Post on 09-Mar-2020
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
Kasper Linnestad Reactor modelling December 3, 2014 2 / 36
Outline
1 TheoryWeighted residuals methodCollocation pointsOrthogonal collocation method
2 Governing equations
3 Implementation
4 Results
5 Conclusion
Kasper Linnestad Reactor modelling December 3, 2014 3 / 36
TheoryWeighted residuals method
General problem
L{f(z, r)
}= g(z, r)
B{fb(zb, rb)
}= gb(z, r)
Approximation
f(z, r) ≈Pz∑
jz=0
Pr∑
jr=0
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) =
P∏
i=0i6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
Kasper Linnestad Reactor modelling December 3, 2014 4 / 36
TheoryWeighted residuals method
General problem
L{f(z, r)
}= g(z, r)
B{fb(zb, rb)
}= gb(z, r)
Approximation
f(z, r) ≈Pz∑
jz=0
Pr∑
jr=0
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) =
P∏
i=0i6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
Kasper Linnestad Reactor modelling December 3, 2014 4 / 36
TheoryWeighted residuals method
General problem
L{f(z, r)
}= g(z, r)
B{fb(zb, rb)
}= gb(z, r)
Approximation
f(z, r) ≈Pz∑
jz=0
Pr∑
jr=0
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) =
P∏
i=0i6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
Kasper Linnestad Reactor modelling December 3, 2014 4 / 36
TheoryWeighted residuals method
General problem
L{f(z, r)
}= g(z, r)
B{fb(zb, rb)
}= gb(z, r)
Approximation
f(z, r) ≈Pz∑
jz=0
Pr∑
jr=0
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) =
P∏
i=0i6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
Kasper Linnestad Reactor modelling December 3, 2014 4 / 36
TheoryWeighted residuals method
General problem
L{f(z, r)
}= g(z, r)
B{fb(zb, rb)
}= gb(z, r)
Approximation
f(z, r) ≈Pz∑
jz=0
Pr∑
jr=0
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) =
P∏
i=0i 6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
Kasper Linnestad Reactor modelling December 3, 2014 4 / 36
TheoryWeighted residuals method
General problem
L{f(z, r)
}= g(z, r)
B{fb(zb, rb)
}= gb(z, r)
Approximation
f(z, r) ≈Pz∑
jz=0
Pr∑
jr=0
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) =
P∏
i=0i 6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
Kasper Linnestad Reactor modelling December 3, 2014 4 / 36
TheoryWeighted residuals method
General problem
L{f(z, r)
}= g(z, r)
B{fb(zb, rb)
}= gb(z, r)
Approximation
f(z, r) ≈Pz∑
jz=0
Pr∑
jr=0
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) =
P∏
i=0i 6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
Kasper Linnestad Reactor modelling December 3, 2014 4 / 36
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
0
2
xLm(x)
Kasper Linnestad Reactor modelling December 3, 2014 5 / 36
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
0
2
xLm(x)
Kasper Linnestad Reactor modelling December 3, 2014 5 / 36
TheoryOrthogonal collocation method
Minimize residuals by∫R(z, r)wi(z, r) dz dr = 0, ∀i
Orthogonal collocation uses
ajz ,jr = f(zjz , rjr)
ln(x) =
P∏
i=0i 6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
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
Kasper Linnestad Reactor modelling December 3, 2014 6 / 36
TheoryOrthogonal collocation method
Minimize residuals by∫R(z, r)wi(z, r) dz dr = 0, ∀i
Orthogonal collocation uses
ajz ,jr = f(zjz , rjr)
ln(x) =
P∏
i=0i 6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
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
Kasper Linnestad Reactor modelling December 3, 2014 6 / 36
TheoryOrthogonal collocation method
Minimize residuals by∫R(z, r)wi(z, r) dz dr = 0, ∀i
Orthogonal collocation uses
ajz ,jr = f(zjz , rjr)
ln(x) =
P∏
i=0i 6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
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
Kasper Linnestad Reactor modelling December 3, 2014 6 / 36
TheoryOrthogonal collocation method
Minimize residuals by∫R(z, r)wi(z, r) dz dr = 0, ∀i
Orthogonal collocation uses
ajz ,jr = f(zjz , rjr)
ln(x) =
P∏
i=0i 6=n
x− xixn − xi
wi(z, r) = δ(z − ziz)δ(r − rir)
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
Kasper Linnestad Reactor modelling December 3, 2014 6 / 36
TheoryLinearisation
Example function
f(ψ) = ψ2 + ψ exp (−ψ)
Fixed-point (Picard-iteration)
f(ψ) ≈(ψ? + exp
(−ψ?
))ψ
solves f(ψ) = 3 in 114 iterations
Taylor (Newton-Raphson-iteration)
f(ψ) ≈ f(ψ?) +∂f
∂ψ
∣∣∣ψ?
(ψ − ψ?
)
solves f(ψ) = 3 in 8 iterations
Kasper Linnestad Reactor modelling December 3, 2014 7 / 36
TheoryLinearisation
Example function
f(ψ) = ψ2 + ψ exp (−ψ)
Fixed-point (Picard-iteration)
f(ψ) ≈(ψ? + exp
(−ψ?
))ψ
solves f(ψ) = 3 in 114 iterations
Taylor (Newton-Raphson-iteration)
f(ψ) ≈ f(ψ?) +∂f
∂ψ
∣∣∣ψ?
(ψ − ψ?
)
solves f(ψ) = 3 in 8 iterations
Kasper Linnestad Reactor modelling December 3, 2014 7 / 36
TheoryLinearisation
Example function
f(ψ) = ψ2 + ψ exp (−ψ)
Fixed-point (Picard-iteration)
f(ψ) ≈(ψ? + exp
(−ψ?
))ψ
solves f(ψ) = 3 in 114 iterations
Taylor (Newton-Raphson-iteration)
f(ψ) ≈ f(ψ?) +∂f
∂ψ
∣∣∣ψ?
(ψ − ψ?
)
solves f(ψ) = 3 in 8 iterations
Kasper Linnestad Reactor modelling December 3, 2014 7 / 36
Outline
1 Theory
2 Governing equationsContinuity equationEnergy equationSpecies mass balanceErgun’s equationInitial and boundary conditions
3 Implementation
4 Results
5 Conclusion
Kasper Linnestad Reactor modelling December 3, 2014 8 / 36
Governing equationsFixed bed reactor
z
r
Kasper Linnestad Reactor modelling December 3, 2014 9 / 36
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
Kasper Linnestad Reactor modelling December 3, 2014 10 / 36
Governing equationsContinuity equation
General form
∂ρ
∂t+∇ · (ρu) = 0
Simplified
∂ρuz∂z
= 0
Linearised
ρ?∂uz∂z
+ uz∂ρ?
∂z︸ ︷︷ ︸⇒Auz
= 0︸︷︷︸⇒buz
Kasper Linnestad Reactor modelling December 3, 2014 11 / 36
Governing equationsContinuity equation
General form
∂ρ
∂t+∇ · (ρu) = 0
Simplified
∂ρuz∂z
= 0
Linearised
ρ?∂uz∂z
+ uz∂ρ?
∂z︸ ︷︷ ︸⇒Auz
= 0︸︷︷︸⇒buz
Kasper Linnestad Reactor modelling December 3, 2014 11 / 36
Governing equationsContinuity equation
General form
∂ρ
∂t+∇ · (ρu) = 0
Simplified
∂ρuz∂z
= 0
Linearised
ρ?∂uz∂z
+ uz∂ρ?
∂z︸ ︷︷ ︸⇒Auz
= 0︸︷︷︸⇒buz
Kasper Linnestad Reactor modelling December 3, 2014 11 / 36
Governing equationsEnergy equation
General form
ρcp
(∂T
∂t+ u · ∇T
)= −∇ · q−∆rxH
Simplified
ρcpuz∂T
∂z=λeff
r
∂
∂r
(r∂T
∂r
)−∆rxH
Linearised
ρ?c?pu?z
∂T
∂z︸ ︷︷ ︸⇒AT
= λeff,?
(1
r
∂T
∂r+∂2T
∂r2
)
︸ ︷︷ ︸⇒AT
−∆rxH?
︸ ︷︷ ︸⇒bT
Kasper Linnestad Reactor modelling December 3, 2014 12 / 36
Governing equationsEnergy equation
General form
ρcp
(∂T
∂t+ u · ∇T
)= −∇ · q−∆rxH
Simplified
ρcpuz∂T
∂z=λeff
r
∂
∂r
(r∂T
∂r
)−∆rxH
Linearised
ρ?c?pu?z
∂T
∂z︸ ︷︷ ︸⇒AT
= λeff,?
(1
r
∂T
∂r+∂2T
∂r2
)
︸ ︷︷ ︸⇒AT
−∆rxH?
︸ ︷︷ ︸⇒bT
Kasper Linnestad Reactor modelling December 3, 2014 12 / 36
Governing equationsEnergy equation
General form
ρcp
(∂T
∂t+ u · ∇T
)= −∇ · q−∆rxH
Simplified
ρcpuz∂T
∂z=λeff
r
∂
∂r
(r∂T
∂r
)−∆rxH
Linearised
ρ?c?pu?z
∂T
∂z︸ ︷︷ ︸⇒AT
= λeff,?
(1
r
∂T
∂r+∂2T
∂r2
)
︸ ︷︷ ︸⇒AT
−∆rxH?
︸ ︷︷ ︸⇒bT
Kasper Linnestad Reactor modelling December 3, 2014 12 / 36
Governing equationsSpecies mass balance
General form
∂ρωi∂t
+∇ · (ρuωi) = −∇ · ji +Ri
Simplified
∂ρuzωi∂z
=Deff
r
∂
∂r
(rρ∂ωi∂r
)−Ri
Linearised
ρ?u?z∂ωi∂z
+ ωiu?z
∂ρ?
∂z+ ρ?ωi
∂u?z∂z︸ ︷︷ ︸
⇒Aωi
= R?i︸︷︷︸⇒bωi
+
Deff,?
(ρ?
r
∂ωi∂r
+∂ρ?
∂r
∂ωi∂r
+ ρ?∂2ωi∂r2
)
︸ ︷︷ ︸⇒Aωi
Kasper Linnestad Reactor modelling December 3, 2014 13 / 36
Governing equationsSpecies mass balance
General form
∂ρωi∂t
+∇ · (ρuωi) = −∇ · ji +Ri
Simplified
∂ρuzωi∂z
=Deff
r
∂
∂r
(rρ∂ωi∂r
)−Ri
Linearised
ρ?u?z∂ωi∂z
+ ωiu?z
∂ρ?
∂z+ ρ?ωi
∂u?z∂z︸ ︷︷ ︸
⇒Aωi
= R?i︸︷︷︸⇒bωi
+
Deff,?
(ρ?
r
∂ωi∂r
+∂ρ?
∂r
∂ωi∂r
+ ρ?∂2ωi∂r2
)
︸ ︷︷ ︸⇒Aωi
Kasper Linnestad Reactor modelling December 3, 2014 13 / 36
Governing equationsSpecies mass balance
General form
∂ρωi∂t
+∇ · (ρuωi) = −∇ · ji +Ri
Simplified
∂ρuzωi∂z
=Deff
r
∂
∂r
(rρ∂ωi∂r
)−Ri
Linearised
ρ?u?z∂ωi∂z
+ ωiu?z
∂ρ?
∂z+ ρ?ωi
∂u?z∂z︸ ︷︷ ︸
⇒Aωi
= R?i︸︷︷︸⇒bωi
+
Deff,?
(ρ?
r
∂ωi∂r
+∂ρ?
∂r
∂ωi∂r
+ ρ?∂2ωi∂r2
)
︸ ︷︷ ︸⇒Aωi
Kasper Linnestad Reactor modelling December 3, 2014 13 / 36
Governing equationsErgun’s equation
General form
dp
dz= −f ρu
2z
dp
Linearised
dp
dz︸︷︷︸⇒Ap
= −f? ρ?(u?z)
2
dp︸ ︷︷ ︸⇒bp
Kasper Linnestad Reactor modelling December 3, 2014 14 / 36
Governing equationsErgun’s equation
General form
dp
dz= −f ρu
2z
dp
Linearised
dp
dz︸︷︷︸⇒Ap
= −f? ρ?(u?z)
2
dp︸ ︷︷ ︸⇒bp
Kasper Linnestad Reactor modelling December 3, 2014 14 / 36
Governing equationsInitial and boundary conditions
Velocity
uz(z = 0, r)︸ ︷︷ ︸⇒Auz
= uz,inlet︸ ︷︷ ︸⇒buz
Temperature
T (z = 0, r)︸ ︷︷ ︸⇒AT
= Tinlet︸ ︷︷ ︸⇒bT
∂T
∂r
∣∣∣∣∣r=0︸ ︷︷ ︸
⇒AT
= 0︸︷︷︸⇒bT
λeff,? ∂T
∂r
∣∣∣∣∣r=R︸ ︷︷ ︸
⇒AT
= −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
Kasper Linnestad Reactor modelling December 3, 2014 15 / 36
Governing equationsInitial and boundary conditions
Velocity
uz(z = 0, r)︸ ︷︷ ︸⇒Auz
= uz,inlet︸ ︷︷ ︸⇒buz
Temperature
T (z = 0, r)︸ ︷︷ ︸⇒AT
= Tinlet︸ ︷︷ ︸⇒bT
∂T
∂r
∣∣∣∣∣r=0︸ ︷︷ ︸
⇒AT
= 0︸︷︷︸⇒bT
λeff,? ∂T
∂r
∣∣∣∣∣r=R︸ ︷︷ ︸
⇒AT
= −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
Kasper Linnestad Reactor modelling December 3, 2014 15 / 36
Governing equationsInitial and boundary conditions
Velocity
uz(z = 0, r)︸ ︷︷ ︸⇒Auz
= uz,inlet︸ ︷︷ ︸⇒buz
Temperature
T (z = 0, r)︸ ︷︷ ︸⇒AT
= Tinlet︸ ︷︷ ︸⇒bT
∂T
∂r
∣∣∣∣∣r=0︸ ︷︷ ︸
⇒AT
= 0︸︷︷︸⇒bT
λeff,? ∂T
∂r
∣∣∣∣∣r=R︸ ︷︷ ︸
⇒AT
= −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
Kasper Linnestad Reactor modelling December 3, 2014 15 / 36
Governing equationsInitial and boundary conditions
Velocity
uz(z = 0, r)︸ ︷︷ ︸⇒Auz
= uz,inlet︸ ︷︷ ︸⇒buz
Temperature
T (z = 0, r)︸ ︷︷ ︸⇒AT
= Tinlet︸ ︷︷ ︸⇒bT
∂T
∂r
∣∣∣∣∣r=0︸ ︷︷ ︸
⇒AT
= 0︸︷︷︸⇒bT
λeff,? ∂T
∂r
∣∣∣∣∣r=R︸ ︷︷ ︸
⇒AT
= −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
Kasper Linnestad Reactor modelling December 3, 2014 15 / 36
Outline
1 Theory
2 Governing equations
3 ImplementationJuliaUnder-relaxationSegregated approachCombined approachCoupled approach
4 Results
5 Conclusion
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ImplementationJulia
Open-source
Just-in-time compilation
Constant global variables
Pass by reference (!)
Easy to call programs written in other languages (Fortran, C, Python)
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ImplementationUnder-relaxation
Trial-and-error
ωi and T
T = γTT + (1− γT )T ?
γT = γω = 5 · 10−2
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ImplementationSegregated approach
T
ωi
||R||2 < tol
uz
p
||R||2 < tol
Update
Initial guess
||R||2 < tol
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ImplementationCombined approach
Coupled mass fractions and temperature
Aω1 0 · · ·0 Aω2 0 · · ·0 · · · . . .
I I · · · I 00 · · · AT
ω1
ω2...
ωCO
T
=
bω1
bω2
...1bT
Coupled velocity, density and pressureAuz 0 0
0 I −diag(
M?
RT?
)
0 0 Ap
uzρp
=
buz0bp
Kasper Linnestad Reactor modelling December 3, 2014 20 / 36
ImplementationCombined approach
Coupled mass fractions and temperature
Aω1 0 · · ·0 Aω2 0 · · ·0 · · · . . .
I I · · · I 00 · · · AT
ω1
ω2...
ωCO
T
=
bω1
bω2
...1bT
Coupled velocity, density and pressureAuz 0 0
0 I −diag(
M?
RT?
)
0 0 Ap
uzρp
=
buz0bp
Kasper Linnestad Reactor modelling December 3, 2014 20 / 36
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(
M?
RT?
)
0 · · · Ap
ω1
ω2...
ωCO
Tuzρp
=
bω1
bω2
...1bTbuz0bp
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Outline
1 Theory
2 Governing equations
3 Implementation
4 ResultsProfilesRun-timeHindsights
5 Conclusion
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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
2
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
3.8
z [m]
uz
[ ms−
1]
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
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ResultsVelocity
0.00
0.01
0.02
0.03
0.04
0.05 01
23
45
67
2
2.5
3
3.5
r [m] z [m]
uz
[ ms−
1]
Velocity in the reactor
Kasper Linnestad Reactor modelling December 3, 2014 23 / 36
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
z [m]
p[Pa]
Pressure in the reactor
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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
800
850
900
950
1,000
1,050
1,100
z [m]
T[K]
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
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ResultsTemperature
0.00
0.01
0.02
0.03
0.04
0.05 01
23
45
67
800
900
1,000
r [m] z [m]
T[K]
Temperature in the reactor
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ResultsHydrogen
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.450
z [m]
xH
2
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
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ResultsMethane
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.040
0.060
0.080
0.100
0.120
0.140
0.160
0.180
0.200
0.220
z [m]
xCH
4
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
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ResultsCarbon dioxide
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.000
0.010
0.020
0.030
0.040
0.050
0.060
z [m]
xCO
2
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
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ResultsSteam
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.350
0.400
0.450
0.500
0.550
0.600
0.650
0.700
0.750
z [m]
xH
2O
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
Kasper Linnestad Reactor modelling December 3, 2014 29 / 36
ResultsCarbon monoxide
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.000
0.010
0.020
0.030
0.040
0.050
0.060
0.070
z [m]
xCO
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
Kasper Linnestad Reactor modelling December 3, 2014 30 / 36
ResultsNitrogen
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.00.031
0.032
0.033
0.034
0.035
0.036
0.037
0.038
0.039
0.040
0.041
z [m]
xN
2
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
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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
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ResultsHindsight
Non-dimensional equations
Sparse matrices
Higher collocation point density near inlet
Linearise with Newton
Iterative matrix solvers (CG, GMRES)
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Outline
1 Theory
2 Governing equations
3 Implementation
4 Results
5 Conclusion
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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
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Questions?
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