# ch14 - Revised: 1-3-04 Chapter 14 14.1 Let GOL(jc) = R + jI...

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14-1 Chapter 14 14.1 L e t G OL ( j ω c ) = R + jI where ω c is the critical frequency. Then, according to the Bode stability criterion | G OL ( j ω c )| = 1 = 2 2 I R + G OL ( j ω c ) = - π = tan –1 ( I/R ) Solving for R and I : R = -1 and I = 0 Substituting s = j ω c into the characteristic equation gives, 1 + G OL ( j ω c ) = 0 I + R + jI = 0 or R = -1 , I = 0 Hence, the two approaches are equivalent. 14.2 Because sustained oscillations occur at the critical frequency 1 2 ω 0.628min 10 min c π == (a) Using Eq. 14-7, 1 = ( K c )(0.5)(1)(1.0) or K c = 2 (b) Using Eq. 14-8, π = 0 + 0 +(- θω c ) + 0 o r θ = 5min ω c = 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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14-2 14.3 (a) From inspection of the Bode diagrams in Tables 13.4 and 13.5, the transfer function is selected to be of the following form G (s) = ) 1 )( 1 ( ) 1 ( 2 1 + τ + τ + τ s s s s K a where τ a , τ 1 , τ 2 correspond to frequencies of ω = 0.1, 2, 20 rad/min, respectively. Therefore, τ a = 1/0.1 = 10 min τ 1 =1/2 = 0.5 min τ 2 = 1/20 = 0.05 min For low frequencies, AR | K / s | = K / ω A t ω = 0.01, AR = 3.2, so that K = ( ω )(AR) = 0.032 Therefore, G ( s ) = ) 1 05 . 0 )( 1 5 . 0 ( ) 1 10 ( 032 . 0 + + + s s s s (b) Because the phase angle does not cross -180 ° , the concept of GM is meaningless. 14.4 The following process transfer can be derived in analogy with Eq. 6-71: 2 11 1 112 2 21 2 2 () ( ) 1 = + ++ + Hs R Qs A R A R s A RA R s F o r R 1 =0.5, R 2 = 2, A 1 = 10, A 2 = 0.8:
14-3 G p (s) = 1 7 8 5 . 0 2 + + s s ( 1 ) For R 2 = 0.5: G p (s) = 1 8 . 5 2 5 . 0 2 + + s s (2) (a) For R 2 = 2 G p = tan -1 ω ω 2 8 1 7 c c , | G p | = 22 2 0.5 (1 8 ) (7 ) −ω + ω cc For G v = K v = 2.5, ϕ v =0, | G v | = 2.5 For G m = 1 5 . 0 5 . 1 + s , ϕ m = -tan -1 (0.5 ω ) , | G m | = 1 ) 5 . 0 ( 5 . 1 2 + ω c K cu and ω c are obtained using Eqs. 14-7 and 14-8: - 1 8 0 ° = 0 + 0 + tan -1 ω ω 2 8 1 7 c c tan -1 (0.5 ω c ) Solving, ω c = 1.369 rad/min. ω + ω = 2 2 2 ) 7 ( ) 8 1 ( 5 . 0 ) 5 . 2 )( ( 1 c c cu K + ω 1 ) 5 . 0 ( 5 . 1 2 c Substituting ω c = 1.369 rad/min, K cu = 10.96, ω c K cu = 15.0 For R 2 =0.5 G p = tan -1 ω ω 2 2 1 8 . 5 c c , | G p | = ω + ω 2 2 2 ) 8 . 5 ( ) 2 1 ( 5 . 0 c c - 1 8 0 ° = 0 + 0 + tan -1 ω ω 2 2 1 8 . 5 c c tan -1 (0.5 ω c ) Solving, ω c = 2.51 rad/min. Substituting ω c = 2.51 rad/min, K cu = 15.93, ω c K cu = 40.0

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14-4 (a) From part (a), for R 2 =2, ω c = 1.369 rad/min, K cu = 10.96 P u = c ω π 2 = 4.59 min Using Table 12.6, the Ziegler-Nichols PI settings are K c = 0.45 K cu = 4.932 , τ I = P u /1.2 = 3.825 min Using Eqs. 13-63 and 13-62 , ϕ c = -tan -1 (-1/3.825 ω ) | G c | = 4.932 1 825 . 3 1 2 + ω Then, from Eq. 14-7 -180 ° = tan -1 ω c 825 . 3 1 + 0 + tan -1 ω ω 2 8 1 7 c c tan -1 (0.5 ω c ) Solving, ω c = 1.086 rad/min. Using Eq. 14-8, A c = AR O L | ω = ω c = = ω + ω + ω 2 2 2 2 ) 7 ( ) 8 1 ( 5 . 0
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## This homework help was uploaded on 04/10/2008 for the course CHE 242 taught by Professor Cummings during the Spring '08 term at Vanderbilt.

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ch14 - Revised: 1-3-04 Chapter 14 14.1 Let GOL(jc) = R + jI...

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