Basis: In the slurry feed, solvent: C = 1 kg H2O, solid: B = 1.667 kg

Transcription

Basis: In the slurry feed, solvent: C = 1 kg H2O, solid: B = 1.667 kg
CENG 3210 Separation Processes
Tutorial 9 Solution
Multistage countercurrent leaching
A multistage countercurrent leaching battery is used to extract the sludge from the reaction
Na2CO3 + CaO + H2O  CaCO3 + 2NaOH
The products from the reaction enter the first unit with no excess reactants but with 1.667 kg
CaCO3/kg H2O. Pure water is used to wash NaOH from the sludge. CaCO3 is assumed to be
completely insoluble in water. The sludge retains solution varying with the concentration as shown
in the following table.
NaOH, wt %
0
5
10
15
20
Kg CaCO3/kg solution 0.667
0.571
0.455
0.370
0.278
If a 20% solution of the NaOH is to be produced, how many stages must be used to recover 97% of
the NaOH?
Basis: In the slurry feed,
solvent: C = 1 kg H2O,
solid: B = 1.667 kg CaCO3.
The solute A (NaOH) is found from the reaction
formula. For 1 mol CaCO3 produced, 2 mol NaOH is
generated.
So A = (2B)(MWNaOH/MWCaCO3)
= 2x1.667(40/100) = 1.333 kg
L0 = A + C = 1.333 + 1 = 2.333 kg
y0 = A/L0 = 1.333/2.333 = 0.571
N0 = B/L0 = 1.667/2.333 = 0.714
xN+1 = 0
x1 = 0.2 (20% solution of NaOH)
1
Since 97% of the NaOH is to be recovered,
NaOH in strong liquor (V1) = S1 = 1.333(0.97) = 1.293 kg
NaOH in washed solid (LN) = AN = 1.333(1-0.97) = 0.04 kg
Water in strong liquor = m1 = (S1/x1)(1-x1)
= (1.293/0.2)(1-0.2) = 5.172 kg
V1 = 1.293 + 5.172 = 6.465 kg
To find yN, do a water balance:
mN + 5.172 = mf + 1
mN = mf – 4.172
yN = AN/( mN + AN) = 0.04/( mN + 0.04)
NN = B/( mN + AN) = 1.667/( mN + 0.04)
NN, yN and mN are obtained by trial from the table.
yN = 0.0153
NN = 0.638
mN = 2.574
LN = mN + AN = 2.574 + 0.04 = 2.614 kg
To determine an intermediate point on the operating curve,
choose yn = 0.1 (assumed), N = 0.455 = B/Ln = 1.667/Ln
Ln = 3.664 kg
Vn1  Ln  V1  L0  3.664  6.465  2.333  7.796 kg
L
V x  L0 y0
x A,n 1  n yn  1 1
Vn 1
Vn 1
3.664
6.465  0.20  2.333  0.571

(0.1) 
 0.042
7.796
7.796
Similarly, with yn = 0.15, xn+1 = 0.0735, with yn = 0.2, xn+1
= 0.114.
2
Using the points of various (yn, xn+1), and (yN, xN+1) to
obtain a slightly curved operating line. By the McCabeThiele method, the number of ideal stages is determined to
be 4.
3