Sample Problem #11

SAMPLE PROBLEM #11 Roasted copper ore containing as CuSO4 is to be extracted in a counter ercu currrent stag age e ex exttract cto or. Each hour, a charge consisting of 10 tons gangue, 1.2 tons CuSO4, and 0.5 ton water is to be treated. The strong solution produced is to consist of   90% water and 10% CuSO4 by we weig igh ht. The recovery of CuSO4 is to be 98% of   that in the ore. Pure water is to be used as fresh solvent. After each stage, one ton of inert gangue retained, 2 tons of water plus the copper sulfate dissolved in that water. Equilibrium is attained at each stag age. e. Calcu cullate the number of stages required.

SOLUTION 98% recovery 10% CuSO4

V1

SLE 1 Lo

n=?

90% water 

ore 10 tons/h gangue 1.2 tons/h CuSO4 0.5 tons/h water 

SLE 2

|| ||

Vn+1

SLE n Ln

retention = 2 tons solution 1 ton gangue

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Feed

Lo: Lo = 10 + 1.2 1 .2 + 0.5 = 11.7

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Solvent Vn+1: solute VAN+1 = solute = 0 solution

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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 @

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XAN = solute solute = 1.3 = 0.06 0.065 5 solution 20 Final

Overflow, V1:  Y1 = 0.1 solute = 0.98 x 1.2 = 1.176

V1 = 1. 1.176 176 = 11.7 11.76 6 0.1

@

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First

underflow, L1: XA1 = solute = YA1 = 0.1 solution

gangue = 10 solution = 20

L1 = 10 + 20 = 30

@

solute = 0.1 x 20 = 2 

Overflow from SLE 2, V 2: 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 + YA2 x 30.06 = 2 + 1.176  YA2 = 0.066

@

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Kremser

Equation for Constant Underflow:

n±1 =

log  YAN+1 ± XAN  YA2 ± XA1 log

n±1 =

 YAN+1 ± YA2 XAN ± XA1

log log

n±1 = @

1.022

n = 2.02 = 2

0 ± 0.065 0.06 0. 066 6 ± 0. 0.1 1 0 ± 0.066 0.06 0. 065 5 ± 0. 0.1 1