Water Quality Management in Rivers

Dissolved Oxygen Depletion

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

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.

Example : Raw Data

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

Example : Time – Concentration Plot

Modeling BOD as a First-order Reaction Organic matter oxidized Organic matter remaining Suppose we imagine a flask with some biodegradable organic waste in it. As bacteria oxidize the waste, the amount of organic material remaining decreases (black line). If the organic material is completely degradable eventually all of it will disappear. Another way to describe to organic matter in the flask is to say the converse: that the amount of organic matter oxidized already increases with time (red line). As the amount of organic matter remaining decreases, the amount of oxygen used to oxidize the waste has increased, but that which is remaining in the bottle decreases. What we often look at is OXYGEN DEMAND - the amount of oxygen needed to degrade a waste - the oxygen used follows the red line, the amount of oxygen demand remaining with time follows the black line - since there is less organic matter remaining there is less oxygen demand. We need to describe these relationships mathematically.

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

Lo BOD exerted BODt Lt L remaining Lo- Lt Lo is the ultimate carbonaceous oxygen demand - amount of oxygen required by microorganisms to oxidize the carbonaceous portion of the waste to simple CO2 and water. Ultimate carbonaceous oxygen demand is the sum of the amount of oxygen already consumed by the waste in the first t days (BODt) plus the amount of oxygen remaining to be consumed.

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) Thus, Lo = BODt + Lt BODt = Lo(1-e-kt)

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? BODt = Lo(1 - e-kt) BOD5 = 102 mg/L BOD20 = 158 mg/L  Lo t = 5 102 mg/L = 158 mg/L(1 - e-k(5 days))

Example (cont)

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

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? Use  = 1.056 since T is between 20 - 30 oC k25 = (0.20)(1.056)25-20 k25 = 0.26 days-1 Use  = 1.135 since T is between 20 - 30 oC k25 = (0.20)(1.135)10-20 k25 = 0.085 days-1

Example

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

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 = When living things die or excrete waste products, the nitrogen that is tied up as complex organic chemicals is converted to ammonia by bacteria and fungi. Aerobic environments: Bacteria Nitrosomonas: converts ammonia to nitrite Nitrobacter converts nitrite to nitrate

Nitrogenous Oxygen Demand

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

Other Measures of Oxygen Demand

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

Water Quality Management in Rivers

Dissolved Oxygen Depletion

Dissolved Oxygen Sag Curve

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

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

Selecting System Boundaries

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)

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

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

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

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

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)

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

5. Calculate Critical time and DO