CHEE 434/821 Process Control II Some Review Material Winter 2006 Instructor:M.GuayTA: V. Adetola

Introduction In the chemical industry, the design of a control system is essential to ensure: Good Process Operation Process Safety Product Quality Minimization of Environmental Impact

Introduction n What is the purpose of a control system? To maintain important process characteristics at desired targets despite the effects of external perturbations. Control Plant Processingobjectives Safety Make $$$ Environment... Perturbations Market Economy Climate Upsets...

Introduction Plant Control What constitutes a control system? Combination of process sensors, actuators and computer systems designed and tuned to orchestrate safe and profitable operation.

Introduction n n Process Dynamics: Study of the transient behavior of processes n n Process Control the use of process dynamics for the improvement of process operation and performance or the use of process dynamics to alleviate the effect of undesirable (unstable) process behaviors

Introduction What do we mean by process? A process, P, is an operation that takes an INPUT or a DISTURBANCE and gives an OUTPUT INPUT: (u) Something that you can manipulate DISTURBANCE: (d) Something that comes as a result of some outside phenomenon OUTPUT: (y) An observable quantity that we want to regulate u d y P Information Flow

Examples n n Stirred tank heater M T in, w Q T, w T in w Q TProcess Inputs Output

Examples n n The speed of an automobile Force of Engine Friction Inputs Output Friction Engine SpeedProcess

Examples e.g. Landing on Mars

Examples e.g. Millirobotics Laparoscopic Manipulators

Introduction n n Process A process, P, is an operation that takes an INPUT or a DISTURBANCE and gives an OUTPUT INPUT: (u) Something that you can manipulate DISTURBANCE: (d) Something that comes as a result of some outside phenomenon OUTPUT: (y) An observable quantity that we want to regulate u d y P Information Flow

Control What is control? n n To regulate of a process output despite the effect of disturbances e.g. Driving a car Controlling the temperature of a chemical reactor Reducing vibrations in a flexible structure n n To stabilize unstable processes e.g. Riding a bike Flight of an airplane Operation of a nuclear plant

Benefits of Control n n Economic Benefits Quality (waste reduction) Variance reduction (consistency) Savings in energy, materials, manpower n n Operability, safety (stability) Performance Efficiency Accuracy l l robotics Reliability Stabilizability l l bicycle l l aircraft l l nuclear reactor

Control n A controller is a system designed to regulate a given process Process typically obeys physical and chemical conservation laws Controller obeys laws of mathematics and logic (sometimes intelligent) e.g. - Riding a bike (human controller) - Driving a car - Automatic control (computer programmed to control) Process Controller What is a controller?

Block representations n n Block diagrams are models of the physical systems Process System Physical Boundary Transfer of fundamental quantities Mass, Energy and Momentum Input variables Output variables Physical Operation Abstract

Control n n A controlled process is a system which is comprised of two interacting systems: e.g. Most controlled systems are feedback controlled systems The controller is designed to provide regulation of process outputs in the presence of disturbances Process Controller OutputsDisturbances ActionObservation monitorintervene

Introduction What is required for the development of a control system? 1. The Plant (e.g. SPP of Nylon) Water Steam Gas Make-up Vent Nylon Blower Dehumidifier Reheater Relief Pot Heater

Introduction What is required? 1. Process Understanding Required measurements Required actuators Understand design limitations 2. Process Instrumentation Appropriate sensor and actuator selection Integration in control system Communication and computer architecture 3. Process Control Appropriate control strategy

Example n n Cruise Control Controller Friction Process Speed Engine Human or Computer

Classical Control n n Control is meant to provide regulation of process outputs about a reference, r, despite inherent disturbances n n The deviation of the plant output, e=(r-y), from its intended reference is used to make appropriate adjustments in the plant input, u ProcessController Classical Feedback Control System d yure + -

Control n n Process is a combination of sensors and actuators n n Controller is a computer (or operator) that performs the required manipulations e.g. Classical feedback control loop yre ACP M d ComputerActuator Process Sensor - +

Examples n n Driving an automobile ye ACP M Driver Automobile - + Steering r Visual and tactile measurement Desired trajectory r Actual trajectory y

Examples n n Stirred-Tank Heater T in, w QT, w Heater TC Thermocouple ye ACP M Controller Tank - + Heater Thermocouple T in, w TRTR

Examples n n Measure T, adjust Q Controller:Q=K(T R -T)+Q nominal whereQ nominal =wC(T-T in ) Q: Is this positive or negative feedback? T in, w Te ACP M Controller Tank - + Heater Thermocouple Feedback control TRTR

Examples n n Measure T i, adjust Q APC M TiTi QiQi QQ Q + + Feedforward Control

Control Nomenclature n n Identification of all process variables Inputs(affect process) Outputs(result of process) n n Inputs Disturbance variables Variables affecting process that are due to external forces Manipulated variables Things that we can directly affect

Control Nomenclature n n Outputs Measured speed of a car Unmeasured acceleration of a car Control variables important observable quantities that we want to regulate can be measured or unmeasured Controller Manipulated Disturbances Process Control Other

Example T L T w i, T i w c, T ci w c, T co w o, T o h Variables w i, w o :Tank inlet and outlet mass flows T i, T o :Tank inlet and outlet temperatures w c :Cooling jacket mass flow P c :Position of cooling jacket inlet valve P o :Position of tank outlet valve T ci, T co :Cooling jacket inlet and outlet temperatures h:Tank liquid level PoPo PcPc

Example Variables InputsOutputs Disturbances Manipulated Measured Unmeasured Control w i T i T ci w c h w o T o P c P o Task: Classify the variables

Process Control and Modeling n n In designing a controller, we must Define control objectives Develop a process model Design controller based on model Test through simulation Implement to real process Tune and monitor Model Controller yure d Process Design Implementation

Control System Development Define Objectives Develop a process model Design controller based on model Test by Simulation Implement and Tune Monitor Performance Control development is usually carried out following these important steps Often an iterative process, based on performance we may decide to retune, redesign or remodel a given control system

Control System Development n n Objectives What are we trying to control? n n Process modeling What do we need? Mechanistic and/or empirical n n Controller design How do we use the knowledge of process behavior to reach our process control objectives? What variables should we measure? What variables should we control? What are the best manipulated variables? What is the best controller structure?

Control System Development n n Implement and tune the controlled process Test by simulation incorporate control strategy to the process hardware theory rarely transcends to reality tune and re-tune n n Monitor performance periodic retuning and redesign is often necessary based on sensitivity of process or market demands statistical methods can be used to monitor performance

Process Modeling n n Motivation: Develop understanding of process a mathematical hypothesis of process mechanisms Match observed process behavior useful in design, optimization and control of process n n Control: Interested in description of process dynamics Dynamic model is used to predict how process responds to given input Tells us how to react

Process Modeling What kind of model do we need? n n Dynamic vs. Steady-state Steady-state Variables not a function of time useful for design calculation Dynamic Variables are a function of time Control requires dynamic model

Process Modeling What kind of model do we need? n n Experimental vs Theoretical Experimental Derived from tests performed on actual process Simpler model forms Easier to manipulate Theoretical Application of fundamental laws of physics and chemistry more complex but provides understanding Required in design stages

Process Modeling n Dynamic vs. Steady-state n Step change in input to observe Starting at steady-state, we made a step change The system oscillates and finds a new steady- state Dynamics describe the transitory behavior 0 50100150200250300 40 45 50 55 60 65 Output Time Steady-State 1 Steady-State 2

Process Modeling n n Empirical vs. Mechanistic models Empirical Models only local representation of the process (no extrapolation) model only as good as the data Mechanistic Models Rely on our understanding of a process Derived from first principles Observing laws of conservation of l l Mass l l Energy l l Momentum Useful for simulation and exploration of new operating conditions May contain unknown constants that must be estimated

Process Modeling n n Empirical vs Mechanistic models Empirical models do not rely on underlying mechanisms Fit specific function to match process Mathematical French curve

Process Modeling n n Linear vs Nonlinear Linear basis for most industrial control simpler model form, easy to identify easy to design controller poor prediction, adequate control Nonlinear reality more complex and difficult to identify need state-of-the-art controller design techniques to do the job better prediction and control n n In existing processes, we really on Dynamic models obtained from experiments Usually of an empirical nature Linear n n In new applications (or difficult problems) Focus on mechanistic modeling Dynamic models derived from theory Nonlinear

Process Modeling n n General modeling procedure Identify modeling objectives end use of model (e.g. control) Identify fundamental quantities of interest Mass, Energy and/or Momentum Identify boundaries Apply fundamental physical and chemical laws Mass, Energy and/or Momentum balances Make appropriate assumptions (Simplify) ideality (e.g. isothermal, adiabatic, ideal gas, no friction, incompressible flow, etc,) Write down energy, mass and momentum balances (develop the model equations)

Process Modeling n n Modeling procedure Check model consistency do we have more unknowns than equations Determine unknown constants e.g. friction coefficients, fluid density and viscosity Solve model equations typically nonlinear ordinary (or partial) differential equations initial value problems Check the validity of the model compare to process behavior

Process Modeling n n For control applications: Modeling objectives is to describe process dynamics based on the laws of conservation of mass, energy and momentum n n The balance equation 1. Mass Balance (Stirred tank) 2. Energy Balance (Stirred tank heater) 3. Momentum Balance (Car speed) Rate of Accumulation of fundamental quantity Flow In Flow Out Rate of Production = - +

Process Modeling n n Application of a mass balance Holding Tank n n Modeling objective: Control of tank level n n Fundamental quantity: Mass n n Assumptions: Incompressible flow h F F in

Process Modeling Total mass in system = Total mass in system = V = Ah Flow in = Flow in = F in Flow out = Flow out = F Total mass at time t = Ah(t) Total mass at time = Total mass at time t+ t = Ah(t t Accumulation = Ah(t t Ah(t) = t( F in - F ),

Process Modeling Model consistency Can we solve this equation? Variables: h, , F in, F, A5 Constants: , A2 Inputs: F in, F2 Unknowns:h1 Equations1 Degrees of freedom0 There exists a solution for each value of the inputs F in, F

Process Modeling Solve equation Specify initial conditions h(0)=h 0 and integrate

Process Modeling n n Energy balance Objective:Control tank temperature Fundamental quantity: Energy Assumptions:Incompressible flow Constant hold-up M T in, w Q T, w

Process Modeling n n Under constant hold-up and constant mean pressure (small pressure changes) Balance equation can be written in terms of the enthalpies of the various streams Typically work done on system by external forces is negligible Assume that the heat capacities are constant such that

Process Modeling After substitution, Since T ref is fixed and we assume constant ,C p Divide by C p V

Process Modeling Resulting equation: Model Consistency Variables: T, F, V, T in, Q, C p, 7 Constants: V, C p, 3 Inputs: F, T in, Q3 Unknown: T1 Equations1 There exists a unique solution

Process Modeling Assume F is fixed where V/F is the tank residence time (or time constant) If F changes with time then the differential equation does not have a closed form solution. Product F(t)T(t) makes this differential equation nonlinear. Solution will need numerical integration.

Process Modeling A simple momentum balance Objective: Objective:Control car speed Quantity Quantity:Momentum Assumption Assumption:Friction proportional to speed MomentumOut = Sum of forces acting on system MomentumIn Rate of Accumulation - + Force of Engine (u) Friction Speed (v)

Process Modeling Forces are:Force of the engine = u Friction = bv Balance: Total momentum = Mv Model consistency Variables:M, v, b, u4 Constants:M, b2 Inputs:u1 Unknownsv1

Process Modeling n n Gravity tank Objectives: Objectives: height of liquid in tank Fundamental quantity: Fundamental quantity: Mass, momentumAssumptions: Outlet flow is driven by head of liquid in the tank Incompressible flow Plug flow in outlet pipe Turbulent flow h L F FoFo

Process Modeling From mass and momentum balances, A system of simultaneous ordinary differential equations results Linear or nonlinear?

Process Modeling Model consistency VariablesF o, A, A p, v, h, g, L, K F, Constants A, A p, g, L, K F, Inputs F o 1 Unknownsh, v2 Equations2 Model is consistent

Solution of ODEs n n Mechanistic modeling results in nonlinear sets of ordinary differential equations n n Solution requires numerical integration n n To get solution, we must first: specify all constants (densities, heat capacities, etc, ) specify all initial conditions specify types of perturbations of the input variables For the heated stirred tank, specify C P, and V specify T(0) specify Q(t) and F(t)

Input Specifications n n Study of control system dynamics Observe the time response of a process output in response to input changes n n Focus on specific inputs 1. Step input signals 2. Ramp input signals 3. Pulse and impulse signals 4. Sinusoidal signals 5. Random (noisy) signals

Common Input Signals 1. Step Input Signal: a sustained instantaneous change e.g. Unit step input introduced at time 1

Common Input Signals 2. Ramp Input: A sustained constant rate of change e.g.

Common Input Signals 3. Pulse: An instantaneous temporary change e.g. Fast pulse (unit impulse)

Common Input Signals 3. Pulses: e.g. Rectangular Pulse

Common Input Signals 4. Sinusoidal input

Common Input Signals 5. Random Input

Solution of ODEs using Laplace Transforms Process Dynamics and Control

Linear ODEs n n For linear ODEs, we can solve without integrating by using Laplace transforms n n Integrate out time and transform to Laplace domain Multiplication Y(s) = G(s)U(s) Integration

Common Transforms Useful Laplace Transforms 1. Exponential 2. Cosine

Common Transforms Useful Laplace Transforms 3. Sine

Common Transforms Operators 1. Derivative of a function f(t) 2. Integral of a function f(t)

Common Transforms Operators 3. Delayed function f(t- )

Common Transforms Input Signals 1. Constant 2. Step 3. Ramp function

Common Transforms Input Signals 4. Rectangular Pulse 5. 5. Unit impulse

Laplace Transforms Final Value TheoremLimitations: Initial Value Theorem

Solution of ODEs We can continue taking Laplace transforms and generate a catalogue of Laplace domain functions. See SEM Table 3.1 The final aim is the solution of ordinary differential equations.Example Using Laplace Transform, solve Result

Solution of Linear ODEs Stirred-tank heater (with constant F) taking Laplace To get back to time domain, we must Specify Laplace domain functions Q(s), T in (s) Take Inverse Laplace

Linear ODEs Notes: The expression describes the dynamic behavior of the process explicitly The Laplace domain functions multiplying T(0), T in (s) and Q(s) are transfer functions + + + T in (s) Q(s) T(0) T(s)

Laplace Transform Assume T in (t) = sin( t) then the transfer function gives directly Cannot invert explicitly, but if we can find A and B such that we can invert using tables. Need Partial Fraction Expansion to deal with such functions

Linear ODEs We deal with rational functions of the form r(s)=p(s)/q(s) where degree of q > degree of p q(s) is called the characteristic polynomial of the function r(s) Theorem: Every polynomial q(s) with real coefficients can be factored into the product of only two types of factors powers of linear terms (x-a) n and/or powers of irreducible quadratic terms, (x 2 +bx+c) m

Partial fraction Expansions 1. q(s) has real and distinct factors expand as 2. q(s) has real but repeated factor expanded

Partial Fraction Expansion Heaviside expansion For a rational function of the form Constants are given by Note: Most applicable to q(s) with real and distinct roots. It can be applied to more specific cases.

Partial Fraction Expansions 3. Q(s) has irreducible quadratic factors of the form where Algorithm for Solution of ODEs Take Laplace Transform of both sides of ODE Solve for Y(s)=p(s)/q(s) Factor the characteristic polynomial q(s) Perform partial fraction expansion Inverse Laplace using Tables of Laplace Transforms

Transfer Function Models of Dynamical Processe Process Dynamics and Control

Transfer Function n Heated stirred tank example e.g. e.g. The block is called the transfer function relating Q(s) to T(s) + + + T in (s) Q(s) T(0) T(s)

Process Control Time Domain Transfer function Modeling, Controller Design and Analysis Process Modeling, Experimentation and Implementation Laplace Domain Ability to understand dynamics in Laplace and time domains is extremely important in the study of process control

Transfer function n n Order of underlying ODE is given by degree of characteristic polynomial e.g. First order processes Second order processes n n Steady-state value obtained directly e.g. First order response to unit step function Final value theorem n n Transfer functions are additive and multiplicative

Transfer function n Effect of many transfer functions on a variable is additive + + + T in (s) Q(s) T(0) T(s)

Transfer Function n Effect of consecutive processes in series in multiplicative n Transfer Function U(s) Y 2 (s)Y 1 (s)

Deviation Variables n n To remove dependence on initial condition e.g. Remove dependency on T(0) Transfer functions express extent of deviation from a given steady-state n n Procedure Find steady-state Write steady-state equation Subtract from linear ODE Define deviation variables and their derivatives if required Substitute to re-express ODE in terms of deviation variables

Example n Jacketed heated stirred tank Assumptions Assumptions: Constant hold-up in tank and jacket Constant heat capacities and densities Incompressible flowModel F, T in F c, T cin F c, T c F, T h

Nonlinear ODEs Q: If the model of the process is nonlinear, how do we express it in terms of a transfer function? A: We have to approximate it by a linear one (i.e.Linearize) in order to take the Laplace. f(x 0 ) f(x) x x0x0

Nonlinear systems n n First order Taylor series expansion 1. Function of one variable 2. Function of two variables 3. ODEs

Transfer function n n Procedure to obtain transfer function from nonlinear process models Find steady-state of process Linearize about the steady-state Express in terms of deviations variables about the steady-state Take Laplace transform Isolate outputs in Laplace domain Express effect of inputs in terms of transfer functions

First order Processes Examples Examples, Liquid storage h F FiFi

First Order Processes Examples Examples: Speed of a Car Stirred-tank heater Note:

First Order Processes Liquid Storage Tank Speed of a car Stirred-tank heater K p / A/ M/b1/b 1/ C p F V/F First order processes are characterized by: 1. Their capacity to store material, momentum and energy 2. The resistance associated with the flow of mass, momentum or energy in reaching their capacity

First order processes Liquid storage: Capacity to store mass : A Resistance to flow : 1/ Car: Capacity to store momentum: M Resistance to momentum transfer : 1/b Stirred-tank heater Capacity to store energy: C p V Resistance to energy transfer : 1/ C p F Time Constant = = (Storage capacitance)* (Resistance to flow)

First order process n n Step response of first order process Step input signal of magnitude M 0.632 y(t)/K p M t/

First order process n What do we look for? Process Gain: Steady-State Response Process Time Constant: = n What do we need? Process at steady-state Step input of magnitude M Measure process gain from new steady-state Measure time constant Time Required to Reach 63.2% of final value

First order process Ramp response: Ramp input of slope a 00.511.522.533.544.55 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 a t/ y(t)/K p a

First order Process Sinusoidal response Sinusoidal input Asin( t) 02468101214161820 -1.5 -0.5 0 0.5 1 1.5 2 AR y(t)/A t/

First order Processes 10 -2 10 10 0 1 2 -2 10 10 0 AR/K p pp Bode Plots 10 -2 10 10 0 1 2 -100 -80 -60 -40 -20 0 pp High Frequency Asymptote Corner Frequency Amplitude RatioPhase Shift

Integrating Processes Example: Liquid storage tank Process acts as a pure integrator h F FiFi

Process Modeling n Step input of magnitude M Output Time Input Time Slope = KM

Integrating processes n Unit impulse response Output Time Input Time KM

Integrating Processes n Rectangular pulse response Output Time Input Time

Second Order Processes Three types of second order process: 1. Multicapacity processes: processes that consist of two or more capacities in series e.g. Two heated stirred-tanks in series 2. Inherently second order processes: Fluid or solid mechanic processes possessing inertia and subjected to some acceleration e.g. A pneumatic valve 3. Processing system with a controller: Presence of a controller induces oscillatory behavior e.g. Feedback control system

Second order Processes n Multicapacity Second Order Processes Naturally arise from two first order processes in series By multiplicative property of transfer functions U(s)Y(s) U(s)

Second Order Processes n Inherently second order process: e.g. Pneumatic Valve Momentum Balance x p

Second order Processes n Second order process: Assume the general form where P = Process steady-state gain = Process time constant = Damping Coefficient n Three families of processes Underdamped =1Critically Damped Overdamped n Note: Chemical processes are typically overdamped or critically damped

Second Order Processes n n Roots of the characteristic polynomial Case 1) >1: Two distinct real roots System has an exponential behavior Case 2) =1:One multiple real root Exponential behavior Case 3) :Two complex roots System has an oscillatory behavior

Second order Processes n Step response of magnitude M 012345678910 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Second order process Observations Responses exhibit overshoot (y(t)/KM >1) when

Second order processes n Example - Two Stirred tanks in series M T in, w Q T 1, w M Q T 2, w Response of T 2 to T in is an example of an overdamped second order process

Second order Processes Characteristics of underdamped second order process 1. Rise time, t r 2. Time to first peak, t p 3. Settling time, t s 4. Overshoot: 5. Decay ratio:

Second order Processes -5% +5% a b c trtr tsts P tptp

Second Order Process n Sinusoidal Response where

Second Order Processes n Bode Plots =1 =1

More Complicated processes Transfer function typically written as rational function of polynomials where r(s) and q(s) can be factored as s.t.

Poles and zeroes Definitions: zeros the roots of r(s) are called the zeros of G(s) poles the roots of q(s) are called the poles of G(s) Poles: Directly related to the underlying differential equation If Re(p i )0 then there is at least one term of the form e p i t - y(t) does not vanish

Poles e.g. A transfer function of the form withcan factored to a sum of A constant term from s A e -t/ from the term ( s+1) A function that includes terms of the form Poles can help us to describe the qualitative behavior of a complex system (degree>2) The sign of the poles gives an idea of the stability of the system

Poles n n Calculation performed easily in MATLAB n n Function ROOTS e.g. ROOTS([1 1 1 1]) ans = 0.0000 + 1.0000i 0.0000 - 1.0000i MATLAB

Poles Plotting poles in the complex plane Roots: -1.0, 1.0j, -1.0j

Poles Process Behavior with purely complex poles

Poles Roots: -0.4368, -0.4066+0.9897j, -0.4066-0.9897j

Poles Process behavior with mixed real and complex poles

Poles Roots: -0.7441, -0.3805+1.0830j, -0.3805-1.0830j, 0.2550

Poles Process behavior with unstable pole

Zeros Transfer function: Let is the dominant time constant

Zeros Observations: Adding a zero to an overdamped second order process yields overshoot and inverse response Inverse response is observed when the zeros lie in right half complex plane, Re(z)>0 Overshoot is observed when the zero is dominant () Pole-zero cancellation yields a first order process behavior In physical systems, overshoot and inverse response are a result of two process with different time constants, acting in opposite directions

Zeros Can result from two processes in parallel If gains are of opposite signs and time constants are different then a right half plane zero occurs U(s)Y(s)

Dead Time Time required for the fluid to reach the valve usually approximated as dead time h FiFi Control loop Manipulation of valve does not lead to immediate change in level

Dead time Delayed transfer functions e.g. First order plus dead-time Second order plus dead-time U(s) Y(s)

Dead time n Dead time (delay) Most processes will display some type of lag time Dead time is the moment that lapses between input changes and process response DD Step response of a first order plus dead time process

Dead Time n n Problem use of the dead time approximation makes analysis (poles and zeros) more difficult n n Approximate dead-time by a rational (polynomial) function Most common is Pade approximation

Pade Approximations n n In general Pade approximations do not approximate dead-time very well n n Pade approximations are better when one approximates a first order plus dead time process n n Pade approximations introduce inverse response (right half plane zeros) in the transfer function n n Limited practical use

Process Approximation n Dead time First order plus dead time model is often used for the approximation of complex processes n Step response of an overdamped second order process 012345678 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 - First Order plus dead time o Second Order

Process Approximation n Second order overdamped or first order plus dead time? n Second order process model may be more difficult to identify -- First order plus dead time - Second order overdamped o Actual process

Process Approximation n n Transfer Function of a delay system First order processes Second order processes G(s) Y(s) U(s)

Process Approximation n n More complicated processes Higher order processes (e.g. N tanks in series) For two dominant time constants and process well approximated by For one dominant time constant process well approximated by Y(s) U(s)

Process Approximation n Example

Empirical Modeling Objective: To identify low-order process dynamics (i.e., first and second order transfer function models) Estimate process parameters (i.e., K p, and Methodologies: 1. Least Squares Estimation more systematic statistical approach 2. Process Reaction Curve Methods quick and easy based on engineering heuristics

Empirical Modeling Least Squares Estimation: Simplest model form Process Description where y vector of process measurement xvector of process inputs 1, 0 process parameters Problem: Find 1, 0 that minimize the sum of squared residuals (SSR)

Empirical Modeling Solution Differentiate SSR with respect to parameters These are called the normal equations. Solving for parameters gives: where

Empirical Modeling Compact form Define Then Problem find value of that minimize SSR

Empirical Modeling Solution in Compact Form Normal Equations can be written as which can be shown to give or In practice Manipulations are VERY easy to perform in MATLAB Extends to general linear model (GLM) Polynomial model

Empirical Modeling Control Implementation: previous technique applicable to process model that are linear in the parameters (GLM, polynomials in x, etc) i.e. such that, for all i, the derivatives are not a function of typical process step responses first order Nonlinear in K p and overdamped second order Nonlinear in K p, and nonlinear optimization is required to find the optimum parameters

Empirical Modeling Nonlinear Least Squares required for control applications system output is generally discretized or, simply First Order process (step response) Least squares problem becomes the minimization of This yields an iterative problem solution best handled by software packages: SAS, Splus, MATLAB (function leastsq)

Empirical Modeling Example Nonlinear Least Squares Fit of a first order process from step response data Model Data

Empirical Modeling Results: Using MATLAB function leastsq obtained Resulting Fit

Empirical Modeling Approximation using delayed transfer functions For first order plus delay processes Difficulty Discontinuity at makes nonlinear least squares difficult to apply Solution 1. Arbitrarily fix delay or estimate using alternative methods 2. Estimate remaining parameters 3. Readjust delay repeat step 2 until best value of SSR is obtained

Empirical Modeling Example 2 Underlying True Process Data

Empirical Modeling Fit of a first order plus dead time Second order plus dead time

Empirical Modeling Process reaction curve method: based on approximation of process using first order plus delay model 1. Step in U is introduced 2. Observe behavior y m (t) 3. Fit a first order plus dead time model GpGp GcGc GsGs M/s D(s) Y(s) Y m (s) Y * (s) U(s) Manual Control

Empirical Modeling First order plus dead-time approximations Estimation of steady-state gain is easy Estimation of time constant and dead-time is more difficult KM

Empirical Modeling Estimation of time constant and dead-time from process reaction curves find times at which process reaches 35.3% and 85.3% Estimate t1t1 t2t2

Empirical Process Example For third order process Estimates: Compare: Least Squares FitReaction Curve

Empirical Modeling Process Reaction Curve Method based on graphical interpretation very sensitive to process noise use of step responses is troublesome in normal plant operations frequent unmeasurable disturbances difficulty to perform instantaneous step changes maybe impossible for slow processes restricted to first order models due to reliability quick and easy Least Squares systematic approach computationally intensive can handle any type of dynamics and input signals can handle nonlinear control processes reliable

Feedback Control n Steam heated stirred tank Feedback control system: Valve is manipulated to increase flow of steam to control tank temperature Closed-loop process: Controller and process are interconnected TT TC IP PsPs Condensate Steam F in,T in F,T IP LT LC

Feedback Control Control Objective: maintain a certain outlet temperature and tank level Feedback Control: temperature is measured using a thermocouple level is measured using differential pressure probes undesirable temperature triggers a change in supply steam pressure fluctuations in level trigger a change in outlet flow Note: level and temperature information is measured at outlet of process/ changes result from inlet flow or temperature disturbances inlet flow changes MUST affect process before an adjustment is made

Examples Feedback Control: requires sensors and actuators e.g. Temperature Control Loop Controller: software component implements math hardware component provides calibrated signal for actuator Actuator: physical (with dynamics) process triggered by controller directly affects process Sensor: monitors some property of system and transmits signal back to controller T in, F Te ACP M Controller Tank - + Valve Thermocouple TRTR

Closed-loop Processes n n Study of process dynamics focused on uncontrolled or Open-loop processes Observe process behavior as a result of specific input signals n n In process control, we are concerned with the dynamic behavior of a controlled or Closed-loop process Controller is dynamic system that interacts with the process and the process hardware to yield a specific behaviour GpGp Y(s)U(s) GcGc GmGm GpGp GvGv + - + + controlleractuator process sensor R(s) Y(s) D(s)

Closed-Loop Transfer Function Block Diagram of Closed-Loop Process G p (s)- Process Transfer Function G c (s)- Controller Transfer Function G m (s)- Sensor Transfer Function G v (s)- Actuator Transfer Function GcGc GmGm GpGp GvGv + - + + controlleractuator process sensor R(s) Y(s) D(s)

Closed-Loop Transfer Function For control, we need to identify closed-loop dynamics due to: - Setpoint changesServo - DisturbancesRegulatory 1. Closed-Loop Servo Response transfer function relating Y(s) and R(s) when D(s)=0 Isolate Y(s)

Closed-Loop Transfer Function 2. Closed-loop Regulatory Response Transfer Function relating D(s) to Y(s) at R(s)=0 Isolating Y(s)

Closed-loop Transfer Function 2. Regulatory Response with Disturbance Dynamics G d (s)Disturbance (or load) transfer function 3. Overall Closed-Loop Transfer Function Regulatory Servo

PID Controllers The acronym PID stands for: P- Proportional I- Integral D- Derivative PID Controllers: greater than 90% of all control implementations dates back to the 1930s very well studied and understood optimal structure for first and second order processes (given some assumptions) always first choice when designing a control system PID controller equation:

PID Control PID Control Equation PID Controller Parameters K c Proportional gain I Integral Time Constant D Derivative Time Constant u R Controller Bias Proportional Action Integral Action Derivative Action Controller Bias

PID Control PID Controller Transfer Function or: Note: numerator of PID transfer function cancels second order dynamics denominator provides integration to remove possibility of steady-state errors

PID Control Controller Transfer Function: or, Note: Many variations of this controller exist Easily implemented in SIMULINK each mode (or action) of controller is better studied individually

Proportional Feedback Form: Transfer function: or, Closed-loop form:

Proportional Feedback Example: Given first order process: for P-only feedback closed-loop dynamics: Closed-Loop Time Constant

Proportional Feedback Final response: Note: for zero offset response we require Possible to eliminate offset with P-only feedback (requires infinite controller gain) Need different control action to eliminate offset (integral) Tracking Error Disturbance rejection

Proportional Feedback Servo dynamics of a first order process under proportional feedback - increasing controller gain eliminates off-set KcKc y(t)/KM t/

Proportional Feedback High-order process e.g. second order underdamped process increasing controller gain reduces offset, speeds response and increases oscillation y(t)/KM

Proportional Feedback Important points: proportional feedback does not change the order of the system started with a first order process closed-loop process also first order order of characteristic polynomial is invariant under proportional feedback speed of response of closed-loop process is directly affected by controller gain increasing controller gain reduces the closed-loop time constant in general, proportional feedback reduces (does not eliminate) offset speeds up response for oscillatory processes, makes closed-loop process more oscillatory

Integral Control Integrator is included to eliminate offset provides reset action usually added to a proportional controller to produce a PI controller PID controller with derivative action turned off PI is the most widely used controller in industry optimal structure for first order processes PI controller form Transfer function model

PI Feedback Closed-loop response more complex expression degree of denominator is increased by one

PI Feedback Example PI control of a first order process Closed-loop response Note: offset is removed closed-loop is second order

PI Feedback Example (contd) effect of integral time constant and controller gain on closed-loop dynamics natural period of oscillation damping coefficient integral time constant and controller gain can induce oscillation and change the period of oscillation

PI Feedback Effect of integral time constant on servo dynamics y(t)/KM 0.01 0.1 0.5 1.0 K c =1

PI Feedback Effect of controller gain affects speed of response increasing gain eliminates offset quicker y(t)/KM 0.1 0.5 1.0 5.0 10.0 I =1

PI Feedback Effect of integral action of regulatory response reducing integral time constant removes effect of disturbances makes behavior more oscillatory y(t)/KM

PI Feedback Important points: integral action increases order of the system in closed-loop PI controller has two tuning parameters that can independently affect speed of response final response (offset) integral action eliminates offset integral action should be small compared to proportional action tuned to slowly eliminate offset can increase or cause oscillation can be de-stabilizing

Derivative Action Derivative of error signal Used to compensate for trends in output measure of speed of error signal change provides predictive or anticipatory action P and I modes only response to past and current errors Derivative mode has the form if error is increasing, decrease control action if error is decreasing, decrease control action Always implemented in PID form

PID Feedback Transfer Function Closed-loop Transfer Function Slightly more complicated than PI form

PID Feedback Example: PID Control of a first order process Closed-loop transfer function

PID Feedback Effect of derivative action on servo dynamics 2.0 1.0 0.5 0.1 y(t)/KM

PID Feedback Effect of derivative action on regulatory response increasing derivative action reduces impact of disturbances on control variable slows down servo response and affects oscillation of process 2.0 1.0 0.5 0.1

Derivative Action Important Points: Characteristic polynomial is similar to PI derivative action does not increase the order of the system adding derivative action affects the period of oscillation of the process good for disturbance rejection poor for tracking the PID controller has three tuning parameters and can independently affect, speed of response final response (offset) servo and regulatory response derivative action should be small compared to integral action has a stabilizing influence difficult to use for noisy signals usually modified in practical implementation

Closed-loop Stability Every control problem involves a consideration of closed-loop stability General concepts: BIBO Stability: An (unconstrained) linear system is said to be stable if the output response is bounded for all bounded inputs. Otherwise it is unstable. Comments: Stability is much easier to prove than unstability This is just one type of stability

Closed-loop Stability Closed-loop dynamics if G OL is a rational function then the closed-loop transfer functions are rational functions and take the form and factor as G OL

Closed-loop stability General Stability criterion: A closed-loop feedback control system is stable if and only if all roots of the characteristic polynomial are negative or have negative real parts. Otherwise, the system is unstable. Unstable region is the right half plane of the complex plane. Valid for any linear systems. Underlying system is almost always nonlinear so stability holds only locally. Moving away from the point of linearization may cause instability.

Closed-loop Stability Problem reduces to finding roots of a polynomial Easy (1990s) way : MATLAB function ROOTS Traditional: 1. Routh array: l l Test for positivity of roots of a polynomial 2. Direct substitution l l Complex axis separates stable and unstable regions l l Find controller gain that yields purely complex roots 3. Root locus diagram l l Vary location of poles as controller gain is varied l l Of limited use

Closed-loop stability Routh array for a polynomial equation is where Elements of left column must be positive to have roots with negative real parts

Example: Routh Array Characteristic polynomial Polynomial Coefficients Routh Array Closed-loop system is unstable

Direct Substitution n n Technique to find gain value that de-stabilizes the system. n n Observation: Process becomes unstable when poles appear on right half plane Find value of K c that yields purely complex poles n n Strategy: Start with characteristic polynomial Write characteristic equation: Substitute for complex pole (s=j ) Solve for K c and

Example: Direct Substitution Characteristic equation Substitution for s=j Real PartComplex Part System is unstable if

Root Locus Diagram Old method that consists in plotting poles of characteristic polynomial as controller gain is changed e.g. K c -0 1

Stability and Performance n n Given plant model, we assume a stable closed-loop system can be designed n n Once stability is achieved - need to consider performance of closed-loop process - stability is not enough n n All poles of closed-loop transfer function have negative real parts - can we place these poles to get a good performance S: Stabilizing Controllers for a given plant P: Controllers that meet performance S P C Space of all Controllers

Controller Tuning Can be achieved by Direct synthesis : Specify servo transfer function required and calculate required controller - assume plant = model Internal Model Control: Morari et al. (86) Similar to direct synthesis except that plant and plant model are concerned Tuning relations: Cohen-Coon - 1/4 decay ratio designs based on ISE, IAE and ITAE Frequency response techniques Bode criterion Nyquist criterion Field tuning and re-tuning

Direct Synthesis From closed-loop transfer function Isolate G c For a desired trajectory (C/R) d and plant model G pm, controller is given by not necessarily PID form inverse of process model to yield pole-zero cancellation (often inexact because of process approximation) used with care with unstable process or processes with RHP zeroes

Direct Synthesis 1. Perfect Control cannot be achieved, requires infinite gain 2. Closed-loop process with finite settling time For 1st order G p, it leads to PI control For 2nd order, get PID control 3. Processes with delay requires again, 1st order leads to PI control 2nd order leads to PID control

IMC Controller Tuning Closed-loop transfer function In terms of implemented controller, G c

IMC Controller Tuning 1. Process model factored into two parts where contains dead-time and RHP zeros, steady-state gain scaled to 1. 2. Controller where f is the IMC filter based on pole-zero cancellation not recommended for open-loop unstable processes very similar to direct synthesis

Example PID Design using IMC and Direct synthesis for the process Process parameters: K=0.3, IMC Design: K c =6.97, I =34.5, d =3.93 Filter 2. Direct Synthesis: K c =4.76, I =30 Servo Transfer function

Example Result: Servo Response IMC and direct synthesis give roughly same results IMC not as good due to Pade approximation y(t) t IMC Direct Synthesis

Example Result: Regulatory response Direct synthesis rejects disturbance more rapidly (marginally) y(t) t IMC Direct Synthesis

Tuning Relations Process reaction curve method: based on approximation of process using first order plus delay model 1. Step in U is introduced 2. Observe behavior y m (t) 3. Fit a first order plus dead time model GpGp GcGc GsGs 1/s D(s) Y(s) Y m (s) Y * (s) U(s) Manuel Control

Tuning Relations Process response 4. Obtain tuning from tuning correlations Ziegler-Nichols Cohen-Coon ISE, IAE or ITAE optimal tuning relations KM

Ziegler-Nichols Tunings - Note presence of inverse of process gain in controller gain - Introduction of integral action requires reduction in controller gain - Increase gain when derivation action is introduced Example: PI:K c = 10 I =29.97 PID: K c = 13.33 I =18 I =4.5

Example Ziegler-Nichols Tunings: Servo response y(t) t

Example Regulatory Response Z-N tuning Oscillatory with considerable overshoot Tends to be conservative

Cohen-Coon Tuning Relations Designed to achieve 1/4 decay ratio fast decrease in amplitude of oscillation Example: PI:K c =10.27 I =18.54 K c =15.64 I =19.75 d =3.10

Tuning relations Cohen-coon: Servo More aggressive/ Higher controller gains Undesirable response for most cases

Tuning Relations Cohen-Coon: Regulatory Highly oscillatory Very aggressive y(t) t

Integral Error Relations 1. Integral of absolute error (IAE) 2. Integral of squared error (ISE) penalizes large errors 3. Integral of time-weighted absolute error (ITAE) penalizes errors that persist ITAE is most conservative ITAE is preferred

ITAE Relations Choose K c, I and d that minimize the ITAE: For a first order plus dead time model, solve for: Design for Load and Setpoint changes yield different ITAE optimum

ITAE Relations From table, we get Load Settings: Setpoint Settings: Example

ITAE Relations Example (contd) Setpoint Settings Load Settings:

ITAE Relations Servo Response design for load changes yields large overshoots for set-point changes

ITAE Relations Regulatory response Tuning relations are based G L =G p Method does not apply to the process Set-point design has a good performance for this case

Tuning Relations n n In all correlations, controller gain should be inversely proportional to process gain n n Controller gain is reduced when derivative action is introduced n n Controller gain is reduced as increases n n Integral time constant and derivative constant should increase as increases n n In general, n n Ziegler-Nichols and Cohen-Coon tuning relations yield aggressive control with oscillatory response (requires detuning) n n ITAE provides conservative performance (not aggressive)

CHE 446 Process Dynamics and Control Frequency Response of Linear Control Systems

First order Process Response to a sinusoidal input signal Recall: Sinusoidal input Asin( t) yields sinusoidal output caharacterized by AR and 02468101214161820 -1.5 -0.5 0 0.5 1 1.5 2 AR y(t)/A t/

First order Processes 10 -2 10 10 0 1 2 -2 10 10 0 AR/K p pp Bode Plots 10 -2 10 10 0 1 2 -100 -80 -60 -40 -20 0 pp High Frequency Asymptote Corner Frequency Amplitude RatioPhase Shift

Second Order Process n Sinusoidal Response where

Second Order Processes Bode Plot =1 =1 Amplitude reaches a maximum at resonance frequency AR

Frequency Response Q: Do we have to take the Laplace inverse to compute the AR and phase shift of a 1st or 2nd order process? No Q: Does this generalize to all transfer function models? Yes Study of transfer function model response to sinusoidal inputs is called Frequency Domain Response of linear processes.

Frequency Response Some facts for complex number theory: i) For a complex number: It follows that where such that Re Im w a b

Frequency Response Some facts: ii) Let z=a-bj and w= a+bj then iii) For a first order process Let s=j such that

Frequency Response Main Result: The response of any linear process G(s) to a sinusoidal input is a sinusoidal. The amplitude ratio of the resulting signal is given by the Modulus of the transfer function model expressed in the frequency domain, G(i ). The Phase Shift is given by the argument of the transfer function model in the frequency domain. i.e.

Frequency Response For a general transfer function Frequency Response summarized by where is the modulus of G(j ) and is the argument of G(j ) Note: Substitute for s=j in the transfer function.

Frequency Response The facts: For any linear process we can calculate the amplitude ratio and phase shift by: i) Letting s=j in the transfer functionG(s) ii) G(j ) is a complex number. Its modulus is the amplitude ratio of the process and its argument is the phase shift. iii) As , the frequency, is varied that G(j ) gives a trace (or a curve) in the complex plane. iv) The effect of the frequency, , on the process is the frequency response of the process.

Frequency Response Examples: 1. Pure Capacitive Process G(s)=1/s 2. Dead Time G(s)=e - s

Frequency Response Examples: 3. n process in series Frequency response of G(s) therefore

Frequency Response Examples. 4. n first order processes in series 5. First order plus delay

Frequency Response n n To study frequency response, we use two types of graphical representations 1. The Bode Plot: Plot of AR vs. on loglog scale Plot of vs. on semilog scale 2. The Nyquist Plot: Plot of the trace of G(j ) in the complex plane n n Plots lead to effective stability criteria and frequency-based design methods

Bode Plot Pure Capacitive Process

Bode Plot G1G1 G2G2 G3G3 G

Example: Effect of dead-time G=G d GGdGd

Nyquist Plot Plot of G(j ) in the complex plane as is varied Relation to Bode plot AR is distance of G(j ) for the origin Phase angle, , is the angle from the Real positive axis Example First order process (K=1, =1)

Nyquist Plot Dead-time Second Order

Nyquist Plot Third Order Effect of dead-time (second order process)

Che 446: Process Dynamics and Control Frequency Domain Controller Design

PI Controller AR

PID Controller AR

Bode Stability Criterion Consider open-loop control system 1. Introduce sinusoidal input in setpoint (D(s)=0) and observe sinusoidal output 2. Fix gain such AR=1 and input frequency such that =-180 3. At same time, connect close the loop and set R(s)=0 Q: What happens if AR>1? GpGp GcGc GsGs D(s) Y(s) Y m (s) R(s) U(s) Open-loop Response to R(s) + - + +

Bode Stability Criterion A closed-loop system is unstable if the frequency of the response of the open-loop G OL has an amplitude ratio greater than one at the critical frequency. Otherwise it is stable. Strategy: 1. Solve for in 2. Calculate AR

Bode Stability Criterion To check for stability: 1. Compute open-loop transfer function 2. Solve for in =- 3. Evaluate AR at 4. If AR>1 then process is unstable Find ultimate gain: 1. Compute open-loop transfer function without controller gain 2. Solve for in =- 3. Evaluate AR at 4. Let

Bode Criterion Consider the transfer function and controller - Open-loop transfer function - Amplitude ratio and phase shift - At =1.4128, =- AR=6.746

Ziegler-Nichols Tuning Closed-loop tuning relation With P-only, vary controller gain until system (initially stable) starts to oscillate. Frequency of oscillation is c, Ultimate gain, K u, is 1/M where M is the amplitude of the open-loop system Ultimate Period Ziegler-Nichols Tunings PK u /2 PI K u /2.2 P u /1.2 PIDK u /1.7 P u /2 P u /8

Nyquist Stability Criterion If N is the number of times that the Nyquist plot encircles the point (-1,0) in the complex plane in the clockwise direction, and P is the number of open- loop poles of G OL that lie in the right-half plane, then Z=N+P is the number of unstable roots of the closed-loop characteristic equation. Strategy 1. Substitute s=j in G OL (s) 2. Plot G OL (j ) in the complex plane 3. Count encirclements of (-1,0) in the clockwise direction

Nyquist Criterion Consider the transfer function and the PI controller

Stability Considerations n n Control is about stability n n Considered exponential stability of controlled processes using: Routh criterion Direct Substitution Root Locus Bode Criterion (Restriction on phse angle) Nyquist Criterion n n Nyquist is most general but sometimes difficult to interpret n n Roots, Bode and Nyquist all in MATLAB n n MAPLE is recommended for some applications. Polynomial (no dead-time)

CHE 446 Process Dynamics and Control Advanced Control Techniques: 1. Feedforward Control

Feedforward Control Feedback control systems have the general form: where U R (s) is an input bias term. n n Feedback controllers output of process must change before any action is taken disturbances only compensated after they affect the process GpGp GcGc GsGs Y(s) Y m (s) R(s) U(s) + + + GDGD GvGv + + D(s) U R (s)

Feedforward Control n n Assume that D(s) can be measured before it affects the process effect of disturbance on process can be described with a model G D (s) Feedforward Control is possible. Feedback/Feedforward Controller Structure GpGp GcGc GsGs Y(s) Y m (s) R(s) U(s) + + + GDGD GvGv GfGf + + D(s) Feedforward Controller

Feedforward Control Heated Stirred Tank Is this control configuration feedback or feedforward? How can we use the inlet stream thermocouple to regulate the inlet folow disturbances Will this become a feedforward or feedback controller? TT TC1 PsPs Condensate Steam F,T in F,T TT

Feedforward Control A suggestion: How do we design TC2? TT TC1 PsPs Condensate Steam F,T in F,T TC2 + + TT

Feedforward Control The feedforward controller: Transfer Function Tracking of Y R requires that GpGp Y(s) U(s) + + GDGD GvGv GfGf + + D(s) U R (s)

Feedforward Control Ideal feedforward controller: Exact cancellation requires perfect plant and perfect disturbance models. Feedforward controllers: very sensitive to modeling errors cannot handle unmeasured disturbances cannot implement setpoint changes Need feedback control to make control system more robust

Feedforward Control GpGp GcGc GsGs Y(s) Y m (s) R(s) U(s) + + + GDGD GvGv GfGf + + D(s) What is the impact of G f on the closed-loop performance of the feedback control system? Feedback/Feedforward Control

Feedforward Control Regulatory transfer function of feedforward/feedback loop Perfect control requires that (as above) Note: Feedforward controllers do not affect closed- loop stability Feedforward controllers based on plant models can be unrealizable (dead-time or RHP zeroes) Can be approximated by a lead-lag unit or pure gain (rare)

Feedforward Control Tuning: In absence of disturbance model lead-lag approximation may be good K f obtained from open-loop data 1 and 2 from open-loop data from heuristics Trial-and-error

Feedforward Control Example: Plant: Plant Model: Feedback Design from plant model: IMC PID tunings

Feedforward Control Possible Feedforward controllers: 1. From plant models: Not realizable 2. Lead-lag unit 3. Feedforward gain controller:

Feedforward Control For Controller 2 and 3 Some attenuation observed at first peak Difficult problem because disturbance dynamic are much faster

Feedforward Control n n Useful in manufacturing environments if good models are available outdoor temperature dependencies can be handle by gain feedforward controllers scheduling issues/ supply requirements can be handled n n Benefits are directly related to model accuracy rely mainly on feedback control n n Disturbances with different dynamics always difficult to attenuate with PID may need advanced feedback control approach (MPC, DMC, QDMC, H 4 -controllers, etc) n n Use process knowledge (and intuition)

CHE 446: Process Dynamics and Control Advanced Control Techniques 2. Cascade Control

Cascade Control Jacketed Reactor: Conventional Feedback Loop: operate valve to control steam flow steam flow disturbances must propagate through entire process to affect output does not take into account flow measurement TT TC1 PsPs Condensate Steam F,T in F,T TT FT

Cascade Control Consider cascade control structure: Note: TC1 calculates setpoint cascaded to the flow controller Flow controller attenuates the effect of steam flow disturbances TT TC1 PsPs Condensate Steam F,T in F,T TT FT FC

Cascade Control Cascade systems contain two feedback loops: Primary Loop regulates part of the process having slower dynamics calculates setpoint for the secondary loop e.g. outlet temperature controller for the jacketed reactor Secondary Loop regulates part of process having faster dynamics maintain secondary variable at the desired target given by primary controller e.g. steam flow control for the jacketed reactor example

Cascade Control Block Diagram G c2 G p1 G p2 G v2 G m2 G m1 G c1 D1D1 D2D2 -- + ++ +

Cascade Control Closed-loop transfer function 1. Inner loop 2. Outer loop Characteristic equation

Cascade Control Stability of closed-loop process is governed by Example

Cascade Control Design a cascade controller for the following system: 1. Primary: 2. Secondary:

Cascade Control 1. 1. PI controller only Critical frequency Maximum gain

AR Bode Plots Cascade Control ln( )

Cascade Control 2. Cascade Control Secondary loop no critical frequency gain can be large Let K c2 =10. Primary loop

Cascade Control Closed-loop stability: Bode Maximum gain K c1 =10.44 Secondary loop stabilizes the primary loop.

Cascade Control Use cascade when: conventional feedback loop is too slow at rejecting disturbances secondary measured variable is available which responds to disturbances has dynamics that are much faster than those of the primary variable can be affected by the manipulated variable Implementation tune secondary loop first operation of two interacting controllers requires more careful implementation switching on and off

CHE 446 Process Dynamics and Control Advanced Control Techniques 3. Dead-time Compensation

Dead-time Compensation Consider feedback loop: Dead-time has a de-stabilizing effect on closed- loop system Presence of dead-time requires detuning of controller Need a way to compensate for dead-time explicitly GcGc GpGp e-se-s R C D

Dead-time Compensation Motivation 0.10.750.50.25

Dead-time Compensation Use plant model to predict deviation from setpoint Result: Removes the de-stabilizing effect of dead-time Problem: Cannot compensate for disturbances with just feedback (possible offset) Need a very good plant model GcGc GpGp e-se-s R C D G pm

Dead-time Compensation Closed-loop transfer function Characteristic Equation becomes Effect of dead-time on closed-loop stability is removed Controller is tuned to stabilize undelayed process model No disturbance rejection

Dead-time Compensation e -0.5s RC D

Dead-time Compensation Include effect of disturbances using model predictions Adding this to previous loop gives GcGc GpGp e-se-s R C D G pm e-se-s + + + - + + + -

Dead-time Compensation Closed-loop transfer function Characteristic Equation Effect of dead-time on stability is removed Disturbance rejection is achieved Controller tuned for undelayed dynamics Fast Dynamics Slow Dynamics

Dead-time Compensation e -0.5s RC D + + + - + + + -

Dead-time Compensation Alternative form Reduces to classical feedback control system with called a Smith-Predictor GcGc GpGp e-se-s R C D G pm (1-e - s ) + + + + + -

Dead-time compensation Smith-Predictor Design 1. Determine delayed process model 2. Tune controller G c for the undelayed transfer function model G pm 3. Implement Smith-Predictor as 4. Perform simulation studies to tune controller and estimate closed-loop performance over a range of modeling errors (G pm and m )

Dead-time Compensation Effect of dead-time estimation errors: e -0.5s RC D e-se-s + + + - + + + -

Documents

CHEE 434/821 Process Control II Some Review Material Winter 2006 Instructor:M.GuayTA: V. Adetola