A single-phase, 230 V, 50 Hz, 8 pole, Induction Motor has the following equivalent circuit parameters:...
A single-phase, 230 V, 50 Hz, 8 pole, Induction Motor has the following equivalent circuit parameters:
|
Z1 = 1.7 + j 2.7 Ω |
ZF = 12 + j 20 Ω |
ZB = 1.2 + j 2.4 Ω |
If the core loss is 84 W, while the friction and windage loss is 100 W at 5% slip. Determine:
- The line current drawn by the machine. (4)
- The power factor. (1)
- The input power. (2)
- The air-gap power. (3)
- The converted power. (2)
- The induced torque. (1)
- The output power. (3)
- The load torque. (2)
- The efficiency. (2)
Homework Answers
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A single-phase, 230 V, 50 Hz, 8 pole,
Induction Motor has the following equivalent circuit
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The input power to a 480V, 3 phase, 6 pole, 50 Hz induction motor is 75kW...
The input power to a 480V, 3 phase, 6 pole, 50 Hz induction motor is 75kW with a line current of 75A and runs at a slip of 2.5%. If the stator core loss is 2 kW, windage and friction loss is 1.2 kW, and the stator resistance per phase is 0.16 ohm. Calculate: a. Power supplied to the rotor (Air gap power) b. Rotor copper loss c. Shaft power d. Efficiency of the motor e. Shaft torque f What...
1. A 20 hp, 230 V, 60 Hz, four-pole, three-phase induction motor operating at rated load...
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A 4 phase, 460 V, 60 Hz, 26.8 hp, 4 pole, Y-connected induction motor draws 25...
A 4 phase, 460 V, 60 Hz, 26.8 hp, 4 pole, Y-connected
induction motor draws 25 A at a power factor of 0.9 lagging. Core
loss is 900 W, stator copper loss is 1100W, friction and winding
losses are 300 W, and stray loss is neglected. The machine shaft
rotates at 1738 rpm. Calculate the air gap power and output load
torque
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A 3-phase, 208 V, 60 Hz, 30 hp, four-pole induction motor has the following equivalent circuit parameters. R1 = 0.2 Ω, , , Xm = 12Ω.The rotational loss is 500 W. For a slip of 5%, calculate (a) The motor speed in rpm and radians per sec (b) The current drawn by the motor from the power supply (c) The stator cu-loss (d) The air gap power (e) The rotor cu-loss (f) The shaft power (g)The developed torque and the shaft torque (h) The efficiency of the motor Use...
6. [Induction Motor - 15 points] A three-phase, 460 V, 60 Hz, 26.8 hp, 4-pole, Y-connected...
6. [Induction Motor - 15 points] A three-phase, 460 V, 60 Hz, 26.8 hp, 4-pole, Y-connected induction motor draws 25 A at a power factor of 0.9 lagging. The core loss is 900 W, stator copper loss is 1100 W, friction and windage loss is 300 W, and stray loss is neglected. The machine shaft rotates at a speed of 1738 rpm. Calculate the air gap power and output load torque.
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Two tests were performed on a 230-V, 60-Hz, 4-pole, Y-connected, three-phase induction motor and the obtained...
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Problem 2. (25 points) For a 120-V, 1/3 hp, four pole, 1-phase induction motor, the parameters of the equivalent circuit, are R,-1.8(2, R,-2.5Ω, X,-X,-2.462, and Xm-60 Ω. At a slip of 500, the motor's rotational losses are 51W. The rotational losses maybe assumed constant over the normal operating range of the motor. If the slip is 0.05, calculate a. Input Power b. Air Gap Power, c. Induced Torque Tin d. Output power e. Load torque. f. Overall Motor efficiency g....
1. A 20 hp, 230 V, 60 Hz, four-pole, three-phase induction motor operating at rated load has rotor copper loss of 331 W, and a combined friction, windage, and stray power loss of 249 W. Determine, (a) Mechanical power developed; (b) Air-gap power; (c) Shaft speed; (d) Shaft torque (10 pts.)
A 4 phase, 460 V, 60 Hz, 26.8 hp, 4 pole, Y-connected
induction motor draws 25 A at a power factor of 0.9 lagging. Core
loss is 900 W, stator copper loss is 1100W, friction and winding
losses are 300 W, and stray loss is neglected. The machine shaft
rotates at 1738 rpm. Calculate the air gap power and output load
torque
- [Induction Motor - 15 points) A three-phase, 460 V, 60 Hz, 26.8 hp, 4-pole, Y-connected induction motor...
A 230 V, 50 Hz, three-phase, Y-connected, 6-pole induction motor has the following per-phase parameters R-0.07 Ro 00 0.08 Ω x, x, 0.3 xm 6.33 At a slip of 2%, determine the followings using approximate equivalent circuit of the motor i) The motor speed in rev/min and rad/s. ii) The rotor current. ii) The stator current. iv) The induced torque. v) The efficiency of the motor.
6. [Induction Motor - 15 points] A three-phase, 460 V, 60 Hz, 26.8 hp, 4-pole, Y-connected induction motor draws 25 A at a power factor of 0.9 lagging. The core loss is 900 W, stator copper loss is 1100 W, friction and windage loss is 300 W, and stray loss is neglected. The machine shaft rotates at a speed of 1738 rpm. Calculate the air gap power and output load torque.
4. A 50-kW, 460-V, 50-Hz, two-pole induction motor has a slip of 5 percent when operating a full-load conditions. At full-load conditions, the friction and windage losses are 700 W, and the core losses are 600 W. Find the following values for full-load conditions: (10 pts.) (a) The shaft speed in rpm (b) The output power in horse power (c) The load torque in lb-ft (d) The induced torque in lb-ft (e) The rotor frequency in hertz
6.12 A three-phase, Y-connected, 460-V (line-line), 37-kW, 60-Hz, four-pole induction motor has the following equivalent-circuit parameters in ohms-per-phase referred to the statr: R10.070 R2 0.152 X 0.743 X2 0.764 Xm 40.1 The total friction and windage losses may be assumed constant at 390 W, and the core loss may be assumed to be equal to 325 W. With the motor connected directly to a 460-V source, compute the speed, output shaft torque and power, input power, and power factor and...
Two tests were performed on a 230-V, 60-Hz, 4-pole, Y-connected, three-phase induction motor and the obtained data are given in the table below. The friction and windage loss is 15 W, and the stator winding resistance between any two lines is 3.96ohm Determine: 1) The equivalent circuit parameters of the motor 3) The motor efficiency at a slip of 8%. No-load test 137 0.43 230 Blocked-rotor test 70 1.18 47 3-phase power input, W Line-line voltage, V
Problem 2. (25 points) For a 120-V, 1/3 hp, four pole, 1-phase induction motor, the parameters of the equivalent circuit, are R,-1.8(2, R,-2.5Ω, X,-X,-2.462, and Xm-60 Ω. At a slip of 500, the motor's rotational losses are 51W. The rotational losses maybe assumed constant over the normal operating range of the motor. If the slip is 0.05, calculate a. Input Power b. Air Gap Power, c. Induced Torque Tin d. Output power e. Load torque. f. Overall Motor efficiency g....
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