CHEE 434/821 Process Control II Some Review Material

Transcription

1 CHEE 434/8 Process Control II Some Review Material Winter 6 Instructor: M.Guay TA: V. Adetola In the chemical industry, Introduction the design of a control system is essential to ensure: Good Process Operation Process Safety Product Quality Minimization of Environmental Impact

2 Introduction What is the purpose of a control system? To maintain important process characteristics at desired targets despite the effects of external perturbations. Perturbations Plant Processing objectives Market Economy Climate Upsets... Safety Make $$$ Environment... Control Introduction What constitutes a control system? Combination of process sensors, actuators and computer systems designed and tuned to orchestrate safe and profitable operation. Control Plant

3 Introduction Process Dynamics: Study of the transient behavior of processes 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 What do we mean by process? Introduction A process, P, is an operation that takes an INPUT or a DISTURBANCE and gives an OUTPUT u d P y Information Flow 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 3

4 Examples Stirred tank heater T in, w M Q T, w Inputs T in w Q Process Output T Examples The speed of an automobile Friction Force of Engine Inputs Friction Engine Process Output Speed 4

5 Examples e.g. Landing on Mars Examples e.g. Millirobotics Laparoscopic Manipulators 5

6 Introduction Process A process, P, is an operation that takes an INPUT or a DISTURBANCE and gives an OUTPUT u d P y Information Flow 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 Control What is control? To regulate of a process output despite the effect of disturbances Driving a car Controlling the temperature of a chemical reactor Reducing vibrations in a flexible structure To stabilize unstable processes Riding a bike Flight of an airplane Operation of a nuclear plant 6

7 Benefits of Control Economic Benefits Quality (waste reduction) Variance reduction (consistency) Savings in energy, materials, manpower Operability, safety (stability) Performance Efficiency Accuracy robotics Reliability Stabilizability bicycle aircraft nuclear reactor Control What is a controller? Process Controller 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) 7

8 Block representations Block diagrams are models of the physical systems Input variables Process Output variables System Physical Boundary Transfer of fundamental quantities Mass, Energy and Momentum Physical Operation Abstract Control A controlled process is a system which is comprised of two interacting systems: e.g. Most controlled systems are feedback controlled systems Disturbances Action intervene Process Controller Outputs Observation monitor The controller is designed to provide regulation of process outputs in the presence of disturbances 8

9 Introduction What is required for the development of a control system?. The Plant (e.g. SPP of Nylon) Nylon Gas Make-up Reheater Dehumidifier Relief Pot Vent Steam Heater Blower Water Introduction What is required?. Process Understanding Required measurements Required actuators Understand design limitations. Process Instrumentation Appropriate sensor and actuator selection Integration in control system Communication and computer architecture 3. Process Control Appropriate control strategy 9

10 Example Cruise Control Friction Engine Process Speed Controller Human or Computer Classical Control Control is meant to provide regulation of process outputs about a reference, r, despite inherent disturbances d r + e - Controller u Process y Classical Feedback Control System The deviation of the plant output, e=(r-y), from its intended reference is used to make appropriate adjustments in the plant input, u

11 Control Process is a combination of sensors and actuators Controller is a computer (or operator) that performs the required manipulations e.g. Classical feedback control loop d r + - e Computer C Actuator A P Process y M Sensor Examples Driving an automobile r + - e Driver C Steering A M P Automobile y Visual and tactile measurement Desired trajectory r Actual trajectory y

12 Examples Stirred-Tank Heater T in, w Heater Q T, w T R + - e Controller C TC Thermocouple Heater A M T in, w P Tank y Thermocouple Examples Measure T, adjust Q T in, w T R + - e Controller C Heater A M P Tank T Thermocouple Feedback control Controller: Q=K(T R -T)+Q nominal where Q nominal =wc(t-t in ) Q: Is this positive or negative feedback?

13 Examples Measure T i, adjust Q T i M C A P + ΔQ Q i + Q Feedforward Control Control Nomenclature Identification of all process variables Inputs Outputs (affect process) (result of process) Inputs Disturbance variables Variables affecting process that are due to external forces Manipulated variables Things that we can directly affect 3

14 Control Nomenclature 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 Disturbances Manipulated Process Other Control Controller Example w i, T i P c L w c, T ci h T w c, T co P o w o, T o Variables w i, w o : T i, T o : w c : P c : P o : T ci, T co : h: T Tank inlet and outlet mass flows Tank inlet and outlet temperatures Cooling jacket mass flow Position of cooling jacket inlet valve Position of tank outlet valve Cooling jacket inlet and outlet temperatures Tank liquid level 4

15 Example Variables Inputs Outputs w i T i T ci w c h w o T o P c P o Disturbances Manipulated Measured Unmeasured Control Task: Classify the variables Process Control and Modeling 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 r e Controller u d Process y Model Design Implementation 5

16 Control System Development Control development is usually carried out following these important steps Define Objectives Develop a process model Design controller based on model Test by Simulation Implement and Tune Monitor Performance Often an iterative process, based on performance we may decide to retune, redesign or remodel a given control system Control System Development Objectives What are we trying to control? Process modeling What do we need? Mechanistic and/or empirical 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? 6

17 Control System Development 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 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 Motivation: Develop understanding of process a mathematical hypothesis of process mechanisms Match observed process behavior useful in design, optimization and control of process Control: Interested in description of process dynamics Dynamic model is used to predict how process responds to given input Tells us how to react 7

18 Process Modeling What kind of model do we need? 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 What kind of model do we need? Process Modeling 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 8

19 Process Modeling Dynamic vs. Steady-state 65 6 Steady-State Output 55 5 Steady-State Time Step change in input to observe Starting at steady-state, we made a step change The system oscillates and finds a new steadystate Dynamics describe the transitory behavior Process Modeling 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 Mass Energy Momentum Useful for simulation and exploration of new operating conditions May contain unknown constants that must be estimated 9

20 Process Modeling Empirical vs Mechanistic models Empirical models do not rely on underlying mechanisms Fit specific function to match process Mathematical French curve.3.. Output Time Process Modeling 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 In existing processes, we really on Dynamic models obtained from experiments Usually of an empirical nature Linear In new applications (or difficult problems) Focus on mechanistic modeling Dynamic models derived from theory Nonlinear

21 General modeling procedure Process Modeling 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) Modeling procedure Process Modeling 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

22 For control applications: Process Modeling Modeling objectives is to describe process dynamics based on the laws of conservation of mass, energy and momentum The balance equation Rate of Accumulation of fundamental quantity Flow = - In Flow Out. Mass Balance (Stirred tank). Energy Balance (Stirred tank heater) 3. Momentum Balance (Car speed) + Rate of Production Process Modeling Application of a mass balance Holding Tank F in h F Modeling objective: Control of tank level Fundamental quantity: Mass Assumptions: Incompressible flow

23 Process Modeling Total mass in system = ρv = ρah Flow in = ρf in Flow out = ρf Total mass at time t = ρah(t) Total mass at time t+δt = ρah(t+δt) Accumulation ρah(t+δt) ρah(t) = Δt(ρF in -ρf ), ρah( t + tδ) ρah( t) = ρ( Fin F), Δt ρah( t + tδ) ρah( t) lim = ρ( Fin F), Δt Δt ρa dh = ρ( Fin F). dt Process Modeling Model consistency Can we solve this equation? Variables: h, ρ, F in, F, A 5 Constants: ρ, A Inputs: F in, F Unknowns: h Equations Degrees of freedom There exists a solution for each value of the inputs F in, F 3

24 Process Modeling Solve equation Specify initial conditions h()=h and integrate t F ht () = h() + in ( τ ) F( τ ) dτ A.5 Fin flow.5 F h Energy balance Process Modeling T in, w M Q T, w Objective: Control tank temperature Fundamental quantity: Energy Assumptions: Incompressible flow Constant hold-up 4

25 Process Modeling 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 dh = H& in H& out + Q+ Ws dt Typically work done on system by external forces is negligible dh = H& in H& out + Q dt Assume that the heat capacities are constant such that H = ρ CPV ( T Tref ) H& in = ρcpw( Tin Tref ) H& = ρc w( T T ) out P ref After substitution, Process Modeling d( ρcpv ( T Tref )) = ρcwt P ( in Tref ) ρcwt P ( Tref ) + Q dt Since T ref is fixed and we assume constant ρ,c p ρcv d ( T T ref ) P = ρcwt P ( in Tref ) ρcwt P ( Tref ) + Q dt Divide by ρ C p V dt dt w V T T Q = ( in ) + ρcv P 5

26 Process Modeling Resulting equation: dt dt Model Consistency F V T T Q = ( in ) + ρvc P Variables: T, F, V, T in, Q, C p, ρ 7 Constants: V, C p, ρ 3 Inputs: F, T in, Q 3 Unknown: T Equations There exists a unique solution Assume F is fixed Process Modeling t t/ τ ( ς t)/ τ Tt () = T() e + e ( Tin( ζ ) + Q( ζ ) ) CV d ζ τ ρ where τ=v/f is the tank residence time (or time constant) p If F changes with time then the differential equation does not have a closed form solution. dt() t Ft () dt V T t T t Qt () = ( in ( ) ( )) + ρ VCP Product F(t)T(t) makes this differential equation nonlinear. Solution will need numerical integration. 6

27 A simple momentum balance Process Modeling Rate of Accumulation = Momentum In - Momentum Out + Sum of forces acting on system Speed (v) Friction Force of Engine (u) Objective: Quantity: Assumption: Control car speed Momentum Friction proportional to speed Process Modeling Forces are: Force of the engine = u Friction = bv Balance: Total momentum = Mv d( Mv( t)) M dv ( t ) = = ut () bvt () dt dt Model consistency Variables: M, v, b, u 4 Constants: M, b Inputs: u Unknowns v 7

28 Process Modeling Gravity tank F o h F Objectives: height of liquid in tank L Fundamental quantity: Mass, momentum Assumptions: Outlet flow is driven by head of liquid in the tank Incompressible flow Plug flow in outlet pipe Turbulent flow Process Modeling From mass and momentum balances, dh F A v = o P dt A A dv hg KFv = dt L ρap A system of simultaneous ordinary differential equations results Linear or nonlinear? 8

29 Process Modeling Model consistency Variables F o, A, A p, v, h, g, L, K F, ρ 9 Constants A, A p, g, L, K F, ρ 6 Inputs F o Unknowns h, v Equations Model is consistent Solution of ODEs Mechanistic modeling results in nonlinear sets of ordinary differential equations Solution requires numerical integration 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, dt F dt V T T Q = ( in ) + ρvc specify ρ, C P, and V P specify T() specify Q(t) and F(t) 9

30 Input Specifications Study of control system dynamics Observe the time response of a process output in response to input changes Focus on specific inputs. Step input signals. Ramp input signals 3. Pulse and impulse signals 4. Sinusoidal signals 5. Random (noisy) signals Common Input Signals. Step Input Signal: a sustained instantaneous change e.g. Unit step input introduced at time.5 Input Time

31 Common Input Signals. Ramp Input: A sustained constant rate of change e.g Input Time Output Time Common Input Signals 3. Pulse: An instantaneous temporary change e.g. Fast pulse (unit impulse) Input Time Output Time 3

32 Common Input Signals 3. Pulses: e.g. Rectangular Pulse.5 Input Time..8.6 Output Time Common Input Signals 4. Sinusoidal input.5.5 Input Time Output Time 3

33 5. Random Input Common Input Signals.5.5 Input Time.6.4. Output Time Solution of ODEs using Laplace Transforms Process Dynamics and Control 33

34 Linear ODEs For linear ODEs, we can solve without integrating by using Laplace transforms st Fs () =I [ f()] t = f() te dt t= Integrate out time and transform to Laplace domain dy() t = ay() t + bu() t dt y( ) = c Integration Multiplication Y(s) = G(s)U(s) Useful Laplace Transforms. Exponential f ()= t e bt Common Transforms bt bt st ( s+ b) t I [ e ] = e e dt = e dt ( s+ b) t bt e I [ e ] = = s+ b s + b. Cosine jωt jωt e + e f () t = cos( ωt) = ( s jω) t ( s+ jω) t I [cos( ωt)] = e dt+ e dt = + s jω s + jω = s s + ω 34

35 Common Transforms Useful Laplace Transforms 3. Sine jωt j t e e ω f () t = sin( ωt) = j ( s jω) t ( s+ jω) t I [sin( ωt)] = e dt e dt j = ω + ω = ω j s j s j s + ω Operators. Derivative of a function f(t). Integral of a function f(t) Common Transforms df () t dt du = df st v = e df st st I [ ] = uv] udv= f( te ) ] ( sf( te ) ) dt dt df st I [ ] = s f ( t) e dt f ( ) = sf( s) f ( ) dt t I = t st Fs () f ( τ) dτ e ( f ( τ) dτ) dt = s 35

36 Common Transforms Operators 3. Delayed function f(t-τ) t < τ gt () = f ( t τ) t τ τ I [ gt] = e st st () ( ) dt+ e f( t τ ) dt τ [ ] I gt = e sτ () Fs () Common Transforms Input Signals. Constant f ()= t a st st I = = [ a] ae dt ( ae ) = a s s. Step t < f ()= t a t st st I = = [ f ( t)] ae dt ( ae ) = a s s t < 3. Ramp function f ()= t at t st e I f t = ate dt = [ ()] st st at ae a + dt = s s s 36

37 Input Signals 4. Rectangular Pulse 5. Unit impulse Common Transforms t < f ()= t a t < tw t tw tw I [ ] = st a t s f () t ae dt = ( e w ) s I [ δ () t ] = lim ( t e ws ) t tws w t se ws I [ δ () t ] = lim = t s w Final Value Theorem Limitations: Initial Value Theorem Laplace Transforms [ yt] = [ sys] lim ( ) lim ( ) t s yt () C, lim [ sy( s) ] exists s Re( s) s [ sy s ] y( ) = lim ( ) s 37

38 Solution of ODEs We can continue taking Laplace transforms and generate a catalogue of Laplace domain functions. See SEM Table 3. The final aim is the solution of ordinary differential equations. Example Using Laplace Transform, solve Result dy 5 + 4y =, y( ) = dt. t yt () = e 8 Solution of Linear ODEs Stirred-tank heater (with constant F) taking Laplace dt F dt V T T Q = ( in ) + ρvcp T( ) = T V dt I F =I I + I dt P τ ( Ts () T()) = T() s Ts () + KQs () [ T () t ] [ T() t ] [ Qt () in ] ρfc in τ KP Ts () = T() + T () s + Qs () in τs + τs + τs + To get back to time domain, we must Specify Laplace domain functions Q(s), T in (s) Take Inverse Laplace P 38

39 Linear ODEs Notes: The expression τ KP Ts () = T() + T () s + Qs () in τs + τs + τs + describes the dynamic behavior of the process explicitly The Laplace domain functions multiplying T(), T in (s) and Q(s) are transfer functions T in (s) Q(s) T() τs + KP τs + τ τs T(s) Laplace Transform Assume T in (t) = sin(ωt) then the transfer function gives directly ω Tin() s = τs + ( s + ω )( τs+ ) Cannot invert explicitly, but if we can find A and B such that A B + s s + + = ω ω τ ( s + ω )( τs+ ) we can invert using tables. Need Partial Fraction Expansion to deal with such functions 39

40 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 +bx+c) m Partial fraction Expansions. q(s) has real and distinct factors expand as n qs ( ) = ( s+ b i ) i= n α rs ()= i i= s+ bi. q(s) has real but repeated factor expanded qs () = ( s+ b) n α rs () = α + α + + n s + b n ( s+ b) Λ ( s+ b) 4

41 Partial Fraction Expansion Heaviside expansion For a rational function of the form ps () ps () n α rs () = = = i qs () n s b s b i ( ( i ) + i ) = + i= Constants are given by α i s b ps () = ( + i ) qs () s = bi 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 qs () = ( s + ds + d) n where d d 4 < 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 4

42 Transfer Function Models of Dynamical Processe Process Dynamics and Control Heated stirred tank example Transfer Function τ KP Ts () = T() + T () s + Qs () in τs + τs + τs + T in (s) Q(s) T() τs + KP τs + τ τs T(s) e.g. The block KP τ s + is called the transfer function relating Q(s) to T(s) 4

43 Process Control Time Domain Laplace Domain Process Modeling, Experimentation and Implementation Transfer function Modeling, Controller Design and Analysis Ability to understand dynamics in Laplace and time domains is extremely important in the study of process control Transfer function Order of underlying ODE is given by degree of characteristic polynomial e.g. First order processes K Ys () = P U() s τs + Second order processes K Ys () = P U() s τ s + ξτs+ Steady-state value obtained directly e.g. First order response to unit step function Kp Ys () = s( τs+ ) Final value theorem [ ] [ ] lim sy( s) = lim G( s) = K P s s Transfer functions are additive and multiplicative 43

44 Transfer function Effect of many transfer functions on a variable is additive τ KP Ts () = T() + Tin() s + Qs () τs + τs + τs + T in (s) Q(s) T() τs + KP τs + τ τs T(s) Transfer Function Effect of consecutive processes in series in multiplicative U(s) KP τ s + Transfer Function Y (s) KP τ s + Y (s) K Y s P () = U() s τs + K Y s P () = Y() s τs + K K Y s P P () = U() s τs + τs + 44

45 Deviation Variables To remove dependence on initial condition e.g. τ KP Ts () = T() + T () s + Qs () in τs + τs + τs + Remove dependency on T() KP T () s = Tin () s + Q () s τs + τs + Transfer functions express extent of deviation from a given steady-state 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 Jacketed heated stirred tank Example F, T in F c, T cin h F c, T c F, T Assumptions: Constant hold-up in tank and jacket Constant heat capacities and densities Incompressible flow Model dt F dt V T T h A = ( in ) + c c ( C V T c T ) ρ P dtc Fc dt V T T ha = ( c c cin c) ( C V T c T ) c ρc Pc c 45

46 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) f(x ) f x ( x ) x x Nonlinear systems First order Taylor series expansion. Function of one variable f ( x f x f x s) ( ) ( s ) + ( x x x s). Function of two variables f ( x f x u f x s u s, us) f ( x u s x x s, s) (, ) (, ) + ( x s) + ( u u u s) 3. ODEs f ( x ) x & = f ( x ) f ( xs ) + s ( x x s) x 46

47 Transfer function 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 Y() s = G() s U () s Ys () = G () s U () s Examples, Liquid storage First order Processes F i h F ρa dh = ρfi ρf = ρfi βh dt ρa dh ρ = Fi h β dt β dh τ + h= KpFi dt dh τ + h = KpF i dt 47

48 Examples: Speed of a Car First Order Processes M dv = u bv dt M dv = b dt b u v dv τ = Kpu v dt Stirred-tank heater v () s Kp = u () s τ s + ρcv dt p = ρcpft + Q dt V dt F dt ρc F Q T T () s Kp = = Q () s τ s + p dt τ = KpQ T dt Note: Tin( t) = First Order Processes Liquid Storage Tank K p ρ/β τ ρa/β Speed of a car M/b /b Stirred-tank heater /ρc p F V/F First order processes are characterized by:. Their capacity to store material, momentum and energy. The resistance associated with the flow of mass, momentum or energy in reaching their capacity 48

49 Liquid storage: Capacity to store mass : ρa Resistance to flow : /β First order processes Car: Capacity to store momentum: M Resistance to momentum transfer : /b Stirred-tank heater Capacity to store energy: ρc p V Resistance to energy transfer : / ρc p F Time Constant = τ = (Storage capacitance)* (Resistance to flow) First order process Step response of first order process K p M Ys ()= τs + s Step input signal of magnitude M.9.8 y(t)/k p M t/τ 49

50 First order process What do we look for? Process Gain: Steady-State State Response K lim s τs Overall Change in y Overall Change in u p K p + = = = Δy Δu Process Time Constant: τ = Time Required to Reach 63.% of final value What do we need? Process at steady-state state Step input of magnitude M Measure process gain from new steady-state state Measure time constant Ramp response: K p a Ys ()= τs + s Ramp input of slope a First order process τ y(t)/k p a a t/τ 5

51 φ Sinusoidal response { } First order Process K Ys P Aω ()= τs + s + ω K lim [ ( ) P A I Ys ] = sin( ωt + φ) t + τ ω Sinusoidal input Asin(ωt).5.5 AR y(t)/a φ t/τ First order Processes AR/K p - Corner Frequency Bode Plots High Frequency Asymptote τ p ω τ p ω Amplitude Ratio K AR = + τ ω Phase Shift φ = tan ( ωτ) 5

52 Example: Liquid storage tank Integrating Processes F i h F ρa dh = ρfi ρf dt A dh = Fi F dt F = F F, F = F F i i is s A dh = F i F dt H () s / A = F () s s Process acts as a pure integrator H () s / A = F () s s i Step input of magnitude M Process Modeling K M KM Ys ()= = s s s Input Output Slope = KM Time Time t < yt ()= KMt t 5

53 Integrating processes Unit impulse response K Ys ()= s M = KM s Input Output KM Time Time t < yt ()= KM t Rectangular pulse response Integrating Processes K M t s KM t s Ys ( ) = ( e w ) = ( e w ) s s s Input Output Time Time KMt t < t yt ()= KMtw t t w w 53

54 Second Order Processes Three types of second order process:. Multicapacity processes: processes that consist of two or more capacities in series e.g. Two heated stirred-tanks in series. 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 Multicapacity Second Order Processes Naturally arise from two first order processes in series U(s) KP τ s + KP τ s + Y(s) U(s) KPKP ( τ s+ )( τ s+ ) Y(s) By multiplicative property of transfer functions K K P P Ys () = U() s ( τ s+ )( τ s+ ) 54

55 Second Order Processes Inherently second order process: e.g. Pneumatic Valve p x Momentum Balance M d dx = pa Kx C dx dt dt dt M d x C dx A + + x = K dt K dt K p A x () s = K p () s M s + C s + K K Second order process: Assume the general form Second order Processes KP YS ( ) = U( s) τ s + ξτs+ where Κ P = Process steady-state state gain τ = Process time constant ξ = Damping Coefficient Three families of processes ξ< Underdamped ξ= Critically Damped ξ> Overdamped Note: Chemical processes are typically overdamped or critically damped 55

56 Second Order Processes Roots of the characteristic polynomial ξτ ± 4ξ τ 4τ τ ξ ± ξ τ τ Case ) ξ>: Two distinct real roots System has an exponential behavior Case ) ξ=: Case 3) ξ<: One multiple real root Exponential behavior Two complex roots System has an oscillatory behavior Second order Processes Step response of magnitude M YS ( ) = τ s KP + ξτs+ M s.8.6 ξ= ξ= ξ=

57 Second order process Observations Responses exhibit overshoot (y(t)/km >) when ξ< Large ξ yield a slow sluggish response Systems with ξ= yield the fastest response without overshoot As ξ (with ξ<) becomes smaller system becomes more oscillatory If ξ<, system oscillates without bounds (unstable) Second order processes Example - Two Stirred tanks in series T in, w M Response of T to T in is an example of an overdamped second order process Q T, w M Q T, w 57

58 Second order Processes Characteristics of underdamped second order process. Rise time, t r. Time to first peak, t p 3. Settling time, t s 4. Overshoot: a OS = = exp b 5. Decay ratio: DR c = = exp b ξ ξ π πξ ξ Second order Processes.8.6 P.4..8 b c +5% -5%.6.4 a. t p t r t s 58

59 Sinusoidal Response Second Order Process Kp Aω Ys ()= τ s + ξτs+ s + ω Kp A yt () = sin( t + ) ω φ [ ( ωτ ) ] + ( ξτ ) where ξωτ φ = tan ( ωτ ) AR n = [ ( ωτ ) ] + ( ξωτ ) Bode Plots Second Order Processes ξ=. ξ= ξ= -5 ξ=. - 59

60 More Complicated processes Transfer function typically written as rational function of polynomials rs () a+ as+ Κ + aρs ρ Gs () = = qs () b+ bs+ Κ + bθ s θ where r(s) and q(s) can be factored as q() s = b( τs + )( τs + ) Λ ( τθ s + ) rs () = a( τas+ )( τas+ ) Λ ( τaθ s+ ) s.t. ( τa s+ ) Λ ( τa s+ ) ρ Gs () = K ( τs+ ) Λ ( τθ s+ ) Poles and zeroes Definitions: the roots of r(s) are called the zeros of G(s) z =, Λ, zρ = τa τ aρ the roots of q(s) are called the poles of G(s) p =, Λ, pθ = τ τθ Poles: Directly related to the underlying differential equation If Re(p i )<, then there are terms of the form e -p i t in y(t) - y(t) vanishes to a unique point If any Re(p i )> then there is at least one term of the form e p i t -y(t)does not vanish 6

61 Poles e.g. A transfer function of the form with K s( τs+ )( τ s + ξτs+ ) ξ < can factored to a sum of A constant term from s A e -t/τ from the term (τ s+) A function that includes terms of the form ξ t e τ sin( ξ t ) τ ξ t e τ cos( ξ t ) τ Poles can help us to describe the qualitative behavior of a complex system (degree>) The sign of the poles gives an idea of the stability of the system Poles Calculation performed easily in MATLAB Function ROOTS e.g. qs ()= s 3 + s + s+» ROOTS([ ]) ans = i. -.i» MATLAB 6

62 Poles Plotting poles in the complex plane Imaginary axis Real axis Roots: -.,.j, -.j qs ()= s 3 + s + s+ Poles Process Behavior with purely complex poles.8 Unit Step Response.6.4. y(t) t 6

63 Poles Imaginary axis Real axis s s + 3s + Roots: , j, j Process behavior with mixed real and complex poles Poles Unit Step Response y(t) t 63

64 Poles.5.5 Imaginary axis Real axis s s 3 + 3s + s 5. Roots: -.744, j, j,.55 Process behavior with unstable pole Poles 6 Unit Step Response 4 8 y(t) t 64

65 Zeros Transfer function: Kp( τas + ) Gs () = ( τs+ )( τs+ ) t t yt ()= Kp M τ + a τ e + a τ τ τ τ e τ τ τ τ Let 3 τ> τ.5 τ is the dominant time constant 6 y(t)/km Time Observations: Zeros 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)> Overshoot is observed when the zero is dominant ( τa > τ ) 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 65

66 Can result from two processes in parallel Zeros U(s) K τ s + K τ s + Y(s) ( τ s Gs () K a + ) = ( τs+ )( τs+ ) K = K+ K τa K τ K τ = + K+ K If gains are of opposite signs and time constants are different then a right half plane zero occurs Dead Time F i Control loop h Time required for the fluid to reach the valve usually approximated as dead time Manipulation of valve does not lead to immediate change in level 66

67 Dead time Delayed transfer functions U(s) e τ d s G() s Y(s) Ys () = e τ d s GsUs () () e.g. First order plus dead-time τ e d s Kp Gs ()= τ s + Second order plus dead-time τ e d s K Gs ()= P τ s + ξτs+ Dead time (delay) Gs ()= e τ d s Dead time Most processes will display some type of lag time Dead time is the moment that lapses between input changes and process response y/km t/tau τ D Step response of a first order plus dead time process 67

68 Dead Time Problem use of the dead time approximation makes analysis (poles and zeros) more difficult τ e d s Kp Gs ()= τ s + Approximate dead-time by a rational (polynomial) function Most common is Pade approximation θ s θ s e G() s = θ + s θ θ s+ s θs e G () s = θ θ + s+ s Pade Approximations In general Pade approximations do not approximate dead-time very well Pade approximations are better when one approximates a first order plus dead time process θ θs e K s p K p Gs ()= τ s + θ + s τs + Pade approximations introduce inverse response (right half plane zeros) in the transfer function Limited practical use 68

69 Process Approximation Dead time First order plus dead time model is often used for the approximation of complex processes Step response of an overdamped second order process First Order plus dead time o Second Order Process Approximation Second order overdamped or first order plus dead time? First order plus dead time - Second order overdamped o Actual process Second order process model may be more difficult to identify 69

70 Process Approximation Transfer Function of a delay system First order processes τ D K e s P Ys () = U() s τs + Second order processes τ D Ke s P YS ( ) = U( s) τ s + ξτs+ U(s) e τ Ds G(s) Y(s) Process Approximation More complicated processes Higher order processes (e.g. N tanks in series) U(s) KPKPΛ KPN ( τs+ )( τs+ ) Λ ( τn s+ ) Y(s) For two dominant time constants τ and τ process well approximated by θs e Kp N Gs () θ = τi ( τs+ )( τs+ ) i= 3 For one dominant time constant τ, process well approximated by θs e Kp N Gs () θ = τi ( τs + ) i= 7

71 Process Approximation Example Gs () = ( s+ )( 5s+ )( s+ ) s e G()= s 5s + s e G() s = ( s+ )( 5s+ ) Objective: Empirical Modeling To identify low-order process dynamics (i.e., first and second order transfer function models) Estimate process parameters (i.e., K p, τ and ξ) Methodologies:. Least Squares Estimation more systematic statistical approach. Process Reaction Curve Methods quick and easy based on engineering heuristics 7

72 Least Squares Estimation: Simplest model form Empirical Modeling Ey [ ]= β + βx Process Description where y x β, β y = β + βx + ε vector of process measurement vector of process inputs process parameters Problem: Find β, β that minimize the sum of squared residuals (SSR) n SSR = ( yi β βxi) i= Empirical Modeling Solution Differentiate SSR with respect to parameters n SSR = ( yi β βxi ) = β i= SSR n = xi( yi β βxi ) = β i= These are called the normal equations. Solving for parameters gives: β = y βx n xy i i nxy β = i= n xi nx i= where n x n y x = i, y = i i= n i= n 7

73 Empirical Modeling Compact form Define y x y x Y = X = = β,, β Μ Μ Μ β y n x n Then y β βx y x E = β β = Y Xβ Μ yn β x β n Problem find value of β that minimize SSR SSR = T E E Empirical Modeling Solution in Compact Form Normal Equations can be written as which can be shown to give T X X T β = X Y or β = E E = β In practice Manipulations are VERY easy to perform in MATLAB Extends to general linear model (GLM) E[ y]= β + βx+ Κ + β pxp Polynomial model Ey [ ]= β + βx+ βx T T ( ) T X X X Y 73

74 Empirical Modeling Control Implementation: previous technique applicable to process model that are linear in the parameters (GLM, polynomials in x, etc ) e i.e. such that, for all i, the derivatives i are not a function of β β typical process step responses first order t/ Eyt [ ( )] = Kp M( e τ ) Nonlinear in K p and τ overdamped second order t/ τ e t/ τ τ e Eyt [ ( )] = Kp M τ τ τ Nonlinear in K p, τ and τ 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 y( t) [ y( t ), y( t ), Κ, y( t n )] y() t [ y, y, Κ, y n ] First Order process (step response) Ey [ i ] t KpM ( e i / = τ ) Least squares problem becomes the minimization of n t SSR = yi K pm e i / τ ( ( )) i= This yields an iterative problem solution best handled by software packages: SAS, Splus, MATLAB (function leastsq) 74

75 Empirical Modeling Example Nonlinear Least Squares Fit of a first order process from step response data Model Eyt [ ( )] = 3. Kp( e t/ τ ) Data 4.5 Step Response y(t) t Empirical Modeling Results: Using MATLAB function leastsq obtained Resulting Fit K p = 343., τ = Step Response y(t) t 75

76 Empirical Modeling Approximation using delayed transfer functions For first order plus delay processes t < θ Ey [ i ] = ( t Kp M( e i θ)/ τ ) t θ Difficulty Discontinuity at θ makes nonlinear least squares difficult to apply Solution. Arbitrarily fix delay or estimate using alternative methods. Estimate remaining parameters 3. Readjust delay repeat step until best value of SSR is obtained Example Underlying True Process Data Empirical Modeling Gs () = ( s+ )( 5s+ )( s+ ) y(t) t 76

77 Empirical Modeling Fit of a first order plus dead time. e s G() s = ( s + ) Second order plus dead time e s G () s = ( s+ )( 9. s+ ) y(t) t Empirical Modeling Process reaction curve method: based on approximation of process using first order plus delay model M/s D(s) Y * (s) G c U(s) G p Y(s) G s Y m (s) Manual Control. Step in U is introduced. Observe behavior y m (t) 3. Fit a first order plus dead time model KMe θs Ym () s = s( τ s+ ) 77

78 Empirical Modeling First order plus dead-time approximations KM. θ τ Estimation of steady-state gain is easy Estimation of time constant and dead-time is more difficult Empirical Modeling Estimation of time constant and dead-time from process reaction curves find times at which process reaches 35.3% and 85.3% τ y(t) t t t Estimate θ = 3. t 9. t τ = 67. ( t t) 78

79 Empirical Process Example For third order process Gs () = ( s+ )( 5s+ )( s+ ) Estimates: t = 3, t = 6. 5 θ = 78., τ = Compare: Least Squares Fit. e s G() s = ( s + ) Reaction Curve. e 78. s G() s = ( 6. 46s + ) Process Reaction Curve Method Empirical Modeling 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 79

80 Steam heated stirred tank Feedback Control F in,t in TT TC IP LT P s Steam LC IP F,T Condensate Feedback control system: Valve is manipulated to increase flow of steam to control tank temperature Closed-loop process: Controller and process are interconnected 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 8

81 Examples Feedback Control: requires sensors and actuators T R e.g. Temperature Control Loop + - e Controller Valve C A M T in, F P Tank T Thermocouple 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 Closed-loop loop Processes Study of process dynamics focused on uncontrolled or Open-loop processes Observe process behavior as a result of specific input signals U(s) G p Y(s) In process control, we are concerned with the dynamic behavior of a controlled or Closed-loop process D(s) controller actuator process R(s) + G c G v G + + Y(s) - p sensor G m Controller is dynamic system that interacts with the process and the process hardware to yield a specific behaviour 8

82 Closed-Loop Transfer Function Block Diagram of Closed-Loop Process D(s) R(s) + - controller actuator process G c G v G p + + Y(s) sensor G m G p (s) G c (s) G m (s) - Process Transfer Function - Controller Transfer Function - Sensor Transfer Function G v (s) - Actuator Transfer Function Closed-Loop Transfer Function For control, we need to identify closed-loop dynamics due to: - Setpoint changes Servo - Disturbances Regulatory. Closed-Loop Servo Response transfer function relating Y(s) and R(s) when D(s)= Y() s = Gp() s V () s Ys () = Gp() sgv() sus () Ys () = Gp() sgv() sgc() ses () Ys () = Gp() sgv() sgc() s Rs () Ym() s YS ( ) = Gp() sgv() sgc() s Rs () Gm() sys () [ ] [ ] Isolate Y(s) Gp() s Gv() s Gc() s Ys () = G s G s G s G s Rs () + p() v() c() m() 8

83 Closed-Loop Transfer Function. Closed-loop Regulatory Response Transfer Function relating D(s) to Y(s) at R(s)= Y() s = D() s + Gp() s V () s Ys () = Ds () + Gp() sgv() sus () Ys () = Ds () + Gp() sgv() sgc() ses () Ys () = Ds () + Gp() sgv() sgc() s[ Ym() s] Ys () = Ds () + Gp() sgv() sgc() s Gm() sys () Isolating Y(s) [ ] Ys () = G s G s G s G s Ds () + p() v() c() m() Closed-loop loop Transfer Function. Regulatory Response with Disturbance Dynamics Ys () = + Gd () s G s G s G s G s Ds () p() v() c() m() G d (s) Disturbance (or load) transfer function 3. Overall Closed-Loop Transfer Function Servo Gp() s Gv() s Gc() s Ys () = G s G s G s G s Rs () + + p() v() c() m() Gd () s G s G s G s G s Ds () + p() v() c() m() Regulatory 83

84 The acronym PID stands for: P - Proportional I - Integral D - Derivative PID Controllers PID Controllers: greater than 9% of all control implementations dates back to the 93s 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: t de ut () = Kc et () + e( ) d + D ur ζ ζ τ τ I dt + PID Control Equation PID Control Proportional Action Derivative Action t de ut () = Kc et () + e( ) d + D ur ζ ζ τ τ I dt + Integral Action Controller Bias PID Controller Parameters K c Proportional gain τ I Integral Time Constant τ D Derivative Time Constant u R Controller Bias 84

85 PID Controller Transfer Function PID Control I[ ut () ur] = U () s = Kc + + τ Ds E() s τ I s or: U () s = P+ I + Ds E() s s Note: numerator of PID transfer function cancels second order dynamics denominator provides integration to remove possibility of steady-state errors Controller Transfer Function: PID Control or, Gc()= s Kc + + τ Ds τ I s G s P I c ()= + + Ds s Note: Many variations of this controller exist Easily implemented in SIMULINK each mode (or action) of controller is better studied individually 85

86 Proportional Feedback Form: u() t ur = Kce() t Transfer function: or, U '( s) = KcE( s) Gc()= s Kc Closed-loop form: Gp() s Gv() s Kc Ys () = G s G s K G s Rs () + + p() v() c m() G s G s K G s Ds () + p() v() c m() Proportional Feedback Example: Given first order process: Kp Gp() s =, Gv() s =, Gm() s = τs + for P-only feedback closed-loop dynamics: K p K c + KpKc Ys () = Rs () τ s + K p K c + τ s + KpKc + KpKc + + Ds () τ s + K p K c + Closed-Loop Time Constant 86

87 Proportional Feedback Final response: K p K c lim y servo ( t) =, lim y reg ( t) = t + K p K c t + K p K c Note: for zero offset response we require lim yservo () t =, lim yreg () t = t t Tracking Error Disturbance rejection Possible to eliminate offset with P-only feedback (requires infinite controller gain) Need different control action to eliminate offset (integral) Proportional Feedback Servo dynamics of a first order process under proportional feedback y(t)/km K c t/τ - increasing controller gain eliminates off-set 87

88 Proportional Feedback High-order process e.g. second order underdamped process.5 y(t)/km increasing controller gain reduces offset, speeds response and increases oscillation 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 closedloop process more oscillatory 88

89 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 t ut () = Kc et () + e( ) d ur ζ ζ τ I + Transfer function model U () s = Kc + s Es () τ I Closed-loop response PI Feedback τ s G s G s K I + p() v() c τ s Ys I () = Rs () + τ s Gp s Gv s K I + + () () c Gm() s τ I s Ds () τ s + Gp() s Gv() s K I + c Gm() s τ I s more complex expression degree of denominator is increased by one 89

90 PI Feedback Example PI control of a first order process Kp Gp() s =, Gv() s =, Gm() s = τs + Closed-loop response τ s Ys I + () = Rs () + τ KK s KK Iτ + c p + τ I s + c p KK c p τiτ KK s + I KK s c p τ c p Ds () τ KK s KK Iτ + c p + τ I s + c p KK c p 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 τ cl = damping coefficient τ Iτ KK c p ξ = K τ τ p KK c p + Kcτ I KK c p integral time constant and controller gain can induce oscillation and change the period of oscillation 9

91 PI Feedback Effect of integral time constant on servo dynamics.8.6. K c = y(t)/km PI Feedback Effect of controller gain y(t)/km τ I = affects speed of response increasing gain eliminates offset quicker 9

92 Effect of integral action of regulatory response PI Feedback y(t)/km reducing integral time constant removes effect of disturbances makes behavior more oscillatory Important points: PI Feedback 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 9

93 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 D de τ D Kc dt if error is increasing, decrease control action if error is decreasing, decrease control action Always implemented in PID form t de ut () = Kc et () + e( ) d + D ur ζ ζ τ τ I dt + Transfer Function Closed-loop Transfer Function PID Feedback U () s = Kc + + τ Ds E() s τ I s τ s s G s G s K Dτ I + τi + p() v() c τ I s Ys () = Rs () + τ s s + Gp() s Gv() s K DτI + τi + c Gm() s τ I s Ds () τ s s Gp s Gv s K DτI + τ + () () I + c Gm() s τ I s Slightly more complicated than PI form 93

94 PID Feedback Example: PID Control of a first order process Kp Gp() s =, Gv() s =, Gm() s = τs + Closed-loop transfer function τ s s Ys DτI + τi + () = Rs () + τiτ KK c p + D I + τ τ s I s KK + c p KK τ + c p τiτ KK s I + τ KK s c p c p Ds () τ Iτ + KK c p + D I τ τ s I s KK + c p KK τ + c p PID Feedback Effect of derivative action on servo dynamics.6.4. y(t)/km

95 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 Important Points: Derivative Action 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 95

96 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 GcGvG p Ys () = GGG G Y * () s + GGG G Ds () + c v p m + c v p m G OL if G OL is a rational function then the closed-loop transfer functions are rational functions and take the form rs () a+ as+ Κ + aρs ρ Gs () = = qs () b+ bs+ Κ + bθ s θ and factor as ( τa s + ) Λ ( τa s + ) ρ Gs () = K ( τs+ ) Λ ( τθ s+ ) 96

97 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 (99s) way : MATLAB function ROOTS Traditional:. Routh array: Test for positivity of roots of a polynomial. Direct substitution Complex axis separates stable and unstable regions Find controller gain that yields purely complex roots 3. Root locus diagram Vary location of poles as controller gain is varied Of limited use 97