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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)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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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--
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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