1. Environmental Modeling Chapter 7: Dissolved Oxygen Sag Curves in Streams Copyright © 2006 by DBS

2. Quote “[Mathematics] The handmaiden of the Sciences” -Eric Temple Bell

3. Concepts Introduction Input sources Mathematical Model Sensitivity analysis Limitations

4. Case Study: Any Stream, Anywhere Every stream has inputs of organic waste –Spreads disease –Consumes DO on decomposition Ancient communities built near flowing water e.g. NY City, London, W. Europe

5. Case Study: Any Stream, Anywhere The Problem: D.O. < BOD Sewage treatment begins Chemical process: MO’s consume DO Physical process: Re-aeration by atmosphere Meadows et al., 2004

6. Introduction Modeling the effects of release of oxidizable organic matter into a flowing body of water –DO = chemical measurement of dissolved oxygen (mg L -1 ) –BOD = total DO needed to oxidize organic matter in a water sample = change from initial DO at saturation to amount after 5 days BOD time

7. Introduction Standard of living ~ adequate water and wastewater treatment Human Risks Challenge of preventing rapid spead of disease e.g. typhoid fever (bacteria), hepatitis (viruses), cryptosporidosis (protozoa) Removed by sand filtration and chlorination/ozonation Aquatic Risks Aerobic organisms depend on DO 8-12 mg L -1 Affected by temperature and salt Inc. salt dec. DO Wipple and Wipple (1911)

8. The Streeter-Phelps Equation without trmt: with trmt: Organic matter is oxidized, stream re-aerates

9. End Review

10. Basic Input Sources Parameters for S-P equation: –Wastewater: Flow rate, temperature, DO, BOD BOD measured in lab – DO measured after several days (flat portion of curve)

11. The following material and model is covered in: CHEM3500/3550

12. Basic Input Sources Sewage Treatment Plants Remove turbidity, oxidizable organic matter, and pathogens –Turbidity – settling tanks and filters –Organic matter – trickling filters, activated sludge –Pathogens – filtration, chloination, ozonation ftp://ftp.wiley.com/public/sci_tech_med/pollutant_fate/

13. Basic Input Sources Sewage Treatment Plants Prelininary - screening of large materials Primary - sedimentation - settling tanks Secondary - biological aeration – trickling filters, activated sludge - metabolizes and flocculates dissolved organics Tertiary – e.g. P removal http://www.waterencyclopedia.com/Tw-Z/Wastewater-Treatment-and-Management.html

14. Basic Input Sources Wastewater Treatment Plant Model

15. Movie 1. Wastewater Treatment and Discharge (2000) 2. Wastewater Generation and Collection (2000) 3. Our Urban Environment: Water Quality (2000)

16. End Review

17. Mathematical Model Take a river: What parameters and processes would be important in developing a model for the oxidation of organic waste? our model river: draw in parameters Ultimate BOD L of mix Stream DO deficit Consumption DO by MO’s Re-aeration by atmosphere Amount DO consumed

18. The Streeter-Phelps Equation D = k’BOD L [exp(-k’(x/v) – exp(-k 2 ’(x/v))] + D 0 exp(-k 2 ’(x/v)) k 2 ’ – k’ where: D = DO concentration deficit (value below saturation) (mg L -1 ), k’ 2 = the re-aeration constant (in d -1 ), BOD L = the ultimate BOD (in mg L -1 ), k’= the BOD rate constant for oxidation (d -1 ), x = distance downstream from the point source (km), v = average water velocity (km d -1 ) D o = initial oxygen deficit of mixed stream and wastewater (mg L -1 ) Consumption by MO’sRe-aeration by atmos. O 2 D is not the remaining DO content but the amount of original DO consumed…must be subtracted from original DO without BOD waste

19. The Streeter-Phelps Equation DO at a given distance below the input:

20. The Streeter-Phelps Equation k 2 ’ = first-order rate constant for re-aeration Eact measurements are difficult, get from tables:

21. The Streeter-Phelps Equation BOD L = ultimate BOD or maximum O 2 required to oxize the waste sample Determined from 5 day BOD test or using equation: BOD L = BOD 5 1 – exp(-k’(x/v)) Where k’ is obtained from a 20 day BOD experiment D 0 = DO level in the stream upstream from input - initial DO of stream-waste mixture

22. The Streeter-Phelps Equation Zone of Clean Water (Zone 1) Zone of Degradation (Zone 2) Zone of Active Decomposition (Zone 3) Zone of Recovery (Zone 4) Zone of Cleaner Water (Zone 5) Algae, fungi, protozoa, worms, larger planst die Gray/black, H 2 S, CH 4, NH 3 productions, Minimum D = critical dissolved oxygen = D c

24. The Streeter-Phelps Equation t c = 1 ln k 2 ’ 1 – D 0 (k 2 ’-k’) k 2 ’ – k’ k’ k’ BOD L and x c = vt c Critical DO concentration, D c : D c = k’ BOD L exp(-k’(x c /v)) k 2 ’

25. Problem 1. Determine D c and its location. 2. Estimate the 20 °C BOD 5 of a sample taken at x c. 3. Plot the curve. Example Problem: A city discharges 25 million gallons per day of domestic sewage into a stream whose typical rate of flow is 250 cubic feet per second. The velocity of the stream is appoximately 3 miles per hour. The temperature of the sewage is 21 °C, while that of the stream is 15 °C. The 20 °C BOD 5 of the sewage is 180 mg/L, while that of the stream is 1.0 mg/L. The sewage contains no DO, but the stream is 90% saturated upstream of the discharge. At 20 °C, k’ is estimated to be 0.34 per day while k 2 ’ is 0.65 per day.

26. 1.Determine DO in stream before discharge (=upstream DO): Saturation conc. at 15 °C = 10.2 mg/L Upstream is 90% saturated = 10.2 mg/L x 0.90 = 9.2 mg/L 2.Determine mixture, T, DO, and BOD using mass balance: Flow rate stream: = 250 ft 3 /s = 612 x 10 6 L/d Flow rate sewage: 25 x 10 6 gallons/d = 94.8 x 10 6 L/d

27. Temperature of mixture:  T = stream input + sewage input – output effect 0 = (stream flow)(stream temp.) + (sewage flow)( sewage temp) – (mix flow)(mix temp) 0 = (612 x 10 6 L/d)(15 °C) + (94.8 x 10 6 L/d)(20 °C) – (612 x 10 6 L/d + 94.8 x 10 6 L/d)T mix T mix = (612 x 10 6 L/d)(15 °C) + (94.8 x 10 6 L/d)(20 °C) = 15.7 °C (612 x 10 6 L/d +94.8 x 10 6 L/d) DO in mixture Net change in DO = Stream input + Sewage output – Output 0 = (stream flow)(stream DO) + (sewage flow)(sewage DO) – (mix flow)(mix DO) 0 = (612 x 10 6 L/d)(9.2 mg/L) + (94.8 x 10 6 L/d)(0.0) - (612 x 10 6 L/d + 94.8 x 10 6 L/d)(Do mix ) DO mix = (612 x 10 6 L/d)(9.2 mg/L) + (94.8 x 10 6 L/d)(0.0 mg/L) (612 x 10 6 L/d + 94.8 x 10 6 L/d) = 7.97 mg/L

28. BOD 5 of mixture: Net change in BOD 5 =  BOD 5 = Stream input + Sewage output – Output 0 = (stream flow)(stream BOD 5 ) + (sewage flow)(sewage BOD 5 ) – (mix flow)(mix BOD 5 ) 0 = (612 x 10 6 L/d)(1.0 mg/L) + (94.8 x 10 6 L/d)(80 mg/L) - (612 x 10 6 L/d + 94.8 x 10 6 L/d)(BOD 5 ) BOD 5mixture = (612 x 10 6 L/d)(1.0 mg/L) + (94.8 x 10 6 L/d)(80 mg/L) = 25.0 mg/L (612 x 10 6 L/d + 94.8 x 10 6 L/d) BOD L of mixture (at 20 °C) BOD L = BOD 5 = 25.0 mg/L= 30.6 mg/L 1 – exp(-k’(x/v) 1 – exp(-0.34/d)(5 d)

29. 3.Correct rate constants to 15.7 °C k’ = 0.34(1.135) 15.7-20 = 0.197 d -1 k 2 ’ = 0.65(1.024) 15.7-20 = 0.587 d -1 4. Determine t c and x c : D 0 = (initial stream O 2 - O 2 of mixture) = (9.2 – 7.97) = 1.23 mg O 2 L -1

30. 4. Determine t c and x c : t c = 1 ln k 2 ’ 1 – D 0 (k 2 ’-k’) k 2 ’ – k’ k’ k’ BOD L = 2.42 d x c = vt c = 3 mi/h x 24 h/d x 2.42 d = 174.2 mi = 280 km 5. Determine D c :

31. V = 3 mi/h = 72 mi/d D c = k’ BOD L exp(-k’(x c /v) k 2 ’ = 0.197 d -1 (30.6 mg/L) exp(-(0.197 d -1 )(174.2 mi / 72 mi d -1 ))) 0.587 d -1 = 6.37 mg L -1 The DO will be depressed 6.37 mg L -1 from saturation. Minimum DO = 9.2 mg L -1 - 6.37 mg L -1 = 2.83 mg L -1

32. 6. Determine BOD 5 at critical point, x c : BOD 5 = BOD L exp(-k’(x/v)) = (30.6 mg L -1 ) exp(-0.197 d -1 )(174.2 mi)/(72 mi d -1 ) = 19.0 mg L -1 20 °C BOD 5 = BOD 5 [1 – exp(-k’)(5)] = 19.0 mg L -1 [1 – exp(-0.34 d -1 )(5 d)] = 15.5 mg L -1 Easier method

33. Use Fate!!! Much easier than by hand

34. End Review

35. Sensitivity Analysis

36. Limitations It uses average re-aeration rates of the stream (problem in alternating riffle and pool areas) Sedimentation is not allowed in the basic model, but can be incorporated with additional experimental data

37. Remediation Problems are:-Eutrophication -Odors -Low/no D.O. -Aquatic death -Microbes/Pathogens Source removal! (install treatment plant) including BOD, NO 3 -, NH 3 /NH 4 +, PO 4 3- removal, but you still will have organic rich sediments for some time Time (flowing aquatic systems can be very resilient) Notice the difference between the recovery of a biodegradable pollutant versus nonbiodegradable!

38. End Review

39. Further Reading Journals and Reports Wipple, G.C. and Wipple, M.C. (1911) Solubility of oxygen in sea water. Journal of the American Chemical Society, Vol. 3 pp 362.

40. Books Craun, G. (1986) Waterborne Diseases in the United States. CRC Press, Boca Raton, FL. Meadows, D., Randers, J., and Meadows, D. (2004) Limits to Growth: The 30-Year Update. Chelsea Gren Publishing Compnay, White River Junction, VT. Metcalf and Eddy Inc. (1991) Wastewater Engineering, 3 rd Ed. McGraw-Hill, New York. Sawyer, C.N. and McCarty, P.L. (1978) Chemistry for Environmental Engineering. McGraw- Hill, New York. Snoeyink, V.L. and Jenkins, D. (1980) Water Chemistry. John Wiley & Sons, New York. Standard Methods for the Examination of Water and Wastewater, 20 th Ed. (1998) American Waterworks Association, Washington D.C. Streeter, H.W. and Phelps, E.B. (1925) A Study of the Pollution and natural Purification of the Ohio River. United States Public Health Service, U.S. Department of Health, Education and Welfare. Tchobanoglous, G. and Burton, F.L. (1991) Wastewater Engineering: Treatment, Disposal, and Reuse. McGraw-Hill, New York.