Modified Ludzack Ettinger Example
3
MODIFIED LUDZACK ETTINGER NITROGEN REMOVAL PROCESS EXAMPLE GIVEN: Wastewater average flow rate, Q = 40,000 m 3 /d Influent concentration Soluble COD (S SO ) 150 mg/l Particulate COD (X SO ) 115 mg/l Soluble ammonia-N (S NHO ) 25 mg-l N Soluble organic N (S NSO ) 6.5 mg/l Particulate organic N (X NSO ) 8.5 mg/l Parameters ˆ H 6 d -1 ˆ A 0.77 d -1 b H 0.18 d -1 b A 0.1 d d -1 K S 20 mg/l COD K NH 1 mg/l NH 4 -N Y H 0.6 g-COD cells/g-substrateCOD Y A 0.24 mg COD cells/g-NH 4 -N f D 0.2 g debrisCOD/g cells COD decayed i NXB 0.087 g-N/g-cell COD Objectives: Effluent COD = S S < 10 mg/l Effluent ammonia nitrogen = S NH < 0.3 mg/l Effluent total inorganic nitrogen (TIN) = 10 mg/l For adequate settling MLSS = X T < 3,000 mg/l or 4,260 mg/l as COD Aerobic Tank S NO,aer Anoxic Tank Q + Q RAS Q RAS MLR Q+Q RAS +MLR Q + Q RAS Q W
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ludzack ettinger example
Transcript of Modified Ludzack Ettinger Example
MODIFIED LUDZACK ETTINGER NITROGEN REMOVAL PROCESS EXAMPLE
GIVEN:
Wastewater average flow rate, Q = 40,000 m3/d
Influent concentration
Soluble COD (SSO) 150 mg/l
Particulate COD (XSO) 115 mg/l
Soluble ammonia-N (SNHO) 25 mg-l N
Soluble organic N (SNSO) 6.5 mg/l
Particulate organic N (XNSO) 8.5 mg/l
Parameters ˆ H 6 d
-1
ˆ A 0.77 d
-1
bH 0.18 d-1
bA 0.1 d d-1
KS 20 mg/l COD
KNH 1 mg/l NH4-N
YH 0.6 g-COD cells/g-substrateCOD
YA 0.24 mg COD cells/g-NH4-N
fD 0.2 g debrisCOD/g cells COD decayed
iNXB 0.087 g-N/g-cell COD
Objectives:
Effluent COD = SS < 10 mg/l
Effluent ammonia nitrogen = SNH < 0.3 mg/l
Effluent total inorganic nitrogen (TIN) = 10 mg/l
For adequate settling MLSS = XT < 3,000 mg/l or 4,260 mg/l as COD
Aerobic Tank
SNO,aer
Anoxic Tank
Q + QRAS
QRAS
MLR
Q+QRAS+MLR
Q + QRAS
QW
Step 1. Find aer for aeration basin – nitrification will control
days).()..(.
.
)bK)bˆ(S(
SK
ANHAANH
NHNHaer 13
1011077030
301
Check Ss.
LCOD/mglCOD/mg.)))(.()((
))(.(
))b(ˆ(
)b(KS
HH
HSS 1090
131801136
13180120
1
1
Nitrogen available for nitrification, NO*/Q * Changed notation from “SNO/Q” to make it less confusing
NO/Q = (SNHO+SNSO+XNSO)-(iNXBYobs(SSO+XSO))-SNH
(assuming both soluble and particulate COD in influent is degraded and neglecting SS << SSO+XSO)
YH,obs = (1+0.2(0.18)13)0.6/(1+0.18(13)) = 0.26 g-cell-COD/g-substrate-COD
NO/Q = (25 + 6.5 + 8.5) – (0.087(115+150)0.26) – 0.3 mg/L N
NO/Q = 33.6 mg/l N
NO = 40,000 m
3/d(33.6 g/m
3) = 1.4 x 10
6 g-N/day available for denitrification
For denitrification design:
Assume anox = 3 d (influent is dilute, maximize available substrate and denitrification by using
generation of COD from decay)
MLE process: sys = 13 + 3 = 16 days
MLSSasCODgx)(.(
))(.)).(.)(d/m,(d
)b(
)SS(Y)bf)(Q(VX
H
SSOHsysHDsyssysT
7104161801
11515060161802010004013
11
3
XTVaer = ( aer/ sys XTVsys = (13/16) 4 x 107 = 3 x 10
7 g MLSS as COD
XTVanox = ( anox/ sys XTVsys = (3/16) 4 x 107 = 7.5 x 10
6 g MLSS as COD
Apply constraint: MLSS < 4,260 mg/l COD and assume MLSS equal in anoxic and aerobic tanks:
Vaer = 3 x 107/4260 = 7,000 m
3
Vanox = 7.5 x 106/4260 = 1,800 m
3
Vsys = 7,000 + 1,800 = 8,800 m3
N/ SS,N = (1+0.18(3)-0.6(1+0.2(0.18)3))/(2.86(1+0.18(3)) = 0.2 g NO3-N/g-COD
N = 0.2(40000)(150) = 1.2 x 106 g NO3-N/d
where N = max amount of nitrate-nitrogen that can be denitrified as limited by available
influent soluble COD
fNO,D = N/(NO)
fNO,D = 1.2 x 106/1.4 x 10
6 = 0.86
= 0.86/(1-0.86) = 6
Let = 0.5
MLR = Q = 5.5(40,000) m3/d = 220,000 m
3/d
QRAS = Q = 0.5(40,000) m3/d = 20,000 m
3/d
Effluent nitrate nitrogen = SNO = (1 - fNO,D)NO/Q= 0.14(33.6) = 4.7 mg/L NO3-N
effluent TIN = NH4-N+NO3-N = 0.3+4.7 = 5 mg-TIN/L (meets permit)
