Separator performance

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If you have residues on {212, 90 , 63, 45, 32} µm for feed, coarse and fine, and if your product is not too fine - as compared to 32µm - , then you could evaluate a Tromp curve.
If you have less data, you have less chances to evaluate your separator, and vice-versa.

However, the main question is: what do you mean by "optimum efficiency"?

On a Tromp curve, you can observe the so-called "bypass level".
The bypass level indicates how much % of fine particles go to the coarse stream.
This measures, of course, and un-wanted behaviour or an "un-efficiency".

If the bypass is 40%, and if the production is 100 t/h, then, naïvely speaking, this means that 40 t/h of good product is returned to the mill.
This could be, naïvely, translated as an efficiency of 100/140 = 70% and would indicate that the production could be increased by 40 t/h if the separator would be 100% efficient. Actually, the impact of the bypass is lower but it increases with the fineness of the product (circulation factor). This can be understood by a model of the complete grinding circuit.

My message is: to evaluate a seperator you do need to evaluate the Tromp curve.
The least you could get from the Tromp curve is the bypass factor.
If you don't have enough information for that, you cannot evaluate the separator.

Michel

my question is what is the normal range of these residues for coarse,feed & product so that i can compare for any improvement in the system raj

There is no answer to this question.
It depends on what you are producing.
For "system improvement", is it not enough to check the quality of the product and the productivity t/h ?
If your focus is on "separator improvement", then it is all about the Tromp curve.

Here is some sample data to illustrate my point:

finish product
3500 Blaines, (P32,P45,P63,P90,P212) = (78%, 90%, 97%, 99.3%, 100%) 
where P=1-R

feed
2450 Blaines, (P32,P45,P63,P90,P212) = (53%, 67%, 80%, 90%, 99.6%) 

reject
1200 Blaines, (P32,P45,P63,P90,P212) = (23%, 39%, 61%, 79%, 98.9%) 

Plant 
Circulating load:             1.8

Separator
bypass:           10%
cut diameter:  40µm  (x50)   (x75= 26µm, x50= 40µm, x25= 63µm)
imperfection:  46%   (=(x75-x25)/2/x50)

Mill
residues reduction factors:  (= residue mill outlet / residue mill inlet)
                                        @32µm   = 40%
                                        @45µm   = 58%
                                        @63µm   = 76%
                                        @90µm   = 87%
                                        @212µm  = 100% (!)
(this is calculated from residues and circulating load)

I don't see how you could use these data and compare them to your own data.
What is your product?
What is the product fineness?
What is you circulating load?

To evaluate your separator, you need the Tromp curve, even with very few points.
The bypass is one point of the Tromp curve that can be evaluated even with few data.