Baixe Resolução Livro Ogata e outras Manuais, Projetos, Pesquisas em PDF para Engenharia Elétrica, somente na Docsity! CHAPTER 4 BAI. Cet m(f-s)dt, Rabo, g=kv7 Thus Fe! cr Sb rh=Rp Note that dane gg Hence tos = A ad . . | Rahe = LEE az ve have . Rm as = Joó Thus, e. Time constant = CR = 5 x 300 = 1500 séc B42. For this system Cef=-Quê, He3r, Gurim =(Lfm : (Eram = —o.c0 sv «dt ntdi) = — 0.008 Tr dt Assume that the head moves dom from H = 2m to x for the 60 sec period. Then 54 Edy =-— a, sos 2) de & (xt 2f) =—cort3z (60-20) vhich can be rewritten as xf —quermais) =— 21480 xf assess /nGO = ISO BP Taking logarithm of both sides of this last equation, we obtain É bg x = ls, ASv Ep X=/6$S2 m B43.- From Figure 4-49 ve obtain the folloving equationsr ME) — did dad as 2,6) di 2,67 = 4,69) -— Ea + 6 6,662 A [2 O ut as For each of the above equations, a block diagram can be dram, as chow next. tals) Hs) TT] 86 [ (6 9 E EA ED) Hals) Beco) Tl MS) 2,6) Tt f Age . + ao eu - 37 - to cl BO Res+l For the bellows and spring, ve have the following equation: Ah =hx The transfer function X(s)/P;(s) is then given by .XO BOLA 1 B6) “RO ROC R Res+r E £o) no = et) E | & ka ia E In this block diagram, Z(s) is the Laplace transform of the small displacement of the diaphragm of the pneumatic relay. The transfer function Po(s)/E(s) is given by 6) =. K k» =k EG)” atb AA " D+ Mika arb & The control action of this controller is proportional.. Thus, the controller is a proportional controller. . . “4-7. Define tho pressure of air in the bellows as Po + Po: Then] Chpegdt, qu fch Hence di R-R, cia = E or Re rh au Define the area of bellows as A and the displacement of the bellows as Y + y; Then, noting that poA = ky, Equation (1) becomes as Re É SG + =A. or d 4 ide +r= A Thus . No RO R6) C Res + A block diagram for this system is shom below. ) R Et) a Xá Ki nt) k (52 a Y Ah A+ Res +! RO. sb St e(s) tb &k A/b É ATE Res+) Assume that KK1. Then RG). b at FE EG) at+b q -( 4 Nes + Thus, the control action is e The controller is a proportional-plus-derivative controller. . B4-8. E(s) EE ? xe) Et x RO Al k 4 a+tb És k Res+tI R6) bb Eis) atb EEE pr Rs + 7 If KjK>p L, then -41- Zels) - Ki ao “+ Kks A RE. (E (+ es) atb & RT The controller is a proportional-plus-integral controller. B4-9. TT E(s) X6) Re) Tê E na k k ] R =2E ad £ Fe EE &xém BRO) + MK Es) atb A ReC 1+k A, 2Cas / k E+E * Relastl Bes+ If KK 1, then RO 5» / E6) atb 4 A Rãs / a+b À Rls+] ASA] (te Eles +17) E Ea) eis +1) dE (1 Sar + Eos TROS) Thus, the control action is proportionai-pius-integral-plus-derivative. The controller is a PID controller. . n B-4-10. Referring to the figure shom on the next page, we can obtain the equations for the system. - 42 — B-4-12. Since the increase of water in the tank during dt seconds is equal to the net inflow to the tank during the same dt seconds, we have Cdh=Chtfa-Bo)dt (1) where h h=% For the feedback lever mechanism, we have =-4 X = a+b 4 Equation (1) can now be written as follows: ah cltenemnohpto Note that dy - q G=kx =k qo h (3) By substituting the given numerical values into Equations (2) ang (3), we obtain Bh 2 =-4+%—2h d de oh Taking the Laplace transforms of the preceding two equations, assuning zero ânitial conditions, we obtain 25 HG) =— YE)+ 04 (3)- 2H69 SYe) = HG) By eliminating Y(s) from the last two equations, we get 25H) = HH) + QUE) -25 HE) Hence (25*+ 25t/)H(9)= s Quis) He) s Qui) 252+25+/ from which ve get B4-13. For the system hA=ktX-E) where À is the area of the bellows and z is the displacemant of the lover end of the spring. . Also, - 45 — g=kfxdt o, g=-E Thus Y)=É xm, Yw=-z6) Hence . AP) = k [X6)-26)] = E [XO + YO) = kl ÃO Therefore, YO KM KA KA Bo Ss Bb) “setor E) — kls+k) B-4-14. Define 69 = amblent temperature 9; = temperature of thermocouple 62 = temperature of thermal well Rj = thermal resistance of thermocouple R2 = thermal resistance of thermal well Cy = thermal capacitance of thermocouple C2 = thermal capacitance of thermal well hy = heat input rate to thermocouple hz = heat input rate to thermal vell Then, the equations for the system can be written as Cdb =h dt CG ds = (ha- A )dé where h; = (02 - 61)/Rj and ho = (60 - 02)/R2. Thus ve have RG SE+O = O dês bola Ba Gage A “Ro R, By eliminating 62 from the last two equations, ve obtain Lib) = / OM) RGRA SAR RC TRaAO)S + Noting that Ry€y = time constant of thermocouple = 2 sec R2€2 = time constant of thermal well = 30 sec RG = Rca É = Jo do = 6 see Hence the denominator of 6,(5)/62(5) becomes as RR S + (RGARO RO) AL = bos pass t/ = (A6S/SLI)CIEIE SAL) Thus, the time constants of the system are Ty = 1.651 sec, To = 36.35 sec - 47 —