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TABLE 34.2
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Response of level for Example 34.2
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TIHE,HII 0.500 L.000 1.500 2.000 2.500 3.000 3.500 LEVEL,PT 0.033 lJ.ll99 0.5OY u.wli L.24b L.!ilJ5
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ll.3~2
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l.Ob7 D. Vr, 0.57L
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9.000 9.500 L0.000
iS.000 h.500
Ct.690 0.899
ll.zl43 L.279 ll.359 ll.3g2
COMPUTERS IN PROCESS CONTROL
5 REH CONTROL OF TRREE-TANK SYSTEH; USE OF RUNGE-KUTTA;EX. 2 10 DEF FNH(M,X,Y,Z) = -KV*KC*Y + KC*KV*(L-X)/TAUI 20 DEF FNX(H,X,Y,Z) = Y 30 DEF FNY(H,X,Y,Z) = Z 90 DEF FNZ(H,X,Y,Z) = -2.5*2 -2*Y -.5*X + .275+H bll KV = -07 70 KC = LO7.2 80 TAUI = 3.7 90 H = KC*KV LOO x = 0 IJIIO Y = 0 II20 z = 0 I,30 T = 0 L40 DT = .I 141 PRINT "TIKE,HIN","LEVEL,FT" I,'#2 K = 0 144 K = K+L 1115 IF K=b TEEN 2117 L4b GOT0 I50 1117 PRINT USING "##.### ";T,X I,'+8 K=L 150 HL = FNH(H,X,Y,Z)*DT Lb0 XL = FNX(M,X,Y,Z)*DT I170 YL = FNY(M,X,Y,Z)*DT LB0 ZL = FNZ(H,X,Y,Z)*DT 190 BP = FNM(H+HL*.5,X+XL*.5,Y+PLc.S,Z+ZLr.S)*DT 200 X2 = FNX(H+HL*.S,X+XL*.S,P+PL+.S,Z+ZLr.S)*DT 2LO Y2 = FNY(M+ML*.5,X+XL*.5,P+YLI.S,Z+ZLI.S)rDT 220 22 = FNZ(H+HL+.S,X+XL*.S,P+PL+.S,Z+ZL+.S)IDT 230 83 = FNH(1+12*.5,X+X2*.5,Y+YP*.5)*DT R&O X3 = FNX(H+MR*.5,X+X2*.5,Y+YP*.5,Z+Z2r.S)rDT 250 Y3 = FNY(H+H2*.5,X+X2*.5,Y+Y, *.5,Z+ZP*.5)*DT 2bO 23 = FNZ(H+H2+.5,X+X2*.5,Y+YP*.5)iDT 270 HIII = FN~I~(K+M~,X+X~,Y+Y~,Z+Z~)*DT 280 X'd = FNX(H+M3,X+X3,Y+Y3,Z+Z3)*DT 290 Ylr = FNY(H+H3,X+X3,Y+Y3,Z+Z3)*DT 300 Zlr = FNZ(H+H3,X+X3,Y+Y3,Z+Z3)*DT 310 II = H+(Z/b)+(H1+82*2+M3*~+Hq) 320 X = X+(l/b)*(Xl+XP*P+X3*P+XQ) 330 Y = Y+(L/b)*(YL+YP*P+Y3*P+Y~) 340 Z = Z+(L/b)*(ZZ+Z2*2+Z3*P+Ztt) 350 T = T+DT 370 IF T>LO.L THEN END 380 GOT0 L'+lr 390 END FIGURE34-6 BASIC computer program for liquid-level control system for Example 34.2 for a step change in set point of 1.0 ft and K, = 107 psi/ft and ~1 = 3.7 min.
DIGITAL COMPUTER
SIMULATION OF CONTROL SYSTEMS
compute y only at discrete times, we must store values of y in an array of computer storage locations, called a stack. The diagram in Fig. 34.8 will help clarify this storage. The array will be used to store past values of y that were computed at the end of each computation interval. At the end of each computation interval the values of y will be moved one position toward the end of the stack and the value of y just computed will be placed in the first storage location of the stack. By this means, a current value of y will not appear at the end of the stack until it has moved through each storage location. The amount of time the current value of y is delayed will depend on the number of storage locations and At. The number of storage locations N is determined-by:, N = 7JA.t Let the values stored in the array be S(i) where i, which represents the array position, will vary from 1 to N + 1. The following terms are now defined for the computer program to be developed. Y = y, present value of y S(i) = stored past values of y S(l) = current value of y, obtained at end of most recent computation interval S(N + 1) = X, the delayed value of y, i.e., X = y(t - s-d)
Present value of Y
*Delayed value of Y (i.e., X)
FIGURE 34-8 Array used to obtain transport lag in Example 34.3.
528 co MF UTERS
IN PROCFiS.5 COhTROL
An outline of the procedure for computing y at discrete values of t is as follows. 1. Let the array for storing values of y be of length N + 1 where N = TdlAt. 2. Initialize the elements of the array to zero. 3. Initialize the time variable, T = 0. 4. Set X = S(N + 1). 5. Print T, Y, and X. 6. Start the Runge-Kutta routine to integrate the differential equation over the first computation interval AT. 7. Rearrange the contents of the array as shown in Fig. 34.8 by shifting the contents of each storage location by one position. Start shifting from the bottom. In this shifting, the oldest value of Y will be discarded and the value of Y just computed will enter the first cell to become S( 1). 8. Store the value of Y just computed into S(I), i.e., S(l) = Y. 9. Increment T by AT and return to step (4) to repeat another cycle of calculation. Using the steps just listed, the BASIC computer program shown in Fig. 34.9 was written for the conditions: r = 1.0, rd = 0.2, R = u(t), K, = 8.4, and At = 0.02. For these conditions, N = Qhit = 0.210.02 = 10 The output from the computer program is shown in Table 34.3. The computer program for simulating a transport lag that has just been presented is quite primitive compared to those provided in commercial software in which the delayed function is not held constant during the time step At, but is allowed to vary by use of an interpolation scheme. Some of the simulation software packages that provide the simulation of transport lag (e.g., TUTSIM, ACSL, and CC) are listed at the end of this chapter. Example 34.4. Simulation of PDD control. The presence of derivative action in a control algorithm, such as PID control, gives some difficulty in the writing of a program for digital computer simulation. Consider the PID control of a first-order system as shown in Fig. 34.10. To obtain a set of first-order differential equations for use with the Runge-Kutta method, we proceed as foIlows. Prom the controller block, we obtain
= $(r,r,s2 + 71s + 1)
(34.42)
Cross-multiplying this expression, solving for sM, and writing the result in the time domain give ti = K,i + (K&)e + K,Td Z (34.43) This expression is not in the form in which the right side is free of derivatives of the variables. To obtain the correct form, we proceed as follows. Since R = 0 for this problem, we may write
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