Material and Energy Balances P2.17

Introduction To Material and Energy Balances by  G.V. Reklaitis

Problem 2.17

A slurry consisting of CaCO3 precipitate in a solution of NaOH and H2O is washed with an equal mass of a dilute solution of 5% (wt) NaOH in H2O. The washed and settled slurry which is withdrawn from the unit contains 2 lbm of solution per 1 lbm of solid (CaCO3). The clear solution withdrawn from the unit can be assumed to have the same concentration as the solution withdrawn with the solids (see the Figure P2.17). If the feed slurry contains equal mass fractions of all components, calculate the concentration of the clear solution.

Answer

(Figure P2.17)

Let's make DOF

(Degree Of Freedom)

Subsidiary Relation
1. A slurry consisting of CaCO3 precipitate in a solution of NaOH and H2O is washed with an equal mass of a dilute solution of 5% (wt) NaOH in H2O.
It’s made F1= F2
2. The washed and settled slurry which is withdrawn from the unit contains 2 lbm of solution per 1 lbm of solid (CaCO3).
(F4,NaOH+F4,H2O) : F4,CaCO3 = 2 : 1
3. The clear solution withdrawn from the unit can be assumed to have the same concentration as the solution withdrawn with the solids.
(F3,NaOH)/(F3,NaOH+F3,H2O) = (F4,NaOH)/(F4,NaOH+F4,H2O)

Use basis 300 lbm/h in feed slurry, it made F1 = 300 lbm/h and F2 = 300 lbm/h respectively because subsidiary relation number 1.
Total Mass Balances
            F1+F2 = F3+F4
     600 lbm/h = F3+F4 …(1)
Independent Balance Equation
NaOH
                         Input = Output
F1,NaOH+F2,NaOH = F3,NaOH + F4,NaOH
100lbm/h+15lbm/h = F3,NaOH + F4,NaOH
         115 lbm/h = F3,NaOH+F4,NaOH …(2)

H2O
                              Input = Output
          F1,H2O+F2,H2O  =   F3,H2O + F4,H2O
   100lbm/h+285lbm/h = F3,H2O + F4,H2O
         385 lbm/h   =   F3,H2O+F4,H2O … (3)

CaCO3
                             Input = Output
                      F1,CaCO3 = F4,CaCO3
                   100 lbm/h   = F4,CaCO3

Subsidiary Relation number 2 can be,
(F4,NaOH+F4,H2O) : F4,CaCO3 = 2 : 1
         (F4,NaOH+F4,H2O) = 2F4,CaCO3
         (F4,NaOH+F4,H2O) = 2(100 lbm/h)
        (F4,NaOH+F4,H2O) = 200 lbm/h …(4)

Subsidiary Relation number 3 can be,
(F3,NaOH)/(F3,NaOH+F3,H2O) = (F4,NaOH)/(F4,NaOH+F4,H2O)
(F3,NaOH)/(F4,NaOH)=(F3,NaOH+F3,H2O)/(F4,NaOH+F4,H2O)…(5)

In stream 4
F4 = F4,NaOH + F4,H2O + F4,CaCO3 , from …(4) and CaCO3 balance we got
F4 = 300 lbm/h
It makes …(1)
F3 = 300 lbm/h

We can explore …(5)
(F3,NaOH)/(F4,NaOH) =  (F3,NaOH+F3,H2O)/(F4,NaOH+F4,H2O)
Recall that F3,NaOH+F3,H2O = F3 = 300 lbm/h and F4,NaOH+F4,H2O = 200 lbm/h based on … (4) and then
(F3,NaOH)/(F4,NaOH) = (300lbm/h)/(200lbm/h)
(F3,NaOH)/(F4,NaOH) = 3/2 …(6)

We can explore …(2)
F3,NaOH+F4,NaOH = 115lbm/h
From …(6) we know that F4,NaOH = (2/3)F3,NaOH and then
F3,NaOH+(2/3)F3,NaOH = 115lbm/h
                  (5/3) F3,NaOH = 115lbm/h
                           F3,NaOH = 69lbm/h
F3,NaOH = 69lbm/h and it makes F3,H2O = 231lbm/h

Concentration formula are,
C = [(mass of a component)/(mass total)]x100%

And then concentration of the clear solution are
C,H2O = [231/300]x100%
C,H2O = 77%
And
C,NaOH = [69/300]x100%
C,NaOH = 23%

So, the concentration of the clear solution are 77% of H2O and 23% of NaOH.

If you find a mistake, comment below.
Thanks.

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