50 L/min of liquid n-pentane is combusted with 100% excess air. The fractional c

Question

50 L/min of liquid n-pentane is combusted with 100% excess air. The fractional conversion of n-pentane is 0.993. There is no incomplete combustion. A partially completed flow chart is provided for you if you want to use it. (HINT: When drawing the flow chart, label the product stream with molar flow rates, not mole fractions.)

a) What is the volume of air that needs to be delivered at 400°C and 2 atm?

b)What is the mole fraction of water in the exit stream?

c) What is the dew point for the product stream if the exit pressure is 1.00 atm?

Explanation / Answer

a)

C5H12 + 8O2   > 5CO2 + 6 H2O

pentane

atNTP gmol weight of npentane = 5*12+ 12*1 =72 g   whichoccupies 22.4 litres at NTP

Moles in 50 L = 50/22.4 = 2.232

From mole ratio O2 required for 8*2.232*.993(conversion factor)

                            = 17.731 moles

           Vol of O2 = 17.731 * 22.4 (molar volume of O2) = 397.17 L

Now P1V1/T1 = P2V2/T2

put values    1* 397.17/273 = V2* 2/673 (400 degree centograde =673 K)

On calculation , V2 = 489.6 L

So this vol of air needs to be delivered

b) In exit stream 5 moles of CO2 and 6 moles of water will be present   so mole fraction of water = 6/5+6 = 6/11

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