1. Lecture 6: Water & Wastewater Treatment
Objectives:
Define primary, secondary, and tertiary treatment
Define BOD
Describe the activated sludge process
Setup and solve a mass balance for an activated sludge system

2. Review Sorption: Settling Kd=Cs/CL CT=(1+KdCss)CL
Fraction sorbed vs. fraction remaining in water
Settling
Settling velocity:
Percent of particles removed: (1-Css/Css,o) x 100%
Where,

3. Well-Mixed Settling Tankvs
Css
Q, Css,o
Q, Css V
Suspended solids remaining:
Define the Overflow Rate:
~ 20 – 100 m/day in treatment plants

4. Wastewater Treatment POTW – Publicly Owned Treatment Works
0.4 – 0.6 m3/person/day
15 million people in Los Angeles  7.5 x 106m3/day or 2000 MGD (million gallons per day)
Hyperion – 450 MGD
Clean Water Act (CWA) – 1977 – Set effluent (what is released by treatment plants into the environment) standards

5. Stages of Water Treatment
Primary
Contaminants (60% of solids and 35% of BOD removed)
Oil & Grease
Total Suspended Solids (Css or TSS) – 60% Removed
Pathogens
BOD – 35% removed
Processes
Screens
Grit Settling
Scum Flotation
Primary Settling

6. Stages (continued) Secondary Contaminants Processes BOD – 90% Removed
TSS – 90% Removed
Processes
Trickling Filter – rotating disk
Activated Sludge – Suspended and mixed
Oxidation ponds – lagoons
(promote contact between microbes and contaminants)

7. Stages (continued) Tertiary Contaminants Processes Nutrients
Dissolved solids (e.g., salt, other ions, etc.)
Processes
Denitrification – bacteria
Phosphorus removal – precipitation
Other chemicals – adsorption and precipitation

8. Primary Sludge (cont’d)

9. Primary Sludge

10. Primary Sludge (cont’d)
Q, Css,o
Q, Css
Given:
Q = 4000 m3/d
Css,o = 200 mg/L and Css = 100 mg/L
Sludge density = 0.05 kg/L
Overflow rate of 50 m/d
Find
Population of town served by this unit
Sludge production rate
Area of settling tank
Settling velocity of particles
Cut-off size of particles (find the particle diameter corresponding to this settling velocity. Assume rs = 2600 kg/m3. All particles larger than this size will settle)

11. Activated Sludge

12. Activated Sludge Components

13. Activated Sludge Components

14. Activated Sludge (cont’d)

15. Definition of BOD
Microorganisms (e.g., bacteria) are responsible for decomposing organic waste. When organic matter such as dead plants, leaves, grass clippings, manure, sewage, or even food waste is present in a water supply, the bacteria will begin the process of breaking down this waste. When this happens, much of the available dissolved oxygen is consumed by aerobic bacteria, robbing other aquatic organisms of the oxygen they need to live. Biological Oxygen Demand (BOD) is a measure of the oxygen used by microorganisms to decompose this waste. If there is a large quantity of organic waste in the water supply, there will also be a lot of bacteria present working to decompose this waste. In this case, the demand for oxygen will be high (due to all the bacteria) so the BOD level will be high. As the waste is consumed or dispersed through the water, BOD levels will begin to decline.

16. Activated Sludge Nomenclature
Q+QR, S, X
Q-Qs, S
Q, So, Xo
Qs+QR, Xs
QR, Xs
Qs, Xs
S stands for conc. of substrate (organic matter, waste, etc.) or BOD
X stands for conc. of microorganisms

17. Activated Sludge Nomenclature (cont’d)
Q, So, Xo
Q+QR, S, X
~Q, S
m, V
Qs+QR, Xs
QR, Xs
Qs, Xs
Assumptions:
Effluent bacteria concentration is 0
Concentration of substrate or BOD in sludge is 0
Sludge flowrate (Qs) is much smaller than Q

18. Decay of BOD and growth of organisms
Substrate or BOD (S) decays with rate k:
Microbes (X) grow at rate m:

19. Activated Sludge Equations
The following equations are derived from conducting mass balances over:
The entire system
The aeration tank
The sedimentation tank
Any good book on wastewater engineering will have the derivations if you are curious!

20. Activated Sludge Equations
Biomass (X) balance over entire system:
Substrate (S) balance over entire system:

21. More AS equations Mass balance over sedimentation tank:
Other equation(s)/rules of thumb:
F/M = QSo/XV - Food-to-microbe ratio: 0.3 – 0.7 d-1
QR ~ 0.25 – 0.50 x Q
X ~ 1000 – 2000 mg/L
Problem types:
Given Q, So, and S (target concentration)
Find QR, Qs, X, m, V, Y

22. Example Find Qs, m, V, Y Given: Q = 1000 m3/d So = 150 mg/L
S = 15 mg/L
QR = 240 m3/d
F/M = 0.3 d-1
X = 2000 mg/L
Xs = 1% or 10,000 mg/L