PSE Summer School 2012
Process Simulation and Optimization of
Chemical Plants
DAAD Summer School 2012, Mexico
Sponsorship:
DAAD
1
1
www.dbta.tu-berlin.de
Process Simulation and Optimization of
Chemical Plants
Dynamic Modeling Approaches
Prof. Dr.-Ing. habil. Prof. h.c. Dr. h.c. G. Wozny
2
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Simulation and Optimization of Chemical Plants
Part II
Aim:
Introduction in Process Dynamic,
Examples
Step responses
Modeling in MOSAIC
Workflow
Basic
Steady state Modeling Approach
Prof. Dr.-Ing. habil. Prof. h.c. Dr. h.c. G. Wozny
3
www.dbta.tu-berlin.de
Process dynamic
Problem solving - Systematic Approach
1
2 Output
Input
?
searched
PFD given, Unit given
given
Workflow- Steady state Modeling (reduced)
0. Aim definition
1. Construction Flow Diagram (Numbering all flows)
2. Declare given Flows (Ströme und Zusammensetzung)
3. Choose suitable Balance volume for Process Unit
4. Formulation of Balance Equations
5. Determine Degree of freedom
6. Test: Number of Variables = Number of Equations?
7. Solve Balance Equation system, …
4
Process analyses aims are given
MOSAIC
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Process dynamic
Input
given
Output
PFD Flowdiagram given
searched
Equation system: Total Material Balances
Component Material Balances
Summation Equations
Equilibrium relationship
Energy Balances
If needed Momentum Balances
Enthalpy Equations
1
NC
NS
Neq
1
+1
NS
Number of Variables:
Const. Parameter
NE = 
NV = NS * ( NC + 4 ) + NK + NP + NQ + NW + …
zi F, T, p, h
Q
W
5
DOF
ND = N V - NE
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?
Steady state Modeling up to now
Process Modelling
Summation Equation
D
yi
Summation Equation
Phase Equilibrium Comp. 1
F
zi
Flash
Phase Equilibrium Comp. 2
xi
Summation Equation Feed
Component Material Balance Comp. 1
Component Material Balance Comp. 2
6
Isothermal Flash
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B
7
MOSAIC – USER Interface
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8
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Process Dynamic
Aims of the next lecture:
- Expenditure of the modeling approach to Dynamic System
- Dynamic Balance Equations
- Solution Approaches
- „Dynamic“ Degree of Freedom
- Introduction additional methods
- Judgment of Process Dynamic
- Process Automation (Control)
- Discussion selected Methods and Examples
9
Introduction
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Process Dynamic
Input
given
PFD Flowdiagram given
Sizing realized
Output
searched
?
Processunit given: Dynamic Total Material Balances
1
Dynamic Component Material Balances NC
Summation Equations
NS
Equilibrium relationship
Neq
Dynamic Energy Balances
1
Thermodynamic Data (e.g. Enthalpy Correlations ,..)
Ns
Dynamic Variables (Hold up, Internal Energy)
Controller Equations
If needed Dynamic Momentum Balances
Number of Equations
10
2
NControl
+1
NE = 
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Formulation of the Model Equations
Description by model equations of conservation (MESH)
System is described by a system of Differential –Algebraic
Equation system (DAE-System) the state variables depends from
input variables and parameter
E.g..:
Balance for changes of inhabitants of Kuantan
Balance : Inhabitants
in Process Sciences (state variable T, p, x,..)
Balance Volume : Berlin
in Process Sciences (Apparatus, Phase,..)
InhabitanB erlin
 movingin  movingout  birth
deaths

















time



Changes
Transport
resevoir
11
* * 0 0 0 T
*
* * * 0 0
p
*
0 * * * 0
x  *
0 0 * * * ..
*
0 0 0 * * 
..


 
x
A
*


b
Matrix formulation
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Formulation of Dynamic Total Material Balances
Material balance
Balance : Mass
Bilanzvolume : Tank
Fin,1
Fin,2
HU
Q
Fout,1
The simple Tank model
12
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Formulation of Dynamic Total Material Balances
Bilanz für Änderung der Masse
Material Balance
Molar basis
d HU
dt
=

k
Fin,k -
Bilanzvolume : Tank
Fin,1
 Fout,m +/- NT
13
NT
m
Konvection
Notation:
HU=Hold up = V*/M
t = time
F = Flow
Q=heat duty
NT =Transport Term
k=No. Input Flows
m=No. Output Flows
Fin,2
Diffusion
HU
Q
Fout,1
The simple Tank model
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Dynamic Energy balance:
Pt
.
m1 ea1
.

u
h1
ea2
m2
HU
h2
.
.
Dynamic Energy Balance
.
TC
Q
Q = k A m
.
msteam
.
d (HU (u+ea) ) /dt = Q + Pt + m1 ( h1 + ea1 ) - m2 ( h2 + ea2 )
Controller Equation:
14
msteam – msteam,0 = kC ( Tset point - Tactuell )
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Process dynamic
Input
given
PFD Flowdiagram given
Sizing realized
Number of Equations
Number of Variables:
Output
searched
?
NE = 
Dynamic Terms
HU U
kC, TN
Const. Parameter
NV = NS * ( NC + 4 ) + NK + NP + NQ + NW + 1 + 1 + Ncontrol.-Param + …
xi F, T, p, h
Q
W
Heat Work
Dynamic
15 DOF
ND = N V - NE
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Process dynamic
Feed
n
F1, x1,1, x1,2
F4
p

n Input
F1
F2
F3
x1,1
p


p
x2,1
Output
t
cooling
Actors:
Valve
Engine speed
F3 (Number of revolution)
Product
Lift of piston
F2 x2,1, x2,2
t
Sensors:
Temperature
Pressure
Concentration
Hagen, J.: Modellierung und Simulation von Chemiereaktoren – Aspekte
einer zeitgemäßen Ingenieurausbildung, CIT 2005, 77, No.1-2, 25-33
16
Introduction
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Process dynamic
Programms - Tools
- gPROMS, ACM
- Aspen Dynamics
- Simusolv
- MOSAIC
MOSAIC
- Hysis …..
ACSL
MATLAB
Mathcad 
Matrixx
Simnon, ....
Knowledge from Lecture one: modeling steady state
Units, Processes – Solution Algebraic Equation Systems
Now: Solution of Differential Algebraic Equation systems:
DAE - Systems
New Tasks: Expenditures of the modeling Approach
Development of general Solution, Methods
17
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Basics of Process dynamic
Literature:
z.B. Luyben, W.L.: Process Modeling, Simulation, and Control for
Chemical Engineers, McGraw-Hill, Inc. 1990, ISBN 0-07-039159-9
Ordys, A.W.; Pike, A.W.; Johnson, M.A.; Katebi, R.M.; Grimble, M.J.:
Modeling and Simulation of Power Generation Plants, Springer Verlag, ISBN 3-540-19907-1
Shelden, R.A.; Dunn, I.J.: Dynamic Simulation - Modeling Processes, The Improvement,
The World CEP December 2001, S. 44-48
Dochain, D.; Vanrolleghem, P.A.: Dynamical Modeling and Estimation in Wastewater
Treatment Processes IWA Publishing, , Alliance House, 12 Caxton Street,
London SW1H 0QS, UK ISBN: 1 900222 50 7
Francis G. Shinskey
Process Control: As Taught vs. as Practiced
Ind. Eng. Chem. Res. 2002, 41,3745 -3750
T. Barz, H. Arellano-Garcia, G. Wozny, Generalization of a Tailored Approach for
Dynamic Simulation and Optimization, AIChE Annual Meeting, Tuesday, November 9, 2010
Dynamic Simulation and Optimization,250 D Room (Salt Palace Convention Center)
Tilman Barz, Stefan Kuntsche, Günter Wozny und Harvey Arellano-Garcia (2011). An
efficient sparse approach to sensitivity generation for large-scale dynamic
optimization Computers and Chemical Engineering. 35. 2053-2065
18
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Process dynamic
Searched: System behavior in dependency of
dependent Variables / independent Variables
(e.g. Design variables, Set Points, Disturbances, Load)
Results: graphical extend of Processvariables (state varaibles) under given
Parameterconditions x = f (t)
Aids: Simulation programs, Tools, lab scale plants, pilot plants,
Production plants
Solutions:
Single DGL (exact Solution , or numerical solution by
Methods like: Runge Kutta, Euler,...)
Systems of DE, DAE (Linearization, Gear, DASSL,.
ODEPACK*, ..., library IMSL, ….)
e.g. : Newton Method:
xk+1=
xk
-
f(xk)
/
f´(xk)
for each time step, after discritization with Euler method
19
Basics
*Hindmarsh A.C.(1983), ODEPACK, a Systematized
Collection of ODE Solvers. In:Scientific Computing
(R.S. Stepleman et al., Eds.), pp. 55-64
North Holland, Amsterdam
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Process dynamic
Solution:
Single DGL (exakt, Runge Kutta, Euler,...)
Systems DGL, DAE (Linearization, Gear, DASSL,...,)
xk+1= xk - f(xk) / f´(xk)
- -k = - k
-xk+1= x- k - f(x
) / f´(x )
Eigenberger, CIT 1992
Num. Methoden
For remembership: Partial DGL are solved e.g. withMethod of Lines
Forward Problem:
Process
Given: input, Process
20
Basics
Searched: Output
e.g.: Destillate purity
Energy demand etc.
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Process dynamic
Backward Problem:
Process
Searched: Input
Given: Process-Model,
Output
e.g..:Disturbance
Inductive Problem:
Process
Given: Input
21
Searched: Process-Model
Or Reason for
observed Behavior
Basics
Given:
Output
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Process dynamic
Question:
- Output Changes for given input changes?
xa(t)
xe(t)
t
t
xe(t)
xa(t)
t
t
22
Basics
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Process dynamic
Chromatography
Optimization des Designs with CFD
Principle of a HPLC Column
Feed
Produkt
Outlet
Inlet
Distributor
(Frit) Verteiler
23
Collector (Frit),
Sammler
A practical Example
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Process dynamic
Chromatography
Optimization of Design with CFD
Data: Brandt et al.
Film1: Simulation und Experiment of a HPLCColumn mit D=50mm without Inlet-Distributer
Verteilerblech.
Film3: Detail of the performance of Film2: Simulation of a HPLC-Column t D=100mm with
the Inlet-Distributer
Inlet- Distributor (Merck Selbstfüllstand NW 100).
24
A practical Example
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Process dynamic
Question :
- Output Changes for a given Input Change?
xa(t)
xe(t)
t
Input: sin , cos
Output: sin , cos
t
Changes of Amplitude
+ Phasen angle
See also lecture Processcontrol
25
Basics
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Process dynamic
Questions:
- Influence of System parameter
z.B. Holdup, Pressure, (Design values - decisions)
but also sometimes mathematical Solution procedure
- Dynamic System output behavior for small/large
disturbances of input variables
e.g. cooling water flow, heating flow (steam)
valve position
- Dynamic behavior of closed loop systems (including controllers)
(Type of controllers, coupling, recycle, heat integration,
Reaction etc.)
- How can the model Parameter are determined (Identification)
26
Basics of Processdynamic
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Process dynamic
Approach in Dynamic Analysis :
- Simulation -> Input changes (larger, smaller, step response investigation)
- Experiment -> Input changes
->Observe Output ,
Challenges: There is a need for a suitable Model, complex, many
investigations are needed,...
- In a real Plant disturbances are undesired,
Investigation at a Pilot plant are difficult, Scale up,
(transfer from small scale to a large scale)
Laplace Domain (Linearization is needed)
Response behavior >
Step response
Transfer behavior
Transfer function G(s)
-> Time domain: many investigations and results, what in general?
normalization, standardization Transformations are needed
27
Basics
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Process dynamic
Column Dynamic
An Industrial example
Dest
C12
C14
C16
C18
Time h
Basics Processdynamic
28
Transferbehavior after Feed step changewww.dbta.tu-berlin.de
Process dynamic
D2
29
Time h
F2
Transfer behavior of a Production plant
With noises , no filtered
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Process dynamic
Concentration
Steam
Time
Step response simulated verses experiments
30
Reboiler duty (steam) changes
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Process dynamic
Steam step change -5%
Systemresponse (Concentration changes)
xB Bottom concentration
xD
xD Distillate concentration
Feed
Steam
xB
31
Time in h
System response steam step change
Step response simulated
Steam step change
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Process dynamic
Steam step change + 5%
System response (Concentration changes)
xD Distillate concentration
xB Bottom concentration
Steam
Time in h
System response steam step change
32
Step response simulated
Steam step change
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Dynamic Flash Model
V = F2
F = F1
HUV
x1,1, x1,2, ..., x1,5
HUL
Variable
Name
Dimension
F
Flow
Kmol/s
x
Molar Fraction
-
T
Temperature
oC
P
Pressure
bar
HU
Hold up
kmol
33
x2,1, x2,2, ..., x2,5
L = F3
x3,1, x3,2, ..., x3,5
Subscript
Superscript
Name
V
Vapor
L
Liquid
i=1,2,3
Flow
j=1,…,5
Component
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Dynamic Flash Model
Example:
F = F1
1

Vapor
HUV
x1,1, x1,2, ... Liquid
HUL
V = F2 2
x2,1, x2,2, ....
L = F3
x3,1, x3,2, …, x3,5
34
Ammonia -Process
5 Components
Nitrogen
Hydrogen
Ammonia
Argon
Methane
1
2
3
4
5
3
Variable
Name
Dimension
F
Flow
Kmol/s
x
Molar
Fraction
-
T
Temperature
oC
P
Pressure
bar
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Dynamic Flash Model
Dynamic Component material balances
F1 x1,1 - F2 x2,1 - F3 x3,1
(1)
d ( HUL x3,2 + HUV x2,2 ) / dt = F1 x1,2 - F2 x2,2 - F3 x3,2
(2)
d ( HUL x3,3 + HUV x2,3 ) / dt = F1 x1,3 - F2 x2,3 - F3 x3,3
(3)
d ( HUL x3,4 + HUV x2,4 ) / dt = F1 x1,4 - F2 x2,4 - F3 x3,4
(4)
d ( HUL x3,5 + HUV x2,5 ) / dt = F1 x1,5 - F2 x2,5 - F3 x3,5
(5)
d ( HUL x3,1 + HUV x2,1 ) / dt =
Summation equations:
1 = x1,1 + x1,2 + x1,3 + x1,4 + x1,5
(6)
1 =
x2,1 + x2,2 + x2,3 + x2,4 + x2,5
(7)
1 =
x3,1 + x3,2 + x3,3 + x3,4 + x3,5
(8)
Phase equilibrium relationships:
35
x2,1 = K1 x3,1
(9)
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Dynamic Flash Model
Phase equilibrium relationships:
x2,1 = K1 x3,1
(9)
x2,2 = K2 x3,2
(10)
x2,3 = K3 x3,3
(11)
x2,4 = K4 x3,4
(12)
x2,5 = K5 x3,5
(13)
T2 = T3
(14)
p2 = p3
(15)
Dynamic Energy balance:
.
d (HUL uL + HUV uV + HUMetal uMetal ) / dt = F1h1 - F2h2 - F3h3 + Q (16)
Enthalpy correlations:
36
h1 = f ( T1, p1, x1,i )
(17)
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Dynamic Flash Model
Dynamic Energy balance:
d
(HUL uL
+
HUV
uV
.
+ HUMetal uMetal ) / dt = F1h1 - F2h2 - F3h3 + Q (16)
Enthalpy correlations:
K-values:
37
h1 = f ( T1, p1, x1,i )
(17)
h2 = f ( T2, p2, x2,i )
(18)
h3 = f ( T3, p3, x3,i )
(19)
K1 = f ( T2, p2, x2,i , x3,i )
(20)
K2 = f ( T2, p2, x2,i , x3,i )
(21)
K3 = f ( T2, p2, x2,i , x3,i )
(22)
K4 = f ( T2, p2, x2,i , x3,i )
(23)
K5 = f ( T2, p2, x2,i , x3,i )
(24)
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Dynamic Flash Model
Momentum balance (pressure loss):
p1 = p2
(25)
Internal Energy:
Hold up:
38
uV = f ( T2, v2, x2,i )
(26)
uL = f ( T3, v3, x3,i )
(27)
uMetal = f ( T3)
(28)
HUV = f ( Geometry-V,T2, p2, x2,i )
(29)
HUL = f ( Geometry-L,T3, p3, x3,i )
(30)
HUMetal = f ( Geometry-Metal,T3)
(31)
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Dynamic Flash Model
Number of variables:
Internal Energy u
K-Values
NV = Ns (Nc + 4 ) + Np + NQ + NHU + Nz + N Geometry=
3 (5+4) +5
+
1
+
3
+ 3
+ 3 = 42
Number specifications (Design variables):
ND = NV - NE = 42 - 31 = 11
.
Design Values choosen: F1, T1, P1, x1,1 , x1,2 , x1,3, x1,4 , Q=0 (adiabat)
Geometry-L, Geometry-V, Geometry Metal
39
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Dynamic Flash Model
Solution of the equation system iteratively .
MOSAIC
40
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Flow driven versus pressure driven
Flow Driven:
• Outlet stream pressures and flow rates of a
block are determined from the inlet conditions
to the block and the block specifications
• Outlet stream pressures or flow rates are
not affected by pressure in the downstream
blocks
Pressure Driven:
• Take into account the effect of downstream
pressures on flow rates in streams
3
Example:
4
F3  F4  K 3 p3  p4
1
2
Holdup = f(T,P,F)
F1  F2  K1 p1  p2
5
41
6
F5  F6  K 5 p5  p6
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Column Process Dynamic
TC
Questions:
PC
LC

TC
L
FC
QC
D
S
FC
F,
zj,iF
TC
xD,i
How many process
variables have to be
controlled?
How can we determine the
control variables?
What is the best pairing?
What is the best place for
the sensors?
TC

LC
B, xB,i = xNS,i
V
Control structure?
Set points?
Controller-Parameter?
Optimization!
Experience: Pressure, quality, level! Are there any more?
42
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Column Process Dynamic
1
PC
LC
Reflux drum

L
2
FC
Balance
volume 1
j-1
jj
F, zj,i
j+1
F
Feed
TC
Balance
volume j
Boil up
n
43
Balance volume n
Vj+1
Lj
Steam
Reboiler
B, xB,i
Vj
HUL(j)
V
Bottom product
Lj-1
D xD,i
Reflux
trap
Francis Weir formula
Lj = 3.33 l hj3/2
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Process Dynamic
1
D, xD
Condenser
Distillate
Choice: Balance volume
Vapor
Interfacial area
Liquid
.
.
.
2
3
Feed
2
j
HUj
Transfer to the model
of a theoretical tray
F, xF
n
44
.
.
.
Reboiler
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Process Dynamic
1
.
.
.
D, xD
Condenser
j-1
Distillate
Vapor phase
Interfacial Area
2
Liquid phase
3
Feed
2
j
j
Vapor phase
HUj
Transfer to the model
of a theoretical plate
F, xF
Liquid phase
j+1
Vapor phase
Liquid phase
n
45
Reboiler
.
.
.
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Complete process model
Yi,j
Lj-1
Vj
hj
V
Xi,j-1
hj-1L
j
URj,l
Tj ; pj
hj
F
Vapor phase
Fj
Uj
HUjV
URk,j
Interfacial area
Liquid phase
Z( Y or X )i,jF
WRn,j
HUjL
Wj
Vj+1
Yi,j+1
with:
•Y = vapor mole fraction
•X = liquid mole fraction
WRj,m
Lj
hj+1V
Xi,j
•V = vapor flow
•L = liquid flow
h jL
•hV = vapor enthalpy
•hL =liquid enthalpy
HU = Hold up
46
Dynamic Model of a theoretical plate
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Complete Process Model
Vj-1
.
.
.
Lj-2
j-1
1. Material balances
2. Component material
Vapor phase
Interfacial Area
Vj
balances
Liquid phase
j
Lj-1
Vapor phase
Feed
3. Phase equilibrium
Liquid phase
Vj+1
j+1
Lj
Vapor phase
Liquid phase
.
.
.
+ pressure loss correlation
Vj+2
4. Summation equation
5. Energy balances
Lj+1
+ tray efficiency
47
Model of a theoretical plate
Steady state
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Complete Process Model
Vj-1
.
.
.
Lj-2
j-1
Vapor phase
1. Dynamic material balance
Interfacial Area
Vj
Plant design:
Diameter, height, area, ….
Liquid phase
j
Lj-1
2. Dynamic component material
Vapor phase
Feed
balance
Liquid phase
Vj+1
j+1
Lj
Vapor phase
4. Summation equation
Liquid phase
.
.
.
+ Pressure loss correlations,
Vj+2
+ Tray efficiency
+ Holdup correlationen, ….
48
3. Phase equilibrium
Lj+1
5. Dynamic energy balances
6. Controller equation
Dynamic Model af a theoretical plate
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Complete Process Model
Yi,j
Lj-1
Vj
hj
V
Xi,j-1
hj-1L
j
Tj ; pj
h jF
Vapor phase
Fj
Z(Y or X)i,j
Without
sidestreams
Interfacial Area
Liquid phase
F
Vj+1
Yi,j+1
Lj
hj+1V
Xi,j
h jL
d HU /dt = Fj + Vj+1 + Lj-1 – Vj - Lj
49
Dynamic material balances
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Complete Process Model
Dynamic component material balances
Yi,j
Lj-1
Vj
hj
V
Xi,j-1
hj-1L
j
Tj ; pj
hjF
Vapor phase
Fj
Z(Y or X)i,j
Assumption:
HUjV<<HUjL
Interfacial Area
Liquid phase
F
Vj+1
Yi,j+1
Lj
hj+1V
Xi,j
h jL
d(HUjL Xi,j )/dt = Fj Zi,jF + Vj+1 Yi,j+1 + Lj-1 Xi,j-1
- Vj Yi,j - Lj Xi,j
50
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Complete Process Model
Phase equilibrium
Yi,j
Lj-1
Vj
hj
V
Xi,j-1
hj-1L
j
Tj ; pj
h jF
Vapor phase
Fj
Z(Y or X)i,j
Interfacial Area
Liquid phase
F
Vj+1
Yi,j+1
Ki,j = Yi,j / Xi,j
Lj
hj+1V
->
Xi,j
h jL
Gi,j = Yi,j – Ki,j Xi,j = 0
with K = equilibrium constant possibly with Murphree tray efficiency
51
Def.: Yi,j= Yi,j+1 + i,j ( Yi,,j – Ki,j+1 Xi,j+1)www.dbta.tu-berlin.de
Summation equation
Yi,j
Lj-1
Vj
hj
V
Xi,j-1
hj-1L
j
Tj ; pj
h jF
Vapor phase
Fj
Interfacial Area
Z(Y or X)i,j
Liquid phase
F
Vj+1
Yi,j+1
Lj
hj+1V
Xi,j
NC
NC
SYj = 1-
 Yi,j = 0
i=1
52
h jL
SXj = 1-
 Xi,j = 0
i=1
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Dynamic energy balance
Yi,j
Lj-1
Vj
hj
V
Xi,j-1
hj-1L
j
Tj ; pj
h jF
Vapor phase
Fj
Z(Y or X)i,j
Interfacial Area
Liquid phase
F
Qj
Lj
Vj+1
Yi,j+1
hj+1V
Xi,j
h jL
d(HUjL ujL + HUjV ujV + Mmetal umetal)/ dt = Fj hjF +
Vj+1 hj+1V + Lj-1 hj-1L - Vj hjV - Lj hjL +/- Qj
53
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Dynamic energy balance
d(HUjL ujL + HUjV ujV + Mmetal umetal)/ dt = Fj hjF +
Vj+1 hj+1V + Lj-1 hj-1L - Vj hjV - Lj hjL +/- Qj
With :
HU = Holdup in kmol
u molar internal energy
for liquids and solids u  h (molar enthalpy)
The term HUjV ujV + Mmetal umetal will be in the following
investigation neglected.
54
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Dynamic model equation for a theoretical plate
. conclusion
Vj-1
.
.
j-1
Lj-2
d(HUjL /dt = Mj
Vapor phase
Interfacial Area
Vj
Vapor phase
Feed
2. Component material balance
Liquid phase
j
Lj-1
j+1
Lj
Vapor phase
.
.
.
Lj+1
+ Holdup Correlation
Pressure loss correlation
Tray efficiency correlations
Flooding, entrainment (constraints)
55
Gi,j = 0
4. Summation equation
Liquid phase
Vj+2
d(HUjL *Xi,j ) / dt = Mi,j
3. Phase equilibrium
Liquid phase
Vj+1
1.Total material balance
SXi,j = 0,
SYi,j =0
5. Energy balances
d( HUjL *hjL ) / dt = Ej
6. Controller equation (e.g. PIDType)
Rl =0 l= Number controller
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Dynamic model equation for a theoretical plate
. conclusion
Vj-1
.
.
j-1
Lj-2
d(HUjL /dt = Mj
Vapor phase
Interfacial Area
Vj
Vapor phase
Feed
2. Component material balance
Liquid phase
j
Lj-1
j+1
Lj
Vapor phase
.
.
.
Lj+1
+ Holdup Correlation
Pressure loss correlation
Tray efficiency correlations
Flooding, entrainment (constraints)
56
Gi,j = 0
4. Summation equation
Liquid phase
Vj+2
d(HUjL *Xi,j ) / dt = Mi,j
3. Phase equilibrium
Liquid phase
Vj+1
1.Total material balance
SXi,j = 0,
SYi,j =0
5. Energy balances
d( HUjL *hjL ) / dt = Ej
6. Controller equation (e.g. PIDType)
Rl =0 l= Number controller
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Solution Methods
Reformulation of the algebraic equations (Summation equations,
Phase Equilibrium Relationships) in Differential Equations
Result: Set of Ordinary Differential Equation system
-> ODE solvers from library can be used (e.g. Runge Kutta, etc.)
Reformulation of the differential equations (Material balances,
Energy Balances) in Algebraic Equations by Euler Method or
Orthogonal Collocation Method
Result: Set of Nonlinear Algebraic Equation system
-> Iterative Methods are needed (Newton Raphson Method)
57
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Dynamic Model equation for a theoretical plate
After transformation
.
1. Total material balance
.
.
Lj-2
Mj - d(HUjL )/dt = 0
V
j-1
j-1
2. Component material balance
Vapor phase
Interfacial area
Vj
j
Vapor phase
Feed
Mi,j -d(HUjL *Xi,j ) / dt = 0
Liquid phase
Lj-1
3. Phase equilibrium
Gi,j
Liquid phase
Vj+1
j+1
Lj
Vapor phase
Vj+2
.
.
.
4. Summation equation
SXj = 0,
Liquid phase
= 0
SYj =0
5. Energy balance
Lj+1
Ej -d(HUjL *hjL ) / dt = 0
6. Controller equation (e.g.:PID Type)
R(l) = 0
58
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Complete Process Model
D = D0 - KC,D (HU1,SetPoint-HU1,actual)
1
PC
TC
LC

L
2
xD
FC
R(1) = 0 = D (D0 - KC,D (HU1,SP-HU1,ac))
D
L1 = L1,0 - KC,L ( T2SP - T2ac)
1
j-1
j
F, z
Feed
R(2)= L1 - L1,0 + KC,L ( T2,SP - T2,ac)
j+1
1
=0
k
TC
TC

LC
n
B, xB
r(l)=(HU1,ac, L1, V(or Q),HUn,ac)
59
V=V0 + KC,V ( Tk,SP – Tk,ac)
V
R(3)=0= V-V0 - KC,V ( Tk,SP – Tk,ac)
B = B0 - KC,B (HUn,SP-HUn,ac)
R(4) = B-(B0 - KC,B (HUn,SP-HUn,ac)) = 0
Integration Controller Equation
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Complete Process Model
D =D0 + KC,D ( Tm,SetPoint – Tm,actual)
1
PC
LC

TC
L
R(1) = ....
2
FC
D xD
m
L1 = L1,0 - KC,L (HU1,SP - HU1,ac)
1
j-1
j
j+1
F, z
R(2)= L1 -(L1,0 - KC,L (HU1,SP - HU1,ac))
1
Feed
k
TC
LC

V= V0 - KC,V (HUn,SP – HUn,ac)
V
TC
n
R(3)= V-(V0 - KC,V (HUn,SP - HUnac))
B, xB
Variable: r(l) = (D, HU1,ac,HUn,ac,B)
60
B=B0 - KC,B ( Tk,SP - Tk,ac)
R(4)= ....
Integration Controller equation
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Dynamic Model equation for a theoretical stage
Solution method
1. Total material balance
The equation system is a so called DAE
(Differential Algebraic Equation System).
In literature you will find powerful
solution methods and programs
e.g. DASSL, Gear –method etc.
Here we will transform the equation system
like the steady state model..
The solution procedure is useful for both
steady state and dynamic simulation.
+ Hold up correlation,
pressure loss correlation,
Tray efficiency correlation,
constraints
61
Mj - d(HUjL )/dt = 0
2. Component material balance
Mj -d(HUjL *Xi,j ) / dt= 0
3. Phase equilibrium:
Gi,j = 0
4. Summation equation:
SXj = 0, SYj =0
5. Energy balance:
Ej -d( HUjL hjL ) / dt = 0
6. Controller equation (e.g. PID Type) :
Rl = 0
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Equation system - Matrix notation
Vj-1
.
.
.
j-1
Lj-2
Vapor phase
Interfacial area
Vj
Liquid phase
j
Vapor phase
Feed
Lj-1
F(X) =
Liquid phase
Vj+1
j+1
Lj
Vapor phase
Liquid phase
Vj+2
.
.
.
Mi,1 –d( HU1L *Xi,1 ) /dt
E1 –d (HU1L *h1L )/dt
Gi,1
...
Mi,j –d(HUjL *Xi,j )/dt
Ej –d(HUjL *hjL )/dt
Gi,j
...
Mi,n –d(HUnL *Xi,n )/dt
En –d(HUnL *hnL )/ dt
Gi,n
Rl =0
Lj+1
F
= (F1 ,...,Fj ,...Fn )T
X (t) = (X1 ,...,Xj ,...Xn )T
62
=0
Xj (t)
j=1,2...n
= (vi,j ,li,j ,Tj )T
r(l)T
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Complete Process Model
Bilinear Terms will be linearized:
NC
l i,j = Xi,j * Lj
with :

l i,j = L j
i=1
NC
v i,j = Y i,j * V j
with:

v i,j = V j)
i=1
Remark: The modeling with component flows has the advantage
for initialization e.g. reactor is cold and empty
63
Selection of the interation variables
Remarks
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Solution Method
The solution of the balance equations (linearized differential equation
with e.g. Euler Approximation) and the nonlinear
equations
is done for each time step with the
-  X +1 =(dF/dX)-1 F 
Newton Raphson method
X +1 =X  +  X +1
 = Iteration Number
X = Variable Vector
F = Function vector
The Jacoby matrix is important!
64
( ) = Jacobi matrix
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Complete Process Model
X = Variable vector, F =Function vector,(...) = Jacoby matrix
B1 C1
P1
A2
B2 C2
P2
A3
B3 C3
P3
A4
B4 C 4
P4
A5 B5 C5
P5
Aj
dF
( ---- ) =
dx
E1
E2
E3
E4
E5
Bj
Cj
..........................
An-2 Bn-2 Cn-2
An-1 Bn-1 Cn-1
An Bn
Ej
En-2 En-1 En
Pj
Pn-2
Pn-1
Pn
G
Aj= ( dFj / dxj-1 ) Bj= ( dFj / dxj ) Cj= ( dFj / dxj+1 ) G=(dR(l)/dr(l))
65
Solution
with Sparse Matrix Method
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Complete Process Model
Transformation für
j=1,2,3,….,n-1, n for Step 1 - 9
-1
j  n-1
-1
jn
-1
Step 1:
Cj = Bj Cj
Step 2:
Pj = Bj Pj
Step 3:
Fj = Bj Fj
jn
Step 4:
Bj+1 = Bj+1 – Aj+1Cj
j  n-1
Step 5:
Pj+1 = Pj+1 - Aj+1Pj
j  n-1
Step 6:
Fj+1 = Fj+1 – Aj+1Fj
j  n-1
Step 7:
Ej+1 = Ej+1 - Ej Cj
j  n-1
66
Solution method
Sparse Matrix Method
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Complete Process Model
Transformation for
Step 8:
Step 9:
j=1,2,3,….,n-1, n for Step 1-9
G = G - Ej Pj
jn
R = R – Ej Fj
jn
Step 10:
r = G-1 R
Step 11:
xn = Fn - Pn r
Step 12:
67
xj = Fj - Cj xj+1 – Pj r
Solution method
Sparse Matrix Method
j = n-1, n-2, …3,2,1
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Complete Process Model
X = Variable vector, F =Function vector,(...) = Jacoby matrix
B1 C1
P1
A2
B2 C2
P2
O*3,n
A3
B3 C3
P3
A4
B4 C 4
P4
A5 B5 C5
P5
..........................
Aj
Bj Cj
Oj,3
Pj
dF
( --- ) =
..........................
dx
An-2 Bn-2 Cn-2
Pn-2
An-1 Bn-1 Cn-1 Pn-1
An Bn Pn
E1 E2
E3 E4
E5 ... Ej ...En-2 En-1 En G
Pj= ( dFj / dr(l)) Ej= ( dR(l) / dxj ) Oj,3=(dFj/dx3) O*3,n=(dF3/dxn)
68
Solution method
Sparse Matrix Method
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o
B1
A2
q
p
C1
B2 C2
A3
B3
A4
r
Oj,3
dF
( --- ) =
dx
E1
69
69
E2
E3
C3
B4 C 4
A5 B5 C5
..........................
Aj j
BBjj CCjj
A
E4
O*3,n
Pj
..........................
An-2 Bn-2 Cn-2
An-1 Bn-1 Cn-1
An Bn
E5 ... Ej ...En-2 En-1 En
Oj,3=(F
Solution
method
j/x3)
Sparse Matrix Method
P1
P2
P3
P4
P5
Pn-2
Pn-1
Pn
G
O*3,n=(F3/xn)
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Complete Process Model
Transformation für Step 1 - 3
Step 1:
j=1,2,3,….,n-1, n
-1
Cj = Bj Cj
-1
O*j,p = Bj O*j,p
-1
Fj = Bj Fj
Step 2:
70
j  n-1
q  j  p-2
jn
Or,j+1 = Or,j Cj
o  j  p-2
Or,p = Or,p – Or,j O*p,j
o  j  p-2 und qj
Or,p = Or,p – Or,j Cj
j = p-1
Or,j+1 = – Or,j Cj
p  j  r-2
Ar = Ar – Or,j Cj
j = r-2
Fr = Fr – Or,j Fj
o  j  r-2
Or,j = 0
Solution method
Sparse Matrix Method
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Complete Process Model
Transformation für Step 1 - 3
Step 3:
71
j=1,2,3,….,n-1, n
Bj+1 = Bj+1 – Aj+1Cj
j  n-1
Fj+1 = Fj+1 – Aj+1Fj
j  n-1
O*j+1,p = O*j+1,p - Aj+1 O*j,p
q  j  p-2
Cj+1 = Cj+1 - Aj+1 O*j,p
j  p-2
Solution method
Sparse Matrix Method
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Complete Process Model
Step 4:
Step 5:
xn = Fn
xj = Fj - Cj xj+1
xj = Fj - Cj xj+1 – O*j,p xp
xj = Fj - Cj xj+1
72
Solution method
Sparse Matrix Method
j = n-1, n-2, …p-2,p-1
j = p-2, p-3, …,q+1, q
j = q-1, q-2, …1
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Complete Process Model
Vj-1
.
.
.
j-1
Lj-2
-  X +1 =(dF/dX)-1 F 
Vapor phase
Interfacial Area
Vj
Liquid phase
j
Vapor phase
Feed
Lj-1
j+1
Lj
Vapor phase
Liquid phase
Vj+2
.
.
.
X +1 =X  +  X +1
For each time step 
Liquid phase
Vj+1
Solution Method
Newton Raphson
X = Variable vector
F =Function vector
Lj+1
( ) = Jacoby matrix
Problems – challenges:
stiffness, Initialization, Index Problem ,...
73
Simultaneous solution for each time step
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Complete Process Model
Vj-1
.
.
.
j-1
Lj-2
Vapor phase
• Flow profile Vj , Lj
Interfacial Area
Vj
Liquid phase
Lj-1
j
Vapor phase
Feed
Liquid phase
Vj+1
• Concentration profile Xi,j, Yi,j
Lj • Temperature profile Tj
j+1
Vapor phase
• Heat duty Q1 , Qn
Liquid phase
Vj+2
Results
.
.
.
Lj+1 • Control variables r(l) (D, V, HU,
PTop)
Simulation results
74
•As a Function of time
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Complete Process Model results
Methanol -Water separation - Feedflow - 10%
20 trays
+ reboiler
+ condenser
D= const.
Qreboiler=const.
75
Simulation results step responses
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Non equilibrium Model
Model : non equilibrium (rate based approach)
Concentrationsprofile
Yi,j
Lj-1
Vj
hjV
Xi,j-1
hj-1L
j
Tj ; pj
Y
Vapor
Film
Vapor phase bulk
Ni,jV
Ni,jL
Fj
X
(Interfacial Area)
Ei,j
Liquid
Film
Liquid phase bulk
hjF
ZF (YF or XF )i,j
Vj+1
76
hj+1V
Yi,j+1
Lj
Xi,j
For complex application
hjL
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Thank you very much for
Your interest!
77
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