3.20 Using base values of 20 kVA and 115 volts in zone 3, rework Example 3.4. EXAMPLE 3.4Per-unit circuit: three-zone single-phase networkThree zones of a single-phase circuit are identified in Figure 3.10(a). The zonesare connected by transformers T, and T, whose ratings are also shown. Usingbase values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and de-termine the per-unit impedances and the per-unit source voltage. Then calculatethe load current both in per-unit and in amperes. Transformer winding resistancesand shunt admittance branches are neglected.FIGURE 3.10Zone 1Zone 2Zone 3Circuit forExample 3.4V, = 220/0° voitsmuwT, Zload = 0.9 + j0.2 NXiine = 2 130 kVA20 kVA240/480 volts460/115 voltsXeg = 0.10 p.u.Xea = 0.10 p.u.(a) Single-phase circuitloou XT1p.u.jXinep.u.įXT2p.u. loadp.u.|j0.1378i p.u.j0.10 p.u. j0.2604 p.u.Zioadp.u. =1.875 + j0.4167 p.u.0.9167/0° p.u.Zone 1Zone 2Zone 3Voase = 240 voltsVosse2 = 480 voitsVoase3 = 120 voltsZoase(240)230,000= 1.92 N Zbase2(480)230.000- 7.68 Ω Zas3(120)230,0000.48 N%3!Spase= 30 kVA30,000120loase3= 250 A%3D(b) Per-unit circuit

Given Data: in zone 3 Vbase3 = 115 V Sbase3 = 20 kVA

Answered: .20 Using base values of 20 kVA and 115…

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.20 Using base values of 20 kVA and 115 volts in zone 3, rework Example 3.4.

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

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.25P: Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows:...

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3.20 Using base values of 20 kVA and 115 volts in zone 3, rework Example 3.4.

EXAMPLE 3.4
Per-unit circuit: three-zone single-phase network
Three zones of a single-phase circuit are identified in Figure 3.10(a). The zones
are connected by transformers T, and T, whose ratings are also shown. Using
base values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and de-
termine the per-unit impedances and the per-unit source voltage. Then calculate
the load current both in per-unit and in amperes. Transformer winding resistances
and shunt admittance branches are neglected.
FIGURE 3.10
Zone 1
Zone 2
Zone 3
Circuit for
Example 3.4
V, = 220/0° voits
muw
T, Zload = 0.9 + j0.2 N
Xiine = 2 1
30 kVA
20 kVA
240/480 volts
460/115 volts
Xeg = 0.10 p.u.
Xea = 0.10 p.u.
(a) Single-phase circuit
loou XT1p.u.
jXinep.u.
įXT2p.u. loadp.u.
|j0.1378
i p.u.
j0.10 p.u. j0.2604 p.u.
Zioadp.u. =
1.875 + j0.4167 p.u.
0.9167/0° p.u.
Zone 1
Zone 2
Zone 3
Voase = 240 volts
Vosse2 = 480 voits
Voase3 = 120 volts
Zoase
(240)2
30,000
= 1.92 N Zbase2
(480)2
30.000
- 7.68 Ω Zas3
(120)2
30,000
0.48 N
%3!
Spase
= 30 kVA
30,000
120
loase3
= 250 A
%3D
(b) Per-unit circuit

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