Transcript Lesson 13 Shunt Connected Dc motor Examples

ET 332a
Dc Motors, Generators and Energy Conversion
Devices
Lesson 13 332a.pptx
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


Explain how changing field excitation of a shunt motor
affects its performance
Explain how the internal feedback inherent in the shunt
motor maintains a nearly constant shaft speed.
Use shunt motor equations and circuit model to compute
motor operating conditions.
Lesson 13 332a.pptx
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1.) Increasing TL causes
motor to slow.
Ea = Ke(n)
Ea
2.) Reduced armature
speed decreases Ea.
TD = KT(Ia)
TD
Ia 
Ia
3. ) More Ia flows in
armature circuit.
More Ia means more
TD.
VT  E a
R acir
4.) When TD matches TL
system stabilizes at
new operating point.
Ia higher:
n
n
Ea
Ke
TL
Time
5.) Ea and n return to
almost the same
values
Lesson 13 332a.pptx
3
Just as in other machines studied up to now, the motor speed,
developed torque and generated emf are all proportional. If an
operating point and a percent increase/decease is known, the new
operating point can be found using proportions.
Ea is proportional to speed
TD is proportional to armature Current
E a1 n 1

Ea2 n 2
TD1 I a1

TD 2 I a 2
Speed is directly proportional to Ea and inversely proportional to field flux
n 1  E a1    p 2 



n 2  E a 2    p1 
Lesson 13 332a.pptx
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A 10 HP 240 volt 1200 rpm motor is operating
at rated conditions. Determine the percent change in
shunt field flux required to lower the speed to 900
rpm. Assume that armature current remains the
same.
Lesson 13 332a.pptx
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Define speeds for two cases
n1 = 1200 rpm n2 = 900 rpm
Set up proportions
V I R
n1  T a1 acir
 p1  k G
VT  Ia 2  R acir
n2 
p2  k G
 VT  I a1  R acir 




k
n1 
  VT  I a1  R acir    p 2  k G  Since Ia1=Ia2 this
p1
G




n 2  VT  I a 2  R acir    p1  k G   VT  I a 2  R acir  simplifies




k
p2
G


n1   p 2 


n 2   p1 
Find the value of
p2 assuming p1
=1
1200 rpm   p 2 

  1.33
900 rpm  1 
  p 2   p1 
1.33  1

100
%

100%  33%




 1 
  p1 
Lesson 13 332a.pptx
Answer
6
A 200 volt shunt motor is rated a 5 HP and 1000 rpm.
At rated output it draws 25 A of line current. The total
armature circuit resistance is 0.5 ohms. The field
resistance is 100 ohms
a.) find the total rotational losses of the motor at rated
conditions
b.) Find the no load speed of the motor at 200 volts
Lesson 13 332a.pptx
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IT = 25 A
Racir
5 HP
1000 rpm
Ia
Find Ia
Rf
200 V
If
Rotational losses are the difference
between Pem and Pshaft
Find Ea
Lesson 13 332a.pptx
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Answer
Lesson 13 332a.pptx
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b.) At no-load the only power absorbed is what is required
to supply Prot.
Power balance on electric side of motor. Neglecting brushes
Put into standard form
Solve quadratic for
Ia
Lesson 13 332a.pptx
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Torques is proportional to armature current Ia. Lower
power requires lower torque and armature current
Speed is proportional to Ea
Lesson 13 332a.pptx
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Solve the previous equation for n2
Answer
Lesson 13 332a.pptx
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Weakening the field increases speed but reduces torque
Example 13-3: A 500 volt 125 HP 1150 rpm shunt motor
operates at rated conditions, driving a constant-torque load.
The line current at rated conditions is 204.3 amps. The total
armature resistance is 0.0343 ohms the field resistance is 96
ohms.
a) Determine the steady-state armature current if a 0.052 ohm
resistor is connected in series with the armature and the
field is weakened by 10% from its rated value.
b) b.) Determine the steady-state speed for conditions in part
a.
Lesson 13 332a.pptx
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R1=0.052 W
Racir
0.0343 W
Rf=96 W
If
IT = 204.3 A
VT=500 V
Find Ea at rated conditions
Lesson 13 332a.pptx
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Switch in additional resistance, R1, and weaken field
TD1 = TD2 Constant torque load
Lesson 13 332a.pptx
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Field weakened by 10% reduces the value of KT by 10% assuming
no magnetic saturation
Lesson 13 332a.pptx
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Answer Part a
b.) Find speed under the above conditions
Lesson 13 332a.pptx
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Answer Part b
Lesson 13 332a.pptx
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ET 332a
Dc Motors, Generators and
Energy Conversion Devices
Lesson 13 332a.pptx
19