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Presentation Slides for
Chapter 17, Part 2of
Fundamentals of Atmospheric Modeling 2nd Edition
Mark Z. JacobsonDepartment of Civil & Environmental Engineering
Stanford UniversityStanford, CA 94305-4020jacobson@stanford.edu
April 1, 2005
Solvation and HydrationSolvation
Bonding between solvent and solute in solution
Hydration
When solvent is liquid water, solvation is hydration
Hydration of cations --> lone pairs of electrons on oxygen atom of water attach to cations
Hydration of anions --> water molecule attaches to anion via hydrogen bonding
Water EquationQuantify amount of hydration with empirical water equation
Zdanovskii-Stokes-Robinson (ZSR) equation
Example with two species, x and y (17.64)
mx,a, my,a = molalities of x and y, alone in solution at given relative humidity
mx,m, my,m = molalities of x and y, when mixed together, at same relative humidity
mx,mmx,a
+my,mmy,a
=1
ZSR Equation
Table 17.2
ZSR equation predictions for a sucrose (species x) - mannitol (species y) mixture at two different water activities.
mx,m/mx,a +
Case mx,a my,a mx,m my,m my,m/my,a
1 0.7751 0.8197 0.6227 0.1604 0.999
2 0.9393 1.0046 0.1900 0.8014 1.000
mx,mmx,a
+my,mmy,a
=1
Water EquationGeneralized ZSR equation (17.64)
Polynomial expression for molality of electrolyte alone in solution at a given water activity (17.66)
mk,mmk,ak
∑ =1
mk,a =Y0,k +Y1,kaw +Y2,kaw2 +Y3,kaw
3 +...
Water Equation
Fig. 17.4a
Water activities of several electrolytes at 298.15 K
0 10 20 30 40 50 60
0
0.2
0.4
0.6
0.8
1
0 10 20 30 40 50 60
Molality
Water activityNaNO
3
HNO
3
H
2
SO
4HCl
Wat
er a
ctiv
ity
Water Equation
Fig. 17.4b
Water activities of several electrolytes at 298.15 K
0 5 10 15 20 25 30
0.4
0.5
0.6
0.7
0.8
0.9
1
0 5 10 15 20 25 30
Molality
Water activity
NH
4
NO
3
NH
4
Cl
NaCl
(NH
4
)
2
SO
4
Na
2
SO
4
Wat
er a
ctiv
ity
Temp. Dependence of Water ActivityTemperature dependence of binary water activity coefficients under
ambient surface conditions is small.
Polynomial for water activity at reference temperature (17.68)
Temperature dependence of water activity (17.67)
lnaw T( ) =lnaw0 −
mvmk,a2
R*TLT0
∂φL∂mk,a
+TC∂φcP∂mk,a
⎛
⎝ ⎜
⎞
⎠ ⎟
lnaw0 =A0 +A1mk,a
12+A2mk,a +A3mk,a
32+...
Temp. Dependence of Water ActivityCombine (17.67), (17.68), (17.54) (17.69-70)
lnaw T( ) =A0 +A1mk,a12
+A2mk,a +E3mk,a32
+E4mk,a2 +...
El =Al −0.5 l −2( )mv
R*TLT0
Ul −2 +TCVl−2⎛
⎝ ⎜
⎞
⎠ ⎟
Example mHCl= 16 m
T = 273 K---> aw = 0.09
T = 310 K---> aw = 0.11
Practical Use of Water EquationRearrange (17.65) (17.71)
mi,j,a = binary molalities of species alone in solution ci,j,m = hypothetical mol cm-3 of electrolyte pair when mixed in solution with all other components
In a model, ion concentrations known but hypothetical electrolyte concentrations unknown --> find hypothetical concentrations
cw =1mv
ci, j,mmi, j,aj =1
NA
∑⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
i=1
NC
∑
Practical Use of Water Equation
6 mol m-3 of H+, 6 mol m-3 Na+
7 mol m-3 of Cl- , 5 mol m-3 of NO3-
Example 17.1:
Combine ions in a way to satisfy mole balance constraintscH+,m=cHNO3,m+cHCl,m
cNa+,m =cNaNO3,m+cNaCl,m
cCl−,m=cHCl,m+cNaCl,m
cNO3- ,m =cHNO3,m+cNaNO3,m
Case cHCl,m cHNO3,m cNaCl,m cNaNO3,m
1 6 0 1 52 4 2 3 3
Concentrations that satisfy mole balance constraints (Table 17.3)
Practical Use of Water Equation
Cation
Automatic method to recombine ions into hypothetical electrolytes
Execute the following three equations, in succession, for each undissociated electrolyte, i,j
Electrolyte (17.72)
Anion
ci, j,m=minci,mνi
,cj,mν j
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
ci,m=ci,m−νici, j,m
cj,m=cj,m−ν jci, j,m
Deliquescence Relative HumidityDeliquescence
Process by which a particle takes up liquid water, lowering its saturation vapor pressure
Deliquescence relative humidity (DRH)The relative humidity at which an initially-dry solid first takes on liquid water during an increase in relative humidity. Above the DRH, the solid may not exist.
Crystallization relative humidity (CRH)The relative humidity at which an initially-supersaturated aqueous electrolyte becomes crystalline upon a decrease in relative humidity.
Deliquescence Relative Humidity
Table 17.4
DRHs and CRHs for several electrolytes at 298 K
In a mixture, the DRH of a solid in equilibrium with the solution is lower than the DRH of the solid alone
Electrolyte DRH(%) CRH(%)
NaCl 75.28 47Na2SO4 84.2 57-59NaHSO4 52.0 <5NH4Cl 77.1 47(NH4)2SO4 79.97 37-40NH4HSO4 40 <5-22NH4NO3 61.83 25-32KCl 84.26 62Oxalic acid 97.3 51.8-56.7
Solid Formation
Consider the equilibrium reaction
Consider the equilibrium reaction
A solid forms when (17.73)
A solid forms when (17.74)
NH4NO3 s( ) NH4++NO3
−
mNH4
+mNO3−γ
NH4+,NO3
−2 >Keq T( )
NH4NO3 s( ) NH3 g( )+HNO3 g( )
pNH3 g( ),spHNO3 g( ),s >Keq T( )
Example Equilibrium ProblemConsider two equilibrium reactions (17.75)
HCl (g)H
+
+ Cl-
HSO4 H
+
+ SO
2-
4
For equilibrium concentrations, solve
equilibrium constant equations
mole balance equations
charge balance equation
water equation
with Newton-Raphson iteration
Example Equilibrium Problem
Equilibrium coefficient equations (17.76)
mH+,eq
mCl- ,eq
γH+,Cl- ,eq2
pHCl,s,eq=Keq T( )
mH+,eq
mSO4
2−,eqγ2H+,SO4
2−,eq3
mHSO4
−,eqγ
H+,HSO4−,eq
2 =Keq T( )
Example Equilibrium Problem
Mole balance equations (17.77)
CHCl(g),eq+cCl-,eq
=CHCl(g),t−h +cCl-,t−h
cHSO4
−,eq+c
SO42−,eq
=cHSO4
−,t−h+c
SO42−,t−h
(17.78)
Example Equilibrium ProblemVapor pressure as a function of mole concentration (17.79)
Charge balance equation (17.80)
Molality as a function of mole concentration
pHCl,s,eq=CHCl(g),s,eqR*T
mCl-,eq
=cCl- ,eq
cw,eqmv
cCl−,eq
+cHSO4
−,eq+2c
SO42−,eq
=cH+,eq
Example Equilibrium ProblemWater equation (17.81)
Hypothetical mole concentration constraints (17.82)
cw,eq =1mv
cH+,Cl- ,m
mH+,Cl- ,a
+cH+,HSO4
−,m
mH+,HSO4
−,a
+c2H+,SO4
2−,m
m2H+,SO4
2−,a
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟
cH+,eq
=cH+,Cl-,m
+cH+,HSO4
−,m+2c
2H+,SO42−,m
cCl−,eq
=cH+,Cl-,m
cHSO4
−,eq=c
H+,HSO4−,m
cSO4
2−,eq=c
2H+,SO42−,m
Mass-Flux Iterative MethodSolve each equation iteratively and iterate over all equations
Initialize species concentrations so that charge is conserved
No intelligent first guess required
Solution mass and charge conserving and always converges
Example solution for one equilibrium equation
Equilibrium equation and coefficient relation
νDD+νEE +... νAA +νBB+...
A{ }νA B{ }νB ...
D{ }νD E{ }νE ...=Keq T( )
Mass-Flux Iterative Method1) Calculate smallest ratio of mole concentration to moles in
denominator and numerator, respectively (17.83)
2) Initialize two parameters
Qd =minCD,0νD
,CE,0νE
⎛
⎝ ⎜
⎞
⎠ ⎟
Qn =mincA,0νA
,cB,0νB
⎛
⎝ ⎜
⎞
⎠ ⎟
z1 =0.5(Qd +Qn) Δx1=Qd −z1
Mass-Flux Iterative MethodAdd mass flux factor (x) to mole concentrations (17.84)
3) Compare ratio of activities to equilibrium coefficient (17.85)
cA,l+1 =cA,l +νAΔxl cB,l+1 =cB,l +νBΔxl
CD,l+1=CD,l −νDΔxl CE,l +1 =CE,l −νEΔxl
F =mA,l+1
νA mB,l +1νB γAB,l+1
νA+νB
pD,l+1νD pE,l+1
νE
1Keq T( )
Mass-Flux Iterative Method4) Cut z in half
5) Check convergence (17.86)
Return to (17.84) until convergence occurs
zl+1 =0.5zl
F =
>1 → Δxl +1=−zl +1<1 → Δxl +1=+zl +1=1 → convergence
⎧
⎨ ⎪
⎩ ⎪
Analytical Equilibrium Iteration MethodSolve most equations analytically but iterate over all equations
Reactions of the form DA
Solve the equilibrium equation (17.87)
Solution for change in concentration (17.88)
Final concentrations
cA,ccD,c
=cA,0+ΔxfincD,0 −Δxfin
=Kr
Δxfin=cD,0Kr −cA,0
1+Kr
cA,c =cA,0+Δxfin cD,c =cD,0 −Δxfin
Analytical Equilibrium Iteration Method
Solve the equilibrium equation (17.89)
Reactions of the form D+EA+B
Solution for change in concentration (17.90)
cA,ccB,ccD,ccE,c
=cA,0 +Δxfin( ) cB,0 +Δxfin( )
cD,0 −Δxfin( ) cE,0−Δxfin( )=Kr
Δxfin=
−cA,0 −cB,0 −cD,0Kr −cE,0Kr
+cA,0 +cB,0 +cD,0Kr +cE,0Kr( )
2
−41−Kr( ) cA,0cB,0−cD,0cE,0( )
2 1−Kr( )
Analytical Equilibrium Iteration Method
Final concentrations
cA,c =cA,0+Δxfin
cB,c =cB,0 +Δxfin
cD,c =cD,0 −Δxfin
cE,c =cE,0 −Δxfin
Analytical Equilibrium Iteration Method
Check if solid can form (17.91)
Reactions of the form D(s)2A+B
If so, solve the equilibrium equation (17.92)
cA,0 +2cD,0( )2
cB,0+2cD,0( )>Kr
cA,c2 cB,c = cA,0 +2Δxfin( )
2cB,0 +Δxfin( )=Kr
Analytical Equilibrium Iteration MethodIterative Newton-Raphson procedure (17.93)
fn x( ) =Δxfin,n3 +qΔxfin,n
2 +rΔxfin,n +s =0
′ f x( ) =3Δxfin,n2 +2qxfin,n +r
q =cA,0 +cB,0
r =cA,0cB,0+0.25cA,02
s =cA,02 cB,0−Kr
Δxfin,n+1=Δxfin,n −fn x( )′ f n x( )
Analytical Equilibrium Iteration Method
Final concentrations
cA,c =cA,0+2Δxfin
cB,c =cB,0 +Δxfin
cD,c =cD,0 −Δxfin
Equilibrium Solver Results
Fig. 17.4
Aerosol composition versus NaCl concentration when the relative humidity was 90%. Other initial conditions were H2SO4(aq) = 10 g m-3, HCl(g) = 0 g m-3, NH3(g) = 10 g m-3, HNO3(g) = 30 g m-3, and T = 298 K.
0
5
10
15
20
25
30
0 5 10 15 20 25 30
Concentration (
g m
-3
)
NO
3
-
H
2
( ) 0.1O aq x
SO
4
2-
NH
4
+
(NaCl concentration g m
-3
)
Cl
-
Con
cent
rati
on (g
m-3)
Equilibrium Solver Results
Fig. 17.5
Aerosol composition versus relative humidity. Initial conditions were H2SO4(aq) = 10 g m-3, HCl(g) = 0 g m-3, NH3(g) = 10 g m-3, HNO3(g) = 30 g m-3, and T = 298 K.
0
5
10
15
20
25
0 20 40 60 80 100
Concentration (
g m
-3
)
NH
4
NO
3
( )s
NO
3
-
( )Relative humidity percent
SO
4
2-
NH
4
+
H
2
( ) 0.1O aq x
(NH
4
)
2
SO
4
( )s
Con
cent
rati
on (g
m-3)
Dissolutional GrowthSaturation vapor pressure of gas q over particle size i (17.95)
Saturation vapor pressure as function of gas mole concentration (17.96)
Molality as function of particle mole concentration (17.97)
pq,s,i =mq,iHq
pq,s,i =Cq,s,i R*T
mq,i =cq,i
mvcw,i
Dissolutional GrowthSubstitute (17.95) and (17.97) into (17.96) (17.98)
where (17.99)
Cq,s,i =pq,s,i
R*T=
mq,i
R*THq=
cq,i
mvcw,i R*THq
=cq,i
′ H q,i
′ H q,i =mvcw,i R*THq
Dissolutional GrowthCondensational growth equations (16.67)
(16.68)
dcq,i,tdt
=kq,i,t−h Cq,t − ′ S q,i,t−hCq,s,i,t−h( )
dCq,tdt
=− kq,i,t−h Cq,t − ′ S q,i,t−hCq,s,i,t( )[ ]i=1
NB
∑
Dissolutional GrowthSubstitute (17.98)
--> Dissolutional growth equations (17.100)
(17.101)
dcq,i,tdt
=kq,i,t−h Cq,t − ′ S q,i,t−hcq,i,t′ H q,i,t−h
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
dCq,tdt
=− kq,i,t−h Cq,t − ′ S q,i,t−hcq,i,t′ H q,i,t−h
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ i=1
NB
∑
Analytical Predictor of DissolutionIntegrate (17.100) for final aerosol concentration (17.102)
Mole balance equation (17.103)
Substitute (17.102) into (17.103) (17.104)
cq,i,t =′ H q,i,t−hCq,t
′ S q,i,t−h+ cq,i,t−h−
′ H q,i,t−hCq,t′ S q,i,t−h
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ exp−
h ′ S q,i,t−hkq,i,t−h′ H q,i,t−h
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
Cq,t + cq,i,ti=1
NB
∑ =Cq,t−h+ cq,i,t−hi =1
NB
∑
Cq,t =
Cq,t−h + cq,i,t−h 1−exp−h ′ S q,i,t−hkq,i,t−h
′ H q,i,t−h
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪ i=1
NB
∑
1+′ H q,i,t−h′ S i,q,t−h
1−exp −h ′ S q,i,t−hkq,i,t−h
′ H q,i,t−h
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪ i=1
NB
∑
Growth During DissociationGrowth equation for hydrochloric acid (17.105)
Total dissolved chlorine (17.106)
Find saturation mole concentration from equilibrium expressions(17.107) HClHCl(aq)
(17.108) HCl(aq)H++Cl-
dcCl,i,tdt
=kHCl,i,t−h CHCl,t − ′ S HCl,i,t−hCHCl,s,i,t( )
cCl,i,t =cHCl aq( ),i,t +cCl-,i,t
Growth During DissociationEquilibrium coefficient relations (17.107)
(17.108)
Equilibrium coefficient relations in terms of mole concentration (17.109)
(17.110)
mHCl aq( ),i
pHCl,s,i=HHCl
molkg atm
mH+,imCl-,iγi,H+ Cl-2
mHCl aq( ),i=KHCl
molkg
CHCl,s,i =cCl,i′ K HCl,i
′ K HCl,i = HHCl 1+KHCl mvcw,i( )
2R*T
cH+,iγi,H+ Cl-2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
Dissolution of Acids/BasesSubstitute saturation mole concentration into growth equation (17.111)
Mole balance equation (17.112)
dcCl,i,tdt
=kHCl,i,t−h CHCl,t − ′ S HCl,i,t−hcCl,i,t′ K HCl,i,t−h
⎛
⎝ ⎜
⎞
⎠ ⎟
CHCl,t + cCl,i,ti =1
NB
∑ =CHCl,t−h + cCl,i,t−hi=1
NB
∑
Dissolution for Dissociating SpeciesIntegrate (17.111) for final aerosol concentration (17.113)
Substitute (17.113) into (17.112) (17.114)
cCl,i,t =′ K HCl,i,t−hCHCl,t
′ S Cl- ,i,t−h
+ cCl,i,t−h −′ K HCl,i,t−hCHCl,t
′ S HCl,i,t−h
⎛
⎝ ⎜
⎞
⎠ ⎟ exp−
hkHCl,i,t−h ′ S HCl,i,t−h′ K HCl,i,t−h
⎛
⎝ ⎜
⎞
⎠ ⎟
CHCl,t =
CHCl,t−h + cCl,i,t−h 1−exp−hkHCl,i,t−h ′ S HCl,i,t−h
′ K HCl,i,t−h
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪ i=1
NB
∑
1+′ K HCl,i,t−h′ S HCl,i,t−h
1−exp−hkHCl,i,t−h ′ S HCl,i,t−h
′ K HCl,i,t−h
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪ i=1
NB
∑
Solve for Ammonia/AmmoniumCharge balance equation (17.115)
where (17.116)
Mole balance equation (17.117)
cNH4+,i,t +cH+,i,t +c±,i,t =0
c±,i,t =−cNO3-,i,t −cCl-,i,t−cHSO4
- ,i,t −2cSO42- ,i,t + z
q∑ cq,i,t−h
CHCl,t =CNH3,t + cNH3 aq( ),i,t +cNH4+,i,t( )
i=1
NB
∑
=CNH3,t−h+ cNH3 aq( ),i,t−h +cNH4+,i,t−h( )
i =1
NB
∑ =Ctot
Solve for Ammonia/AmmoniumEquilibrium expressions (17.118) NH3(g)NH3(aq)
(17.119) NH3(aq)+H+NH4+
Equilibrium coefficient expressions (17.118)
(17.119)
mNH3 aq( ),i
pNH3
=HNH3 mol
kg atm
mNH4+,iγi,NH4
+
mNH3 aq( ),imH+,iγi,H+=KNH3
kgmol
Solve for Ammonia/AmmoniumNH4
+/H+ activity coefficient relationship (17.120)
Equilibrium coefficient relations in terms of mole concentration (17.121,2)
γi,NH4+
γi,H+=
γi,NH4+γi,NO3
−
γi,H+γi,NO3−
=γ
i,NH4+ NO3
−2
γi,H+ NO3
−2
=γi,NH4
+γi,Cl-
γi,H+γi,Cl-=
γi,NH4
+ Cl-2
γi,H+ Cl-2
cNH3 aq( ),i
CNH3
= ′ H NH3,imolmol ′ H NH3,i =HNH3R*Tmvcw,i
cNH4+,i
cNH3 aq( ),icH+,i= ′ K NH3,i
cm3
mol′ K NH3,i =KNH3
1mvcw,i
γi,H+
γi,NH4+
Solve for Ammonia/AmmoniumIon concentration in each size bin (17.124)
Substitute into mole-balance equation (17.125)
cNH4+,i,t =
−c±,i,tCNH3,t ′ H NH3,i,t−h ′ K NH3,i,t−h
CNH3,t ′ H NH3,i,t−h ′ K NH3,i,t−h +1
CNH3,t +
CNH3,t ′ H NH3,i,t−h
−c±,i,tCNH3,t ′ H NH3,i,t−h ′ K NH3,i,t−hCNH3,t ′ H NH3,i,t−h ′ K NH3,i,t−h +1
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ i =1
NB
∑ −Ctot=0
Solve for Ammonia/AmmoniumIterate for ammonia gas concentration (17.126)
where (17.128)
CNH3,t,n+1=CNH3,t,n−fn CNH3,t,n( )
′ f n CNH3,t,n( )
′ f n CNH3,t,n( ) =
1+
′ H NH3,i,t−h −c±,i,t ′ H NH3,i,t−h ′ K NH3,i,t−h
CNH3,t,n ′ H NH3,i,t−h ′ K NH3,i,t−h +1
+c±,i,tCNH3,t,n ′ H NH3,i,t−h ′ K NH3,i,t−h( )
2
CNH3,t,n ′ H NH3,i,t−h ′ K NH3,i,t−h +1( )2
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥ ⎥
i=1
NB
∑
Simulations of Growth/Dissociation
Fig. 17.7
Initial distributions for simulation
10
-1
10
0
10
1
10
2
10
-2
10
-1
10
0
10
1
10
2
10
3
10
4
0.1 1 10
dM (
g m
-3
) / d log
10
D
p
( . dN No cm
-3
) / d log
10
D
p
(Particle diameter D
p
, )m
Soildust
( )NaCl s
.Number conc
( ) a Initial
dM (g
m-3)
/ dlo
g 10 D
pdN
(No. cm
-3) / dlog10 D
p
Simulations of Growth/Dissociation
0
5
10
15
20
25
30
0 2 4 6 8 10 12
Summed concentration (
g m
-3
)
( )Time from start h
H
2
0.1O x
NH
4
+
NO
3
-
Cl
-
( )S VI
( ) =5b h s
Na
+
Aerosol concentrations, summed over all sizes, during nonequilibrium growth plus internal aerosol equilibrium at RH=90 percent when h=5 s.
Sum
med
con
cent
rati
on (g
m-3)
Simulations of Growth/Dissociation
0
5
10
15
20
25
30
0 2 4 6 8 10 12
Summed concentration (
g m
-3
)
( )Time from start h
H
2
0.1O x
NH
4
+
NO
3
-
Cl
-
( )S VI
( ) =300c h s
Na
+
Sum
med
con
cent
rati
on (g
m-3)
Same as previous slide, but h=300 s
Nonequilibrium Growth of SolidsGas-solid equilibrium reactions (17.129)
NH4NO3(s)NH4(g)+HNO3(g)
Solids can form when (17.131)
NH4Cl(s)NH4(g)+HCl(g) (17.130)
(17.132)
pNH3pHNO3 >KNH4NO3
pNH3pHCl >KNH4Cl
Nonequilibrium Growth of Solids
Gas-solid equilibrium coefficient relation (17.133)
(17.134)
CNH3,s,tCHNO3,s,t =KNH4NO3 R*T( )−2
CNH3,s,tCHCl,s,t =KNH4Cl R*T( )−2
Nonequilibrium Growth of SolidsGrowth equations for gases that form solids (solids formed during
operator-split equilibrium calculation)
dcNO3−,i,t
dt=kHNO3,i,t−h CHNO3,t − ′ S HNO3,i,t−hCHNO3,s,t( )
dcCl−,i,tdt
=kHCl,i,t−h CHCl,t − ′ S HCl,i,t−hCHCl,s,t( )
Simulations of Solid GrowthTime-dependent aerosol concentrations, summed over all sizes, during nonequilibrium growth
plus internal aerosol equilibrium at RH=10 percent when h=5 s.
Fig. 17.8
1
10
0 2 4 6 8 10 12
Summed concentration (
g m
-3
)
( )Time from start h
NH
4
NO
3
( )s
NaNO
3
( )s
(NH
4
)
2
SO
4
( )s
Na
2
SO
4
( )s
NH
4
( )Cl s
( ) =5a h s
( )NaCl s
Sum
med
con
cent
rati
on (g
m-3)
Simulations of Solid Growth
Fig. 17.8
1
10
0 2 4 6 8 10 12
Summed concentration (
g m
-3
)
( )Time from start h
NH
4
NO
3
( )s
NaNO
3
( )s
(NH
4
)
2
SO
4
( )s
Na
2
SO
4
( )s
NH
4
( )Cl s
( ) =300b h s
( )NaCl s
Same as previous slide, but h=300 sS
umm
ed c
once
ntra
tion
(g
m-3)