Water Quality Management in Rivers

1. Water Quality Management in Rivers

2. Dissolved Oxygen Depletion Dissolved Oxygen Depletion

3. Biochemical Oxygen Demand Measurement • Take sample of waste; dilute with oxygen saturated water; add nutrients and microorganisms (seed) • Measure dissolved oxygen (DO) levels over 5 days • Temperature 20° C • In dark (prevents algae from growing) • Final DO concentration must be > 2 mg/L • Need at least 2 mg/L change in DO over 5 days

4. Example • A BOD test was conducted in the laboratory using wastewater being dumped into a Lake. The samples are prepared by adding 3.00 mL of wastewater to the 300.0 mL BOD bottles. The bottles are filled to capacity with seeded dilution water.

5. Example : Raw Data

6. Example : Calculations • What is the BOD5 of the sample? • Plot the BOD with respect to time.

7. Example : Time – Concentration Plot

8. Organic matter oxidized Organic matter remaining Modeling BOD as a First-order Reaction

9. Modeling BOD Reactions • Assume rate of decomposition of organic waste is proportional to the waste that is left in the flask.

10. Lo Lo- Lt BOD exerted BODt Lt L remaining

11. Ultimate Biochemical Oxygen Demand Lt = amount of O2 demand left in sample at time, t Lo = amount of O2 demand left initially (at time 0, no DO demand has been exerted, so BOD = 0) At any time, Lo = BODt + Lt (that is the amount of DO demand used up and the amount of DO that could be used up eventually) Assuming that DO depletion is first order BODt = Lo(1 - e-kt)

12. Example • If the BOD5 of a waste is 102 mg/L and the BOD20 (corresponds to the ultimate BOD) is 158 mg/L, what is k?

13. Example (cont)

14. Biological Oxygen Demand: Temperature Dependence • Temperature dependence of biochemical oxygen demand • As temperature increases, metabolism increases, utilization of DO also increases kt = k20T-20  = 1.135 if T is between 4 - 20 oC  = 1.056 if T is between 20 - 30 oC

15. Example The BOD rate constant, k, was determined empirically to be 0.20 days-1 at 20 oC. What is k if the temperature of the water increases to 25 oC? What is k if the temperature of the water decreases to 10 oC?

16. Example

17. Nitrogenous Oxygen Demand • So far we have dealt only with carbonaceous demand (demand to oxidize carbon compounds) • Many other compounds, such as proteins, consume oxygen • Mechanism of reactions are different

18. Nitrogenous Oxygen Demand • Nitrification (2 step process) 2 NH3 + 3O2 2 NO2- + 2H+ + 2H2O 2 NO2- + O2  2 NO3- • Overall reaction: NH3 + 2O2 NO3- + H+ + H2O • Theoretical NBOD =

19. Nitrogenous Oxygen Demand

20. Nitrogenous oxygen demand • Untreated domestic wastewater ultimate-CBOD = 250 - 350 mg/L ultimate-NBOD = 70 - 230 mg/L Total Kjeldahl Nitrogen (TKN) = total concentration of organic and ammonia nitrogen in wastewater: 15 - 50 mg/L as N Ultimate NBOD  4.57 x TKN

21. Other Measures of Oxygen Demand

22. Chemical Oxygen Demand • Chemical oxygen demand - similar to BOD but is determined by using a strong oxidizing agent to break down chemical (rather than bacteria) • Still determines the equivalent amount of oxygen that would be consumed • Value usually about 1.25 times BOD

23. Water Quality Management in Rivers

24. Dissolved Oxygen Depletion

25. Dissolved Oxygen Sag Curve

26. Mass Balance Approach • Originally developed by H.W. Streeter and E.B. Phelps in 1925 • River described as “plug-flow reactor” • Mass balance is simplified by selection of system boundaries • Oxygen is depleted by BOD exertion • Oxygen is gained through re-aeration

27. Steps in Developing the DO Sag Curve • Determine the initial conditions • Determine the re-aeration rate from stream geometry • Determine the de-oxygenation rate from BOD test and stream geometry • Calculate the DO deficit as a function of time • Calculate the time and deficit at the critical point

28. Selecting System Boundaries

29. Initial Mixing Qw = waste flow (m3/s) DOw = DO in waste (mg/L) Lw = BOD in waste (mg/L) Qr = river flow (m3/s) DOr = DO in river (mg/L) Lr = BOD in river (mg/L) Qmix = combined flow (m3/s) DO = mixed DO (mg/L) La = mixed BOD (mg/L)

30. Determine Initial Conditions • Initial dissolved oxygen concentration • Initial dissolved oxygen deficit where D = DO deficit (mg/L) DOs = saturation DO conc. (mg/L)

31. 1. Determine Initial Conditions DOsat is a function of temperature. Values can be found in Table. • Initial ultimate BOD concentration

32. 2. Determine Re-aeration Rate • O’Connor-Dobbins correlation where kr = reaeration coefficient @ 20ºC (day-1) u = average stream velocity (m/s) h = average stream depth (m) • Correct rate coefficient for stream temperature where Θ = 1.024

33. Determine the De-oxygenation Rate • rate of de-oxygenation = kdLt where kd = de-oxygenation rate coefficient (day-1) Lt = ultimate BOD remaining at time (of travel downstream) t • If kd (stream) = k (BOD test) and

34. 3. Determine the De-oxygenation Rate • However, k = kd only for deep, slow moving streams. For others, where η = bed activity coefficient (0.1 – 0.6) • Correct for temperature where Θ = 1.135 (4-20ºC) or 1.056 (20-30ºC)

35. 4. DO as function of time • Mass balance on moving element • Solution is

36. 5. Calculate Critical time and DO