4lecture;wastewater treatment

8/17/2019 4Lecture;Wastewater Treatment

1/50

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W

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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!

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– T ( .. 40 L/ .

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C 15,000 EP 150 L/EP.

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ADWF = 15,000(EP) 150(L/EP.) 1(L)/1000 (L)

= 2,250 L/

P

, .

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, (

) 19

U

.

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• 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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12/5/20

• S D M L

– A BOD S

S P

E (PE)

– D

,

.

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• WW

.

•

• T WW :

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P

C (

CFSTR)

C

A

P

F22

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• U , / .

• R .

• A , , .

• A .

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D

•W .

1. P : , , , .

2. C : ,, , .

3. B : ( , ); ( )

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:

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C31

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.

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(; RW)

REACTORVESSEL

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(; )

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T .

T , .

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F

.

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.

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.

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.

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A

• A

.

37

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.

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I , . 38

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• T

, :

– T

– T

– T

– A C

– T .

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Q, C0 Q, C

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40


11/50

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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

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I

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12/5/20

C T

. T :

A

T

45

P (P T)

46

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.

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()

.

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I

.

T .

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H

H

. T

.

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.

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.

I

,

. 48


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12/5/20

PT

, .. F

, ,

.

49

MM

TP

TP

G

G

FOG

FOG

SS

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N,

B

H, ,

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50

H,

( ),

, ( )

( ).

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A

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H

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A

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C (,

)

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F (

)

F

F

C ( )

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14/50

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R SG:• F .

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, , ,

• S ( )

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.., ,

.

P .

P .

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15/50

12/5/20

C : O (, )

C ( )

M (,

, )

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I ,

.

C S• C (

)

. C

.

B 5.08 10. 16

0.64 5.08 .

58

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,

.

F S

• F

.

•F S .

59

M

.

S

.

A

.

60


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12/5/20

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

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S

.

70

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. 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

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, , , ,

:

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 ,

.

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, ).

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) M ( ).

78

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0.3 /.

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(

),

Q/A

.

80

H F G C D C


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81

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= ( − )

= (/) = (/)

= (/) = (/) = () = (//)

82

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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


35/50

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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

4lecture;wastewater treatment

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