The

Mass Balance Equation

Flux in = Flux out

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

Lake Surface Area = 1,000,000 m2

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

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.

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

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.

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

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

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

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 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)

Steady-State:Xw = Input/V