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sludge thickening, equalization basins

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I am feeling pretty good about the exam right now, but am having trouble with sludge thickening and equalization basins.

Could someone please give me a "for dummies" refresher on these topics? I've done lts of digging and am not getting it...

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I am feeling pretty good about the exam right now, but am having trouble with sludge thickening and equalization basins.

Could someone please give me a "for dummies" refresher on these topics? I've done lts of digging and am not getting it...

Share on other sites
I am feeling pretty good about the exam right now, but am having trouble with sludge thickening and equalization basins.

Could someone please give me a "for dummies" refresher on these topics? I've done lts of digging and am not getting it...

I'm having the same problem!

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I am feeling pretty good about the exam right now, but am having trouble with sludge thickening and equalization basins.

Could someone please give me a "for dummies" refresher on these topics? I've done lts of digging and am not getting it...

Can you post some examples of the type that is giving you problems?

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I am feeling pretty good about the exam right now, but am having trouble with sludge thickening and equalization basins.

Could someone please give me a "for dummies" refresher on these topics? I've done lts of digging and am not getting it...

Can you post some examples of the type that is giving you problems?

Like the attached problem.

sludge_problem.pdf

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Like the attached problem.

The thickener sees 2 sources - primary and secondary sludge.

PRIMARY

Solids removal = 135 mg/L x 20 MGD x 8.3454 (conversion factor) = 22533 lb/day solids mass removed

Sludge mass = 22533/0.035 = 643788 lb/day

Assuming sludge is near water in sp gr, sludge volume = 643788/8.3454 = 77,193 gal/day

SECONDARY

Solids removal = 10000 lb/day (given)

Secondary sludge volume = 10000/7500x8.3454 = 0.16 MGD = 160,000 gal/day

AT THICKENER

Incoming solids mass = 22533+10000 = 32533 lb/day

Outgoing solids = 32533 lb/day @4.5% solids content

Outgoing sludge volume = 32533/(0.045x8.3454) = 86,629 gal/day

At the thickener, incoming volume = 237,193 gpd, outgoing volume = 86629 gpd (63.5% volume reduction)

THIS IS A LOT OF WORK FOR A 5 MINUTE QUESTION!

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yes, these are the types of questions that i'm dealing with. Is this above and beyond what we would see on the afternoon exam?

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Like the attached problem.

The thickener sees 2 sources - primary and secondary sludge.

PRIMARY

Solids removal = 135 mg/L x 20 MGD x 8.3454 (conversion factor) = 22533 lb/day solids mass removed

Sludge mass = 22533/0.035 = 643788 lb/day

Assuming sludge is near water in sp gr, sludge volume = 643788/8.3454 = 77,193 gal/day

SECONDARY

Solids removal = 10000 lb/day (given)

Secondary sludge volume = 10000/7500x8.3454 = 0.16 MGD = 160,000 gal/day

AT THICKENER

Incoming solids mass = 22533+10000 = 32533 lb/day

Outgoing solids = 32533 lb/day @4.5% solids content

Outgoing sludge volume = 32533/(0.045x8.3454) = 86,629 gal/day

At the thickener, incoming volume = 237,193 gpd, outgoing volume = 86629 gpd (63.5% volume reduction)

THIS IS A LOT OF WORK FOR A 5 MINUTE QUESTION!

OK...I have a lot of questions. The solids removed makes sense. But on the sludge mass why is that divided by 3.5%. Seems like it should be multiplied.

Also, the 7500 doesn't have any units. What is that?

And again...why is the sludge volume divided by 4.5% and where did the 237,193 come from?

Thanks so much! As you can see....i'm pretty lost.

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OK...I have a lot of questions. The solids removed makes sense. But on the sludge mass why is that divided by 3.5%. Seems like it should be multiplied.

Also, the 7500 doesn't have any units. What is that?

And again...why is the sludge volume divided by 4.5% and where did the 237,193 come from?

Thanks so much! As you can see....i'm pretty lost.

When a sludge comes out with 3.5% solids, then the solids removed (per day) represents 3.5% by weight of the weight of the liquid sludge. So, if we have the weight of solids removed, then we divide that by 0.035 to get the weight of the wet sludge. And since at 3.5% solids the sludge is nearly water, we can divide by the specific gravity of water 8.3454 lb/gal to get the volume of sludge.

The 7500 is the concentration of suspended solids in the secondary (recycled) sludge. It is mg/L

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I am feeling pretty good about the exam right now, but am having trouble with sludge thickening and equalization basins.

Could someone please give me a "for dummies" refresher on these topics? I've done lts of digging and am not getting it...

Can you post some examples of the type that is giving you problems?

Like the attached problem.

I tried to solve this problem on my own but was getting nowhere because I was multiplying the solids load by the fraction of percent solids instead of dividing. I recalculated the answer with a sludge specific gravity of 1.2 which gives a sludge mass of 10 pounds per gallon, and you end up with the same answer. Can anyone else out there please post onto this thread another problem similar to this one?