Standard Methods, 21st Ed., Method 2540D TSS, end of Section 3.c.: "Duplicate determinations should agree within 5% of their average weight." Does this mean that the two final weights should be averaged, and each weight compared to that average must be within 5% of that average? How can this critera best be handled?

If I recall correctly, this statement appears in all of the SM Solids analyses. I'm working with TSS first and will make adjustments to other analyses if needed.

Any assistance to clarify this criteria would be appreciated.

Thanks!

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Analysis =R1

Duplicate =R2

(R1-R2)/(average) * 100

You can't really compare the weights and need to compare the TSS value. Unless of course, you filter the same volume for both analysis, then comparing the weights should be fine.

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If the sample and duplicate volumes are not exactly the same, convert the Weight of the Dry Residues from mg/L to g/L:

Weight of Dry Residue mg/L/(Sample volume mLs/1000).

Then compare the two Weight of Dry Residues in g/L to 5% of the average of the two.

This should be indicated somewhere on the benchsheet.

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