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Unformatted text preview: ECE 430 Exam #2, Fall 2009 Name: 50 / OL 7‘7 ‘am
90 Minutes Section (Check One) MWF 10am MWF 2pm
1. / 35 2. / 35
3. / 30 Total / 100 Useful information sin(x)=cos(x—90°) V =21 S =V1‘ yo = 47: 10‘7 H/m [CHd1=[SJnda [CEd1=—%jSBnda SR=—L MMF=Ni=¢ER ,uA 91:1. szH ¢=BA 1=N¢ 1=Li(ifunear) N ®
2. i I
0 0 6x 6x fe_>Te
b I
. _ e _ _6Wm .__6Wm
Egg—W law—1M EaEbE+€Ii¥~Wme % a "6,1 1. (35 points) For the structure drawn in the ﬁgure below, the movable mass M is constrained to move left and right
only as indicated in the ﬁgure where “x” is the distance from the left edge of the stationary member to
the right edge of the movable mass. The depth into the page for both members is 2cm. The width shown
is 4 cm. The gap g is 1mm, and the number of turns N =100. Find: a) Flux linkage 7L (deﬁned for the voltage polarity shown) in terms of current i and distance x. b) An expression for the voltage v in terms of the current i, distance x, resistance R, the velocity (v)
of the mass in the positive x direction, and time t. c) An expression for the force of electric origin on the movable mass M in the positive x direction. d) A numerical value for the energy stored in the coupling ﬁeld (in Joules) when the current is 10
Amps and the distance x is 3cm You may neglect fringing in the gap, and you may assume the iron is inﬁnitely permeable. 4/3‘ =1 ‘ . V; LY
M W , t
m/Wm '063ﬂ03)/0a : 0187 U 2. (35 points) An electric machine (1 = stator, 2 = rotor) has the following linear ﬂux linkage vs current
characteristic: M = 0.2i1 + 0.1sin6 l2
X2 = 0.lsin6i1+ 0.3 l2 a) What is the energy stored in the coupling ﬁeld when 6 = 90 degrees, i1 = 3 Amps, and i2 = 5
Amps? b) How much energy is given to the coupling ﬁeld by the mechanical system if 9 is changed
from 90 degrees to 60 degrees while the two currents remain constant? c) How much energy is given to the coupling ﬁeld by the electrical system during that same path
from 9 equals 90 degrees to 60 degrees while the two currents remain constant? ’ l, ,L 2 t.
4\ WM 7 Wm .. L422., 2/ +045mm/LZ iii/3221 M (9V70 .. i
‘ a an ,Ir J a. . a
a” Z 7* “(*2 3’25“— 34/573 3. (30 points)
Consider the following nonlinear equations 0
X1
0
X2 X1 ' XIXZ
0.5 xlxz  2 x; H II Assume the initial conditions for this system are
_ X1(0)] _ [ 2 ]
I‘m) ‘ Low) ‘ 0.5
a) Using Euler’s method with a time step At = 0.1, determine the value of x(0.2). b) Determine two separate equilibrium points for this system. X (03):— 514/04) ixiﬁjw :: c:
2.? ’53 #072; ...
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 Spring '08
 Kim
 Energy, Mass, Electromagnet, coupling field, movable mass

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