Assignment 4 Solutions 2009

Why solution thetransfer funct

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Unformatted text preview: s of the system. q i h 1 q 1 R 1 h 2 q 2 R 2 5 3 (a) What is the response q (t ) to a step input of magnitude 0.5 m /min in q (t ) if the system is 2 i init ially at steady­state corresponding to qi = q1 = q = 1 m 3 / min . 2 (b) The head­flow relat ions for the tanks are 5 3 / min m 2 3 / min m q1 = h , q = h 1 2 2 m m 3 What are the ult imate values o f each tank level aft er 1 m of liquid is suddenly added to the first tank? Why? Solution: The transfer funct ion between the input flowrate and the exit flowrate is given by, Q’ (s) i 1 5 + 1 s Q’ (s) 1 1 4 + 1 s Q’ (s) 2 Figure: A two­tank surge process transfer funct ion model Q 2 ( ) ' s 1 = Q i ( ) (5 + 1 ( s + 1) ' s s ) 4 If there is a step change in the input flowrate, the exit flowrate will be given as, Q 2 ( ) = ' s 1 0.5 (5 + 1 ( s + 1 s s ) 4 ) Either using the partial fract ion expansio ns, or the formulas for second­order process responses, we can find the time­do main response o f the exit flowrate. I will use the latter, æ t 1e - t / t1 - t 2 e - t / t 2 ö ç ÷ q ( t = KM 1 - ') t 1 - t 2 è ø Where KM=0.5, and t 1 = 5 t 2 = 4 , as this is a slight ly overdamped process ( z = 1.006 ). The result is, , 6 q ) = 1 + 0.5(1 - 5 - t / 5 + 4 - t / 4 ) ( t e e For part (b), we need to first find the transfer functio ns between the inlet flowrate and the levels in each tank. H 1 ( ) H ' ( ) Q 1 ( ) 1 1 ' s s ' s =1 = Q i ( ) Q 1 ( ) Q i ( ) 5 5 + 1 ' s ' s ' s s H 2 ( ) H 2 ( ) Q 2 ( ) 1 ' s ' s ' s 1 = = Q i ( ) Q 2 ( ) Q i ( ) 2 (5 + 1 ( s + 1) ' s ' s ' s s ) 4 The disturbance for this case is an impulse of magnitude 1. Hence the level responses will look like, 1 1 (1 ) 5 5 + 1 s 1 1 H ' ( ) = (1) 2 s 2 (5 + 1 ( s + 1 s ) 4 ) H ' ( ) = 1 s Using the Final Value Theorem, we can find the ultimate values of these levels, æ 1 1 ö lim( H ' ( )) = lim s s 1 s ç (1 = 0 )÷ è 5 5 + 1 ø s 0 ® s 0 ® s æ 1 1 ö lim( H ' ( )) = lim s s 2 s (1 = 0 )÷ ç s 0 ® s 0 2 (5 + 1 ( s + 1 ®è s ) 4 ) ø This means that the levels go back to their original height after a transient deviation, as a result of the fact that the levels act like “self­regulating” process 7 8...
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