In a high-pressure separation process, a gas having a mass composition of 50% be

Question

In a high-pressure separation process, a gas having a mass composition of 50% benzene (C6H6), 30% toluene (C7H8), and 20% xylene (C8H10) is fed into the process at a rate of 483 m3/h at 607 K and 26.8 atm. One exit stream is a vapor with a mole composition of 91.2% benzene, 7.2% toluene, and 1.6% xylene. A second exit stream is liquid with a mass composition 6.0% benzene, 9.0% toluene, and 85.0% xylene. What is the composition of the third exit stream if it is liquid flowing at a rate of 9800 kg/h, and the ratio of benzene to xylene in the stream is 3 kg benzene to 2kg xylene? C6H6:MW=78,Tc =562.6K,pc =48.6atm C7H8: MW = 92, Tc = 593.9 K, pc =40.3 atm C8H10:MW=106,Tc =619K,pc =34.6atm

Explanation / Answer

Solution:

The composition of third exit stream is,

Benzene = 1-(0.912+0.06) = 0.028

Toluene = 1-(0.072+0.09) = 0.838

Xylene = 1-(0.016+0.85) = 0.134

Flow rate = 9800 kg/h

Benzene = 0.028*9800=274.4 kg/h

Toluene = 8212.4 kg/h

Xylene = 0.134*9800=1313.2 kg/h

Asper the ratio of 3:2 for benzene:xylene the composition will be,

Benzene = 0.0972

Toluene = 0.838

Xylene = 0.0648

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