Question

A mixture, F, of 620 lbmol/h of 46 mol% benzene (B) and 54 mol% toluene (T) is separated continuously into two products. One product, the distillate , is 98 mol% benzene. The other product, the bottoms , is 95 mol% toluene. Compute the flow rates of these two streams in: (a) lbmol/h, and (b) kmol/h. Calculate the fraction of the benzene fed to the unit that is recovered in the distillate.

Answer 1

F = 620

xb = 0.46

xt = 0.54

P = ?

xb = 0.98

xt = 0.02

B = ?

xb= 0.95

xt = 0.05

Calculate

A) B and P (lbmol/h and kmol/h)

F = B + P

620 = B + P

balance Toluene

620*0.54 = B*0.02 + 0.95*P

Solve for P

620 = B + P .... P = 620-B

Then substitute in 620*0.54 = B*0.02 + 0.95*P

620*0.54 = B*0.02 + 0.95*(620-B)

Solve for B

334.8 = 0.02B + 589 - 0.95B

-0.93B = 334.8 -589

B = (334.8 -589 )/(-0.93 ) = 273.333 lbmolh = 273.33*0.454 = 124.09 kg/h

Solve for P

620 = B + P

620 = 273.333+ P

P = 620 - 273.333 = 346.667 lbmol/h = 346.667 *0.454 = 157.38 kg/h

calculate the fraction of benzene being fed

balance benzene

benzene in destillate = 0.98*273.33= 267.8634 lbmol/h (from those 285 lbmol/h being fed)