CHMT2023A – ASSIGNMENT 2
REMOVAL OF COPPER
SULPHATE FROM A
POLLUTION CONTROL
RESIDUE
13 MAY 2024
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GROUP 40:
❖ DIVYA VENKETSAMY
❖ LEHLONOLO MASILO
❖ JESSICA TIFO
❖ KGOTHATSO MAKWELA
❖ SMANGALISO PRINCE
GAMA
University of the Witwatersrand, Johannesburg
School of Chemistry
SENATE PLAGIARISM POLICY
Declaration by Students
I, DIVYA VENKETSAMY (2695022), LEHLONOLO MASILO (2674273), JESSICA
TIFO (2340874), KGOTHATSO MAKWELA (2662416), SMANGALISO PRINCE
GAMA (2700408), am a student registered for CHMT2021A in the year 2024. I hereby
declare the following:
•
♣
I am aware that plagiarism (the use of someone else’s work without their
permission and/or without acknowledging the original source) is wrong.
•
♣
I confirm that ALL the work submitted for assessment for the above course
is my own unaided work except where I have explicitly indicated otherwise.
•
♣
I have followed the required conventions in referencing the thoughts and
ideas of others.
•
♣
I understand that the University of the Witwatersrand may take disciplinary
action against me if there is a belief that this in not my own unaided work or that I
have failed to acknowledge the source of the ideas or words in my writing.
PROOF OF CONTRIBUTION OF GROUP MEMBERS:
Signature: _________________________ Date: 13/05/24 (D. VENKETSAMY)
Signature:
Date: 13/05/24 (L. MASILO)
Signature:
Date: 13/05/24 (J. TIFO)
Signature:
Date: 13/05/24 (K. MAKWELA)
Signature:
Date: 13/05/24 (S.P GAMA)
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GIVEN:
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➢
Pulp density of slurry = 2.2
Mass fraction of solid in the slurry = 61.54% = 0.6154kg
Discharge slurry = 1% of incoming amount of Cu
Concentration of CuSO4 in the slurry solution = 0.16
Volume fraction of liquid in the slurry is consistent throughout
Volumetric flow rate of liquid for washing = volume of wash water
Density of solids in the slurry = 5kg/m3
Approximate specific gravity: p (CuSO4 solution) = 1 + wCuSO4
BASIS: 1000L of feed slurry
 volume(total) = 1000L
p = m/v
mass(total) = p x v
= (2.2kg/L) (1000L)
= 2200 kg
MASS BALANCE OF FEED SLURRY:
1. mass(liquid) + mass(solid) = 2200kg
VOLUME BALANCE OF FEED SLURRY:
2. volume(liquid) + volume(solid) = 1000L
FEED SLURRY (mass and volume):
mass(solid) = (0.6154) (2200kg) = 1354 kg
mass(liquid) = 2200 – m(solid) = 2200kg – 1354kg = 846 kg
p(solid) = m(solid) / v(solid)
v(solid) = 1354kg / 5kg/L = 271L
v(liquid) + 271L = 1000L
v(liquid) = 729L
w(liquid) = 1 – w(solid)
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= 1 - 0.6154kg = 0.3846
Table 1 showing mass and volume of each phase of the feed slurry.
FEED SLURRY
Solid
Liquid
Total
MASS (kg)
1354
846
2200
VOLUME (L)
271
729
1000
MASS FRACTION
1.0 solid
0.16 CuSO4
0.3846 liquid
FEED SLURRY (STREAM A):
FA = 2200kg = 1000L
given wA(solid) = 0.6154
wA/L(CuSO4) = 0.16
mass of CuSO4 in liquid = (0.16) (846) = 135.4kg
mass of CuSO4 entering = 135.4 – 1.4 = 134kg
mass of H2O in liquid = 135.4 + 846.4kg = 711kg
WASHED / DISCHARGED SLURRY (STREAM D):
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Total volume is unchanged (1000L)
Volume of weight and solids and liquids is constant throughout the process.
Volume of solids (stream D) = volume of solids (stream A) = 1354kg
Volume of liquids (stream D) = volume of liquids (stream A) = 729L
assume density of slurry is 1.002.
mD(liquid) = v x p (H2O) = 729 x 1.002 = 730 kg
xD/L(CuSO4) = 1.4/729 = 0.0019
FD = total mass (stream D) = mass (solids) + mass (solution) = 1354kg + 729kg = 2083kg
•
Total CuSO4 in stream D = 1% CuSO4 in stream A
= (0.01) (135.4) = 1.4kg
wD/L(CuSO4) = 1.4kg / 729 = 0.00192
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wD/L (H2O) = 1 - 0.00192 = 0.9981
wD/LCuSO4 x wD/L x FD = mass of FD CuSO4
(0.00192) (wD/L) (2083) = 1.4
wD/L = 0.35
wD(solid) = 1-0.35 = 0.65
FEED SLURRY (stream A):
FA = 2200kg (1000L)
wA(solid) = 0.6154
wA/L(CuSO4) = 0.16
WASHED / DISCHARGED SLURRY (stream D):
FD = 2083kg (1000L)
wD(solid) = 0.65
wD/L(CuSO4) = 0.00192
Figure 1 showing block flow diagram for washing leached CuSO4 from insoluble residue.
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INDIVIDUAL MASS BALANCES:
CuSO4 MASS BALANCE:
▪
M/S I: 135.4 + wB/L(CuSO4) (Fy) = wB/L (CuSO4) (Fz +Fb)
▪
M/S II: 0.001918(Fx) + wB/L(CuSO4) (Fb) = wB/L (CuSO4) (Fc + Fy)
▪
M/S III: wB/L (CuSO4) (Fc) = 0.001918(Fx) + 1.4
▪
OVERALL: 135.4 = 1.4 + wB/L (CuSO4) (Fx)
OVERALL MASS BALANCE:
▪
W + 846 = 730 + Fz
Let F = Vp = V (1+ wCuSO4)
Where V = 729L in each slurry
stream
Flow rate of stream X and Y = W
M/S I: 135.4 + wB/L (CuSO4) (W) (V (1+ wCuSO4)) = wB/L (CuSO4) (Fz)
M/S II: (0.00192) (W) (1.00192) + wB/L (CuSO4) (729) (1+ wCuSO4) = wB/L(CuSO4) (W + 729) (1+
wCuSO4)
M/S III: wB/L (CuSO4) (729) (1+ wCuSO4) = 1.4 + (0.00192) (W) (1.00192)
OVERALL CuSO4: 135.4 = 1.4 + wB/L (CuSO4) (Fx)
OVERALL: W + 116 = Fz
Let A = mass (volume) of wash water
B = wI/L (CuSO4)
C = wII/L (CuSO4)
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M/S I: 135.4 + A (C + C2) = B (A+116) + 729(B + B2)
M/S II: (0.00192) (A) + 729(B + B2) = (A + 729) (C + C2)
M/S III: 729(C + C2) = (0.001918) (A) + 1.4
OVERALL CuSO4: 134 = B (A + 116)
Solved using solver function in excel.
A = 3063L – volume of streams W, X, Y
B = 0,04215
C = 0,0099
➢ Mass of Wash Liquid (out):
M/S I: 3050L x (1 + 0,04215) = 3179kg
M/S II: 3063L x (1 + 0.0099) = 3093kg
M/S III: 3063L x 0.00192 = 3069kg
➢ Mass of Wash Liquid (in):
M/S I: 3179 x 0,04215 = 134kg
M/S II: 3093 x 0.0099 = 30.6kg
M/S III: 3069 x 0.00192 = 5.9kg
➢ Mass of Slurry Liquid (out):
M/S I: 729L x (1 + 0,04215) kg/L = 760kg
M/S II: 729L x (1 + 0.0099) kg/L = 736kg
M/S III: 729L x 0.00192kg/L = 730kg
➢ Mass of Slurry Liquid (in):
M/S I: 760 x 0,04215 = 32kg
M/S II: 736 x 0.0099 = 7.3kg
M/S III: 730 x 0.00192 = 1.4kg
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Table 2 showing mass balance for CuSO4 washing circuit (kg)
Mass Fraction of
CuSO4
Mass of Wash Liquid
(out)
Mass of Wash Liquid
(in)
Mass of Slurry Liquid
(out)
Mass of Slurry Liquid
(in)
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M/S I
0.04215
M/S II
0.00990
M/S III
0.00192
3179
3090
3069
134
30.6
5.9
760
736
730
32.0
7.3
1.4