determine the per-unit impedances and the per-unit source voltage. Then cal- culate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected. V₂ Zone 1 = 220/0° volts Three zones of a single-phase circuit are identified in Figure 3.10(a). The zones are connected by transformers T₁ and T₂, whose ratings are also shown. Using base values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and مع و T₁ 30 kVA 240/480 volts = 0.10 p.u. Xa Zone 2 Xune 202 उह T₂ 20 KVA 460/115 volts = 0.10 p.u. Xea Zone 3 کند Zload = 0.9 +10.20 ทะ

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 3.53P: The ratings of a three-phase, three-winding transformer are Primary: Y connected, 66kV,15MVA...
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HOME WORK: Per-unit circuit: three-zone single-phase network
determine the per-unit impedances and the per-unit source voltage. Then cal-
culate the load current both in per-unit and in amperes. Transformer winding
resistances and shunt admittance branches are neglected.
Zone 1
Vs = 220/0° volts i
Three zones of a single-phase circuit are identified in Figure 3.10(a). The zones
are connected by transformers T₁ and T₂, whose ratings are also shown. Using
base values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and
T₁
30 kVA
240/480 volts
Xeq = 0.10 p.u.
V sp.u. =
0.9167/0° p.u.
Zoase
Zone 2
Xune 20
(a) Single-phase circuit
I'spu i Xtio.u
Zone 1
Vbase1 = 240 volts
(240)²
30,000
Spase = 30 kVA
= 1.92
T₂ Zload = 0.9 +0.2
20 KVA
460/115 volts
Xea = 0.10 p.u.
jXunepu
j0.10 p.u. j0.2604 p.u.
Zone 2
Vbase2 = 480 volts
Zbase2 =
Zone 3
(480)²
30,000
¡XT2pu hoadp.u.
= 7.68
(b) Per-unit circuit
/0.1378
p.u.
Zone 3
Vbase3= 120 volts
Zbase3 =
base3 =
Zoadp.u. =
1.875+ 0.4167 p.u.
(120)²
30,000
30,000
120
= 0.48 Ω
€10
= 250 A
Transcribed Image Text:HOME WORK: Per-unit circuit: three-zone single-phase network determine the per-unit impedances and the per-unit source voltage. Then cal- culate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected. Zone 1 Vs = 220/0° volts i Three zones of a single-phase circuit are identified in Figure 3.10(a). The zones are connected by transformers T₁ and T₂, whose ratings are also shown. Using base values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and T₁ 30 kVA 240/480 volts Xeq = 0.10 p.u. V sp.u. = 0.9167/0° p.u. Zoase Zone 2 Xune 20 (a) Single-phase circuit I'spu i Xtio.u Zone 1 Vbase1 = 240 volts (240)² 30,000 Spase = 30 kVA = 1.92 T₂ Zload = 0.9 +0.2 20 KVA 460/115 volts Xea = 0.10 p.u. jXunepu j0.10 p.u. j0.2604 p.u. Zone 2 Vbase2 = 480 volts Zbase2 = Zone 3 (480)² 30,000 ¡XT2pu hoadp.u. = 7.68 (b) Per-unit circuit /0.1378 p.u. Zone 3 Vbase3= 120 volts Zbase3 = base3 = Zoadp.u. = 1.875+ 0.4167 p.u. (120)² 30,000 30,000 120 = 0.48 Ω €10 = 250 A
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