Hello lucas ,

As far as I know , BOD is a better representative value of food , but the test of BOD take 3-5 days , so they use COD as an alternative.

volatile suspended solid is a good estimate of microorganisms count.

check (Metcalf and Eddy/wastewater engineering ) for values of f/m.

Regards .

Rami

Dear Lucas,

COD includes intractable compounds not readily degraded biologically but determined by chemical oxidation. Hence BOD is used almost exclusively to represent food in the F/M ratio. Typically the COD value is 2.2 - 2.8 times the BOD value.

M represents the mass of micro-organisms and is usually measured as the mixed liquor total suspended solids. TDS typically includes 20 to 30% non-organic matter and therefore using VSS is a little more accurate. However, VSS contains organic matter other than micro-organisms (and also dead and inactive micro-organisms) so the correction may be misleading and not worth the effort.

A reference range of 0.04 to 0.10 kgBOD/kgTSS would be for a treatment plant capable of both carbonaceous and nitrogen oxidation. An F/M ratio of 0.5 would be typical of a carbonaceous oxidation only plant. Both ratios are affected by temperature.

Regards

Michael Lever

Dear Lucas,

You have already gotten very good and valuable answers. It is just to add that when talking about both COD and BOD it important to consider that both have different fraction and to have a very accurate measure of F you need to consider the readily available fraction that is not very straightforward to get it. The main problem can come when BOD has its major fraction as not readily available, or persistent, or recalcitrant. It can also be the opposite case and if COD has its major part as readily available then you can use it as F. So this is a bit complex and as the colleagues already mentioned there are already established values for that. But the tricky thing is to be really sure that the values you use are readily available. a very good method to understand the effluent you are working with and evaluate F/M ratio that could work best for your case is to run some respirometric tests and analyse the graphs of oxygen uptake rate in relation to time.

My best,

Fabio

F/M means Food to microorganism ratio/ d You may consider BOD/MLSS= 0.2 t0 0.6 however for oxidation ditch or extended aeration it should be 0.05-0.15 and mean cell residence time 20-30 days for extended aeration or 5-15for complete mix aeration.

You may take 3000-6000 mg/L as MLSS , 18-36 hrs of aeration period for extended aeration, 2-5 hrs in case of complete mix aeration or high rate aeration

i suggest you should start experiment with complete mix aeration first.

I will interact and follow your work in case if you think so

I go with Dr Sheth

Generally BOD is used for F and TSS for M. If other parameters are used (COD or TOC for F resp. VSS for M) the range is different and you have to indicate which parameters you have used (this should be done in any case).

I have always thought it more logical to consider (Q*BOD5)/(V*MLVSS). In systems with very long SRT values, and/or where alum or ferric are being added, the fixed SS can be considerable in the ML.

Wright, Mr. Doran. F/M can be starting point of technological calculation, or only derived data on the end of calculation if you use different (more sophisticated) approach.

Dear all,

Thank you very much for the interesting answers.

I got the point that BOD for F and VSS for M represents better the food-to-microorganisms ratio.

However, most of the ranges you can find in literature and in the answers posted here are expressed in TSS. That is why my initial confusion.

For example: for activated sludge with extended aeration and both carbonaceous and ammonia oxidation, the ranges posted here were:

0.04 to 0.10 kgBOD/kgTSS (Michael Lever)

0.05 to 0.15 kgBOD/kgTSS (Dr K N Sheth)

Regards,

For practical purposes, TSS is MLSS. Ratio of MLVSS to MLSS is normally 0.8

The influent BOD5 and biodegradable biomass=68% of BODL.

The biological solids TSS are 65% biodegradable.

So F/M = kg BOD5 applied/kg MLVSS per day. This MLVSS are the bacterial cells biomass

So you consider F/M= 0.05 to 0.15 kg BOD/kg MLSS x 0.8

Hope this simplification resolves your doubt

As per Dr Sheth' s answer

Thank you very much sir.

Be cautious with the 0.8 assumption, as longer SRTs and/or the addition of a metal coagulant for P removal can reduce the volatile fraction considerably.