Question

Q2) 8000 kg/hr of feed contains 21% acetic acid ( C) and 79% water (A ) is to be counter current extracted with isopropyl ether ( B) to remove 87% of the solute (C) with totally immiscible between two layers. Determine three of the following. 1-The number of theoretical stages required if 10000 kg/hr of solvent B (free of C) is used. 2-What is the concentration of C in the extract in the third stage? 3-What is the minimum solvent required ( B min. ) 4- What is the concentration of acetic acid in the raffinate at the fourth stage. Given the equilibrium data: X( C/A ) Y( C/B) (25 degrees) 0.05 0,1 0.15 0.2 0.25 0.04 0.08 0.12 0.16 0.2

Answer 1

F = 8000 Kg/h

Counter current extraction takes place

C- acetic acid

B- isopropyl alcohol

A- water

Solute is acetic acid

x_f = 0.21

X_f = 0.21/(1-0.21) = 0.2658

Water present in feed = (8000) (1-0.21)

= 6320 Kg/h= A

B = 10000 Kg/h

Slope of operating line of extraction = (-A/B) =

(-6320/10000) = -0. 632

The equilibrium data given

Y	0.04	0.08	0.12	0.16	0.20	0.24	0.28
X	0.05	0.1	0.15	0.20	0.25	0.30	0.35

87% of C is to be extracted

X_N = (1-0.87) (X_F)

X_N = (1-0.87) (0.2658) = 0.03455

The equation for counter current extraction is

(-A/B) = (Y_s- Y₁) /(X_F - X_N)

Solvent is pure

Ys = 0

-0. 632= (0-Y1) /(0.2658-0.03455)

Y₁ = 0.14615

Operating line coordinates are

(X_F, Y₁) and (X_N, Y_s)

(0.2658, 0.14615) and (0.03455, 0)

The graph is given below

From graph number of stages = 4.7~ 5 stages

D)

Concentration of raffinate at fourth stage from top = 0.05

C)

To find minimum solvent rate

We should find extract in equilibrium with feed concentration

At X= 0.2658 from graph

Y_max = 0.212

Equation

-(A/B) = (Y_s- Y₁) /(X_F-X_N)

-(6320/B) = (0-0.212) /(0.2658-0.03455)

Bmin = 6893.867 Kg/h

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