An ore containing copper sulphate (CuSO4) is extracted in a counter-current stag
ID: 532414 • Letter: A
Question
An ore containing copper sulphate (CuSO4) is extracted in a counter-current staged leaching cascade
operated at atmospheric pressure and 25 °C. Each hour, a solid charge containing 10 tonne of gangue,
1.2 tonne CuSO4 and 0.5 tonne of water is treated. The target extract solution concentration is 10
wt.% CuSO4. The recovery of CuSO4 is to be 98 % of that originally in the ore. Pure water is used as
the fresh solvent. After each stage, 1 tonne of inert gangue entrains 2 tonnes of solution in the
underflow (raffinate) stream.
a) Calculate the number of stages required to achieve this separation.
b) Determine the theoretical minimum solvent (water) rate that could be used to obtain a 98 %
recovery of CuSO4.
Explanation / Answer
98% recovery
10% CuSO4
90% water
N=?
V1
SLE 1
-<---à
SLE 2
-<---à
SLE n
Lo
ore
10 tons/h gangue
1.2 tons/h CuSO4
0.5 tons/h water
retention = 2 tons solution/1 ton gangue
Feed Lo:
Lo= 10 + 1.2 + 0.5 = 11.7
SOLVENT Vn+1 = solute / solution =0
Final underflow, Ln:
gangue = 10 tons
solution = 2 tons solution
1 ton gangue
solution = 20 tons solution
Ln= 10 + 20 = 30
solute = 0.1 + 1.2 = 1.3
XAN= solute/ solution = 1.3 /20= 0.065
Final Overflow, V1:
Y1= 0.1
solute = 0.98 x 1.2 = 1.176
V1= 1.176/0.1= 11.76
First underflow, L1:
XA1== solute/ solution=YA1= 0.1
gangue = 10
solution = 20
L1= 10 + 20 = 30
solute = 0.1 x 20 = 2
Overflow from SLE 2, V2:
material balance at SLE 1,
Lo+ V2= L1+ V1
11.7 + V2= 30 + 11.76
V2= 30 + 11.76 ±11.7
V2= 30.06
solute balance at SLE 1,
XAoLo+ YA2V2= XA1L1+ YA1V1
1.2 + YA2x 30.06 = 2 + 1.176
YA2= 0.066
kremser Equation for Constant Underflow:
n-1 = log (YAN+1 – XAN) /YA2 - XA1 ] / [ log ( YAN+1 ±Y A2 /XAN-XA1)]
n-1 = log[ (0 – 0.065) /0.066 – 0.1 ] / log ( 0 -0.066 /0.065 -0.1)
N-1=1.022
n = 2.02 = 2---------------------answer
90% water
N=?
V1
SLE 1
-<---à
SLE 2
-<---à
SLE n
Lo
ore
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