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Activated Sludge Problem


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#1 jenni179

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Posted 05 October 2010 - 03:13 PM

I'm having difficulty with this activated sludge problem. I have a few questions about it:

1.) Why does the Food equation (mass of BOD entering the aeration tank) have 1 ppm = 1* 10^6 mg/L ?? I thought 1 ppm = 1 mg/L.

Also, when you cancel the units out, you get lbm and ppm leftover. However, the answer just has 8,340 lbm. Where did the ppm go?

2.) When I calculated the values in the MLVSS equation, I get 0.003053. In order to get their answer of 3053 mg/L, I have to multiply by 10^6. I am to assume the 0.003053 was unitless and if I wanted mg/L then I needed to multiply by 10^6? I thought that the units for MLVSS were already in mg/L.

3.) Finally, the problem states that the plant has an 85% BOD removal efficiency. Why is this value not used for this problem? I understand why the 67% efficiency was used for the primary tank, but why should I ignore the 85%?

Any help would be greatly appreciated. The units in this problem are driving me crazy!

Thanks,
Jen

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#2 IlPadrino

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Posted 05 October 2010 - 04:15 PM

Let me help you on 1)

0.67 x 150 mg/L x 10,000,000 gal/day x 8.35 lbm/gal x (1 L / 1,000,000 mg) = 8350 lbm/day (all other units cancel out)

It looks like the solution incorrectly wrote "ppm" in place of "L". You are correct that 1 ppm = 1 mg/L. You'll need this to get your final answer into ppm.

#3 jenni179

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Posted 05 October 2010 - 06:58 PM

QUOTE (IlPadrino @ Oct 5 2010, 12:15 PM) <{POST_SNAPBACK}>
Let me help you on 1)

0.67 x 150 mg/L x 10,000,000 gal/day x 8.35 lbm/gal x (1 L / 1,000,000 mg) = 8350 lbm/day (all other units cancel out)

It looks like the solution incorrectly wrote "ppm" in place of "L". You are correct that 1 ppm = 1 mg/L. You'll need this to get your final answer into ppm.


I was wondering if the answer was incorrect. I figured it was just me. Thank you!! Does anyone know the answer for #2 or 3? smile.gif

Jen


#4 Octave

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Posted 05 October 2010 - 08:56 PM

QUOTE (jenni179 @ Oct 5 2010, 02:58 PM) <{POST_SNAPBACK}>
I was wondering if the answer was incorrect. I figured it was just me. Thank you!! Does anyone know the answer for #2 or 3? smile.gif


#2 is similar to #1. It's all in the units. Usually people get confused when converting between lbm/gal and mg/L.

By the way, you may find F/M expressed in different ways, sometimes dimensionless, sometimes with dimension of 1/time.

#3. The 85% removal efficiency would apply to the whole process, including the secondary or final clarifiers. Since the questions address only the aeration tank, this information is not needed.


#5 IlPadrino

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Posted 06 October 2010 - 11:10 AM

QUOTE (Octave @ Oct 5 2010, 04:56 PM) <{POST_SNAPBACK}>
By the way, you may find F/M expressed in different ways, sometimes dimensionless, sometimes with dimension of 1/time.


The Food to Microorganism Ratio should always be dimensionless. If it's expressed in dimensions that don't cancel directly then they can be canceled with a unity conversion factor. Do you have an example to the contrary?

#6 Octave

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Posted 06 October 2010 - 12:34 PM

QUOTE (IlPadrino @ Oct 6 2010, 07:10 AM) <{POST_SNAPBACK}>
The Food to Microorganism Ratio should always be dimensionless. If it's expressed in dimensions that don't cancel directly then they can be canceled with a unity conversion factor. Do you have an example to the contrary?


Yep, a quick Google search yields several:

F/M = 1,669 lbs/day / 3,552 lbs = 0.47 day-1
CE 356: Fundamental (sic) of Environmental Engineering

For conventional aeration tanks the F/M ratio is 0.2-0.5 lb BOD5/day/lb MLSS...
Environmental microbiology
By Raina M. Maier, Ian L. Pepper, Charles P. Gerba


F/M ratio, kg BOD/kg MLVSS/d (Table 6.2)
Biological Treatment Processes By Lawrence K. Wang, Norman C. Pereira, Yung-Tse Hung

F: M = Ns = QLa / XV (KgBOD5/KgMLSS * d)
Settlement than in the activated sludge sewage treatment operation and management of the guiding role of the (must be a computer-translated paper)

F / M (kg BOD / kg MLSS . day) Food to microorganism ratio
Abbreviations


And the list goes on. The point is, F/M is not always expressed in dimensionless form. It can also be expressed in terms of BOD or CBOD, MLSS or MLVSS, etc.


#7 IlPadrino

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Posted 06 October 2010 - 03:34 PM

QUOTE (Octave @ Oct 6 2010, 08:34 AM) <{POST_SNAPBACK}>
And the list goes on. The point is, F/M is not always expressed in dimensionless form. It can also be expressed in terms of BOD or CBOD, MLSS or MLVSS, etc.


I appreciate the google search and I'm not trying to argue or debate needlessly, but I think this is an important point.

F/M is a ratio of masses - it is, by its very definition, dimensionless. I won't go through every example you provide... but picking just the last one:

QUOTE (Octave @ Oct 6 2010, 08:34 AM) <{POST_SNAPBACK}>
F / M (kg BOD / kg MLSS . day) Food to microorganism ratio


Do you see how this is unitless? If you want to pick the mass of BOD and MLSS, you have to pick some time period, but the units of the F/M ratio are *not* "per day". The F/M ratio is dimensionless and the value can be based on a day's masses.

Bottom line: If you always pay strict attention to units, everything should cancel properly to give you the answer listed. If the units don't work out to the answers, you should check your work as there's a mistake somewhere (even if it's an implicit assumption).

Again, thanks for the dialog (and search time)!

#8 Octave

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Posted 06 October 2010 - 11:32 PM

QUOTE (IlPadrino @ Oct 6 2010, 11:34 AM) <{POST_SNAPBACK}>
F/M is a ratio of masses - it is, by its very definition, dimensionless.

Well, here's where we disagree. I have seen F/M expressed with dimension 1/T in more than one occasion. I do believe that many years ago it was always expressed as mass BOD5/lb mass of MLSS, but I've also seen it expressed with different units, as I explained in my other message. I don't see why it would have to be dimensionless, it's just a ratio.

I mean, if F/M is always dimensionless, how do you explain its usage in the references I cited?

Anyways, my objective was to point to the original poster (Jen) that it is very important to watch the units, and in that I completely agree with you. Good discussion.

#9 Jacob_PE

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Posted 02 September 2011 - 12:32 AM

I incorrectly tried to solve this probem by including the RAS as Food. For the CERM equation 30.3 the book states that So is usually taken as the incoming BOD. Isn't RAS measurable as BOD?  I guess the way I need to remember it is that the concentration coming from the primary clarifier is the Food and the MLVSS in the RAS is going to chow down on that food.

#10 civilized_naah

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Posted 02 September 2011 - 02:39 PM

Attached File  ActivatedSludge.jpg   131.71KB   22 downloads
QUOTE (Jacob @ Sep 1 2011, 08:32 PM) <{POST_SNAPBACK}>
I incorrectly tried to solve this probem by including the RAS as Food. For the CERM equation 30.3 the book states that So is usually taken as the incoming BOD. Isn't RAS measurable as BOD?  I guess the way I need to remember it is that the concentration coming from the primary clarifier is the Food and the MLVSS in the RAS is going to chow down on that food.

See the attached solution. A lot of the data can be used to calculate other stuff but are not relevant to the question asked.

#11 Jacob_PE

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Posted 02 September 2011 - 05:54 PM

QUOTE (civilized_naah @ Sep 2 2011, 09:39 AM) <{POST_SNAPBACK}>
Attached File  ActivatedSludge.jpg   131.71KB   22 downloads
QUOTE (Jacob @ Sep 1 2011, 08:32 PM) <{POST_SNAPBACK}>
I incorrectly tried to solve this probem by including the RAS as Food. For the CERM equation 30.3 the book states that So is usually taken as the incoming BOD. Isn't RAS measurable as BOD?  I guess the way I need to remember it is that the concentration coming from the primary clarifier is the Food and the MLVSS in the RAS is going to chow down on that food.

See the attached solution. A lot of the data can be used to calculate other stuff but are not relevant to the question asked.

Thanks for posting that write-up. Something I'm not 100% clear on is the flow of return activated sludge and waste activated sludge, do the flow (MGD), concentraion (mg/l) or loading in (lb/day) always have to be given in order to determine those parameters. Can they be found knowing the MLSS/MLVSS in the aeration tank and knowing the solids removal in the secondary clarifier?

#12 civilized_naah

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Posted 02 September 2011 - 06:18 PM

QUOTE (Jacob @ Sep 2 2011, 01:54 PM) <{POST_SNAPBACK}>
Thanks for posting that write-up. Something I'm not 100% clear on is the flow of return activated sludge and waste activated sludge, do the flow (MGD), concentraion (mg/l) or loading in (lb/day) always have to be given in order to determine those parameters. Can they be found knowing the MLSS/MLVSS in the aeration tank and knowing the solids removal in the secondary clarifier?

No, the best way to solve these problems is to draw a closed boundary and examine what goes into and out of that 'box' and equate IN = OUT. The boundary can enclose a single node or an entire region. You should 'cut through' those branches of the process which have either given data or desired parameters. So, if you internalize the paths with unknown parameters, those variables are not even part of the resulting equation.

#13 SneadFrank

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Posted 28 November 2011 - 08:23 AM

Durable and resistant to corrosion, they are especially used to remove sludge or other unwanted materials like slurries and mud, oil spill, waste water etc. There are various manufacturers of displacement stainless pumps, yet it is very important that you get genuine quality ones for the best results.




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