# ch05.pdf - 1234567898 5.1 a xDP(t = hS(t 2hS(t-tw hS(t-2tw...

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5-1 ±²³´µ¶·¸¹¸ 5.1 a) x DP ( t ) = hS ( t ) 2 hS ( t-t w ) + hS ( t- 2 t w ) x DP ( s ) = s h ( 1 - 2 e -t w s + e - 2 t w s ) b) Response of a first-order process, s h s K s Y + τ = 1 ) ( ( 1 - 2 e -t w s + e - 2 t w s ) or Y ( s ) = ( 1 - 2 e -t w s + e - 2 t w s ) + τ α + α 1 2 1 s s Kh s Kh s = + τ = α = 0 1 1 τ - = = α τ - = Kh s Kh s 1 2 Y ( s ) = ( 1 - 2 e -t w s + e - 2 t w s ) + τ τ - 1 s Kh s Kh Kh ( 1 - e -t/ τ ) , 0 < t < t w y ( t ) = Kh ( 1 – e -t/ τ + 2 e - ( t-t w ) / τ ) , t w < t < 2 t w Kh ( –e -t/ τ + 2 e - ( t-t w ) / τ - e - ( t- 2 t w ) / τ ) , 2 t w < t Response of an integrating element, s h s K s Y = ) ( ( 1 - 2 e -t w s + e - 2 t w s ) Kht , 0 < t < t w y ( t ) = Kh ( -t + 2 t w ) , t w < t < 2 t w 0 , 2 t w < t c) This input gives a response, for an integrating element, which is zero after a finite time. 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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5-2 5.2 a) For a step change in input of magnitude M y ( t ) = KM ( 1- e -t/ τ ) + y (0) We note that KM = y ( ) – y (0) = 280 – 80 = 200 ° C Then K = Kw C 5 . 0 200 = 400 ° C/Kw At time t = 4 , y (4) = 230 ° C Thus τ - - = - - / 4 1 80 280 80 230 e or τ = 2.89 min 1 89 . 2 400 ) ( ) ( + = s s P s T [ ° C/Kw] a) For an input ramp change with slope a = 0.5 Kw/min Ka = 400 × 0.5 = 200 ° C/min This maximum rate of change will occur as soon as the transient has died out, i.e., after 5 × 2.89 min 15 min have elapsed. 0 1 2 3 4 5 6 7 8 9 10 0 500 1000 1500 time(min) T' Fig S5.2. Temperature response for a ramp input of magnitude 0.5 Kw/min.
5-3 5.3 The contaminant concentration c increases according to this expression: c ( t ) = 5 + 0.2 t Using deviation variables and Laplace transforming, ( ) 0.2 c t t = or 2 2 . 0 ) ( s s C = Hence 2 2 . 0 1 10 1 ) ( s s s C m + = and applying Eq. 5-21 /10 ( ) 2( 1) 0.2 t m c t e t - = - + As soon as ( ) 2 ppm m c t the alarm sounds. Therefore, t = 18.4 s (starting from the beginning of the ramp input) The time at which the actual concentration exceeds the limit ( t = 10 s) is subtracted from the previous result to obtain the requested t . t = 18.4 - 10.0 = 8.4 s 0 2 4 6 8 10 12 14 16 18 20 0 0.5 1 1.5 2 2.5 time c' m Fig S5.3. Concentration response for a ramp input of magnitude 0.2 Kw/min.

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5-4 5.4 a) Using deviation variables, the rectangular pulse is 0 t < 0 F c = 2 0 t < 2 0 2 t Laplace transforming this input yields ( s F e s s c 2 1 2 ) ( - - = The input is then given by ) 1 2 ( 8 ) 1 2 ( 8 ) ( 2 + - + = - s s e s s s c s and from Table 3.1 the time domain function is ) 1 ( 8 ) 1 ( 8 ) ( 2 / ) 2 ( 2 / - - - - - - = t t e e t c S ) 2 ( - t (1) 0 2 4 6 8 10 12 14 16 18 20 0 1 2 3 4 5 6 time C' Fig S5.4. Exit concentration response for a rectangular input. b) By inspection of Eq. 1, the time at which this function will reach its maximum value is 2, so maximum value of the output is given by
5-5 ) 1 ( 8 ) 1 ( 8 ) 2 ( 2 / 0 1 - - - - - = e e c S ) 0 ( (2) and since the second term is zero, 057 . 5 ) 2 ( = c c) By inspection, the steady state value of ) ( t c will be zero, since this is a first-order system with no integrating poles and the input returns to zero.

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• Spring '16
• Mark Darby
• dt, τS

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