ME422 Mechanical Control Systems Modeling Fluid Systems


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1 Cal Poly San Luis Obispo Mecanical Engineering ME422 Mecanical Control Systems Modeling Fluid Systems Owen/Ridgely, last update Mar 2003 Te dynamic euations for fluid flow are very similar to te dynamic euations for most oter systems. We begin wit te concept of resistance. Tis is analogous to te linear friction present in some mecanical systems, were we ave F bv, wit F being force, v being velocity and b te friction coefficient. In electrical systems, we ave Om s Law, V IR wic works te same way. Eac of tese rules relates a generalized force variable (wic may be force or voltage or some oter motivating potential) to a generalized flow variable (suc as velocity or current). Fluid systems ave a similar relationsip relating te pressure difference, or ead, across a fluid restriction to te flow troug te restriction. Unfortunately te relationsip isn t linear: K were is te rate of flow and is te ead, a function of te pressure drop: p/ρg. Wen tis relationsip is graped, it looks someting like tis: Tis is clearly not a linear relationsip (it s a parabola) so we can t use it in linear control system design. However, we can set up a linear approximation wic is good enoug for use in designing control systems. Te process of setting up tis approximation is referred to as linearization; it is described in Section 2.11 of te Nise text. Te most important ting to remember about linearization is tat it only works in a limited range about a given operating point. To begin linearizing te fluid flow vs. pressure function, we first coose our operating point. Tis is some point (, ) on te grap at wic te system can operate in steadystate euilibrium. We ten draw a tangent line troug te grap at tat point:
2 Te line is a reasonably good approximation to te nonlinear function witin te area sown by te dotted rectangle. Outside tat region, te line is not a useful approximation; we would ave to coose a different operating point and draw a new line (resulting in a different system euation) in order to linearize te system out tere. Next, we define variables and wic represent te deviations of flow and ead from teir steadystate euilibrium values. Tis means tat if te steadystate value of is 10 gal/min, a value of 2 gal/min means tat te actual flow is 12 gal/min. We can use tese deviations to write a linear euation: [ ] d d, Tis linear euation is useful in designing a control system wic maintains te flow of fluid at some value close to te operating point (, ). Now we need to find te value of te slope of te line, d/d. K 1 K 2 2 d d 2 K 2 K 2 2 d 2 d ( 2 /) 2 We can evaluate tis expression at te operating point to get [ ] d d, 2 2 Tis relationsip between ead and flow is useful for many fluid devices wose pressureflow relationsip is uadratic, including orifices, valves, and even te odd ole in a tank.
3 In te ME422 lab we will be working wit a tank wose top is open to atmosperic pressure (wic is treated as a zero gauge pressure) and wic as a flow restriction at te bottom, as sown below. Te eigt is te vertical distance from te surface of te liuid to te flow restriction (valve, orifice, or watever). Tere will be some flow i into te top of te tank and some flow o out of te bottom of te tank. By volumetric continuity, te rate of cange of te volume V of fluid in te tank must be te difference between wat s going in and wat s going out: i o dv Ad Tis euation assumes tat te tank as a constant cross section A. Te preceding euation was written for te actual values of te flows and ead. Now we write te euation for te deviations from te steadystate operating values i, o, and. Note tat te derivatives of te steady state values ave to be zero, as tose values are constants for a given operating point. Our linearized euation was i o dv 2 o, o 2 Ad ( ) Plugging tis into te continuity euation, we ave A d ( ) i ( ) Ad i d ( ) or ( ( ) 1 i A) 2A
4 Now tis is te familiar form for a firstorder linear differential euation. Te state of te system is and te input to te system is i. Tis euation is transformed into te Laplace domain to get ( s + ) H(s) Q i(s) 2A A H(s) 1/A s + ( /2A ) Q i(s) Tere are two system outputs in wic we migt be interested. If te eigt of fluid in te tank is of interest, we use tis euation as given. If te flow o out of te tank is of interest, we note tat te output flow as already been found to be o 2 Tis is of course a static relationsip, and it can be multiplied by te transfer function found above to give us Q o / Q i. Take a moment to tink about tese euations must be used. Te linearization tecniue only works for a system wose flow deviates sligtly from some steadystate value. It does not work for a system wose flow is turned fully on and off, for example. If you re going to do a step response test of a fluid system, you cannot start te system wit zero pressure and zero flow. At te start of te test, te system must be running in steadystate euilibrium wit ead and flow. It migt seem strange tat fluid flow systems ave different dynamic euations depending on ow muc fluid is flowing. But we can deal wit tis. Often, fluid systems are operated as process systems in wic our goal is to keep te fluid moving at a constant rate all te time. Cemical processing plants are an example of suc systems. Only one operating model is needed because te flow remains constant wenever te plant is in operation. At oter times, systems are designed to work at different flow rates. Tis may necessitate tat te design of te control system is repeated many times, wit different controllers being designed for different operating points.
5 Fluid Tank Example i o Consider te system sown above. Te tank is cylindrical and its diameter is 15 inces. Te steadystate operating point is 25 in 3 /sec, 27 in. Note tat te units are all in inces and seconds if answers are needed in oter units suc as feet or gallons, te mat sould still be done in inces and te answers converted to oter units afterward. We want to find te response of te system to a step input cange i of 5 in 3 /sec, wic means te flow canges from 25 to 30 in 3 /sec. A π 4 d2 177 in 2 H(s) Q i (s) H(s) Q i (s) 1/A s + ( /2A ) s + 1/(177 in 2 ) (25 in3 /sec) 2(177 in 2 )(27 in) s in in 3 /sec Te time constant of tis system is 382 seconds, or 6 minutes 22 seconds. A sanity ceck suggests tat tis makes sense; consider ow slowly te water level would cange in a kitcen sink wen you cange te flowrate. We can find te new steadystate value of fluid eigt in te tank by using te Final Value Teorem, wit te input s Laplace transform being Q i (s) (5in 3 /sec)(1/s). ss lim s 0 s H(s) ss lim s 0 s (1/A)( Q i(s)) s + ( /2A ) ss (5 in3 /sec)/(177 in 2 ) (25 in 3 /sec) 2(177 in 2 )(27 in) 10.8 in
6 Tis means tat te fluid level in te tank will rise to 37.8 inces. If we need to find (t), we could use partial fraction expansion. But we could also kludge togeter an answer by noting tat for a simple first order system, we know wat te response must look like. It s an exponential rise from te initial value to te final value. Te formula for te time response is ) (t) 10.8 in (1 e sec Te time response of te system can also be found using numerical simulation suc as tat done by Simulink. A simple model wic implements tis system is sown ere. Step s Fluid Tank Scope simout To Workspace Te results of running tis simulation are sown below. Note tat te settling time is rater long. One of te advantages of simulation is tat we can analyze systems reasonably uickly instead of aving to perform tests wic can last many minutes or even ours. 38 Tank Simulation Results Heigt [in] Time [sec]
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