Keeping this in mind and the fact that the question is oversimplified, I just need to confirm if my assumption was correct.

I'm a bit unsure about the inclusion of the BOD in the daily weight of sludge produced. Maybe I should only use the SS removed in primary sedimentation, and then calculate a weight of excess solids from the BOD content ratio provided. In that case I remove the section 1 of my sum.

Maxim

Hi Christoph,

Thank you for your answer. I understand your reaction, I had the same one when trying to wrap my head around the problem. I did more detailed calcs on this process in the past but it looks like things are oversimplified in this question, so I need to understand where the prof wants us to go and get there.

I attach a scanned copy of the question, you can see the details of the equation I mentioned. I think we have to keep in mind that the question is simple and we need to make assumptions on the conditions that are not given. The % provided for the mass of excess biological solids is probably only there to make the calculations easier, even if it is not representative of reality.

Maxim

Hello Maxim,

I´m a wastewater engineer since more than 20 years, I never used the "R=(t/(a+bt)) equation".

Could you explain the formula? Maybe put your calcs.

In regard to the follow up.

"Following biological oxidation, the mass of excess biological solids that are settled out is equal to 44% of the BOD content remaining after primary sedimentation. Calculate the daily weight of sludge solids removed during both primary and secondary sedimentation."

This makes not even sense as a brain twister and has no relation at all to reality.

It says (from my understanding): mass of excess biological solids that are settled out – in exact words that would be, “the mass of sludge that settles in the clarifier”. – but that makes no sense, as the return sludge settles as well and the number has absolutely nothing to do with the primary treatment, and you do not end up with the total sludge ammount.

There is no way to calculate the excess sludge (sludge formed due to BOD removal in the secondary treatment) by the data you gave.

You have to have:

BOD influent concentration, BOD effluent concentration or efficiency, sludge age….as minimum to be able to predict a number. For domestic wastewater this would be ok, for industrial wastewater you have to determine more parameters as sludge build up and sludge decomposition.

Christoph

Hello SuSanA members,

I know there is a great wealth of knowledge in wastewater treatment on the forum and figured someone might be able to offer some advice. I am a water engineer working for a consulting firm specialized in international development, and on the side I am currently completing a MSc at WEDC. I am working on an assignment and inputs would be greatly appreciated. It's a simple question but I want to make sure I get it right, wastewater treatment is not my specialty.

We have a simple sludge calculation question for wastewater treatment works. I have the total flow in m3/day, the SS and BOD concentrations at the influent. There is a primary sedimentation tank and we have the data to estimate the removal using the R=(t/(a+bt)) equation with parameters a and b for BOD and SS.

In a follow-up question, we have:
"Following biological oxidation, the mass of excess biological solids that are settled out is equal to 44% of the BOD content remaining after primary sedimentation. Calculate the daily weight of sludge solids removed during both primary and secondary sedimentation."

I would like to confirm my key assumptions. The daily weight of sludge solids would be equal to the sum of 1) BOD removed after primary sedimentation (already calculated), 2) SS removed after primary sedimentation (already calculated) and 3) mass of excess biological solids that settle after the 2nd stage, biological oxidation.

To calculate 3), I use the mass of BOD removed after primary sedimentation and calculate 44% of that.

What are your thoughts on this?

Regards,

Maxim