iPad 8:57 PM Problem 4.40 Sequential Distillation A liquid mixture containing 30
ID: 718082 • Letter: I
Question
iPad 8:57 PM Problem 4.40 Sequential Distillation A liquid mixture containing 30.0 mole% benzene (B), 25.0% toluene (T), and the balance xylene (X) is fed to a distillation The bottoms product contains 98.0 mole% x and no B, and 90.0% of the x in the feed is recovered in this stream. The overhead product is fed to a second column. The overhead product from the second column contains 95.5% of the B in the feed to this column. The composition of this stream is 94.0 mole% B and the balance T. Still1 You should be able to solve for one variable at a time. Write three balances and use one constraint for the performance of Still 1 to solve for the following quantities for a process feed (n) of 300.0 mol/h First Still Bottoms flow rate (nz) nol/h Overhead fow rate (n3) nol/h Benzene mole fraction in overhead (X3) Toluene mole fraction in overhead (X37) LINK TO TEXT LINK TO TEXT By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Attempts: 1 of 8 used SAVE FOR LATER Still 2Explanation / Answer
column 1
Stream 1
Feed rate
F = 300 moles/h
Molar flow of Benzene in feed = 0.30 x 300 = 90 moles/h
Molar flow of toluene in feed = 0.25 x 300 = 75 moles/h
Molar flow of xylene in feed = (1-0.30-0.25) x 300 = 135 moles/h
Stream 2
90% of the xylene = 0.90 x 135 = 121.5 moles/h
Molar flow of xylene in bottom product of column 1
= 121.5 moles/h
xylene in bottom product of column 1 = 98%
Toluene in bottom product of column 1 = 100 - 98 =2 %
Molar flow of bottom product of column 1 = 121.5/0.98
= 123.979 mol/h
Molar flow of Toluene in bottom product of column 1
= 123.979 x 0.02 = 2.479 mol/h
Stream 3
Molar flow rate of overhead in column 1 = feed rate - bottom rate
= 300 - 123.979 = 176.021 mol/h
Benzene rate = benzene in feed - benzene in bottom
= 90 - 0 = 90 mol/h
Toluene rate = Toluene in feed - Toluene in bottom
= 75 - 2.479 = 72.521 mol/h
xylene rate = xylene in feed - xylene in bottom
= 135 - 121.5 = 13.5 mol/h
First still
Molar flow of bottom product of column 1
n2 = 123.979 mol/h
Molar flow rate of overhead in column 1 =
n3 = 176.021 mol/h
Benzene mol fraction in overhead x3b = 90/176.021 = 0.511
Toluene mol fraction in overhead x3t = 72.521/176.021= 0.412
Column 2
Stream 5 overhead
95.5% of B in column 2 feed = 95.5% of B in overhead of column 1
= 0.955 x 90 = 85.95 mol/h
Mol% B = 94%
Mol% T = 100 - 94 = 6 %
Molar flow rate of overhead stream 5 = 85.95/0.94 = 91.43 mol/h
Molar flow of Toluene = 91.43*0.06= 5.48 mol/h
Stream 4
Molar flow of bottom = feed - overhead
= 176.021 - 91.43 = 84.59 mol/h
Benzene rate = B in ( feed - overhead) = 90 - 85.95 = 4.05 mol/h
Toluene rate = T in ( feed - overhead)
= 72.521 - 5.48 = 67.041 mol/h
Xylene rate = X in (feed - overhead) = 13.5 - 0 = 13.5 mol/h
Second still
n5 = 91.43 mol/h
n4 = 84.59 mol/h
X4b = 4.05/84.59 = 0.0479
X4T = 67.041/84.59 = 0.793
% of B recovered = 85.95/90 = 0.955 = 95.5%
% of T recovered = 67.041/75 = 0.894 = 89.4%
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