Spring2009_Exam2

# Spring2009_Exam2 - ECE 430 Exam#2 Spring 2009 Name g0 I...

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Unformatted text preview: ECE 430 Exam #2, Spring 2009 Name: g0 I ‘A ‘HOV‘ 90 Minutes Section (Check One) MWF TTH 1. /25 2. /25 3. /25 4. /25 Total /100 Useﬁll information #0 = 47:.10'7 H/m fCH-d1=jSJ.nda ICE-dl=—a—at [SB-nda 92:1; MMF:N2’=¢ER B=,uH ¢=BA A=N¢ 2=Li(if1inear) ,u :_ m 6x 6x Wm=idi Wm’2Iﬂtdf Wm+W,,’,=/tz' fe—an 6W x—w 0 fe_9Te b I 15%;“: idl EFM=— fedx €§§+EFM=erWm 1:6Wm - 6W»: l : a—>b 4'95 a 61' 8/7. 1. (25 points) A 2-phase machine, with two stator coils and one rotor coil, has following ﬂux linkage-current relationship 13,, L5 0 McosB isa 2,3,, = 0 LS Msine x isb A, Moose MsinG Lr ir a) Compute the co—energy W "'1 (10 points) b) Compute the energy Wm (5 points) 0) Find the torque of electrical origin Te (10 points) x 1/ ’ J '7' . . u .. . t A\ Wm :% 365’“, ‘l' lljcs‘s 4' m/dﬁQg‘F/ I’mg‘ﬂaggcr fillrc/ // a l ( {PM a 5757377“) 2. (25 points) The ﬁgure below shows a magnetic strucuture composed of an iron core and a moveable iron plunger. The plunger is connected to a solid structure through a spring with stiffness coefﬁcient K. The spring exerts no force when the plunger is located at x = C. The plunger is separated from the core by a constant gap of length g above and below the plunger. The horizontal length of the plunger is W. The plunger and the core both have a depth of D. The iron core and plunger both have inﬁnite permeability. You should neglect fringing in the air gaps. You may neglect the force of gravity and all friction. Depth: D [1—900 For V = det, a) Find the ﬂux linkage )L in terms of current i and position x plus no, N, V/, g, and D. b) Find the force of electric origin either in terms of ﬂux linkage 9t and position x, or in terms of current i and position x (whichever you prefer). c) Select either ﬂux linkage 9» or current i as an independent state and write the three differential equations that describe all of the dynamics of this electromechanical system with voltage v as the only input. P2 . N1,“ (w... 150 . 4) C\ 9 «1%» 2 M : “We 11/“ 2 ﬂ “"1. {g 1 WWW” #35270 F" 1 ,,. ‘2‘ 0 [ii *‘O’V'xm My): Mil—w“ is z , (this page is extra space for showing work) Niad/wv-XWAJ ’ NEWQEZ (Vcaglé-f‘ 15 cur ’Lj 513% {hng .2». MK,()<~-CS ’- W M «a 0K . K A)? M» «312/70 ’r 75" “ N/40(W”‘ \$4 13;; g :4 ,ﬂ. m4," ..: z—de ’ 1;? kowymwm a“ Z 6 ; 4d7<rc8 2 DH?” (Wad 3. (25 points) The co—energy of a device is given by 51' 6/ z 6 Z ‘ .5 .3 . _ A ‘ W / "a = L + L. + .1. C {—— ‘l’ ‘f— + max) 24x 6x x (A iii 6* X Find: i a) The force of electrical origin f e (i, x) b) The energy stored in the coupling ﬁeld Wm as a function of i and x. (,,CC ,3”! M\ ’F ’ mL 41" X7" 4. (25 points) An electromechanical system has a nonlinear ﬂux linkage-current relationship: 12 1':— x a) Find the energy stored in the coupling ﬁeld when i = 2.0 Amps, x = 0.02 m, k = 0.2 Wan b) Find the force of electric origin when i = 2.0 Amps, x = 0.02 In, 9» = 0.2 Wan c) Find the energy transferred from the mechanical system into the coupling ﬁeld (EFM) as the position x changes from 0.02 m to 0.01 m along a constant ﬂux linkage (0.2 Wan) path. VA A .g A 2’: '73 , L) y’ﬁv-ta’ﬂf 7c 3'01 8 ,,7/ "'" 3 3 A] Otl%: {dd A '3 L5 ‘00 r , 9\ 5" “WM-"2. ’L 7“/-" c M 7[ r . Bx 3x .u 3mg.” Bacon-é x:«.02 j’d‘ ’ 2’0 ’0} [ 3‘7” .0. 1 31V ‘3' 3 n -’ x , 3.. 3 3 0 Wm : l’f/‘ﬁ ‘ '( 3* :él’vf"; M , , ' ll ’6" A b 0?.— ;ﬁoll 3",. 3#0 .0L «068. . é’ , X .. : WW * ~33” ~ 3 6% T 'p}v'_ .OC ...
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## This note was uploaded on 01/24/2012 for the course ECE 330 taught by Professor Petersauer during the Spring '12 term at University of Illinois, Urbana Champaign.

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Spring2009_Exam2 - ECE 430 Exam#2 Spring 2009 Name g0 I...

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