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May 23rd, 2013, 12:29pm
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Sludge age calculation problem (Read 824 times)
wasted
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Sludge age calculation problem
May 30th, 2012, 6:44am
 
I have this book which says to calculate the sludge age aka SRT=(X*Va + X*Vc) / (Qw*Xu)

Qw=(X*Va + Xu*Vc) / (SRT*Xu)

SRT - sludge age
Va - aeration tank volume (m3)
Vc - the volume of mud at secondary clarifier (m3)
X - activated sludge MLSS (mg/l)
Xu - mud concentration coming from secondary clarifier (mg/l)
Qw - sludge to be removed from proccess (m3/d)

So i calculate mine:
SRT - lets say i want 7 days
Va - with anoxic,anaerobic and aeration chamber 659m3
X - lab said 4,8 g/l aka 4800mg/l ? or its 4,8mg/l ? (confused)
Vc - 28,25m3
Xu - 6,4g/l aka 6400mg/l ? or its 6,4mg/l instead ? (confused)


So i want to know how much should i remove if i want the SRT to be 7 days.

What would be the correct answer ?
Mine seem to come really wierd or insane...probably doing smthing wrong...

Would appriciate if someone would calculate it for me and give some tips...or is this formula wrong ?

thnx in advance
Ardi
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Re: Sludge age calculation problem
Reply #1 - May 30th, 2012, 7:08am
 
still get really wierd answer

Qw= (4,8kg/m3*659m3) + (6,4kg/m3*28,25m3) / 7*6,4kg/m3 =  74,64m3

Is that answer acceptable ? Since aeration is 659m3 and if i remove 74,64m3 daily....should i believe that ?

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Re: Sludge age calculation problem
Reply #2 - May 30th, 2012, 7:32am
 
I got no clue about these formulas, but this is what i see now:
SRT=(X*Va + X*Vc) / (Qw*Xu)

Qw=(X*Va + Xu*Vc) / (SRT*Xu)

Wierd thing here is before you do X*va + X*Vc and when you start to write it differently there is X and Xu at the first part. Beside that mistake the formule is rewritten perfect.
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Re: Sludge age calculation problem
Reply #3 - May 30th, 2012, 7:49am
 
yeh its like that written at the book also..noticed that also before..but other than that the answer is correct ?
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Re: Sludge age calculation problem
Reply #4 - May 30th, 2012, 7:56am
 
SRT = (4,8kg/m3*659m3)+(6,4kg/m3*28,25m3) / 14m3*6,4kg/m3 = 37,3d

So if i remove 14m3 then my sludge age is 37,3d ?
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Re: Sludge age calculation problem
Reply #5 - May 30th, 2012, 9:58am
 
wasted wrote on May 30th, 2012, 7:56am:
SRT = (4,8kg/m3*659m3)+(6,4kg/m3*28,25m3) / 14m3*6,4kg/m3 = 37,3d

So if i remove 14m3 then my sludge age is 37,3d ?


Well I really hope someone else can help with this, since this isnt my expertise at all. But im wondering, if it tells you how much sludge in cubic meter you dispose of, then shouldnt it know the concentration of it? And we only have the concentration of activated sludge and Xu out going sludge concentration, arent we missing from Vc so the concentration of the medium in the secondary clarifier.

So what i actually mean is Qw=(X*Va + Xc*Vc) / (SRT*Xu)

SRT - sludge age
Va - aeration tank volume (m3)
Vc - the volume of mud at secondary clarifier (m3)
X - activated sludge MLSS (mg/l)
Xc - sludge concentration secondary clarifier (kg/m3)
Xu - mud concentration coming from secondary clarifier (mg/l)
Qw - sludge to be removed from proccess (m3/d)

So then this would become:
Qw=(X*Va + Xc*Vc) / (SRT*Xu)


Could someone else confirm this, since this isnt my expertise Cry

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Re: Sludge age calculation problem
Reply #6 - May 30th, 2012, 12:24pm
 
sometimes engineers like to make things really complicated Wink

Take the mass of your solids inventory in the plant/ the mass of solids leaving the plant per day to get the sludge age.
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Re: Sludge age calculation problem
Reply #7 - May 30th, 2012, 4:23pm
 
well, some sources do not include the volume of secondary clarifiers.
SRT=X*Va/Qw*Xu
which one is correct?
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Re: Sludge age calculation problem
Reply #8 - Jun 4th, 2012, 5:22am
 
"sometimes engineers like to make things really complicated Wink

Take the mass of your solids inventory in the plant/ the mass of solids leaving the plant per day to get the sludge age."

So it should be:

When i take SV test and get result of 50% in 1 litre. I have 659m3 of total in the aeration basin. 50% of 659m3 = 329,5m3 and activated sludge MLSS is 4800mg/l aka 4,8kg/m3 then i have total of 1581,6 kg sludge in aeration tank.

but the mass of solids leaving the plant per day....i dont know how to calculate that...i only know how much mud i take out and press...but i dont know how much leaves from clarifier with suspended solids...and i only can measure incoming flow not outgoing.
Or should i leave out the suspended solids count and just calculate mud taken out from process*calrifier mud MLSS ?

Correct me if i misunderstood smthing
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Re: Sludge age calculation problem
Reply #9 - Jun 4th, 2012, 5:27am
 
to: srknk

If i calculate it with your formula...i get 35,3days of sludge age
My formula gets me 37,3 days

So i guess both are almoust correct Smiley
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Re: Sludge age calculation problem
Reply #10 - Jun 4th, 2012, 10:29am
 
People will measure this different ways. Personally I like to ignore the solids in the clarifier and leaving in the effluent TSS unless there's very large clarifiers or a lot of TSS. To figure out the mass of solids wasted you need to know the concentration of these solids and you can calculate that lbs. (WAS concentration * WAS flow rate in MGD * 8.34 lb/gal)

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Re: Sludge age calculation problem
Reply #11 - Jun 4th, 2012, 2:18pm
 
"The sludge judge"  Surprised you have been not drummed out of the Wastewater industry.  I am being sarcastic..  In my career I argued your point  of view  till I was blue in face.   How can one determine the amount of solids in the clarifier.... Why not also include the RAS contents.  

The big issue is that the SRT formula is used by certification exams, which confuses the operations staff.
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