AR-9161 B.Tech. VI Sem. Chemical Engineering Process Dynamics &Control Model Answer

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1 AR-9161 B.Tech. VI Sem. Chemical Engineering Process Dynamics &Control Model Answer Ans (1) Section A i. (A) ii. iii. iv. (B) (B) (B) v. (D) vi. vii. viii. ix. (C) (B) (B) (C) x. (A) Section B

2 (1) (i) Servo Problem versus Regulator Problem Fig. Block diagram of a simple control system. The control system of Fig. can be considered from the point of view of its ability to handle either of two types of situations. In the first situation, which is called the servomechanism-type (or servo) problem, we assume that there is no change in load Ti and that we are interested in changing the bath temperature according to some prescribed function of time. For this problem, the set point TR would be changed in accordance with the desired variation in bath temperature. If the variation is sufficiently slow, the bath temperature may be expected to follow the variation in TR very closely. There are occasions when a control system in the chemical industry will be operated in this manner. For example, one may be interested in varying the temperature of a reactor according to a prescribed time-temperature pattern. However, the majority of problems that may be described as the servo type come from fields other than the chemical industry. The tracking of missiles and aircraft and the automatic machining of intricate parts from a master pattern are well-known examples of the servo-type problem. The other situation will be referred to as the regulator problem. In this case, the desired value TR is to remain fixed and the purpose of the control system is to maintain the controlled variable at TR in spite of changes in load Ti. This problem is very common in the chemical industry, and a complicated industrial process will often have many self-contained control systems, each of which maintains a particular process variable at a desired value. These control systems are of the regulator type. In considering control systems in the following chapters, we shall frequently discuss the response of a linear control system to a change in set point (servo problem) separately from the response to a change in load (regulator problem). However, it should be realized that this is done only for convenience. The basic approach to obtaining the response of either type is essentially the same, and the two responses may be superimposed to obtain the response to any linear combination of set-point and load changes.

3 Fig. Block diagram of a simple control system. (1) (ii)feedback Control The control system shown in Fig. is called a closed-loop system or a feedback system because the measured value of the controlled variable is returned or fed back to a device called the comparator. In the comparator, the controlled variable is compared with the desired value or set point. If there is any difference between the measured variable and the set point, an error is generated. This error enters a controller, which in turn adjusts the final control element in order to return the controlled variable to the set point. Negative Feedback The feedback principle, which is illustrated by Fig. 9.2, involves the use of the controlled variable T to maintain itself at a desired value TR. The arrangement of the apparatus of Fig. 9.2 is often described as negative feedback to contrast with another arrangement called positive feedback. Negative feedback ensures that the difference between TR and T, is used to adjust the control element so that the tendency is to reduce the error. For example, assume that the system is at steady state and that T = T,,, = TR. If the load Ti should increase, T and T,,, would start to increase, which would cause the error E to become negative. With proportional control, the decrease in error would cause the controller and final control element to decrease the flow of heat to the system with the result that the flow of heat would eventually be reduced to a value such that T approaches T R. Positive Feedback If the signal to the comparator were obtained by adding TR and T,, we would have a positive feedback system, which is inherently unstable. To see that this is true, again assume that the system is at steady state and that T = T,,, = TR. If Ti were to increase, T and T,,, would increase, which would cause the signal the comparator (E in Fig) to increase, with the result that the heat to the system would increase. However, this action, which is just the opposite of that needed, would cause T to increase further. It should be clear that this situation would cause T to run away and control would not be achieved. For this reason, positive feedback would never be used intentionally in the system of Fig. However, in more complex systems it may arise naturally.

4 Example: Feedback Level Control

5 (2) Solve x(0) = x (0) = 0 Taking Laplace transform of both sides, Inserting the initial conditions and rearranging gives Equating the numerators on each side gives We now equate the equation of like power of s to obtain A+B=0 2A+C=0 2A=2 Solving these equations gives A = 1, B = - 1, and C = -2. Equation of x(s) now becomes Express the quadratic term in the form to match the following transform

6 Equating the coefficient of like power of s gives a=1, k=1 Using the table, we get (3) Two interacting systems (tanks) are connected in series. Obtain a transfer function that relates H 2 to Q. Fig. Two interacting tanks connected in series

7 Our problem is to find a transfer function that relates h 2 to q, that is, H 2 (s)/q(s).consider a case of two liquid level system connected in series as shown in fig. Where, q, q 1, h 1, A 1, R 1, are the input flow rate, output flow rate, liquid head, cross sectional area and resistance for tank one respectively. q 2, h 2, A 2, R 2, are the output flow rate, liquid head, cross sectional area and resistance for tank two respectively. Taking the material balance for tank one (1) (2) (3) At steady state From equation (1) and (5) (4) (5) (6) (7) From equation (2) and (6) (8) From equation (3) and (4) (9) (10) Transforming equation (7) through (10) gives (11) (12) (13) (14) (15)

8 Substituting the value of H1(s) from equation 15 into equation 13 gives Substituting the value of Q1(s) from equation (16) and Q2(s) from equation (14) into equation (12) gives (16) (17) (4) A mercury thermometer having a time constant of 0.1 min is placed in a temperature bath at 100 F and allowed to come to equilibrium with the bath. At time t = 0, the temperature of the bath begins to vary sinusoidally about its average temperature of l00 o F with an amplitude of 2 F If the frequency of oscillation is 10/π cycles/min, plot the ultimate response of the thermometer reading as a function of time. What is the phase lag? Solution Time constant τ = 0.1 min x s = 100 F A = 2 F

9 Consider a thermometer to be in equilibrium with a temperature bath at temperature x S. At some time t = 0, the bath temperature begins to vary according to the relationship x = x s +A sinwt (1) where, x = temperature of bath x s = temperature of bath before sinusoidal disturbance is applied A = amplitude of variation in temperature w = radian frequency, radkime In anticipation of a simple result we shall introduce a deviation variable X which is defined as X = x - x s, ( 2) X = x - x s = A sin wt ( 3) By referring to a table of transforms, the transform of Eq. (3) is (4) (5) (6) (7) (8) (9) (10) Equating the numerators on each side gives (11) We now equate the equation of like power of s to obtain C 1 + C 2 =0 (12) C 1 +( C 3 /τ)=0 (13) Putting s= -1/ τ gives

10 (14) Taking inverse Laplace transform gives (15) (16) At steady state t, the first term of the equation (16) vanishes (17) Time constant τ = 0.1 x s = 100 F A = 2 F Frequency of oscillation f = 10/π cycles/min Radian frequency w =2π f = 2π(10/π) = 20 rad/min Amplitude of the response τ Phase angle ϴ= --tan -1 2 = or Phase lag = The response of the thermometer is therefore Y(t) = sin (20t ) or y(t) = sin (20t ) To obtain the lag in terms of time rather than angle, we proceed as follows: A frequency of 10/π cycles/min means that a complete cycle (peak to peak) occurs in (10/π) -1 min. Since one cycle is equivalent to and the lag is , the time corresponding to this lag is min The response of the thermometer reading and the variation in bath temperature are shown in Fig.

11 Fig. Response of thermomete

12 (5) (a) Give one example each of a system with first order and second order dynamics. Derive the transfer function of a second order system. Example of first order system : A mercury thermometer Example of second order system : Damped vibrator A second-order transfer function will be developed by considering a classical example from mechanics. This is the damped vibrator, which is shown in Fig. A block of mass W resting on a horizontal, frictionless table is attached to a linear spring. Fig. Damped vibrator A viscous damper (dashpot) is also attached to the block. Assume that the system is free to oscillate horizontally under the influence of a forcing function F(t). The origin of the coordinate system is taken as the right edge of the block when the spring is in the relaxed or unstretched condition. At time zero, the block is assumed to be at rest at this origin. * Positive directions for force and displacement are indicated by the arrows in Fig. Consider the block at some instant when it is to the right of Y = 0 and when it is moving toward the right (positive direction). Under these conditions, the position Y and the velocity dy/dt arc both positive. At this particular instant, the following forces are acting on the block: 1. The force exerted by the spring (toward the left) of -KY where K is a positive constant, called Hooke s constant. 2. The viscous friction force (acting to the left) of -C dy/dt, where C is a positive constant called the damping coefficient. 3. The external force F(t) (acting toward the right). Newton s law of motion, which states that the sum of all forces acting on the mass is equal to the rate of change of momentum (mass X acceleration), takes the form (1) Rearrangement gives (2) where W = mass of block, lb, g c = 32.2(lb,)(ft)/(lbf)(sec2) C = viscous damping coefficient, lbf/(ft/sec) K = Hooke s constant, lbf/ft F(t) = driving force, a function of time, lbf Dividing Eq. (2) by K gives (3)

13 For convenience, this is written as (4) τ (5) τ (6) (7) Solving for τ and ϕ from Eqs. (5) and (6) gives τ sec (8) dimensionless (9) By definition, both τ and ϕ must be positive. Equation (4) is written in a standard form that is widely used in control theory. Notice that, because of superposition, X(t) can be considered as a forcing function because it is proportional to the force F(t). If the block is motionless (dyldt = 0) and located at its rest position (Y = 0) before the forcing function is applied, the Laplace transform of Eq. (4) becomes τ τ (10) From this, the transfer function follows: τ τ (11) 5 (b) A control system having transfer function is expressed as The radian frequency for the control system is 1.9 rad/min. The time constant is 0.5 min. The control system is subjected to a step change of the magnitude 2. Calculate : (i) Rise time (ii) Decay ratio (iii) Maximum value of Y(t) (iv) Response time

14 Fig. Terms used to describe an underdamped second-order response. Given X(s) = 2/s Time constant τ=0.5 min Radian frequency w= 1.9 rad/min Φ=0.312 (i)rise time ϴ= min Decay ratio=0.127

15 Ultimate value of the response Y ultimte (B) at t X(s)= 2/s = 10 Y ultimte (B)=10 Overshoot (B/A)=0.356 Response time ts=3 τ/ ϕ =4.76 min for % of ultimate value Response time ts=4 τ/ ϕ = 6.33 min for of ultimate value

16 (6) Determine the transfer function Y(s)/X(s) for the block diagram shown in Figure 1. From figure (1) (2) (3) (4) (5) (6) From equation (4) &(6) (7) From equation (3) &(4)

17 (8) From equation (2),(5)&(8) (9) From equation (1) &(7) (10) From equation (9) &(10) (11)

18 (7) Open loop transfer function of a negative feedback control system is given as: Sketch the root locus diagram. The open loop transfer function is given as: Negative feedback control becomes positive feedback control system. The open loop poles are P 1 = -1-j P 2 = -1+j P 3 =-5 The open loop zeros is, Z 1 =5 No. of open loop poles= b=3 No. of open loop poles= a=1 The BIP can be calculated as:

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21 (8) A control system is represented by means of a block diagram shown in figure 2. R(s) + - C Figure 2 Determine the value of k c gain of controller which just causes instability. Use Routh criterion. Also determine the location of the pair of roots lie on the imaginary axis for the control system. Solution : The open loop transfer function for the control system is given as The characteristics equation of the control system is given as 1+G(s) = 0 Constructing the Routh Array. There are three numbers of roots therefore the number of rows are four Row r 1 2 (kc+1) m-1 r 2 3 2kc m r 3 A 1 A 2 m+1 r 4 B 1 B 2

22 The unknown coefficients A 1, A 2, B 1 & B 2 can be calculated as and A 2 =0 and B 2 =0 The Routh Array becomes, Row r 1 2 (kc+1) m-1 r 2 3 2kc m r 3 (3-k c )/3 0 m+1 r 4 2k c 0 The control system becomes just unstable when all the element of m th row become zero

23 (9) Multicapacity non-interacting three CSTRS in series having time constants τ=1 minute for two CSTRS & τ=0.4 minute for third CSTRS. The system is controlled by PD controller having τ D = 0.1 minute transportation lag τ d = 0.4 minute & gain of controller is k c. Sketch the asymptotic Bode diagram. Obtain the design value of gain of controller. Solution: The individual transfer functions are The open loop transfer function is kc is the gain of the controller. 1. LFA W 0 AR 1 =1 LFA W AR 1 =w This is the equation of straight line with slope =1 passing through AR 1 =1, 0.1w=1, w=10. Corner frequency 2.

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27 (10) There are two non-interacting liquid-level system in series. The time constants for liquid level systems are 1 minute and 0.1 minute. The overall gain process is 0.1. The transportation lag parameter of the measuring elements is 0.1 minute. Sketch the asymptotic Bode diagram and determine : a. Ultimate gain. b. Ultimate period. c. Optimum controller setting. d. The design value of gain proportional controller. Solution: From given data The open loop transfer function is Where K is gain of the process K c is gain of the controller 1. LFA W 0 AR 1 =1 LFA W AR 1 =1/w This is the equation of straight line with slope =-1 passing through AR 1 =1, w=1.

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