CHE412 Process Dynamics and Control BSc (Engg) Chemical Engineering (7 th Semester) Week 2/3 Mathematical Modeling Luyben (1996) Chapter 2-3 Stephanopoulos (1984) Chapter 4, 5 Seborg et al (2006) Chapter 2 Dr Waheed Afzal Associate Professor of Chemical Engineering Institute of Chemical Engineering and Technology University of the Punjab, Lahore [email protected]1
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CHE412 Process Dynamics and ControlBSc (Engg) Chemical Engineering (7th Semester)
• Stability • Block diagram • Transducer • Final control element • Mathematical model• Input-out model,
transfer function • Deterministic and
stochastic models• Optimization• Types of Feedback
Controllers (P, PI, PID)
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(t)
• Proportional: c(t) = Kc Є(t) + cs
• Proportional-Integral:
• Proportional-Integral-Derivative:
Nomenclature actuating output , error , gain , time constant Consult your class notes onmodelling of stirred tank heater
Types of Feedback Controllers
(Stephanopoulos, 1984)
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• Mathematical representation of a process (chemical or physical) intended to promote qualitative and quantitative understanding
• Set of equations• Steady state, unsteady state (transient) behavior
• Model should be in good agreement with experiments
Mathematical Modeling
Experimental Setup
Set of Equations(process model)
Inputs Outputs
Outputs
Compare
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1. Determine objectives, end-use, required details and accuracy
2. Draw schematic diagram and label all variables, parameters3. Develop basis and list all assumptions; simplicity Vs reality 4. If spatial variables are important (partial or ordinary DEs)5. Write conservation equations, introduce auxiliary equations 6. Never forget dimensional analysis while developing
equations 7. Perform degree of freedom analysis to ensure solution 8. Simplify model by re-arranging equations 9. Classify variables (disturbances, controlled and manipulated
variables, etc.)
Systematic Approach for Modelling (Seborg et al 2004)
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• To understand the transient behavior, how inputs influence outputs, effects of recycles, bottlenecks
• To train the operating personnel (what will happen if…, ‘emergency situations’, no/smaller than required reflux in distillation column, pump is not providing feed, etc.)
• Selection of control pairs (controlled v. / manipulated v.) and control configurations (process-based models)
• To troubleshoot • Optimizing process conditions (most profitable
scenarios)
Need of a Mathematical Model
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• Theoretical Models based on principal of conservation- mass, energy, momentum and auxiliary relationships, ρ, enthalpy, cp, phase equilibria, Arrhenius equation, etc)• Empirical model based on large quantity of experimental data) • Semi-empirical model (combination of theoretical
and empirical models) Any available combination of theoretical principles and empirical correlations
Classification of Process Models based on how they are developed
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Theoretical Models Physical insight into the process Applicable over a wide range of conditions Time consuming (actual models consist of large
number of equations) Availability of model parameters e.g. reaction rate
coefficient, over-all heart transfer coefficient, etc. Empirical model Easier to develop but needs experimental data Applicable to narrow range of conditions
Advantages of Different Models
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State variables describe natural state of a process Fundamental quantities (mass, energy, momentum)
are readily measurable in a process are described by measurable variables (T, P, x, F, V)
State equations are derived from conservation principle (relates state variables with other variables)
(Rate of accumulation) = (rate of input) – (rate of output) + (rate of generation) - (rate of consumption)
State Variables and State Equations
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BasisFlow rates are volumetric Compositions are molar A → B, exothermic, first order Assumptions Perfect mixingρ, cP are constant
Perfect insulation Coolant is perfectly mixedNo thermal resistance of jacket
Modeling ExamplesJacketed CSTR
Coolant
Fi, CAi, Ti
F, CA, T
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Overall Mass Balance (Rate of accumulation) = (rate of input) – (rate of output) Component Mass Balance (Rate of accumulation of A) = (rate of input of A) – (rate of output of A) + (rate of generation of A) – (rate of consumption of A)
Modeling of a Jacketed CSTR (Contd.)
CoolantFci,Tci
VCA
T
Fi, CAi, Ti
F, CA, T
CoolantFco,Tco
Energy Balance (Rate of energy accumulation) = (rate of energy input) – (rate of energy output) - (rate of energy removal by coolant) + (rate of energy added by the exothermic reaction)
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Overall Mass Balance
Component Mass Balance
Energy Balance
Modeling of a Jacketed CSTR (Contd.)
CoolantFci,Tci
VCA
T
Fi, CAi, Ti
F, CA, T
CoolantFco,Tco
Input variables: CAi, Fi, Ti, Q, (F) Output variables: V, CA, T
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Nf = Nv - NE
Case (1): Nf = 0 i.e. Nv = NE (exactly specified system)
We can solve the model without difficulty
Case (2): f > 0 i.e. Nv > NE (under specified system), infinite number of solutions because Nf process variables can be fixed arbitrarily. either specify variables (by measuring disturbances) or add controller equation/s
Case (3): Nf < 0 i.e. Nv < NE (over specified system) set of equations has no solution
remove Nf equation/s
We must achieve Nf = 0 in order to simulate (solve) the model
Degrees of Freedom (Nf) Analysis
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Basis/ Assumptions Perfectly mixed, Perfectly insulated ρ, cP are constant
Stirred Tank Heater: Modeling and Degree of Freedom Analysis
Stirred Tank Heater: Modeling and Degree of Freedom Analysis
SteamFst
ACV MV
h F
T Q
Nf = 2 - 2 = 0 Can you draw these control loops?
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F
Z
mB
mD
VB
DxD
BxB
R xD
Reboiler
Condenser
Reflux Drum
(Stephanopoulos, 1984)
Basis/ Assumptions 1. Saturated feed2. Perfect insulation of column3. Trays are ideal4. Vapor hold-up is negligible 5. Molar heats of vaporization of A
and B are similar 6. Perfect mixing on each tray7. Relative volatility (α) is constant 8. Liquid holdup follows Francis weir
formulae 9. Condenser and Reboiler dynamics
are neglected10. Total 20 trays, feed at 10 2, 4, 5 → V1 = V2 = V3 = … VN
Mass balance (total and component) around 6 segments of a distillation column: reflux drum, top tray, Nth tray, feed tray, 1st tray and column base.
Solution of ODE for total mass balance gives liquid holdups (mN) Solution of ODE for component mass balance gives liquid
compositions (xN) V1 = V2 = … = VN = VB (vapor holdups) How to calculate y (vapor composition) and L (liquid flow rate) Recall αij is constant throughout the column Use αij = ki/kj , xi + xj =1, yi + yj = 1, and k = y/x to prove