In a glycerol plant a 10 mass basis aqueous glycerin solutio

In a glycerol plant, a 10% (mass basis) aqueous glycerin solution containing 3% NaCl is treated with butyl alcohol as illustrated in Figure 4.28. The alcohol fed to the tower contains 2% water on a mass basis. The raffinate leaving the tower contains all the original salt, 1.0% glycerin and 1.0% alcohol. The extract from the tower is sent to a distillation column. The distillate from this column is the alcohol containing 5% water. The bottoms from the distillation column are 25% glycerin and 75% water. The two feed streams to the extraction tower have equal mass flow rates of 1000 lb_m per hour. Determine the output of glycerin in pounds per hour from the distillation column.

Solution

stream 1

glycerien = 10% = 100 Lbm/hr

Salt = 3% 0f total stream = 30 lbm/hr

water = 1000-100-30= 870 lbm/hr

stream 2

Alcohol = 98% of total stream =980 lbm/hr

water =2% of total stream =20 lbm/hr

total mass flow including both stream = 2000 lbm/hr

suppose X percentage is going to Raffinate

then raffinate stream composition

salt = 30 lbm/hr (all salt goes to raffinate stream)

glycerin =.01 X ( 1 percent)

alcohol = .01 X ( 1 percent)

extract stream

mass flow rate = (2000-X) lbm/hr

suppose it

overhead stream contains x1 part

and bottom stream contains (1-X1) part

overhead of distillation column 95% alcohol

(2000-X) X1x95/100=(980-.01X) ---------1)

bottom stream contains 25 of glycerin

(2000-X)(1-X1)X25/100=(100-.01X) -----2)

by solving both