1 / 90

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.

Download Presentation

Environmental Partitioning in Evaluative Environments Merits:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related