Question

A distillation column is separating 150.0 kmol/h of a saturated liquid mixture that is 30.0 mol% methanol and 70.0 mol% water. The column operates at 1.0 atm pressure. Reflux ratio is 2.0, and reflux is returned as a saturated liquid. We desire a 97.0% recovery of methanol in the distillate and a methanol distillate mole fraction of 0.990. Find distillate ow rate D, bottoms flow rate B, methanol mole fraction in the bottoms xM, bot, and the fractional recovery of water in the bottoms.

Answer 1

Answer:

Explanation:

From the information given:

Feed F = 150.0 kmol/hr

The saturated liquid mixture of the distillation column is $X_F$ = 30%

Reflux ration  = 2.0%

methanol distillate mole fraction $X_D$ = 0.990

recovery of methanol in the distillate = 97.0%

The distillate flow rate D can be determined by using the formula;

$D = 0.97 (F* X_F)$

D = 0.97 × 150 × 0.3

D = 43.65 kmol/h

The bottom flow rate Balance B  on the column is:

F = D + B

150 = 43.65 + B

B = ( 150 - 43.65 )kmol/h

B = 106.35 kmol/h

The methanol mole fraction in the bottom $x_M$ can be computed by using the formula:

$F*X_F = DX_D + BX_B$

150(0.3) = 43.65(0.999) + 106.3($X_B$)

45 = 43.60635 + 106.3($X_B$)

45 -  43.60635 = 106.3($X_B$)

1.39365 = 106.3($X_B$)

$X_B$ = 1.39365 /  106.3

$X_B$ =  0.013

the fractional recovery of water in the bottoms f is calculated as:

$f = \dfrac{B(1-X_B)}{F_X_F}$

$f = \dfrac{106.35(1-0.013)}{150\times 0.7}$

$f = \dfrac{106.35(0.987)}{150\times 0.7}$

f = 0.99969