Introduction to Process Control


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1 Introduction to Process Control For more visit : By: M. P. Girnari (SSEC, Bhavnagar) For more visit: 1
2 Contents: Introduction Process control Dynamics Stability The Development of Mathematical Model What is Mathematical Model? Types of Mathematical Models Model Derivation : For different Process control, Mechanical & Electrical system Dynamic Behavior of Linear and Nonlinear system First order system Second order system Dynamics and Stability of Controlled System GTU Previous Year Questions Tutorial:1 For more visit: 2
3 Process Control Process control is an engineering discipline that deals with architectures, mechanisms and algorithms for maintaining the output of a specific process within a desired range. In other words, process control refers to Regulation and manipulation of variables influencing the conduct of process in a such a way as to obtain a product of desired quality and quantity in an efficient manner. For Example? For more visit: 3
4 Examples: Process Control Level control Temp control Pressure control For more visit: 4
5 Dynamics Dynamics means time varying behaviour. Process dynamics means time varying behaviour of a regulated or unregulated process when it is subjected to disturbances. For example : Car Driving Driving a car around curves in road, avoiding potholes and other obstructions, starting and stopping at traffic lights etc. Speeding the car or slowing it down. It takes a finite time for the car to accelerate or decelerate to the new constant speed. To predict dynamic operation of car we would have to accumulate the information in terms of mass, energy and momentum balance. We would have to specify All system inputs and how you would manipulate speed. For more visit: 5
6 Stability When a system is said to be stable? A system is said to be stable if the controlled variables converges to a steady state after the disturbance, else the system is unstable. In unstable system the response of the controlled variable grows without any limit. Example? Stable system : Gravity Driven Pendulum Always returns to its vertical (down) state, also known as zero state or equilibrium state. So it is an example of stable system. UnStable system : Now, consider the same system in vertical (up) state. It is theoretically in an equilibrium state. But the slightest deviation away from the equilibrium state causes it to move sideways till it hit the horizontal ground. So it is an example of unstable system. For more visit: 6
7 Types of Dynamic Process 1. Instantaneous process Y(t) = K. X(t) 2. Integral Process dy dt = K. X(t) 3. Frist Order Process τ dy dt + y = K p f(t) 4. Second Order Process τ 2 d2 y dt + 2ζτ dy dt + y = K p f(t) 5. Higher order Process For more visit: 7
8 Instantaneous process The output follows the change in inputs quickly as it can be represented by the following equation Y t = K. X t Where, y(t) = output, X(t) = input, and K= static gain. Also known as zero order system as there are no derivatives in the describing equation. This type of process represents the ideal or perfect dynamic response. Example..? Unloaded potentiometer Output voltage is proportional to the movement of wiper. For more visit: 8
9 Integral process An integral process can be represented by the following equation dy t dt = K. X t Where, y(t) = output, X(t) = input, and K= static gain. The output will continue to increase until limit of the system is reached. Example..? Liquid level container with only inflow or more inflow than outflow The level in the tank will increases continuously until the tank is full or overflow. Positive slope Liquid level container with only outflow or more outflow than inflow The level in the tank will decreases continuously until the tank is empty. Negative slope For more visit: 9
10 First Order system A first order system is one whose output y t is modelled by a first order differential equation. For linear (or linearized) system, a 1 dy dt + a 0y = b f(t) Where, f(t) is the input (forcing function). If a 0 then above equation can be rewritten as ( a 1 a 0 ) dy dt + y = ( b a 0 ) f(t) Now, let a 1 a 0 = τ (time constant) and b a 0 = K p (steady state gain / static gain) τ dy dt + y = K p f(t) For more visit: 10
11 First Order system Equation of first order system is τ dy + y = K dt p f(t) Now, assume that initial conditions is y(0) = 0 and f(0) = 0 By taking a Laplace transform, the transfer function of a first order system can be given by G s = Y s F s = K p (sτ+1) The transfer function of a first order system having a time delay / lag is given by G s = Y s F s = e st K p (sτ+1) is also known as FOPDT process. For more visit: 11
12 Second Order system A second order system is one whose output is described by the solution of second order differential equation. For example, second order system can be described by following ODE a 2 d 2 y dt + a 1 If a 0 0, then τ 2 d2 y dt dy dt + a 0y = b f(t) + 2ζτ dy dt + y = K p f(t) Where, τ 2 = a 2 a 0, 2ζτ = a 1 a 0, K p = b a 0. τ = natural period of osscilation of the system ζ = damping factor K p = steady state or static gain of the system For more visit: 12
13 Mathematical Model For design and analysis of control system we need a mathematical model that describes the dynamics behavior of the system. A mathematical model of a process is a description of the characteristics of a number of phenomena that take place in a process. The mathematical modelling is a way of describing a process with the help of mathematical equations. A model could be a set of mathematical equations or a graph, or a series of data. For more visit: 13
14 Types of mathematical models For more visit: 14
15 Theoretical Modelling Model construction is based on the physical properties of the system (based on physicochemical law). Also known as fundamental modelling. With the use of Material (mass) balances equations Energy balances equations Momentum balance equations And depending on the levels of accuracy of the assumptions we will have different models ranging from simple to rigorous models. The model derived from this method do not require experimental data, but data is required for validation of the model. For more visit: 15
16 Theoretical Modelling Advantage : Global validity Accuracy Complete process understanding Disadvantages : Complex method Huge computation & simplification is needed Time consuming Useful for extrapolation, scaling etc. For more visit: 16
17 Empirical Modelling For some processes, the mathematical modelling is too complex to perform in order to develop process model. Empirical modelling is based on the operating data. Means it utilizes the input & output data from the operation of plant to build the relationship between input & output. The experimental data can be fitted into some equations, empirical relationship. Also known as Black Box modeling. Black box modeling bcz you don t need to worry about what is inside the process means the inner dynamics. For more visit: 17
18 Empirical Modelling Advantages : Easy to develop Quite simple No need of inner dynamics Reduction in computation. Disadvantage : Well designed and highly accurate experimental data is required Result from one plant may not be applied to other plant. Result from the same plant may not be applied to the same plant if the experimental condition changes. Dominantly used in industries For more visit: 18
19 SemiEmpirical Modelling (Hybrid modelling) Combination of theoretical (fundamental) and empirical modeling. Model equations are derived from the basic conservation laws and certain model parameters (such as reaction rate and heat transfer coefficient) evaluated from the physical experiments or from the process operating data. Also known as Gray Box modeling. Gray box modeling bcz?. Utilizes the advantage of individual methods. But, the critical thing is that we need to decide for which part of the model to use theoretical (fundamental) modeling and for which part to use empirical data. For more visit: 19
20 Uses of Mathematical Model in process control To improve understanding of the process. Process model can be used to investigate the process behavior without the expense and unexpected hazards of operation of real process. To train operating personnel. Plant operator can be trained to operate a complex process and how to deal with emergency situations by working on the process simulator. To design a control strategy for the process. Before designing a control strategy you can see that control strategy will work efficiently for your plant or not. To optimize process design and operating condition. Can be used to determine most profitable operating condition. For more visit: 20
21 Dynamics Behavior : Linear First Order system Single Tank Level Control system Assumptions : Density of liquid is constant Crosssectional area of tank is uniform Disturbance MV LC SP Mass Balance equation : Accumulation = inlet flow outlet flow A dh dt d + f i f o (1) At steady state: Inlet Flow LT A dh s dt = f ds + f is f os h Outlet Flow For more visit: 21
22 Dynamics Behavior : Linear First Order system f o = h R 2 (2) Putting value of equ.(2) into equ.(1). A dh = f dt i h R 2 Putting the above equation in variable form A dh = F dt i H R 2 Taking Laplace on both sides we get AsH s = F i s H s Rearranging above equation we get H s = R 2 F i s R 2 As+1 = K R 2 (sτ+1) Where, K = R 2 and τ = R 2 A Disturbance For more visit: h MV Inlet Flow LT LC SP Outlet Flow
23 Dynamics Behavior : Linear First Order system Single Tank Temperature Control Assumptions : Density of liquid is constant Crosssectional area of tank is uniform Mixer speed constant. Energy Balance equation : Accumulation = inlet flow outlet flow Vρc P dt dt = Q = UA (T st T) (1) Where, c P = heat capacity U = Overall heat transfer coefficient. T st = Temp. of saturated steam. A = Total heat transfer area For more visit: 23
24 Dynamics Behavior : Linear First Order system At steady state 0 = UA T st,s T s (2) Subtracting eq. (2) from (1) and Writing the terms in deviation variable form dt Vρc P = Q = UA T dt st T.. (3) Where, T = T T s and T st = T st T st,s Tanking Laplace transform G s = T s T = st s Where, 1 Vρc P UA s+1 = K (sτ+1) τ = time constant = Vρc P UA K p = steady state gain = 1 For more visit: 24
25 Dynamics Behavior : Linear Second Order system Noninteracting tank system For tank1 > τ 1 dy 1 dt + y 1 = K p1 f 1 t G 1 s = For tank2 > τ 2 dy 2 dt + y 2 = K p2 f 2 (t) G 2 s = K p1 (sτ 1 +1) K p2 (sτ 2 +1) First system affect the second by its output G 1 s = Y 1 s and G F 1 s 2 s = Y 2 s Y 1 s The overall transfer function is given by G s = Y 2 s F 1 s G s = K p2 = Y 2 s Y 1 s = K p1 * Y 1 s F 1 s (sτ 1 +1) (sτ 2 +1) K P τ 2 s 2 +2ζτ s+1 Where, τ = τ p1 τ p2, 2ζτ = τ p1 +τ p2 and K P = K P1 K P2 For more visit: 25
26 Dynamics Behavior : Linear Second Order system Interacting tank system For tank1 > τ 1 dy 1 dt + y 1 = K p1 f 1 t G 1 s = For tank2 > τ 2 dy 2 dt + y 2 = K p2 f 2 (t) G 2 s = How? K p1 (sτ 1 +1) K p2 (sτ 2 +1) F 1 = h 1 h 2 R 1 and F 2 = h 2 R 2 Putting the values of F 1 and F 2 in respective above equations we get H 2 s F i s = R 2 τ 1 τ 2 s 2 +(τ 1 +τ 2 +A 1 R 2 )s+1 Where, τ 1 = A 1 R 1 and τ 2 = A 2 R 2 For more visit: 26
27 Mathematical model of RLC circuit According to KVL V t = V L + V R + V C Using Ohm's law to solve for the voltage across each element of the circuit, the equation for this circuit can be written as V t = L di + Ri + 1 i t dt dt C Changing variables from current to charge using i t = dq yields dt V i t = L d2 dt q + R d dt q + 1 C (q) Take Laplace transform V i s = Ls 2 Q(s) + RsQ(s) + 1 C Q(s) For more visit: 27
28 Mathematical model of RLC circuit Output voltage is the voltage across capacitor can be written as V C t = 1 C q t Take Laplace transform V o s = 1 C Q(s) For transfer function G s = (V o s /V i s ) = 1 C Q(s)/(Ls2 + Rs + 1 C ) Q(s) After simplification G s = 1/(LCs 2 + CRs + 1 ) For more visit: 28
29 2nd Order system : An Undamped Pendulum Consider figure: the pendulum can only move in two directions in the plane of figure. Notation: l = length of pendulum, m = weight of mass h = vertical position of the center of mass θ = angle of swing away from a vertical position F = force acting on the suspension poin negative opposite direction For more visit: 29
30 2nd Order system : An Undamped Pendulum On the application of suspension system, Newton s second law yield Two component of force Horizontal : my = F sinθ (1) Vertical : mh = Fcosθ + mg (2) Here, y = 2nd order time derivatives of y (acceleration in horizontal direction) h = 2nd order time derivatives of h (acceleration in vertical direction) Now assume that Swing of the pendulum is moderate so that the angle θ is always small. The pendulum then moves very little in vertical direction and we can assume h = 0 For more visit: 30
31 2nd Order system : An Undamped Pendulum From equation (1) and (2), we can write y + g tan θ = 0 (3) From figure we can write tan θ = y u h y u l (4) For final equation of system, put the value of eq.(4) into eq. (3) y + g l y = ( g l )u For more visit: 31
32 2nd Order system : Spring mounted mass system See the figure. (springmassdamper system) At equilibrium point y = u = 0 According to the Newton s second law, my + by + ky = u t (1) Where, m= mass of the body k = spring constant b= damping constant u(t) = force applied on body y(t) = distance of movement The gravitational force mg is cancelled out bcz we have considered equilibrium condition. For more visit: 32
33 Dynamics Behavior : First Order system First order system transfer function G s = K p (sτ+1) Step response: Y s = K p (sτ+1) A Take inverse Laplace y t = K p A (1 e t τ) Impulse response: Y s = s K p (sτ+1) A Take inverse Laplace y t = K pa τ (e t τ) For more visit: 33
34 Dynamics Behavior : First Order system First order system transfer function G s = K p (sτ+1) Ramp response: Y s = K p A (sτ+1) s 2 Take inverse Laplace y t = K p A(t τ + τe t τ) Parabolic response: Y s = K p (sτ+1) A s 3 Take inverse Laplace y t = For more visit: 34
35 Dynamics Behavior : Second Order system First order system transfer function G s = K τ 2 s 2 +2ζτs+1 = ω2 n s 2 +2ζω n s+ω n 2 Step response of second order system Response specification Delay time (t d ) Rise time (t r ) Peak time (t p ) Settling time t s Peak overshoot (M P ) Steadystate error (e ss ) For more visit: 35
36 Stability of controlled system Let us summarize our definitions of stability for linear, timeinvariant systems. Using the natural response: A system is stable if the natural response approaches zero as time approaches infinity. A system is unstable if the natural response approaches infinity as time approaches infinity. A system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates. Using the total response (BIBO): A system is stable if every bounded input yields a bounded output. A system is unstable if any bounded input yields an unbounded output. For more visit: 36
37 Stability Analysis Methods for stability analysis RouthHurwitz Criteria Jurystability criterion Direct Substitution method Rootlocus (timedomain) Bodeplot (frequencydomain) Nyquist plot For more visit: 37
38 Stability Analysis using RouthHurwitz criteria Test the stability of closed loop system Characteristic equation will be s s s From Routh table? Sign change in first column So, System is unstable. For more visit: 38
39 Stability Analysis using Bode Plot Frequency domain technique Graphical representation of phase and gain contribution Stability can be determined from Gain margin : Allowable gain GM = G jw H jw Phase margin : Allowable phase lag PM = 180+ < G jw H jw 1 System Gain Margin Phase Margin Frequency Stable System +ve +ve ω gc > ω pc Unstable System ve ve ω gc < ω pc Marginally Stable System 0 0 ω gc = ω pc For more visit: 39
40 Stability Analysis using Root Locus Root locus is a graphical representation of a path traced by closed loop poles as a system parameters (usually gain) varies. The closedloop system response is nonoscillatory for points on the loci of the real axis. The roots are real and therefore after inversion the response will be nonoscillatory. When the loci branch is away from the real axis, the closed loop response becomes oscillatory. The converse of the statement given above. The closed loop response is marginally stable when one of the branches intersects the real axis. in going from the left half plane to the right half plane. For more visit: 40
41 GTU Previous Year Questions 1. Compare black, white and gray box models for process control problems with suitable examples. For more visit: 41
42 Tutorial : Explain in detail. Dynamics and Stability of Controlled system. 2. Discuss about the dynamic behavior of first order system with suitable example. 3. Discuss about the dynamic behavior of second order system with suitable example. 4. What do you mean by mathematical modelling? Explain different types of mathematical modelling in detail. For more visit: 42
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