Treatment Plants for Phosphorus Removal from Wastewater
For the phosphorus sludge it normally applies that the amount of suspended
phosphorus is much larger than the amount of dissolved phosphorus, that is,
03 · X TP,3 > 03 · STP,3
hence
Example 10.1
A treatment plant treats 11 ,000 m 3 of municipal wastewater per day. The content of
phosphorus is 12 g P/m 3 .
The plant is an activated sludge plant with a surplus sludge production of 4,000 kg
SS/d with a content of dry solids of 20 per cent.
Phosphorus is to be removed by simultaneous precipitation to an effluent concentration of 1 g P/m 3 .
What is the content of phosphorus in the surplus sludge production?
Phosphorus in the surplus sludge, a 3 · CTRS· is found from the mass balance, Expression (1 0.1 ):
a, · CTP,1 + a4 · CTP.4 = a2 · CTR2 +as· CTR3
(10.1)
It is assumed that there is no phosphorus in the precipitant, that is, CTR4 = 0, and that
a4 = 0 («a,)
Phosphorus in the surplus sludge, CTRS· is found from the expression
as· CTR3 =a, · CTR1- a2 · CTR2
CTR3 =(a, · CTR1- a2 · CTR2)/as
Here we know
a, = 11 ,000 m 3 /d
CTR1 = 12 g P/m 3
CTRZ = 1 g P/m 3
a 3 and a 2 must be calculated.
The surplus sludge, a 3, is found from
as · Xss.s = 4,000 kg SS/d
Xss.s = 20% 88, that is, 200 kg SS/m 3
a 3 = (4,000 kg SS/d)/(200 kg SS/m 3 ) = 20 m 3 /d
From a water balance we find a 2
a, =Oz +as
az =a, -as= 11 ,000 m 3 /d- 20 m 3 /d = 10,980 m 3 /d
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For the phosphorus sludge it normally applies that the amount of suspended
phosphorus is much larger than the amount of dissolved phosphorus, that is,
03 · X TP,3 > 03 · STP,3
hence
Example 10.1
A treatment plant treats 11 ,000 m 3 of municipal wastewater per day. The content of
phosphorus is 12 g P/m 3 .
The plant is an activated sludge plant with a surplus sludge production of 4,000 kg
SS/d with a content of dry solids of 20 per cent.
Phosphorus is to be removed by simultaneous precipitation to an effluent concentration of 1 g P/m 3 .
What is the content of phosphorus in the surplus sludge production?
Phosphorus in the surplus sludge, a 3 · CTRS· is found from the mass balance, Expression (1 0.1 ):
a, · CTP,1 + a4 · CTP.4 = a2 · CTR2 +as· CTR3
(10.1)
It is assumed that there is no phosphorus in the precipitant, that is, CTR4 = 0, and that
a4 = 0 («a,)
Phosphorus in the surplus sludge, CTRS· is found from the expression
as· CTR3 =a, · CTR1- a2 · CTR2
CTR3 =(a, · CTR1- a2 · CTR2)/as
Here we know
a, = 11 ,000 m 3 /d
CTR1 = 12 g P/m 3
CTRZ = 1 g P/m 3
a 3 and a 2 must be calculated.
The surplus sludge, a 3, is found from
as · Xss.s = 4,000 kg SS/d
Xss.s = 20% 88, that is, 200 kg SS/m 3
a 3 = (4,000 kg SS/d)/(200 kg SS/m 3 ) = 20 m 3 /d
From a water balance we find a 2
a, =Oz +as
az =a, -as= 11 ,000 m 3 /d- 20 m 3 /d = 10,980 m 3 /d
313