Nitrogen gas and alcohol are combined in a boiler where the alcohol is fully vap

Question

Nitrogen gas and alcohol are combined in a boiler where the alcohol is fully vaporized. The combined output is carried out on the outlet stream. 111 moles per minute of alcohol are input to the boiler. The output is combined with a second nitrogen stream flowing at 4.23 m^3/min at a pressure of 2 atm, and the combined stream is fed to a compressor. The compressor has only one product stream. There is no reaction and the process is at steady state. All gases can be considered ideal. A partially labeled flow diagram for the process is: a) What is the value of n_3N, the molar flow rate of N_2 in the second nitrogen input stream? b) Find the mol fraction of A on the compressor output stream given that n_IN = 700 mole N_2/c) Write it the other way.

Explanation / Answer

Ans a) Given V3N = 4.23m3/min

T=298K

P=2atm

R=0.0821

n3N =RT/PV3N

      = 0.0821x298 /2x4.23

      = 2.891 molN2/min

Now molar flow rate = n3N x V3N

                             = 2.891x4.23

   =12.23m3 /min

Ans b) From gas equation we get

PN/Pt = n1N / nt

Here PN = 3.01atm

Pt =3.3atm

n1N = 700mol N2 /min

nt= mole fraction of A

Therefore 3.01/3.3 = 700 /nt

nt = 700x3.3/3.01

    = 767.4molN2/min

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