# ch04.pdf - 1234567898 4.1 a b c d iii iii v v a 5 b 10 4.2...

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4-1 ±²³´µ¶·¸¹¸ 4.1 a) iii b) iii c) v d) v 4.2 a) 5 b) 10 c) ) 1 10 ( 10 ) ( + = s s s Y From the Final Value Theorem, y ( t ) = 10 when t d) y ( t ) = 10(1 - e - t /10 ) , then y (10) = 6.32 = 63.2% of the final value. e) s e s s Y s ) 1 ( ) 1 10 ( 5 ) ( - - + = From the Final Value Theorem, y ( t ) = 0 when t →∞ f) 1 ) 1 10 ( 5 ) ( + = s s Y From the Final Value Theorem, y ( t )= 0 when t →∞ g) ) 9 ( 6 ) 1 10 ( 5 ) ( 2 + + = s s s Y then y ( t ) = 0.33e -0.1 t - 0.33cos(3 t ) + 0.011sin(3 t ) The sinusoidal input produces a sinusoidal output and y ( t ) does not have a limit when t →∞ . 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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4-2 By using Simulink-MATLAB, above solutions can be verified: 0 5 10 15 20 25 30 35 40 45 50 0 1 2 3 4 5 6 7 8 9 10 time y(t) 0 5 10 15 20 25 30 35 40 45 50 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 time y(t) Fig S4.2a. Output for part c) and d) Fig S4.2b. Output for part e) 0 5 10 15 20 25 30 35 40 45 50 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 time y(t) 0 2 4 6 8 10 12 14 16 18 20 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 time y(t) Fig S4.2c. Output for part f) Fig S4.2d. Output for part g) 4.3 a) The dynamic model of the system is given by ) ( 1 w w dt dV i - ρ = (2-45) C V Q T T V w dt dT i i ρ + - ρ = ) ( (2-46) Let the right-hand side of Eq. 2-46 be f( w i ,V,T ), ( 29 T T f V V f w w f T V w f dt dT s s i s i i + + = = , , (1)
4-3 ) ( 1 T T V w f i s i - ρ = 0 1 ) ( 2 2 = - = ρ - - ρ - = s i i s dt dT V C V Q T T V w V f ρ - = V w T f i s = dt dT i i w T T V - ρ ) ( 1 T V w i ρ - , dt T d dt dT = Taking Laplace transform and rearranging 1 / ) ( ) ( ) ( + ρ - = s w V w T T s W s T i i i i (2) Laplace transform of Eq. 2-45 gives s s W s V i ρ = ) ( ) ( (3) If s V f were not zero, then using (3) 1 1 ) ( ) ( ) ( + ρ + - = s w V s V f w V w T T s W s T i s i i i i (4) Appelpolscher guessed the incorrect form (4) instead of the correct form (2) because he forgot that s V f would vanish. b) From Eq. 3, s s W s V i ρ = 1 ) ( ) (

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4-4 4.4 1 K Y( s ) G( s )X( s ) s( s ) = = τ + G ( s ) Interpretation U ( s ) Interpretation of G ( s ) of u ( t ) ) 1 ( + τ s s K 2 nd order process * 1 δ (0) [ Delta function] 1 + τ s K 1 st order process s 1 S (0) [Unit step function] s K Integrator 1 + τ s K τ - τ / 1 t e [Exponential input] K Simple gain ) 1 ( 1 + τ s s τ - - / 1 t e (i.e no dynamics) [Step + exponential input] * 2 nd order or combination of integrator and 1 st order process 4.5 a) 2 dt d 1 y = -2y 1 – 3y 2 + 2u 1 (1) dt d 2 y = 4y 1 – 6y 2 + 2u 1 + 4u 2 (2) Taking Laplace transform of the above equations and rearranging, (2 s +2)Y 1 ( s ) + 3Y 2 ( s ) = 2U 1 ( s ) (3) -4 Y 1 ( s ) + ( s +6)Y 2 (s)=2U 1 ( s ) + 4U 2 ( s ) (4) Solving Eqs. 3 and 4 simultaneously for Y 1 (s) and Y 2 (s),
4-5 Y 1 ( s ) = ) 4 )( 3 ( 2 ) ( U 12 ) ( U ) 3 ( 2 24 14 2 ) ( U 12 ) ( U ) 6 2 ( 2 1 2 2 1 + + - + = + + - + s s s s s s s s s s Y 2 ( s ) = ) 4 )( 3 ( 2 ) ( U ) 1 ( 8 ) ( U ) 3 ( 4 24 14 2 ) ( U ) 8 8 ( ) ( U ) 12 4 ( 2 1 2 2 1 + + + + + = + + + - + s s s s s s s s s s s s Therefore, 4 1 ) ( U ) ( Y 1 1 + = s s s , ) 4 )( 3 ( 6 ) ( U ) ( Y 2 1 + + - = s s s s 4 2 ) ( U ) ( Y 1 2 + = s s s , ) 4 )( 3 ( ) 1 ( 4 ) ( U ) ( Y 2 2 + + + = s s s s s 4.6

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• Spring '16
• Mark Darby
• Trigraph, dt, dt Vρ

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