- To participate in the 911Metallurgist Forums, be sure to JOIN & LOGIN
- Use Add New Topic to ask a New Question/Discussion about Flotation.
- OR Select a Topic that Interests you.
- Use Add Reply = to Reply/Participate in a Topic/Discussion (most frequent).
Using Add Reply allows you to Attach Images or PDF files and provide a more complete input. - Use Add Comment = to comment on someone else’s Reply in an already active Topic/Discussion.
Laboratory Batch Flotation TestsRecovery (6 replies)
Any comments regarding quality control around lab testing programs and use of raw and adjusted data, e.g., should you consider discarding / repeating a test if change / error is greater than X %?
Since many years now we utilise on routine the BILCO software to make mass and metallurgical balances in lab flotation test. Or any other lab or pilot plant as well. Very often we do have an excess of data so such software is extremely powerful.
Would you care to comment on the size of the error (or default rate) that you use to screen out problematic tests and avoid arriving at false conclusions?
First of all I want to say that these notes are based on readings from the book Barry A. Wills, Mineral Processing Technology, ISBN 0-7506-2838-3 (I recommend to buy this book) and from my experience with actual plant and laboratory data
• If you would like to see the opinion about the recovery equations please see section 1 of these notes
• If you want to see how to determine errors please skip this section and go to section 2 of these notes
• If you want to see an example please go to section 3 of these notes
Section 1: Recovery equations in a separator that produces two products
All the Equations shown on
are equivalent and belong to the same system mass balance where a weight in the feed F is fed to a separation system at a certain grade f to produce a concentrate mass C at a grade c and a tail weight T at a grade t; where c, f and t are grades of the same element.
This can be demonstrated by knowing that the recovery of a certain element is calculated using
R = 100 (Cc)/(Ff) (1)
And that, in a perfect world, the weight F in the feed to the separation system is equal to the weight of the concentrate C plus the weight of the tail T.
F = C + T (2)
By developing simple algebraic substitutions between equations 1 and 2, a person can determine that equation 3 and 4 corresponds to different algebraic arrangements based on equations 1 and 2.
R = 100 Cc/(Cc + Tt) (3)
R = 100 (c (f-t)) / (f (c-t)) (4)
Hence there is no difference between calculating the recovery of an element using either equations 3 o or 4.
The feed sample fed to a separation system is the one that produced the concentrate and the tails, hence that is the feed assay that needs to be used in equations 3 and 4. The researcher would want that this world were perfect and that a subsample of the feed to the separation system would have been obtained in such a way that it has the same chemical and physical characteristics of the samples that produced the tails and concentrates under study. This is never going to be true because it is impossible to have a physical system that would allow to split two samples of the same lot to produce two (or more) samples with identical composition, hence the calculated feed to the separation system is going to be always different to the external reference feed sample.
The problem gets worse when we find that metallurgical accounting errors are produced once a person (or a robotic system) starts handling the concentrates and tails. In summary, a researcher should not think that either equations 3 or 4 are best suited for a particular case. This problem gets multiplied when a 3 product system is studied.
Limitations of the Two Products Recovery Equations
(These notes are taken from Barry A. Wills, Mineral Processing Technology)
The two products recovery equations assume the following:
The system is under steady state conditionsconcentrate and tails are produced starting from the same feed there are not sampling errors and chemical assay errors the fundamental error doesn’t exist.
Equations 3 and 4 are very sensitive to the value of the tails assays t. If we differentiate equation 4 with respect to f and c we obtain the following two equations:
"Where VR, Vf, Vc and Vt are the variances in R, f, c and t, respectively"
Replacing equations 5, 6 and 7 on equation 9 gives
At 95% confidence the Recovery will be between R = Mean Rec ± 1.96 *SD
R = 84.2% ± 2.6%
As can be seen with 5% relative standard deviation a researcher should expect that its Cu recovery is located between 81.6 and 86.9, i.e. a margin of 5.3 points in recovery.
When management is presented with these values a decision needs to be made whether to invest more money in training and in better sampling and chemical assays techniques to reduce the relative standard deviation of the data.
Here the reader has been presented with calculations to deal with only one element but in reality a software, such as Bilco or Bilmat, should be used to determine a multi-element mass balance. In order to do this the researcher needs to have an estimate of the standard deviation for each element, hence a program to collect this data needs to be developed.
Standard Deviation can be calculated for every stream in a mill if a sample is split in let's say 10 subsamples. The average, the standard deviation and the relative standard deviation can be calculated. This can be done continuously until a database is formed and the researcher now will know the sensitivity of its information.
These calculations have a major impact on business revenue.
This is a very interesting conversation!
Let me weigh in by saying that it is vitally important to get the right (representative) sample before worrying about how many tests to do on it. This is the old story around accuracy vs. precision - you can have a very precise answer, with very low variation in the results, but it can be totally different to the 'true' answer.
Having said that, and assuming all the sample collection and preparation stages have been done correctly, the number of batch flotation tests performed per sample required is reliant on the degree of accuracy you would like. If you are just looking at trends, then it is maybe not as necessary to do triplicate tests. However, if you are trying to use these results to design a plant, you would want a higher degree of confidence and obviously more tests are required.
A major mining company used to have as their standard procedure triplicate batch tests for everything, providing the standard deviation across all results. It's a cost-benefit thing that needs to be weighed up.
A few years ago a wise mentor in the industry said to me - the answer to every question in metallurgy is 'it depends’!
The sample selection is vital to have an answer on what a researcher is trying to find. I agree with that.
It is also very important to understand the system sampling and assay errors. Every system is different because of the ore characteristics and because of employee training and company standards.
Remember, there is no perfect sampling and assay system; hence the error can be minimized but not eliminated (think fundamental error). The role of a researcher (engineer, student, etc.) is to measure the sampling and assay error. Once this is measured then that person needs to understand what those numbers mean to give his/her results the power to make informed decisions.
For example, if someone finds that the Cu recovery can be between 80 and 85%, is that acceptable? What are the implications on concentrate production, on the economics of the project?
There are steps that can be followed to answer the question that was initially asked here.
Increasing the amount of flotation tests will not increase precision or repeatability if the sampling and assay errors are not understood and minimized.
If this doesn’t happen then the explanation I provided before, on how to proceed to develop these calculations is useless.
Laboratory batch flotation testing generally involves the determination of recovery for a given component of the system, for example copper or gold. The method used to calculate recovery is not as straight forward as you may.