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Exam2_solutions[1]

# Exam2_solutions[1] - Exam II ECE321 Fall 2010 First Name 2C...

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Unformatted text preview: Exam II ECE321 Fall, 2010 First Name: 2C) W 1’ Last Name: Student ID: Work gr'oblems and gravide answers in mace gr'ovided - do not unstagle gages 1-page crib allowed (to be submitted with exam). No calculators, you may express answers in terms of 11: , J2- , e , oo , including fractions thereof. Simpiify to extent possible. Some answers may be 0 or 00. (25) 1. Consider the foliowing multi—stack variable—reluctance stepper motor (VRSM). {i-ii] eMr: E’r'm The maximum self inductance =5 H. The minimum self—inductance = l H. (a) ias m ics : 0, ibs = 4 A. TL 2 0.The stable equilibrium position closestto that shown above "‘1 0 mm: eﬁ (b) The rotor is at the equilibrium position in (a). Then, ibs is stepped to 0 and £68 to 4 A with ias remaining 0. The rotor settles to the new equilibrium position at em = /20 0. (C)Lbsbs(9rm)=W§m+~Lcos[ (ermﬂLm H ,., 9 .m, m- / m- (M (9 *3“ + cost em» (em—Miamin CSCS rm) 2 (e)ib3;4A,i =i =0. (18 CS T _______ w 6/2? ) i“ 2 Vi,” [Z {6W wéofi/ 5:" (25) 2. The static torque—versus-position characteristics of a VRSM are depicted below for one—phase-on-at-a- [i-ii] time excitation. as bs cs as bs cs Tmax = 1 N—m (I: 1 A) 9: arm (a) RT = 8 (b) TL = 0.5, ics = 1A, ids = ibs = 0. The stable equilibrium closest to em = 0 is 6rm :H o (c) The rotor is at the equilibrium position in (b). Then, it,S is stepped to 0 and ibs to 1 A with ias re— maining 0. Sketch the trajectory in the ﬁgure above assuming no overshoot. After transients subside, 8 = d 75/4 ° . rm (d) The rotor is at the equilibrium position in (b) [TL 2 0.5, ics = l A, ias = ibs = 0]. Then, ias is stepped to l A with ics remaining at l A and ibs at 0. Sketch the trajectory in the figure below as- suming no overshoot. After transient subside, em = ‘3' 75 °. as _ bs cs as _ bs cs ll 55 (v .2 B (e) If the current in an excited phase is decreased to 0.5 A, the corresponding Tmax (a)Laf= (2 + (2,5 cos( / 9r+ O °)H. (b) Assuming there is no commutator, if = 2 A , coil a-a' is open circuited, and 9,, = 605. Mm’ tﬁl >kaf = ﬁﬂqwtfj: ijééﬂ —" .p— 0,5 5an (67/) : 6 O va_a.(t) = C) + 0/63 sin( 50 t+ 4590 °) Sketch va_a, vs. 9,. in space below. Ua~a', max = [email protected] 9, = “£0 270° Ua-a',min - #60 [email protected] e: _ mo (c) Assume a commutator connects the rotor coil to stationary terminals through brushes as shown below. Sketch 116 vs. 9,, for conditions in (b). —270° —-130° —-90° C: .9 s a I [g < ® 10: u Ht 4) CD H. N \3 Q 0 C I Q < ® CD 11 O a \ m Q: ‘2) (10) 4. A PM dc machine has parameters: kv = 0.05 V~s/rad, ra : 4 Q,and LAA = i mH. If 00 m 20 V iii, iv . . . . [ ] (constant), sketch Te versus 03,. ludicatmg numcncal values for Te when 03,. r 0 and 03,. when Te = 0. “P c - um 3 [\$3. J ’ ‘ M. a" (fng 1'" my”, (15) 5. The dc machine in problem 4 is supplied by a 2—quadrant chopper. [iv, V] is w?- (MI—a: g A :1, 5M CW; 42;“ m“? — 2 P296?) . u, (0) car [ad/s if ﬂ fr ﬂu diff,» Exam II ECE321 Fall, 2010 (a; f: 1‘ “fag; Last Name: “0.31:3 {J7 {ﬂieﬂofigzs We}! ‘3 First Name: Student ID: Work )mblems and )mvide answers in smce )mvided ~ do not unstaﬂc )(1 es 1~page crib allowed (to be submitted with exam). No calculators, you may express answers in terms of rc, ﬂ , e , oo , including fractions thereof. Simpiify to extent possible. Some answers may be 0 or 00 . (25) 1. Consider the following multi—stack variable-reluctance stepper motor (VRSM). [Hi] 9’77! (O 1% 3) (b) The rotor is at the equilibrium position in (a). Then, ibs is stepped to 0 and ics to 2 A with ias remaining 0. The rotor settles to the new equilibrium position at Gm = j .2 :1} 0. (C)Lbsbs(9rm)= + 2 ces[ Z (BWW6C) °)]H ,3 w/zo <d>Lcscs(8,-m) = 3 + ‘e costﬂLte,‘me;)i H @HM:ZAJ =i :0 as CS Tewrm): + 8 sin[24 (9. +30 °)]N—m (25) 2. The static torque-versus-position characteristics of a VRSM are depicted below for one-phase-on-at-a- [i-ii] time excitation. (1)) TL = 0.5, lies 2 1A, ias 2 £58 = 0. The stable equilibrium closest to 6 = 0 is rm. 9rm zﬁ—ﬁk I ' 0 a hu‘ we 80,”: 0.5" -» 8am: ~30° rt, (2.”: #3750 (c) The rotor is at the equilibrium position in (b). Then, ics is stepped to O and ribs to l A with ias re- maining 0. Sketch the trajectory in the figure above assuming no overshoot. After transients subside, #75 o em: 45" 5/? = “75/4 :- #973 (d) The rotor is at the equilibrium position in (b) [TL = 0.5, ics = 1 A, ias = ibs = 0]. Then, ias is stepped to 1 A with ics remaining at l A and i bs at 0. Sketch the trajectory in the figure below as— suming no overshoot. After transient subside, em = 335‘ °. — 3.75 1’" 7:5 S 3.73 as _ bs _ cs as _ bs _ cs (e) If the current in an excited phase is increased to 2 A, the corresponding Tmax = if N-m. aa Laﬁ max = 0.5 H (await: 0 + 0,6‘cos(/ 9r+ 0 °)H. (b) Assuming there is no commutator, if = 4 A, coil a~a’ is open circuited, and Br = 402?. 7/6102" : \$264! : [1‘ro fjoéZaJ'JCg??? :’(f(40) 0:5.5/7} 6/“ :.—804,‘040f' va_a.(t) = 0 + 50 sin(f0t+ /90 °) Sketch va‘ a. vs. 0,, In space below. va-a',max = (at) [email protected] 9r =d—90 2 7O 0 ll -80 [email protected] = Cfo —2?0 o Ua-a’, min (0) Assume a commutator connects the rotor coil to stationary terminals through brushes as shown below. Sketch va vs. 9,, for conditions in (b). —180° —90° (10) 4. A PM do machine has parameters: I30 2 0,05 V—s/rad, rq = 2 Q, and LAA 2 1 mil . If ya = 20 V li-ii ivl . . . . ’ ‘ (constant), sketch Te versus (0,. indicating numencai values for TB when (D). = 0 and 0),. when G i}. s. {M} (“Jr 53 um will n 39:33:”? 541,213 (15) 5. The dc machine in problem 4 is supplied by a 2-quadrant chopper. [i~ii, iv] 55 t, ms If the load torque TL = 0.1 N—m, (a) E m v (sex. b I— ; A m ( ) a m 53:" all 5;" (€7.53: {3% (C) 631‘ 3 1'ad/S 2:: E. r“ -- r“ ;~ , a o. if: S: r9 {:4 m4: <3»- "3' MW 5/ Lee?" M ‘g’ {:3 7'5“ {7. Qg'ﬁefk’" ...
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