1. Water Resources Planning and Management Daene C. McKinney Water Quality

2. Water Quality Management Critical component of overall water management in a basin Water bodies serve many uses, including – Transport and assimilation of wastes – Assimilative capacities of water bodies can be exceeded WRT intended uses Water quality management measures – Standards Minimum acceptable levels of ambient water quality – Actions Insure pollutant load does not exceed assimilative capacity while maintaining quality standards – Treatment

3. Water Quality Management Process Identify – Problem – Indicators – Target Values Assess source(s) Determine linkages – Sources  Targets Allocate permissible loads Monitor and evaluate Implement

4. Physical Processes Controlling Flux Advection – Solutes carried along by flowing water Diffusion – Transport by molecular diffusion Dispersion – Transport by mechanical mixing

5. Models Advection + dispersion - major processes by which dissolved matter is distributed throughout a water body (e.g., river) C = concentration (M/L 3 ) V = Average velocity in reach (L/T) D = Longitudinal dispersion coefficient (L 2 /T) t= time x = longitudinal distance Advection term Dispersion term Source term Reaction term Eq. 10.10 in text

6. Steady-state Model Steady-state –Where k = decay rate (1/T) –Solution is –W = loading (M/T) at x = 0 Eq 10.12 in text

7. Advection - Dispersion

8. Advection Dominated Flow

9. Water Quality Example W 1, W 2 = Pollutant loads (kg/day) x 1, x 2 = Waste removal efficiencies (%) P 2 max, P 3 max = Water quality standards (mg/l) P 2, P 3 = Concentrations (mg/l) Q 1, Q 2, Q 3 = Flows (m 3 /sec) a 12, a 13, a 23, = Transfer coefficients

10. Water Quality Example

11. Paramet er UnitsValue Q1Q1 m 3 /s10 Q2Q2 m 3 /s12 Q3Q3 m 3 /s13 W1W1 kg/day250,000 W2W2 kg/day80,000 P1P1 mg/l32 P 2 max mg/l20 P 3 max mg/l20 a 12 -0.25 a 13 -0.15 a 23 -0.60

12. Water Quality Example

13. Cost of treatment at 1 greater than cost at 2 (bigger waste load at 1) Marginal cost at 1 greater than marginal cost at 2, c 1 > c 2 for same level of treatment

14. Water Quality Example Cost of treatment at 1 >= cost at 2 marginal cost at 1, c 1, >= marginal cost at 2, c 1, for the same amount of treatment.

15. Water Quality Example

16. Example Irrigation project –1800 acre-feet of water per year Decision variables –x A = acres of Crop A to plant? –x B = acres of Crop B to plant? 1,800 acre feet = 2,220,267 m 3 400 acre = 1,618,742 m 2 Crop ACrop B Water requirement (Acre feet/acre)32 Profit ($/acre)300500 Max area (acres)400600

17. Example 2 4 6 8 10 2468 x A (hundreds acres) x B (hundreds acres) x B < 600 x A > 0 x A < 400 3x A +2 x B < 1800 x B > 0

18. Example 2 4 6 8 10 2468 x A (hundreds acres) x B (hundreds acres) x B < 600 x A > 0 x A < 400 x B > 0 Z=3600=300x A +500x B Z=2000=300x A +500x B Z=1000=300x A +500x B (200, 600)

19. GAMS Code POSITIVE VARIABLES xA, xB; VARIABLES obj; EQUATIONS objective, xAup, xBup, limit; objective.. obj =E= 300*xA+500*xB; xAup.. xA =L= 400.; xBup.. xB =L= 600.; limit.. 3*xA+2*xB =L= 1800; MODEL Calibrate / ALL /; SOLVE Calibrate USING LP MAXIMIZING obj; Display xA.l; Display xB.l; Marginal, Lagrange multiplier, shadow price, dual variable

20. GAMS Output LOWER LEVEL UPPER MARGINAL ---- EQU objective... 1.000 ---- EQU xAup -INF 200.000 400.000. ---- EQU xBup -INF 600.000 600.000 300.000 ---- EQU limit -INF 1800.000 1800.000 100.000 LOWER LEVEL UPPER MARGINAL ---- VAR xA. 200.000 +INF. ---- VAR xB. 600.000 +INF. ---- VAR obj -INF 3.6000E+5 +INF. Marginal

21. Marginals Marginal for a constraint = Change in the objective per unit increase in RHS of that constraint. –i.e., change x B –Objective = 360,000 –Marginal for constraint = 300 –Expect new objective value = 360,300

22. New Solution LOWER LEVEL UPPER MARGINAL ---- EQU objective... 1.000 ---- EQU xAup -INF 199.333 400.000. ---- EQU xBup -INF 601.000 601.000 300.000 ---- EQU limit -INF 1800.000 1800.000 100.000 LOWER LEVEL UPPER MARGINAL ---- VAR xA. 199.333 +INF. ---- VAR xB. 601.000 +INF. ---- VAR obj -INF 3.6030E+5 +INF. Note: Adding 1 unit to x B adds 300 to the objective, but constraint 3 says and this constraint is “tight” (no slack) so it holds as an equality, therefore x A must decrease by 1/3 unit for x B to increase by a unit.

23. Unbounded Solution 2 4 6 8 10 2468 x A (hundreds acres) x B (hundreds acres) x A > 0 x A < 400 x B > 0 unbounded Take out constraints 3 and 4, objective can Increase without bound

24. Infeasibility 2 4 6 8 10 2468 x A (hundreds acres) x B (hundreds acres) x B < 600 x A > 0 x A < 400 3x A +2 x B > 3000 x B > 0 Change constraint 4 to >= 3000, then no intersection of constraints exists and no feasible solution can be found

25. Multiple Optima Change objective coefficient to 200, then objective has same slope as constraint and infinite solutions exist 2 4 6 8 10 2468 x A (hundreds acres) x B (hundreds acres) x B < 600 x A > 0 x A < 400 x B > 0 Z=1800=300x A +200x B Infinite solutions on this edge 3x A +2 x B < 1800