Design of denitrifying plants
By means of the nitrogen balance over the plant, with the use of symbols in Fig 7.1a,
the amount of nitrogen to be denitrified, MN, can be calculated:
(7.27)
are given assumptions for the composition of the raw wastewater,
is a determined, acceptable effluent concentration,
is the nitrogen in the exess sludge production(= fx,NFEsr).
When MN is known, it can be checked whether the C/N ratio is big enough, or
whether it is necessary to reduce the estimated denitrification rate, rx,s due to a too
low C/N ratio, or possibly add an external carbon source. Subsequently the necessary tank volume for denitrification, V2, may be determined on the basis of:
260
MN=Vz·rv,s
Vz= MN/rv,s
Example 7.7
Design of plant for denitrification with activated sludge.
(7.28)
How big should the denitrification tank be when a volume of wastewater of 5,000 m 3 /d
is to be denitrified?
The amount of nitrogen to be denitrified, MN. is found from a nitrogen balance, see
Fig 7.1a.
MN = a, . c, - a3 . c3- as . Cs
The following is known:
C, = 50 g TN/m 3
C3 =7gTN/m 3
(7.1S)
Furthermore, a, = 5,000 m 3 /d is known here. It is reasonable to assume that a 3 - a,,
that is, a3 = 5,000 m 3 /d.
Nitrogen in the sludge production, fx.N. is estimated at 0.05 kg Nlkg SS. The sludge
production is assumed to be 700 kg SS/d. This means that a5 · C5 is 700 kg SS/d ·
0.05 kg N/kg SS = 35 kg N/d.
By substitution we find:
MN = 5,000 m 3 /d · 50 g N/m 3 - 5,000 m 3 /d · 7 g N/m 3 -35 kg N/d = 1SO kg N/d
The lowest temperature in the denitrification tank is estimated to be soc. From Fig
3.11 we find a denitrification rate, rx.s. of approx. 0.5 g N/(kg VSS · h) (for raw wastewater at S 0 C}. The sludge concentration in the denitrification tank, X2, is set at 4 kg
VSS/m 3 . By substitution in Expression (7.26) we find:
By means of the nitrogen balance over the plant, with the use of symbols in Fig 7.1a,
the amount of nitrogen to be denitrified, MN, can be calculated:
(7.27)
are given assumptions for the composition of the raw wastewater,
is a determined, acceptable effluent concentration,
is the nitrogen in the exess sludge production(= fx,NFEsr).
When MN is known, it can be checked whether the C/N ratio is big enough, or
whether it is necessary to reduce the estimated denitrification rate, rx,s due to a too
low C/N ratio, or possibly add an external carbon source. Subsequently the necessary tank volume for denitrification, V2, may be determined on the basis of:
260
MN=Vz·rv,s
Vz= MN/rv,s
Example 7.7
Design of plant for denitrification with activated sludge.
(7.28)
How big should the denitrification tank be when a volume of wastewater of 5,000 m 3 /d
is to be denitrified?
The amount of nitrogen to be denitrified, MN. is found from a nitrogen balance, see
Fig 7.1a.
MN = a, . c, - a3 . c3- as . Cs
The following is known:
C, = 50 g TN/m 3
C3 =7gTN/m 3
(7.1S)
Furthermore, a, = 5,000 m 3 /d is known here. It is reasonable to assume that a 3 - a,,
that is, a3 = 5,000 m 3 /d.
Nitrogen in the sludge production, fx.N. is estimated at 0.05 kg Nlkg SS. The sludge
production is assumed to be 700 kg SS/d. This means that a5 · C5 is 700 kg SS/d ·
0.05 kg N/kg SS = 35 kg N/d.
By substitution we find:
MN = 5,000 m 3 /d · 50 g N/m 3 - 5,000 m 3 /d · 7 g N/m 3 -35 kg N/d = 1SO kg N/d
The lowest temperature in the denitrification tank is estimated to be soc. From Fig
3.11 we find a denitrification rate, rx.s. of approx. 0.5 g N/(kg VSS · h) (for raw wastewater at S 0 C}. The sludge concentration in the denitrification tank, X2, is set at 4 kg
VSS/m 3 . By substitution in Expression (7.26) we find: