1. Wastewater Treatment: Characteristics and Systems

2. Significance of Wastewater Contaminants
Suspended solids – can cause sludge deposits and anaerobic conditions in the environment
Biodegradable organics – can cause anaerobic conditions in the environment
Pathogens – transmit disease
Nutrients – can cause eutrophication
Heavy metals – toxicity to biota and humans
Refractory organics – toxicity to biota and humans
Dissolved solids – interfere with reuse

3. Characteristics of Wastewater
Physical
Colour – grey
Septic sewage is black (due to precipitation of iron sulfide)
Chemical
Number of chemicals found in w/w is limitless
Measure of w/w strength often use COD and BOD
COD – Chemical Oxygen Demand
BOD – Biological Oxygen Demand

4. Characteristics of Domestic Wastewater

5. On-Site Disposal Systems
In locations where sewers and a centralized wastewater treatment system are not available, on site disposal must be used
Septic systems most common for individual residences
“Engineered systems” used for unfavorable site conditions
Larger systems required for housing clusters, rest areas, commercial and industrial facilities

6. Septic Systems

7. Septic Systems
Septic Tank – settling, flotation and anaerobic degradation

8. Septic Systems
Drain field (cross-section) – aerobic degradation

9. Septic Systems Soil must pass percolation test Design specifications
soil type
rate of water infiltration
depth to water table
Design specifications
Tank volume and number of chambers
Drain field size
Drain field materials
Basis for design is empirical
Tank must be “pumped” to remove solids every 1-3 years
Drain field replacement may be required

10. Engineered Systems
Engineered system required for soils that don’t “perc” – common in many areas, especially lakefront property in MI
Typical approach is to design and build something equivalent to a drainfield
Mound systems most common in MI
Other types of “filters” may be used
Typically about 3 times more expensive than drain field

11. Engineered Systems
Mound System

12. Engineered Systems Intermittent sand filter can be designed for
pulsed dosing
even distribution
high treatment efficiency
leakage protection

13. Municipal Wastewater Treatment Systems
Pretreatment – removes materials that can cause operational problems, equalization optional
Primary treatment – remove ~60% of solids and ~35% of BOD
Secondary treatment – remove ~85% of BOD and solids
Advanced treatment – varies: 95+ % of BOD and solids, N, P

14. Pretreatment of Industrial Wastewaters
Industrial wastewaters must be pretreated prior to being discharged to municipal sewer system
Approach is to remove materials that will not be treated by municipal system
Local authority must monitor and regulate industrial discharges
Pretreatment requirements set by U.S. EPA

15. Bar racks Purpose Solid material stored in hopper and sent to landfill
remove larger objects
Solid material stored in hopper and sent to landfill
Mechanically or manually cleaned

16. Grit Chambers
Purpose: remove inert dense material, such as sand, broken glass, silt and pebbles
Avoid abrasion of pumps and other mechanical devices
Material is called “grit”

17. Grit Chambers: Velocity Controlled

18. Type I Settling -- Stokes’ Law
where
νs = settling velocity
ρs = density of particle (kg/m3)
ρ = density of fluid (kg/m3)
g = gravitational constant (m/s2)
d = particle diameter (m)
μ = dynamic viscosity (Pa·s)

19. Example: Grit Chamber Design
Design a grit chamber to remove sand particles (p = 2650 kg/m3) with a mean diameter of 0.21 mm. Assume the sand is spherical and the temperature of the wastewater is 20 oC. The wastewater flow is 10,000 m3/d. A velocity of 0.3 m/s will be automatically maintained, and the depth must be 1.5 times the width at maximum flow.
Dynamic viscosity:
mPa-s
Pa-s=10-3mPa-s
Pa=N/m2
N=kg-m/s2
Therefore,
Pa-s = (kg-m/s2)(1/m2)(s) = kg/(m-s)

20. Example
Calculate settling velocity

21. Example
Calculate the cross-sectional area

22. Example
Calculate the width and depth

23. Example
Determine the detention time required for a particle to fall the entire tank depth
Determine the length to achieve this detention time

24. Example Thus, the tank must have dimensions W = 0.51 m D = 0.76 m
L = 5.8 m