Apresenta

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EQE038 Simulação e Otimização de Processos
Químicos
CSTR Multiplicidade de Soluções e Análise de
Estabilidade
Argimiro R. Secchi Programa de Engenharia Química
COPPE/UFRJ Rio de Janeiro, RJ
EQ/UFRJ março de 2014
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System Analysis

• Multiplicity of steady states
• Linearization
• System stability
• Complex dynamic behaviors (limit cycles, strange
  attractors)
• Parametric sensitivity and input sensitivity

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Multiplicity of Steady States
Non-isothermal CSTR
Fe , CAf , CBf , Tf
Fws , Tw
V , T
Fwe , Twe
Fs , CA , CB , T
A
B
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Multiplicity of Steady States
Process description
In a non-isothermal continuous stirred tank
reactor, with diameter of 3.2 m and level
control, pure reactant is fed at 300 K and 3.5
m3/h with concentration of 300 kmol/m3. A first
order reaction occur in the reactor, with
frequency factor of 89 s-1 and activation energy
of 6 x 104 kJ/kmol, releasing 7000 kJ/kmol of
reaction heat. The reactor has a jacket to
control the reactor temperature, with constant
overall heat transfer coefficient of 300
kJ/(h.m2.K). Assume constant density of 1000
kg/m3 and constant specific heat of 4 kJ/(kg.K)
in the reaction medium. The fully-open output
linear valve has a constant of 2.7 m2.5/h.
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Multiplicity of Steady States
Model assumptions

• perfect mixture in the reactor and jacket
• negligible shaft work
• (-rA) k CA
• constant density
• constant overall heat transfer coefficient
• constant specific heat
• incompressible fluids
• negligible heat loss to surroundings
• ?(internal energy) ? ?(enthalpy)
• negligible variation of potential and kinetic
  energies
• constant volume in the jacket
• thin metallic wall with negligible heat capacity.

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Multiplicity of Steady States
CSTR modeling
Mass balance in the reactor Overall
(1)
Component
(2)
(3)
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Multiplicity of Steady States
CSTR modeling
Energy balance in the reactor
where
(4)
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Multiplicity of Steady States
CSTR modeling
(5)
where
q U At (T Tw)
qr (-?Hr) V (-rA)
(6)
(-rA) k CA
(7)
k k0 exp(E/RT)
(8)
A ? D2/4
(9)
V A h
(10)
At A ? D h
(11)
Fs x Cv ? h
(12)
x f(h)
Level control (13)
Tw f(T)
Temperature control (14)
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Multiplicity of Steady States
Consistency analysis

• variable units of measurement
• Fe, Fs m3 s-1
• V m3
• t, ? s
• CA, CAf kmol m-3
• rA kmol m-3 s-1
• kg m-3
• Cp kJ kg-1 K-1
• T, Tf, Tw K
• qr, q kJ s-1
• U kJ m-2 K-1 s-1
• At, A m2
• h, D m
• Cv m2.5 h-1
• x
• ?Hr, E kJ kmol-1
• R kJ kmol-1 K-1
• k, k0 s-1

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Multiplicity of Steady States
Consistency analysis
variables Fe, Fs, V, t, CA, CAf, rA, ?, Cp, T,
Tf, Tw, qr, q, U, At, A, h, D, Cv, x, ?Hr, E, R,
k, k0,? ? 27 constants ?, Cp, U, D, Cv, ?Hr,
E, R, k0 ? 9 specifications t ? 1 driving
forces Fe, Tf, CAf ? 3 unknown variables Fs,
V, CA, rA, T, Tw, qr, q, A, At, h, x, k, ? ?
14 equations 14 Degree of Freedom variables
constants specifications driving forces
equations unknown variables equations 27
9 1 3 14 0 Dynamic Degree of Freedom
(index lt 2) differential equations 3 ? Needs
3 initial condition h(0), CA(0), T(0) ? 3
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Multiplicity of Steady States

• Running EMSO

Open MSO file
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Multiplicity of Steady States
The CSTR example at the steady state satisfy
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Multiplicity of Steady States
Rewriting the energy balance
stable
unstable
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Multiplicity of Steady States
Path Following
Newton-Raphson
Homotopic Continuation
affine homotopy
Newton homotopy
Multiples solutions can be obtained by
continuously varying the parameter p
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Multiplicity of Steady States
Path Following
Parametric Continuation
where s is some parameterization, e.g., path arc
length
Frechet derivative
a point (xo, po) is
- Regular if is
non-singular
reparameterization
- Turning point if is
singular and DF has rank n
- Bifurcation if is
singular and DF has rank lt n
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Multiplicity of Steady States
Example a) execute flowsheet in file
CSTR_noniso.mso with initial condition of 578 K
and compare with result changing the initial
condition to 579 K b) find the three steady
states using file CSTR_sea.mso by changing the
initial guess for T and CA (use the section
GUESS).
Solutions 1) CA 13,13 kmol/m3 and T
659,46 K 2) CA 132,87 kmol/m3 and T
523,01 K 3) CA 299,86 kmol/m3 and T
332,72 K
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Linearization
Generate linearized model at given operating
point. Implicit DAE Considering the
specification as input, u(t), (SPECIFY section in
EMSO) And identifying the algebraic variables
as y(t)
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Linearization
Differentiating F and extracting The
partition Define the linearized system
(index lt 2)
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Linearization
Test example for a linear model exact solution!
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Linearization
Verifying the results for the linear model
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Non-isothermal CSTR linearization
Example execute the flowsheet in file
CSTR_linearize.mso with the option Linearize
true and evaluate the characteristic values of
the Jacobian matrix (matrix A). Repeat the
example with the value of Cp 10 times smaller,
i.e., 0.4 kJ / (kg K). Compare the ratio between
the greater and the smaller characteristic values
in module.
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Stability Analysis
Liapunov Stability is stable (or
Liapunov stable) if, given e gt 0, there exists a
d d(e) gt 0, such that, for any other solution,
y(t), of
, then
satisfying
for t gt t0.
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Stability Analysis
Asymptotic Stability is asymptotic
stable if Liapunov stable and there exists a
constant b gt 0 such that, if
then
Defining deviation variables
Expanding in Taylor series
Linearization
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Stability Analysis
For an equilibrium point x, the
stability is characterized by the characteristics
values of the Jacobian matrix J(x) A
? x is a hyperbolic point if none
characteristics values of J(x) has zero real
part. ? x is a center if the characteristics
values are pure imaginary. Fixed point
non-hyperbolic. ? x is a saddle point,
unstable, if some characteristics values have
real part gt 0 and the remaining have real part lt
0. ? x is stable or attractor or sink point if
all characteristics values have real part lt 0. ?
x is unstable or repulsive or source point if
at least one characteristic value have real part
gt 0.
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Stability Analysis
For a second-order linear system
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Stability Analysis
Considering the CSTR example with constant volume
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Stability Analysis
1) Stable node
2) Saddle Point, unstable
3) Stable Node
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Stability Analysis
file CSTR_nla/traj_cstr.m
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Complex Dynamic Behavior
CSTR example
stable solutions
CA
unstable solutions
Tw
T
Hopf point Tw 200,37 K
Tw
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Complex Dynamic Behavior
unstable limit cycle
file CSTR_auto/cstr_bif.mso
t (h)
A limit cycle is stable if all characteristics
values of exp(J p) (Floquet multipliers) are
inside the unitary cycle, where J is the Jacobian
matrix in the cycle, p 2 ? / ? is the
oscillation period and ? ?Hopf.
t (h)
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Interface EMSO-AUTO
parameters
Equation system Jacobian matrix First
steady-state solution
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Interface EMSO-AUTO
p 0 x (0, 0) ?(J) (-1, -3)
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Interface EMSO-AUTO
Parameter p Eigenvalues Phase plane
p lt 0.06361 Real negatives eigenvalues stable node p 0.05 ? -1.13, -2.06
p 0.06361 Repeated real negatives eigenvalues stable node (star) ? -1.4372, -1.4372
0.06361 lt p lt 0.0889 Complex eigenvalues with negative real part - stable focus p 0.085 ? -1.095 ? 0.565 i
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Interface EMSO-AUTO
p 0,0889 unstable node gives rise to two points unstable node and saddle point p gt 0.0889 Turning point (fold) One stable solution (focus) and other unstable (node). (point 3 in figure below) ? -1.009 ? 0.605 i ? 0, 3.432
0.0889 lt p lt 0.0933 One stable solution (focus) and two unstable (saddle and node) p 0.09 ? -0.982 ? 0.614 i ? -0.213, 3.332 ? 0.364, 3.151
0.0933 lt p lt 0.10574 at p 0.105738931 the first point goes from stable focus to stable node ? -0.055, -0.046 One stable solution (focus) and two unstable (saddle and focus) p 0.10 ? -0.652 ? 0.651 i ? -0.439, 1.953 ? 1.431 ? 1.851 i
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Interface EMSO-AUTO
p 0.10574 the stable node gives rise to two points stable node and saddle for p lt 0.10574 Turning point (fold) One stable solution (node), other unstable (focus), and one sable limit cycle. (point 2 in figure below) ? -0.097, 0 ? 1.186 ? 2.478 i
0.10574 lt p lt 0.1309 One unstable solution (focus) and one stable limit cycle p 0.12 ? 0.528 ? 3.487 i
p 0.1309 Hopf bifurcation pure imaginary eigenvalues. (point 4 in figure below) ? ? 4.008 i
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Interface EMSO-AUTO
p gt 0.1309 Complex eigenvalues with negative real part - stable focus p 0.15 ? -0.952 ? 4.627 i
x2
p
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Interface EMSO-AUTO
Hopf
2nd turning point
1st turning point
Trajectories stable point saddle point unstable
point
Hopf
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Interface EMSO-AUTO
Example copy files auto_emso.exe and r-emso.bat
(Windows) or _at_r-emso (linux) in bin folder of
EMSO to the folder CSTR_auto and execute the
command below in a prompt of commands
(shell) Windows r-emso cstr_bif Linux
./_at_r-emso cstr_bif The results are stored in file
fort.7. In Linux the graphic tool PLAUT can be
used to plot the results using the command _at_p.
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Sensitivity Analysis
Objective determine the effect of variation of
parameters (p) or input variables (u) on the
output variables.
Steady-state simulation
local
Sensitivity analysis
(case study)
global bifurcation diagram, surface response
Normalized form
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Sensitivity Analysis
Dynamic simulation
where
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Sensitivity Analysis
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Sensitivity Analysis
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References

• DAE Solvers
• DASSL Petzold, L.R. (1989), http//www.enq.ufrgs.
  br/enqlib/numeric/numeric.html
• DASSLC Secchi, A.R. (1992), http//www.enq.ufrgs.
  br/enqlib/numeric/numeric.html
• MEBDFI Abdulla, T.J. and J.R. Cash (1999),
  http//www.netlib.org/ode/mebdfi.f
• PSIDE Lioen, W.M., J.J.B. de Swart, and W.A. van
  der Veen (1997), http//www.cwi.nl/cwi/projects/PS
  IDE/
• SUNDIALS Serban, R. et al. (2004),
  http//www.llnl.gov/CASC/sundials/description/desc
  ription.html