# Environmental Modeling Chapter 6: Fate and Transport of Pollutants in Rivers and Streams Copyright © 2006 by DBS.

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Environmental Modeling Chapter 6: Fate and Transport of Pollutants in Rivers and Streams Copyright © 2006 by DBS

Quote “[Mathematics] The handmaiden of the Sciences” -Eric Temple Bell

Concepts Descriptive aspects of rivers General mathematical concepts Pulse and step models Chemical aspects Limitations Professional modeling Remediation

Case Study: Rhine River Major river passing through dense population Pollution spread was well documented Capel et al., 1988

Case Study: Rhine River Explanatory modeling – reverse fit of data to computer model Pulse release

Case Study: Rhine River Disulfoton – thiophosphoric acid insecticide Diluted and degraded downstream

Introduction Rivers and streams make up ~ 0.014 % of terrestrial water

Introduction Humans settle near running surface water because wastes can be quickly removed Pulse and step releases Models assume complete mixing (no cross-sectional gradient) Models include – longitudinal dispersion (spreading in the direction of flow), advection (transport in the direction of flow at velocity of water), and first-order removal (biological, chemical, or radioactive decay)

Examples of Rivers Depth, flow rate and water velocity needed for modeling

Examples of Rivers Few streams achieve the length, drainage area, or discharge rate of the major rivers

Input Sources Point – well-defined source e.g. industrial, feedlot, sewage plant effluent, Non-point - less well defined, cannot be pinpointed e.g. runnoff from farming, leachate from landfills, outboard motors, resuspension (bioturbation, storms, dredging) (Bioturbation is a continuous process, resuspension via storms or dredging would be considered pulse events.)

End Review

Terms for Modeling SA = the cross sectional area of the stream (width, w, multiplied by average depth, d) (m 2 ) Q i = the flow rate at the beginning of the section of the stream to be modeled (m 3 /yr) Q e = the flow rate at the end of the section of the stream to be modeled (m 3 /yr) (We usually assume that Q i is equal to Q e and we represent both by simply Q) C(x) = the pollutant concentration profile as a function of distance (at a fixed time) downstream from the pollutant addition (kg/m 3 or similar units) C(t) = the pollutant concentration in the stream at a fixed distance as a function of time (kg/m 3 or similar units) k = the first-order removal rate of pollutant from the river/stream (yr -1 or in the Fate ® model s -1 ) t = time (years or in the Fate ® model, seconds) M o = the total mass of pollutant in the river (kg or Ci) W = rate of continuous discharge of the waste (mass/time in kg/s or Ci/s)

Modeling of Streams The Stream Channel Q = flow rate, m 3 /yr E = longitudinal dispersion v = velocity of water m/s k = rate constant Physical characteristics of the channel determine these terms

Modeling of Streams Mixing and Dispersion Advective-dispersive transport equations Dispersion is a direct result of advection – results in dilution of pollutant by unpolluted water Cost of remediation increases due to larger volume to deal with Dispersion – hydrodynamic vs diffusion Turbulent mixing (hydrodynamic) dominates

Modeling of Streams Mixing and Dispersion Dispersion is characterized by the longitudinal dispersion coefficent (E) E = 0.011 v 2 w 2 du Where u = √ gds, v = average water velocity (ms -1 ), w = average stream width (m), d = average stream depth (m), g = 9.81 m s -2, s = slope of stream bed Stream velocity and width increase longitudinal dispersion

Modeling of Streams Mixing and Dispersion Mixing of two rivers separated by a concrete barrier just below the surface

Modeling of Streams Mixing and Dispersion E can be determined experimentally using tracers (fluorescent dyes)

Modeling of Streams Mixing and Dispersion Increased dilution Pulse: increased peak width Step: increased spreading front of pollution

Modeling of Streams Removal Mechanisms Degradation: Photochemical, biological, abiotic (chemical) and nuclear All first-order Add rate constants (k) together C t = C 0 e -kt Determined in lab, plot of ln(C/C 0 ) vs. t, slope is k Applies as long as processes are 1 st order

Modeling of Streams Removal Mechanisms Sorption Depends on (a) K D, (b) suspended solids concentration, (c) rate of turbulence Streams tend to be highly turbulent, less time for particles to settle Storm events may increase and redistribute suspended particles Modeling is complicated – generally left out

End Review

Simple Transport Models Start with mass balance: Accumulation = inputs - outputs + sources - sinks What terms do we need?

Simple Transport Models Most are one dimensional Again based on a mass balance Are a function of time (and distance) Pulse: Step:

Pulse Input River Model Pulse: Where C(t) = pollutant concentration (mg L -1 ), t = time, M 0 = pollutant mass released (mg), d = average stream depth (m), w = average stream width (m), E = longitudinal dispersion coefficient (m 2 s -1 ), x = distance downstream from input (m), v = average water velocity (m s -1 ), k = 1 st order degradation rate constant (s -1 )

Pulse Input River Model Example Problem Solution: 1. Calculate the average stream velocity, v = discharge/cross-sectional area = 40 m 3 s -1 / 80 m 2 = 0.50 m s -1 2. Calculate the rate constant, k ln (C/C 0 ) = -kt  k = ln2 / t 1/2 = 1.07 x 10 -8 s -1 3. Calculate the dispersion coefficient, E Slope = 1 m / 10000 m = 10 -4 u = √(gds) = √(9.81)(2)(10 -4 ) = 0.044 m s -1  E = 0.011 v 2 w 2 / du = 50 m 2 s -1 One curie of 134 Cs is accidentally released into a small stream. The stream channel has an average width of 40 m and a depth of 2 m. The average water discharge (Q) in the stream is 40 m s -1, and the stream channel drops 1 m in elevation over a distance of 10 km. Assuming that the 134 Cs is evenly distributed across the stream channel, estimate the distribution of 134 Cs as a function of distance downstream (using a max. distance of 30 km) at 1, 3, 6, and 12 hours. Also estimate the 134 Cs activity at a distance of 10 km at 6 hours after the release. 134 Cs has a half-life of 2.07 years.

Pulse Input River Model Example Problem 4. Arrange data in proper units M 0 = 1 curie = 1 x 10 6 μCi w = 40 m d = 2 m E = 50 m 2 s -1 t = variable (s) x = variable (m) v = 0.50 m s -1 k = 1.07 x 10-8 s -1 6. Calculate C(t) at 10 km after 6 hr Plot of concentration at 12 hr Peak concentration at 21.6 km 5.

Pulse Input River Model Example Problem 6. Calculate C(t) at 10 km after 6 hr

Step Input River Model Step Where C(x, t) = pollutant concentration (mg L -1 ), at distance x and time t, W = rate of continuous discharge of the waste (kg s -1 ), Q = stream flow (m 3 s -1 ), E = longitudinal dispersion coefficient (m 2 s -1 ), x = distance downstream from input (m), v = average water velocity (m s -1 ), and k = first order decay constant (s -1 ) The positive root is the upstream direction (-x), negative root is the downstream direction (+x)

Step Input River Model When there is no degradation set k = 0 (use very long t 1/2 ) Equation reduces to: Step input, concentration varies with distance not time

Step Input River Model Example Problem Solution: 1. Calculate the mass input, W W = (1500 L / min)(min / 60 s)(500 mg / L)(kg / 10 6 mg) = 0.0125 kg s -1 2. Calculate the flow rate of the stream, Q Q = width x depth x velocity = (20)(2.3)(0.85)= 39.1 m 3 s -1 3. Calculate the rate constant, k k = ln2 / t1/2 = 0.277 d -1 = 3.21 x 10 -6 s -1 An abandoned landfill leaches water into an adjacent stream at a rate of 1500 L min -1. The concentration of 2-chlorophenol in the water is 500 mg L -1. The stream is 20 m wide and 2.3 m deep and has a water velocity of 0.85 m s-1. The regional slope of the stream channel is 1 ft per 1500 ft distance, and the first-order half-life of 2-chlorophenol is 2.5 days. Construct the concentration profile of 2-chlorophenol in the stream. What is the concentration 25 km downstream from the point source?

Step Input River Model Example Problem 4. Calculate E Slope = 1 / 1500 = 6.67 x 10 -4 u = √(gds) = √ (9.81)(2.3)(6.67 x 10 -4 ) = 0.12 m s -1 E = 0.011 v 2 w 2 / du = 11 m 2 s -1 5. Arrange data in proper units: Q = 39.1 m3 s -1 W = 0.0125 kg s -1 w = 20 m d = 2.3 m v = 0.85 m s -1 E = 11.24 m2 s -1 x = variable (m) k = 3.21 x 10 -6 s -1 7. Calculate C(x) at 25 km 6.

Step Input River Model Example Problem 7. Calculate C(x) at 25 km

Step Input River Model What two processes account for decrease in concentration with distance?

End Review

Sensitivity Analysis Sensitivity of Concentration (C) to Water velocity (v) Water velocities of 0.25, 0.50, 0.85, 1.00, 1.25, and 1.50 m s -1 were tested at 500 km downstream Higher v causes polutants to be transported before they degrade Higher C downstream Exp since v is in exp term

Limitations One-dimensional (large rivers may require two-dimensional models) but we can change input function to account for incomplete mixing in the beginning –In real-world pollutant not evenly distributed Volatilization must be handled as a first-order process (probably o.k.) No sorption and particle settling directly built into the models (you can do this in models based on numerical methods)

Remediation of Rivers Source removal Booms and barriers for LNAPLs Activated carbon (tons) Clay capping (does not work very well for rivers, especially turbulent rivers) Dredging of polluted sediments (\$100-\$1000 per cubic yard) – becomes hazardous waste

End Review

Further Reading Journals and Reports Capel, P.D., Giger, W., Reichart, R., and Warner, O. (1988) Accidental release of pesticides into the Rhine River. Environmental Science and Technology, Vol. 22, pp. 992-997. Szestay, K. (1982) River basin development and water management. Water Quality Bulletin, Vol. 7, pp. 155-162.

Books Fischer, H.B.E., List, E.J., Koh, R.C.Y., Imberger, I., and Brooks, N.H. (1979) Mixing in Inland and Coastal Waters, Academic Press, New York. Wetzel, R.G. (2001) Limnology: River and Lake Ecosystems, Academic Press, New York. Wetzel, R.G. and Likens, G.E. (2000) Limnological Analysis, 3 rd Edition, Springer-Verlag, New York.