hw_7_w06_soln - HFH—Eb—EUUb 1U=4H MtEH & HtHU...

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Unformatted text preview: HFH—Eb—EUUb 1U=4H MtEH & HtHU tNUthtHlNU them HomeWorio Set #7: MAE 3600 '— System Dynamics, Winter 2,, 06. 5T5 HH4 DUHU . ” ~ Due date: Friday, April 21, 2006 Only Problems 1-3 will be graded Problem. 1. Considerga simple pneumatic system that consists of a supply-tattle filled with compressed air at 3310‘ deg C, an inlet pipe with valve (resistance R), and rigid vessel with volume 0.1 mill,“ For time t i O, the entire system is at steady-state ,‘pressure I) = 6(10)5 hl/m2 (absolute). At time t = 0, the supply tank pressure is stepped {to absolute pressurefi+ p,, where the incremental step change is p,- =3(1t))‘l mez. The laminar flow resistance of the valve is R = 1.5(10)3 N—s/kg-mz. Assume the expansion process through the valvefis isothennal. Q? a) Compute the tilnfiefrto reach steady state after the step pressure change.:, V. I. "Hi b) Sketch theabsdhiie pressure response of the rigid vessel, and labelcalldmportant fie parameters ops-{flielsketch « J I 0 c) Verify part bytbyzsimulating the response using Matlab or Simulink. Hand in a ____,... plot of the vessel» pressure response vs. time. ‘ Problem 2. A rflmtiflmltIMCChHMCfli system has the following system model: fate + at? +1366 = T(r) The angular positionlgfiérfl (in rad) is measured from equilibrium. The system is initially in equilibrium (zero , " and at rest). A step torque is applied at time t= 0,4 torque To) = 6.5N—m f?! at: a, o. ‘ £3. a) Compute dream natural frequency and damping ratio 4 for tingsyzstem. to b) Sketch the total asthma so) and show the important characteristicsoffl'te transient “r- and steady-state response on your sketch (show numerical values). ‘ (O c) Verify your sketch with a Matlab or Simulink simulation. Hand in a‘ plot of the --‘ computer simulation, and explain what method you used. Pgoblem 3. The roptglocations are given for three different second-order‘.systems. In each ease, deterniinelthe corresponding characteristic equation. In additionytcompute the following values (only applicable): 4‘ , a)", a), , time to reach steady state; and period of oscillation. \ i 8 a) s = —2 i j5 f" b) s = as: 12 F. U1 HI—‘H—Eb—Elfillfilb lkll4tl P‘llzL'H E: HtHLI IzNLillelelNLi b'i’d titl4 51115111 Problem 4. An engineer wants to model a mechanical system as a simple mass— spring-damper. The? system has mass m = 2.5 kg. The system is subjeeted to an impulsive force input“ (impulse hammer), and the response is shown below: “ 0.5 1 ‘1 .5 2 2.5 3 3.5 4 Time, a a) Compute, thelgmnping ratio, unclamped natural frequency, spring Constant, and friction coeffith for this system. ' ~ b) Derive. a transfer'function for this mechanical system. Textbook emanates: graded): 3-8-1, 3-3-2, 3-3-3, 13-3—7, B-8-‘8«,;B:i8¥"llii-,fB-S-ll, 3-3-18 1 :3: - '1. |-'. [£13 a No. 93? B11E 9mm“ Engineer's Computation Pad HFH—Eb—EUUb 1U=4H MtEH & HtHU tNUthtHlNU 5T5 HH4 DUHU F.Ud «‘ \ ' Pfiofi I‘ 1:.) \ : :Wm‘ Maid c‘ “9&4 A -'-’ 7’5 79$ 1:. inarmewm Presjum a-F VESFef (gfljgle‘r-essure, WW +~ *p' > 77: E " i fan pressure. mt \ @931} fink. , '3 ( w") N/ML Tc: V CQFaCCHAM :.~ n R T D (:1 = 1-!5 (WM Lag—m7” $ Trill-4E T =. RC. = at.) 3am”: 3-3 in “t5. __ [ res/mu) 1» 848 w + m, : * (41:5,- Egr£5$figé\;fifj i rpfi. _+ 73 HI—‘H—Eb—Elfllflb 11:1: 45' MtL'H E: HtHLI tNLillelelNLi b'r’d HES-4 DUHU |-‘ . [£14 ‘Vl‘i \ r e : .—-""-_-_'—-_ ‘r > ~H€Wifi7i PM? é ____.._.____ x 105 _‘ Abs pressure of vessel forstep inPUt ,’ r f‘ 5.5 I 6.45 5.4 5.35 51" w Abs vessel pressure. Pa m M U'I 6.15 6.1 6.05 .900 ‘ \ L Time, s T ' r A ‘ Ub F. 1U=4H HFH—Eb—EUUb 5T5 HH4 DUHU MtEH & HtHU tNUthtHlNU wmumnmfioz... E. E393 ommEsmcm Sufi whammfln EEmEEuE nun”... , emmfidmwgmi . ,, ,F 5&3; , manic? Ema ms... Engineer's Camputafion Pad ND. 93? ENE 95mm“ HI—‘H—Eb—Elfllflb 11:1: 45' MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUHU |-‘ . [db PMS z. g + M3 +13¢9 = T .---—"'—_'__ i: I :3) BMW ; 9p+ wané + wing??? an d: 6:: lg 5: 7-1.; ' myr- : ‘ E? 2'5"?" . DE :1 ‘ :1 0 II 'I] H Time "h: 154. = i ‘ " ' ' V 3-Way I“.- Z‘ ma graff/Zu-Mq .berflééfl . <‘ I“ .l I.- /. a 6.417 ‘ x V 31+»? % 0. S, = 38,93 .A' : (I-3593).@I =,~,;7’42;.e_:peé¢ HFH—Eb—EUUb 0.07 0. 0664- ‘ ‘* 0.06 0.05 0.04 0.03 anguiar displacement. rad 0.02 0.01 1U=4H MtEH & HtHU tNUthtHlNU w f-gxéiH w it? 7 -. Peg. 2 / Rotationei response for step torque input 0.6 0.0 1 1.2 1.4 5T5 HH4 DUHU e 1.6 ‘ p m .‘,V . . HA“ ' 1.8 F. HK :—a?;.-;:v;c- 3:" .. aruu v.7; UH F. MtEH & HtHU tNUthtHlNU 1U=bU HFH—Eb—EUUb 5T5 HH4 DUHU h- Fmomfiutog D... s smwfiwimflohw Emfihm fig: Enos. mmimmdmmfi w EnE mun—EB nmfi .mfifium \ N. FEE; Hg: Sam “2:5; can NE...— Engineers Computafion Pad No. 93? B11E 35mm” HI—‘H—Eb—Elfllflb llfllblfl MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUHU |-‘ S); m PfloB 3 c, 15465; .F X ,‘.i:_ +- 2173 2...J5 [1) Roof“: ': UH. Jerqu :. (s+3+j$)(s+2_-j§) : “(3 Villa“) SL+43+Z9 = flu 31+ armnfi' -\" Lu?” =* h 539;; rad/S =9 119.22% rad/J a. ,‘ 4 r: . ,1 F .‘ ; . .’ A ‘ \ 2' ‘ . ~ H . \ ,1 1 v ’ J L : ¢ 1' A x \ . .111‘3' HFH—Eb—EUUb lUlbU MtEH & HtHU tNUthtHlNU 5T5 HH4 DUHU . \ Male 3 F__-w—-h-ll-F .. ,2 {Mali/HFM) CM, (5+E)(§+3) : C) 31+Es+ég = (3 J C) SCIhafi‘Hfin fig!) .3" f Engineer's Computation Pad ND. 93? 31 ‘I E 775mg 4: 1!: z 47‘ 3 N U! {A «Blames? :;I"7Hme. can 37914!- =» £5“? 47- : 2. 9mm° F. 1U HFH—Eb—EUUb 1m=bm MtEH & HtHU tNUthtHlNU Pflofi 4 (12,) L09 défir-«emm‘f MEf-haa‘ s; -35"; K. n H “ 3"um Z:%Eflvpfitfl ‘ rfififib’ : " Mi SJ' - M 0J3! M)? + 5%,;4- kr = 1c 5T5 HH4 DUHU F.11 :C(53 Em 5(5) = i)‘ 1 \ x I :3? ‘ ' ‘F-gbgg‘Z-‘s‘ 5" + 7,43: + loo-F ) ———.——«————I-,—;—-————'*——‘—"u ‘ . _.—.——_’—WII—‘rl—-""_-—'—H~‘—'———— 4H4 ,P 5" + +4 40.2; .J.>; _ ,__H__—— TDTHL P.11 , "/0 “V V}: .V: “4.”, er‘w, “L, ,._ 14-” a," ...
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This note was uploaded on 02/11/2010 for the course MAE 3600 taught by Professor Pai during the Summer '09 term at Missouri (Mizzou).

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hw_7_w06_soln - HFH—Eb—EUUb 1U=4H MtEH & HtHU...

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