Essentials of Process Control phần 2 ppsx

The linal solulioli is The ramp response is skc~chctl in f;ig. 2.5. n I1 is 1’rcc~uc1~t~y uscl’ul to bc abic to dc~crminc the time constant of a first-order system from cxpcrimcntal step rcspotisc d:Ua. This is easy to do. When time is equal to T,, in Eq. (2.53). the Icrm ( I - 0 “Tlt) hecomcs ( I - c ’ ) = 0.623. This means that the output variable has undergone 62.3 percent of the total change it is going to make. Thus, the time constant of a first-order system is simply the time it takes the step response to reach 62.3 percent of its new final steady-state value. 2.3.2 Second-Order Linear ODES with Constant Coefficients The first-order sysfem considered in the previous section yields well-behaved expo- nential responses. Second-order systems can be much more exciting since they can give an oscillatory or underdmlpecl response. The first-order linear equation [Eq. (2.44)] could have a time-variable coeffi- cient; that is, Pt,) could be a function of time. We consider only linear second-order ODES that have constant coefficients (T,, and 5 are constants). ,d’X CIX 7-0 __ clt’ + 25rq + x = m(,) (2.62) Analytical methods are available for linear ODES with variable coefficients, but their solutions are usually messy infinite series, and we do not consider them here. The solution of a second-order ODE can be deduced from the solution of a first- order ODE. Equation (2.45) can be broken up into two parts: X(t) = (2.63) The variable xc is called the complementary solution. It is the function that satisfies the original ODE with the forcing function Q(,, set equal to zero (called the homo- geneous differential equation): dX dt + P(,)X = 0 The variable x,’ is called the prricuiur solution. It is the function that satisfies the original ODE with a specified Q,,,. One of the most useful properties of linear ODES is that the total solution is the sum of the complementary solution and the particular solution. Now we are ready to extend the preceding ideas to the second-order ODE of Eq. (2.62). First we obtain the complementary solution x,. by solving the homogeneous 42 PART ONE: Time Domain Dynamics and Control equation (2.65) Then we solve for the particular solution x,’ and add the two to obtain the entire solution. A. Complementary solution Since the complementary solution of the first-order ODE is an exponential, it is reasonable to guess that the complementary solution of the second-order ODE is also of exponential form. Let us guess that XC = ce”t (2.66) where c and s are constants. Differentiating xc with respect to time gives dxc - II= (-seS’ dt and d2Xc dt2 = c&~t Now we substitute the guessed solution and its derivatives into Eq. (2.65) to find the values of s that satisfy the assumed form [Eq. (2.66)]. 7,2(cs2es’) + 257-O(cseSt) + (ceSf) = 0 I 1 7,2s2 + 257,s + 1 = 0 (2.67) This equation, called the characteristic equation, contains the system’s most impor- tant dynamic features. The values of s that satisfy Eq. (2.67) are called the roots of the characteristic equation (they are also called the eigenvalues of the system). Their values, as we will shortly show, dictate if the system is fast or slow, stable or unsta- ble, overdamped or underdamped. Dynamic analysis and controller design consist of finding the values of the roots of the characteristic equation of the system and changing their values to obtain the desired response. Much of this book is devoted to looking at roots of characteristic equations. They represent an extremely important concept that you should fully understand. Using the general solution for a quadratic equation, we can solve Eq. (2.67) for its two roots S= 27; 5, = 70 (2.68) Two values of s satisfy Eq. (2.67). There are two exponentials of the form given in Eq. (2.66) that are solutions to the original homogeneous ODE [Eq. (2.65)]. The sum of these solutions is also a solution since the ODE is linear. Therefore, the comple- mentary solution is (for sI # ~2) x, = cle31t + c2e”*’ (2.69) c‘ttwtw 2. Time Domain Dynamics 43 where cl and c2 are constants. The two roots .SI and s2 are (2.70) (2.7 1) The shape of the solution curve depends strongly on the values of the physical parameter 5, called the damping coefficient. Let us now look at three possibilities. 5 > I (overdamped system). If the damping coefficient is greater than unity, the quantity inside the square root is positive. Then SI and s2 will both be real numbers, and they will be different (distinct roots). EXAMPLE 2.8. Consider the ODE (2.72) Its characteristic equation can be written in several forms: s2+5s+6=0 (2.73) (s + 3)(S + 2) = 0 (2.74) ($s’+2(4(&)s+l =o (2.75) All three are completely equivalent. The time constant and the damping coefficient for the system are The roots of the characteristic equation are obvious from Eq. (2.74), but the use of Eq. (2.68) gives ,=-&- A”-1 = -5 + L 70 70 2-2 St = -2 s2 = -3 The two roots are real, and the complementary solution is XC = cte -21 + c2e-31 The values of the constant cl and c2 depend on the initial conditions. (2.76) n 44 twcr ONI;: Time Domain Dynamics ant1 Control l=l( crl KU y ‘t’ 11 damped system). If the damping coefficient is equal to unity, the term inside the square root of Eq. (2.68) is zero. There is only one value of s that satisfies the characteristic equation. (2.77) The two roots are the same and are called reputed roots. This is clearly seen if a value of J = 1 is substituted into the characteristic equation [Eq. (2.67)]: T(yS2 + 2r,,s + I = 0 = (7,s + l)(T,S + I) (2.78) The complementary solution with a repeated root is XC = (c, + c*t)e”t = (c, + c*t)e -t/r,, (2.79) This is easily proved by substituting it into Eq. (2.65) with 5 set equal to unity. EXAMPLE 2.9. If two CSTRs like the one considered in Example 2.6 are run in series, two first-order ODES describe the system: Differentiating the second-order ODE: (2.80) $$ +(; +k,)C,z = (-$A, (2.8 1) second equation with respect to time and eliminating CA! give a d2G2 +(;+k,+;+k~)~+(~+k,)(~+k+~= dt2 If temperatures and holdups are the same in both tanks, the specific holdup times T will be the same: k, = k2=k 3-1 = 72 = 7 The characteristic equation is (2.82) reaction rates k and s2+2(;+k)r+(l+Pr =O (s+;+k)(s+;+k)=O The damping coefficient is unity and there is a real, repeated root: The complementary solution is (CA2)c = ((.I + Qf).c (k+ I/T)/ 5 < 1 (underdampedsystem). Things begin to get interesting when the damp- ing coefficient is less than unity. Now the term inside the square root in Eq. (2.68) (2.83) (2.84) 1 ~YIAIWK I. Time Domain Dynamics 45 is negative, giving an imaginary number in the roots. 5 2-I $ t Jr__=- i,;J-T 70 7,) 7,) 70 The roots are complex numbers with real and imaginary parts. (2.85) (2.86) (2.87) To be more specific, they are complex conjugnfes since they have the same real parts and their imaginary parts differ only in sign. The complementary solution is xc = clesl’ + c2es2’ = cl exp ii -i+iF)l} +Czexp[(-& iy)t] = e-!Jf/To {c, eIp(+i yt)+ C:exp(-i yt)] Now we use the relationships e ix = cos x + isinx cos( -x) = cos x sin(-x) = - sinx Substituting into Eq. (2.88) gives XC = e+“~~(+ [ cos( ,/I,t)+ isin( J’-t)] + cz[ cos( il_t)- isin( yt)l) = epcfiTo[ (cl + q)cos( yt The complementary solution consists o 1 + i(cl - c2)sin f oscillating sinus (2.88) (2.89) (2.90) (2.91) (2.92) ;oidal terms multiplied by an exponential. Thus, the solution is oscillatory or underdamped for [ < 1. Note that as long as the damping coefficient is positive (c > 0), the exponential term will decay to zero as time goes to infinity. Therefore, the amplitude of the oscillations decreases to zero, as sketched in Fig. 2.6. If we are describing a real physical system, the solution xc must be a real quan- tity and the terms with the constants in Eq. (2.92) must all be real. So the term cl + Q and the term i(cl - c2) must both be real. This can be true only if cl and c2 are com- plex conjugates, as proved next. 46 PARTONE: Time Domain Dynamics and Control I+ J , Sinusoidd terms FIGURE 2.6 Complementary solution for 5 < 1. Let z, be a complex number and 7 be its complex conjugate. z = x + iy and 7 = x - iy Now look at the sum and the difference: z + -2 = (x + iy) + (x - iy) = 2x a real number z - Z = (x + iy) - (x - iy) = 2yi a pure imaginary number i(z-2) = -2y a real number So we have shown that to get real numbers for both ct + c2 and i(q - 4, the numbers cl and c2 must be a complex conjugate pair. Let ct = cR + ic’ and c2 = cR - ic’. Then the complementary solution becomes Xc(r) = e -@To{ (2cR)cos( yt) - (2c’)sin( Tl)] (2.93) EXAMPLE 2.10. Consider the ODE d2x dx dt2+dt+X=0 Writing this in the standard form, We see that the time constant TV = 1 and the damping coefficient 5 = 0.5. The charac- teristic equation is s2+s+1 =o Its roots are I 1 r I t C tl (WWITR z: Time Domain Dynamics 47 The complementary solution is (2.95) 5 = 0 (undamped system). The complementary solution is the same as Eq. (2.93) with the exponential term equal to unity. There is no decay of the sine and cosine terms, and therefore the system oscillates forever. This result is obvious if we go back to Eq. (2.65) and set 5 = 0. You might remember from physics that this is the differential equation that describes a harmonic oscillator. The solution is a sine wave with a frequency of I/T,. We dis- cuss these kinds of functions in detail in Part Three, when we begin our “Chinese” lessons covering the frequency domain. 5 < 0 (unstable system). If the damping coefficient is negative, the exponential term increases without bound as time becomes large. Thus, the system is unstable. This situation is extremely important because it shows the limit of stability of a second-order system. The roots of the characteristic equation are s= -<kiJ1-52 70 70 If the real part of the root of the characteristic equation (-l/r,) is a positive number, the system is unstable. So the stability requirement is: A system is stable if the real parts of all the roots of the characteristic equation are negative. We use this result extensively throughout the rest of the book since it is the foundation upon which almost all controller designs are based. B. Particular solution Up to this point we have found only the complementary solution of the homo- geneous equation This corresponds to the solution for the unforced or undisturbed system. Now we must find the particular solutions for some specific forcing functions m(,). Then the total solution will be the sum of the complementary and particular solutions. Several methods exist for finding particular solutions. Laplace transform meth- ods are probably the most convenient, and we use them in Part Two. Here we present the method c~undetermined coefficients. It consists of assuming a particular solution 48 PART ONI;: Time Domain Dynamics and Control with the same form as the forcing function. The method is illustrated in the following examples. EX A MPLK 2. I I . The ova-damped system of Example 2.8 is forced with a unit step func- tion. -+&+6x= I d2X dt2 dt (2.97) Initial conditions are X(O) = 0 and dX i 1 dt (0) = 0 The forcing function is a constant, so we assume that the particular solution is also a constant: xP = ~3. Substituting into Eq. (2.97) gives 0 + 5(O)+ 6~3 = 1 -$ ~3 = i (2.98) Now the total solution is [using the complementary solution given in Eq. (2.76)] x = xc + xp = c,c2’ + c*e-3’ + ; (2.99) The constants are evaluated from the initial conditions, using the total solution. A com- mon mistake is to evaluate them using only the complementary solution. X(0) = 0 = Cl + c2 + ; dx c-i dt (0) = 0 = (-2c.,e-2’ - 3c2ep3’)(,=0) = -2cr - 3c2 = 0 Therefore q = -; and c2 = f The final total solution for the constant forcing function is X(f) = +-2' + +-3r + ; (2.100) w EXAMPLE 2.12. A general underdamped second-order system is forced by a unit step function: (2.101) Initial conditions are X(0) = 0 and Since the forcing function is a constant, the particular solution is assumed to be a con- stant, giving x, = 1. The total solution is the sum of the particular and complementary solutions [see Eq. (2.93)]. ( (2cx)cos(~t) - (2c’)sinj vt)l (2.102 ) CHN~~:.H 2, Time Domain Dynamics 49 Using lhe initial conditions to CVillllntC COllSlillllS, -t-(o) = 0 = I + [2P(l) - 2c'(O)] (1X i 1 - tit (0) = () = (@) + -Q t ( :I,) Solving for the constants gives 2cR = -1 and 2~’ = s JF-p I.5 0 2 FIGURE2.7 Step responses of 4 6 8 Time O&J a second-order underdamped system. IO SO mcrorw Time Domain Dynamics and Conlrol The total solution is This step response is sketched in Fig. 2.7 for several values of the damping coefficient. Note that the amount the solution overshoots the final steady-state value increases as the damping coefficient decreases. The system also becomes more oscillatory. In Chapter 3 we tune feedback controllers so that we get a reasonable amount of overshoot by selecting a damping coefficient in the 0.3 to 0.5 range. n It is frequently useful to be able to calculate damping coefficients and time con- stants for second-order systems from experimental step response data. Problem 2.7 gives some very useful relationships between these parameters and the shape of the response curve. There is a simple relationship between the “peak overshoot ratio” and the damping coefficient, allowing the time constant to be calculated from the “rise time” and the damping coefficient. Refer to Problem 2.7 for the definitions of these terms. EX A M PLE 2.13 The overdamped system of Example 2.8 is now forced with a ramp input: d*x -+5%+6x=r dt* (2.104) Since the forcing function is the first term of a polynomial in f, we will assume that the particular solution is also a polynomial in t. xI’ = b. + b,t + b2t2 + b3t3 f ’ (2. I OS) where the b; are constants to be determined. Differentiating Eq. (2.105) twice gives dx, = b, + 2bg + 3b3t2 + *. . dt d2X, = 2b2 + 6b3t + *-a dt2 Substituting into Eq. (2.104) gives (2b2 + 6b3t + . . .) + 5(b, + 2b2t + 3b7t2 + . . .) + 6(bo + b,t + b2t2 + b3t3 + . ) = t Now we rearrange the above expression to group together all terms with equal powers of t. . . . + t3(6b3 + . . .) + t2(6b2 + 15b3 + . . -) + t(6b3 + lob2 + 6b,) + (2b2 + 5b, + 6bo) = t Equating like powers oft on the left-hand and right-hand sides of this equation gives the simultaneous equations 6b3 + . . . = 0 6b2 + 15b3 + . * * = 0 6b3 + IOh:! + 6b, = 1 2h2 + SO, + 6bo = 0 [...]... solve for the openloop and closedloop responses of the two-heatedtank process using a proportional temperature controller with K, values of 0, 2, 4, and 8; T2 is controlled by Ql 2. 23 Use MATLAB to solve for the openloop and closedloop responses of the two-heatedtank process using a PI temperature controller with T/ = 0 I hr and K, values of 0, 2, 4, and 8 2. 24 A reversible reaction occurs in an isothermal... Chapter 12) (2. 127 ) ~=CX+DU -_ - - - (2. 128 ) where x = vector of the three temperatures T1, T2, and T3 u = vector of the two inputs To and Qr y = vectoro f measured variables (in our case just the scalar quantity T3) A, B, C, and D are matrices of constants === = (YIAIWK F V F v A == 0 F v 0 0 c -= 0 Time Domain Dynamics 5 5 0 F -v 2 (2. 129 ) lvpq, 0 0 i! 0 0 1 (2. 130) D = [o o] (2. 131) Table 2. 1 gives... K(h)3 12 2 .2 A fluid of constant density p is pumped into a cone-shaped tank of total voluine HTR~/~ The flow out of the bottom of the tank is proportional to the square root of the height h of liquid in the tank Derive the nonlinear ordinary differential equation describing this system Linearize the ODE FO I t H 1 F=KJ7; FIGURE P2 .2 2.3 Solve the ODES: (a) d$+5g+4x =2 cl X(O) =(Jg i 1(0) (b) d$ +2g +2x... and s2 must be a complex conjugate pair if they are complex This is exactly what we found in Eq (2. 85) in the previous section For a third-order system with three roots sl , ~2, and ~3, the roots could all be real: st = cyt, s2 = cy2, and s3 = CY~ Or there could be one real root and two complex conjugate roots: SI =a{’ (2. 117) s2 = s3 = t2y2 - iw2 cy2 + (2. 118) io2 (2. 119) where ak = real part of sk... real part of sk = Re[sk] Ok = imaginary part of sk = lm[.sk] These are the only two possibilities We cannot have three complex roots The complementary solution would be either (for distinct roots) x, = clesIf + c2es2’, + cjes3t or x, = C@ + e”2f[(c2 + c3) cos(02t) + i(c2 - c3) (2. 120 ) sin(ozr)] (2. 121 ) wwrf:.R 2: Time Domain Dynamics 53 where the constants ~2 and cj must also hc complex conjugates in... temperature controller is used to control T2 by manipulating Q, Conventional Control Systems and Hardware In this chapter we study control equipment, controller performance, controller tuning, and general control system design concepts Questions explored include: How do we decide what kind of control valve to use? What type of sensor can be used, and what are some of the pitfalls we should be aware of that... publications of the control valve manufacturers [for example, the Masonielan Handbook for Control Valve Sizing, 6th ed (1977), Dresser Industries] that handle flows of gases, flashing liquids, and critical flows with either English or SI units Sizing of control valves is one of the more controversial subjects in process control The sizing of control valves is a good example of the engineering trade-off that... complementary solution is N xc = (c, + c2t + (2. 123 ) c3t2)enif $ 2 ckeSAf k=4 The stability of the system is dictated by the values of the real parts of the roots ‘The system is stable if the real parts of all roots are negative since the exponential terms go to zero as time goes to infinity If the real part of nrzv one of the roots is positive, the system is unstable The roots of the characteristic equation... equivalent control signal Figure 3.46 shows a temperature transmitter that accepts thermocouple input signals and is set up so that its current output goes from 4 to 20 mA as the process temperature varies from 50 to 25 0°F The range of the temperature transmitter is 50 to 25 0°F, its span is 20 0°F, and its zero is 50°F The gain of the temperature transmitter is 20 mA-4mA =16 mA 25 0°F - 50°F 20 0°F (3 .2) As... interface with the process at the other end of the control loop is made by the final control element In a vast majority of chemical engineering processes the final control element is an automatic control valve that throttles the flow of a manipulated variable In mechanical engineering systems the final control element is a hydraulic actuator or an electric servo motor Most control valves consist of a plug on . Differentiating Eq. (2. 105) twice gives dx, = b, + 2bg + 3b3t2 + *. . dt d2X, = 2b2 + 6b3t + *-a dt2 Substituting into Eq. (2. 104) gives (2b2 + 6b3t + . . .) + 5(b, + 2b2t + 3b7t2 + roots) x, = clesIf + c2es2’, + cjes3t or x,. = C@ + e”2f[(c2 + c3) cos(02t) + i(c2 - c3) sin(ozr)] (2. 120 ) (2. 121 ) wwrf:.R 2: Time Domain Dynamics 53 where the constants ~2 and cj must. complex conjugate roots: SI =a{’ (2. 117) s2 = cy2 + io2 (2. 118) s3 = t2y2 - iw2 (2. 119) where ak = real part of sk = Re[sk] Ok = imaginary part of sk = lm[.sk] These are the