4Lecture;Wastewater Treatment
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Transcript of 4Lecture;Wastewater Treatment
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CE 4312 W
W E
ECE 4
. , ,
F E,
830@.
1
:
U .
U : , , .
A .
2
C
W
•• I TI T
•• W W
•• W W
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• W
•• P P 3
W S ?W S ?• W
(, , , ), ( );D ( , .) (, ).
• W ,, .
• B , .
4
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W
I J T R
E
5
Where does it all go!
Where does thewater from thewasher go?
When you flush thetoilet where doesthe contents go?
By gravity flow, the waste is on its wayto your local wastewater treatment plant!
6
W ?
• C (
DO )
• A ( )
• I
( )
7
W T
G :
B
.
R
P
A
..
8
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WWTreatmentWWTreatmentWWTreatmentWWTreatment---- ObjectivesandregulationsObjectivesandregulationsObjectivesandregulationsObjectivesandregulations D ,
.
T
.
BOD,
.
S 1980, (..
). R
(: ) ..
9
I ,
, ,
VOC
.
10
A F AEAE EAE
W
.
11
T GOAL W T :
T
,
.
12
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K W T P
13
W T P D
P
.
C (F) E
, ,
.
D S ,
.
C
.
14
B :
F
I .. TSS, COD,
BOD5, , TKN, T P,
O.
M .. ,
.
15
• E
• E (EP)
• E I , , /.
• T .
16
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• S D F
– D: . V
– I: B
– I/E: %
– P F: D
17
• F
• A
() .
• E :
– T ( .. 40 L/ .
, 40 L/ .
)
– E , ,
– N /.
18
• E :
C 15,000 EP 150 L/EP.
E ADWF:
ADWF = 15,000(EP) 150(L/EP.) 1(L)/1000 (L)
= 2,250 L/
P
, .
P
, (
) 19
U
.
E: T N 50 /L, 6.5 ML/.
• M = 50 106/L 6.5 106L/
= 325 /
E EP.
E: P BOD5 55 /EP.
28,000 EP
BOD5 = 55 (/EP.) 28,000 (EP)
= 1,540,000 /
= 1,540 / 20
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• S D M L
– A BOD S
S P
E (PE)
– D
,
.
21
• WW
.
•
• T WW :
B
P
C (
CFSTR)
C
A
P
F22
23
K D C
24
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25 26
27
• U , / .
• R .
• A , , .
• A .
28
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D
•W .
1. P : , , , .
2. C : ,, , .
3. B : ( , ); ( )
29
B B
30
A , B C
:
C C
B B
C31
R V
32
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R V T
.
I
(; RW)
REACTORVESSEL
E
(; )
33
B R
T
.
T .
T .
T , .
M .
E: BOD .
34
P , ( ) R
F
.
T
.
T .
35
C ( )
(CSTR) C
.
C
.
36
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A
• A
.
37
C
T
.
I ,
.
I , . 38
P
R ,
, , .
W ,
( ) ( ).
39
• T
, :
– T
– T
– T
– A C
– T .
A
Q, C0 Q, C
C V, C
40
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M E
/
IO+ G/ = A
ORI S
41
S :M B
= − + (−)
= , 3 = ℎ ℎ ℎ , −3−1
= ℎ ,3−1 0 = ℎ ,−3 = ,−3
= − , −1 42
E F EAE F AEAE
AEBC AAEBC• AEROBIC TREATMENT TAKES PLACE IN
PRESENCE OF AIR.
• THE ORGANIC POLLUTANTS ARECONVERTED INTO CARBON DIOXIDE
AND WATER
• THERE IS NO SMELL IN THESURROUNDING AREA
• IT CONSUMES MORE ELECTRICAL
ENERGY.
• THE SLUDGE PRODUCTION IS HIGH
• ANAEROBI C TREATMENT TAKESPLACE IN ABSENCE OF AIR.
• THE ORGANIC POLLUTANTS ARECONVERTED BY ANAEROBICMICROORGANISMS TO A GASCONTAINING METHANE ANDCARBON DIOXIDE.
• THERE IS SMELL ALL AROUND THEPLANT
• THE ENERGY CONSUMPTION ISLOW
• THE SLUDGE PRODUCTION IS LOW
43
T
W T P
S P
SA
S
S
D
T
F
A
D
L D
I
C
R
S
T
44
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C T
. T :
A
T
45
P (P T)
46
PT F
.
S ()
()
.
T :
I
.
T .
47
T
H
H
. T
.
W
.
T
.
I
,
. 48
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PT
, .. F
, ,
.
49
MM
TP
TP
G
G
FOG
FOG
SS
B
H, ,
N,
B
H, ,
N,
50
H,
( ),
, ( )
( ).
I
A
O
H
O
P
F
P
O
A
A
K
R S
C (,
)
G C
F (
)
F
F
C ( )
52
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R SG:• F .
• S .
• S
, , ,
• S ( )
P
.
O:
R ,
.., ,
.
P .
P .
53
S E M S 10 30
54
S S 3 6
55
R D S 3 6
56
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C : O (, )
C ( )
M (,
, )
F
57
S C: T .
I ,
.
C S• C (
)
. C
.
B 5.08 10. 16
0.64 5.08 .
58
•S
,
.
F S
• F
.
•F S .
59
M
.
S
.
A
.
60
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D S
V
O = 0.6 / (
)
M = 0.751.0 / (
.
M = 0.4 /
T = 0.6 1.0 /
61
H L
(a) Head loss (hL) through bar racks as a function of:•
•
62
L = ()0.7 =
.
V = V
(/)
U =
(/)
= (2)
N:
63
() H (L)
•A
•B
•A
: ;
.
L = ()
W = ()
= ( ) ()
V = (/)
= .
= (2)
β= ( = 2.42;= 1.79)
64
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( ) R
H0 =
P0 =
P=
= 0.050.15 .
= 0.10.4 ().
• H .
• R
.
65
C
A 45 60 75 85
M ) 80 80
T
. T
.
66
67 68
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S
C
F
:
B O,
T S
C W
C C
T C
R, 3.5 35 L 1000 3
.
S 1020% 6401100 /3. 69
W D S
S
.
70
T
. T
.
T
.
71
D
G
I T
.T
,
.
C
72
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73
E 1: F :
– P (Q) = 80 X 106 L/
– C = 1 ; 5
– I = 60°
– C = 0.75
C;
– C
– M
– C
C
C
C
610 ,
.
C
.
74
G ()
, , , ,
:
S G;
• S
• D ( , ,
.)
• B
GRIT REMOVAL
75
G ;
U
C ,
B
T
T 70
85%.
M 10 70%.
T 0.004 0.1
3/ML . T
.76
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O
P
.
R ,
.
R .
T
B .
U S ( )
H
, ( 1
, ).
77
D
) C (T ) .
) A .
) M ( ).
78
C V G C ()
C )T ,
)V .79
• S
0.3 /.
• L
(
),
Q/A
.
80
H F G C D C
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W S
81
S L :
= ( − )
= (/) = (/)
= (/) = (/) = () = (//)
82
H F G C D C
F = 0.2 ,
=2650 /3 , =1.518× 103 //
= ( − ) 2
18
= 9.81 (2650 −1000) 0.00022
18 1.518 10−3
= 0.021 /
83
W
84
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85
Differentiationbetweenhorizontalcrosssection
andtransversalcrosssection
Horizontalcrosssectionarea=WXL
QQ
Transversalcrosssection
area=WXH
R , V
H
WL
C
(W X H);
, = (3 )
ℎ ( )
, = max ℎ (0.3 )
86
C
(W X L);
, = (3 )
( )
, = max (0.021 )
87
R :
C ,
At maximum flow, ℎ
= ℎ
= ℎ
= G.C.
V = / 0.2 = 0.021
/ (S )L = L G.C.
V = F (0.3 /)
88
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T H F G C
• D 4590
• H 0.3 0.45 /
• S 0.61.2 /
• L . A
1.52.0 .
• I ,
, 20%
.
L 14 ; 1.2 1.5 L > = 18
• H 3040%
, .
89
A G C
90
A G C
91
• T
. T
.
• A
.
• T .
N 0.15 0.45 3/
.
• L
.
• L 2.5 :1 5:1 2 5 .
D
R T
D:
D, 2:5
L, 7.5:20
W, 2.5:7.0
W D 1:1 5:1 2:1
D
,
25 3
A ,
3/ .
0.150.45 0.3
G :
G, 3/103 30.0040.200 0.015
92
S: M & E, I. [536]
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93
P :• H
• R
• U
• A
T ,
3 %.
Q :
• D
• R
• H
( )
E 1: D
:
= 10600 L/
= 17070 L/.
E
.
E 2: A
0.2 ,
2.65. S
0.016 0.022 /,
. A 0.3 /
. D
10 0003/. 94
F .
I , (FOG).
I
T .
T
. T . T .
FLOTATION
95
FLOW BALANCINGT
D 6
ADWF
T
A 96
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P
97
P
.
T
(),
() .
P
(≈ 2)
( )
.
98
T B :
R
R BOD5
R (WAS)
R
P
99
E
.
.
100
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L T P
101
T
T :
()R ;
()C, ;
()H,
102
C,
H, 103
T C P S T
104
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• T 2 ( 20
3 ). I ,
.
• T
, B
S (SOR). T SOR 3//2.
T 2 .
105
T C ST
K
B (S 3//2)
D
S G
H
I ( )
I
R :
F
S
W
106
T . T
.
R
.
S BOD5 30
.
S
.
A ,
. F ,
2
.
E
. I
(.. L )
.
H, , 50 70
25 40 BOD . F .
107
S O R P
R
108
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R T
D , 1.52.5 2.0
O , 3/2.
A 3050 40
P 80120 100
D,
R
D 34.9 4.3
L 1590 2440
W 324 4.99.8
C
D 34.9 4.3
D 3.060 1245
B , / 1/161/6 1/12
Designcriteriaforprimarysedimentationtank
109
ExampleExampleExampleExample :::: A WW
5000 3/. D
60 % .
110
S
111
S W T
• P
– C (60% 35% BOD )
• O & G
• T S S (TSS) 60% R
• P
• BOD 35%
– P
• S
• G S
• S F• P S
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• S
– C
• BOD 90% R
• TSS 90% R
– P
• T F
• A
• O
( )
S ()W ?
114
S T O
115
Aerationand rapid
mixing
Settlingcollects sludge
on bottom
S
airdiffuser
F r o m p
r i m a r y
p r o c e s s
T o t e r t i a r y
p r o c e s s
116
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I
117
• T
. .. BOD, COD
.
• T ,
,
.
• T .
• I .
118
• A . I
.
• T
.
• I ,
, .
• B WW ,
, , .
• T
.
Q L
119
B
H G?
?
D?
B
120
M
• T
,
• O
C
• O
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121
• H
• A
:
O ,
B
M C:
122
A
.
H (4.0
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B B
125
M
C A
A
.
O + O2 CO
2 + H
2O + NO
3
+
A
.
T
R +
C (M) G F U
126
T
.
A
.
I S (/L) X
(/L),
F.
B
R
WW
B
B U
127
C /
E
B
F
T
()
E
128
( 1)
T /
.
T () 1.
( 2)
• O , .
• B ; ,
.
• T ,
, .
• S , 2 .
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129
M . T
, .
S 3 ,
.
E
• A
, 4
.
• I , ,
.
N .
T .
• B M
• L
L .
=0 , X=X , S=S
T
/ /
∝
R
M
=
S
U µ ( 1)
T µ .
µ .
S ,
Aµ .
L : .
= () S
µ
L .
= . + D :
Wµ = µ /2
µ
S
µ :
S µ
A S µ : µ
S= 0
µµµµ
S S :
A (BOD)
=0
µµµµ=0 .
A µµµµ
S
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µµµµ :
O .
µµµµ = µµµµ /2
S µ = µ /2
K . =
=
.+
X M
S
C (
)
G
E & .
L
L C 1
W S (. P .)
. V S
T .
>>>>
∴ + ≈
∴ = .
= .
S .
= 0 =
=0
=
µµµµ
.
= 0 max I
+ + +
T
(.. 2)
XS 2
C 2
W , ..
S .
S
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I 2, , .S ∴ + ≈
N
=
= . +
=
. +
.
=
.
S C
C
.
, .(R : W ).
S & X
A S X & S
S X .
S .
= ( )
=
=
(−)
C =
= − 00 − C
= 0C
=
C
= C
= 0
Y T (Y) Y
140
E
• I ,
.
• D
.
• T, (X/, )
.
T .
( ) = X
= , (1)
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B T P
• T :
– S
– A
141
A A A F
B
O2
B
O2
B
O2
M/O
O2
A
S A
M/O
WW
M/O
,
D
142
SG P
A
AL
AG P
T
R
B
143
S G P
B
.
A
.
S
.
144
S G P C:
C
E
C
E
C
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145
A S P B• T
. T A,L F D S W M.
• S .
• M .
• O .
• O CO2 .
146
A S P B
147
A S P
• A B
• A
• F • R
• W
C
:
• C
• P
• O
• S
148
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A S P• W
• B
• O
• F (BOD)
• N
• C
• T
• A BOD,
• T • B ,
• P ,
• R .150
Q , X, SQ, X0, S0
Q , X
Q , X, S
Q , X, SQ, X0, S0
Q , X, S
Q , X
Figure:Figure:Figure:Figure: Schematic diagramofASP(a)withwastingfromtheaerationtankand(b)withwastingfromthesludgereturnline(recycleline)
()
()
Biological waste treatment with ASP is typically accomplished using a flow
diagram such as that shown in the figure.
A S P P
151
A S ()
• B () /.
• A .
• A , . T
. T .
152
T
S:
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A T () C:
• O
.
• T M L.
• I ,
.
153
:
• E :
154
A T () C:
• COHNS WW.
• I , ,
BOD 1.42
.
• T
,
.
• A ,
, WW.
2. F A T SS (SST, S C)
• R .
• I , .
• C L
, N .
• T .
• L A ( 20%)
RETURNED
ACTIVATED SLUDGE (RAS). T
.
• I ,
80%
.155
S S T (SST)
• W M
L :
.
• MLSS: M T
2000 4000 /L
. T MLSS RAS
10,000 20,000 /L. T
MLSS
> 2000 /L, O,
.
• MLVSS : M : 80% MLSS
MLVSS. 156
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• T
.
• A:
– C
– N M/O
– W M/O
157
M
– VT = Volume of the reactor + Volume of the settling tank
– Q = Influentflow rate
– Vr = Volume of the reactor
– Vs = Volumethe settlingtank
158
P D P
H T (HRT)
θ
159
= () = units of time ( ) − 0 ( ) −
N: Ф HRT.
160
M C R T (θ )
• A S A
• T
• R
.
• I
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161
θ = /
.
X = C MLSS ,
/L
Q = T , ML/
X = S .
, /L
V = A , ML
X = MLSS , /L
Q = V , ML/
M C R T (θ )
162
, ,
, ,
, ,
, ,
A S N
. ( , , .) BD
.
163
DIAGRAM OF ACTIVATED SLUDGE
Q
164
A T E D• C
;
aerationunder MLVSS of massTotal
daychamber aerationentering BODof Mass M
F /=
MLVSS kg
day BODkg /=
Note
MLSS is simpler to measure than MLVSS and because the ratio of MLVSS
to MLSS tends to be relatively constant for a given system (75 – 80%) F/M is
often expressed in terms of MLSS concentration
/ = 3 (
3 )
3 (3)
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165
FM R
• C .
• S M L S SC (MLSS), (MLVSS)
• F/M .
• C
166
/ = 3 (
3 )
3 (3)
/ = 00
/ =
0 = (3) 0 = (
3)
= . . , (3) = (3) = ()
A T E DI :
167
A T V
• R :
• V F/M X
/ = 00
= 00 (/)
168
A T E D
P :
• N
.
• D M
C R T
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C M R• T
.
• T .
• A .
• T .
• T F/M 0.040.07.
• T 1 BOD5//3.
• D O
2 /L.
• G (RAS)
.
• A RAS
.169
C M R A
• A .
• D F/M .
• G .
• T, .
170
air or
oxygen
RAS
WAS
aeration basin clarifier
to tertiary treatmentor surface discharge
C M R C:
• T
I
A
S
S
S B
171
T C S
172
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W
• Q= , 3/
• S = S , /3 /L
• X = B ( ) /L
• A T V 3
173
C M R C: A E
• T
:
– T
– T
– T
• A
!
U M B (
175
F .
. 176
Activated Sludge Wastewater
Treatment Plant
TertiaryFiltration(Optional)
Bar Rack/Screens
InfluentForceMain
Screenings
Grit
GritTank
PrimarySettling Tank
PrimarySludge
Activated SludgeAeration Basin
Air or Oxygen
Diffusers
SecondarySettling Tank
Return ActivatedSludge (RAS)
Waste Activated Sludge (WAS)Cl2
Chlorine Contact Basin
(optional)
toreceivingstream
wastewater flow
residuals flow
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Completely Mixed Activated Sludge (CMAS) Conventional (plug flow)
Activated Sludge (CAS)
plan viewPrimary effl.
to secondary clarifierRAS
C A S
180
• ExampleExampleExampleExample1111 :Design a complete-mix ASP and secondary settling
facilities to treat 0.25 m3/s of settled WW with 250 mg/L of
BOD5. The effluent is to have 20 mg/L of BOD 5 or less. Assume
that the temperature is 20 ◦ C and that the following conditions
are applicable:
– Influent VSS to reactor are negligible.
– Ratio of MLVSS to MLSS = 0.8.
– Return sludge concentration= 10000 mg/L of SS.
– MLVSS = = 3500 mg/L.
– D esig n = 1 0 d .
– Hydraulic regime of reactor = complete mix.
– Kinetic coefficients, Y = 0.65 lb cells/lb BOD5 utilized;kd = 0.06 d-1
– It is estimated that the effluent will contain about 22 mg/L of biological
solids,of which 65 % is biodegradable. – BOD5 = 0.68 X BODL.
– WW contains adequate nitrogen, phosphorus and other trace nutrients
for biologicalgrowth.
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181
• ExampleExampleExampleExample 2222 :::: An organic waste having a soluble BOD 5 of 250mg/L is to be treated with a complete-mix ASP. The effluent
BOD5 is to be equal to or less than 20 mg/L. Assume that the
temperature is 20 ◦ C, the flow rate is 5.0 Mgal/d, and that the
following conditions are applicable.
– Influent VSS to reactor are negligible.
– Return sludge concentration = 10000 mg/L of suspended solids = 8000
mg/L VSS.
– Mix ed-liquor VSS (MLVSS) = 3500 mg/L = 0.8 X total MLSS.
– = 10 d.
– Hydraulic regime of reactor = complete mix.
– Kinetic coefficients, Y = 0.65 lb cells/lb BOD5 utilized;kd = 0.06 d-1
– It is estimated that the effluent will contain about 20 mg/L of biological
solids, of which 80 % is volatile and 65 % is biodegradable. Assume thatthe biodegradable biological solids can be converted from ultimate BOD
demand to a BOD5 demand using the factor 0.68 ( e.g. BOD k value = 0.1
d-1, base 10).
– Waste contains adequate nitrogen, phosphorus and other trace nutrientsfor biologicalgrowth. 182
A G PM
.
E
B (T )
R
B T
183
A G P C:
()
C
E
E
C
()
184
B (T) F
L
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185
B F C:C :
• W
• S ; T M/O WW . T .
• U ; T WW .
• E C
• S ; T WW.
• R
186
O:
• W .
• M .
• O
B F C:
187
• O .
• W
• C
B F C:
188
• A
.
• O2
.
• P
, .
B F C:
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189
B FT
190
TF
S
R
S
A
A
A
CO2
O2D
A ,
191
B F C
• M 1.8 2.0
( )
• M 50 100
( )
• S 40 60 3/2
80 120 3/2
192
D G
BOD Loading(Kg/m3
media.d)
HydraulicLoading(m3 /m2
media.d)
Sludge Yield(kg/kg BOD)
Single pass 0.1 – 0.2 0.3 – 0.8 0.4 – 0.6
High rate
filtration(plasticmedia)
Up to 0.6 Up to 20 0.8 – 1.0
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ROTATING BIOLOGICAL CONTACTOR
A
. T
(, PVC,
)
40% . T
. A
,
.
ROTATING BIOLOGICAL CONTACTOR
B
• D .
• L,
.
• A,
.
• D
,
.• F E;
195
• S = E (/L)
• S = I (/L)
• = T
, (1 )
• D = F ()
• Q = H (3/2.)
• = C
()
196
W:
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197
F. S
B
S
E
I
P
D
S
E
A
S
P
P
• W ;
• O U M
CO2, H2O .
• S,
(.. M ).
• N .
198
S .............
SECONDARY SEDIMENTATION TANK
• M .
• T
.
• T
.
• D Z
S .
(R .)
199
S S T D
200