3.2 Models for technical systems

Transcription

1 onrol Laboraory 3. Mahemacal Moelng 3. Moels for echncal sysems 3.. Elecrcal sysems Fg. 3. shows hree basc componens of elecrcal crcs. Varables = me, = volage [V], = crren [A] omponen parameers R = ressance [Ω], = capacance [F], L = ncance [H] Relaonshps ressance capacor ncor Fgre 3.. Basc componens n a elecrcal crc. Ressance Ohm s law: = R 3. apacor: = 0 + τ τ 0 = L Incor: KEH Dynamcs an onrol 3-9

2 3.. Elecrcal sysems onrol Laboraory Example 3.. A passve analog low-pass fler. Fgre 3. shows a passve analog low-pass fler. How he volage on he op se epens on he volage n on he np se f he crc s ncharge a he op? Fgre 3.. A passve analog low-pass fler. Noaon: R = volage across he ressor, R = crren hrogh he ressor = volage across he capacor, = crren hrogh he capacor If we con all he volages volage rop as posves when applyng Krchhoff's secon law aron he rgh loop, we oban n = R + = When he op s ncharge, here s no crren o from he fler, an we have R = 3 KEH Dynamcs an onrol 3-0

3 3.. Elecrcal sysems onrol Laboraory The combnaon of an an sbson wh 3. gves = n R R 4 Frhermore, combnng an 3. gves = = 0 + τ τ 0 The ervave of boh ses of 5 wh respec o he me gves = τ = R 5 6 where he las eqaly s gven by 3. ombnng 4 an 6 gves R + = n 7 Ths s a lnear frs orer fferenal eqaon. The crc s a low-pass fler ha flers.e., reces he ample of hgh freqences n n. In pracce, we also have an amplfer on he op se, whch allows s o charge he crc so ha 3 sll hols approxmaely. KEH Dynamcs an onrol 3-

4 3.. Elecrcal sysems onrol Laboraory Example 3.. Smple RL crc. Fgre 3.3 shows a smple RL crc rven by a crren sorce. How oes he volage across he capacor epen on he crren from he crren sorce? Fgre 3.3. Smple RL crc. Noaon: R = volage across he ressor, R = crren hrogh he ressor = volage across he capacor, = crren hrogh he capacor L = volage across he ncor, L = crren hrogh he ncor Krchhoff s laws gve = R + L = R + R = L 3 KEH Dynamcs an onrol 3-

5 3.. Elecrcal sysems Sbson of 3. an 3.3 n : 4 Elmnaon of R an L : 5 Accorng o eq. 6 n Ex. 3.: 6 Sbson of 6 n 5: Afer rearrangemen: 7 where s he np sgnal an s he op sgnal. Ths s a lnar secon orer fferenal eqaon. KEH Dynamcs an onrol 3-3 onrol Laboraory L R L R + = L R + = = L R + = L R R L + = + +

6 3. Moel for echncal sysems onrol Laboraory 3.. Mechancal sysems The moelng of mechancal sysems are manly base on Newon s secon law F = ma 3.4 where F s he force acng on he mass m an a s he acceleraon of he mass. Example 3.3. Unampe penlm. Fgre 3.4 shows an nampe swngng penlm. The penlm can only move n wo recons n he plane of he fgre. Is pon of sspenson s a a sance an s cener of mass he ron wegh a he lower en of he penlm s a a sance y from he vercal lne o he lef. How oes he poson y of he cener of mass epen on he poson of he sspenson pon? Fgre 3.4. Swngng penlm. Noaons: l = penlm s lengh, θ = angle he penlm swngs away from a vercal poson m = wegh of mass, h = vercal poson of he cener of mass F = force acng n he negave recon on he sspenson pon of he penlm KEH Dynamcs an onrol 3-4

7 3.. Mechancal sysems onrol Laboraory When he penlm s affece by he sspenson force F an he gravaonal force mg, accorng o Newon s secon law, we oban horzonal force componens : mÿ = F sn θ vercal force componens: mḧ = F cos θ + mg ÿ an ḧ are secon orer me ervaves of y an h, respecvely,.e. he acceleraon n he respecve recons. Assme ha he penlm s swng s moerae so ha he angle θ s always small. The penlm hen moves harly a all n he vercal recon an we can assme ha ḧ 0. The elmnaon of F hen gves ÿ + g an θ = 0 3 The angel θ s gven by he rgonomerc eny y y anθ = 4 h l where he las eqaly ses he fac ha h l when θ s small. By combnng 3 an 4 we oban he moel y + g l y Noce ha he approxmaons ḧ 0 an θ small lm he valy of he moel. = g l 5 KEH Dynamcs an onrol 3-5

8 3.. Mechancal sysems onrol Laboraory Example 3.4. Sspenson sysem n an aomoble. Fgre 3.5. a Sprng-mone mass wh ampng; b ar sspenson sysem. KEH Dynamcs an onrol 3-6

9 3.. Mechancal sysems onrol Laboraory a How oes he vercal evaon y from an eqlbrm poson epen on a force acng on he sprng-mone mass m? An eqlbrm poson apples when y = = 0 apar from he ns. If he ownwar recon s he posve vercal recon, Newon s secon law for he sprng force an he ampng force of he cylner gves mÿ = ky bẏ +.e. mÿ + bẏ + ky = where b an k are consans. The gravaonal force mg s no ncle; becase also affecs he eqlbrm poson, s cancelle o when he evaon from he eqlbrm poson s moele. b How o he evaons y an y n a car sspenson sysem epen on, whch enoes he roghness of he gron? m s he mass of he car, m s he mass of he wheels an he axles, b an k escrbe he ynamcs of he car shock absorber an k enoes he elascy of he res. In he eqlbrm poson, y = y = = 0. If he pwar recon s he posve recon, we ge m ÿ = k y y + b ẏ ẏ m ÿ = k y y + b ẏ ẏ + k y 3 These are wo cople n orer fferenal eqaons, ha escrbe he car boy an he vercal moon of he wheels as fncon of he vercal roghness of he roa. KEH Dynamcs an onrol 3-7

10 3. Moel for echncal sysems onrol Laboraory 3..3 engneerng sysems engneerng sysems are ypcally moele wh flow balances mass an energy balances an consve relaonshps. Example 3.5. Lq conaner wh free oflow. A volmerc flow rae s spple connosly o he conaner an a volmerc flow rae q flows o freely by gravy, case by he hegh of he lq h n he conaner. The conaner has a consan crossseconal area A, an he ole be has he effecve cross-seconal area a. How oes he level of he lq epen on he nflow? We assme ha he lq has consan ensy ρ. Fgre 3.6. onaner wh free oflow. Mass balance: ρah = ρ ρq Becase he ensy an he cross-seconal area are consan, can be smplfe o h A = q KEH Dynamcs an onrol 3-8

11 3..3 engneerng sysems onrol Laboraory Accorng o Bernoll s law, he followng consve relaonshp apples for he oflow of a lq v = gh where v s he velocy of he oflow an g s he acceleraon of gravy. De o he conracon a he begnnng of he oflow be, he volme flow rae q s hen gven by q = av = a gh 3 4 where a s he effecve cross-seconal area of he oflow be, whch s slghly smaller han he acal cross-secon. ombnaon of an 4 fnally gves h a g = h + A A.e. a nonlnear fferenal eqaon ha escrbes how he level h epens on he nflow. 5 KEH Dynamcs an onrol 3-9

12 3..3 engneerng sysems onrol Laboraory Example 3.6. Mxng ank. Two volmerc flow raes F an F, wh he concenraons mass/volme c an c, respecvely, of some nflowng componen X. They are mxe connosly n a conaner an a volmerc flow rae F 3, wh concenraon c 3, s scharge from he conaner. The lq n he conaner, whch has a consan cross-seconal area A, reaches he hegh h. The concenraon n he conaner of he componen X s c. The srrng n he conaner s assme o be perfec. FLOW FLOW Fgre 3.7. Mxng ank. FLOW 3 How o he level h an he concenraon c an c 3 epen on oher varables? I s reasonable o assme ha he lq ensy n he fferen flows s consan an he same f he lq emperare s consan an he concenraon of he componens s moerae. Analogosly o ex. 3.5, we oban afer cancellng o he ensy h Toal mass balance: A = F + F F3 We canno elmnae he oflow F 3 becase we o no know wha epens on. KEH Dynamcs an onrol 3-0

13 3..3 engneerng sysems onrol Laboraory We can also se p a mass balance for each componen X, paral mass balance: Ahc = F c + F c If he srrng n he conaner s perfec, we have complee mxng whch means ha he concenraon s he same all over he conaner a a gven me nsan. Ths also means ha he concenraon n he oflow ms be he same as he concenraon n he conaner,.e. we have he consve relaonshp F c 3 3 c 3 = c 3 The evelopmen of he ervave n accorng o he proc rle an conserng eqaon 3 gves h c Ac + Ah = F c + Fc F3c 4 Then, combnaon of 4 wh gves c Ah = F c c + F c c Ths s a lnear fferenal eqaon wh n general non-consan parameers. 5 KEH Dynamcs an onrol 3-

14 3..3 engneerng sysems onrol Laboraory Example 3.7. Waer-heaer. The nflow of waer s a mass flow ṁ wh emperare T an he oflow s a mass flow ṁ wh emperare T. The waer, wh mass M n he heaer, s heae p o a emperare T wh a heang flow rae Q. The srrng n he heaer s assme o be perfec. How o he waer volme an he emperare n he heaer epen on oher varables? FLOW Fgre 3.8. Waer-heaer. FLOW Mass balance: M = m m Energy balance: E = E E + Q where Ė an Ė are energy flows ha are spple by he nflow an he oflow, respecvely. KEH Dynamcs an onrol 3-

15 3..3 engneerng sysems onrol Laboraory The energy n a sbsance s proporonal o s mass or mass flow rae, an for lqs apples wh goo accracy ha he energy s also proporonal o he emperare, Ths gves onsve relaonshps: E = c p TM, Ė = c p T ṁ, Ė = c p T ṁ 3 where c p s he specfc hea capacy for waer n hs case assme o be consan. ombnng an 3 an he evelopmen of he ervave accorng o he chan rle gves T M + M T = T m T m + Q c p 4 The consve relaonshp T = T apples base on he assmpon of perfec mxng. The elmnaon of M / wh gves M T = m + T T Q c p 5 Eqaon an 5 ncae how he mass an he emperare n he heaer epen on he nflow an he heang effcency Q. KEH Dynamcs an onrol 3-3

16 3..3 engneerng sysems onrol Laboraory If we wan o se ns of volme nsea of ns of mass, we can now nser M = ρah an ṁ = ρ F n eqaon 5 o ge T Q ρah = ρ 6 F T T + c p Noe ha n eqaon 6 he ensy s no assme o be a consan. Eqaon expresse n ns of volme becomes more complex when we have a varable ensy. We can, however, show ha even f he emperare epenence of he ensy s no neglgble, he effecs of a varable ensy on en o cancel o. A compleely aeqae form of expresse n ns of volme s hen h A = F F 7 KEH Dynamcs an onrol 3-4

17 3..3 engneerng sysems onrol Laboraory Example 3.8. Gas n a close ank. Fgre 3.9 llsraes a close gas-ank wh he volme V, sbsance amon molar amon n, pressre p an emperare T. The nflow o he ank s he molar flow rae ṅ a he pressre p, he oflow s he molar flow rae ṅ a he pressre p. Valve can be se for conrol by ajsng he valve poson. Fgre 3.9. Gas n a close ank. How oes he pressre p n he ank epen on oher varables? n Sbsance amon balance: = n n The molar flow rae hrogh a valve wh a gven openng poson s proporonal o he sqare roo of he pressre fference across he valve. Moreover, s assme ha he facor of proporonaly s proporonal o he sqare of he lnear valve poson. The molar flow raes are hen gven by consve relaonshps: n = k p p, n = k p p KEH Dynamcs an onrol 3-5

18 3..3 engneerng sysems onrol Laboraory Frhermore, we can assme ha he eal gas law hols,.e. pv = nrt 3 apples. Here R s he general gas consan an T s he emperare expresse n Kelvn. If he emperare T s consan, hen sbson of an 3 n gves p = RT V n = RT V k p p k p p 4 whch s a relavely complex nonlnear fferenal eqaon, even f s of frs orer. KEH Dynamcs an onrol 3-6