Question

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

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Answer #1

4) Given in a Induction Motor

V2 = 460 volts

Stator frequency, f = 50 Hz.

Poles = 2

Slip = 5% = 0.05.

Ns = 120f / P = (120 ) (50 ) / 2 = 3000 RPM = Sychronous frequency.

also given, friction and windage losses (or) mechanical losses

Pme  = 700 w.

core losses = 600 w.

(a) Shaft speed in RPM.

Shaft speed (or) Rotor speed, Nr = Ns ( 1 - s)

= 3000 ( 1 - 0.04 )

Nr = 2850 RPM

(b) Output power in horse power.

we have , 1 British horse Power = 0.746 kw.

1 Metric horse Power = 0.785 kw (mostly preferred)

Out put power , point = 50 kw

= 50 / 0.735

Pout = 68 HP

in MHP

Out put power , point = 50 kw

Pout = 50/ 0.746

Pout = 67 HP in BHP.

Ne 禾峯体. Poles 1.356 stp-5C405 ous requer cat los 3000 (t-o.05 산초 5D o:335

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