modelling steam methane reforming in a fixed bed reactor ...folk.ntnu.no/andersty/5....

Modelling steam methane reforming in a fixed bedreactor

using orthogonal collocation

Kasper Linnestad

Department of Chemical EngineeringNorwegian University of Science and Technology

December 3, 2014

Kasper Linnestad Reactor modelling December 3, 2014 1 / 36

Outline

1 Theory

2 Governing equations

3 Implementation

4 Results

5 Conclusion

Outline

1 TheoryWeighted residuals methodCollocation pointsOrthogonal collocation method

3 Implementation

4 Results

5 Conclusion

TheoryWeighted residuals method

General problem

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(z, r)

Approximation

f(z, r) ≈Pz∑

ajz ,jr ljz(z)ljr(r)

Residual

R(z, r) = L{f(z, r)

}− g(z, r)

Minimize residuals by∫R(z, r)wi(z, r) dz dr = 0, ∀i

Orthogonal collocation uses

ajz ,jr = f(zjz , rjr)

ln(x) =

i=0i6=n

x− xixn − xi

wi(z, r) = δ(z − ziz)δ(r − rir)

General problem

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(z, r)

Approximation

f(z, r) ≈Pz∑

Residual

R(z, r) = L{f(z, r)

}− g(z, r)

ln(x) =

i=0i6=n

x− xixn − xi

General problem

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(z, r)

Approximation

f(z, r) ≈Pz∑

Residual

R(z, r) = L{f(z, r)

}− g(z, r)

ln(x) =

i=0i6=n

x− xixn − xi

General problem

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(z, r)

Approximation

f(z, r) ≈Pz∑

Residual

R(z, r) = L{f(z, r)

}− g(z, r)

ln(x) =

i=0i6=n

x− xixn − xi

General problem

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(z, r)

Approximation

f(z, r) ≈Pz∑

Residual

R(z, r) = L{f(z, r)

}− g(z, r)

ln(x) =

i=0i 6=n

x− xixn − xi

General problem

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(z, r)

Approximation

f(z, r) ≈Pz∑

Residual

R(z, r) = L{f(z, r)

}− g(z, r)

ln(x) =

i=0i 6=n

x− xixn − xi

General problem

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(z, r)

Approximation

f(z, r) ≈Pz∑

Residual

R(z, r) = L{f(z, r)

}− g(z, r)

ln(x) =

i=0i 6=n

x− xixn − xi

TheoryCollocation points

Legendre polynomials∫ 1

−1Ln(x)Lm(x) dx = 0, m 6= n

Collocation points

Lm(xi) = 0, ∀i ∈ {0, . . . ,m}

−1 −0.5 0 0.5 1−2

xLm(x)

TheoryCollocation points

Legendre polynomials∫ 1

−1Ln(x)Lm(x) dx = 0, m 6= n

Collocation points

Lm(xi) = 0, ∀i ∈ {0, . . . ,m} −1 −0.5 0 0.5 1−2

xLm(x)

TheoryOrthogonal collocation method

ln(x) =

i=0i 6=n

x− xixn − xi

Inserted

R(ziz , rir) = 0, ∀ iz ∈ {0, . . . ,Pz}ir ∈ {0, . . . ,Pr}

System of equations

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(zb, rb)

Linearisation

Af = b

ln(x) =

i=0i 6=n

x− xixn − xi

Inserted

System of equations

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(zb, rb)

Linearisation

Af = b

ln(x) =

i=0i 6=n

x− xixn − xi

Inserted

System of equations

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(zb, rb)

Linearisation

Af = b

ln(x) =

i=0i 6=n

x− xixn − xi

Inserted

System of equations

L{f(z, r)

}= g(z, r)

B{fb(zb, rb)

}= gb(zb, rb)

Linearisation

Af = b

TheoryLinearisation

Example function

f(ψ) = ψ2 + ψ exp (−ψ)

Fixed-point (Picard-iteration)

f(ψ) ≈(ψ? + exp

(−ψ?

solves f(ψ) = 3 in 114 iterations

Taylor (Newton-Raphson-iteration)

f(ψ) ≈ f(ψ?) +∂f

∣∣∣ψ?

(ψ − ψ?

TheoryLinearisation

Example function

f(ψ) = ψ2 + ψ exp (−ψ)

f(ψ) ≈(ψ? + exp

(−ψ?

f(ψ) ≈ f(ψ?) +∂f

∣∣∣ψ?

(ψ − ψ?

TheoryLinearisation

Example function

f(ψ) = ψ2 + ψ exp (−ψ)

f(ψ) ≈(ψ? + exp

(−ψ?

f(ψ) ≈ f(ψ?) +∂f

∣∣∣ψ?

(ψ − ψ?

Outline

1 Theory

2 Governing equationsContinuity equationEnergy equationSpecies mass balanceErgun’s equationInitial and boundary conditions

3 Implementation

4 Results

5 Conclusion

Governing equationsFixed bed reactor

Governing equationsSteam methane reforming

Assumptions

Pseudo-homogeneousEfficiency factor = 10−3

Reactions

CH4 + H2O = CO + 3 H2

CO + H2O = CO2 + H2

CH4 + 2 H2O = CO2 + 4 H2

Governing equationsContinuity equation

General form

∂t+∇ · (ρu) = 0

Simplified

∂ρuz∂z

Linearised

ρ?∂uz∂z

+ uz∂ρ?

∂z︸︷︷︸⇒Auz

= 0︸︷︷︸⇒buz

General form

∂t+∇ · (ρu) = 0

Simplified

∂ρuz∂z

Linearised

ρ?∂uz∂z

+ uz∂ρ?

= 0︸︷︷︸⇒buz

General form

∂t+∇ · (ρu) = 0

Simplified

∂ρuz∂z

Linearised

ρ?∂uz∂z

+ uz∂ρ?

= 0︸︷︷︸⇒buz

Governing equationsEnergy equation

General form

∂t+ u · ∇T

)= −∇ · q−∆rxH

Simplified

ρcpuz∂T

∂z=λeff

(r∂T

)−∆rxH

Linearised

ρ?c?pu?z

∂z︸︷︷︸⇒AT

= λeff,?

∂r+∂2T

︸︷︷︸⇒AT

−∆rxH?

︸︷︷︸⇒bT

General form

∂t+ u · ∇T

)= −∇ · q−∆rxH

Simplified

ρcpuz∂T

∂z=λeff

(r∂T

)−∆rxH

Linearised

ρ?c?pu?z

∂z︸︷︷︸⇒AT

= λeff,?

∂r+∂2T

︸︷︷︸⇒AT

−∆rxH?

︸︷︷︸⇒bT

General form

∂t+ u · ∇T

)= −∇ · q−∆rxH

Simplified

ρcpuz∂T

∂z=λeff

(r∂T

)−∆rxH

Linearised

ρ?c?pu?z

∂z︸︷︷︸⇒AT

= λeff,?

∂r+∂2T

︸︷︷︸⇒AT

−∆rxH?

︸︷︷︸⇒bT

Governing equationsSpecies mass balance

General form

∂ρωi∂t

+∇ · (ρuωi) = −∇ · ji +Ri

Simplified

∂ρuzωi∂z

(rρ∂ωi∂r

)−Ri

Linearised

ρ?u?z∂ωi∂z

+ ωiu?z

∂ρ?

∂z+ ρ?ωi

∂u?z∂z︸︷︷︸

⇒Aωi

= R?i︸︷︷︸⇒bωi

Deff,?

∂ωi∂r

+∂ρ?

∂ωi∂r

+ ρ?∂2ωi∂r2

︸︷︷︸⇒Aωi

General form

∂ρωi∂t

+∇ · (ρuωi) = −∇ · ji +Ri

Simplified

∂ρuzωi∂z

(rρ∂ωi∂r

)−Ri

Linearised

ρ?u?z∂ωi∂z

+ ωiu?z

∂ρ?

∂z+ ρ?ωi

∂u?z∂z︸︷︷︸

⇒Aωi

= R?i︸︷︷︸⇒bωi

Deff,?

∂ωi∂r

+∂ρ?

∂ωi∂r

+ ρ?∂2ωi∂r2

︸︷︷︸⇒Aωi

General form

∂ρωi∂t

+∇ · (ρuωi) = −∇ · ji +Ri

Simplified

∂ρuzωi∂z

(rρ∂ωi∂r

)−Ri

Linearised

ρ?u?z∂ωi∂z

+ ωiu?z

∂ρ?

∂z+ ρ?ωi

∂u?z∂z︸︷︷︸

⇒Aωi

= R?i︸︷︷︸⇒bωi

Deff,?

∂ωi∂r

+∂ρ?

∂ωi∂r

+ ρ?∂2ωi∂r2

︸︷︷︸⇒Aωi

Governing equationsErgun’s equation

General form

dz= −f ρu

Linearised

dz︸︷︷︸⇒Ap

= −f? ρ?(u?z)

dp︸︷︷︸⇒bp

Governing equationsErgun’s equation

General form

dz= −f ρu

Linearised

dz︸︷︷︸⇒Ap

= −f? ρ?(u?z)

dp︸︷︷︸⇒bp

Governing equationsInitial and boundary conditions

Velocity

uz(z = 0, r)︸︷︷︸⇒Auz

= uz,inlet︸︷︷︸⇒buz

Temperature

T (z = 0, r)︸︷︷︸⇒AT

= Tinlet︸︷︷︸⇒bT

∣∣∣∣∣r=0︸︷︷︸

= 0︸︷︷︸⇒bT

λeff,? ∂T

∣∣∣∣∣r=R︸︷︷︸

= −U?(T ?(r = R)− Ta)︸︷︷︸⇒bT

Pressure

p(z = 0, r)︸︷︷︸⇒Ap

= pinlet︸︷︷︸⇒bp

Mass fractions

ωi(z = 0, r)︸︷︷︸⇒Aωi

= ωi,inlet︸︷︷︸⇒bωi

∂ωi∂r

∣∣∣∣∣r=0︸︷︷︸

⇒Aωi

= 0︸︷︷︸⇒bωi

∂ωi∂r

∣∣∣∣∣r=R︸︷︷︸

⇒Aωi

= 0︸︷︷︸⇒bωi

Velocity

uz(z = 0, r)︸︷︷︸⇒Auz

Temperature

T (z = 0, r)︸︷︷︸⇒AT

∣∣∣∣∣r=0︸︷︷︸

= 0︸︷︷︸⇒bT

λeff,? ∂T

∣∣∣∣∣r=R︸︷︷︸

= −U?(T ?(r = R)− Ta)︸︷︷︸⇒bT

Pressure

p(z = 0, r)︸︷︷︸⇒Ap

Mass fractions

ωi(z = 0, r)︸︷︷︸⇒Aωi

∂ωi∂r

∣∣∣∣∣r=0︸︷︷︸

⇒Aωi

= 0︸︷︷︸⇒bωi

∂ωi∂r

∣∣∣∣∣r=R︸︷︷︸

⇒Aωi

= 0︸︷︷︸⇒bωi

Velocity

uz(z = 0, r)︸︷︷︸⇒Auz

Temperature

T (z = 0, r)︸︷︷︸⇒AT

∣∣∣∣∣r=0︸︷︷︸

= 0︸︷︷︸⇒bT

λeff,? ∂T

∣∣∣∣∣r=R︸︷︷︸

= −U?(T ?(r = R)− Ta)︸︷︷︸⇒bT

Pressure

p(z = 0, r)︸︷︷︸⇒Ap

Mass fractions

ωi(z = 0, r)︸︷︷︸⇒Aωi

∂ωi∂r

∣∣∣∣∣r=0︸︷︷︸

⇒Aωi

= 0︸︷︷︸⇒bωi

∂ωi∂r

∣∣∣∣∣r=R︸︷︷︸

⇒Aωi

= 0︸︷︷︸⇒bωi

Velocity

uz(z = 0, r)︸︷︷︸⇒Auz

Temperature

T (z = 0, r)︸︷︷︸⇒AT

∣∣∣∣∣r=0︸︷︷︸

= 0︸︷︷︸⇒bT

λeff,? ∂T

∣∣∣∣∣r=R︸︷︷︸

= −U?(T ?(r = R)− Ta)︸︷︷︸⇒bT

Pressure

p(z = 0, r)︸︷︷︸⇒Ap

Mass fractions

ωi(z = 0, r)︸︷︷︸⇒Aωi

∂ωi∂r

∣∣∣∣∣r=0︸︷︷︸

⇒Aωi

= 0︸︷︷︸⇒bωi

∂ωi∂r

∣∣∣∣∣r=R︸︷︷︸

⇒Aωi

= 0︸︷︷︸⇒bωi

Outline

1 Theory

3 ImplementationJuliaUnder-relaxationSegregated approachCombined approachCoupled approach

4 Results

5 Conclusion

ImplementationJulia

Open-source

Just-in-time compilation

Constant global variables

Pass by reference (!)

Easy to call programs written in other languages (Fortran, C, Python)

ImplementationUnder-relaxation

Trial-and-error

ωi and T

T = γTT + (1− γT )T ?

γT = γω = 5 · 10−2

ImplementationSegregated approach

||R||2 < tol

Update

Initial guess

||R||2 < tol

ImplementationCombined approach

Coupled mass fractions and temperature

Aω1 0 · · ·0 Aω2 0 · · ·0 · · · . . .

I I · · · I 00 · · · AT

ω2...

...1bT

Coupled velocity, density and pressureAuz 0 0

0 I −diag(

0 0 Ap

buz0bp

ImplementationCombined approach

Coupled mass fractions and temperature

Aω1 0 · · ·0 Aω2 0 · · ·0 · · · . . .

I I · · · I 00 · · · AT

ω2...

...1bT

Coupled velocity, density and pressureAuz 0 0

0 I −diag(

0 0 Ap

buz0bp

ImplementationCoupled approach

Fully coupled system of equations

Aω1 0 · · ·0 Aω2 0 · · ·0 · · · . . .

I I · · · I 0 · · ·0 · · · AT 0 · · ·0 · · · Auz 0 0

0 · · · I −diag(

0 · · · Ap

ω2...

Tuzρp

...1bTbuz0bp

Outline

1 Theory

3 Implementation

4 ResultsProfilesRun-timeHindsights

5 Conclusion

ResultsVelocity

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.01.8

[ ms−

Axial velocity in the reactor

r = 0,00m

r = 1,01 · 10−3m

r = 5,19 · 10−3m

r = 1,21 · 10−2m

r = 2,08 · 10−2m

r = 3,02 · 10−2m

r = 3,89 · 10−2m

r = 4,58 · 10−2m

r = 5,00 · 10−2m

r = 5,10 · 10−2m

ResultsVelocity

0.05 01

r [m] z [m]

[ ms−

Velocity in the reactor

ResultsPressure

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.02.76 · 106

2.78 · 106

2.80 · 106

2.82 · 106

2.84 · 106

2.86 · 106

2.88 · 106

2.90 · 106

Pressure in the reactor

ResultsTemperature

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0750

Temperature in the reactor

r = 0,00m

r = 1,01 · 10−3m

r = 5,19 · 10−3m

r = 1,21 · 10−2m

r = 2,08 · 10−2m

r = 3,02 · 10−2m

r = 3,89 · 10−2m

r = 4,58 · 10−2m

r = 5,00 · 10−2m

r = 5,10 · 10−2m

ResultsTemperature

0.05 01

r [m] z [m]

Temperature in the reactor

ResultsHydrogen

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.000

Mole fraction of H2 in the reactor

r = 0,00m

r = 1,01 · 10−3m

r = 5,19 · 10−3m

r = 1,21 · 10−2m

r = 2,08 · 10−2m

r = 3,02 · 10−2m

r = 3,89 · 10−2m

r = 4,58 · 10−2m

r = 5,00 · 10−2m

r = 5,10 · 10−2m

ResultsMethane

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.040

Mole fraction of CH4 in the reactor

r = 0,00m

r = 1,01 · 10−3m

r = 5,19 · 10−3m

r = 1,21 · 10−2m

r = 2,08 · 10−2m

r = 3,02 · 10−2m

r = 3,89 · 10−2m

r = 4,58 · 10−2m

r = 5,00 · 10−2m

r = 5,10 · 10−2m

ResultsCarbon dioxide

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.000

Mole fraction of CO2 in the reactor

r = 0,00m

r = 1,01 · 10−3m

r = 5,19 · 10−3m

r = 1,21 · 10−2m

r = 2,08 · 10−2m

r = 3,02 · 10−2m

r = 3,89 · 10−2m

r = 4,58 · 10−2m

r = 5,00 · 10−2m

r = 5,10 · 10−2m

ResultsSteam

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.350

Mole fraction of H2O in the reactor

r = 0,00m

r = 1,01 · 10−3m

r = 5,19 · 10−3m

r = 1,21 · 10−2m

r = 2,08 · 10−2m

r = 3,02 · 10−2m

r = 3,89 · 10−2m

r = 4,58 · 10−2m

r = 5,00 · 10−2m

r = 5,10 · 10−2m

ResultsCarbon monoxide

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.000

Mole fraction of CO in the reactor

r = 0,00m

r = 1,01 · 10−3m

r = 5,19 · 10−3m

r = 1,21 · 10−2m

r = 2,08 · 10−2m

r = 3,02 · 10−2m

r = 3,89 · 10−2m

r = 4,58 · 10−2m

r = 5,00 · 10−2m

r = 5,10 · 10−2m

ResultsNitrogen

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.031

Mole fraction of N2 in the reactor

r = 0,00m

r = 1,01 · 10−3m

r = 5,19 · 10−3m

r = 1,21 · 10−2m

r = 2,08 · 10−2m

r = 3,02 · 10−2m

r = 3,89 · 10−2m

r = 4,58 · 10−2m

r = 5,00 · 10−2m

r = 5,10 · 10−2m

ResultsRun-time

Solution method Run-time [s]

Finite-difference & ode15s (Matlab) 1.8Collocation – segregated 129.9Collocation – combined 51.1Collocation – coupled 52.0Collocation – coupled & SparseMatrixCSC 14.8

ResultsHindsight

Non-dimensional equations

Sparse matrices

Higher collocation point density near inlet

Linearise with Newton

Iterative matrix solvers (CG, GMRES)

Outline

1 Theory

3 Implementation

4 Results

5 Conclusion

Conclusion

FDM with ode15sFastSimpleAutomatic step-size controlNo support for boundary conditions in z

Orthogonal collocation

SlowNot as simple (but not that hard)Supports boundary conditions in zEasy differentiation (Kronecker product)Trial and error for relaxation factorsGlobal index notationCollocation points must be given beforehand

Questions?

modelling steam methane reforming in a fixed bed reactor ...folk.ntnu.no/andersty/5....

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