ece320 exam3b 2010 solution

7/26/2019 Ece320 Exam3b 2010 Solution

1/5

ECE320, Spring 2010, Exam 3b March 19, 2010

Name

PID Number

Problem Max Point Point

1 20

2 !0

3 !0

"otal 100

Intruction#

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.

$rite %our olution an& tep neatl% an& or&erl%' No cre&it (or ran&om i&e

)riting'

*ox %our an)er, $here nee&e&'


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Problem 1 +20 point. A DC motor is powered from a !! " source. #he motor rotates at

!!! rpm at no load and at $%!! rpm at half load. &hat is the motor speed at full load'

kRIV aaa += (e). $*

rpmkRa !!!++,!-!! += No -oa& Cae

==

rpm

Vk $!.!

!!!

!!

rpmkRI arateda $%!!++

$!! += .al( -oa& Cae

rpmRI arateda $%!!+$!.!+

$!! +=

$%!+

$!! += arateda RI

,-/!

$

$%!!!+. VRI arateda =

=

0rom (e). $* ,-$ aaa RIVk =

/ull -oa& Cae

rpm$1!!,/!!!-$!.!

$==


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Problem 2' +!0 point

#hree 2-turn stator windings are placed in space $! apart from each other as shown in the

figure to produce a rotating magnetic field, B with a constant magnitude. Determine the

following things4

a i3t 5 ,,-$!+6!cos-$!! At o+ 78 ,,-/!+6!cos-$!! At o . $! pts

b #he rotating speed of B is 6! 9: 9: or rpm, circle one and its initial position

at t5! is ;=?. $! pts

c B coincides with a=is >=? when t5 !.!!6 , !.!6 , !.!/6 , and so on. $!pts

give first three consecutive times of the coincidence.

Note# the initial position of B is ;=?, i.e. @ of a cycle, where # is the period.

,-!!6.!!!

$

6!

$

/

$

/

$sT ==

= s the first occurrence

,-!6.!/

$sTT =+ 0or the second occurrence

,-!/6.!/

$sTT =+ s the third occurrence

d &hatBs the rotating direction of B ' Clockwise and


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how to reverse the direction of B ' wap any two of the three currents. $! pts

Problem 3' +!0 point lease read the entire problem and sort out your thoughts before start writing.

+aE=plain the principle of a s)uirrel-cage F you need to convince the grader that you

understand the principle by using your own words andGor sketches. #he flow of your

e=planation has to be easy to follow. #he following criteria will be used for grading4 what isthe stator and its function + ptH how voltage is induced in the rotor + ptH how slip

fre)uency affects the voltage induced + pt and tor)ue produced + pt in the rotorH and

e=plain why no tor)ue is produced when the rotor is rotating at the synchronous speed + pt.+bAn FIs e)uivalent circuit is similar to a transformerIs. An FIs rotor can be locked i.e.,

the rotor speed is :ero or an F can be operating at its synchronous speed. &hat is the

e)uivalent transformer operating condition of an F that its rotor is locked and why' +' pt

&hat is the e)uivalent transformer operating condition of an F that is operating at itssynchronous speed and why' +' pt

n example olution i a (ollo)#

+a'

tator and its function4#he stator is the stationary part of the motor. t is usually made up of 3 outer

windings.

#he function of the stator is to create a rotating magnetic field and to transfer

power.

9ow voltage is induced in the rotor4

&hen os the stator flu= is moving with respect to the rotor bars. #hemoving flu= then cuts across the rotor bars. #his creates a changing magnetic

field

dt

d, as seen by the rotor bars, which induces a voltage on the rotor bars.


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9ow slip fre)uency affects voltage induced4t is the slip that causes the rotor bars to e=perience a magnetic field that

changes in time. As slip increases,dt

dincreases and thereby increases the

induced voltage. As slip decreases, dt

d

decreases and thereby decreases theinduced voltage.

9ow slip fre)uency affects tor)ue produced4

#he lower slip results in a lower rotor bar voltage and therefore a lower rotor

bar current. #he rotor current is directly related to tor)ue

&hy no tor)ue is produced when rotor is rotating at synchronous speed4

&hen the rotor is rotating at synchronous speed, the rotor bars are rotating at the

same speed as the stator field. #herefore there is not a time rate of change inthe magnetic field from the perspective of the rotor bars and therefore a voltage

is not induced. &ithout an induced voltage, rotor current is :ero and therefore

tor)ue is :ero as shown in e)uation 1.J above.

+b E)uivalence of operating conditions4

nduction Fotor #ransformer 8eason

i Kocked 8otor hort the secondary windings Kot of rotor current

without outputpower

ii ynchronous peed 7pen the secondary windings 2o rotor current

ece320 exam3b 2010 solution

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