ch18 - 123456789 8 18.1 McAvoy has reported the PI...

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18-1 ±²³´µ¶·¸¹º¸ 18.1 McAvoy has reported the PI controller settings shown in Table S18.1 and the set-point responses of Fig. S18.1a and S18.1b. When both controllers are in automatic with Z-N settings, undesirable damped oscillations result due to the control loop interactions. The multiloop tuning method results in more conservative settings and more sluggish responses. Controller Pairing Tuning Method K c τ I (min) T 17 - R Single loop/Z-N -2.92 3.18 T 4 - S Single loop/Z-N 4.31 1.15 T 17 - R Multiloop -2.59 2.58 T 4 - S Multiloop 4.39 2.58 Table S18.1. Controller Settings for Exercise 18.1 0 5 10 15 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Time(min) Single loop tuning (one loop in manual) Single loop tuning (both loops in automatic) Multiloop tuning T 17 Figure S18.1a. Set point responses for Exercise 18.1. Analysis for T 17 Solution Manual for Process Dynamics and Control, 2 nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
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18-2 0 5 10 15 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 Time(min) Single loop tuning (one loop in manual) Single loop tuning (both loops in automatic) Multiloop tuning T 4 Figure S18.1b. Set point responses for Exercise 18.1. Analysis for T 4 18.2 The characteristic equation is found by determining any one of the four transfer functions Y 1 ( s )/ Y sp 1 ( s ), Y 1 ( s )/ Y sp2 ( s ), Y 2 ( s )/ Y sp 1 ( s ) and Y 2 ( s )/ Y sp2 ( s ), and setting its denominator equal to zero. In order to determine, say, Y 1 ( s )/ Y sp 1 ( s ), set Y sp2 = 0 in Fig 18.3 b and use block diagram algebra to obtain 12 1 11 1 1 1 1 ( ) [ ( ) ( )] ( ) P C P C s G G R s C s G M s = - + (1) 2 21 22 1 1 1 1 1 ( ) ( [ ( ) [ ( ) ( )]]) C P P C M s G G M s G G R s C s = - + - (2) Simplifying (2), 2 22 1 2 21 1 1 1 ( ) [ ( ) ( )] 1 C P C C P G G G M s R s C s G G - = - + (3) Substituting (3) into (1) and simplifying gives 1 12 2 21 1 2 11 22 1 12 2 21 1 2 11 22 1 1 ( )(1 ) ( ) ( ) (1 )(1 ) C P C P C C P P C P C P C C P P G G G G G G G G C s R s G G G G G G G G + - = + + -
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18-3 Therefore characteristic equations is (1 + G c 1 G p 12 ) (1 + G c 2 G p 21 ) G c 1 G c 2 G p 11 G p 22 = 0 If either G p 11 or G p 22 is zero, this reduces to (1 + G c 1 G p 12 ) = 0 or (1 + G c 2 G p 21 ) = 0 So that the stability of the overall system merely depends on the stability of the two individual feedback control loops in Fig. 18.3 b since the third loop containing G p 11 and G p 22 is broken. 18.3 Consider the block diagram for the 1-1/2-2 control scheme in Fig.18.3 a but including a sensor transfer function ( G m 1 , G m 2 ) for each output ( y 1 , y 2 ). The following expressions are easily derived, Y ( s ) = G p ( s ) U ( s ) or 11 12 1 1 21 22 2 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) p p p p G s G s Y s U s G s G s Y s U s = (1) U ( s ) = G c (s) E (s) or 1 1 1 2 2 2 ( ) 0 ( ) ( ) 0 ( ) ( ) ( ) c c G s U s E s G s U s E s = (2) E ( s )= Y sp ( s )- G m ( s ) Y ( s ) or 1 1 1 1 2 2 2 2 ( ) ( ) 0 ( ) ( ) ( ) 0 ( ) ( ) ( ) sp m sp m Y s G s E s Y s Y s G s E s Y s = - (3) If Eqs. 1 through 3 are solved for the response of the output to variations of set points, the result is Y ( s ) = G p ( s ) G c ( s ) [ I + G p ( s ) G c ( s ) G m ( s )] -1 Y sp ( s ) = where I is the identity matrix.
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ch18 - 123456789 8 18.1 McAvoy has reported the PI...

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