Water, Part 2
Wastewater Treatment
Primary Chapter: 11
Supplemental Chapters: 8, 9
1
Activity
What’s the problem with direct discharge of
untreated wastewater?
Write your ideas.
Share with a partner.
Share with the class.
2
Treatment Processes
Treatment process = f(amount, type/source)
Discharge limits = f(type/source, discharge
location, time of year)
3
CENTRALIZED TREATMENT
4
Types of Wastewater
(Section 11.1.1)
From where does it all come?
5
Transport of WW
6
Characteristics of WW
(Section 11.1.2)
Parameter
BOD5
SS
COD
Total-N
NH3-N
Organic-N
NO3P-Total
Ortho-P
Organic-P
Typical Concentration
(mg/L)
250
250
500
40
30
10
0
10
6
4
7
Activity
Why is it important to obtain local data on
wastewater composition and flow rates
rather than using typical data when
designing a new or an expansion to a
treatment plant?
Write your ideas.
Share with a partner.
Share with the class.
8
Regulations
(Sections 9.3.2 and 9.3.3)
What’s the primary law for WWT?
9
Activity
Who must obtain an NPDES permit?
A.
B.
C.
D.
E.
Manufacturers
Point source dischargers
Farmers
Septic tank owners
Municipalities
10
Typical Municipal WWTP
11
Degrees of Treatment - Example
Raw Sewage
Bar Rack
Grit Chamber
Preliminary
Treatment
Equalization Basin
Solids
Handling
Primary
Pump
Treatment
Primary Clarifier
Biological Treatment
Advanced or
Tertiary Treatment
Secondary Clarifier
Secondary
Treatment
Disinfection
Receiving Body
12
Influent – How much is removed?
Solids
BOD
13
After Primary Treatment
14
After Secondary Treatment
15
Preliminary Treatment
(Section 11.2.1)
Screens
Grit Chamber
Comminutors
(Grinders)
Bar
Racks
16
Primary Treatment
(Section 11.2.2)
17
Secondary Treatment
(Section 11.3)
18
Section 9.1.2
OXYGEN DEMAND
19
Oxygen Demand
Amount of oxygen required to oxidize a
waste
 Methods

 Theoretical
oxygen demand (ThOD)
 Biochemical oxygen demand (BOD)
 Chemical oxygen demand (COD)
20
Theoretical Oxygen Demand
Total ThOD = C-ThOD + N-ThOD


C-ThOD = stoichiometric amount of O2 required
to convert an organic substance to CO2, H2O,
and NH3
N-ThOD = stoichiometric amount of O2 required
to convert NH3 and organic N to NO321
Example 1
What is the total ThOD to oxidize completely
25 mg/L of ethanol (CH3CH2OH)?
CH3CH2OH + a O2  b CO2 + c H2O
22
Determine Volume of O2 or Air
23
Example 2
What is the ThOD to oxidize completely 25
mg/L of serine (CH2OHCHNH2COOH)?
CH2OHCHNH2COOH + a O2  b CO2 + c
H2O + d NH3
NH3 + a O2  b HNO3 + c H2O
24
Biochemical Oxygen Demand
(BOD)
25
Lab: Unseeded BOD
DOi  DO f
BOD t 
 DOi  DO f DF
Vs
Vb
BODt = BOD at t days (mg/L)
 DOi = initial dissolved oxygen (mg/L)
 DOf = final dissolved oxygen (mg/L)
 Vs= sample volume (mL)
 Vb = bottle volume (mL) = 300 mL
 DF = dilution factor = Vb/Vs

26
Lab: Seeded BOD
V

DOi  DOf   Bi  B f 1  s Vb 


BODt 
Vs
Vb
Bi = initial DO of blank (mg/L)
 Bf = final DO of blank (mg/L)

27
In-Class Activity


10 mL of a wastewater sample are placed in a 300-mL
BOD bottle with unseeded nutrient broth. The initial DO
of the sample is 8.5 mg/L. The DO is 3 mg/L after 5
days. What is the 5-day BOD?
A 17oC sample is initially saturated with oxygen.
Saturated seeded dilution water is used to obtain a 1:25
dilution. The final DO of the seeded dilution water is 8.2
mg/L while the final DO of the diluted sample is 2.8
mg/L. What is the 5-day BOD?
28
In-Class Activity
You received the results of a BOD test of the influent to a municipal WWTP
run with 300-mL bottles. The initial DOs of the samples and seeded dilution
water were at saturation (9.07 mg/L). All samples were run at a dilution
factor of 40:1. The 5-day DOs are shown in the table below. The client is on
the phone with your boss wanting to know why he hasn’t gotten a report yet.
Justify why you threw out this data and made the lab redo the test.
Bottle Number
10
14
18
22
28
1B
Type1
B
B
S
S
S
DO (mg/L)
8.91
8.89
7.85
8.15
8.52
= blank (seeded dilution water), S = sample with seeded dilution water
29
Rate of BOD Removal
Relate BOD exerted (BODt or Lt) to total,
or ultimate, BOD (BODu or L)
 Assume that the BOD reduction rate
(dC/dt) is proportional to the BOD
remaining (C):

dC
dt
 kC
30
Rate of BOD Removal cont.
Integration yields:

y  L 1 e
 k1t


BODt  BODu 1  e
y
 k1t

= BOD exerted in t days = BODt
 L = ultimate BOD = BODu
 k1 = BOD degradation rate constant = deoxygenation
constant
31
Ultimate BOD
32
In-Class Activity continued
10 mL of a wastewater sample are placed in
a 300-mL BOD bottle. The initial DO of the
sample is 8.5 mg/L. The DO is 3 mg/L after
5 days. What is the 5-day BOD? 165 mg/L
What is the 3-day BOD if the reaction rate
constant is 0.23/d?
33
Chemical Oxygen Demand (COD)
34
Secondary Treatment
(Section 11.3)
35
Activated Sludge: Aeration Basin
(Sections 11.3.2 - 11.3.4)
36
Aeration Basin Design


Kinetics
Mean cell residence time & hydraulic detention time
37
Kinetics: Logistic Growth
max
^
dX
 X
dt
 (d1)
max/
2

S
Ks  S
 ^

X
X  S 
rs 

Y
Y  K s  S 


Ks
S (mg/L)
38
Aeration Basin Design
Mean cell residence time & hydraulic detention time
39
MCRT from a Reactor without
Recycle
V, S, X
Q, So, Xo
Q, S, X
40
Wasting from Recycle Line
Qe = Q-Qw,
S, Xc
Q, So, Xo
V, S, X
Vc
Qr , S, Xr
Qw , S, Xr
41
Example
A conventional WWTP receives 2 MGD with
an average BOD of 165 mg/L to the aeration
basin. The aeration basin is 100,000 ft3.
The MLSS is 2,800 mg/L and the effluent SS
is 25 mg/L. The WAS is 38,000 gpd from
the recycle line. The SS of the recycle flow
is 9,000 mg/L. What is the mean cell
residence time?
42
General Equation for
Mean Cell Residence Time
VX
VX
c 

Qw X w  Qe X c Qw X w
43
Secondary Clarifier
(Section 11.3.5)
44
Sludge Volume Index (SVI)
45
Other Secondary Treatment
Options
46
Sequencing Batch Reactor
(SBR)
47
Aerated Lagoons
48
Oxidation Ditch
•
49
Trickling Filters
(Section 11.3.1)
50
Rotating Biological Contactors
(Section 11.3.1)
51
UV Generator and Lamps
Section 11.3.6
DISINFECTION
Chlorine Contact Basin
52
Section 11.4
TERTIARY TREATMENT
53
Wetlands
(Section 11.4.3)
54
EFFLUENT DISCHARGE
55
Rapid Infiltration
•
56
Slow-Rate Land Application
57
Overland Flow
58
Discharge to a Stream
(Section 8.2.3)
Effects:
 Nitrogen
species
 Biodiversity
 DO
59
Discharge to a Stream: DO
9
Initial Deficit
Saturation DO
8
Deficit
DO Recovery
DO (mg/L)
7
6
5
4
Minimum DO
Actual DO
3
2
1
0
Critical
0
2
Discharge Point
Point
4
6
8
10
12
Time (d) or Distance (km)
60
Streeter-Phelps Model


k1 Lo
 k1t
k2t
k2 t
D
e e
 Do e
k 2  k1
where D = oxygen deficit
= DOsat - DOactual
61
Critical Point
Obtain from dD/dt = 0:
t crit
 k2 
Do k 2  k1  
1


ln  1 
k 2  k 1  k1 
k1 Lo

xcrit  utcrit
62
Evaluation of Model


Very simple to use
But not like nature
 Assumes
steady state
 Assumes a single discharge
 Assumes no upstream dispersion
 Assumes complete mixing
 Assumes all the BOD is soluble
 Doesn’t include scouring
 Doesn’t include DO from algae
63
Discharge to a Lake
(Section 8.2.2)
Cougar
Lake
64
Effect on a Lake
Cougar
Lake
65
Section 11.5
SLUDGE MANAGEMENT
66
Purpose
Reduce/inactivate pathogens
 Increase solids content
 Reduce odor & putrescence

67
Sludge Treatment
68
Sludge Disposal
How can we get rid of the sludge???
And how do we choose which option
is best?
69
OTHER DESIGN ISSUES
70
ONSITE TREATMENT
71
Conventional Septic System
72
Anaerobic Septic Tank
73
Absorption Fields
Shallow
Trench
Drop Box
74
Septic Tank with Sand Filter
Single Pass (Intermittent)
Recirculating
75
Mound System
Distribution Lateral
Topsoil and
Vegetation Cover
Marsh Hay
Sand
Fill
Bed of Coarse
Pipe from Pump
Aggregate (0.5 - 2”) (pressure distribution line) Plowed Layer
(Ground Surface)
76
Aerobic Treatment Units (ATU)
77
Lagoon/Waste Stabilization Pond
78