A 400-hp, 1100-V, 3-phase induction motor gave the following data during light-load tests, the figures representing line values. 1100 V 100 A 22:2 kW total input; mechanical loss 8 kW No load 230 V 200 A 18 kW total input Locked rotor Resistance between any two stator terminals 0-144 . LUsing this resistance, satisfy yourself that the stator copper loss for any given line urrent is the same whether the stator is assumed to be star or delta connected. Cu Estimate the stator line current and the ratio of stator to rotor m.m.f. at full load. (a Estimate the increase of stator resistance per phase required to reduce the starting current to three times the full-load value and what reduction in starting torque will this cause? Express the additional resistance in terms of R,.

A 400-hp, 1100-V, 3-phase induction motor gave the following data during light-load tests, the figures representing line values. 1100 V 100 A 22:2 kW total input; mechanical loss 8 kW No load 230 V 200 A 18 kW total input Locked rotor Resistance between any two stator terminals 0-144 . LUsing this resistance, satisfy yourself that the stator copper loss for any given line urrent is the same whether the stator is assumed to be star or delta connected. Cu Estimate the stator line current and the ratio of stator to rotor m.m.f. at full load. (a Estimate the increase of stator resistance per phase required to reduce the starting current to three times the full-load value and what reduction in starting torque will this cause? Express the additional resistance in terms of R,.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Problem 3Calculate the converted, air gap, output powers and the efficiency of a three-phase induction motor connected at 678.82 cos (377 t) V, which is supplied with 60 A at a lagging power factor of 0.85.R1 is 0.185 Ω. Losses in the core are 1800 W and due to friction and friction are 600 W. Miscellaneous losses are negligible and slip corresponds to 1.81%.
Consider a 480-V, 50-Hz, three-phase induction motor that consumes 80 A at 0.85 PF lagging The stator and rotor copper losses are 2 kW and 800 W. The friction and windage losses are 600 W. The core loses are 1.6 kW. The stray losses are negligible Find: • The air-gap power PAG • The converted power Pconv • The output power Pout • The efficiency, η, of the motor
A 440V, 60Hz, 10 pole delta connected induction motor is rated at 100KW. The equivalent parameter for the motor are: Rs = 0.15 Ω R2` = 0.15 Ω Xs = 0.3 Ω X2` = 0.2 Ω Xm = 6 Ω R c = 120 Ω At full load condition, the friction and windage losses are 600W. The slip of the motor is 0.04. Calculate: The stator line current Is ? The input power ? Stator copper losses ? The air gap power PG ? The converted power Pconv? The torque induced by the motor ? The load torque ? The starting torque? The efficiency of the motor ?
What will be the operating power factor of a 3-phase induction motor, whose ratings are 230 V, 50 Hz, 120 hp, and 4-poles? Motor is operating at rated conditions with an efficiency of 92% and draws a line current of 260 A. The core loss and stator copper losses are 1710 W and 2830 W respectively. (a) 0.85 (c) 0.93 (b) 0.90 (d) 0.80
230 V, 50 Hz, 4 – pole single-phase induction motor has the following equivalent circuit impedances:R1 = 2.22, R2 = 4.52, X1 = 3.12, X2 = 2.62, XM = 802, Friction, windage and core loss = 40 W. For a slip of 0.03pu, calculation (a) input current, (b) power factor, (c) developed power, (d) output power, (e) efficiency.
A three-phase Y-connected 230-V (line-to-line) 7.5-kW 60-Hz six-pole induction motor has the following parameter values in Ω/phase referred to the stator: R1 = 0.294, R2 = 0.144 X1 = 0.503, X2 = 0.209, Xm = 13.25 The total friction, windage, and core losses may be assumed to be constant at 403 W, independent of load. For a slip of 4 percent, determine: (a) stator current, power factor, (b) output torque and power,(c) efficiency when the motor is operated at rated voltage and frequency, (d) the load component 12 of the stator current, the induced torque Tind, and the converted power Pconv for a slip s= 0.04,(e) the maximum torque and the corresponding speed; and (f) the starting torque Tst and the corresponding stator load current 12 start
An approximate per phase equivalent circuit of a 3-phase, 220-V, wye-connected, 6 pole, 60 Hz induction motor is shown in figure below for which R1 = 0.3 ohm; X1 = 0.4 ohm; R’2 = 0.4 ohm; X’2 = 0.6 ohm; Xm = 15 ohms. For the given numerical values calculate the total developed(electromagnetic) torque in (kg-cm) .The motor speed is 1080 rpm.
We have a 50 hp Y-connected three-phase induction motor at √3*138.56 V line voltage. Find the slip, the converted, air gap, and output powers, and the efficiency, if the frequency is 60 Hz and it is fed with 35 A at a power factor of 0.85 lagging. The losses are: in the stator copper 2.5 kW; on the rotor 750 W; by friction 650 W and in the core 1850 W. Note: Miscellaneous losses are negligible.
QI: The parameters of a 4 kW, 220V. 2-phase induction motor, in ohms per phase, are: r1=0.53, x1 =2.4, XM= 70, r2 = 0.96, x2 = 3.0. The motor is operated from unbalanced 2- phase source whose voltages are 230V and 210V, the smaller voltage is lagging the larger by 800. For a slip of 5% and rotational losses of 30 W, find the positive- and negative- sequence components of: (a) the applied voltages. (b) Find the values of phase currents (c) the input current. (d) Find the developed mechanical power. (e) Find the efficiency.
single-phase induction motor employed for a mechanical work at a rate of 45 HP. Its electrical power will be 33.57 kW. A)Estimate the amount of line current necessary to power the motor if the line voltage is 460 Volts assuming 100% efficiency and a power factor of 1. ( B)Now, estimate the necessary line current for this motor if its power factor drops to 0.65. Assume the efficiency and mechanical power as the same. C)What do these calculations indicate about the importance of maintaining a high-power factor value in an AC circuit?
The current drawn by 230V, 60Hz, 4-pole, star-connected 3-phase induction motor at full load is 15∠0°A. The equivalent circuit parameters of the motor are as follows. R₁=5Ω , R2=3 Ω, X1=0, X2=0, Xm=∞, RFE=15 Ω. friction loss = 4W. Since the slip is 2.5%, find the efficiency and motor torque.
A 208 V, 60 Hz, 2-pole, Y-connected induction motor is rated at 16 hp. Its equivalentcircuit components have the following values:(R1 = 0.200 ohm, X1 =0.410 ohm); (R2 = 0.120 ohm, X2 = 0.410 ohm); Xm = 15.0 ohm.Rotor slip is 4.5% at the rated voltage and frequency. Mechanical and core losses are 230Wand 200W respectively. Find (a) rotor speed (b) line current and motor p.f (c) Air-gap andconverted power (d) Induced torque.