Design of biological phosphorus removal
The amount available of easily degradable organic matter can be estimated on the
basis of 3 contributions:
influent:
Ql · SHAc,l
hydrolysis/
fermentation:
rv,HAc' exp{lC{T -20)) .Yz
denitrification: - VN03,HAc' Qs · SNo3,5
Using the following assumptions:
rv;HAc
0.25 kg COD(S) / (m 3 · d)
1C
0.1 oc-l
VN03,HAc
5 kg COD /kg N03-N
and that 0.1 kg P /kg COD can be removed. Hence the maximum possible biological
phosphorus removal can be calculated:
Ql (CP,l- C.r,4) = 0.1 (Ql · SHAc,l +
0.25 · exp (0.1(T-20)) · Vz- 5 · Qs · SNo3,s)
(8.7)
Hence it is assumed that all organic matter is removed so that the effluent concentration of easily degradable organic matter, SHAc,2, from the anaerobic tank is 0.
280
Example 8.2
Calculate the maximum possible biological phosphorus removal at 20°C and at 8°C for
the plant described in Example 8.1. Additional information is that the nitrate concentration
in the return sludge flow, 3,000 m 3 /d, is 0.003 kg N03-N/m 3 .
Maximum phosphorus removal corresponds to SHAc,2 = SHAc,s = 0.
By substitution into Expression (8.7) we find for 20°C
3,600 m 3 /d (CP. 1 - CP.4) = 0.1 kg P/kg COD(S) (3,600 m 3 /d · 0.060 kg COD(S)/m 3 +
0.25 kg COD(S)/(m 3 · d) · exp (0.1 (20 -20) 200 m 3 -
5 kg COD(S)/kg N03-N · 3,000 m 3 /d · 0.003 kg N03-N/m 3 )
CP.1 - CP.4 = 0.1 (216 +50- 45)/3,600 = 0.006 kg P/m 3
At 8°C, all terms, with the exception of the hydrolysis, are unchanged. The hydrolysis
is 0.25 kg COD(S)/(m 3 · d) · exp (0.1 (8-20)) . 200m 3 =
15 kg COD(S)/d
Cp, 1 - Cp,4 = 0.1 (216 + 15 - 45)/3,600 = 0.005 kg P/m 3
Hence a maximum of respectively 6 and 5 g P/m 3 can be removed by biological phosphorus removal at 20°C and 8°C.