Modeling Surface Water Contamination

Transcription

1 Modeling Surface Water Contamination A municipal wastewater treatment facility is planning to locate upstream on a popular fishing river. The wastewater facility will continuously discharge wastewater into the river. The effluent will have have high biochemical oxygen demand (BOD) and low dissolved oxygen (DO) concentrations We want to determine the impact of the effluent on the DO concentration of the river (and ultimately the fish population in the river) We also want to evaluate the potential impact of corrective measures the facility can take to address potential DO problems

2 Flows in this System Objective: Predict DO levels in the river when organic waste is added from the treatment plant What pieces of information do we need to do this? We need all the inflows and outflows of O 2 : 1. DO of river water before organic waste is added 2. DO of effluent containing organic waste 3. BOD of river water before organic waste is added 4. BOD of effluent containing organic waste 5. Volume of river flow before effluent is added 6. Volume of effluent flow 7. The DO sat level of the river before mixing with the effluent

3 Processes and Model Structure We will model the resulting DO concentration in the river using a two-step approach: 1. We will establish initial values of DO and BOD immediately after the river water and effluent are mixed using a mass-balance approach 2. We will predict the expected downstream DO levels, taking into account some processes that will occur as the water moves downstream that change the DO: Reoxygenation (or aeration) will absorb O 2 from the atmosphere as the water moves downstream Deoxygenation will consume DO in the river through the consumption of the organic waste

4 Initial Stage: River and Effluent Mixing DO a = (DO s V s + DO p V p ) / (V s + V p ) where: DO a is the DO conc. in the mixed water (mg/l) DO s is the DO conc. in the river water (mg/l) V s is the volume of river water before mixing (L) DO p is the dissolved oxygen concentration in the effluent (mg/l) is the volume of effluent (L) V p For flowing water, we can use flow rates (Q in L/sec) instead of volumes, yielding: DO a = (DO s Q s + DO p Q p ) / (Q s + Q p ) We can apply the very same mass-balance principles to calculate the BOD values immediately after mixing the river water and the effluent

5 Predicting Downstream DO Levels Inflows Organic material enters the river naturally at a constant rate BOD in (t) = A Biochemical Oxygen Demand Outflows BOD (and linked DO) is consumed by decomposition BOD out (t) = k 1 BOD(t) DO is recharged by O 2 from the atm. up to DO sat DO in (t) = k 2 [DO sat (t) DO(t)] Dissolved Oxygen DO (linked to BOD) is consumed by decomposition DO out (t) = k 1 BOD(t)

6 Predicting Downstream DO Levels Difference Equations To model this system, we will need to come up with difference equations for both the BOD and DO portions of the system: R(t+ t) = R(t) + {sum of all inflows sum of all outflows}* t For DO, we must take the difference between the reoxygenation and deoxygenation processes: DO(t+ t) = DO(t) + k 2 * [DO sat -DO(t)] * t k 1 * BOD(t) * t For BOD, we must take the difference between the BOD in and BOD out processes: BOD(t+ t) = BOD(t) + A* t k 1 * BOD(t) * t

7 Predicting Downstream DO Levels Difference Equations DO(t+ t) = DO(t) + k 2 * [DO sat -DO(t)] * t k 1 * BOD(t) * t BOD(t+ t) = BOD(t) + A* t k 1 * BOD(t) * t Recall that the reoxygenation process (the second term in the DO equation above) operates according to the size of the oxygen deficit (DO sat - DO), such that: The reoxygenation term is large when (DO sat DO) is large The reoxygenation term is small when (DO sat DO) is small Immediately after adding the DO poor (and BOD rich) effluent, we expect reoxygenation to proceed rapidly, driving the DO level back towards DO sat, then this inflow scales back to zero as the level is near DO sat

8 Predicting Downstream DO Levels Difference Equations DO(t+ t) = DO(t) + k 2 * [DO sat -DO(t)] * t k 1 * BOD(t) * t BOD(t+ t) = BOD(t) + A* t k 1 * BOD(t) * t The deoxygenation process links the two difference equations together (via the k 1 term), as we have defined BOD consumption to equal the reduction of DO The rate of deoxygenation proceeds according to how much BOD/ DO remains: The deoxygenation term is large when BOD/DO is large The deoxygenation term is small when BOD/DO is small Once the organic waste is consumed, the rate of BOD and DO reduction decreases correspondingly

9 DO Sag Curve (Streeter & Phelps, 1925) If we decompose the net change of of the dissolved oxygen level in time into two components (one from deoxygenation and the other from reoxygenation) we can describe these components in terms of two plots: Due to Reoxygenation Due to Deoxygenation C DO Oxygen Deficit DO sat C DO DO sat Time Time The rate of change (slope) is largest initially and slows down as the BOD and DO are consumed; the actual C DO combines the effects of both

10 Predicting Downstream DO Levels Rate Equations We can derive rate equations from the difference equations in the usual fashion, starting with DO: DO(t+ t) = DO(t) + k 2 * [DO sat -DO(t)] * t k 1 * BOD(t) * t DO(t+ t) - DO(t) = k 2 * [DO sat -DO(t)] * t k 1 * BOD(t) * t lim t 0 DO(t+ t) - DO(t) t = k 2 * [DO sat -DO(t)] k 1 * BOD(t) ddo dt = k 2 * [DO sat -DO(t)] k 1 * BOD(t) Thus the rate of change of DO is a function of the reoxygenation and deoxygenation coefficients, DO sat (which is a f(x) of temp.) and the DO and BOD levels

11 Predicting Downstream DO Levels Rate Equations We can also derive the rate equation for BOD in the usual fashion, starting with the difference equation: BOD(t+ t) = BOD(t) + A* t k 1 * BOD(t) * t BOD(t+ t) - BOD(t) = A* t k 1 * BOD(t) * t lim t 0 BOD(t+ t) - BOD(t) t = A k 1 * BOD(t) dbod dt = A k 1 * BOD(t) Thus the rate of change of BOD is a function of the deoxygenation coefficient (k 1 ), the BOD level, and the constant rate BOD enters the system naturally (A)

12 Predicting Downstream DO Levels Steady-State Conditions This system has a steady-state solution when the rate of change of both the DO and BOD stocks is equal to zero (i.e. inflows equal outflows in both halves of the system) simultaneously. Mathematically, this occurs when the following two conditions are true: ddo dt dbod dt = 0 = A k 1 * BOD(t), and = 0 = k 2 * [DO sat -DO(t)] k 1 * BOD(t) From these conditions, we can derive effective steadystate levels for the BOD and DO reservoirs: BOD = A k 1 DO = DO sat - A k 2

13 Predicting Downstream DO Levels Steady-State Conditions BOD = A k 1 DO = DO sat - Note that the DO level will never reach DO sat as long as there is some BOD naturally entering the system at rate A, as is usually the case However, if the reoxygenation coefficient (k 2 ) is a sufficiently large value when compared to A, then the denominator of the term is big enough to make the overall term quite small, which results in the steadystate condition for the DO reservoir only being slightly smaller than the DO sat value A k 2

14 Modeling Surface Water Contamination A municipal wastewater treatment facility is planning to locate upstream on a popular fishing river. The wastewater facility will continuously discharge wastewater into the river. The effluent will have have high biochemical oxygen demand (BOD) and low dissolved oxygen (DO) concentrations We want to determine the impact of the effluent on the DO concentration of the river (and ultimately the fish population in the river) We also want to evaluate the potential impact of corrective measures the facility can take to address potential DO problems

15 Modeling Surface Water Contamination We will use the DO model as we have described it in Lab #6 in a couple of ways: First we will model the system s DO function as it occurs before the introduction of the wastewater treatment plant and associated effluent Then, we will do further simulations that include the impact of building the treatment plant etc. For the natural condition, p.124 provides the initial parameters for the system: Initial BOD = 3.33 mg/l Initial DO = 8.5 mg/l DO sat = 11 mg/l k 1 = 0.3 k 2 = 0.4 A = 1 mg/l

16 Modeling Surface Water Contamination After constructing the model from scratch, you ll make use of the provided natural parameters, and perform a standard 3-run sensitivity analysis on each of them Next, we need to introduce the effluent from the wastewater treatment plant into the system We will use the assumptions described in the previous lecture and the following values to come up with the characteristics of the river/effluent mixed water: Q r = 300,00 L/min BOD r = 3.33 mg/l DO r = 8.5 mg/l Q e = 100,000 L/min DO e = 2 mg/l BOD e = 40 mg/l V r = 20 m/min d = 40 km

17 Initial Stage: River and Effluent Mixing Q r = 300,00 L/min BOD r = 3.33 mg/l DO r = 8.5 mg/l Q e = 100,000 L/min DO e = 2 mg/l BOD e = 40 mg/l DO a = (DO r Q r + DO e Q e ) / (Q r + Q e ) where: DO a is the DO conc. in the mixed water (mg/l) DO r is the DO conc. in the river water (mg/l) Q r is the flow rate of the river before mixing (L) DO e is the dissolved oxygen concentration in the effluent (mg/l) is the flow rate of effluent (L) Q e The remaining values (V r & d) will be used to translate elapsed time in the simulation to distance down the river

18 Modeling Other Pollutants We have focused almost exclusively on DO in the surface water system what about other pollutants? What can we take from what we have learned about modeling DO in the surface water system to use in the context of modeling another pollutant? What kind of information would we need to model some other pollutant that is introduced into a river at a point source? Initial concentrations of the pollutant in the river and in the effluent The set of processes that cause the concentration of that pollutant to increase and decrease in surface water For example, lead compounds from a factory