Water Quality Management in Rivers Dissolved Oxygen Depletion

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 m. L of wastewater to the 300. 0 m. L BOD bottles. The bottles are filled to capacity with seeded dilution water.

Example : Raw Data

Example : Calculations • What is the BOD 5 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

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

Lo Lo- Lt BOD exerted BODt Lt L remaining

Ultimate Biochemical Oxygen Demand Lt = amount of O 2 demand left in sample at time, t Lo = amount of O 2 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)

Example • If the BOD 5 of a waste is 102 mg/L and the BOD 20 (corresponds to the ultimate BOD) is 158 mg/L, what is k?

Example (cont)

Biological Oxygen Demand: Temperature Dependence • Temperature dependence of biochemical oxygen demand !As temperature increases, metabolism increases, utilization of DO also increases kt = k 20 T-20 = 1. 135 if T is between 4 - 20 o. C = 1. 056 if T is between 20 - 30 o. C

Example The BOD rate constant, k, was determined empirically to be 0. 20 days-1 at 20 o. C. What is k if the temperature of the water increases to 25 o. C? What is k if the temperature of the water decreases to 10 o. C?

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 NH 3 + 3 O 2 2 NO 2 - + 2 H+ + 2 H 2 O 2 NO 2 - + O 2 2 NO 3– Overall reaction: NH 3 + 2 O 2 NO 3 - + H+ + H 2 O • Theoretical NBOD =

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 1. Determine the initial conditions 2. Determine the re-aeration rate from stream geometry 3. Determine the de-oxygenation rate from BOD test and stream geometry 4. Calculate the DO deficit as a function of time 5. Calculate the time and deficit at the critical point

Selecting System Boundaries

Initial Mixing Qw = waste flow (m 3/s) DOw = DO in waste (mg/L) Lw = BOD in waste (mg/L) Qr = river flow (m 3/s) DOr = DO in river (mg/L) Lr = BOD in river (mg/L) Qmix = combined flow (m 3/s) DO = mixed DO (mg/L) La = mixed BOD (mg/L)

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

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

2. Determine Re-aeration Rate a. O’Connor-Dobbins correlation where kr = reaeration coefficient @ 20ºC (day-1) u = average stream velocity (m/s) h = average stream depth (m) b. Correct rate coefficient for stream temperature where Θ = 1. 024

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

3. Determine the De-oxygenation Rate c. However, k = kd only for deep, slow moving streams. For others, where η = bed activity coefficient (0. 1 – 0. 6) d. 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