enters a furnace at a rate of 90.0 L/min where it reacts with 11.0% excess air. Determine the volumetric flow rate of air to the furnace 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?

*What is the molar flow rate of fuel entering the furnace? (in mole fuel/min)

Explanation / Answer

CH4 (g) + 2O2 (g) ----> CO2 (g) + 2H2O (g)
nCH4= 0.53 mol
nO2 = 2 * 0.53 =1.06 moles

C3H8 (g) + 5O2 (g) ----> 3CO2 (g) + 4H2O (g)
nC3H8 = 0.320 mol
nO2 = 5 * 0.320 = 1.600 moles

C5H12 (g) + 9O2 (g) ----> 5CO2 (g) + 8H2O (g)
nC3H8 = 0.150 mol
nO2 = 9 * 0.150 = 1.350 moles

nO2 = 1.06 + 1.6 + 1.35 = 4.01 moles

mole fraction of O2 in air = 0.21
nair = nO2/xO2 * 1.11= 4.01*1.11/0.21 = 21.1957 moles
For 3 moles of fuel, 21.957 moles of air are required.

Volumetric flow rate of fuel = 90L/min
For 1 minute,
n = PV/RT = (1.23/1.0135)*(90)/(0.0821*327) = 4.068

molar flow rate of fuel = 4.068 mol/min

molar flow rate of fuel = 4.068 * 21.957/3 = 29.774 mol/min

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