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ECE320, Spring 2010, Exam 3b
March 19, 2010
Name
PID Number
Problem
1
2
3
Total
Max Points
20
40
40
100
Points
Instructions:

Closed book, but you are allowed to have one-page double-sided handwritten notes.

You can have 3 pages of blank papers for calculation purpose.

Write your solutions and steps neatly and orderly. No credits for random side
writings.

Box your answers, Where needed.
Problem 1 (20 points). A DC motor is powered from a 200 V source. The motor rotates at
2000 rpm at no load and at 1800 rpm at half load. What is the motor speed at full load?
Va  I a Ra  k
[eq. 1]
200  (0) * Ra  k * 2000rpm
No Load Case
k
 V 
200

 0.10
2000
 rpm 
200 
1
I a _ rated * Ra  k * 1800rpm
2
Half Load Case
1
I a _ rated * Ra  0.10 * 1800rpm
2
1
200  I a _ rated * Ra  180
2
200  180
I a _ rated * Ra 
 40(V )
1
2
200 
From [eq. 1]  

1
(Va  I a Ra )
k
1
(200  40)  1600rpm
0.10
Full Load Case
Problem 2. (40 points)
Three (N-turn) stator windings are placed in space 120º apart from each other as shown in the

figure to produce a rotating magnetic field, B with a constant magnitude. Determine the
following things:
Dot “•” Indicates flux
direction (flowing out)
when applying a positive
current to the coil
i1(t) = 100*cos(50*2pt) A
X
i3(t)
i2(t) = 100*cos(50*2pt120o) A
a) i3(t) =
100 cos(50 * 2p  t  120 o )( A) OR

b) The rotating speed of B is
t=0 is
+90o
100 cos(50 * 2p  t  240 o )( A)
. (10 pts)
50 Hz
(Hz or rpm, circle one) and its initial position at
degrees from axis “x”. (10 pts)

c) B coincides with axis “x” when t = 0.005 ,
0.025 ,
0.045 , and so on. (10 pts)
(give first three consecutive times of the coincidence.)

Note: the initial position of B is +90o from “x”, i.e. ¼ of a cycle, where T is the period.
1
1
 1  1 
T     
 0.005( s ) Is the first occurrence
4
 4  50  200
1
T  T  0.025( s ) For the second occurrence
4
1
T  2T  0.045( s ) Is the third occurrence
4

d) What’s the rotating direction of B ?

how to reverse the direction of B ?
Clockwise
and
Swap any two of the three currents__. (10 pts)
Problem 3. (40 points) Please read the entire problem and sort out your thoughts before start writing.
(a) Explain the principle of a squirrel-cage IM (you need to convince the grader that you
understand the principle by using your own words and/or sketches. The flow of your
explanation has to be easy to follow). The following criteria will be used for grading: what is
the stator and its function (5 pts); how voltage is induced in the rotor (5 pts); how slip
frequency affects the voltage induced (5 pts) and torque produced (5 pts) in the rotor; and
explain why no torque is produced when the rotor is rotating at the synchronous speed (5 pts).
(b) An IM's equivalent circuit is similar to a transformer's. An IM's rotor can be locked (i.e.,
the rotor speed is zero) or an IM can be operating at its synchronous speed. What is the
equivalent transformer operating condition of an IM that its rotor is locked and why? (7.5 pts)
What is the equivalent transformer operating condition of an IM that is operating at its
synchronous speed and why? (7.5 pts)
An example solution is as follows:
(a.)
Stator and its function:
The stator is the stationary part of the motor. It is usually made up of 3 outer
windings.
The function of the stator is to create a rotating magnetic field and to transfer
power.
How voltage is induced in the rotor:
When s  o the stator flux is moving with respect to the rotor bars. The
moving flux then cuts across the rotor bars. This creates a changing magnetic
 d 
field   , as seen by the rotor bars, which induces a voltage on the rotor bars.
 dt 
How slip frequency affects voltage induced:
It is the slip that causes the rotor bars to experience a magnetic field that
d
changes in time. As slip increases,
increases and thereby increases the
dt
d
induced voltage. As slip decreases,
decreases and thereby decreases the
dt
induced voltage.
How slip frequency affects torque produced:
The lower slip results in a lower rotor bar voltage and therefore a lower rotor
bar current. The rotor current is directly related to torque
Why no torque is produced when rotor is rotating at synchronous speed:
When the rotor is rotating at synchronous speed, the rotor bars are rotating at the
same speed as the stator field. Therefore there is not a time rate of change in
the magnetic field from the perspective of the rotor bars and therefore a voltage
is not induced. Without an induced voltage, rotor current is zero and therefore
torque is zero as shown in equation 6.7 above.
(b)
Equivalence of operating conditions:
Induction Motor
(i)
Transformer
Reason
Locked Rotor
Short the secondary windings
Lot of rotor current
without output
power
(ii) Synchronous Speed
Open the secondary windings
No rotor current