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Region: Saskatchewan (prairie) Intake rates for Part 3, question 2:

Region: Saskatchewan (prairie) Intake rates for Part 3, question 2:. Exercise 3. REM 610 January 27 and February 3, 2010. Part I: Level II Models Revisited. Equilibrium between compartments Steady state over entire environment Transport and transformation is included

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Region: Saskatchewan (prairie) Intake rates for Part 3, question 2:

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  1. Region: Saskatchewan (prairie) • Intake rates for Part 3, question 2:

  2. Exercise 3 REM 610 January 27 and February 3, 2010

  3. Part I: Level II Models Revisited • Equilibrium between compartments • Steady state over entire environment • Transport and transformation is included • Flux = D.f D(total) = Di D(reaction) = k.V.Z k = reaction rate (d-1) D(advection) = G.Z G = flow rate (m3.d-1)

  4. Part I: Fugacity 2 - Naphthalene1. What is the flux of Naphthalene in & out of the environment? Form II – Spreadsheet: Flux = f * D = 0.016665 Pa * 6.66 E+03 mol/Pa day = 111 mol/day Level II Model – Volumes are different Inputs = Outputs = Flux Results: Outputs = 1.20E6 mol/hr*24h/d = 2.88E7 mol/day

  5. 2. What is the most important source of naphthalene into the environment? A) Point Source Emissions = 100 mol/day B) Inflow Water Conc. (Cw) = 1E-03 mol/m3 Advective Water Flow (Gw) = 1E+03 m3/day Water Inflow = Cw * Gw = 1 mol/day C) Inflow Air Conc. (Ca) = 1E-06 mol/m3 Advective Air Flow (Ga) = 1E+07 m3/day Air Inflow = Ca * Ga = 10 mol/day  Point Source Emissions are largest

  6. 3. Which medium contains the greatest mass of naphthalene? • Air - 98.6% - contains the largest mass of chemical • Water - 1% • Other media - less than 1%

  7. 4. What is the residence time and half-life time of naphthalene? Residence Time (total) = 369 days = total mass (mol) / total loss rate (mol/day) = 40916 (mol) / 111 (mol/day) Half-Life Time = ln 2 * Res. Time = 0.693 * 369 = 256 days

  8. 5. What is the most important route of naphthalene removal? Loss Routes (mol/day) 1. Air advection = 67 mol/day - most important loss route 2. Air reaction = 38 mol/day 3. Sediment reaction = 2.5 mol/day

  9. 6. In which medium does naphthalene degrade the fastest? Slowest? • Fastest degradation rate = biota (t1/2 = 10 days) • Slowest degradation rate = air (t1/2 = 730 days)

  10. 7. Which medium is responsible for most of the degradation for naphthalene? - Most of the chemical undergoes reaction in the AIR, although it has the slowest rate of reaction (Que. 6) - WHY? - Air contains most of the mass of the chemical - The loss rate is a function of both the half-life and the amount of chemical in the medium - T1/2 is not a good indication of persistence by itself Reactive loss rates (mol/day); Half-life time (day); % Mass

  11. 8. In which medium can you expect the highest concentrations? Highest chemical concentrations are found in biota - Que. 6 - biota has the highest reaction rate - Results are not consistent

  12. 9. Evaluate the relationship between half-life and chemical concentration in biota • Expectation: Relatively low half-life = relatively fast reaction rate therefore you would expect relatively lower concentrations due to metabolism. • However, - Fugacity 2 model assumes equal fugacities, therefore concentrations are only a function of the fugacity capacity of each medium (i.e., C = f*Z) - Biota has the highest Z value and therefore the highest chemical concentration • - Fugacity 2 model does not incorporate metabolism

  13. Part II – DDT Level II & III Models • Input data “New Chemical” in Level II & III Models • Reaction half-lives (h): • Soil = 1663.2 • Sediment = 166320 • Suspended Sediment = 24000 • Biota = 24000 • Water = 16632 • Air = 2.4E9

  14. Part II – Input Data DDT • Chemical Properties: • Molecular Weight = 200 g/mol • Temperature = 27.5 ºC • Solubility = 20.0 g/m3 • Vapour Pressure = 1.00 Pa • Low Kow = 4 • Melting Point = 109 ºC • Emissions and Inflows: • Input = 0.833 kg/h • Inflow concentration in water = 2E6 ng/L • Inflow concentration in air = 2E5 ng/m3

  15. 1. How and at what rate does DDT enter and leave the environment? • Steady state assumed in both models • Look at total outputs (loss) = total inputs in both Level II & III Models • Keep in mind default environmental parameters are different for Level II & III

  16. 2. What is the fugacity of DDT in the media of the evaluative environments • In Level II equilibrium is assumed • Fugacity = ratio of Ci/Zi in any medium • Fugacities are equal in all media in Level II • Level III – Look at “fugacity” tab under results • In Level III system is at steady state but NOT equilibrium

  17. 3. What is the relationship between the emission and the fugacity of each of the environments? • Use Level III model to adjust emissions in a given medium • Notice how fugacity increases LINEARLY with emissions

  18. 4. a) What is the effect of changing the medium in which the emission occurs? • Changes in the total loss rates from each medium with different mode of entry • Overall residence time or “Environmental Persistence” (see Webster paper) changes b) How do you explain the results? • Overall residence time is a function of both the loss rates through advection and reaction as well as the mass of chemical in each medium

  19. 5. a) What are the relative concentrations and amounts of chemical in each of the phases? • See results sections of both Level II and Level III models or summary diagram b) How do they compare to with those from the level I model? • Level II and III models include LOSS & TRANSPORT • Level I models do not • Concentrations in Level I models tend to be lower as a result

  20. 6. In which phase does most of the DDT react? • See Results of Level II and III models b) Is this what you would expect from the half-life times? Explain your observation • Medium with shortest half-life is not necessarily the phase where most of the chemical reaction • Depends on where most of the mass of chemical partitions

  21. 7. What is the relative importance of advection and reaction as loss mechanisms of the chemical in each of the two models? • Note that degradation of DDT occurs very slowly due to relatively long half lives in all media • Advective removal in air accounts for the majority of DDT loss

  22. 8. What is the overall persistence or average residence time of DDT in this unit world? • See Overall Residence Time in 1st Tab of Results in Level II & Level III models • Will depend on the ecosystems parameters and emissions inputs

  23. 9. What is the main mechanism of transport of DDT from the site of origin to neighboring environments? • Advective removal through air

  24. 10. Is it possible to apply the Level II and III models to estimate the concentrations of DDT in real environments? • Good Question • Can adjust environmental parameters to better reflect real conditions • May still be an oversimplification of reality • Personal decision/compromise as a modeler • Note CHEMCAN Model is a “real” environment

  25. Part III: ChemCan4 - Benzo(a)pyrene 1. Forecast the concentrations of BaP in water, soil and fish in Saskatchewan (prairie)? Emissions = 100 tonnes/yr = 100,000 kg/yr • Water = 8.92 E-10 mol/m3 or 2.25 E-07 g/g • Soil = 1.25 E-05 mol/m3 or 2.11 E-03 g/g • Fish = 5.55 E-03 g/g

  26. 2. Regional Environmental Fatea) What are the predominant routes of removal? 1. Air advection 2. Soil reaction 3. Air reaction

  27. b) What are the main routes for human exposure? Possible routes for human exposure: • Air inhilation • Water intake • Dietary intake (fish, terrestrial animals & plants)

  28. b) What are the main routes for human exposure?  Intake from dietary sources results in greatest exposure • Assumptions - intake rates are valid or in the right proportion - concentrations in the media are homogeneous throughout the Saskatchewan prairie - water not treated

  29. 3. What is the residence time and half-life time of BaP in Saskatchewan? Residence Time (total) = 272 days Half-Life Time = ln 2 * Res. Time = 0.693 * 272 = 188 days

  30. 4. What is an acceptable emission level to meet a fish residue guideline of 2ppm? Method 1: vary emissions in ChemCan to determine the resulting concentration in fish or Method 2: set up a ratio, since emissions and fish concentrations are linearly related: E1/E2 = C1/C2 100 tonnes / E2 = (0.00555g/g) / (2g/g) E2  36,036 tonnes

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