A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase equivalent circuit parameters as follows: Locked rotor, Nominal code letter kVA/hp R = 0.15 Q R2 = 0.10 Q Xị = 0.362 X2 = 0.36 2 C 3.55 – 4.00 XM =8.362 D 4.00 – 4.50 E 4.50 – 5.00 F 5.00 – 5.60 The core-losses are 1.7 G 5.60 – 6.30 kW, mechanical losses are 1.8 kW, and stray losses are 500 W. a) For a slip of 4% find the line current, the rotor copper losses, the load torque, and the efficiency of the motor under the given conditions.
A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase equivalent circuit parameters as follows: Locked rotor, Nominal code letter kVA/hp R = 0.15 Q R2 = 0.10 Q Xị = 0.362 X2 = 0.36 2 C 3.55 – 4.00 XM =8.362 D 4.00 – 4.50 E 4.50 – 5.00 F 5.00 – 5.60 The core-losses are 1.7 G 5.60 – 6.30 kW, mechanical losses are 1.8 kW, and stray losses are 500 W. a) For a slip of 4% find the line current, the rotor copper losses, the load torque, and the efficiency of the motor under the given conditions.
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Question
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![A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase
equivalent circuit parameters as follows:
Nominal code
Locked rotor,
letter
kVA/hp
R = 0.15 Q
R2 = 0.10 2
X1 = 0.362
X2 = 0.36 2
C
3.55 – 4.00
XM =8-362
D
4.00 – 4.50
E
4.50 – 5.00
F
5.00 – 5.60
The core-losses are 1.7
G
5.60 – 6.30
kW, mechanical losses are 1.8 kW, and stray
losses are 500 W.
a) For a slip of 4% find the line current, the rotor copper losses, the load torque, and the
efficiency of the motor under the given conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc176afbe-1520-4dca-a8b8-3efd3b2e5db4%2F19e61a30-de78-4daa-a11b-75ed48e1f528%2F4sxebf9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase
equivalent circuit parameters as follows:
Nominal code
Locked rotor,
letter
kVA/hp
R = 0.15 Q
R2 = 0.10 2
X1 = 0.362
X2 = 0.36 2
C
3.55 – 4.00
XM =8-362
D
4.00 – 4.50
E
4.50 – 5.00
F
5.00 – 5.60
The core-losses are 1.7
G
5.60 – 6.30
kW, mechanical losses are 1.8 kW, and stray
losses are 500 W.
a) For a slip of 4% find the line current, the rotor copper losses, the load torque, and the
efficiency of the motor under the given conditions.
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