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

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

Introductory Circuit Analysis (13th Edition)
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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**3.20** Using base values of 20 kVA and 115 volts in zone 3, rework Example 3.4.
Transcribed Image Text:**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 \( T_1 \) and \( T_2 \), whose ratings are also shown. Using base values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and determine 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:** Circuit for Example 3.4 **(a) Single-phase circuit** - **Zone 1:** - \( V_s = 220 \angle 0^\circ \text{ volts} \) - Transformer \( T_1 \): 30 kVA, 240/480 volts, \( X_{\text{eq}} = 0.10 \text{ p.u.} \) - **Zone 2:** - Line impedance: \( X_{\text{line}} = 2 \, \Omega \) - **Zone 3:** - Transformer \( T_2 \): 20 kVA, 460/115 volts, \( X_{\text{eq}} = 0.10 \text{ p.u.} \) - Load impedance: \( Z_{\text{load}} = 0.9 + j0.2 \, \Omega \) **(b) Per-unit circuit** - **Per-unit Source Voltage (\( V_{s, p.u.} \)):** - \( 0.9167 \angle 0^\circ \text{ p.u.} \) - **Impedances:** - Zone 1 Impedance: \( j0.10 \text{ p.u.} \) - Line Impedance: \( j0.2604 \text{ p.u.} \) - Zone 3 Impedance: \( j0.1378 \text{ p.u.} \) - **Load Impedance (\( Z_{\text{load}, p.u.} \)):** - \( 1.875 + j0.4167 \text{ p.u.} \) - **Base Values:** - Zone 1: -
Transcribed Image Text:**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 \( T_1 \) and \( T_2 \), whose ratings are also shown. Using base values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and determine 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:** Circuit for Example 3.4 **(a) Single-phase circuit** - **Zone 1:** - \( V_s = 220 \angle 0^\circ \text{ volts} \) - Transformer \( T_1 \): 30 kVA, 240/480 volts, \( X_{\text{eq}} = 0.10 \text{ p.u.} \) - **Zone 2:** - Line impedance: \( X_{\text{line}} = 2 \, \Omega \) - **Zone 3:** - Transformer \( T_2 \): 20 kVA, 460/115 volts, \( X_{\text{eq}} = 0.10 \text{ p.u.} \) - Load impedance: \( Z_{\text{load}} = 0.9 + j0.2 \, \Omega \) **(b) Per-unit circuit** - **Per-unit Source Voltage (\( V_{s, p.u.} \)):** - \( 0.9167 \angle 0^\circ \text{ p.u.} \) - **Impedances:** - Zone 1 Impedance: \( j0.10 \text{ p.u.} \) - Line Impedance: \( j0.2604 \text{ p.u.} \) - Zone 3 Impedance: \( j0.1378 \text{ p.u.} \) - **Load Impedance (\( Z_{\text{load}, p.u.} \)):** - \( 1.875 + j0.4167 \text{ p.u.} \) - **Base Values:** - Zone 1: -
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