EE 255 - Chapter 7c

# EE 255 - Chapter 7c - 197(d To determine the starting code...

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Unformatted text preview: 197 (d) To determine the starting code letter, we must find the locked-rotor kVA per horsepower, which is equivalent to finding the starting kVA per horsepower. The easiest way to find the line current (or armature current) at starting is to get the equivalent impedance F Z of the rotor circuit in parallel with M jX at starting conditions, and then calculate the starting current as the phase voltage divided by the sum of the series impedances, as shown below. 0.33 Ω j 0.42 Ω +- V φ I A, start R 1 jX 1 R F jX F The equivalent impedance of the rotor circuit in parallel with M jX at starting conditions ( s = 1.0) is: ,start 2 1 1 0.167 0.415 0.448 68.1 1 1 1 1 30 0.172 0.42 F M Z j jX Z j j = = = + = ∠ ° Ω + + Ω + The phase voltage is 460/ 3 = 266 V, so line current ,start L I is ,start 1 1 266 0 V 0.33 0.42 0.167 0.415 L A F F R jX R jX j j φ ∠ ° = = = + + + Ω + Ω + Ω + Ω V I I ,start 274 59.2 A L A = = ∠ - ° I I Therefore, the locked-rotor kVA of this motor is ( )( ) ,rated 3 3 460 V 274 A 218 kVA T L S V I = = = and the kVA per horsepower is 218 kVA kVA/hp 4.36 kVA/hp 50 hp = = This motor would have starting code letter D , since letter D covers the range 4.00-4.50. 7-20. Answer the following questions about the motor in Problem 7-19. (a) If this motor is started from a 460-V infinite bus, how much current will flow in the motor at starting? (b) If transmission line with an impedance of 0.35 + j 0.25 Ω per phase is used to connect the induction motor to the infinite bus, what will the starting current of the motor be? What will the motor’s terminal voltage be on starting? (c) If an ideal 1.4:1 step-down autotransformer is connected between the transmission line and the motor, what will the current be in the transmission line during starting? What will the voltage be at the motor end of the transmission line during starting? S OLUTION (a) The equivalent circuit of this induction motor is shown below: 198 0.33 Ω j 0.42 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 0.172 Ω j 0.42 Ω j 30 Ω I 2 The easiest way to find the line current (or armature current) at starting is to get the equivalent impedance F Z of the rotor circuit in parallel with M jX at starting conditions, and then calculate the starting current as the phase voltage divided by the sum of the series impedances, as shown below. 0.33 Ω j 0.42 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 at starting conditions ( s = 1.0) is: 2 1 1 0.167 0.415 0.448 68.0 1 1 1 1 30 0.172 0.42 F M Z j jX Z j j = = = + = ∠ ° Ω + + Ω + The phase voltage is 460/ 3 = 266 V, so line current L I is 1 1 266 0 V 0.33 0.42 0.167 0.415 L A F F R jX R jX j j φ ∠ ° = = = + + + Ω + Ω + Ω + Ω V I I 273 59.2 A L A = = ∠ - ° I I (b) If a transmission line with an impedance of 0.35 + j 0.25 Ω per phase is used to connect the induction motor to the infinite bus, its impedance will be in series with the motor’s impedances, and the starting...
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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 7c - 197(d To determine the starting code...

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