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

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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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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