A mixture containing 45% benzene (B) and 55% toluene (T) by mass is fed to a dis

Question

A mixture containing 45% benzene (B) and 55% toluene (T) by mass is fed to a distillation column. An overhead stream of 95 wt% B is produced, and 8% of the benzene fed to the column leaves in the bottom stream. The feed rate is 2000 kg/h. Determine the overhead flow rate and the mass flow rates of benzene and toluene in the bottom stream Basis: Given Feed Rate The labeled flowchart is as follows. D kg/h 0.95 kg B/kg 0.05 kg T/kg 2000 kg/h 0.45 kg B/kg 0.55 kg T/kg Contains 8% of the B in the feed wa kg B/kg WT kg T/kg

Explanation / Answer

determine overhead flow rate... D

mass flow of benzene + toluene in B

overall mass balance

2000 = D + B

mass balance of B:

0.45*2000 = 0.95*D + wb*B

note that we know that

8% of B

that is ...

0.45*2000 --> 100% of benzene

wb*B --> 8% of b

wb*B = 8/100*0.45*2000 = 72

2000 = D + B

900 = 0.95*D + 72

now... we have 2 equations + 2 unkown

solv efor D

D = (900-72)/(0.95) = 871.57

get B

B = 2000-D = 2000-871.57 = 1128.43

now...

we have D = 871.57 kg of overhead rate

now... get

flow of toluene in bottoms = 1128.43 - 72 = 1056.43 kg of toluen ( bottoms)

flow of benzene in bottoms = 8% of inlet --> 72 kg of benzene ( bottoms)

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