Fugacity Models Level 1: Equilibrium

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


“Chemical properties control”

fugacity of chemical in medium 1 =



…..

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)

Fugacity Models





Level 2 :

Steady-state over the entire environment & Equilibrium between compartment

Flux in = Flux out




…..

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)

Fugacity Models





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

20

40

60

80

100

120

0 200 400 600 800 1000

Time (days)

Xw

(g

)

Xw ater (g)

Xw ater (g)

Numerical Integration:

No assumption regarding input overtime.


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

20000

40000

60000

80000

100000

120000

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76

Time (days)

Mas

s in

Lak

e W

ater

(g

ram

s)

Steady-State:Xw = Input/V

Evaluative Models vs. Real Models

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.

6. If steady-state does not apply, solve by numerical simulation

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?

dXwater /dt = Input - Output

dXwater /dt = Input - (Flow x Cwater)

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

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

20

40

60

80

100

120

0 200 400 600 800 1000

Time (days)

Xw

(g

)

Xw ater (g)

Xw ater (g)

Numerical Integration:

No assumption regarding input overtime.


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


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

20000

40000

60000

80000

100000

120000

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76

Time (days)

Mas

s in

Lak

e W

ater

(g

ram

s)

Steady-State:Xw = Input/V

Documents

Fugacity Models Level 1: Equilibrium