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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
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Level 1 : Equilibrium
“Chemical properties control”
fugacity of chemical in medium 1 =
fugacity of chemical in medium 2 =
fugacity of chemical in medium 3 =
…..
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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)
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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
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Level 2 :
Steady-state over the entire environment & Equilibrium between compartment
Flux in = Flux out
fugacity of chemical in medium 1 =
fugacity of chemical in medium 2 =
fugacity of chemical in medium 3 =
…..
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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)
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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
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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
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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
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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))
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0
20
40
60
80
100
120
0 200 400 600 800 1000
Time (days)
Xw
(g
)
Xw ater (g)
Xw ater (g)
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Numerical Integration:
No assumption regarding input overtime.
dXwater /dt = Input - k. Xwater
Xwater /t = Input - k. Xwater +
If t then
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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
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After the 2nd time stept = t = 2 d
Xwater = (1 - 0.01. Xwater).1
at t=1, Xwater = 1
Xwater = (1 - 0.01. 1).1 = 0.99
Xwater = 1 + 0.99 = 1.99
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After the 3rd time stept = t = 3 d
Xwater = (1 - 0.01. Xwater).1
at t=2, Xwater = 1.99
Xwater = (1 - 0.01. 1.99).1 = 0.98
Xwater = 1.99 + 0.98 = 2.97
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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
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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
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Evaluative Models vs. Real Models
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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
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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.
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What is the difference between
Equilibrium & Steady-State?
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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
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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))
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0
20
40
60
80
100
120
0 200 400 600 800 1000
Time (days)
Xw
(g
)
Xw ater (g)
Xw ater (g)
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Numerical Integration:
No assumption regarding input overtime.
dXwater /dt = Input - k. Xwater
Xwater /t = Input - k. Xwater +
If t then
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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
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After the 2nd time stept = t = 2 d
Xwater = (1 - 0.01. Xwater).1
at t=1, Xwater = 1
Xwater = (1 - 0.01. 1).1 = 0.99
Xwater = 1 + 0.99 = 1.99
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After the 3rd time stept = t = 3 d
Xwater = (1 - 0.01. Xwater).1
at t=2, Xwater = 1.99
Xwater = (1 - 0.01. 1.99).1 = 0.98
Xwater = 1.99 + 0.98 = 2.97
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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
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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