Acrylonitrile vapors frequently occur in specialty chemicals industry. However,

Question

Acrylonitrile vapors frequently occur in specialty chemicals industry. However, acrylonitrile is highly flammable and when mixed with air can ignite if the concentration of acrylonitrile is between 3 and 9 mol%, which are the lower flammability limit (LFL) and upper flammability limit (UFL), respectively. In one pipeline, a gas mixture of air and acrylonitrile is flowing such that the acrylonitrile content is 11.7 mol% and the flow rate of the stream is 2328 lbm/h. It is desired to dilute this stream by mixing a pure, inert, nitrogen stream at steady state such that the mixed output gas mixture has an acrylonitrile content of 2.7 mol%, temperature of 135 degree F, and a pressure of 20 psig. Air composition of 79 mol% nitrogen and 21 mol% oxygen and atmospheric pressure of 1 atm can be assumed. Determine the flow rate of required input nitrogen for dilution, the composition of output stream and thereby, its average molecular weight, and the volumetric flow rate of the output gas stream in cubic feet per minute (CFM). For density of output gas mixture, ideal gas behavior can be used. Then choose the correct option. required input nitrogen flow rate in Ib-mol/hr is 294 average molecular weight (g/mol) of output gas mixture is 29.5 volumetric flow rate in CFM of output gas mixture is 976.2 Two choices, B and C, are correct The three A-B-C choices are all correct

Explanation / Answer

Molar mass of Acrylonitrile = 53. The feed contains 11.7 mole% Acrylonitrile and rest air

Molar mass of air = 0.79*28+0.21*32= 28.84

Mass of 1 mole of mixture = 0.117*53+28.84*(1-0.117)= 31.66lb

Molar mass of mixture = 31.66 lb/mole

Molar flow rate of mixture = 2328/31.66 lbmoles/h=73.5 lb moles/hr

Molar flow rate of nitrogen in the feed = 73.5* (1-0.117)*0.79=51.3 lbmoles/hr

Molar flow rate of oxygen = 73.5*(1-0.117)*0.21= 13.6 lbmoles/hr

Molar flow rate of Acrylonitrile = 73.5*0.117=8.6moles/hr

When mixed with pure nitrogen, moles of oxygen or Acrylonitrile does not change.

It is given that after mixing, mole % of Acrylonitrile = 2.7%. Hence molar flow rate of diluted mixture = 8.6*100/2.7=318 moles/hr

This stream contains 8.6 moles/hr Acrylonitrile and 13.6 lbmoles/hr oxygen. Rest is N2. Some nitrogen is there in the inlet of mixture. So balance has to be supplied from pure N2 stream

Hence flow rate of nitrogen in the pure stream = 318-13.6-8.6 -51.3=244.5 lbmoles/hr

Products contains (composition): Acrylonitrile = 2.7%, N2=100* (244.5+51.3)/318 =93% and balance is oxygen = 100-93-2.7= 4.3%.Molar mass of product = 53*0.027+0.93*28+0.043*32=28.847 lbmoles/hr. This is as good as 28.847 g/mole

Flow rate of gas = 318 lbmoles/hr, T1=135F= 135+460= 595R, P1= 20psig= 20+14.7 psia= 34.7 psia =34.7/14.7= 2.4 atm

One moles of any gas at STP occupies 359 ft3

Hence 318 lbmoles/hr at STP occupies 359*318=114162 ft3/hr

V2= 114162 ft3/hr, T2=460R, P2= 1 atm, V1= (P2V2/T2)*(T1/P1)=(1*114162/460)*(595/2.4)=61527ft3/hr=1025ft3/min.

All the answers are close to calculated values except A. Hence B and C are correct

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