akm metallurgical balances

Metallurgical BalancesOverview of 2-Product and 3-Product Formulas

Metallurgical Balances

Usessteady-state accounting of mass flows in a systemevaluation of metallurgical testworkcomparison of two different mills or circuitsprocess control of an operation plant

Properties of the Balancerequires samples for assay and weights/flowratesaccuracy of the assays usedturnaround time of the assays

Metallurgical BalancesThe method relies on equations and tablesEquations2-Product Formula

F = C + TFf = Cc + Tt

whereF = 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 FormulaThere are 6 variables Total MassSpecies AnalysisFandfCandcTandtStep 1:Reduce number of variables to 5 by setting F = 100Step 2:Obtain the value of three variablesStep 3:Calculate the remaining two variables

Two-Product FormulaThe 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

Metallurgical BalancesC = 100 * (f-t)/(c-t)

%Recovery = 100 * c(f-t) /f(c-t)2-Product Formula SolutionThe Metallurgical Balance TableWeight%Assay (%)Units%RecoveryProductConcentrateTailingFeedT100CcftfTt100f100CcCc/fTt/f

Two-Product FormulaProduct WeightWeight%%Cu Cu Units %Recovery (tpd)Concentrate 1,135 4.5426.9 122.126 94.67Tailing 23,865 95.460.072 6.873 5.33------------------------------------------------------------------------------------------------------------------Feed 25,000 100.00 1.29 128.999 100.00Given the following three variables:All AssaysCalculate the Weight% of C: C = F * (f t) / (c t)C = 100(1.29-0.072) / (26.9-0.072)

Two-Product FormulaProduct WeightWeight%%Cu Cu Units %Recovery (g)Concentrate 45.3 4.54226.9 122.180 94.67Tailing 952.1 95.4580.072 6.873 5.33------------------------------------------------------------------------------------------------------------------Feed 997.4 100.00 (1.291) 129.053 100.00Given the following four variables:Product WeightsProduct Assays

Two-Product FormulaProduct WeightWeight%%Cu Cu Units %Recovery (tpd)Concentrate 1,135 4.54026.9 122.124 94.67Tailing 23,865 95.4600.072 6.876 5.33------------------------------------------------------------------------------------------------------------------Feed 25,000 100.00 1.29 129.00 100.00Given the following three variables:%Recovery%Cu in Concentrate%Cu in Feed

Metallurgical BalancesThe method relies on equations and tablesEquationsFC13-Product Formula

F = C1 + C2 + TFf1 = C1c11 + C2c21 + Tt1Ff2 = C1c12 + C2c22 + Tt2

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 = stream assay for element 1 (%, g/t, ppm, etc.) f2, c12 , c22 , t2 = stream assay for element 2 (%, g/t, ppm, etc.)

Metallurgical BalancesLets examine application of the 2-product formulaEquationsFC1

2-Product Formulato solve in stages

We need the assays of the intermediate product T1 (t11 and t12)Step 1 Step 2Ff1 = C1c11 + T1t11 then T1t12 = C2c22 + Tt2100 = C1 + T1 and T1 = C2 + T

Check that the assays of element 2 in Circuit 1 are balancedCheck that the assays of element 1 in Circuit 2 are balancedT1f1 = 1.29 %Cuf2 = 4.32 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 1.10 %Cuc22 = 57.7 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn

Metallurgical BalancesLets examine application of the 2-product formulaEquationsFC1First step:

Ff1 = C1c11 + T1t11100 = C1 + T1So C1 = 100(1.29 - 0.104) / (26.9 - 0.104)

Stream Weight% %Cu Cu units%Recovery C1 4.426 26.9 119.059 92.29 T1 95.574 0.104 9.940 7.71 F 100.00 1.29 128.999 100.00

T1f1 = 1.29 %Cuf2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.104 %Cu t12 = 4.15 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 1.10 %Cuc22 = 57.7 %Zn


Check assays of T1 balance

1. Zn assays in Circuit 1

StreamWeight%%Zn Zn units%Recovery C1 4.426 9.25 40.940 9.477 T1 95.574 (4.092) 391.060 90.523 F 100.00 4.32 432.000 100.000So the Zn assay of T1 must be adjusted by -0.058 %ZnT1f1 = 1.29 %Cuf2 = 4.32 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 2.23 %Cuc22 = 57.7 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.104 %Cu t12 = 4.15 %Zn

Metallurgical BalancesLets examine application of the 2-product formulaEquationsFC1Second step:

T1t12 = C2c22 + Tt2 T1 = C2 + TSo C2 = 95.574(4.092 - 0.342) / (57.7 - 0.342)

StreamWeight%%Zn Zn units%Recovery* C2 6.249 57.7 360.570 83.465 T 89.325 0.342 30.550 7.072 T1 95.574 4.092 391.120 90.537 (391.090) (90.530)* with respect to F (4.32 %Zn)T1f1 = 1.29 %Cuf2 = 4.32 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 1.10 %Cuc22 = 57.7 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.104 %Cu t12 = 4.15 %Znt12 = 4.092 %Zn

t11 = 0.104 %Cu t12 = 4.40 %ZnMetallurgical BalancesLets examine application of the 2-product formulaEquationsFC1


2. Cu assays in Circuit 2

StreamWeight%%Cu Cu units%Recovery C2 6.249 1.10 6.870 5.326 T 89.325 0.072 6.430 4.984 T1 95.574(0.139) 13.300 10.310So the Cu assay of T1 must be adjusted by +0.035 %CuT1f1 = 1.29 %Cuf2 = 4.32 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 1.10 %Cuc22 = 57.7 %Zn t1 = 0.072 %Cu t2 = 0.342 %Znt12 = 4.092 %Zn

Metallurgical BalancesLets examine application of the 2-product formulaEquationsFC1Return to Circuit 1:

Ff1 = C1c11 + T1t11100 = C1 + T1So C1 = 100(1.29 - 0.139) / (26.9 - 0.139)

Stream Weight% %Cu Cu units%Recovery C1 4.301 26.9 115.700 89.69 T1 95.699 0.139 13.300 10.31 F 100.00 1.29 129.000 100.00

T1f1 = 1.29 %Cuf2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.139 %Cu t12 = 4.09 %Znt11 = 0.139 %Cuc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 1.10 %Cuc22 = 57.7 %Znt12 = 4.092 %Zn


Recheck assays of T1 balance

1. Zn assays in Circuit 1

StreamWeight%%Zn Zn units%Recovery C1 4.301 9.25 39.780 9.208 T1 95.699 (4.098) 392.220 90.792 F 100.00 4.32 432.000 100.000

So the Zn assay of T1 must be adjusted by +0.006 %Zn T1f1 = 1.29 %Cuf2 = 4.32 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 2.23 %Cuc22 = 57.7 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.139 %Cu t12 = 4.092 %Zn

Metallurgical BalancesLets examine application of the 2-product formulaEquationsFC1Return to Circuit 2:

T1t12 = C2c22 + Tt2 T1 = C2 + TSo C2 = 95.699(4.098-0.342) / (57.7 - 0.342)

StreamWeight%%Zn Zn units%Recovery* C2 6.267 57.7 361.610 83.706 T 89.432 0.342 30.590 7.081 T1 95.699 4.098 392.200 90.787 (392.175) (90.781)* with respect to F (4.32 %Zn)T1f1 = 1.29 %Cuf2 = 4.32 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 1.10 %Cuc22 = 57.7 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.104 %Cu t12 = 4.15 %Zn t11 = 0.139 %Cu t12 = 4.098 %Zn

t11 = 0.104 %Cu t12 = 4.40 %ZnMetallurgical BalancesLets examine application of the 2-product formulaEquationsFC1


2. Cu assays in Circuit 2

StreamWeight%%Cu Cu units%Recovery C2 6.267 1.10 6.890 5.341 T 89.432 0.072 6.440 4.994 T1 95.699(0.139) 13.330 10.330So the Cu assay of T1 requires no further adjustmentT1f1 = 1.29 %Cuf2 = 4.32 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 1.10 %Cuc22 = 57.7 %Zn t1 = 0.072 %Cu t2 = 0.342 %Znt12 = 4.092 %Zn t11 = 0.139 %Cu


Final Results:

Stream Weight% %Cu %Zn Units %Recovery Cu Zn Cu Zn C1 4.301 26.9 9.25 115.70 39.78 89.69 9.21 C2 6.2671.10 57.7 6.89 361.61 5.34 83.71 T 89.432 0.072 0.342 6.44 30.59 4.99 7.08 F 100.00 1.29 4.32 129.03 431.98 100.03 100.00 T1f1 = 1.29 %Cuf2 = 4.32 %Znc11 = 26.9 %Cuc12 = 9.25 %Znc21 = 1.10 %Cuc22 = 57.7 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.104 %Cu t12 = 4.15 %Zn t11 = 0.139 %Cu t12 = 4.098 %Zn


Three Product Results:



Difference between2-Product and 3-Product:


C1C2Metallurgical BalancesThe method relies on equations and tablesEquationsBulkConcTF3-Product Formula

F = C1 + C2 + TFf1 = C1c11 + C2c21 + Tt1Ff2 = C1c12 + C2c22 + Tt2

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 = stream assay for element 1 (%, g/t, ppm, etc.) f2, c12 , c22 , t2 = stream assay for element 2 (%, g/t, ppm, etc.)

Three-Product FormulaFeed = Conc1 + Conc2 + Tailing

Total Mass: F = C1 + C2 + T 4 unknowns

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 FormulaSolution

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 FormulaProblem: Equations for 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 minerals) 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

The reason an answer may be obtained is because of measurement errors from

Sampling, Sample preparation, Contamination, and Non-steady-state conditions in the process during sampling .

Analytical lab results are usually very accurate, although mistakes do occur and "strange" assays can occur.


Process Disturbances leading to variation in 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


First step

What is the purpose of the sample and the balance?Evaluation of lab testworkEvaluation of plant testwork Accounting purposesProcess control

Two important sampling issues:

Accuracy (representativeness and processing)Turn-around time


Second step

How should the sample be taken?Grab samplesComposite samplesMethod usedWater versus solids

Two sampling issues:

Manual techniques proper trainingAutomated methods proper maintenance


Third step

How should the sample be prepared?Sub-sampling (riffle-splitting)Cone & quarteringDewateringWeighingSize reduction

Two sampling issues:

Retention of sample make-upAvoiding contamination


Fourth step

How should the sample be assayed and stored?Assay tolerancesDuplicates or triplicatesAutomated (self-assayed)Use of an assay labMethod of assay (A.A., XRF, GC, fire-assay, etc.)

Two assaying issues:

Sample retention for future examinationSample degradation (oxidation/moisture pick-up)


Fifth step

How should the results be reported?

Qualified person (public release)Chain of custody issues (samples and data)Some samples submitted as blanks and surrogates

Two reporting issues:

Security of dataReliability of results and interpretation


Analytical Errors:

Sampling ErrorsSample Preparation ErrorsAssay ErrorsHuman Communication ErrorsWeighing ErrorsNoisy Data ErrorsUnstable Process ErrorsTime Delay ErrorsParticle Size Errors


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



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 is taken- Mistakes on how sample is prepared- Mistakes in reporting results- Rush samples can lead to poor quality



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



Particle Size Errors- Low-grade gold oresthe 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

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akm metallurgical balances

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