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7) Id 8) f cIwrf:K 7: Laplace-Domain Dynamics 237 B = lim K,‘tT: 4) ___ = Therefore, I l/T, 1 ; - (s + 1/T,)2 - s + l/T” Inverting term by term yields 7.4 TRANSFER FUNCTIONS Our primary use of Laplace transformations in process control involves representing the dynamics of the process in terms of “transfer functions.” These are output-input ~~~ relationships and are obtained by Laplace-transforming algebraic and differentktl equations. In the following discussion, the output variable of the process is ~(~1. The ~~~~ input variable or the forcing function is ~(~1. 7.4.1 Multiplication by a Constant Consider the algebraic equation Y(t) = KU(t) (74lj~~ mm Laplace-transforming both sides of the equation gives I 02 cc o Ywe -dt = K i tqt)P’ dt 0 Y(Y) = KU(s) (7.42) where YtS, and Ut,) are the Laplace transforms of y& and ~(~1. Note that U(Q is an arbitrary function of time. We have not specified at this point the exact form of the input. Comparing Eqs. (7.41) and (7.42) shows that the input and output variables ~~ are related in the Laplace domain in exactly the same way as they are related in the time domain. Thus, the English and Russian words describing this situation are the same. Equation (7.42) can be put into transfer function form by finding the outputiinput ratio: Y(.d _ K __ - U(s) (7.43) 238 PARTTWO: Laplace-Domain Dynamics and Control %) -v(f) * ‘Time domain * Laplace domain FIGURE 7.2 Gain transfer function. For any input UcS) the output Y(,, is found by simply multiplying Ucs) by the constant K. Thus, the transfer function relating YtS, and Uc,y, is a constant or a “gain.” We can represent this in block-diagram form as shown in Fig. 7.2. 7.4.2 Differentiation with Respect to Time Consider what happens when the time derivative of a function ‘ytt) is Laplace trans- formed. Integrating by parts gives (7.44) u = p dv = 2dt dt du = -sedsf dt v=y Therefore, -st dt = [yC’]::; + sye ” dt I- a = 0 - Y(t=O) + s Y(t)e- dt o The integral is, by definition, just the Laplace transformation of ycr), which we call Y(s). = sq,, - Y(f=O) (7.45) The result is the most useful of all the Laplace transformations. It says that the op- eration of differentiation in the time domain is replaced by multiplication by s in the Laplace domain, minus an initial condition. This is where perturbation variables become so useful. If the initial condition is the steady-state operating level, all the initial conditions like Y(~=o) are equal to zero. Then simple multiplication by s is equivalent to differentiation. An ideal derivative unit or a perfect differentiator can be represented in block diagram form as shown in Fig. 7.3. The same procedure applied to a second-order derivative gives 2 = s2 Y,,) - sy(, =()) - (7.46) 1 > CIIAWI:.K 7: Laplace-IXtnain Dynatnics 239 Y(s) SY(,, + Laplace domain 42 Time domain Y(s) 2Y(s) * Laplace domain FIGURE 7.3 Differential transfer function. Thus, differentiation twice is equivalent to multiplying twice by S, if all initial con- ditions are zero. The block diagram is shown in Fig. 7.3. The preceding can be generalized to an Nth-order derivative with respect to time. In going from the time domain into the Laplace domain, dNxldtN is replaced by sN. Therefore, an Nth-order differential equation becomes an Nth-order algebraic equation. dNY dN-’ y ‘IN dtN - + aN-l- dY dtN-’ + . . . + al dt + a0y = U(t) (7.47) aNsN Y(s) + aN- 1s N-‘Y@) + **- +‘a&,) + a0Y(,) = &) (7.48) (aNsN + aN-lsNel + ” ’ + als + aO>Y(,) = u(s) (7.49) Notice that the polynomial in Eq. (7.49) looks exactly like the characteristic equa- tion discussed in Chapter 2. We return to this not-accidental similarity in the next section. 7.4.3 Integration Laplace-transforming the integral of a function yet) gives Integrating by parts, u = J ydt dv = e ” dt du = ydl 1 ” = e s’ s 240 PAW TWO: Laplace-Domain Dynamics and Control Y(l) IY(,#f + Time domain t I ; Y(.s, s =- Laplace dolnain FIGURE 7.4 Integration transfer function. Therefore, 2 if 1 y(,)dt = ‘Y + A s w s (7.50) The operation of integration is equivalent to division by s in the Laplace domain, using zero initial conditions. Thus integration is the inverse of differentiation. Fig- ure 7.4 gives a block diagram representation. The l/s is an operator or a transfer function showing what operation is performed on the input signal. This is a completely different idea from the simple Laplace trans- formation of a function. Remember, the Laplace transform of the unit step function is also equal to l/s. But this is the Laplace transformation of a function. The l/s operator discussed above is a transfer function, not a function. 7.4.4 Deadtime Delay time, transportation lag, or deadtime is frequently encountered in chemical en- gineering systems. Suppose a process stream is flowing through a pipe in essentially plug flow and that it takes D minutes for an individual element of fluid to flow from the entrance to the exit of the pipe. Then the pipe represents a deadtime element. If a certain dynamic variable fit,, such as temperature or composition, enters the front end of the pipe, it will emerge from the other end D minutes later with exactly the same shape, as shown in Fig. 7.5. pp, FIGURE ,.5 t=O t=D Effect of a dead-time element. cItAfTER 7: Laplace-Domain Dynamics 241 - Time domain Laplace domain FIGURE 7.6 Deadtime transfer function. Let us see what happens when we Laplace-transform a function h,-o, that has been delayed by a deadtime. Laplace transformation is defined in Eq. (7.5 1). af,,,1 = I Y (@I dt = F(s) o (7.5 1) The variable t in this equation is just a “dummy variable” of integration. It is inte- grated out, leaving a function of only s. Thus, we can write Eq. (7.5 1) in a completely equivalent mathematical form: (7.52) where y is now the dummy variable of integration. Now let y = t - D. F(s) = I m fit-& SO D) d(t - 0) = p m fit-D,e-“’ dt 0 f 0 (7.53) F(,) = eD”3Lf&d Therefore, %ht-o)l = e-DsFcs) (7.54) Thus, time delay or deadtime in the time domain is equivalent to multiplication by eeDS in the Laplace domain. If the input into the deadtime element is ~(~1 and the output of the deadtime element is y(+ then u and y are related by Y(f) = q-D) And in the Laplace domain, Y(,) = e -Ds 4) (7.55) Thus, the transfer function between output and input variables for a pure deadtime process is epDS, as sketched in Fig. 7.6. 7.5 EXAMPLES Now we are ready to apply all these Laplace transformation techniques to some typ- ical chemical engineering processes. I 242 PARTTWO: Laplace-Domain Dynamics and Control E x A M P I , E 7.3. Consider the isothermal CSTR of Example 2.6. The equation describing the system in terms of perturbation variables is - + L + k CiCn dt i 1 7 CA(,) = L CAOW 7 (7.56) where k and r are constants. The initial condition is CA(o) = 0. We do not specify what Cnocr, is for the moment, but just leave it as an arbitrary function of time. Laplace- transforming each term in Eq. (7.56) gives scA(s) - CA(t=O) + (7.57) The second term drops out because of the initial condition. Grouping like terms in CA($) gives Thus, the ratio of the output to the input (the “transfer function” Cc,,) is %s) Gtsj zz - = l/r CAO(s) s+k+ l/r (7.58) The denominator of the transfer function is exactly the same as the polynomial in s that was called the characteristic equation in Chapter 2. The roots of the denominator of the transfer function are called the poles of the transfer function. These are the values of s at which Gc,, goes to infinity. The roots of the characteristic equation are equal to the poles of the transfer function. This relationship between the poles of the transfer function and the roots of the charac- teristic equation is extremely important and useful. The transfer function given in Eq. (7.58) has one pole with a value of -(k + l/r). Rearranging Eq. (7.58) into the standard form of Eq. (2.51) gives G(,) = s+l 6 =- 7,s + 1 (7.59) where K, is the process steady-state gain and r,, is the process time constant. The pole of the transfer function is the reciprocal of the time constant. This particular type of transfer function is called a jrst-order lag. It tells us how the input CAO affects the output C A, both dynamically and at steady state. The form of the transfer function (polynomial of degree 1 in the denominator, i.e., one pole) and the numerical values of the parameters (steady-state gain and time constant) give a com- plete picture of the system in a very compact and usable form. The transfer function is a property of the system only and is applicable for any input. We can determine the dynamics and the steady-state characteristics of the system without having to pick any specific forcing function. If the same input as used in Example 2.6 is imposed on the system, we should be able to use Laplace transforms to find the response of CA to a step change of magni- - tude CAM. : :- ). U le kv ,f ke l- In le ‘Y )e i- CtIAtTtiK 7: Laplace-Domain Dynamics 243 We take the Laplace transform of Cncy,,, substitute into the system transfer function, solve for CA(,~), and invert back into the time domain to find CA(,). ~e[cAO,,,l = CA(@) = r,O; (7.61) (7.62) Using partial fractions expansion to invert (see Example 7.1) gives C n(t) = K,C,q) (1 - e-y This is exactly the solution obtained in Example 2.6 [Eq. (2.53)]. n EXAMPLE 7.4. The ODE of Example 2.8 with an arbitrary forcing function uct) is d2y dy - + 5& + 6y = u(t) dt2 with the initial conditions (7.63) (7.64) Laplace transforming gives s2 Y(s) + 5q,, + 6Y(,, = u(s) YG)(s3 + 5s + 6) = U(s) The process transfer function Gcs) is s = GcS) = s2 + ;, + 6 = 1 4s) (s + 2)(s + 3) Notice that the denominator of the transfer function is again the same polynomial in s as appeared in the characteristic equation of the system [Eq. (2.73)]. The poles of the transfer function are located at s = -2 and s = -3. So the poles of the transfer function are the roots of the characteristic equation. If uct) is a ramp input as in Example 2.13, (7.66) Y(s) = G(s) U(s) = (,+:,ih)(~)= s2(s+:)(s+3) Partial fractions expansion gives AB C Y(s) = p + ; + - s i- 2 A = li$ ,s2 YW i i = lim e-+0 B = lim s-o = lim c +o ;(s2y(v,j] = ;5&$2+:J+fj)] -(2x + 5) 1 5 (s’ + 5s + 6)’ = - 36 244 rAr<TTwo: Laplace-Domain Dynamics and Control f Therefore, Yt,s,] = lim .s+-2 Y(,s)] = lim s-r-3 5 I Y(,.) = h - E + -s - - 9 5 s2 s s + 2 s+3 (7.68) Inverting into the time domain gives the same solution as Eq. (2.109). jr(,) = it - 6 + $!-2f - $-3f (7.69) EXAMPLE 7.5. An isothermal three-CSTR system is described by the three linear ODES dCAr dt dcii2 dt The variables can be either total or perturbation variables since the equations are linear (all k’s and r’s are constant). Let us use perturbation variables, and therefore the initial conditions for all variables are zero. CA l(0) = CA*(O) = cA3(0) = 0 (7.71) Laplace transforming gives (3 + kr + -+(sj = -$AO(,, (S + k2 + -$A,,,, = $-A,(s) (S + k3 + +(s, = ;cA2(s) These can be rearranged to put them in terms of transfer functions for each tank. G CAl(s) I(J) = - = l/T, c AC’(s) s + k, + UT, G c _ AXE) 1172 2(s) _ - CA I($) s + k2 + l/~~ CA3(s) G3(s) = - = l/T3 C A2(s) s + k3 + l/~~ (7.72) (7.73) CtlAI’I‘I;R 7: Laplace-Domain Dynamics 245 CAOW CAI(,) GZ(.s) &3(s) ___L G(s) - G2(s) + G(s) L GO(s) CAR(s) cc I(s) G2(s) . G3(s) t- FIGURE 7.7 Transfer functions in series. If we are interested in the total system and want only the effect of the input CAO on the output CAM, the three equations can be combined to eliminate CA, and CA*. CA3(s) = G3CA2(s) = G3 (G2CA,(s)) = ‘GG2 (G CAOW) (7.74) The overall transfer function Go) is G(s) = C’A3(s) - = G(.&s)G(s) CAO(s) (7.75) This demonstrates one very important and useful property of transfer functions. The total effect of a number of transfer functions connected in series is just the product of all the individual transfer functions. Figure 7.7 shows this in block diagram form The overall transfer function is a third-order lag with three poles. G(s) = l/TtQT3 (s + kt + l/~t)(s + k2 + 1/r2)(s + k3 + 11~~) (7.76) Further rearrangement puts the above expression in the standard form with time con- stants roi and a steady-state gain K,. 1 1 1 G= (1 + klT1) (1 + km) (1 + km) i 71 s+l ji 72 s+l 73 1 + k,q 1 + k2r2 1 + k3r3 s+l (7.77) G(s) = K?J (701s + 1x702s + 1)(703s + 1) Let us assume a unit step change in the feed concentration CAO and solve for the response of cA3. We will take the case where all the T,~‘s are the same, giving a repeated root of order 3 (a third-order pole at s = - l/7,). C AW) = &r(l) +’ CAO(s) = f C = G(.FJCAO(.X) = K, 1 -= K/Jr,3 AX(s) (7,s + I>3 s s(s + l/7,)3 (7.78) 246 PARTTWO: Laplace-Domain Dynamics and Control Applying partial fractions expansion, = K,, J I 1 2K& -___ Inverting Eq. (7.79) with the use of Eq. (7.18) yields 1 z-z 1 2, s3 1 - KP (7.80) m EXAMPLE 7.6. A nonisothermal CSTR can be linearized (see Problem 2.8) to give two linear ODES in terms of perturbation variables. dCA - = a,,C,‘j + czl;?T + Q,jCAO + a15F dr dT - = a?lC~ + n22T + a2JTo + a2SF + az6TJ dt where F a,, = -_ -k a12 = -Z?;AEk F V RT2 aI3 = - V a15 = c,, - c, -AT; -hkE& F V a*1 = - a22 = PC, pC,RT2 v F a24 = - 7,) - ;7; VA V a2.j = ____ V a26 = vpc, (7.81) UA (7.82) VPC, The variables CAo, TO. F. and T, are all considered inputs. The output variables are CA and T. Therefore, eight different transfer functions are required to describe the sys- tem completely. This multivariable aspect is the usual situation in chemical engineering systems. [...]... fleet, under the command of sinister Admiral von Dietrich, consists of 200 U-boats at the beginning of the battle The British destroyer fleet, under the command of heroic Admiral Steadman (a direct descendant of the intelligence officer responsible for the British victory at the Battle of Trafalgar), consists of 150 ships at the beginning of the battle The rate of destruction of submarines by destroyers... l/2160 “F/Btu/min (s + 1)(5s + 1) = I( 50 0,000 Btu/min WC) 16 mA ! I( (8.7) l + l/2160 “F/Btu/min 5 0 0 , 0 0 0 Btu/min 16 mA )(Kc)(%) (s + 1)(5s + 1) [ + 50 ,000K,/216/16 “F/mA = 5s2 6s + I + 5OOKJ216 If we look at the closedloop transfer function between PV and SP, we must multiply the above by Gr PV 50 0 Kc/2 16 mAImA sp= 5s2 + 6s + 1 + 50 0KJ2 I6 (5. 8) Notice that the denominators of all these closedloop... the error signal E through a feedback controller transfer function Gets), and sends out a controller output signal CO The controller output signal changes the position of a control valve, which changes the flow rate of the manipulated variable M Figure 8.26 gives a sketch of the feedback control system and a block diagram for the two-heated-tank process with a controller Let us use an analog electronic... two-heated-tank process can be calculated from the openloop process transfer functions and the feedback controller transfer function We choose a proportional controller, so Gco) = Kc Note that the dimensions of the gain of the controller are mA/mA, i.e., the gain is dimensionless The controller looks at a milliampere signal (PV) and puts out a milliampere signal (CO) I“F /OF GLw = (s + I) (57 + 1) l/2160... tank process has a steadystate gain with units of “F /OF The GMtsj transfer function has a steady-state gain with units of OF/Btu/min 8.1.2 Closedloop Characteristic Equation and Closedloop Transfer Functions Now let us put a feedback controller on the process, as shown in Fig 8.2~ The controlled variable is converted to a process variable signal PV by the sensor/transmitter element GT(~J The feedback controller... Laplace-Domain Dynamics and Control chemical engineering), for a prediction of the outcome of the upcoming battle Steadman has been working with the new engineering officers in the fleet, Lt Moquin and Lt Walsh, who have replaced the retired Lt Scott These innovative officers have been able to increase the firepower of half of the vessels in Kirk’s fleet by a factor of 2 over the firepower of the Klingon vessels,... roots of the denominator of the system transfer function (poles) are exactly the same as the roots of the characteristic equation Thus, for the system to be stable, the poles of the transfer function must lie in the left half of the s plane (LHP) This stability requirement applies to any system, openloop or closedloop The stability of an openloop process depends on the location of the poles of its... stable or unstable, depending on the values of the controller parameters We will show that any real process can be made closedloop unstable by making the gain of the feedback controller large enough There are some processes that are openloop unstable We will show that these systems can usually be made closedloop stable by the correct choice of the type of controller and its settings The most useful... which all have the same firepower The firepower of the rest of Kirk’s fleet is on a par with that of the Klingons But these officers have also been able to improve the defensive shields on this second half of the fleet The more effective shields reduce by 50 percent the destruction rate of these vessels by the Klingon firepower Thus, there are two classes of starships: eight vessels are Class Et with... Openloop process The dynamics of this openloop system depend on the roots of the openloop characteristic equation, i.e., on the roots of the polynomials in the denominators of the openloop transfer functions These are the poles of the openloop transfer functions If all the roots lie in the left half of the s plane, the system is openloop stable For the two-heated-tank example shown in Fig 8.lb, the poles of . l/r). Rearranging Eq. (7 .58 ) into the standard form of Eq. (2 .51 ) gives G(,) = s+l 6 =- 7,s + 1 (7 .59 ) where K, is the process steady-state gain and r,, is the process time constant. The pole of the transfer. yields 7.4 TRANSFER FUNCTIONS Our primary use of Laplace transformations in process control involves representing the dynamics of the process in terms of “transfer functions.” These are output-input ~~~ relationships. +o ;(s2y(v,j] = ;5& amp;$2+:J+fj)] -(2x + 5) 1 5 (s’ + 5s + 6)’ = - 36 244 rAr<TTwo: Laplace-Domain Dynamics and Control f Therefore, Yt,s,] = lim .s+-2 Y(,s)] = lim s-r-3 5 I Y(,.)