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Metallurgical Balances

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)

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Metallurgical Balances

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  1. Metallurgical Balances Overview of 2-Product and 3-Product Formulas

  2. 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

  3. 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

  4. Concentration Ratio • Tonnes of ore / tonnes of concentrate or product Commodity C.R. Coal 1.15-1.30 Iron Ore 1.02-2.50 Lead/Zinc 10-50 Nickel 20-100 Copper 20-200 Gold/Silver/PGMs 50-500,000 Diamonds 1,0000-1,000,000

  5. 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

  6. 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

  7. Separation Circuit T F C Metallurgical Balances The method relies on equations and tables Equations 2-Product Formula F = C + T Ff = Cc + Tt 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.) >>> C = F (f - t) / (c - t)

  8. 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

  9. 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.

  10. Metallurgical Balances 2-Product Formula Solution C = 100 * (f-t)/(c-t) %Recovery = 100 * c(f-t) /f(c-t) The Metallurgical Balance Table f Units Assay (%) %Recovery Product Weight% c C Cc Cc/f Concentrate t T Tt Tt/f Tailing 100f f 100 100 Feed

  11. Two-Product Formula Given the following three variables: All Assays Calculate the Weight% of C: C = F * (f – t) / (c – t) 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 C = 100(1.29-0.072) / (26.9-0.072)

  12. 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

  13. 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

  14. Circuit 2 T C2 Metallurgical Balances The method relies on equations and tables Equations Circuit 1 F 3-Product Formula F = C1 + C2 + T Ff1 = C1c11 + C2c21 + Tt1 Ff2 = 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 = element1 stream assay (%, g/t, ppm, etc.) f2, c12 , c22 , t2 = element2 stream assay (%, g/t, ppm, etc.) C1

  15. 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

  16. 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

  17. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn 2-Product Formula to solve in stages We need the assays of the intermediate product T1 (t11 and t12) Step 1 Step 2 Ff1 = C1c11 + T1t11 then T1t12 = C2c22 + Tt2 100 = C1 + T1 and 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 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn C1 T1

  18. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.104 %Cu t12 = 4.15 %Zn First step: Ff1 = C1c11 + T1t11 100 = C1 + T1 So 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 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn C1

  19. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn Check assays of T1 balance 1. Zn assays in Circuit 1 Stream Weight% %Zn Znunits %Recovery C1 4.426 9.25 40.940 9.477 T1 95.574 (4.092) 391.060 90.523 F 100.00 4.32 432.000 100.000 So the Zn assay of T1 must be adjusted by -0.058 %Zn t11 = 0.104 %Cu t12 = 4.15 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  20. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.104 %Cu Second step: T1t12 = C2c22 + Tt2 T1 = C2 + T So C2 = 95.574(4.092 - 0.342) / (57.7 - 0.342) Stream Weight% %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 * with respect to F (4.32 %Zn) t12 = 4.092 %Zn c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn C1

  21. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn Check assays of T1 balance 2. Cu assays in Circuit 2 Stream Weight% %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 10.310 So the Cu assay of T1 must be adjusted by +0.035 %Cu t11 = 0.104 %Cu t12 = 4.092 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  22. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.139 %Cu t12 = 4.09 %Zn Return to Circuit 1: Ff1 = C1c11 + T1t11 100 = C1 + T1 So 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 t11 = 0.139 %Cu t12 = 4.092 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  23. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn Recheck assays of T1 balance 1. Zn assays in Circuit 1 Stream Weight% %Zn Znunits %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 t11 = 0.139 %Cu t12 = 4.092 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  24. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn Return to Circuit 2: T1t12 = C2c22 + Tt2 T1 = C2 + T So C2 = 95.699(4.098-0.342) / (57.7 - 0.342) Stream Weight% %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 * with respect to F (4.32 %Zn) t11 = 0.139 %Cu t12 = 4.098 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  25. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn Check assays of T1 balance 2. Cu assays in Circuit 2 Stream Weight% %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 10.330 So the Cu assay of T1 requires no further adjustment t11 = 0.139 %Cu t12 = 4.098 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  26. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn Final Results: Stream Wt% Assays Units %Recovery %Cu %Zn Cu Zn Cu Zn C1 4.301 26.9 9.25 115.70 39.78 89.699.21 C2 6.267 1.10 57.7 6.89 361.61 5.3483.71 T 89.432 0.072 0.342 6.44 30.59 4.997.08 F 100.00 1.29 4.32 129.03 431.98 100.03100.00 t11 = 0.139 %Cu t12 = 4.098 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  27. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn Three Product Results: Stream Wt% Assays Units %Recovery %Cu %Zn Cu Zn Cu Zn C1 4.300 26.9 9.25 115.67 39.77 89.669.21 C2 6.268 1.10 57.7 6.89 361.64 5.3483.71 T 89.433 0.072 0.342 6.44 30.59 4.997.08 F 100.00 1.29 4.32 129.00 432.00 100.00100.00 t11 = 0.139 %Cu t12 = 4.098 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  28. Circuit 2 T C2 Metallurgical Balances Let’s examine application of the 2-product formula to a 3-product circuit T1 Equations Circuit 1 F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn Difference between 2-Product and 3-Product: Stream Wt% Assays Units %Recovery %Cu %Zn Cu Zn Cu Zn C1 0.001 0.0 0.00 0.03 0.01 0.03 0.00 C2 0.001 0.00 0.0 0.00 0.03 0.00 0.00 T 0.001 0.000 0.000 0.00 0.00 0.00 0.00 F 0.00 0.00 0.00 0.03 0.02 0.03 0.00 t11 = 0.139 %Cu t12 = 4.098 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  29. Metallurgical Balances The method relies on equations and tables Equations Bulk T F 3-Product Formula F = C1 + C2 + T Ff1 = C1c11 + C2c21 + Tt1 Ff2 = 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 = element1 stream assay (%, g/t, ppm, etc.) f2, c12 , c22 , t2 = element2 stream assay (%, g/t, ppm, etc.) Bulk Conc Selective C2 C1

  30. Three-Product Formula Feed = 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 -------------------------------------------------------------------------------------

  31. 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

  32. 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.

  33. Three-Product Formula An answer may be obtained because of Measurement Errors from: • 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.

  34. 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

  35. 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)

  36. Cross-Belt Sample Splitter

  37. Moving Gang Sampler

  38. Two-stage Continuous Flow Thief Sampler

  39. Rotary Sample Splitter

  40. 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

  41. 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

  42. 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

  43. 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)

  44. 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

  45. Metallurgical Balances - Sampling • Analytical Errors: • Sampling Errors • Sample Preparation Errors • Assay Errors • Human Communication Errors • Weighing Errors • Noisy Data Errors • Unstable Process Errors • Time Delay Errors • Particle Size Errors

  46. 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

  47. 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

  48. 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

  49. 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

  50. Questions?

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