Chap12FAMII

Presentation Slides for

Chapter 12of

Fundamentals of Atmospheric Modeling 2nd Edition

Mark Z. JacobsonDepartment of Civil & Environmental Engineering

Stanford UniversityStanford, CA [email protected]

March 29, 2005

Initial Value Problems

Vector of concentrations at time t (12.1)

Concentration of species i is known at time t-h=0 (12.2)

ˆ N t = N1,t,N2,t,...,Ni,t,...,NK,t[ ]

Ni,t−h =Ni,0

Values are known at an initial time and desired at a final time

Properties of Good ODE SolversAccuracy

Mass conservation

Positive-definiteness

Speed

Normalized gross error (12.3)

NGE =1

Ntim

1Ks,tj

Ni,tj−Ei,tj

Ei,tji=1

Ks,t j

∑⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ j=1

Ntim

∑ ×100%

Analytical Solution to ODEsNitrogen dioxide photolysis (12.4)

Time rate of change of NO2 concentration (12.5)

Analytical solution (12.6)

Example:[NO2]t-h = 1010 molec. cm-3

J = 0.02 s-1

---> [NO2]t = 1010 e-0.02t

NO2

+ h ν + NO O

d NO2[ ]dt

=−J NO2[ ]

NO2[ ]t = NO2[ ]t−he−Jh

Taylor Series Solution to ODEs

Taylor series expansion for NO2 (12.8)

Explicit Taylor series expansion for one species (12.7)

Ni,t =Ni,t−h+hdNi,t−h

dt+

h2

2

d2Ni,t−h

dt2+

h3

6

d3Ni,t−h

dt3+...

NO[ ]t = NO[ ]t−h +hd NO[ ]t−h

dt+

h2

2

d2 NO[ ]t−hdt2

+h3

6

d3 NO[ ]t−hdt3

+...

Taylor Series Example

Assume

(12.8-12.12)

d NO[ ]dt

=−kb NO[ ] O3[ ]

d2 NO[ ]

dt2=−kb

d NO[ ]dt

O3[ ]−kb NO[ ]d O3[ ]

dt

d3 NO[ ]

dt3=−kb

d2 NO[ ]

dt2O3[ ]−2kb

d NO[ ]dt

d O3[ ]dt

−kb NO[ ]d2 O3[ ]

dt2

NO[ ]t = NO[ ]t−h +hd NO[ ]t−h

dt+

h2

2

d2 NO[ ]t−hdt2

+h3

6

d3 NO[ ]t−hdt3

+...

NO + O3

NO2

+ O2

Taylor Series Solution

(12.13)

DisadvantagesUnstable unless small time step is usedHigh derivatives computationally demanding

d O3[ ]dt

=−d NO2[ ]

dt=

d NO[ ]dt

d2 O3[ ]

dt2=−

d2 NO2[ ]

dt2=

d2 NO[ ]

dt2

d3 O3[ ]

dt3=−

d3 NO2[ ]

dt3=

d3 NO[ ]

dt3

NO + O3

NO2

+ O2

Forward Euler Solution to ODEsExplicit forward Euler solution for an individual species (12.14)

Example set of three equations (12.15-17)

Ni,t =Ni,t−h+hdNi,t−h

dt

NO + O3

NO2

+ O2

NO2

+ O3

NO3

+ O2

NO2

+ h ν + NO O

(k1)

(k2)

(J)

d NO2[ ]t−hdt

=k1 NO[ ]t−h O3[ ]t−h −k2 NO2[ ]t−h O3[ ]t−h −J NO2[ ]t−h

First derivative of NO2 (12.18)

Forward Euler Solution to ODEsFirst derivative of NO2 (12.18)

Forward Euler equation (12.19)

d NO2[ ]t−hdt

=k1 NO[ ]t−h O3[ ]t−h −k2 NO2[ ]t−h O3[ ]t−h −J NO2[ ]t−h

NO2[ ]t = NO2[ ]t−h +h k1 NO[ ]t−h O3[ ]t−h−k2 NO2[ ]t−h O3[ ]t−h−J NO2[ ]t−h( )

Forward Euler Solution to ODEs

Rewrite in terms of production and loss (12.20-22)

AdvantagesMass conserving, easy to calculate derivatives

DisadvantageUnconditionally unstable

NO2[ ]t = NO2[ ]t−h +h Pc,NO2,t−h−Lc,NO2,t−h( )

Pc,NO2,t−h =k1 NO[ ]t−h O3[ ]t−h

Lc,NO2,t−h =k2 NO2[ ]t−h O3[ ]t−h+J NO2[ ]t−h

NO2[ ]t = NO2[ ]t−h +h k1 NO[ ]t−h O3[ ]t−h−k2 NO2[ ]t−h O3[ ]t−h−J NO2[ ]t−h( )

Forward Euler equation (12.19)

Mass Conservation of Forward EulerRequirement for mass conservation of nitrogen (12.28)

Forward Euler concentrations of nitrogen species (12.24,6)

NO2[ ]+ NO[ ]( )t = NO2[ ]+ NO[ ]( )t−h

NO[ ]t = NO[ ]t−h −hk1 NO[ ]t−h O3[ ]t−h

NO2[ ]t = NO2[ ]t−h +hk1 NO[ ]t−h O3[ ]t−h

Mass Conservation of Forward EulerRequirement for mass conservation of oxygen (12.29)

2 O2[ ]+3 O3[ ]+2 NO2[ ]+ NO[ ]( )t= 2 O2[ ]+3O3[ ]+2 NO2[ ]+ NO[ ]( )t−h

Forward Euler concentrations (12.24-7)

NO[ ]t = NO[ ]t−h −hk1 NO[ ]t−h O3[ ]t−h

O3[ ]t = O3[ ]t−h−hk1 NO[ ]t−h O3[ ]t−h


O2[ ]t = O2[ ]t−h+hk1 NO[ ]t−h O3[ ]t−h

Backward Euler Solution to ODEsLinearized backward Euler solution for one species (12.30)

Example set of three equations (12.15-17)

NO + O3

NO2

+ O2

NO2

+ O3

NO3

+ O2

NO2

+ h ν + NO O

(k1)

(k2)

(J)

First derivative of NO2 (12.31)

d NO2[ ]t,t−hdt

=k1 NO[ ]t−h O3[ ]t−h −k2 NO2[ ]t O3[ ]t−h−J NO2[ ]t

Ni,t =Ni,t−h+hdNi,t,t−h

dt

Backward Euler Solution to ODEsFirst derivative of NO2 (12.30)

d NO2[ ]t,t−hdt


NO2[ ]t = NO2[ ]t−h +h k1 NO[ ]t−h O3[ ]t−h−k2 NO2[ ]t O3[ ]t−h −J NO2[ ]t( )

Linearized backward Euler equation (12.32)

Backward Euler Solution to ODEs

Rewrite in terms of production and loss (12.36)

Linearized backward Euler equation (12.32)

NO2[ ]t = NO2[ ]t−h +h k1 NO[ ]t−h O3[ ]t−h−k2 NO2[ ]t O3[ ]t−h −J NO2[ ]t( )

NO2[ ]t = NO2[ ]t−h +h Pc,NO2,t−h−Λc,NO2,t−h NO2[ ]t( )

Production term (12.33)

Loss term (12.34)

Implicit loss coefficient (12.35)

Pc,NO2,t−h =k1 NO[ ]t−h O3[ ]t−h

Lc,NO2,t,t−h =k2 NO2[ ]t O3[ ]t−h +J NO2[ ]t

Λc,NO2,t−h =Lc,NO2,t,t−h

NO2[ ]t=k2 O3[ ]t−h +J


Rearrange (12.37)

AdvantagesAlways gives positive solutions to chemical ODEs

DisadvantagesRequires a small time step for accuracyNot mass conserving

NO2[ ]t 1+hΛc,NO2,t−h( )= NO2[ ]t−h +hPc,NO2,t−h

NO2[ ]t =NO2[ ]t−h +hPc,NO2,t−h

1+hΛc,NO2,t−h

NO2[ ]t = NO2[ ]t−h +h Pc,NO2,t−h−Λc,NO2,t−h NO2[ ]t( )


Example of mass non-conservation of nitrogen (12.39)

NO[ ]t = NO[ ]t−h −hk1 NO[ ]t O3[ ]t−h


Requirement for mass conservation of nitrogen (12.28)

NO2[ ]+ NO[ ]( )t = NO2[ ]+ NO[ ]( )t−h

Simple Exponential Solution to ODEsObtained by integrating a linearized first derivative.

AdvantagesAlways gives positive solutions to chemical ODEsEasy to implement

DisadvantagesRequires a small time step for accuracyNot mass conserving

Simple Exponential Solution to ODEsLinearized first derivative from backward Euler example (12.31)

Rewrite in terms of production and loss (12.43)

Integrate (12.44)

d NO2[ ]t,t−hdt


d NO2[ ]tdt

=Pc,NO2,t−h−Λc,NO2,t−h NO2[ ]t

NO2[ ]t = NO2[ ]t−he−hΛc,NO2,t−h +Pc,NO2,t−h

Λc,NO2,t−h1−e−hΛc,NO2,t−h( )

Simple Exponential Solution to ODEsGeneric solution (12.45)

When the implicit loss coefficient is zero (no loss term),

When the production term is zero,

When the implicit loss coefficient is large (short-lived species),

Ni,t =Ni,t−he−hΛc,i ,t−h +Pc,i,t−hΛc,i,t−h

1−e−hΛc,i ,t−h( )

Ni,t =Ni,t−h+hPc,i,t−h

Ni,t =Ni,t−he−hΛc,i ,t−h

Ni,t =Pc,i,t−hΛc,i,t−h

(steady state solution)

Quasi-Steady State ApproximationQSSA (12.46)

Use forward Euler for long-lived speciesUse simple exponential for medium-lived speciesUse steady-state solution for short-lived species

Ni,t =

Ni,t−h+h Pc,i,t−h −Lc,i,t−h( ) hΛc,i,t−h <0.01

Ni,t−he−hΛc,i ,t−h +Pc,i,t−hΛc,i,t−h

1−e−hΛc,i ,t−h( ) 0.01≤hΛc,i,t−h ≤10

Pc,i,t−h Λc,i,t−h hΛc,i,t−h >10

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

Multistep Implicit-Explicit (MIE) Solution

Premises:The forward Euler and linearized backward Euler solutions converge to each other upon iteration.

Since the converged backward Euler is always positive, so is the converged forward Euler.

Since the converged forward Euler always conserves mass, so does the converged backward Euler

--> The MIE solution, which requires iteration, conserves mass and is positive definite.

Multistep Implicit-Explicit Solution

1) Set initial estimates to initial concentrations (12.49)

2) Set initial maximum estimates to initial values (12.50)Maximum estimates are used to bound the backward Euler during the initial iteration steps.

3) Estimate reaction rates from backward-Euler values (12.51)These equation apply for any iteration number m

Ni,B,1 =Ni,t−h

Ni,MAX,1=Ni,t−h

Rc,n,B,m =knNi,B,mNj,B,m

Rc,n,B,m =J nNi,B,m

Multistep Implicit-Explicit Solution4) Sum production, loss, implicit loss terms (12.53)

(12.55)

(12.57)

Pc,i,B,m= Rc,nP (l,i),B,ml =1

Nprod,i

∑

Lc,i,B,m= Rc,nL (l,i),B,ml =1

Nloss,i

∑

Λc,i,B,m=

Lc,i,B,mNi,B,m

Multistep Implicit-Explicit SolutionExample reactions (12.47)

(12.48)

Estimate reaction rates. (12.52)

NO + O3

NO2

+ O2

O3

+ h ν O2

+ O λ > 310 nm

Rc,1,B,m=k1 NO[ ]B,m O3[ ]B,m

Rc,2,B,m =J O3[ ]B,m

k1

J

Multistep Implicit-Explicit SolutionSummed production terms (12.54)

Summed loss terms (12.56)

Pc,O,B,m=J O3[ ]B,m

Pc,NO2,B,m=k1 NO[ ]B,m O3[ ]B,m

Lc,NO,B,m=k1 NO[ ]B,mO3[ ]B,m

Lc,O3,B,m=k1 NO[ ]B,m O3[ ]B,m+J O3[ ]B,m


Summed implicit loss coefficients (12.58)

Λc,NO,B,m=k1 O3[ ]B,m

Λc,O3,B,m=k1 NO[ ]B,m+J

Multistep Implicit-Explicit Solution5) Estimate concentrations with backward Euler (12.59)

6) Estimate concentrations with forward Euler (12.60)

Ni,B,m+1=

Ni,t−h+hPc,i,B,m1+hΛc,i,B,m

Ni,F,m+1=Ni,t−h+h Pc,i,B,m−Lc,i,B,m( )

Multistep Implicit-Explicit Solution7) Check convergence (12.61)

ˆ N F,m+1≥0→ nP =nP +1

Ni,F,m+1<0→ nP =0

⎧ ⎨ ⎪

⎩ ⎪

Method 1: Each iteration, check whether all concentrations from forward Euler ≥ zero. If so, update a counter, nP by 1. If one concentration < 0, reset nP to 0.

When nP=NP (all concentrations from forward Euler exceed 0 for NP iterations in a row), convergence has occurred

NP = 5 for large sets of equations = 30 for small sets of equations

Multistep Implicit-Explicit Solution7) Check convergence (12.62)

ˆ N F,m+1≥0 for specieswith hΛc,i,B,m<LT → nP =nP +1

Ni,F,m+1<0 and hΛc,i,B,m<LT → nP =0

⎧

⎨ ⎪

⎩ ⎪

Method 2: Each iteration, check whether all concentrations of medium- and long-lived species (those for which hLc,I,B,m<LT) from forward Euler ≥ zero. If so, update the counter nP by 1. If one such concentration < 0, reset nP to 0.

When nP=NP, convergence has occurred

LT = 102-106


8) Upon convergence, set final concentrations (12.63)

9a) Recalculate maximum for next iteration (12.64)

9b) Bound backward Euler for next iteration (12.65)Note that this equation uses the MAX concentration from the previous iteration, not from the value just calculated.

Ni,t =Ni,B,m+1 backwardEuler( ) hΛc,i,B,m≥LT

Ni,F,m+1 forwardEuler( ) hΛc,i,B,m<LT

⎧ ⎨ ⎪

⎩ ⎪

Ni,MAX,m+1=maxNi,B,m+1, Ni,t−h( )

Ni,B,m+1=min Ni,B,m+1,Ni,MAX,m{ }


Converged backward Euler equation (12.66)

Demonstration that iterated forward Euler solutions converge to iterated backward Euler solutions and to

positive numbers.

Converged forward Euler equation (12.64)

Thus, convergence of forward to backward Euler occurs when

where

Ni,B,m+1 =Ni,t−h +h Pc,i,B,m−Λc,i,B,mNi,B,m+1( )

Ni,F,m+1=Ni,t−h+h Pc,i,B,m−Λc,i,B,mNi,B,m( )

Λc,i,B,mNi,B,m=Lc,i,B,m

Ni,B,m+1=Ni,B,m

which occurs upon iteration of backward Euler (12.68)

Convergence of Forward Euler (FE) to Backward Euler (BE) Upon Iteration

Fig. 12.1

h = 10 seconds.

-1x10

15

-5x10

14

1x10

7

5x10

14

1x10

15

1000

10

5

10

7

10

9

10

11

10

13

10

15

1 10 100

NO

2

and O

3

(molecules cm

-3

)

O (molecules cm

-3

)

Iteration number

O (FE)

O (BE)

NO

2

(FE)

NO

2

(BE)

O

3

(FE)

O

3

(BE)

NO

2 and

O3 (

mol

ec. c

m-3)

O (m

olec. cm-3)

Comparison of MIE With Exact Solution

Fig. 12.2

MIE solution (circles); exact solution (lines); h = 10 seconds.

1x10

0

5x10

11

1x10

12

1.5x10

12

2x10

12

2.5x10

12

3x10

12

0 4 8 12 16 20 24

Concentration (molecules cm

-3

)

Hours from start of simulation

HNO

3

O

3

Con

cent

rati

on (

mol

ec. c

m-3)

Effects of Time Step on MIE Solution

Fig. 12.3

Time step sizes of 2, 10, 100, 500, and 1000 seconds are compared. Results in all cases are shown after 10,000 s.

2 10

4

4 10

4

6 10

4

8 10

4

1 10

5

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

1 10 100 1000

No. of iterations required

Average error per species (%)

Time step (s)

No.

of

iter

atio

ns r

equi

red A

verage error per species

Gear’s MethodDiscretize ODE time derivative over many past steps (12.69)

Rearrange for an individual species (12.70)

Set up predictor equation (12.71)

and for a set of species (12.70)

dNi,tdt

≈Ni,t −αs,1Ni,t−h −αs,2Ni,t−2h...−αs,sNi,t−sh

hβs=

Ni,t − αs, jNi,t−jhj=1

s

∑

hβs

0 =−ˆ N t + αs, jˆ N t−jh

j =1

s

∑ +hβsdˆ N tdt

Ptˆ N t,m+1− ˆ N t,m( ) =−ˆ N t,m+ αs, j

j =1

s

∑ ˆ N t−jh +hβsd ˆ N t,m

dt

0 =−Ni,t + αs, j Ni,t−jhj =1

s

∑ +hβsdNi,tdt

Gear’s MethodSimplify to (12.72)

where

Predictor matrix (12.77)

(12.73)

PtΔˆ N t,m= ˆ B t,m

Δ ˆ N t,m= ˆ N t,m+1− ˆ N t,m

ˆ B t,m=−ˆ N t,m+ αs, jj =1

s

∑ ˆ N t−jh +hβsd ˆ N t,m

dt

Pt ≈I −hβsJ t

Jacobian matrix of partial derivatives (12.78)

J t =∂2 ˆ N i,t,m∂Nk,t,m∂t

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ i,k=1

K,K

Gear’s Method

Iterate until

Solve

Δ ˆ N t,m

ˆ N t,m+1 = ˆ N t,m+Δ ˆ N t,m


then update concentration with (12.74)

for

Δ ˆ N t,m≈0

Error TestsLocal error test (12.75)

Global error test (12.76)

NRMSt,m=1K

ΔNi,t,mRtolNi,t,1+Atol,t

⎛

⎝ ⎜

⎞

⎠ ⎟

2

i=1

K

∑

NRMSt =1K

ΔNi,t,mm∑RtolNi,t,1+Atol,t

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

i=1

K

∑

Example Reactions and 1st Derivatives

Example reactions (12.79)

(12.80)

(12.81)

NO + O3

NO2

+ O2

O + O2

+ M O3

+ M

NO2

+ h ν + NO O

Example Reactions and 1st Derivatives

First derivatives of NO, NO2, O, and O3 (12.82)

d NO[ ]dt

=J NO2[ ]−k1 NO[ ] O3[ ]

d NO2[ ]dt

=k1 NO[ ] O3[ ]−J NO2[ ]

d O[ ]dt

=J NO2[ ]−k2 O[ ] O2[ ] M[ ]

d O3[ ]dt

=k2 O[ ] O2[ ] M[ ]−k1 NO[ ]O3[ ]

Partial DerivativesPartial derivatives of NO (12.83)

Partial derivatives of NO2 (12.84)

∂2 NO[ ]∂ NO[ ]∂t

=−k1 O3[ ]∂2 NO[ ]

∂ NO2[ ]∂t=J

∂2 NO[ ]∂ O3[ ]∂t

=−k1 NO[ ]

∂2 NO2[ ]∂ NO[ ]∂t

=k1 O3[ ]∂2 NO2[ ]∂ NO2[ ]∂t

=−J

∂2 NO2[ ]∂ O3[ ]∂t

=k1 NO[ ]

Partial DerivativesPartial derivatives of O (12.85)

Partial derivatives of O3 (12.86)

∂2 O[ ]∂ NO2[ ]∂t

=J∂2 O[ ]∂ O[ ]∂t

=−k2 O2[ ] M[ ]

∂2 O3[ ]∂ NO[ ]∂t

=−k1 O3[ ]∂2 O3[ ]∂ O[ ]∂t

=k2 O2[ ] M[ ]

∂2 O3[ ]∂ O3[ ]∂t

=−k1 NO[ ]

Predictor Matrix

(12.87)

1−hβs∂2 NO[ ]∂ NO[ ]∂t

−hβs∂2 NO[ ]

∂ NO2[ ]∂t−hβs

∂2 NO[ ]∂ O[ ]∂t

−hβs∂2 NO[ ]∂ O3[ ]∂t

−hβs∂2 NO2[ ]∂ NO[ ]∂t

1−hβs∂2 NO2[ ]∂ NO2[ ]∂t

−hβs∂2 NO2[ ]∂ O[ ]∂t

−hβs∂2 NO2[ ]∂ O3[ ]∂t

−hβs∂2 O[ ]

∂ NO[ ]∂t−hβs

∂2 O[ ]∂ NO2[ ]∂t

1−hβs∂2 O[ ]∂ O[ ]∂t

−hβs∂2 O[ ]

∂ O3[ ]∂t

−hβs∂2 O3[ ]∂ NO[ ]∂t

−hβs∂2 O3[ ]

∂ NO2[ ]∂t−hβs

∂2 O3[ ]∂ O[ ]∂t

1−hβs∂2 O3[ ]∂ O3[ ]∂t

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

NO NO2 O O3

NO

NO2

O

O3

Substitute Partial Derivatives Into Predictor Matrix

(12.88)

NO NO2 O O3

NO

NO2

O

O3

1−hβs −k1O3[ ]( ) −hβs J( ) 0 −hβs −k1 NO[ ]( )

−hβs k1 O3[ ]( ) 1−hβs −J( ) 0 −hβs k1 NO[ ]( )0 −hβs J( ) 1−hβs −k2 O2[ ] M[ ]( ) 0

−hβs −k1 O3[ ]( ) 0 −hβs k2 O2[ ] M[ ]( ) 1−hβs −k1 NO[ ]( )

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

Matrix Equation

(12.89)


1−hβs −k1O3[ ]( ) −hβs J( ) 0 −hβs −k1 NO[ ]( )

−hβs k1 O3[ ]( ) 1−hβs −J( ) 0 −hβs k1 NO[ ]( )0 −hβs J( ) 1−hβs −k2 O2[ ] M[ ]( ) 0

−hβs −k1 O3[ ]( ) 0 −hβs k2 O2[ ] M[ ]( ) 1−hβs −k1 NO[ ]( )

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

Δ NO[ ]t,mΔ NO2[ ]t,m

Δ O[ ]t,mΔ O3[ ]t,m

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

=

BNO,t,mBNO2,t,mBO,t,mBO3,t,m

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

Reordered Matrix

(12.90)

1−hβs −k2 O2[ ] M[ ]( ) −hβs J( ) 0 0

0 1−hβs −J( ) −hβs k1 O3[ ]( ) −hβs k1 NO[ ]( )

0 −hβs J( ) 1−hβs −k1 O3[ ]( ) −hβs −k1 NO[ ]( )

−hβs k2 O2[ ] M[ ]( ) 0 −hβs −k1 O3[ ]( ) 1−hβs −k1 NO[ ]( )

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

Δ O[ ]t,mΔ NO2[ ]t,mΔ NO[ ]t,mΔ O3[ ]t,m

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

=

BO,t,mBNO2,t,mBNO,t,mBO3,t,m

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

Place species with most partial derivatives at the bottom and fewest partial derivatives at the top of the matrix

Effect of Sparse-Matrix Reductions on Matrix Fill in and Number of Multiplies

1427 Gases, 3911 Reactions

Table 12.2

Without With PercentQuantity Reductions Reductions ReductionOrder of matrix 1427 1427 0Initial fill-in 2,036,329 14,276 99.30Final fill-in 2,036,329 17,130 99.16Decomp. 1 967,595,901 47,596 99.995Decomp. 2 1,017,451 9,294 99.09Backsub. 1 1,017,451 9,294 99.09Backsub. 2 1,017,451 6,409 99.37

Vectorization of Inner Nested LoopVectorized inner loop

Nested Loop ADO 105 NK = 1, NBIMOLEC

JSP1 = JPROD1(NK)

JSP2 = JPROD2(NK)

DO 100 K = 1, KTLOOP

TRATE(K,NK) = RRATE(K,NK) * CONC(K,JSP1)

* CONC(K,JSP2)

100 CONTINUE

105 CONTINUE

Vectorization of Inner Nested LoopNon-vectorized inner loop

Nested Loop BDO 105 K = 1, KTLOOP

DO 100 NK = 1, NBIMOLECJSP1 = JPROD1(NK)JSP2 =JPROD2(NK) TRATE(K,NK) = RRATE(K,NK) * CONC(K,JSP1) * CONC(K,JSP2)

100 CONTINUE

105 CONTINUE

Stiffness Predictor(12.91)Sp =

1K

dNi,t dt

Ni,t+Atol,t

⎛

⎝ ⎜

⎞

⎠ ⎟

i=1

K

∑2

Number of backsubstitution calls as a function of the stiffness predictor. The more calls, the greater the stiffness of equations.

Fig. 12.40

50

100

150

200

10

-14

10

-12

10

-10

10

-8

10

-6

10

-4

0.01 1

No. of backsubstitution calls

Predictor value (s

-2

)

No.

of

back

subs

titu

tion

cal

ls

Family MethodOrganize short- and long-lived species into families, which are long-lived, solve for family concentration, then partition family concentrations into individual species concentrations

Odd chlorine family (12.95)

Odd hydrogen family (12.93)

Odd nitrogen family (12.94)

Odd oxygen family (12.92)

[OT] =[O]+[O(1D)]+[O3]+[NO2]

[HOT ]=[OH]+[HO2]+[H2O2]

[NOT ]=[NO]+[NO2]+[NO3]

[ClT ]=[Cl]+[ClO]+[ClO2]

Family Method

Example of species cycling within odd oxygen family (12.96)

O3

+ h νO

2 + (O

1

D )

O2

+ h ν O + O

O(

1

D ) O

M

O + O2

+ M O3

+ M

Solution with Family Method(1) Calculate reaction rates and first derivatives of all species

(2) Sum production and loss terms of each family

(3) Apply forward Euler to family concentration

(4) Repartition species concentrations in the family

Example with 4 total species and 2 families (12.97)

[Fam1]=[A]+[B]+[C]

[Fam2]=[D]

A → B

B → C

C → D

Reactions in the system (12.98)

ka

kb

kc

Solution with Family MethodFirst derivatives of A, B, C, and D (12.99)

d A[ ]dt

=−ka A[ ]

d B[ ]dt

=ka A[ ]−kb B[ ]

d C[ ]dt

=kb B[ ]−kc C[ ]

d D[ ]dt

=kc C[ ]

Example Family SolutionFirst derivatives of each family as a whole (12.100)

Apply forward Euler to family concentration with (12.101)

d Fam1[ ]dt

=d A[ ]dt

+d B[ ]dt

+d C[ ]dt

=−kc C[ ]

d Fam2[ ]dt

=d D[ ]dt

=kc C[ ]

Fam[ ]t = Fam[ ]t−h +hd Fam[ ]t−h

dtFor example, (12.102)

Fam1[ ]t = Fam1[ ]t−h−hkc C[ ]t−h

Fam2[ ]t = Fam2[ ]t−h +hkc C[ ]t−h

Partition Species in Family

Relate species to family concentration (12.103)

Concentration of D equals second family concentration

D[ ]t = Fam2[ ]t =hkc C[ ]t−h

Fam1[ ]t = A[ ]t + B[ ]t + C[ ]t = A[ ]t 1+B[ ]tA[ ]t

+C[ ]tA[ ]t

⎛

⎝ ⎜

⎞

⎠ ⎟

Repartitioning Species

Solve for A, B, and C, respectively (12.104)

Many ways to obtain partitioning ratios, and .

A[ ]t =Fam1[ ]t

1+B[ ]tA[ ]t

+C[ ]tA[ ]t

B[ ]t = A[ ]tB[ ]tA[ ]t

C[ ]t = A[ ]tC[ ]tA[ ]t

B[ ]tA[ ]t

C[ ]tA[ ]t

Estimating Partitioning Ratios1) Assume each species in steady state (12.105)

---> Partitioning ratios of B and C are (12.106)

d B[ ]dt

=ka A[ ]−kb B[ ]=0

d C[ ]dt

=kb B[ ]−kc C[ ] =0

B[ ]tA[ ]t

≈B[ ]A[ ]

=kakb

C[ ]tA[ ]t

≈C[ ]A[ ]

=C[ ]B[ ]

B[ ]A[ ]

=kbkc

kakb

=kakc

Estimating Partitioning Ratios---> Final species concentrations (12.107)

A[ ]t =Fam1[ ]t

1+kakb

+kakc

B[ ]t =

Fam1[ ]tkakb

1+kakb

+kakc

C[ ]t =

Fam1[ ]tkakc

1+kakb

+kakc

Estimating Partitioning Ratios2) First linearize the ODE of each species -- (12.108)

-- then integrate with simple exponential method -- (12.109)

d A[ ]tdt

=−ka A[ ]td B[ ]t

dt=ka A[ ]t−h −kb B[ ]t

d C[ ]tdt

=kb B[ ]t−h −kc C[ ]t

B[ ]t ≈ B[ ]t−he−kbh+ka A[ ]t−h

kb1−e−kbh( )

C[ ]t ≈ C[ ]t−he−kch+kb B[ ]t−h

kc1−e−kch( )

A[ ]t ≈ A[ ]t−he−kah

---> plug results into partitioning ratios

Estimating Partitioning Ratios3) First write each species ODE in finite difference form -- (12.110)

A[ ]t − A[ ]t−hh

=−ka A[ ]t

B[ ]t − B[ ]t−hh

=ka A[ ]t −kb B[ ]t

C[ ]t − C[ ]t−hh

=kb B[ ]t −kc C[ ]t

Estimating Partitioning Ratios-- then rearrange finite difference equations into a matrix -- (12.111)

1+hka 0 0

hka 1+hkb 0

0 hkb 1+hkc

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

A[ ]tB[ ]tC[ ]t

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

=

A[ ]t−hB[ ]t−hC[ ]t−h

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

---> then solve the matrix for [A]t, [B]t, and [C]t and plug results into partitioning ratios

ExampleAdd bimolecular reaction to the system

Linearized, finite difference forms of A, B, and C (12.112)

(12.113)

(12.114)

A +C → B+D

A[ ]t − A[ ]t−hh

=−ka A[ ]t −kac A[ ]t C[ ]t−h

B[ ]t − B[ ]t−hh

=ka A[ ]t −kb B[ ]t +0.5kac A[ ]t C[ ]t−h + A[ ]t−h C[ ]t( )

C[ ]t − C[ ]t−hh

=kb B[ ]t −kc C[ ]t −kac A[ ]t−h C[ ]t

Example With Bimolecular Reaction

Resulting matrix (12.115)

---> then solve the matrix for [A]t, [B]t, and [C]t and plug results into partitioning ratios

1+hka +hkac C[ ]t−h 0 0

−hka −0.5 hkac C[ ]t−h( ) 1+hkb −0.5 hkac A[ ]t−h( )0 −hkb 1+hkc +hkac A[ ]t−h

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

A[ ]tB[ ]tC[ ]t

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ =

A[ ]t−hB[ ]t−hC[ ]t−h

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

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Chap12FAMII