1. INTRODUCTION TO Physical-chemical Treatment CE 523
J.(Hans) van Leeuwen

2. Instructor (33%) Professor J. (Hans) van Leeuwen
from/of the Lions
Born in Gouda, Netherlands
Grew up in South Africa
Lived in Australia for 7 years
Lived in Ames for 11 years
Specialty: Environmental and Bioengineering
Industrial wastewater treatment and
product development based on waste materials

3. Research activities
Beneficiation of biofuel co-products by cultivating fungi

4. Ozonation applications
Selective disinfection
Selective oxidation
Alcohol purification

5. Keeping exotic aliens out of our portsET
Zebra mussels

6. Human technological development
From scavengers…

7. …to use of fire… but, this ultimately led to…
Use of fire was the turning
point in the technological
development of humans
leading to extended diet
food preservation
better hunting
agriculture
industry
Top of the food chain!
but, this ultimately led to…

8. …Overpopulation…

9. P o l u t i n .

10. Pollution of a small stream

11. Consequence of pollution

12. …and ecological disasters

13. …and more disasters

14. Distribution of Earth’s water

15. Dangers lurking in water

16. Pollution from informal housing

17. Waterborne diseases
Map by Lord John Snow of the cholera outbreak in London in 1854 – the Broad Street Epidemic.
This is considered the root of epidemiology.

18. Spread of Cholera in London 1854
1-3 September: 127 dead
By 10 September: 500
Ultimately: 616 dead

19. Cholera – the rapid killer
SEM micrograph of Vibrio cholerae, a Gram-negative bacterium that produces cholera toxin, an enterotoxin, which acts on the mucosal epithelium lining of the small intestine This is responsible for the disease's most salient characteristic, exhaustive diarrhea.
Bottom: cholera toxin

20. Examples of organisms secreting enterotoxins
Bacterial
Escherichia coli O157:H7
Clostridium perfringens
Vibrio cholerae
Yersinia enterocolitica
Shigella dysenteriae
Staphylococcus aureus (pictured)
Viral
Rotavirus (NSP4) 
(Institute for Molecular Virology. WI)

21. Dissolved oxygen Importance Why is oxygen in water important?
Dissolved oxygen (DO) analysis measures the amount of gaseous oxygen (O2) dissolved in an aqueous solution. Oxygen gets into water by diffusion from the surrounding air, by aeration (rapid movement), and as a product of photosynthesis.
DO is measured in standard solution units such as milligrams O2 per liter (mg/L), millilitres O2 per liter (ml/L), millimoles O2 per liter (mmol/L), and moles O2 per cubic meter (mol/m3).
DO is measured by way of its oxidation potential with a probe that allows diffusion of oxygen into it.
The saturation solubility of oxygen in wastewater can be expressed as
Cs =  (0.99)h/88 x /(T )
For example, in freshwater in Ames at 350m and 20°C, O2 saturation is 8.8 mg/L. (Check for yourself, with  = 1)

22. BOD
Biochemical oxygen demand or BOD is a procedure for determining the rate of uptake of dissolved oxygen by the organisms in a body of water
BOD measures the oxygen uptake by bacteria in a water sample at a temperature of 20°C over a period of 5d in the dark. The sample is diluted with oxygen saturated de-ionized water, inoculating it with a fixed aliquot of microbial seed, measuring the (DO) and then sealing the sample to prevent further oxygen addition. The sample is kept at 20 °C for five days, in the dark to prevent addition of oxygen by photo-synthesis, and the dissolved oxygen is measured again.
The difference between the final DO and initial DO is the BOD or, BOD5.
Once we have a BOD5 value, it is treated as just a concentration in mg/L
BOD can be calculated by:
Diluted: ((Initial DO - Final DO + BOD of Seed) x Dilution Factor
BOD of seed (diluted activated sludge) is measured in a control: just deionized water without wastewater sample.
Significance: BOD is a measure of organic content and gives an indication on how much oxygen would be required for microbial degradation.

23. Oxygen depletion in streams

24. DO sag definitions

25. Cumulative oxygen supply + demand
Plotting the two kinetic equations separately on a cumulative basis and adding these graphically produce the DO sag curve

26. Streeter-Phelps Model*
Mass Balance for the Model
Not a Steady-state situation
rate O2 accum. = rate O2 in – rate O2 out + produced – consumed
rate O2 accum. = rate O2 in – – rate O2 consumed
Kinetics
Both reoxygenation and deoxygenation are 1st order
* Streeter, H.W. and Phelps, E.B. Bulletin #146, USPHS (1925)

27. Kinetics* for Streeter-Phelps Model
Deoxygenation
L = BOD remaining at any time
dL/dt = Rate of deoxygenation equivalent to rate of BOD removal
dL/dt = -k1L for a first order reaction
k1 = deoxygenation constant, f’n of waste type and temp.
*See Kinetics presentation if unfamiliar with the mathematical processing

28. Developing the Streeter-Phelps
Rate of reoxygenation = k2D
D = deficit in D.O.
k2 = reoxygenation constant*
Where
T = temperature of water, ºC
H = average depth of flow, m
ν = mean stream velocity, m/s
D.O. deficit
= saturation D.O. – D.O. in the water
There are many correlations for this.
The simplest one, used here, is from O’Connor and Dobbins, 1958
Typical values for k2 at 20 °C, 1/d (base e) are as follows:
small ponds and back water
sluggish streams and large lakes
large streams with low velocity
large streams at normal velocity
swift streams
rapids and waterfalls > 1.15

29. Combining the kinetics
Net rate of change of oxygen deficiency, dD/dt
dD/dt = k1L - k2D
where L = L0e-k1t OR
dD/dt = k1L0e-k1t - k2D

30. Integration and substitution
The last differential equation can be integrated to:
It can be observed that the minimum value, Dc is achieved when dD/dt = 0:
, since D is then Dc
Substituting this last equation in the first, when D = Dc and solving for t = tc:

31. Example: Streeter-Phelps
Wastewater mixes with a river resulting in a
BOD = 10.9 mg/L, DO = 7.6 mg/L
The mixture has a temp. = 20 C
Deoxygenation const.= 0.2 day-1
Average flow = 0.3 m/s, Average depth = 3.0 m
DO saturated = 9.1 mg/L
Find the time and distance downstream at which the oxygen deficit is a maximum
Find the minimum value of DO

32. Solution…some values needed
Initial Deficit 
Do = 9.1 – 7.6 = 1.5 mg/L
(Now given, but could be calculated from proportional mix of river DO, presumably saturated, and DO of wastewater, presumably zero)
Estimate the reaeration constant:
k2 = 3.9 v½ (1.025T-20)½
H3/2  
k2 = 3.9 x (0.3m/s)½ ( )½
(3.0m)3/2  
= d-1

33. Solution…time and distance
Note that the effects will be maximized almost 70 km downstream

34. Solution…maximum DO deficiency
Note that this BOD could have been calculated from mixing high-BOD wastewater with zero or near-zero BOD
The minimum DO value is = 6 mg/L
Implication: DO probably not low enough for a fishkill, but if continued could lead to species differentiation and discourage sensitives species like trout.