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Environmental Partitioning in Evaluative Environments Merits:

Environmental Partitioning in Evaluative Environments Merits: Provides assessments of the environmental distribution of chemicals based on chemical properties: Relative Concentrations Mass distribution Can be used for comparing/ranking chemicals.

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Environmental Partitioning in Evaluative Environments Merits:

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  1. Environmental Partitioning in Evaluative Environments • Merits: • Provides assessments of the environmental distribution of chemicals based on chemical properties: • Relative Concentrations • Mass distribution • Can be used for comparing/ranking chemicals

  2. Environmental Partitioning in Evaluative Environments • Limitations: • Closed System • Describes an end-situation, achieved after a long time when equilibrium is reached. • Absolute values of concentrations are irrelevant • Well mixed environment • Assumes chemical losses (through transformation and transport) do not occur

  3. REACTION & TRANSPORT RATES dX/dt : Moles/day (Flux) d(C.V) : Moles/day dC/dt : Moles/L.day if V is constant X : Mass of chemical (moles) C : Concentration (mol/m3) V : Volume of medium in which the chemical resides (m3)

  4. 1-st Order Kinetic Process • e.g. dechlorination of PCBs in sediments • applies to : • reacting substrate is present in small quantities • reacting medium is present in large amounts • dC/dt = - k. C • k : first order rate constant (1/day)

  5. Integrate: C = Co.exp(-k.t) ln C = ln Co - k.t Half-life time : t1/2 = 0.693/k ln C - ln Co = - k . t ln (C/ Co) = - k . t ln (0.5) = - k . t ln (1/0.5) = k . t ln 2 = 0.693 = k . t t = 0.693/k Ln C Time

  6. 0-order Kinetics • e.g. photolysis of oil in oil slicks • applies to : • reacting substrate is present in very high amounts • reacting agent (light, micro-organism, substance B) is present at low levels • dC/dt = - k . Coil = Constant • But: Coil is constant over time

  7. Coil dCoil/dt = - Constant = - k Integrate Coil = - k.t +Constant Units of rate constant: mol/m3.d Time

  8. 2nd Order Kinetic Process • e.g. NO + O3 ===> NO2 + O2 • applies to : • Both reacting substrates are present in small quantities • d[NO]/dt = - k . [NO] . [O3] • k : 2nd order rate constant (m3/mol.day)

  9. Integrate: 1/C = k.t + 1/Co Half-life time : t1/2 = 1/(2.C0.k)

  10. Michaelis-Menten Kinetics e.g. Many enzymatic/Biological degradation processes Vmax . Corg . Cs -dCs/dt = ----------------------- Km + Cs

  11. If Cs is very high (compared to Km): -dCs/dt = Vmax . Corg : 0-order If Cs is very low (compared to Km): -dCs/dt = Vmax . Corg .Cs/Km = k.Cs : 1-order

  12. ADVECTIVE TRANSPORT Piggy-Backing • Rainfall • Dry deposition (dust fall) • Sediment deposition • Resuspension • Soil-run-off • Food ingestion

  13. Water Flow = 10,000 m3/day Lake Volume = 1,000,000 m3 Concentration = 1 g/m3 Flux = Flow x Concentration = 10,000 g/d Rate Constant k = Flow / Volume = 0.01 d-1 Residence Time = Volume / Flow = 1/ k = 100 d

  14. Water Flow = 10,000 m3/day Lake Volume = 1,000,000 m3 Concentration = 1 g/m3 Flux = Flow x Concentration = 10,000 g/d Rate Constant k = Flux / Total Mass in Lake Rate Constant k = 10,000 g/d / 1,000,000 g Rate Constant k = 0.01 1/d

  15. Diffusive Transport One-Medium C1 C2 Flux = k.A.(C1 - C2) A : Area of Diffusion (m2) k : permeability / velocity / mass transfer coefficient (m/day)

  16. Diffusive Transport Two-Media C2 Air Water C1 Flux = k.A.(C1 - K12C2) A : Area of Diffusion (m2) k : permeability / velocity / mass transfer coefficient (m/day)

  17. Question? What is the half-life time of Benzene in Lake X given that: Benzene is transformed at a rate of 0.01 1/day and the lake water flows out of the lake at a flow rate 2,000,000 m3/day and diffuses to the air which is characterized with a mass transfer coefficient of 0.3 m/day

  18. Volatilisation = 0.3 m/d Outflow = 2,000,000 m3/d Reaction = 0.01 1/d Spill of benzene : 1 tonne Lake Volume = 100,000,000 m3 Lake Surface Area = 1,000,000 m2

  19. Reaction: k = 0.01 1/day Flow: k = flow/volume = 2,000,000 m3/day /100,000,000 m3 = 0.02 1/day Diffusion: k = k.Area/Volume = 0.3 m/d x 1,000,000 m2/100,000,000 m3 = 0.003 1/day

  20. k (TOTAL) = 0.01 + 0.02 + 0.003 = 0.033 1/day Half-life Time = 0.693/0.033 = 21 days

  21. Fugacity Format Transport or Transformation: Flux (mol/day) = D.f D : Transport Parameter (mol/d.Pa) f : Fugacity (Pa)

  22. Reaction Rate: Flux = k.V.C Flux = D.f D = k.V.Z k = Reaction Rate Constant (1/d) V = Reaction Volume (m3) C = Concentration of reacting substrate (mol/m3)

  23. Advective Transport Rate: Flux = G.C Flux = D.f D = G.Z G = Flow Rate Constant (m3/d) C = Concentration of reacting substrate (mol/m3)

  24. Diffusive Transport Rate: Flux = k.A.C Flux = D.f D = k.A.Z k= Mass Transfer Coefficient (m/d) A = Area of diffusion (m2) C = Concentration of reacting substrate (mol/m3)

  25. D (TOTAL) = Sum (D values) D (TOTAL) = S Di

  26. 0.3 10-12 mol/Pa.d 2 10-12 mol/Pa.d 10-12 mol/Pa.d Z = 100 mol/m3.Pa Spill of benzene : 1 tonne Lake Volume = 100,000,000 m3 Lake Surface Area = 1,000,000 m2

  27. D (TOTAL) = Sum (D values) = S Di = 3.3.10-12 mol/Pa.d

  28. The Mass Balance Equation

  29. Question : What is the concentration of chemical X in the water (fish kills?) Tool : Use steady-state mass-balance model Lake Volatilisation Emission Outflow CW=? Reaction Lake Volume = 100,000,000 m3 Lake Surface Area = 1,000,000 m2 Sedimentation

  30. Question : What is the concentration of chemical X in the water (fish kills?) Tool : Use steady-state mass-balance model 0.001 1/d 1 mol/day 0.002 1/d CW=? 0.003 1/d Lake Volume = 100,000,000 m3 Lake Surface Area = 1,000,000 m2 0.004 1/d

  31. Concentration Format dMW/dt = E - kV.MW - kS.MW - kO.MW - kR.MW dMW/dt = E - (kV + kS+ kO+ kR).MW 0 = E - (kV + kS+ kO+ kR).MW E = (kV + kS+ kO+ kR).MW MW = E/(kV + kS+ kO+ kR) & CW = MW/VW MW : Mass in Water (moles) t : time (days) E : Emission (mol/day) kV: Volatilization Rate Constant (1/day) kS: Sedimentation Rate Constant (1/day) kO: Outflow Rate Constant (1/day) kR.: Reaction Rate Constant (1/day)

  32. Concentration Format dMW/dt = 1 - 0.001.MW - 0.004.MW - 0.002.MW - 0.003.MW dMW/dt = 1 - (0.001+ 0.004+ 0.002+ 0.003).MW 0 = 1 - (0.001+ 0.004+ 0.002+ 0.003).MW 1 = (0.001+ 0.004+ 0.002+ 0.003).MW MW = 1/(0.001+ 0.004+ 0.002+ 0.003) = 1/0.01 CW = 0.01/100,000,000 = 1.10-10 mol/m3

  33. Fugacity Format d(VW ZW.fW )/dt = E - DV.fW - DS.fW - DO.fW - DR.fW VW ZW.dfW/dt = E - (DV + DS+ DO+ DR).fW 0 = E - (DV + DS+ DO+ DR).fW E = (DV + DS+ DO+ DR).fW fW = E/ (DV + DS+ DO+ DR) & CW = fW.ZW VW : Volume of Water (m3) ZW : Fugacity Capacity in water (mol/M3.Pa) fW : Fugacity in Water (Pa) t : time (days) E : Emission (mol/day) DV: Transport Parameter for Volatilization (mol/Pa. day) DS: Transport parameter fro Sedimentation (mol/Pa.day) DO: Transport Parameter for Outflow (mol/Pa.day) kR.: Transport Parameter for Reaction (mol/Pa.day)

  34. Steady-state mass-balance model: 2 Media Volatilisation Emission Outflow Settling CW=? Reaction Resuspension CS=? Burial

  35. Water: dMw/dt = Input + ksw.Ms - kw.Mw - kws.Mw = 0 Sediments: dMs/dt = kws.Mw - kb.Ms - ksw.Ms = 0

  36. From : Eq. 2 kws.Mw = kb.Ms + ksw.Ms Ms = kws.Mw / (kb + ksw) Substitute in eq. 1 Input + ksw.{kws.Mw / (kb + ksw)} = kw.Mw + kws.Mw Input = kw.Mw + kws.Mw - ksw.{kws.Mw / (kb + ksw)}

  37. In Fugacity Format Water: dMw/dt = Input + Dsw.fs - Dw.fw - Dws.fw = 0 Sediments: dMs/dt = Dws.fw - Db.fs - Dsw.fs = 0

  38. From : Eq. 2 Dws.fw = Db.fs + Dsw.fs fs = Dws.fw / (Db + Dsw) Substitute in eq. 1 Input + Dsw.{Dws.fw / (Db + Dsw)} = Dw.fw + Dws.fw Input = Dw.fw + Dws.fw - Dsw.{Dws.fw / (Db + Dsw)}

  39. Recipe for developing mass balance equations 1. Identify # of compartments 2. Identify relevant transport and transformation processes 3. It helps to make a conceptual diagram with arrows representing the relevant transport and transformation processes 4. Set up the differential equation for each compartment 5. Solve the differential equation(s) by assuming steady-state, i.e. Net flux is 0, dC/dt or df/dt is 0.

  40. Level III fugacity Model: Steady-state in each compartment of the environment Flux in = Flux out Ei + Sum(Gi.CBi) + Sum(Dji.fj)= Sum(DRi + DAi + Dij.)fi For each compartment, there is one equation & one unknown. This set of equations can be solved by substitution and elimination, but this is quite a chore. Use Computer

  41. Fugacity Models Level 1 : Equilibrium Level 2 : Equilibrium between compartments & Steady-state over entire environment Level 3 : Steady-State between compartments Level 4 : No steady-state or equilibrium / time dependent

  42. LEVEL I

  43. Mass Balance Total Mass = Sum (Ci.Vi) Total Mass = Sum (fi.Zi.Vi) At Equilibrium : fi are equal Total Mass = M = f.Sum(Zi.Vi) f = M/Sum (Zi.Vi)

  44. LEVEL II GA.CA GA.CBA E GW.CBW GW.CW

  45. Level II fugacity Model: Steady-state over the ENTIRE environment Flux in = Flux out E + GA.CBA + GW.CBW = GA.CA + GW.CW All Inputs = GA.CA + GW.CW All Inputs = GA.fA .ZA + GW.fW .ZW Assume equilibrium between media : fA= fW All Inputs = (GA.ZA + GW.ZW).f f = All Inputs / (GA.ZA + GW.ZW) f = All Inputs / Sum (all D values)

  46. Reaction Rate Constant for Environment: Fraction of Mass of Chemical reacting per unit of time : kR (1/day) kR = Sum(Mi.ki) / Mi Reaction Residence time: tREACTION = 1/kR

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