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Metallurgical Balances

Metallurgical Balances. Overview of 2-Product and 3-Product Formulas. 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 operation plant

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Metallurgical Balances

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

  2. 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 operation plant • Properties of the Balance • requires samples for assay and weights/flowrates • accuracy of the assays used • turnaround time of the assays

  3. 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.)

  4. 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 by setting F = 100 Step 2: Obtain the value of three variables Step 3: Calculate the remaining two variables

  5. 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

  6. 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

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

  8. 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

  9. 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

  10. T C2 Metallurgical Balances The method relies on equations and tables Equations 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 = stream assay for element 1 (%, g/t, ppm, etc.) f2, c12 , c22 , t2 = stream assay for element 2 (%, g/t, ppm, etc.) C1

  11. T C2 Metallurgical Balances Let’s examine application of the 2-product formula Equations 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

  12. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations 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

  13. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations 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 Zn units %Recovery C1 4.426 9.25 40.9409.477 T1 95.574 (4.092) 391.06090.523 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 = 2.23 %Cu c22 = 57.7 %Zn

  14. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations 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 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 (391.090) (90.530) * with respect to F (4.32 %Zn) t12 = 4.092 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  15. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations 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 = 4.40 %Zn t12 = 4.092 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  16. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations 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 t11 = 0.139 %Cu 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 t12 = 4.092 %Zn c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn C1

  17. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations 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 Zn units %Recovery C1 4.301 9.25 39.7809.208 T1 95.699 (4.098) 392.22090.792 F 100.00 4.32 432.000100.000 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 = 2.23 %Cu c22 = 57.7 %Zn

  18. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations F f1 = 1.29 %Cu f2 = 4.32 %Zn t1 = 0.072 %Cu t2 = 0.342 %Zn t11 = 0.139 %Cu t12 = 4.098 %Zn t11 = 0.104 %Cu t12 = 4.15 %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 (392.175) (90.781) * with respect to F (4.32 %Zn) C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  19. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations 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 t11 = 0.104 %Cu t12 = 4.40 %Zn t12 = 4.092 %Zn C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  20. T C2 Metallurgical Balances Let’s examine application of the 2-product formula T1 Equations 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 t11 = 0.139 %Cu t12 = 4.098 %Zn 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 6.267 1.10 57.7 6.89 361.61 5.34 83.71 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 C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  21. T C2 Metallurgical Balances Let’s examine application of the 3-product formula T1 Equations 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 t11 = 0.139 %Cu t12 = 4.098 %Zn Three Product Results: Stream Weight% %Cu %Zn Units %Recovery Cu Zn Cu Zn C1 4.300 26.9 9.25 115.67 39.77 89.66 9.21 C2 6.268 1.10 57.7 6.89 361.64 5.34 83.71 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 C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  22. T C2 Metallurgical Balances Let’s examine application of the 3-product formula T1 Equations 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 t11 = 0.139 %Cu t12 = 4.098 %Zn Difference between 2-Product and 3-Product: Stream Weight% %Cu %Zn Units %Recovery 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 C1 c11 = 26.9 %Cu c12 = 9.25 %Zn c21 = 1.10 %Cu c22 = 57.7 %Zn

  23. Metallurgical Balances The method relies on equations and tables Equations 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 = stream assay for element 1 (%, g/t, ppm, etc.) f2, c12 , c22 , t2 = stream assay for element 2 (%, g/t, ppm, etc.) Bulk Conc C2 C1

  24. 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 -------------------------------------------------------------------------------------

  25. 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

  26. Three-Product Formula Problem: 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.

  27. 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.

  28. Metallurgical Balances • 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

  29. Metallurgical Balances • 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

  30. Metallurgical Balances • Second step • How should the sample be taken? • Grab samples • Composite samples • Method used • Water versus solids • Two sampling issues: • Manual techniques – proper training • Automated methods – proper maintenance

  31. Metallurgical Balances • 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

  32. Metallurgical Balances • 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)

  33. Metallurgical Balances • 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 data • Reliability of results and interpretation

  34. Metallurgical Balances • 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

  35. Metallurgical Balances 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

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

  37. Metallurgical Balances 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

  38. Metallurgical Balances 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

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