Assignment 4 Solutions 2009

Why solution thetransfer funct

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online