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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 1page 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). {iii] 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—versusposition characteristics of a VRSM are depicted below for one—phaseonata [iii] 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 aa' 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 = V@ 9, = “£0 270°
Uaa',min  #60 V@ 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
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<
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10:
u
Ht
4)
CD
H.
N
\3
Q
0 C
I
Q
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®
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 variablereluctance 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 torqueversusposition characteristics of a VRSM are depicted below for onephaseonata
[iii] 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 Nm. 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. vaa',max = (at) V@ 9r =d—90 2 7O 0 ll 80 [email protected] = Cfo —2?0 o Uaa’, 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 liii 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 2quadrant 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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This note was uploaded on 11/29/2010 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue.
 Spring '08
 Staff

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