Answered: A 760-kVA, 2300-V, three-phase,…

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

Chapter12: Power System Controls

Section: Chapter Questions

Problem 12.5P

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Question

A 760-kVA, 2300-V, three-phase, Y-connected, synchronous generator is operated at its rated speed to obtain its rated no-load voltage. Later, another test finds that the short-circuit phase current is 128A. The average resistance of each phase is 0.56 ohm.

1) Determine the synchronous reactance per phase.
2) Determine the voltage regulation when the generator delivers the rated load at its rated

voltage and 0.75 PF lagging.

Expert Solution

Step 1

For given generator

$S = VA rating = 760 \times 10^{3} V = line voltage = 2300 V V_{ϕ} = per phase voltage = \frac{V}{\sqrt{3}} R_{a} = per phase resistance = 0.56 Ω X_{s} = synchronous reactance per phase I_{sc} = short circuit current = 128 A Z_{s} = synchronous impedance I = line current = phase current$

Step 2

$V_{ϕ} = \frac{2300}{\sqrt{3}} = 1328 V$

$Z_{s} = \frac{V_{ϕ}}{I_{s c}} = \frac{1328}{128} = 10.37 Ω$

$X_{s} = \sqrt{Z_{s}^{2} - R_{a}^{2}} = \sqrt{{(10.37)}^{2} - {(0.56)}^{2}} X_{s} = 10.355 Ω . . synchronous reactance per phase$

Step 3

$I = \frac{S}{\sqrt{3} V} = \frac{760 \times 10^{3}}{\sqrt{3} \times 2300} = 190.77 A power factor is 0.75 lagging \emptyset = \cos^{- 1} (0.75) I = 190.77 ∠ - 41.40 A$

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