3. Mathematical Modelling

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1 3. Mahemaical Moelling 3.1 Moelling principles Moel ypes Moel consrucion Moelling from firs principles 3.2 Moels for echnical sysems Elecrical sysems Mechanical sysems engineering sysems 3.3 Moel linearizaion Moivaion Linearizaion of ODEs KEH Dynamics an 3 1

2 3. Mahemaical Moelling 3.1 Moelling principles Moel ypes For esign an analysis of a conrol sysem we nee a mahemaical moel ha escribes he ynamical behaviour of he sysem. The ynamics can be escribe by ifferenial equaions for coninuous-ime ynamics ifference equaions for iscree-ime ynamics Mos processes are ime coninuous, bu some processes are inherenly ime iscree (e.g. raioacive ecay) compuer algorihms (i.e. conrollers) an many measuring evices prouce oupus a iscree ime insans» o esign such conrollers, we someimes use iscree-ime moels o escribe coninuous-ime processes In his course, we will consier boh ypes of moels conroller esign boh in coninuous an iscree ime However, he major par of he course eals wih coninuous ime. KEH Dynamics an 3 2

3 3. Mahemaical Moelling 3.1 Moelling principles Moel consrucion There are wo main principles for consrucion of mahemaical moels Moelling from firs principles: we erive moels using physical laws an oher known relaionships (moels). Sysem ienificaion: we use observaions (measuremens) of he sysem o fin a moel empirically. Usually, esigne ienificaion experimens are carrie ou o generae suiable aa. Ofen boh mehos are combine: we erive he basic moel from firs principles an eermine uncerain parameers by sysem ienificaion. I is imporan o realize ha all moels have a limie valiiy range, even he physical laws (e.g. Newon s laws of moion o no apply close o he spee of ligh). I is especially imporan o noe ha moels eermine hrough sysem ienificaion shoul no be use ousie he experimenal range. KEH Dynamics an 3 3

3. Mahemaical Moelling 3.1 Moelling principles 3.1.3 Moelling from firs principles In he following we consier moelling from firs principles.

4 3. Mahemaical Moelling 3.1 Moelling principles Moelling from firs principles In he following we consier moelling from firs principles. Because real echnical sysems en o be complex we canno or o no wan o inclue all eails of he sysem in he moel. We ry o make a goo compromise beween he following wo requiremens. The moel shoul be sufficienly accurae for is inene purpose simple enough o use e.g. for sysem analysis an conrol esign In moelling from firs principles, wo ypes of mahemaical relaionships are use: conservaion laws consiuive relaionships KEH Dynamics an 3 4

5 3.1 Moelling principles Moelling from firs principles onservaion laws onservaion laws apply o aiive quaniies of he same ype in a sysem. There are wo general kins of conservaion laws: flow balances effor balances A flow balance for a given quaniy in a sysem has he general form accumulaion / ime uni = inflow ouflow + proucion / ime uni where accumulaion an proucion occurs insie he sysem inflow an ouflow cross he sysem bounaries Flow balances apply o conserve quaniies (uner normal coniions). If no chemical or nuclear reacions ake place, he proucion is zero. Examples of flow balance quaniies: mass paricles (moles) energy curren (Kirchhoff s firs law) Noe ha volume is no a conserve quaniy for compressive fluis. KEH Dynamics an 3 5

6 3.1 Moelling principles Moelling from firs principles An effor balance for a given quaniy has he general form change / ime uni = forcing quaniy loaing quaniy where change refers o a sysem propery riving an loaing refer o ineracion wih he surrouning Generally, effor balances are applicaions of Newon s laws of moion an Kirchhoff s secon law. Examples of effor balance quaniies: force momenum angular momenum volage (Kirchhoff s secon law) KEH Dynamics an 3 6

7 3.1 Moelling principles Moelling from firs principles onsiuive relaionships onsiuive relaionships are saic relaionships ha relae quaniies of ifferen kins in a sysem. Examples of consiuive relaionships: Ohm s law: relaes he curren o he volage over a resisance valve characerisics: relaes he flow rae o he pressure rop over a valve Bernoulli s law: relaes he velociy of he flow ou of a ank o he liqui level in he ank he ieal gas law: relaes he emperaure o he pressure of a gas in a close conainer KEH Dynamics an 3 7

8 3.1 Moelling principles Moelling from firs principles The general moelling proceure 1. Formulae balance equaions. 2. Inrouce consiuive relaionships o relae variables o each oher; possibly o inrouce new variables in he balance equaions. 3. Do a correcness check by a leas checking ha all aiive erms in an equaion have he same uni; he lef an righ han sie of an equaion have he same uni. KEH Dynamics an 3 8

3. Mahemaical Moelling 3.2 Moels for echnical sysems 3.2.1 Elecrical sysems Fig. 3.1 shows hree basic componens of elecrical circuis.

9 3. Mahemaical Moelling 3.2 Moels for echnical sysems Elecrical sysems Fig. 3.1 shows hree basic componens of elecrical circuis. Variables = ime, u = volage [V], i = curren [A] omponen parameers R = resisance [Ω], = capaciance [F], L = inucance [H] resisor capacior inucor Figure 3.1. Basic componens in a elecrical circui. Relaionships Resisor (Ohm s law): u() = R i() (3.1) 1 apacior: u ( ) = u(0) + i( ) 0 (3.2) τ τ i( ) Inucor: u( ) = L (3.3) KEH Dynamics an 3-9

3.2 Moels for echnical sysems 3.2.1 Elecrical sysems Example 3.1. A passive analog low-pass filer. Figure 3.2 shows a passive analog low-pass filer.

10 3.2 Moels for echnical sysems Elecrical sysems Example 3.1. A passive analog low-pass filer. Figure 3.2 shows a passive analog low-pass filer. How oes he volage u u () on he oupu sie epen on he volage u in () on he inpu sie if he circui is uncharge a he oupu? Figure 3.2. A passive analog low-pass filer. Noaion: u R () = volage across he resisor, i R () = curren hrough he resisor u () = volage across he capacior, i () = curren hrough he capacior Accoring o Kirchhoff's secon law, he volages over he componens saisfy u in () = u R () + u () (1) u u () = u () (2) When he oupu is uncharge, here is no curren ou from he filer, an we have i R () = i () (3) KEH Dynamics an 3-10

11 3.2.1 Elecrical sysems Example 3.1. A passive analog low-pass filer The combinaion of (1) an (2) an subsiuion of (3.1) give u u () = u in () R i R () (4) Furhermore, combining (2) an (3.2) gives 1 uu ( ) = u ( ) = u (0) + i ( τ ) τ 0 The erivaive of boh sies of (5) wih respec o he ime gives (5) u ( ) u 1 = i ( τ ) = 1 i R ( ) (6) where he las equaliy is given by (3). ombining (4) an (6) gives uu ( ) R + uu ( ) = u in ( ) (7) This is a linear firs-orer ifferenial equaion. The circui is a low-pass filer ha filers (i.e., reuces he ampliue of) high frequencies in u in (). In pracice, we also have an amplifier on he oupu sie, which allows us o charge he circui so ha (3) sill hols (approximaely). KEH Dynamics an 3-11

3.2 Moels for echnical sysems 3.2.1 Elecrical sysems Example 3.2. Simple RL circui. Figure 3.3 shows a simple RL circui charge by a curren source.

12 3.2 Moels for echnical sysems Elecrical sysems Example 3.2. Simple RL circui. Figure 3.3 shows a simple RL circui charge by a curren source. How oes he volage across he capacior epen on he curren from he curren source? Figure 3.3. Simple RL circui. Noaion: u R () = volage across he resisor, i R () = curren hrough he resisor u () = volage across he capacior, i () = curren hrough he capacior u L () = volage across he inucor, i L () = curren hrough he inucor Kirchhoff s laws give u () = u R () + u L () (1) i() = i R () + i () (2) i R () = i L () (3) KEH Dynamics an 3-12

13 3.2.1 Elecrical sysems Subsiuion of (3.1) an (3.3) ino (1): (4) Eliminaion of i R () an i L (): (5) Accoring o eq. (6) in Ex. 3.1: (6) Subsiuion of (6) ino (5): Afer rearrangemen: (7) where i () is he inpu signal an u () is he oupu signal. This is a linar secon-orer ifferenial equaion. KEH Dynamics an 3-13 Example 3.2. Simple RL circui i L i R u L R ) ( ) ( + = ( ) ( ) i i L i i R u ) ( ) ( ) ( ) ( ) ( + = u i ) ( = u i L u i R u ) ( ) ( ) ( + = i L i R u u R u L ) ( ) ( = + +

3. Mahemaical Moelling 3.2 Moels for echnical sysems 3.2.2 Mechanical sysems The moeling of mechanical sysems are mainly base on Newon s secon law F = ma (3.

14 3. Mahemaical Moelling 3.2 Moels for echnical sysems Mechanical sysems The moeling of mechanical sysems are mainly base on Newon s secon law F = ma (3.4) where F is he force acing on he mass m an a is he acceleraion of he mass. Example 3.3. Unampe penulum. Figure 3.4 shows an unampe swinging penulum. The penulum can only move in wo irecions in he plane of he figure. Is poin of suspension is a a isance u an is cener of mass (he roun weigh a he lower en of he penulum) is a a isance y from he verical line o he lef. How oes he posiion y of he cener of mass epen on he posiion u of he suspension poin? Figure 3.4. Swinging penulum. Noaion l = penulum s lengh, θ = angle he penulum swings away from a verical posiion m = weigh of mass, h = verical posiion of he cener of mass F = force acing in he negaive irecion on he suspension poin of he penulum KEH Dynamics an 3-14

15 3.2 Moels for echnical sysems Mechanical sysems When he penulum is affece by he suspension force F an he graviaional force mg, accoring o Newon s secon law, we obain horizonal force componens : mÿ = F sin θ (1) verical force componens: mḧ = F cos θ + mg (2) ÿ an ḧ are secon-orer ime erivaives of y an h, respecively, i.e. he acceleraion in he respecive irecions. Assume ha he penulum s swing is moerae so ha he angle θ is always small. The penulum hen moves harly a all in he verical irecion an we can assume ha ḧ 0. The eliminaion of F hen gives The angel θ is given by he rigonomeric ieniy y u y u anθ = h l ÿ + g an θ = 0 (3) where he las equaliy uses he fac ha h l when θ is small. By combining (3) an (4) we obain he moel g g y + y = u l l (5) (4) Noice ha he approximaions ḧ 0 an θ small limi he valiiy of he moel. KEH Dynamics an 3-15

3.2 Moels for echnical sysems 3.2.2 Mechanical sysems Example 3.4.

16 3.2 Moels for echnical sysems Mechanical sysems Example 3.4. Suspension sysem in a car. Figure 3.5. a) Spring-moune mass wih amping; b) ar suspension sysem. KEH Dynamics an 3-16

17 3.2.2 Mechanical sysems Example 3.4. Suspension sysem in a car a) How oes he verical eviaion y() from an equilibrium posiion epen on a force u() acing on he spring-moune mass m? An equilibrium posiion applies when y = u = 0 (apar from he unis). If he ownwar irecion is he posiive verical irecion, Newon s secon law for he spring force an he amping force of he cyliner gives mÿ = bẏ ky + u() i.e. mÿ + bẏ + ky = u() (1) where b an k are consans. The graviaional force mg is no inclue; because i also affecs he equilibrium posiion, i is cancelle ou when he eviaion from he equilibrium posiion is moele. b) How o he eviaions y 1 () an y 2 () in a car suspension sysem epen on u(), which enoes he roughness of he groun? m 1 is he mass of he car, m 2 is he mass of he wheels an he axles, b 1 an k 1 escribe he ynamics of he car shock absorber an k 2 enoes he elasiciy of he ires. In he equilibrium posiion, y 1 = y 2 = u = 0. If he upwar irecion is he posiive irecion, we ge m 1 ÿ 1 = k 1 ( y 2 y 1 ) + b 1 ( ẏ 2 ẏ 1 ) (2) m 2 ÿ 2 = k 1 ( y 1 y 2 ) + b 1 ( ẏ 1 ẏ 2 ) + k 1 ( u y 2 ) (3) These are wo couple 2 n orer ifferenial equaions, ha escribe he car boy an he verical moion of he wheels as funcion of he verical roughness of he roa. KEH Dynamics an 3-17

3. Mahemaical Moelling 3.2 Moels for echnical sysems 3.2.3 engineering sysems engineering sysems are ypically moele wih flow balances (mass an energy balances) an consiuive relaionships. Example 3.5.

18 3. Mahemaical Moelling 3.2 Moels for echnical sysems engineering sysems engineering sysems are ypically moele wih flow balances (mass an energy balances) an consiuive relaionships. Example 3.5. Liqui conainer wih free ouflow. A volumeric flow rae u is supplie coninuously o he conainer an a volumeric flow rae q flows ou freely by graviy, cause by he heigh of he liqui h in he conainer. The conainer has a consan crosssecional area A, an he oule ube has he effecive cross-secional area a. How oes he level of he liqui epen on he inflow u? We assume ha he liqui has consan ensiy ρ. Figure 3.6. onainer wih free ouflow. Mass balance: (1) ( ρah) = ρu ρq Because he ensiy an he cross-secional area are consan, (1) can be simplifie o h A = u q (2) KEH Dynamics an 3-18

19 3.2.3 engineering sysems Ex Liqui ank wih free ouflow Accoring o Bernoulli s law, he following consiuive relaionship applies for he ouflow of a liqui v = 2gh (3) where v is he velociy of he ouflow an g is he acceleraion of graviy. Due o he conracion a he beginning of he ouflow ube, he volume flow rae q is given by q = av = a 2gh where a is he effecive cross-secional area of he ouflow ube, which is slighly smaller han he acual cross-secion. ombinaion of (2) an (4) finally gives h a 2g 1 = h + u A A i.e. a nonlinear ifferenial equaion ha escribes how he level h epens on he inflow u. (4) (5) KEH Dynamics an 3-19

3.2 Moels for echnical sysems Example 3.6. Mixing ank. Two volumeric flow raes F 1 an F 2, wih he concenraions (mass/volume) c 1 an c 2, respecively, of some inflowing componen X.

20 3.2 Moels for echnical sysems Example 3.6. Mixing ank. Two volumeric flow raes F 1 an F 2, wih he concenraions (mass/volume) c 1 an c 2, respecively, of some inflowing componen X. They are mixe coninuously in a conainer an a volumeric flow rae F 3, wih concenraion c 3, is ischarge from he conainer. The liqui in he conainer, which has a consan crosssecional area A, reaches he heigh h. The concenraion in he conainer of he componen X is c. The sirring in he conainer is assume o be perfec engineering sysems FLOW 1 FLOW 2 FLOW 3 Figure 3.7. Mixing ank. How o he level h an he concenraion c (an c 3 ) epen on oher variables? I is reasonable o assume ha he liqui ensiy in he ifferen flows is consan an he same if he liqui emperaure is consan an he concenraion of he componens is moerae. Analogously o Ex. 3.5, we obain afer cancelling ou he ensiy Toal mass balance: h A = F (1) 1 + F2 F3 We canno eliminae he ouflow F 3 because we o no know wha i epens on. KEH Dynamics an 3-20

21 3.2.3 engineering sysems Example 3.6. Mixing ank We can also se up a mass balance for each componen X, parial mass balance: ( Ahc) = F (2) 1c1 + F2c2 F3c 3 If he sirring in he conainer is perfec, we have complee mixing which means ha he concenraion is he same all over he conainer a a given ime insan. This also means ha he concenraion in he ouflow mus be he same as he concenraion in he conainer, i.e. we have he consiuive relaionship c 3 = c (3) The evelopmen of he erivaive in (2) accoring o he prouc rule an consiering equaion (3) gives h c Ac + Ah = F1 c1 + F2c2 F3c Then, combinaion of (4) wih (1) gives c Ah = F1 ( c1 c) + F2 ( c2 c) This is a linear ifferenial equaion wih (in general) non-consan parameers. (4) (5) KEH Dynamics an 3-21

3.2 Moels for echnical sysems 3.2.3 engineering sysems Example 3.7. Waer heaer. The inflow of waer is a mass flow ṁ 1 wih emperaure T 1 an he ouflow is a mass flow ṁ 1 wih emperaure T 2.

22 3.2 Moels for echnical sysems engineering sysems Example 3.7. Waer heaer. The inflow of waer is a mass flow ṁ 1 wih emperaure T 1 an he ouflow is a mass flow ṁ 1 wih emperaure T 2. The waer, wih mass M in he heaer, is heae up o a emperaure T wih a heaing flow rae Q. The sirring in he heaer is assume o be perfec. How o he waer volume an he emperaure in he heaer epen on oher variables? FLOW 1 Figure 3.8. Waer-heaer. FLOW 2 Mass balance: M = m 1 m 2 (1) Energy balance: E = E E + Q 1 2 (2) where Ė 1 an Ė 2 are energy flows ha are supplie by he inflow an he ouflow, respecively. KEH Dynamics an 3-22

23 3.2.3 engineering sysems Example 3.7 Waer heaer The energy in a subsance is proporional o is mass or mass flow rae, an for liquis i applies wih goo accuracy ha he energy is also proporional o he emperaure, This gives onsiuive relaionships: E = c p TM, Ė 1 = c p T 1 ṁ 1, Ė 2 = c p T 2 ṁ 2 (3) where c p is he specific hea capaciy for waer (in his case assume o be consan). ombining (2) an (3) an he evelopmen of he erivaive accoring o he chain rule gives M T Q T + M = T1m 1 T2m 2 + (4) c p The consiuive relaionship T 2 = T applies base on he assumpion of perfec mixing. The eliminaion of M / wih (1) gives T Q M = m 1 ( T1 T ) + (5) c Equaion (1) an (5) inicae how he mass an he emperaure in he heaer epen on he inflow an he heaing efficiency Q. p KEH Dynamics an 3-23

24 3.2.3 engineering sysems Example 3.7 Waer heaer If we wan o use unis of volume insea of unis of mass, we can now inser M = ρah an ṁ 1 = ρ 1 F 1 in equaion (5) o ge T Q ρah = ρ (6) 1 F1 ( T1 T ) + c p Noe ha in equaion (6) he ensiy is no assume o be a consan. Equaion (1) expresse in unis of volume becomes more complex when we have a variable ensiy. We can, however, show ha even if he emperaure epenence of he ensiy is no negligible, he effecs of a variable ensiy on (1) en o cancel ou. A compleely aequae form of (1) expresse in unis of volume is hen h A = F 1 F 2 (7) KEH Dynamics an 3-24

3.2 Moels for echnical sysems Example 3.8. Gas in a close ank. Figure 3.9 illusraes a close gas-ank wih he volume V, subsance amoun (molar amoun) n, pressure p an emperaure T.

25 3.2 Moels for echnical sysems Example 3.8. Gas in a close ank. Figure 3.9 illusraes a close gas-ank wih he volume V, subsance amoun (molar amoun) n, pressure p an emperaure T. The inflow o he ank is he molar flow rae ṅ 1 a he pressure p 1, he ouflow is he molar flow rae ṅ 2 a he pressure p 2. Valve 2 can be use for conrol by ajusing he valve posiion u engineering sysems Figure 3.9. Gas in a close ank. How oes he pressure p in he ank epen on oher variables? n Molar balance: = n 1 n 2 (1) The molar flow rae hrough a valve wih a given opening posiion is proporional o he square roo of he pressure ifference across he valve. Moreover, i is assume ha he facor of proporionaliy is proporional o he square of he linear valve posiion u. The molar flow raes are hen given by 2 consiuive relaionships: n = k p p, n = k u p p (2) KEH Dynamics an 3-25

26 3.2.3 engineering sysems Example 3.8. Gas in close ank Furhermore, we can assume ha he ieal gas law hols, i.e. pv = nrt (3) applies. Here R is he general gas consan an T is he emperaure expresse in Kelvin. If he emperaure T is consan, hen subsiuion of (2) an (3) in (1) gives p = RT V n = RT V 2 ( k p p k u p p ) (4) which is a relaively complex nonlinear ifferenial equaion, even if i is of firs orer. KEH Dynamics an 3-26

27 3. Mahemaical Moelling 3.3 Moel linearizaion Moivaion We have in a number of examples shown how o erive ynamic moels for many ypes of echnical sysems. In all cases, he obaine moels are orinary ifferenial equaions. We noe ha he ifferenial equaions (DEs) are ofen nonlinear even if hey are linear, he coefficiens are generally no consan because hey epen on some physical ime-varying quaniy i is ifficul, maybe impossible, o fin general soluions o hese kins of DEs Therefore, we nee o suy special cases an/or o simplifying assumpions Frequenly use simplificaions is o assume ha some quaniies are consan, even if hey are (slighly) ime-varying inpu signals change in some ieal (bu reasonable) way KEH Dynamics an 3 27

28 3.3 Moel linearizaion Moivaion In pracice, i is ofen enough o know he sysem behaviour in a limie region close o a known operaing poin. Then, he moel simplificaion may be o linearize he moel equaions a he operaing poin. The avanage of his is ha efficien analysis, synhesis, an esign mehos base on linear algebra can be use. If he sysem is very nonlinear, or he operaing region very large, one can use several linear moels ha are linearize a ifferen operaing poins. Because of he reasons menione above, moelling from firs principles is ofen followe by a linearizaion of he moel. In his course we are only consiering moels obaine from orinary ifferenial equaions, no parial DEs. KEH Dynamics an 3 28

29 3. Mahemaical Moelling 3.3 Moel linearizaion Linearizaion of ODEs A general ODE onsier an n:h orer ODE, which we can formally wrie as f y n,, y, y, u = 0 (3.5) To simplify, ime erivaives of u are no inclue; hey can be hanle in he same way as he ime erivaives of y. Usually he ime erivaives appear linearly in (3.5); however, he linearizaion applies also when hey o no appear linearly. We can linearize (3.5) by means of a firs-orer Taylor series expansion a he nominal operaing poin y n,, y, y, u, enoe by f : f y n,, y, y, u = f y n,, y, y, u + + y f y y + y f y (n) y y + f u f y n y n + u u (3.6) Usually he operaing poin is a saic (seay-sae) poin wih all ime erivaives equal o zero, bu (3.6) hols even if his is no so. KEH Dynamics an 3 29

30 3.3 Moel linearizaion Linearizaion of ODEs We inrouce he variables y (n) y n y n,, y y y, y y y, u u u (3.7) which enoe eviaions from he nominal operaing poin. We call such variables eviaion variables, or simply, -variables. ombinaion of (3.5), (3.6) an (3.7) wih he fac ha he operaing poin saisfies (3.5), gives y (n) f y (n) + + y f y + f y + f u = 0 (3.8) This is a linear n:h orer ODE wih consan coefficiens. Noe: If he ODE conains ime erivaives of u, hey appear as - variables in (3.8) in he same way as he ime erivaives of y appear. KEH Dynamics an 3 30

31 3.3 Moel linearizaion Linearizaion of ODEs ODEs linear in ime erivaives If he ime erivaives appear linearly in (3.5), we can formally wrie f n y, u y n + + f 1 y, u y + f 0 y, u = 0 (3.9) We can apply (3.6) o linearize every erm separaely. We ge f 0 y, u = f 0 y, u + f 0 an for he erm wih he i:h erivaive: f 0 y + f 0 f i y, u y i = f i y, u y i + f i y, u y i f 0 u + f i f i y i y + f i u f i y i u Subsiuion ino (3.9), which also hols for he operaing poin, gives f n y, u y n + + f 1 y, u y + f 0 n f 0 y + f 0 f 0 u = f (3.10) where f = y i f i i=1 y + u (3.11) Noe ha f = 0 if he operaing poin is a seay-sae wih all y i = 0. KEH Dynamics an 3 31 f i f i f i

32 3.3 Moel linearizaion Linearizaion of ODEs onsiuive relaionships Nonlinear consiuive relaionships also nee o be linearize. Such a relaionship can be formally wrien g z, y, u = 0 (3.12) where z is a new variable ha is relae o y an u accoring o (3.12). Linearizaion using a firs-orer Taylor series expansion as in (3.6) gives g z + g y + z g y g g u g u = 0 (3.13) If he nominal operaing poin is a seay-sae wih all ime erivaives zero, iffereniaion wih respec o ime gives for he i:h ime erivaive g z (i) + g y (i) + z g y g g u g u (i) = 0 (3.14) If esire, he variable z can be inrouce as epenen variable insea of y in (3.8) or (3.10) by means of (3.13) an (3.14). KEH Dynamics an 3 32

33 3.3 Moel linearizaion Linearizaion of ODEs Example 3.9. Linearizaion of a firs-orer DE. In Example 3.5 we erive he nonlinear DE Ah + a 2gh u = 0 We wan o linearize his DE a he seay-sae operaing poin h, u. Applicaion of (3.10) gives A h + a 2gh u h + h h,u a 2gh u u = u h,u which gives A h + a 2g h h A h + a h 2g 2 h h u u u u = 0 h = u KEH Dynamics an 3 33

3.3 Moel linearizaion 3.3.2 Linearizaion of ODEs Exercise 3.1. Linearizaion of consiuive relaionship. onsier a conrol valve in a pipeline, schemaically illusrae in he figure.

34 3.3 Moel linearizaion Linearizaion of ODEs Exercise 3.1. Linearizaion of consiuive relaionship. onsier a conrol valve in a pipeline, schemaically illusrae in he figure. A a given pressure, he flow rae q hrough he valve epens on he posiion x of he valve plug (sem) accoring o he valve characerisic q = (α x 1)/(α 1) where an α are consan parameers epening on he size an consrucion of he valve. The valve is close when x = 0 an fully open when x = 1. The posiion x is ajuse by an acuaor responing o a conrol signal u. Because of ineria, x follows u accoring o he ynamic relaionship Tx + x = KK where T an K are consan parameers (ime consan an saic gain). Derive a linear ynamic moel ha shows how q epens on u close o an operaing poin q = q. KEH Dynamics an 3 34

Homework 2 Solutions

Homework 2 Solutions Mah 308 Differenial Equaions Fall 2002 & 2. See he las page. Hoework 2 Soluions 3a). Newon s secon law of oion says ha a = F, an we know a =, so we have = F. One par of he force is graviy, g. However,