CE3503 Expectations – Wastewater Treatment
Legislative Mandate
-
-
Clean Water Act
o Fishable-swimmable
o Work toward zero discharge
o Technology-based treatment standards
National Pollutant Discharge Elimination System
o Issues discharge permits
o Water quality-based treatment standards
Wastewater Sources for Publicly-Owned Treatment Works (POTWs)
-
wastewater from homes and commercial premises
infiltration (know definition)
inflow (know definition)
problems caused by infiltration and inflow
-
industrial sources
o priority pollutants
o problems associated with industrial wastes
 pass-through pollutants
 harm to plant
industrial waste options
o treatment handled by industry
o industry pre-treats and sends to POTW
-
Wastewater Collection Systems
-
connection to house, 6”
collecting sewers, gather from houses, 8-12”
interceptor or trunk sewers, to plant, 15-27” or larger
-
designed for gravity flow and to –
o prevent in-line sedimentation
o rapid transport, avoid anaerobic conditions
o avoid scour to pipe
Wastewater Characteristics
-
BOD and SS, 200 mg/L in, 30 mg/L out
Ammonia, 25 mg/L in, 2 mg/L out
Phosphorus, 10 mg/L in, <1 mg/L out
Coliforms, 108 cells/100 mL in, 100 cells/100 mL out
Daily variation in wastewater flow
Preliminary Treatment
-
know purpose and method for each
o screening (bar racks and screens)
o comminutors
o grit chambers [design problem, Stokes’ Law]
o floatation
o flow equalization
Primary Treatment
-
objective and approach
basis of method
handling of floatables, sludge
primary clarifier [design problem, SOR]
Secondary Treatment
-
-
microbiological components
suspended growth, activated sludge
o aeration tank
o mixed liquor
o mixed liquor suspended solids
o secondary clarifier
o return activated sludge
o waste activated sludge
o F/M ratio [design problem, F/M]
 growth phases
 settleability
 BOD removal
o Sludge age or solids retention time [design problem, SRT]
o Process variations
 high rate
 conventional
 extended aeration
o Process modifications
 plug flow, tapered aeration
 plug flow, step aeration
 completely mixed, conventional
 completely mixed, contact stabilization
fixed film
o trickling filters [design problem, TF]
o rotating biological contactors
Tertiary Treatment
-
a.k.a. advanced waste treatment or AWT
suspended solids, filtration
dissolved BOD, adsorption
phosphorus, precipitation with alum or ferric chloride
nitrogen, nitrification to eliminate ammonia
denitrification, to remove all nitrogen
Disinfection
-
chlorination
ozonation
UV
Sludge Treatment and Disposal
-
-
-
stabilization to reduce odor, putrescibility and pathogens
dewatering to make sludge easier to handle in disposal
aerobic digestion; long term endogenous phase aeration
anaerobic digestion
o acid formers
o methane formers
dewatering
o sand drying beds (small plants)
o belt filter
o centrifuge
disposal
o landfill
o land application
o re-use (Millorganite)
o incineration
 fluidized bed
 multiple hearth furnace
 co-generation
Summary
Be able to sketch out a schematic of a wastewater treatment plant.
Design Problem 1. Grit Chamber.
A wastewater treatment plant will receive a flow of 35000 m3·d-1 (~10 MGD).
Calculate the surface area (m2), volume (m3), and detention time (s) of a 3m deep
horizontal flow grit chamber which will remove grit with a specific gravity of >1.9 and a
size >0.2 mm.
Calculate the particle settling velocity (vs) using Stokes’ Law.
vs 
g  ( p  w )  d 2
18
9.81  (19
.  10
. )  998  (0.2 x103 ) 2
vs 
18  1002
.
x103
vs  0.02 m  s1 or 1728 m  d 1
The design particle (and all larger, denser particles) will be removed if the horizontal
velocity within the tank is less than the settling velocity (vh < vs).
Calculate the required surface area of the tank.
vh 
A
Q
or
A
A
Q
vh
35000
 20.25 m2
1728
Calculate the tank volume.
V  A  H  20.25  3  60.75 m3
Calculate the detention time.

V 60.75

 0.0017 d  2.5 min
Q 35000
Design Problem 2. Primary Clarifier.
A wastewater treatment plant will receive a flow of 35000 m3·d-1 (~10 MGD).
Calculate the surface area (m2), diameter (m), volume (m3), and retention time of a 3m
deep, circular, primary clarifier which would achieve 50% suspended solids removal.
What BOD removal efficiency would be expected?
Calculate the required surface area.
Q
35000
m3  d 1
A

 3 2 1  583 m2
SOR
60
m m d
Calculate the tank diameter.
A    r2; r 
A


583

 13.6 m ; D  2r  27.2 m
Calculate the tank volume.
V  A  H  583  3  1749 m3
Calculate the retention time.

V 1749

 0.05 d  12
. h
Q 35000
From the design plot, a BOD removal efficiency of ~25% could be expected.
Design Problem 3. Activated Sludge - F/M Ratio.
A wastewater treatment plant will receive a flow of 35000 m3·d-1 (~10 MGD)
with a raw wastewater CBOD5 of 250 mg·L-1. Primary treatment removes ~25% of the
BOD. Calculate the volume (m3) and approximate retention time (d) of the activated
sludge aeration basin required to run the plant as a “high rate” facility (F/M = 2 d -1). The
aeration basin MLSS concentration will be maintained at 2000 mgMLSS·L-1.
Calculate the CBOD5 feed to the aeration tank.
So  250 mg  L1  250 g  m3  0.75103 kg  g 1  01875
.
kg  m3
Calculate the MLSS concentration of the aeration tank.
X  2000 mg  L1  2000 g  m3 103 kg  g 1  2 kg  m3
Use the definition of F/M and the data above to calculate V.
F Qo  So 35000 m3  d 1  01875
.
kg  m3


 2.0 d 1
3
M VX
V  2 kg  m
V  1640 m3
Calculate the retention time.

V 1640

 0.047 d  11
. h
Q 35000
This retention time is approximate, because the flow estimate does not include return
activated sludge.
Design Problem 4. Activated Sludge - SRT.
A wastewater treatment plant will receive a flow of 35000 m3·d-1 (~10 MGD). Calculate
the mass of sludge wasted each day (Qw·Xw, kg·d-1) for an activated sludge system
operated at a solids retention time (SRT or c) of 5 days. Assume an aeration tank
volume of 1640 m3 and a MLSS concentration of 2000 mg·L-1. What fraction of the
solids leaving the aeration tank is this?
Set up the SRT design equation.
c 
XV
Xw Qw
Rearrange and solve for Qw·Xw.
Qw  Xw 
X  V 1640  2000

 656,000 g  d 1  656 kg  d 1
c
5
Calculate the % of the solids leaving the aeration tank.
Waste sludge 
656
 1%
35000  2
Design Problem 5. Trickling Filter.
A wastewater treatment plant will receive a flow of 35000 m3·d-1 (~10 MGD) with a raw
wastewater CBOD5 of 250 mg·L-1. Primary treatment removes ~25% of the BOD.
Calculate the diameter (m) of the 3m deep trickling filter(s) which would accommodate a
hydraulic loading of 5 m3·m-2·d-1 and an organic loading of 250 gCBOD·m-2·d-1.
Calculate the CBOD5 load to the filter.
CBOD Load  250 mg  L1  0.75  35000 m3  d 1  6562500 g  d 1
Calculate the required filter area based on CBOD loading.
CBOD Load
6562500 g  d 1
2
A

2
1  26250 m
Design Organic Load 250 g  m  d
Calculate the diameter.
A
A    r2; r 


26250

 91 m ; D  2r  182 m
Thus, ~9 filters with an area of 2831 m2 (60m diameter) would be required.
Calculate the required filter area based on the hydraulic loading.
Hydraulic Load
35000 m3  d 1
2
A

3
2
1  7000 m
Design Hydraulic Load 5 m  m  d
Calculate the diameter.
A    r2; r 
A


7000

 47 m ; D  2r  94 m
Thus, ~3 filters with an area of 2831 m2 (60m diameter) would be required.
The design must be made to accommodate both the BOD and hydraulic loads, therefore
the BOD design criterion governs here.