Solution_to_problem_5.14 - 2 2 '( ) '( ) 0.127 0.167 '( )...

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5.14 – 1 Solution to problem 5.14 a) 1) From the initial and final steady states we can obtain the gain, knowing that the step change is from 15 psi to 31 psi. 2) From the overshoot we can obtain the damping coefficient ζ . 3) From the period (once the damping coefficient is known) we can obtain the time constant τ . See pages 118 and 119 in Seborg et al. 2004 for more information. We get: b) For this exercise we need to follow the steps of Figure 4.5. in Seborg et al. 2004, but moving from the bottom of the flow diagram towards the top, in an opposite way to what we have usually done until now in the exercises. 1) The denominator of the transfer function 2 '( ) 0.2 '( ) 0.127 0.167 1 R s P s s s = + + is due to the derivative terms of R . Therefore we rearrange: ( ) 2 '( ) 0.127 0.167 1 0.2 '( ) R s s s P s + + = And if we apply the inverse Laplace Transform we obtain the derivatives in the time domain:
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Unformatted text preview: 2 2 '( ) '( ) 0.127 0.167 '( ) 0.2 '( ) d R t dR t R t P t dt dt + + = These are deviation variables, and we are interested in the original variables R ( t ) and P ( t ). Recalling the definition of the deviation variables: '( ) ( ) SS R t R t R =-'( ) ( ) SS P t P t P =-And consequently: 5.14 – 2 '( ) ( ) dR t dR t dt dt = And: 2 2 2 2 '( ) ( ) d R t d R t dt dt = Substituting in the differential equation we obtain: ( ) 2 2 ( ) ( ) 0.127 0.167 ( ) 0.2 ( ) SS SS d R t dR t R t R P t P dt dt + +-=-Assuming that we start from the steady state in Figure E5.14, i.e. 15 SS P psi = and 8 SS R mm = , we get: ( ) 2 2 ( ) ( ) 0.127 0.167 ( ) 8 0.2 ( ) 15 d R t dR t R t P t dt dt + +- =-Which is equivalent to: 2 2 ( ) ( ) 1.31 7.88 ( ) 1.57 ( ) 39.5 d R t dR t R t P t dt dt + + = +...
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This note was uploaded on 11/20/2009 for the course CHME DTU-abroad taught by Professor Rafiqulgani during the Spring '09 term at Rensselaer Polytechnic Institute.

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Solution_to_problem_5.14 - 2 2 '( ) '( ) 0.127 0.167 '( )...

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