Fall2010_Exam2 - ECE 430 Exam#2 Fall 2010 Name M 90 Minutes...

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Unformatted text preview: ECE 430 Exam #2, Fall 2010 Name: M 90 Minutes Section (Check One) MWF 10am MWF 2pm 1. /25 2. /25 3. /25 4. /25 Total / 100 Useful information sin(x)=cos(x—90°) 17:27 §=V_I* ,u0 : 47:.10‘7 H/m a z . [CH-dlszJ-nda [CE-dl=—EJ‘SB-nda 92:1; WF=N2=¢91 912—114. BzyH ¢=BA x1=N¢ x1=Li(if1inear) ,u m‘ 1. i I . A I_ 1‘ ,_ . e_an __ Wm=£zdl Wm 1M: Wm+Wm—/12 f _ ax _— ax x—)6 fe_)Te b b I . g _an ._an 121:1?ij @2249;de €E§+€i¥=Wme 1- a,- 1m Problem 1. (25 points) Assume the two stator coils (sa and sb) shown below are identical and the rotor coil has a resistance of 1.5 Ohms. a) Write down Xsa(isa,isb,ir,6), ksb(isa,isb,ir,6), and N(isa,isb,i,,6) by inspection using inductance parameters LS, L, and M. You may use sinusoidal approximations for position effects as appropriate and you should assume the positive polarity for the flux linkages and voltages is on the “X” mark of each coil. (9 points) b) Find an expression for the torque of electrical origin in the positive 9 direction in terms of the ~ currents, 6, and the mutual inductance between stator and rotor (M). (8 points) Let if = 8 A (DC). If de/dt = 1207: rad/s, what are the voltages vsa(t) and vsb(t) when isa=isb= 0 in terms of the mutual inductance M and the angle 9? (8 points) fl\ ASa: L5 L;a+/Y)[659&’/ (extra space on next page) < , . . . T : “MS/lflfltgflc, fn’lefiLgbtf 5) Wk: 54:. «8mx/7o-r5m9 ngngbg gmxizofiwsé Problem 2. (25 points.) a) Explain in your own words why it is usually easier to use coenergy (Wm’) rather than the energy stored in the coupling field (Wm) to compute the force or torque of electrical origin. 9 .r f j 3 M; a; filo “flue M¢7nC~FfC may?» if [0mg J’iwflflj jarng [Ia/St A I; 79“. turn ij‘fluw) 70a. "Huffy/V4. fir» L [fl firm; 59f b) Show graphically why the energy stored in the coupling field and the coenergy are equal for magnetically linear systems. c) Why is the integral for energy stored in the coupling field done along a constant x (or theta) path while the EFE integral might not be? winch [am/2994117 Wm} gym mail '1'? thfitjflwfé 7%! {ant at!!! X wfifié flwé [mun-r the! {(642 Ime-yt A,” yfg’ Tklf‘lvlv’f W‘cn you t‘n'fcfuwfi‘ twirtm‘f fiV‘gf ikoL‘jfx )vu 4‘559‘ All,“ '2" If“ [‘13 )[fmf Vdéqlfi B 7/44 Witty-“f 01‘ I: avu 2 Alan} IL fflééflflfz ’flfi‘fifl warm; éczmii thewéew‘f tLILMerj X, Problem 3. (25 points) An electromechanical system has the following flux linkage-current relationship: 0.08 —— z‘ (0.02+x) In the following questions, EF E stands for “Energy From the Electrical system into the coupling field”, and EFM stands for “Energy From the Mechanical system into the coupling field”. Consider the following to be point a: x = O, i = 0, 7» = 0. a) Find the EFE and EF M as the system is moved along constant x from point a to point b which has i = 5 Amps. b) Find the EFE and EFM as the system is moved along constant 9» from point b to point c which has x = 0.02 meters. 0) Find Wm at point c d) Find the EFE and EFM as the system is moved along constant x from point c to point d which has 9» = i = 0. e) For this cycle, is this a motor or a generator? Explain why. 9(50 7"”2 é) Wmc=izauoaloar , 0 EFE- : :; ——] ' 3* D Wing/’Wmc; “[003.— A) C, l [24; 003' V; J . 2 0r 6) EFE :— {0+0 4400 = “57/5 ‘9 “wmflr (EFM 2501 cycle: a“; a) c) ,0? . Filer/yr firearm 3), A 3., WW L ‘L. Problem 4. ’02 ' 3: (25 points) The movable mass of the electromechanical system of problem 3 has a mass of 0.5 kg. The electrical system is producing a current of 4 Amps (DC) and the wires have a total resistance of 3 Ohms. There is an external force equal to 200 Newtons being applied in the positive x direction and the mass is stationary under these conditions. a) What is the equilibrium value of x for the condition described above (constrained to be positive)? b) Describe as best you can what happens if the applied voltage is suddenly changed from 12 Volts to 24 Volts (DC). c) Solve the electrical and mechanical dynamic equations of motion using Euler’s method for one time step of 0.1 seconds to see what happens after the moment of suddenly changing the voltage. —Z g I f All}: 0 zi- Zae as» L I, (hung) frtz’fl‘e) but“ 4 6., g 7n.- :J fiteiw 00244.1() .fi 2w X¢7 0 WC (cu/auf WIN [n[rt45< 'f'p Qh'm/f The (6 win Phat/age WlnFZK “(I/t )MIT'“ ‘flw" M‘ug Mat/3"} m m naga.r;v< x gamma, It: We we fire/9 .‘MH/ x warm saw), 3 ' ., 1—0.3...- JL’ ,, M N": [K 4» 2'7, 1 L 4' [Jab/rue) 1:1.“ {3014?le As! .JV L ’2 _.. wiy 015,3200“ ’0/0 V/Olwm x/o)=.0364m ' 610/)" H+ ’a2+’£§i(27,3x7+0)x0./= LMLM A f "D we 7C/Del31,03éc + om: :. ,0344 m ‘ Jr ’0‘”quny = 0 y/fl/I) ‘" 0 +5315'(200 ’/,oz.;,0}((l ...
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