Question

Acetic acid is to be extracted from benzene into water at 25C. 100 kg/h of a feed that is 0.0092 wt. fraction acetic acid and 0.9908 wt. fraction benzene is fed to a column. The inlet water is pure and flows at 25 kg/h. The desired outlet weight fraction is 0.00040 acetic acid in benzene. Assume total flow rates are constant, water and benzene are completely immiscible, eqilibrium is linear.

K=y/x = wt. fraction solute in extract/wt. fraction solute in raffinate = 30.488

Determine the weight fraction of acetic acid in the water and the required number of stages.

Answer 1

Assuming countercurrent Extraction

solvent is immiscible that is given in the problem

Using all usual notations

X=mass of C/mass of A in the raffinate phase

Y=mass of C/mass of B in the extract phase

Y_s=mass of C/mass of B in the solvent used

where C is solute i.e. Acetic acid

A=solvent in feed or raffinate i.e.benzene

B=solvent used for extraction i.e. Water in given problem

material balance of C over stage one

material balance of C over entire unit

(note that Y_n+1=Y_s)

after rearrenging & simplifying above equation

orY

this is equation of straight line of slope A/B

this the operating line passing through (X_F,Y₁) & (X_n,Y_n+1)

the g_raphical construction of the no. of stages required to pass X_F to X_n.

equilibrium curve is given as

K=y/x =30.488

solution can found by this way

but in this problem equilibrium lines lope is given 30.488 which practicaly imposible to plote

tan^-1(30.488)= 88.121 degree

which Y axis we can say so on plotting this problem

X_F=mass of acetic acid / mass of benzene in feed= 100*0.0092/100*0.9908 =9.28 *10^-3

Y₁=mass acetic acid /Mass of water in final extract phase =0.035

X_n=mass of acetic acid / mass of benzene in final raffinate =0.0004/0.9996=0.4 *10^-3

Y_n+1=Y_s=0   as pure water is used

0 =100*0.9908/25 (0.4 *10^-3-9.28 *10^-3)+Y₁

therefore

Y₁=0.035

now we can plot it graphically

so there will be only one stage as our equilibrium curve is Y axis