Problem 4.40 Sequential Distillation lead mixture cortaring 30..0 mle% benaene (

Question

Problem 4.40 Sequential Distillation lead mixture cortaring 30..0 mle% benaene (. 2S0% toluene CT), and the balance ryle. CX) "%d to adstilation eolumn. hebottorns pro t contains 980mo'es x and no B, and 90.0% of the Kinthe feed is recovered n this stren. ht overhead product is fed to second column. The overhead product from the second couren cotars 95.5% of the B th·feed to this eok mnThe composition of this stream is 94.0 mole% 8 and the balance T. Still 1 nite three balances and use one contraire nor the performance ofsun 1 tososve for . fviwng Oarenies for . process feed cm)of300.0 meth First Still Overhead fow rabe- molt Toluene mole fraction in overhead () newer *1: the tolerance is 6.2% newer .2: ehe tolerance"/-2% Answer .3: the tolerance is +/-2% Attempts: 0 of 8 userd

Explanation / 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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