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# ch07 - 1234567898 7.1 In the absence of more accurate data...

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7-1 ±²³´µ¶·¸¹¸ 7.1 In the absence of more accurate data, use a first-order transfer function as '( ) '( ) 1 s i T s Ke Q s s = τ + o ( ) (0) (124.7 120) F 0.118 540 500 gal/min i T T K q ∞ - - = = = - θ = 3:09 am – 3:05 am = 4 min Assuming that the operator logs a 99% complete system response as “no change after 3:34 am”, 5 time constants elapse between 3:09 and 3:34 am. 5 τ = 3:34 min - 3:09 min = 25 min τ = 25/5 min = 5 min Therefore, 4 '( ) 0.188 '( ) 5 1 s i T s e Q s s - = + To obtain a better estimate of the transfer function, the operator should log more data between the first change in T and the new steady state. 7.2 Process gain, 2 (5.0) (0) (6.52 5.50) min 0.336 30.4 0.1 ft i h h K q - - = = = × a) Output at 63.2% of the total change = 5.50 + 0.632(6.52-5.50) = 6.145 ft Interpolating between h = 6.07 ft and h = 6.18 ft 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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7-2 (0.8 0.6) 0.6 (6.145 6.07)min 0.74min (6.18 6.07) - τ = + - = - b) 0 (0.2) (0) 5.75 5.50 ft ft 1.25 0.2 0 0.2 min min t dh h h dt = - - = = - Using Eq. 7-15, 0 t KM dh dt = τ = = min 84 . 0 25 . 1 ) 1 . 0 4 . 30 ( 347 . 0 = × × c) The slope of the linear fit between t i and - - - ) 0 ( ) ( ) 0 ( ) ( 1 ln h h h t h z i i gives an approximation of (-1/ τ ) according to Eq. 7-13. Using h ( ) = h (5.0) = 6 .52, the values of z i are t i z i t i z i 0.0 0.00 1.4 -1.92 0.2 -0.28 1.6 -2.14 0.4 -0.55 1.8 -2.43 0.6 -0.82 2.0 -2.68 0.8 -1.10 3.0 -3.93 1.0 -1.37 4.0 -4.62 1.2 -1.63 5.0 - Then the slope of the best-fit line, using Eq. 7-6 is 2 13 1 13 ( ) tz t z tt t S S S slope S S - = - = τ - (1) where the datum at t i = 5.0 has been ignored. Using definitions, 0 . 18 = t S 4 . 40 = tt S 5 . 23 - = z S 1 . 51 - = tz S Substituting in (1), 1 1.213 - = - τ 0.82min τ =
7-3 d) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.4 5.6 5.8 6 6.2 6.4 6.6 6.8 Experimental data Model a) Model b) Model c) Figure S7.2. Comparison between models a), b) and c) for step response. 7.3 a) 1 1 1 ( ) ( ) 1 T s K Q s s = τ + 2 2 1 2 ( ) ( ) 1 T s K T s s = τ + 2 2 1 2 1 2 1 2 1 ( ) ( ) ( 1)( 1) ( 1) s T s K K K K e Q s s s s = τ + τ + τ + (1) where the approximation follows from Eq. 6-58 and the fact that τ 1 > τ 2 as revealed by an inspection of the data. 667 . 2 82 85 0 . 10 0 . 18 ) 0 ( ) 50 ( 1 1 1 = - - = - = q T T K 75 . 0 0 . 10 0 . 18 0 . 20 0 . 26 ) 0 ( ) 50 ( ) 0 ( ) 50 ( 1 1 2 2 2 = - - = - - = T T T T K Let z 1 , z 2 be the natural log of the fraction incomplete response for T 1 , T 2 , respectively. Then,

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7-4 - = - - = 8 ) ( 18 ln ) 0 ( ) 50 ( ) ( ) 50 ( ln ) ( 1 1 1 1 1 1 t T T T t T T t z - = - - = 6 ) ( 26 ln ) 0 ( ) 50 ( ) ( ) 50 ( ln ) ( 2 2 2 2 2 2 t T T T t T T t z A graph of z 1 and z 2 versus t is shown below. The slope of z 1 versus t line is –0.333 ; hence (1/- τ 1 )=-0.333 and τ 1 =3.0 From the best-fit line for z 2 versus t , the projection intersects z 2 = 0 at t 1.15. Hence τ 2 =1.15. 1 3 667 . 2 ) ( ' ) ( ' 1 + = s s Q s T (2) 1 15 . 1 75 . 0 ) ( ' ) ( ' 1 2 + = s s T s T (3) Figure S7.3a. z 1 and z 2 versus t b) By means of Simulink-MATLAB, the following simulations are obtained 0 2 4 6 8 10 12 14 16 18 20 22 10 12 14 16 18 20 22 24 26 28 time T 1 , T 2 T 1 T 2 T 1 (experimental) T 2 (experimental) Figure S7.3b. Comparison of experimental data and models for step change -8.0 -7.0 -6.0 -5.0 -4.0 -3.0 -2.0 -1.0 0.0 0 5 10 15 20 time,t z 1 ,z 2
7-5 7.4 s s s s s X s G s Y 5 . 1 ) 1 )( 1 3 )( 1 5 ( 2 ) ( ) ( ) ( × + + + = = Taking the inverse Laplace transform ( ) -75/8*exp(-1/5*t)+27/4*exp(-1/3*t)-3/8*exp(-t)+3 y t = (1) a) Fraction incomplete response

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