EE 255 - Chapter 7b

# EE 255 - Chapter 7b - 185 0.075 Ω j 0.17 Ω V φ I A R 1...

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Unformatted text preview: 185 0.075 Ω j 0.17 Ω +- V φ I A R 1 jX 1 R 2 g184 g185 g183 g168 g169 g167- s s R 1 2 jX 2 jX M 0.065 Ω j 0.17 Ω j 7.2 Ω 1.56 Ω (a) The easiest way to find the line current (or armature current) is to get the equivalent impedance F Z of the rotor circuit in parallel with M jX , and then calculate the current as the phase voltage divided by the sum of the series impedances, as shown below. 0.075 Ω j 0.17 Ω +- V φ I A R 1 jX 1 R F jX F The equivalent impedance of the rotor circuit in parallel with M jX is: 2 1 1 1.539 0.364 1.58 13.2 1 1 1 1 7.2 1.625 0.17 F M Z j jX Z j j = = = + = ∠ ° Ω + + Ω + The phase voltage is 440/ 3 = 254 V, so line current L I is 1 1 254 0 V 0.075 0.17 1.539 0.364 L A F F V I I R jX R jX j j φ ∠ ° = = = + + + Ω + Ω + Ω + Ω 149.4 18.3 A L A I I = = ∠ - ° (b) The stator power factor is ( ) PF cos 18.3 0.949 lagging = ° = (c) To find the rotor power factor, we must find the impedance angle of the rotor 1 1 2 2 0.17 tan tan 5.97 / 1.625 R X R s θ-- = = = ° Therefore the rotor power factor is PF cos5.97 0.995 lagging R = ° = (d) The stator copper losses are ( ) ( ) 2 2 SCL 1 3 3 149.4 A 0.075 1675 W A P I R = = Ω = (e) The air gap power is 2 2 2 AG 2 3 3 A F R P I I R s = = 186 (Note that 2 3 A F I R is equal to 2 2 2 3 R I s , since the only resistance in the original rotor circuit was 2 / R s , and the resistance in the Thevenin equivalent circuit is F R . The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit.) ( ) ( ) 2 2 2 2 AG 2 3 3 3 149.4 A 1.539 103 kW A F R P I I R s = = = Ω = (f) The power converted from electrical to mechanical form is ( ) ( ) ( ) conv AG 1 1 0.04 103 kW 98.9 kW P s P =- =- = (g) The synchronous speed of this motor is ( ) sync 120 50 Hz 120 3000 r/min 2 e f n P = = = ( ) sync 2 rad 1 min 3000 r/min 314 rad/s 1 r 60 s π ω = = Therefore the induced torque in the motor is ( ) AG ind sync 103 kW 327.9 N m 2 rad 1 min 3000 r/min 1 r 60 s P τ π ω = = = ⋅ (h) The output power of this motor is OUT conv mech core misc 98.8 kW 1.0 kW 1.1 kW 150 W 96.6 kW P P P P P =--- =--- = The output speed is ( ) ( ) ( ) sync 1 1 0.04 3000 r/min 2880 r/min m n s n =- =- = Therefore the load torque is ( ) OUT load 98.8 kW 327.6 N m 2 rad 1 min 2880 r/min 1 r 60 s m P τ π ω = = = ⋅ (i) The overall efficiency is OUT OUT IN 100% 100% 3 cos A P P P V I φ η θ = × = × ( )( ) ( ) 96.6 kW 100% 89.4% 3 254 V 149.4 A cos 18.3 η = × = ° (j) The motor speed in revolutions per minute is 2880 r/min. The motor speed in radians per second is ( ) 2 rad 1 min 2880 r/min 301.6 rad/s 1 r 60 s m π ω = = 7-15. For the motor in Problem 7-14, what is the pullout torque? What is the slip at the pullout torque? What is the rotor speed at the pullout torque?...
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## This note was uploaded on 10/15/2009 for the course EE 255 taught by Professor Shaban during the Spring '05 term at Cal Poly.

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EE 255 - Chapter 7b - 185 0.075 Ω j 0.17 Ω V φ I A R 1...

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