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3 years ago

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A steady state and continuous separator has a total feed rate of 100. kg/h of a 55.0 wt. % benzene mixture. The balance is tolue

ne (i.e., 45 wt. % toluene). A vapor stream leaves the process and has a benzene concentration of 85.0% by mass and a liquid stream leaving the process has a benzene concentration of 10.6% by mass. a. Draw and label the process flow chart. Develop a consistent variable naming system. b. Simplify the general balance equation. c. Write and solve the total mass balance and the benzene mass balance.

1 answer:

Tju [1.3M]3 years ago

7 0

Answer:

Part a : The flow chart is as given in the attached file

Part b: Overall Mass balance is given as $m_1=m_2+m_3$

Part c: The mass flow rate of Vapor and Liquid are 59.667 kg/hr

and 40.323 kg/hr respectively.

Explanation:

Part a

The flow chart is as given in the attached file.

The labeling is as

Feed Rate is given as m_1 which is 100 kg/h

Vapour Rate is given as m_2 which is not known

Liquid Rate is given as m_3 which is not known

Fraction of Benzene at Feed is given as x_1B which is 55% wt.

Fraction of Toluene at Feed is given as x_1T which is 45% wt.

Fraction of Benzene at Vapour is given as x_2B which is 85% wt.

Fraction of Toluene at Vapour is given as x_2T which is 15% wt.

Fraction of Benzene at Liquid is given as x_3B which is 10.6% wt.

Part b

Overall Mass balance is given as

Total Mass in=Total Mass out

m_1=m_2+m_3

100=m_2+m_3 -----------------1

Part c

Benzene Mass Balance is given as

Mass of Benzene in Feed=Mass of Benzene in Vapor+Mass of Benzene in Liquid

x_1B *m1=x_2B*m2+x_3B*m3

0.55*m1=0.85*m2+0.106*m3 -----------------------2

Solving the 2 equations simultaneously gives

m_2=59.667 kg/hr

m_3=40.323 kg/hr

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