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Lesson 13 Shunt Connected Dc Motor Examples ET 332a Dc Motors, Generators and Energy Conversion Devices 1 Lesson 13 332a.pptx

ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

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Page 1: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

Lesson 13 Shunt Connected Dc Motor

ExamplesET 332a

Dc Motors, Generators and Energy Conversion Devices

1

Page 2: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

2

Learning Objectives

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.

Page 3: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

3Lesson 13 332a.pptx

Dynamic Response of Shunt Motors

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

Page 4: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

Lesson 13 332a.pptx4

Dc Shunt Motor Solution MethodsJust 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

Page 5: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

Lesson 13 332a.pptx5

Example 13-1:Shunt Motor Speed Control Using Field Weakening

A 10 HP 240 volt 1200 rpm motor is operatingat 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.

Page 6: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

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Example 13-1 Solution (1)

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

kRIV

kRIV

n

n Since Ia1=Ia2 this simplifies

1p

2p

2

1

n

n Find the value of Fp2 assuming Fp1 = 1

33.11rpm 900

rpm 1200 2p

%33%1001

133.1%100

1p

1p2p

Answer

Page 7: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

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Example 13-2A 200 volt shunt motor is rated a 5 HP and 1000 rpm.At rated output it draws 25 A of line current. The totalarmature 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

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Example 13-2 Solution (1)

200 V

IT = 25 A

Racir Rf

5 HP1000 rpm

If

Ia

Find Ia

Rotational losses are the differencebetween Pem and Pshaft

Find Ea

Page 9: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

Lesson 13 332a.pptx9

Example 13-2 Solution (2)

Answer

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Example 13-2 Solution (3)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 forIa

Page 11: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

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Example 13-2 Solution (4)Torques is proportional to armature current Ia. Lower power requires lower torque and armature current

Speed is proportional to Ea

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Example 13-2 Solution (5)Solve the previous equation for n2

Answer

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Lesson 13 332a.pptx13

Example 13-3: Effects of Changing Field ExcitationWeakening 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.

Page 14: ET 332a Dc Motors, Generators and Energy Conversion Devices 1Lesson 13 332a.pptx

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Example 13-3: Solution (1)

VT=500 V

Racir

0.0343 W

R1=0.052 W

Rf=96 WIT = 204.3 A

If

Find Ea at rated conditions

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Example 13-3: Solution (2)

Switch in additional resistance, R1, and weaken field

TD1 = TD2 Constant torque load

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Example 13-3: Solution (3)

Field weakened by 10% reduces the value of KT by 10% assuming no magnetic saturation

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Example 13-3: Solution (4)

Answer Part a

b.) Find speed under the above conditions

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Example 13-3: Solution (5)

Answer Part b

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Lesson 13 332a.pptx

End Lesson 13ET 332aDc Motors, Generators and Energy Conversion Devices

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