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

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

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)

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)

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

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

dCoil/dt = - Constant = - k

Integrate

Coil = - k.t +Constant

Units of rate constant: mol/m3.d

Coil

Time

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)

Integrate: 1/C = k.t + 1/Co

Half-life time : t1/2 = 1/(2.C0.k)

Michaelis-Menten Kinetics

e.g. Many enzymatic/Biological degradation processes

Vmax . Corg . Cs

-dCs/dt = -----------------------

Km + Cs

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

ADVECTIVE TRANSPORT

Piggy-Backing

•Rainfall

•Dry deposition (dust fall)

•Sediment deposition

•Resuspension

•Soil-run-off

•Food ingestion

Lake Volume = 1,000,000 m3

Concentration = 1 g/m3

Water Flow = 10,000 m3/day

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

Lake Volume = 1,000,000 m3

Concentration = 1 g/m3

Water Flow = 10,000 m3/day

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

Diffusive Transport

One-Medium

Flux = .A.(C1 - C2)

A : Area of Diffusion (m2)

: permeability / velocity / mass transfer coefficient (m/day)

C1 C2

Diffusive Transport

Two-Media

Flux = .A.(C1 - K12C2)

A : Area of Diffusion (m2)

: permeability / velocity / mass transfer coefficient (m/day)

C1

AirC2

Water

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

Spill of benzene : 1 tonne

Lake Volume = 100,000,000 m3

Lake Surface Area = 1,000,000 m2

Volatilisation = 0.3 m/d

Reaction = 0.01 1/d

Outflow = 2,000,000 m3/d

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 = .Area/Volume = 0.3 m/d x 1,000,000 m2/100,000,000 m3 = 0.003 1/day

k (TOTAL) = 0.01 + 0.02 + 0.003 = 0.033 1/day

Half-life Time = 0.693/0.033 = 21 days

Fugacity Format

Transport or Transformation:

Flux (mol/day) = D.f

D : Transport Parameter (mol/d.Pa)

f : Fugacity (Pa)

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)

Advective Transport Rate:

Flux = G.C

Flux = D.f

D = G.Z

G = Flow Rate Constant (m3/d)


Diffusive Transport Rate:

Flux = .A.C

Flux = D.f

D = .A.Z

= Mass Transfer Coefficient (m/d)

A = Area of diffusion (m2)


D (TOTAL) = Sum (D values)

D (TOTAL) = Di

Z = 100 mol/m3.Pa

Spill of benzene : 1 tonne

Lake Volume = 100,000,000 m3


0.3 10-12 mol/Pa.d

10-12 mol/Pa.d

2 10-12 mol/Pa.d

D (TOTAL) = Sum (D values)

= Di

= 3.3.10-12 mol/Pa.d

The

Mass Balance Equation

Question : What is the concentration of chemical X in the water (fish kills?)

Tool : Use steady-state mass-balance model

Lake

CW=?

Volatilisation

Emission

Sedimentation

Reaction

Outflow

Lake Volume = 100,000,000 m3


Question : What is the concentration of chemical X in the water (fish kills?)

Tool : Use steady-state mass-balance model

CW=?

0.001 1/d

1 mol/day

0.004 1/d

0.003 1/d

0.002 1/d

Lake Volume = 100,000,000 m3


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)

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

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)

Steady-state mass-balance model: 2 Media

Burial

CW=?

Volatilisation

Emission

Settling

Reaction

Outflow

CS=?

Resuspension

Water:

dMw/dt = Input + ksw.Ms - kw.Mw - kws.Mw = 0

Sediments:

dMs/dt = kws.Mw - kb.Ms - ksw.Ms = 0

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)}

In Fugacity Format

Water:

dMw/dt = Input + Dsw.fs - Dw.fw - Dws.fw = 0

Sediments:

dMs/dt = Dws.fw - Db.fs - Dsw.fs = 0

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)}

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.

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

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

LEVEL I

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)

E

GA.CBAGA.CA

GW.CBW

GW.CW

LEVEL II

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)

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

Removal Rate Constant for Environment:

Fraction of Mass of Chemical removed per unit of time by advection: kA 1/day

kA = Sum(Gi.Ci) / Vi.Ci

tADVECTION = 1/kA

Total Residence Time in Environment:

ktotal = kA + kR = E/M

tRESIDENCE = 1/kTOTAL = 1/kA + 1/kR

1/tRESIDENCE = 1/tADVECTION + 1/tREACTION

LEVEL III

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

dXwater /dt = Input - Output

dXwater /dt = Input - (Flow x Cwater)

dXwater /dt = Input - (Flow . Xwater/V)

dXwater /dt = Input - ((Flow/V). Xwater)

dXwater /dt = Input - k. Xwater

k = rate constant (day-1)

Time Dependent Fate Models / Level IV

Analytical Solution

Integration:

Assuming Input is constant over time:

Xwater = (Input/k).(1- exp(-k.t))

Xwater = (1/0.01).(1- exp(-0.01.t))

Xwater = 100.(1- exp(-0.01.t))

Cwater = (0.0001).(1- exp(-0.01.t))

0

10

20

30

40

5060

70

80

90

100

0 200 400 600 800 1000

Time (days)

Xw

(g) Xw ater (g)

Xw ater (g)

Numerical Integration:

No assumption regarding input overtime.

dXwater /dt = Input - k. Xwater

Xwater /t = Input - k. Xwater +

If t then

Xwater = (Input - k. Xwater).t

Split up time t in t by selecting t : t = 1

Start simulation with first time step:Then after the first time step

t = t = 1 d

Xwater = (1 - 0.01. Xwater).1

at t=0, Xwater = 0

Xwater = (1 - 0.01. 0).1 = 1

Xwater = 0 + 1 = 1

After the 2nd time stept = t = 2 d


at t=1, Xwater = 1

Xwater = (1 - 0.01. 1).1 = 0.99

Xwater = 1 + 0.99 = 1.99

After the 3rd time stept = t = 3 d


at t=2, Xwater = 1.99

Xwater = (1 - 0.01. 1.99).1 = 0.98

Xwater = 1.99 + 0.98 = 2.97

then repeat last two steps for t/t timesteps

Analytical Num. IntegrationTime Xwater Xwater

(days) (g) (g)0 0 01 0.995017 12 1.980133 1.993 2.955447 2.97014 3.921056 3.9403995 4.877058 4.9009956 5.823547 5.8519857 6.760618 6.7934658 7.688365 7.7255319 8.606881 8.648275

10 9.516258 9.561792

Mass of contaminant in water of lake vs time

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77

Time (days)

Mas

s in

Lak

e W

ater

(g

ram

s)

Application of the Models

•To assess concentrations in the environment

(if selecting appropriate environmental conditions)

•To assess chemical persistence in the environment

•To determine an environmental distribution profile

•To assess changes in concentrations over time.

What is the difference between

Equilibrium & Steady-State?

Cwater

Steady-State & Equilibrium

GILL UPTAKE

GILL ELIMINATION

Steady-State Flux Equation:

“Mass Balance Equation”

dM/dt = DWF.fW - DFW.fF = 0

DWF.fW = DFW.fF

fF/fW = DWF/DFW = 1.0

CF/CW = fF.ZF/fW.ZW = ZF/ZW = KFW

Equilibrium

Cwater


GILL UPTAKE

GILL ELIMINATION

METABOLISM


“Mass Balance Equation”

dM/dt = DWF.fW - DFW.fF - DM.fF = 0

DWF.fW = DFW.fF + DM.fF

fF/fW = DWF/(DFW + DM) < 1.0

CF/CW = (ZF/ZW). DWF/(DFW + DM) < KFW

Steady-State

Diffusive Transport Rate between

2 media:

Air

Water

Transport in Series:

Rtot = R1 + R2


Rtot = RAir + RWater

1/Dtot = 1/DA + 1/DW

1/Dtot = 1/kA.A.ZA + 1/kW.A.ZW

D from water-to-air =

D from air-to-water


1/Dtot = 1/kWA.A.ZW

1/Dtot = (ZW/kA.A.ZA + 1/kW.A).1/ZW

1/Dtot = (1/kA.A.KAW + 1/kW.A).1/ZW

1/kWA = (1/kA.KAW + 1/kW).1/ZW


1/Dtot = 1/kAW.A.ZA

1/Dtot = (1/kA.A + ZA/kW.A.ZW).1/ZA

1/Dtot = (1/kA.A + KAW/kW.A).1/ZA

1/kAW = (1/kA + KAW/kW).1/ZA


KAW = kAW/kWA

The mass transfer coefficient from air to water is not equal to the mass

transfer coefficient from water to air

ADVECTIVE TRANSPORT


Air

Water

DAW DWA


Flux (Water-to-Air) = Flux (Air-to-Water)

DAW.fW = DAW.fA

fW/fA = DAW/DAW = 1.0

Equilibrium