Draw the circle diagram for a 5.6 kW, 400-V, 3-φ, 4-pole, 50-Hz, slip-ring induction motor from the following data : No-load readings : 400 V, 6 A, cos φ0 = 0.087 : Short-circuit test : 100 V, 12 A, 720 W. The ratio of primary to secondary turns = 2.62, stator resistance per phase is 0.67 Ω and of the rotor is 0.185 Ω. Calculate (i) full-load current (ii) full-load slip (iii) full-load power factor
Draw the circle diagram for a 5.6 kW, 400-V, 3-φ, 4-pole, 50-Hz, slip-ring induction motor from the following data : No-load readings : 400 V, 6 A, cos φ0 = 0.087 : Short-circuit test : 100 V, 12 A, 720 W. The ratio of primary to secondary turns = 2.62, stator resistance per phase is 0.67 Ω and of the rotor is 0.185 Ω. Calculate (i) full-load current (ii) full-load slip (iii) full-load power factor
Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Question
Draw the circle diagram for a 5.6 kW, 400-V, 3-φ, 4-pole, 50-Hz, slip-ring
induction motor from the following data :
No-load readings : 400 V, 6 A, cos φ0
= 0.087 : Short-circuit test : 100 V, 12 A, 720 W.
The ratio of primary to secondary turns = 2.62, stator resistance per phase is 0.67 Ω and of the
rotor is 0.185 Ω. Calculate
(i) full-load current (ii) full-load slip
(iii) full-load power factor
(iv) maximum torque
full - load torque
(v) maximum power.
![A:PA
2_5231136897..
Computations and Circle Diagrams
1325
Since the two Cu losses are equal, point H is the mid-point of A B.
Line MK represents the maximum torque of the motor in synchronous watts
MK = 7.3 cm (by measurement) = 7.3×7,120=51,980 synch. watt.
Example 35.7. Draw the circle diagram for a 5.6 kW, 400-V, 3-̟, 4-pole, 50-Hz, slip-ring
induction motor from the following data :
No-load readings : 400 V, 6 A, cos 0, = 0.087 : Short-circuit test : 100 V, 12 A, 720 W.
The ratio of primary to secondary turns = 2.62, stator resistance per phase is 0.67 N and of the
rotor is 0.185 N. Calculate
(i) full-load current
(iii) full-load power factor
(ii) full-load slip
(iy) тахітит torque
full - load torque
(у) тахітит power.
Solution. No-load condition
Po
= cos
(0.087) = 85°
M
¡R
P
Output Line
N.
69°40¢
T
85°
Torque Line
A
-X¢
D
Fig. 35.15
Short-circuit condition
Short-circuit current with normal voltage = 12 × 400/100 = 48 A
Total input = 720 × (48/12)² = 11.52 kW
720
cos o, =
3x100×12
= 0.347
0 = 69° 40'
or
Current scale is,
1 cm = 2 A
In the circle diagram of Fig. 35.15, OA = 3 cm and inclined at 85° with O V. Line OB represents
short-circuit current with normal voltage. It measures 48/2 = 24 cm and represent 48 A. BD is
perpendicular to OX.
For Drawing Torque Line
K = 2.62
R, = 0.67 Q
R,= 0.185 N
(in practice, an allowance of 10% is made for skin effect)
rotor Cu loss
0.185
2.62 x
rotor Cu loss
:.
total Cu loss
..
= 1.9
1.9
= 0.655
2.9
stator Cu loss
0.67
Now
BD = 8.25 cm and represents 11.52 kW
power scale = 11.52/8.25 = 1.4 kW/cm
--t--](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1e93bd3-7a88-4831-b350-d9c66e8543f4%2F7fb1cc8d-e545-42f6-bbb8-e7ab89bfdd6f%2Fmj0v5zf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A:PA
2_5231136897..
Computations and Circle Diagrams
1325
Since the two Cu losses are equal, point H is the mid-point of A B.
Line MK represents the maximum torque of the motor in synchronous watts
MK = 7.3 cm (by measurement) = 7.3×7,120=51,980 synch. watt.
Example 35.7. Draw the circle diagram for a 5.6 kW, 400-V, 3-̟, 4-pole, 50-Hz, slip-ring
induction motor from the following data :
No-load readings : 400 V, 6 A, cos 0, = 0.087 : Short-circuit test : 100 V, 12 A, 720 W.
The ratio of primary to secondary turns = 2.62, stator resistance per phase is 0.67 N and of the
rotor is 0.185 N. Calculate
(i) full-load current
(iii) full-load power factor
(ii) full-load slip
(iy) тахітит torque
full - load torque
(у) тахітит power.
Solution. No-load condition
Po
= cos
(0.087) = 85°
M
¡R
P
Output Line
N.
69°40¢
T
85°
Torque Line
A
-X¢
D
Fig. 35.15
Short-circuit condition
Short-circuit current with normal voltage = 12 × 400/100 = 48 A
Total input = 720 × (48/12)² = 11.52 kW
720
cos o, =
3x100×12
= 0.347
0 = 69° 40'
or
Current scale is,
1 cm = 2 A
In the circle diagram of Fig. 35.15, OA = 3 cm and inclined at 85° with O V. Line OB represents
short-circuit current with normal voltage. It measures 48/2 = 24 cm and represent 48 A. BD is
perpendicular to OX.
For Drawing Torque Line
K = 2.62
R, = 0.67 Q
R,= 0.185 N
(in practice, an allowance of 10% is made for skin effect)
rotor Cu loss
0.185
2.62 x
rotor Cu loss
:.
total Cu loss
..
= 1.9
1.9
= 0.655
2.9
stator Cu loss
0.67
Now
BD = 8.25 cm and represents 11.52 kW
power scale = 11.52/8.25 = 1.4 kW/cm
--t--
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