A ternary liquid mixture containing 35 mol% methanol, 45 mol% water, and remaind

Question

A ternary liquid mixture containing 35 mol% methanol, 45 mol% water, and remainder acetone is sent to a distillation column as feed for separation. The distillate liquid from the column contains 16 mol% acetone. It is found that the bottoms liquid output contains 25% of the methanol fed to the column and 92% of the water fed to the column. It is desired to have a bottoms stream flow at 3700 L/hr at steady state. Calculate the volume flow rates of feed and distillate as well as the mass % composition and molar flow rate of the bottoms stream. Then choose correct option below. Assume that the S.G. values in Appendix B1 can be used and that mixture densities for liquid solution can be estimated by weight average approximation.

Explanation / Answer

Data Given:

Feed:

Methanol zmeth:0.35

Water zwat:0.45

Acetone zac:0.20

F=feed flow rate

Distillate:

D=distillate flow rate

yace = 0.16

Bottom:

B= Bottom flow rate = 3700 l/hr

We know that mole fraction and volume fraction same.

Methanol =0.20 of feed=0.20*zmeth *F=0.2*0.35*F=0.07F=xmeth*B

Water= 0.90 of feed =0.92*zwat*F=0.92*0.45*F=0.414F=xwat*B

So acetone = B-methanol-water=B-0.07 F-0.414F=B-0.484F

Solution:

Now take overall balance

F=D+B

F=D+3700

Now take methanol balance

zmethF=ymeth*D+xmeth*B

0.35F=ymeth*D +0.07 F

0.28 F =ymeth*D

Now take acetone balance

zaceF=yaceD+xaceB

0.20F=0.16D+(3700-0.484F)

0.684F=0.16D+3700

And we know that

F=D+3700

so solving this equation:

F=5931.29 lit/hr

D = 2231.29 lit/hr =2231.9*0.004403=9.827 GPM

So

In bottom

methanol =0.07*5931.29=415.1903 lit/hr

Water = 0.414*5931.29=2455.55 lit/hr

Acetone = 829.2597 lit/hr

So mole fraction:(mole fraction and volume fraction same)

xmeth = 0.1122

xwater = 0.6632

xace =0.2246

In terms of mass:

Mass = volumetric flow rate * density

Specific gravityof methanol=0.791

density =s.g*water density =0.791*1000=791 kg/lit

So methanol =415.1903 lit/hr *791=328415.52 kg/hr

Acetone=829.2597*792=656773.68 kg/hr

water=2455.55*1000=2455550 kg/hr

Now weight fraction = mass/total mass

Total mass =2455550+656773.68+328415.52=3440739.2 kg/hr

Weight fraction:

methanol=328415.52/3440739.2=0.095=9.5%

acetone=656773.68/3440739.2=0.19088=19.088%

water=2455550/3440739.2=0.7136=71.36%

Density mix=939.92 kg/lit

B= 3700 lit/hr*939.92 kg/lit=3.4*106 kg /hr

Molecular weight of mix = sum(mol fraction*molecular weight)=28.57 gm/mol=28.57 kg/kmol

B = 3.4*106/28.57=119.005 kmol/hr

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