A 20-pole alternator is having 180 identical stator slots with 6conductors in each slot. All the coils of a phase are in series. If thecoils are connected to realize single-phase winding, the generatorvoltage is V. If the coils are reconnected to realize three-phase star-connected winding, the generated phase voltage is V₂. Assuming fullpitch, single-layer winding, the ratio V₁/V₂ is

Given- Alternator poles P=20, Number of slots in stator = 180, Number of conductors in each slot=6,…

A 20-pole alternator is having 180 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are connected to realize single-phase winding, the generator voltage is V₁. If the coils are reconnected to realize three-phase star- connected winding, the generated phase voltage is V₂. Assuming full pitch, single-layer winding, the ratio V₁/V₂ is

A 20-pole alternator is having 180 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are connected to realize single-phase winding, the generator voltage is V₁. If the coils are reconnected to realize three-phase star- connected winding, the generated phase voltage is V₂. Assuming full pitch, single-layer winding, the ratio V₁/V₂ is

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.16P: A single-phase, 120V(rms),60Hz source supplies power to a series R-L circuit consisting of R=10 and...

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A three-phase star-connected alternator delivers at rated load a current I = 200 A under a voltage of 5 kV-50 Hz (between phases). The load is inductive of power factor equal to 0.87, the resistance of each stator winding is 0.292 and the speed of the rotor is n = 750 r.p.m. The sum of the constants losses and the stator Joule's losses and the rotor Joule's losses is 55 kW. The no load test, at rated frequency, gave: E-4200 V (between phases) for an excitation current lexc = 40 A. The short-circuit test, for an excitation current lexc = 40 A, gave a short-circuit current in the stator windings Ise =2.5 kA. 1- Calculate the numbers of poles of the alternator. 2- Calculate the synchronous reactance (X). 3- Calculate the electromotive force of the alternator (E) between phases for the same excitation current. 4- Calculate the power absorbed by the alternator (Pa).
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i. A 2300 V, With 2500 kVA three phase delta-connected alternator has a resistance of 0.1ohm/phase and a synchronous reactance of 1.5 ohms/phase. The alternator is adjustedto rated voltage at no load. Calculate its terminal voltage when carrying ratedcurrent at no load power factor of 0.6 lagging.
A three-phase, star-connected alternator has 8 number of poles, armature with 160 slots and 2 conductors per slot. The speed of rotation is 1000 rpm. The flux per pole is 0.06 weber. If the winding factor is 0.94 and all the conductors in each phase are in series. Calculate a) Frequency of generation b) Number of turns c) EMF generated (phase voltage) d) EMF generated (Line voltage)
P:11) Solve all parts.
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3.The voltage generated across one inductor of an alternator is 8.11V. There are 120 slots and 4 turns in every coil. The conductors are all in series and connected in double layer lap windings distributed sinusoidally in several slots. The coil pitch is 10 slots, frequency is 60 Hz and synchronous speed is 720 rpm. Calculate the line-to-line generated voltage if the alternator windings are connected wye. a. 2400Vb. 4160Vc. 4800Vd. 856V
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