ECE 4411 Complete Equivalent Circuit for the Shunt Motor R acir = resistance of the armature circuit Interpoles and Compensating Windings.

ECE 441 1

Complete Equivalent Circuit for the Shunt Motor

Racir = resistance of the armature circuit

Interpoles and Compensating Windings

ECE 441 2

General Speed Equation For a DC Motor

a p G

a

p G

T a acir

p G

E n k

En

k

V I Rn

k

ECE 441 3

Example 10.8

• A 25-hp, 240-V shunt motor operating at 850 r/min draws a line current of 91 A when operating at rated conditions. A 2.14-Ω resistor inserted in series with the armature causes the speed to drop to 634 r/min. The respective armature-circuit resistance and field-circuit resistance are 0.221-Ω and 120-Ω. Determine the new armature current.

ECE 441 4

Solution

2402

12091 2 89

f

a T f

VI A

I I I A

ECE 441 5

1

1 1

22 2

2

1

1

2

2

[ ][ ]

[ ( )] [ ( )]

[ ( )]

( )634

[240 (240 89 0.221)]850 32.050.221 2.14

T a acir

p G

T a acir

p G T a acir

T a acir x T a acir x

p G

T T a acir

a

acir x

a

V I Rn

k

V I Rkn V I R

V I R Rn V I R Rk

nV V I R

nI

R R

I A

ECE 441 6

Example 10.9

• A shunt motor rated at 10hp, 240-V, 2500 r/min, draws 37.5 A when operating at rated conditions.

• Ra=0.213Ω, RCW=0.065Ω, RIP=0.092Ω, Rf=160Ω

ECE 441 7

Example 10.9 (continued)

• Determine the steady-state armature current if a rheostat in the shunt field reduces the flux in the air gap to 75% of its rated value, a 1.0-Ω resistor is placed in series with the armature, and the load torque on the shaft is reduced to 50% rated

ECE 441 8

At rated conditions

2401.5

160

37.5 1.5 36

f

f

f

a T f

VI A

R

I I I A

ECE 441 9

At the new conditions

1 11

2 2 2

1 12 1

2 1

1 2 1 1

[ ] [ ][ ] [ ]

0.536 24

0.75

D p a M

p p a p a

p p a p a

p p

a a

p p

T B I k

B I ITT B I I

T TI I A

T T

ECE 441 10

1 11

2 2 2

1 12 1

2 1

1 2 1 1

[ ] [ ][ ] [ ]

0.536 24

0.75

D p a M

p p a p a

p p a p a

p p

a a

p p

T B I k

B I ITT B I I

T TI I A

T T

ECE 441 11

Determine the new steady-state speed

2

2 1

1

2

2

[ ] [ ]

240 24 (1 0.370)2500 [ ] [ ]

0.75 240 36 0.370

3046 / min

T a acir

p G

p GT a acir

p G T a acir

p G

p G

V I Rn

k

kn V I Rn k V I R

kn

k

n r

Check page 421 for = sign

ECE 441 12

Speed Control of DC Motors

• Armature Control– Insert a resistor or

rheostat in series with the armature

– Reduce speed below the base speed

• Shunt Field Control– Insert a resistor or

rheostat in series with the shunt field

– Increase speed above the base speed

ECE 441 13

For speed reduction

Armature current decreases

Armature current increases

Torque decreases

Torque recovers (increases)

Speed decreases

Speed settles to new value

Counter-emf decreases

ECE 441 14

For speed increase

Decreasing field current reduces the flux

Reduction in flux causes cemf to decrease

Motor accelerates, cemf increases

Armature current decreases

Armature current increases

Torque increases, machine accelerates

Torque settles down Motor runs at new speed

Speed increases

ECE 441 15

Mechanical Power and Developed Torque

Pmech = Total Power Input to the Armature – Copper Losses in the Armature

Pmech = VTIa – Ia2Racir

Racir = Ra + RIP + RCW

Pmech = EaIa

ECE 441 16

Power-flow Diagram – DC Motor

Ploss = Pacir + Pb + Pcore + Pfcl + Pf,w + Pstray

Pb = VbIa

ECE 441 17

Power-flow Diagram – DC Generator

Ploss = Pacir + Pb + Pcore + Pfcl + Pf,w + Pstray

Pb = VbIa

ECE 441 18

Starting a DC Motor

• At “locked-rotor”, or “blocked-rotor”

,

0

T a

a

cir

T T

a lr

acir acir

V EI

RV V

IR R

ECE 441 19

Manually-Operated DC Motor Starter

“Start” with all of the rheostat resistance in the circuit

“Off” position

“Run” position

All resistance is cut out when motor reaches full-speed

“Holds” the lever in “Run” position

(A “break” in the field circuit de-energizes the coil, shutting the motor down)

ECE 441 20

Example 10.12Motor Starting

• A 15-hp, 230-V, 1750 r/min shunt motor with a compensating winding draws 56.2A when operating at rated conditions. The motor parameters are Racir = 0.280 Ω and Rf = 137 Ω.

ECE 441 21

Example 10.12 (continued)

• Determine• (a) the rated torque

5252525215 5252

45.01750

rated

rated

Tn PP T

n

T lb ft

ECE 441 22

Example 10.12 (continued)

• (b) the armature current at locked-rotor if no starting resistance is used

,

230 0821.4

0.280T a

a lr

cir

V EI A

R

ECE 441 23

Example 10.12 (continued)• (c) the external resistance required in the

armature circuit that would limit the current and develop 200% rated torque when starting

ECE 441 24

, ,

1 1

2 2

2 1

2 1

1 1

2301.68

137

56.2 1.68 54.52

254.52 109.0

T

f

f

a rated T rated f

a

a

a a

VI A

R

I I I A

T IT I

T TI I A

T T

ECE 441 25

( )

230 02.80 1.83

109.0

T a a acir x

T a

x acir

a

x

V E I R R

V ER R

I

R

ECE 441 26

Example 10.12 (continued)

• (d) Assuming that the system voltage drops to 215 V, determine the locked-rotor torque using the external resistor in (c).– Assume that the flux density is proportional to the

field current

D p a M

p f

f a

T B I k

B I

T I I

ECE 441 27

Example 10.12 (continued)

11

2 2

2

2 1

1

2

[ ][ ]

[ ][ ]

1.57 101.945.0 78.6

1.68 54.52

f aD

D f a

f a

D D

f a

D

I ITT I I

I IT T

I I

T lb ft