ET 332a Dc Motors, Generators and Energy Conversion Devices · p1 p2 2 1 n n Find the value of p2...
Transcript of ET 332a Dc Motors, Generators and Energy Conversion Devices · p1 p2 2 1 n n Find the value of p2...
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ET 332a
Dc Motors, Generators and Energy Conversion Devices
1 Lesson 13 332a.pptx
Lesson 13 332a.pptx 2
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.
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Lesson 13 332a.pptx 3
TD = KT(Ia)
Time
TD
Ia
n
TL
Ea
Ea = Ke(n)
1.) Increasing TL causes motor to slow.
2.) Reduced armature
speed decreases Ea. 3. ) More Ia flows in
armature circuit. More Ia means more
TD.
4.) When TD matches TL
system stabilizes at new operating point. Ia higher:
5.) Ea and n return to
almost the same values
e
a
K
En
acir
aTa
R
EVI
Lesson 13 332a.pptx 4
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
2
1
2a
1a
n
n
E
E
TD is proportional to armature Current
2a
1a
2D
1D
I
I
T
T
1p
2p
2a
1a
2
1
E
E
n
n
Speed is directly proportional to Ea and inversely proportional to field flux
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Lesson 13 332a.pptx 5
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 6
n1 = 1200 rpm n2 = 900 rpm
Define speeds for two cases
Set up proportions
G1p
acir1aT1
k
RIVn
G2p
acir2aT2
k
RIVn
acir2aT
G2p
G1p
acir1aT
G2p
acir2aT
G1p
acir1aT
2
1
RIV
k
k
RIV
k
RIV
k
RIV
n
n Since Ia1=Ia2 this simplifies
1p
2p
2
1
n
nFind the value of
p2 assuming p1 = 1
33.11rpm 900
rpm 1200 2p
%33%1001
133.1%100
1p
1p2p Answer
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Lesson 13 332a.pptx 7
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 8
200 V
IT = 25 A
Racir Rf
5 HP 1000 rpm
If
Ia
Find Ia
Rotational losses are the difference between Pem and Pshaft
Find Ea
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Lesson 13 332a.pptx 9
Answer
Lesson 13 332a.pptx 10
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
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Lesson 13 332a.pptx 11
Torques is proportional to armature current Ia. Lower power requires lower torque and armature current
Speed is proportional to Ea
Lesson 13 332a.pptx 12
Solve the previous equation for n2
Answer
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Lesson 13 332a.pptx 13
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 14
VT=500 V
Racir
0.0343
R1=0.052
Rf=96
IT = 204.3 A
If
Find Ea at rated conditions
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Lesson 13 332a.pptx 15
Switch in additional resistance, R1, and weaken field
TD1 = TD2 Constant torque load
Lesson 13 332a.pptx 16
Field weakened by 10% reduces the value of KT by 10% assuming no magnetic saturation
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Lesson 13 332a.pptx 17
Answer Part a
b.) Find speed under the above conditions
Lesson 13 332a.pptx 18
Answer Part b
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ET 332a
Dc Motors, Generators and Energy Conversion Devices
Lesson 13 332a.pptx 19