ECE 441 1

Problem # 5-7

• A 3-phase, 8-pole, induction motor, rated at 874rpm, 30hp, 60Hz, 460V, operating at reduced load, has a shaft speed of 880 rpm. The combined stray power loss, windage loss, and friction loss is 350W. The rotor parameters in Ω/phase are– R1=0.1891 R2=0.191 XM=14.18

– X1=1.338 X2=0.5735 Rfe=189.1

ECE 441 2

Equivalent Circuit for an Induction Motor with all parameters referenced to the stator

ECE 441 3

Problem # 5-7 continued

• The motor is NEMA Design C and Y-connected.

• Determine – the motor input impedance/phase

22 2

RZ jX

s

Need to determine the value of s.

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120 (120)(60)

8900

900 874

900

0.0222

s

s

s r

s

fn

Pn rpm

n ns

n

s

22 2

2

2

0.1910.5735

0.022

8.623 3.814 8.604 0.5735

RZ jX

s

Z j

Z j

ECE 441 5

0

0

2 0

2 0

(189.1)(14.18)

189.1 14.18

14.14 85.71 1.058 14.1

8.623 3.814 14.14 85.71

8.604 0.5735 1.058 14.1

6.940 32.887 5.828 3.768

fe M

fe M

p

p

R jX jZ

R jX j

Z j

Z ZZ

j jZ Z

Z j

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10.1891 1.338 5.828 3.768

6.017 5.106 7.96 39.899

in p

in

Z Z Z j j

Z j

The Input Impedance per Phase

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Solution continued

• Determine the line current

1

4600

3 33.364 39.8997.96 39.899

T

in

VI A

Z

ECE 441 8

3-Φ Delta and Wye Connections

• (a) shows the sources (phases) connected in a wye (Y).– Notice the fourth

terminal, known as Neutral.

• (b) shows the sources (phases) connected in a delta (∆).– Three terminals

ECE 441 9

Solution continued

• Determine the active, reactive, and apparent power and power factor

* 4603 3( )(33.364 39.899 )

3

26,582.55 39.899 20,393.5 17,051.0

26,582.55

20,393.5

17,051

cos(39.899 ) 0.767 76.7%p

S V I

S VA j VA

S VA

P W

Q VARS

F

Solution continued

• Determine the equivalent rotor current

ECE 441 10

82.10853.26814.3623.8

012.754.231

012.754.231

887.32940.6)899.39364.33(

2

22

2

12

Z

EI

VE

ZIE p

ECE 441 11

Solution continued

• Determine the stator copper loss

21 1

2

3

3(33.364) (0.1891)

631.5

scl

scl

scl

P I R

P

P W

ECE 441 12

Solution continued

• Determine the rotor copper loss

22 2

2

3

3(26.853) (0.191)

413.2

rcl

rcl

rcl

P I R

P

P W

ECE 441 13

Solution continued

• Determine the core loss

22

2

3

(231.55)3

189.1

850.5

corefe

core

core

EP

R

P

P W

ECE 441 14

Solution continued

• Determine the air gap power

413.2

0.022218,781.8

rclgap

gap

gap

PP

sW

P

P W

ECE 441 15

Solution continued

• Determine the mechanical power developed

18,781.8 413.2

18,368.6

mech gap rcl

mech

mech

P P P

P

P W

ECE 441 16

Solution continued

• Determine the developed torque

22 2

2

21.12

(21.12)(26.853) (0.191)

(0.0222)(900)

146.9

Ds

D

D

I RT

sn

T

T lb ft

ECE 441 17

Solution continued

• Determine the shaft horsepower

,

746

74620,393.5 631.5 850.5 413.2 350

74624.33

in lossshaft

in scl core rcl f w strayshaft

shaft

shaft

P PP hp

P P P P P PP

P

P hp

ECE 441 18

Solution continued

• Determine the shaft torque

(5252)

5252

(24.33)(5252)

880145.19

r

r

Tn HPHP T

n

T

T lb ft

ECE 441 19

Solution continued

• Determine the efficiency

(24.33 )(746 / )

20,393.5

0.890 89%

out

in

P

P

hp W hp

W

ECE 441 20

Solution continued

• Sketch the power flow diagram and enter all data

631.5W 850.5W 413.2W 350W

20,393.5W 24.33hp

18,368.6W

18,781.8W

ECE 441 21

Solution continued

• Determine the expected minimum locked-rotor torque, breakdown torque, and pull-up torque

• From Table 5-1, Tlr=200%Trated

– Tlr = (2)(180.27) = 360.54 lb-ft

• From Table 5-3, Tbreakdown = 190%Trated

– Tbreakdown = (1.9)(180.27) = 342.52 lb-ft

• From Table 5-6, Tpull-up = 140%Trated

– Tpull-up = (1.4)(180.27) = 252.37 lb-ft

ECE 441 22

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