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Water Resources Planning and Management Daene C. McKinney Water Quality

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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

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

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

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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

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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

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Advection - Dispersion

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Advection Dominated Flow

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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

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Water Quality Example

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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

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Water Quality Example

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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

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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.

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Water Quality Example

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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

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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

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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)

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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

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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

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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

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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.

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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

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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

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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

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3-1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Linear Programming: Computer Solution and Sensitivity Analysis Chapter 3.

3-1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Linear Programming: Computer Solution and Sensitivity Analysis Chapter 3.

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