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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
Steady state analysis
Paweł Ptaszek
Cracov University of Technology
February 26, 2013
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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
We will analyze situation when t →∞, derivative ∂x∂t = 0.
Our chemical engineerig objects will be in steadystate. Forlumped objects like CSTR’s we obtain nonlinear algebraicsystem of equestions in the folowing form (good example forfurther analysis):
1ταA − rA(αA,T ) = 0
1τ
(T0 − T ) + (∆Tad) · rA(αA,T )− U(T − Tq) = 0
We can’t solute this system of equations analytically but wehave numerical methods ...
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Steady stateanalysis
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Process Dynamics
Newton method:
f1 =1ταA − rA(αA,T )
f2 =1τ
(T0 − T ) + (∆Tad) · rA(αA,T )− U(T − Tq)
x =
[αAT
]f =
[f1f2
]∂f∂x
=
[∂f1∂αA
∂f1∂T
∂f2∂αA
∂f2∂T
]iteration scheme (staring from aproximation solution x0):
xi+1 = xi −(∂f∂x
)−1· f
This method is not comfortable because on each iteration stepwe must invert the Jacobi matrix.
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Steady stateanalysis
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Process Dynamics
More flexible is Newton-Krylov method:
xi+1 = xi − c
on each iteration we must solve system of equations, wherevector c is unknown: (
∂f∂x
)· c = f
This modification is much faster than classical Newton methodand is very good for sparse problems.
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Process Dynamics
Using described methods we obtain only one solution. Sometimes we need the family of solutions as a function of theselected parameter (λi ). In our case (CSTR) vector of controlparameters looks like as folow:
λ =
τT0
∆TadUTq
Then we tabulated our solutions as parameter function. Itcould happen that it exists more solutions for one valueparameter. Then we have a problem ...
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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
Continuation methods:The simplest case implicit function derivative. We shouldreduce our CSTR model to one equation: We must expressmass balance as a function αA = g(T , τ)and heat balance:
1τ
(T0 − T ) + (∆Tad) · rA(αA,T )− U(T − Tq) = 0
in compact form:Ψ(T , λi ) = 0
Now we calculate implicit function Ψ(T , λi ) derivative (fromdefinition):
dλidT
= −∂Ψ∂T∂Ψ∂λi
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Process Dynamics
Last equation can be regarded as nonlinear initial valueproblem and can be integrate well known methods: Euler, Gear,Runge-Kutta, etc.The initial condition is any steadystate λi (T̄ ) = λ̄i0.Drawback of presented method is singularity when thederivations (denominator) ∂Ψ
∂λi= 0.
To solve this problem there are several methods:Predictors:
ODE-methods, including tangent (DERPAR); doi:10.1145/355666.355674polynominal extrapolation
Parametrizations:the parametrization as additional equationarclength and pseudoarclenght (AUTO):http://indy.cs.concordia.ca/autolocal parametrization (PITCON); doi:10.1145/357456.357461
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Process Dynamics
Using continuation method we obtain bifurcation diagram:
300
350
400
450
500
550
300 310 320 330 340 350 360 370 380 390 400
T,K
T0
Turning point
Turning point
Steady state curve
Steady state diagram
, K
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Process Dynamics
Analysis of bifurcation diagram gives information about:
structure of steady states,
range of occurrence of multiple steady states,
number of steady states,
location of turning points (static bifurcations).
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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
Analysis of bifurcation diagram gives information about:
structure of steady states,
range of occurrence of multiple steady states,
number of steady states,
location of turning points (static bifurcations).
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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
Analysis of bifurcation diagram gives information about:
structure of steady states,
range of occurrence of multiple steady states,
number of steady states,
location of turning points (static bifurcations).
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Steady stateanalysis
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Process Dynamics
Analysis of bifurcation diagram gives information about:
structure of steady states,
range of occurrence of multiple steady states,
number of steady states,
location of turning points (static bifurcations).
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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
Analysis of bifurcation diagram gives information about:
structure of steady states,
range of occurrence of multiple steady states,
number of steady states,
location of turning points (static bifurcations).
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Steady stateanalysis
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Process Dynamics
When we made continuation as a function of another controlparameter, we obtained the structure called: fold.
4060
80100
120140
160180
Tad, K
300320
340360
380400
T0, K
0
100
200
300
400
500
600
700
T, K
thre
e st
eady
stat
es
... but this procedure is very computer time consuming, so wemake continuation of the turning points (static bifurcations)and obtain catastrophic cross section
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Steady stateanalysis
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Process Dynamics
More complex example: cascade of n nonisothermal CSTR withrecycle and A→B reaction:
(ξαn − αi ) + τi ri (αi , θi ) = 0
(ξθn − θi ) + τiηri (αi , θi )− τiwi (θi − θq) = 0
ri = (1− αi )k0 exp(− γ
1 + θi
)i = 1, · · · , n − 1
Now: n = 4, η = 0.333, γ = 28, k0 = 109s−1, wi = 0.0055,θq = 0, ξ = 0.05, τi = τ
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CSTR-1 CSTR-2 CSTR-3 CSTR-4
, s
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Process Dynamics
0.325
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20 40 60 80 100 120 140 160
, s
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, s
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Analysis of steady states in distributed objects (BVP problems).Nonisothermal tubular reactor with longitudinal dispersion(A→ B):
1Pem
d2αdz2− dαdz
+ τ r(α, θ) = 0
1Peq
d2θdz2− dθdz
+ τηr(α, θ)− Q(θ − θQ) = 0
α(0)− 1Pem
dα(0)
dz= 0,
dα(1)
dz= 0
θ(0)− 1Peq
dθ(0)
dz= 0,
dθ(1)
dz= 0
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Process Dynamics
After substitutions α = u1, dαdz = u2, θ = u3, dαdz = u4:
du1dz
= u2
du2dz
= Pem[u2 − τ r(u1, u3)]
du3dz
= u4
du4dz
= Peq[u4 − τηr(u1, u3)] + Q(u3 − θQ)]
Bondary conditions:
Pemu1(0)− u2(0) = 0, u2(1) = 0
Pequ3(0)− u4(0) = 0, u4(1) = 0
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Backward shooting method: Assume values of u1(1), u3(1)
u2(1) = 0
u4(1) = 0
Backward integrate model of our reactor from z = 1 to z = 0and On the begining we have system of nonlinear algebraicequations conjugated with initial value problem:
f1 = Pemu1(0)− u2(0) = 0
f2 = Pequ3(0)− u4(0) = 0
If f1 and f2 cerates system of algebraic equations we will usecontinuation methods.
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Process Dynamics
0.6
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, s
3 - stady states
1 - stady state
1 - stady state
5 - stady states
3 - stady states
3 - stady states
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Process Dynamics
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0
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stable
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able
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�(1)
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Process Dynamics
0
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stable
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Other aplications of continuations methods in chemicalengineerig:
homothopy method for finding azeotropy points,
polymer-polymer-solvent equlibria,
least squares method for parameter estimations.
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Paweł Ptaszek
Process Dynamics
Other aplications of continuations methods in chemicalengineerig:
homothopy method for finding azeotropy points,
polymer-polymer-solvent equlibria,
least squares method for parameter estimations.
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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
Other aplications of continuations methods in chemicalengineerig:
homothopy method for finding azeotropy points,
polymer-polymer-solvent equlibria,
least squares method for parameter estimations.
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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
Other aplications of continuations methods in chemicalengineerig:
homothopy method for finding azeotropy points,
polymer-polymer-solvent equlibria,
least squares method for parameter estimations.
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Steady stateanalysis
Paweł Ptaszek
Process Dynamics
Homotopy method for finding azeotropes points.Azeotropes of an Nc component system satisfy the folowingsystem of nonlinear equations:
xo − y = 0
feq(xo , y ,T ,P) = 0
Nc∑i=1
yi − 1 = 0
T ,P, xoi , yi 0
where xo is the liquid composition, y is vapor composition, andfeq is some equlibrium relationship between the liquid and thevapor.
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Process Dynamics
Now we entroduce homotopy:
h(x , λ) = λ(xo − y) + (1− λ)(xo − y id)
where: λ ∈ [0, 1] is the homotopy parameter,first part of this sum is describing real (nonideal) vapor-liquidequlibrium:
xo − y = 0
second part describing ideal vapor-liquid equlibrium:
xo − y id = 0
and from Raoult’s law: y idi =PsatjP xoj
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Process Dynamics
By writing out the vapor-liquid equlibrium and sumation ofmole fractions explicitly, this system of equations can bereformulated as:
xo − y∗ = 0
y∗j − [λK oj + (1− λ)PsatjP
]xoj = 0
Nc∑i=1
y∗i − 1 = 0
where the nonideal vapor-liquid equlibrium is described by thefolowing expresion: K oj (y∗, xo ,T ,P). We have 2Nc + 1equations with 2Nc + 3 variables (xo , y∗,T ,P, λ). When wefixed pressure we obtain 2Nc + 2 variables. When λ iscontinuation parameters we have 2Nc + 1 equations with2Nc + 1 variables. y∗ is ”perturbed” vapor composition.
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Process Dynamics
We calculate the K oj using UNIFAC method (activitycoefficients), and saturated vapor pressure using Antoine’aequation in form:
log(Psati ) = A− BC + T
Our system: water, acrylic acid, methanol, methyl acrylate.
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Steady stateanalysis
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Process Dynamics
Polymer-polymer-solvent equlibria (Flory-Huggins theory):
∆µ1 = RT[
ln(φ1) +
(1− 1x2
)φ2 +
(1− 1x3
)φ3 + ζ1
]ζ1 = φ2φ3(χ12 + χ13 − χ23 + χ12φ
22 + χ13φ
23)
∆µ2 = RT[
ln(φ2) +
(1− x2x3
)φ3 + (1− x2)φ1 + ζ2
]ζ2 = x2(χ23 + φ23 + χ12φ
21 + φ1φ3(χ12 + χ23 − χ13)
∆µ3 = RT[
ln(φ3) +
(1− x3x2
)φ3 + (1− x3)φ1 + ζ3)
]ζ3 = x3(χ23 + φ22 + χ13φ
21 + φ1φ2(χ13 + χ23 − χ12)
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Process Dynamics
where: φ1 - volume fraction of solvent, φ2 - volume fraction ofpolymer 1, φ3 - volume fraction of polymer 2, χ12, χ13, χ23 areFlory -Huggins coefficients, and
xi =MiνiM1ν1
i = 1, 2, 3
νi - specific volume, Mi - molar massPhase equlibria conditions:
∆µα1 −∆µβ1 = 0
∆µα2 −∆µβ2 = 0
∆µα3 −∆µβ3 = 0
α and β are phases.27 / 30
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Process Dynamics
We have three equations with six variables (φαi and φβi ,i = 1, 2, 3). But φα,β3 = 1− (φα,β1 + φα,β2 ), and φα1 iscontinuation parameter. Finaly we obtain three equations withthree variables.The problem is to find one solutions to start our continuationprocedure.Numerical example:
x2 = 4 · 103
x3 = 103
χ12 = 0.4
χ13 = 0.048
χ23 = 0.00199
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Solvent
Polymer 1 Polymer 2
II - phases
I - phaseI- phase
0.5
1
0.5
I- phase
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Steady stateanalysis
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Process Dynamics
Least squares method for parameter estimations.Homework !!
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