Metallurgical Balances
Overview of 2-Product and 3-Product Formulas
Mineral Processing
• Goals (quality–quantity trade-off)
– To beneficiate an ore economically
i.e., to recover as much of the value as
possible to suitable products (or concentrates)
– To reduce the weight of material to be shipped
– To meet the customer quality requirements
Size of Mining Operations
- Production Rate Levels
•
•
•
•
Tonnes per day of ore
Tonnes per day of concentrate
Tonnes per day of contained metal
Terms used: high tonnage operation,
low tonnage operation
• Terms are different for different commodities
Concentration Ratio
• Tonnes of ore / tonnes of concentrate or product
Commodity
Coal
Iron Ore
Lead/Zinc
Nickel
Copper
Gold/Silver/PGMs
Diamonds
C.R.
1.15-1.30
1.02-2.50
10-50
20-100
20-200
50-500,000
1,0000-1,000,000
Metallurgical Balances
Massin + Accumulationprior = Massout + Accumulationnew
• What goes in must come out
–
–
–
–
Assume any material accumulation = 0
Steady-state mass balance
Account for accumulation
Dynamic analysis or non-steady-state balance
• You can't recover all the values
• You can't produce a perfect product
Metallurgical Balances
• Uses
– steady-state accounting of mass flows in a system
– evaluation of metallurgical testwork
– comparison of two different mills or circuits
– process control of an operating plant
• Properties of the Balance
– requires samples for assay and weights/flowrates
– accuracy of the assays used
– turnaround time of the assays
Metallurgical Balances
The method relies on equations and tables
Equations
F
Separation
Circuit
T
2-Product Formula
F=C+T
Ff = Cc + Tt
C
>>> C = F (f - t) / (c - t)
where F = feed tonnage rate or 100%
C = concentrate tonnage or weight%
T = tailing tonnage or weight%
and f, c, t = assay of each respective stream (%, g/t, ppm, etc.)
Two-Product Formula
There are 6 variables
Total Mass
Species Analysis
F
and
f
C
and
c
T
and
t
Step 1: Reduce number of variables to 5: set F = 100
Step 2: Measure the value of 3 variables
Step 3: Calculate the remaining two variables
Two-Product Formula
• The calculation can be done using the
formula or by simply filling in a Table.
• If f, c, and t are the three measured
variables, then the formula is used.
• If other variables are given, then the
Table is simply filled in one step at a time.
Metallurgical Balances
2-Product Formula Solution
C = 100 * (f-t)/(c-t)
%Recovery = 100 * c(f-t) /f(c-t)
f
The Metallurgical Balance Table
Product
Weight%
Assay (%)
Tailing
C
T
c
t
Feed
100
f
Concentrate
Units
Cc
Tt
100f
%Recovery
Cc/f
Tt/f
100
Two-Product Formula
Given the following three variables: All Assays
Calculate the Weight% of C: C = F * (f – t) / (c – t)
C = 100(1.29-0.072) / (26.9-0.072)
Product
Weight
Weight%
%Cu
Cu Units %Recovery
(tpd)
Concentrate 1,135
4.54
26.9
122.126
94.67
Tailing
23,865
95.46
0.072
6.873
5.33
------------------------------------------------------------------------------------------------------Feed
25,000
100.00
1.29
128.999
100.00
Two-Product Formula
Given the following four variables:
Product Weights
Product Assays
Product
Weight
Weight%
%Cu
Cu Units %Recovery
(g)
Concentrate 45.3
4.542
26.9
122.180
94.67
Tailing
952.1
95.458
0.072
6.873
5.33
-----------------------------------------------------------------------------------------------------Feed
997.4
100.00
(1.291)
129.053
100.00
Two-Product Formula
Given the following three variables:
%Recovery
%Cu in Concentrate
%Cu in Feed
Product
Weight
Weight%
%Cu
Cu Units %Recovery
(tpd)
Concentrate
1,135
4.540
26.9
122.124
94.67
Tailing
23,865
95.460
0.072
6.876
5.33
-----------------------------------------------------------------------------------------------------------------Feed
25,000
100.00
1.29
129.00
100.00
Metallurgical Balances
The method relies on equations and tables
Equations
F
Circuit 1
Circuit 2
T
3-Product Formula
C1
F = C1 + C2 + T
Ff1 = C1c11 + C2c21 + Tt1
Ff2 = C1c12 + C2c22 + Tt2
where
C2
F = feed tonnage rate or 100%
C1 and C2 = concentrate1 and 2 tonnage or weight%
T = tailing tonnage or weight%
and f1, c11 , c21 , t1 = element1 stream assay (%, g/t, ppm, etc.)
f2, c12 , c22 , t2 = element2 stream assay (%, g/t, ppm, etc.)
Three-Product Formula
Matrix Solution
C1
C2
T
F
1.0
1.0
1.0
100
c11
c21
t1
f1
c12
c22
t2
f2
http://www.agdconsulting.ca/MatrixMethod.pdf
Three-Product Formula
Solution
C1 = 100(f1 - t1) (c22 - t2) - (f2 - t2) (c21 - t1)
(c11 - t1)(c22 - t2) - (c12 - t2)(c21 - t1)
C2 = 100(f1 - t1) (c12 - t2) - (f2 - t2) (c11 - t1)
(c21 - t1)(c12 - t2) - (c22 - t2)(c11 - t1)
T = 100 - C1 - C2
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
2-Product Formula
to solve in stages
Circuit 1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
C1
Circuit 2
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
We need the assays of the intermediate product T1 (t11 and t12)
Step 1
Ff1 = C1c11 + T1t11
100 = C1 + T1
Step 2
then
and
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
T1t12 = C2c22 + Tt2
T1 = C2 + T
Check that the assays of element 2 in Circuit 1 are balanced
Check that the assays of element 1 in Circuit 2 are balanced
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
Ff1 = C1c11 + T1t11
100 = C1 + T1
So C1 = 100(1.29 - 0.104) / (26.9 - 0.104)
Weight%
4.426
95.574
100.00
Circuit 2
t11 = 0.104 %Cu
t12 = 4.15 %Zn
First step:
Stream
C1
T1
F
T1
C1
%Cu
26.9
0.104
1.29
C2
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Cu units %Recovery
119.059
92.29
9.940
7.71
128.999
100.00
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
Check assays of T1 balance
1. Zn assays in Circuit 1
Stream
C1
T1
F
Weight%
4.426
95.574
100.00
T1
Circuit 2
t11 = 0.104 %Cu
t12 = 4.15 %Zn
C1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
%Zn
9.25
(4.092)
4.32
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Zn units %Recovery
40.940
9.477
391.060
90.523
432.000
100.000
So the Zn assay of T1 must be adjusted by -0.058 %Zn
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
C1
T1t12 = C2c22 + Tt2
T1 = C2 + T
So C2 = 95.574(4.092 - 0.342) / (57.7 - 0.342)
Weight%
6.249
89.325
95.574
* with respect to F (4.32 %Zn)
Circuit 2
t11 = 0.104 %Cu
t12 = 4.092 %Zn
Second step:
Stream
C2
T
T1
T1
%Zn
57.7
0.342
4.092
C2
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Zn units %Recovery*
360.570
83.465
30.550
7.072
391.120
90.537
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
Check assays of T1 balance
2. Cu assays in Circuit 2
Stream
C2
T
T1
Weight%
6.249
89.325
95.574
T1
Circuit 2
t11 = 0.104 %Cu
t12 = 4.092 %Zn
C1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
%Cu
1.10
0.072
(0.139)
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Cu units %Recovery
6.870
5.326
6.430
4.984
13.300
10.310
So the Cu assay of T1 must be adjusted by +0.035 %Cu
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
Ff1 = C1c11 + T1t11
c
100 = C1 + T1
c
So C1 = 100(1.29 - 0.139) / (26.9 - 0.139)
C1
= 26.9 %Cu
12 = 9.25 %Zn
11
Weight%
4.301
95.699
100.00
Circuit 2
= 0.139%Cu
%Cu
11 0.139
t11t=
t = 4.09 %Zn
t12 12
= 4.092 %Zn
Return to Circuit 1:
Stream
C1
T1
F
T1
%Cu
26.9
0.139
1.29
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Cu units %Recovery
115.700
89.69
13.300
10.31
129.000
100.00
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
Recheck assays of T1 balance
1. Zn assays in Circuit 1
Stream
C1
T1
F
Weight%
4.301
95.699
100.00
T1
Circuit 2
t11 = 0.139 %Cu
t12 = 4.092 %Zn
C1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
%Zn
9.25
(4.098)
4.32
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Zn units %Recovery
39.780
9.208
392.220
90.792
432.000 100.000
So the Zn assay of T1 must be adjusted by +0.006 %Zn
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
C1
T1t12 = C2c22 + Tt2
c = 26.9 %Cu
T1 = C2 + T
c = 9.25 %Zn
So C2 = 95.699(4.098-0.342) / (57.7 - 0.342)
11
12
Weight%
6.267
89.432
95.699
* with respect to F (4.32 %Zn)
Circuit 2
%Zn
57.7
0.342
4.098
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
t11 = 0.139 %Cu
t12 = 4.098 %Zn
Return to Circuit 2:
Stream
C2
T
T1
T1
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Zn units %Recovery*
361.610
83.706
30.590
7.081
392.200
90.787
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
Check assays of T1 balance
2. Cu assays in Circuit 2
Stream
C2
T
T1
Weight%
6.267
89.432
95.699
T1
Circuit 2
t11 = 0.139 %Cu
t12 = 4.098 %Zn
C1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
%Cu
1.10
0.072
(0.139)
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Cu units %Recovery
6.890
5.341
6.440
4.994
13.330
10.330
So the Cu assay of T1 requires no further adjustment
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Circuit 1
T1
Circuit 2
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
t11 = 0.139 %Cu
t12 = 4.098 %Zn
Final Results:
Stream
C1
C2
T
F
Wt%
4.301
6.267
89.432
100.00
C1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Assays
Units
%Recovery
%Cu %Zn
Cu
Zn
Cu
Zn
26.9 9.25 115.70 39.78 89.69
9.21
1.10 57.7
6.89 361.61 5.34
83.71
0.072 0.342 6.44 30.59 4.99
7.08
1.29 4.32 129.03 431.98 100.03 100.00
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Three Product Results:
Circuit 1
C1
C2
T
F
Wt%
4.300
6.268
89.433
100.00
Circuit 2
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
t11 = 0.139 %Cu
t12 = 4.098 %Zn
C1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
Stream
T1
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Assays
Units
%Recovery
%Cu %Zn
Cu
Zn
Cu
Zn
26.9 9.25 115.67 39.77 89.66
9.21
1.10 57.7
6.89 361.64 5.34
83.71
0.072 0.342 6.44 30.59 4.99
7.08
1.29 4.32 129.00 432.00 100.00 100.00
Metallurgical Balances
Let’s examine application of the 2-product formula
to a 3-product circuit
Equations
F
f1 = 1.29 %Cu
f2 = 4.32 %Zn
Difference between
2-Product and 3-Product:
Stream
C1
C2
T
F
Wt%
0.001
0.001
0.001
0.00
Circuit 1
T1
Circuit 2
t11 = 0.139 %Cu
t12 = 4.098 %Zn
C1
c11 = 26.9 %Cu
c12 = 9.25 %Zn
C2
c21 = 1.10 %Cu
c22 = 57.7 %Zn
Assays
Units
%Cu %Zn
Cu
Zn
0.0 0.00 0.03 0.01 0.03
0.00 0.0
0.00 0.03 0.00
0.000 0.000 0.00 0.00 0.00
0.00 0.00 0.03 0.02 0.03
T
t1 = 0.072 %Cu
t2 = 0.342 %Zn
%Recovery
Cu
Zn
0.00
0.00
0.00
0.00
Metallurgical Balances
The method relies on equations and tables
Equations
F
Bulk
T
3-Product Formula
F = C1 + C2 + T
Ff1 = C1c11 + C2c21 + Tt1
Ff2 = C1c12 + C2c22 + Tt2
Bulk
Conc
Selective
C2
C1
where
F = feed tonnage rate or 100%
C1 and C2 = concentrate1 and 2 tonnage or weight%
T = tailing tonnage or weight%
and f1, c11 , c21 , t1 = element1 stream assay (%, g/t, ppm, etc.)
f2, c12 , c22 , t2 = element2 stream assay (%, g/t, ppm, etc.)
Three-Product Formula
Feed = Conc1 + Conc2 + Tailing
Total Mass: F = C1
unknowns
+ C2
+T
4
Element 1: Ff1 = C1c11 + C2c21 + Tt1
+ 4 unknowns
Element 2: Ff2 = C1c12 + C2c22 + Tt2
+ 4 unknowns
= 12 unknowns
Let F = 100
- 1 variable
Measure: f1, c11, c21, t1, f2, c12, c22, t2
- 8 variables
------------------------------------------------------------------------------------= 3 unknowns
-------------------------------------------------------------------------------------
Three-Product Formula
Solution
C1 = 100(f1 - t1) (c22 - t2) - (f2 - t2) (c21 - t1)
(c11 - t1)(c22 - t2) - (c12 - t2)(c21 - t1)
C2 = 100(f1 - t1) (c12 - t2) - (f2 - t2) (c11 - t1)
(c21 - t1)(c12 - t2) - (c22 - t2)(c11 - t1)
T = 100 - C1 - C2
Three-Product Formula
Problem: The three equations must be independent.
Element 1 & Element 2 and total mass must be independent.
So:
if f1 is similar to c11 is similar to c21 is similar to t1
then Element 1 equation is same as Total Mass equation.
And: if Element 1 is associated with Element 2 (Ag dissolved in Cu mineral)
then Element 1 equation is same as Element 2 equation.
Association may be due to interlocked minerals.
The 3 equations are reduced to 2 and an incorrect solution is obtained.
An answer may be obtained, but it is likely wrong.
Three-Product Formula
An answer may be obtained because of Measurement Errors from:
1.
2.
3.
4.
Sampling;
Sample preparation;
Contamination; and/or
Non-steady-state conditions in the process plant during sampling.
Analytical lab results (assays) are usually very accurate, although
mistakes do occur and "strange" assays can occur.
It is O.K. to question the lab results, but not too often.
Metallurgical Balances
Process Disturbances can cause variations in sampling results
• Mineralogy changes (quality & quantity)
• Liberation changes (locking characteristics)
• Particle size changes (coarse and ultra-fines)
• Water chemistry changes (pH and ions and S.S.)
• Process control of flow rates
• Reagent addition control (quantity & quality)
• Poor house-keeping issues
• Equipment mal-functions
• Planned maintenance interruptions
• Temperature and pressure changes
• Moisture changes
What is a sample?
• All of the stream / part of the time
– Moving gang sample splitter
– Voting in an election
• Part of the stream / all of the time
– Thief sample
– Conducting regular polls (daily, monthly, yearly)
Cross-Belt Sample Splitter
Moving Gang Sampler
Two-stage Continuous Flow
Thief Sampler
Rotary Sample Splitter
Metallurgical Balances - Sampling
First step
What is the purpose of the sample and the balance?
• Evaluation of lab testwork
• Evaluation of plant testwork
• Accounting purposes
• Process control
Two important sampling issues:
• Accuracy (representativeness and processing)
• Turn-around time
Metallurgical Balances - Sampling
Second step
How should the sample be taken?
• Grab samples
• Composite samples
• Method used to obtain the sample
• Sampling pulp or water vs. sampling solids
Two sampling issues:
• Manual techniques – proper training
• Automated methods – proper maintenance
Metallurgical Balances - Sampling
Third step
How should the sample be prepared?
• Sub-sampling (riffle-splitting)
• Cone & quartering
• Dewatering
• Weighing
• Size-reduction
Two sampling issues:
• Retention of sample make-up
• Avoiding contamination
Metallurgical Balances - Sampling
Fourth step
How should the sample be assayed and stored?
• Assay tolerances
• Duplicates or triplicates
• Automated (self-assayed)
• Use of an assay lab
• Method of assay (A.A., XRF, GC, fire-assay, etc.)
Two assaying issues:
• Sample retention for future examination
• Sample degradation (oxidation/moisture pick-up)
Metallurgical Balances - Sampling
Fifth step
How should the results be reported?
• Qualified person (public release)
• Chain of custody issues (samples and data)
• Some samples submitted as blanks & surrogates
Two reporting issues:
• Security of data
• Reliability of results and interpretation
Metallurgical Balances - Sampling
Analytical Errors:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Sampling Errors
Sample Preparation Errors
Assay Errors
Human Communication Errors
Weighing Errors
Noisy Data Errors
Unstable Process Errors
Time Delay Errors
Particle Size Errors
Metallurgical Balances - Sampling
Minimize the Impact of Errors
Sampling
- sample part of the stream, all of the time
- sample all of the stream, part of the time
- ensure cross sample contamination cannot occur
- ensure pulp sampler does not overflow
- ensure that segregation of particles does not occur
Assaying
- different labs may produce different results
- a well-run lab does not make many mistakes
- assay involves at least three sub-samples
- agreement must meet rigid variance standards
Metallurgical Balances - Sampling
Minimize the Impact of Errors
Sample Preparation (for the Assay Lab)
- Samples must be filtered and dried and recovered
- Samples must be "bucked"
- Samples must be less than 100 microns in size
- Samples must be bagged and properly labelled
- Most cross-contamination occurs at this stage
Human Communication
- Mistakes on where sample has been taken
- Mistakes on how sample was prepared
- Mistakes in reporting results
- "Rush" samples can lead to poor quality
Metallurgical Balances - Sampling
Minimize the Impact of Errors
Weighing Errors
- part of sample is lost during processing and/or testing
- calibration of instruments not done well
- in lab, tare weights must be properly accounted for
- improper dewatering
- improper compositing
Process Issues
- unbalanced dynamic effects
- steady-state balances can be done, but are meaningless
- inaccurate sampling may result
- on-line assays are timely, but less-accurate
Metallurgical Balances - Sampling
Minimize the Impact of Errors
Particle Size Errors
- Low-grade gold ores…the “Nugget” effect
- Coarse size distributions lead to settling and segregation
- Non-representative samples
- Samples must be reduced in size for assaying
- Ultra-fines may require Cyclosizer analysis
Reporting Issues
- In production accounting, material must be written-off
- Errors accumulate due to moisture pick-up and losses
- Stockpiles must be accurately measured and sampled
- Sampling railcars is an art-form
Questions?
Metallurgical Balances