ch15 - Revised: 1-3-04 Chapter 15 15.1 For Ra=d/u Kp = Ra d...

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15-1 Chapter 15 15.1 F o r R a = d /u 2 u d u R K a p = = which can vary more than K p in Eq. 15-2, because the new K p depends on both d and u . 15.2 By definition, the ratio station sets ( u m u m0 ) = K R ( d m d m0 ) Thus 2 1 2 2 1 2 2 0 0 = = = d u K K d K u K d d u u K m m m m R (1) For constant gain K R , the values of u and d in Eq. 1 are taken to be at the desired steady state so that u / d = R d , the desired ratio. Moreover, the transmitter gains are 2 1 mA ) 4 20 ( d S K = , 2 2 mA ) 4 20 ( u S K = Substituting for K 1 , K 2 and u/d into (1) gives: 2 2 2 2 = = u d d d d u R S S R R S S K Solution Manual for Process Dynamics and Control, 2 nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp Revised: 1-3-04
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15-2 15.3 (a) The block diagram is the same as in Fig. 15.11 where Y H 2 , Y m H 2 m , Y sp H 2 sp , D Q 1 , D m Q 1 m , and U Q 3 . b) (A steady-state mass balance on both tanks gives 0 = q 1 q 3 or Q 1 = Q 3 (in deviation variables) (1) From the block diagram, at steady state: Q 3 = K v K f K t Q 1 From (1) and (2), K f = 1 vt KK ( 2 ) c) (No, because Eq. 1 above does not involve q 2 . 15.4 (b) From the block diagram, exact feedforward compensation for Q 1 would result when Q 1 + Q 2 = 0 Substituting Q 2 = K V G f K t Q 1 , G f = 1
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15-3 (c) Same as part (b), because the feedforward loop does not have any dynamic elements. (d) For exact feedforward compensation Q 4 + Q 3 = 0 ( 1 ) From the block diagram, Q 2 = K V G f K t Q 4 (2) Using steady-state analysis , a mass balance on tank 1 for no variation in q 1 gives Q 2 Q 3 = 0 ( 3 ) Substituting for Q 3 from (3) and (2) into (1) gives Q 4 + K V G f K t Q 4 = 0 or G f = 1 vt KK For dynamic analysis , find G p 1 from a mass balance on tank 1, 1 11 2 1 1 dh A qqC h dt =+−
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15-4 Linearizing (4), noting that 1 q = 0, and taking Laplace transforms: 1 12 1 1 2 C dh A qh dt h ′′ =− or 11 1 2 11 1 (2 / ) () (2 / ) 1 hC Hs Qs A hCs = + (5) Since 31 1 1 1 2 qC h C h = = or 3 1 1 1 2 C h = (6) From (5) and (6), 1 3 2 1 (2 / ) 1 P G AhCs == + (7) Substituting for Q 3 from (7) and (2) into (1) gives 44 1 0 (2 / ) 1 vft QK G K Q += + o r 1 1 [(2 / ) 1] f vt GA h C s KK + 15.5 (a) For a steady-state analysis: G p =1, G d =2, G v = G m = G t =1 From Eq.15-21, G f = 2 ) 1 )( 1 )( 1 ( 2 = = p t v d G G G G (b) Using Eq. 15-21,
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15-5 G f = 1 5 2 1 1 ) 1 )( 1 ( ) 1 5 )( 1 ( 2 + = + + + = s s s s G G G G p t v d (c) Using Eq. 12-19, + = + = = G G s G G G G m p v ~ ~ 1 1 ~ where 1 1, 1 GG s +− == + ±± For τ c =2, and r =1, Eq. 12-21 gives f = 1 2 1 + s From Eq. 12-20 *1 11 (1 ) 21 +  + =  + +  ± c s fs ss From Eq. 12-16 s s s s s G G G G c c c 2 1 1 2 1 1 1 2 1 ~ 1 + = + + + = = (d) For feedforward control only, G c =0. For a unit step change in disturbance, D ( s ) = 1/ s. Substituting into Eq. 15-20 gives Y ( s ) = ( G d +G t G f G v G p ) s 1 For the controller of part (a) Y ( s ) = s s s s 1 1 1 ) 1 )( 2 )( 1 ( ) 1 5 )( 1 ( 2 + + + +
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15-6 Y ( s ) = 5 / 1 5 . 2 1 5 . 2 1 5 2 / 25 1 2 / 5 ) 1 5 )( 1 ( 10 + + = + + + = + + s s s s s s or y ( t ) = 2.5 (e - t – e - t /5 ) For the controller of part (b) Y ( s ) = 0 1 1 1 ) 1 ( 1 5 2 ) 1 ( ) 1 5 )( 1 ( 2 = +
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ch15 - Revised: 1-3-04 Chapter 15 15.1 For Ra=d/u Kp = Ra d...

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