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From Example 2.1 the maximum diversified kW demands were computed. Using the 0.9 lagging power factor, the maximum diversified kW and kVA demands for the line segments and transformers are Segment N1-N2: P12 = 92.8 kW S12 = 92.8 + j45.0 kVA S3 = 72.6 + j35.2 kVA Su = 48.9 + j23.7 kVA S = 30.3 + j14.7 kVA Segment N2-N3: P23 = 72.6 kW %3D Segment N3-N4: P4 = 48.9 kW Transformer T1: P = 30.3 kW %3D Transformer T2: Pr2 = 35.5 kW S2 = 35.5 + j17.2 kVA %3D Transformer T3: Pr3 = 48.9 kW ST3 = 48.9 + j23.7 kVA %3!

From Example 2.1 the maximum diversified kW demands were computed. Using the 0.9 lagging power factor, the maximum diversified kW and kVA demands for the line segments and transformers are Segment N1-N2: P12 = 92.8 kW S12 = 92.8 + j45.0 kVA S3 = 72.6 + j35.2 kVA Su = 48.9 + j23.7 kVA S = 30.3 + j14.7 kVA Segment N2-N3: P23 = 72.6 kW %3D Segment N3-N4: P4 = 48.9 kW Transformer T1: P = 30.3 kW %3D Transformer T2: Pr2 = 35.5 kW S2 = 35.5 + j17.2 kVA %3D Transformer T3: Pr3 = 48.9 kW ST3 = 48.9 + j23.7 kVA %3!

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
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Q)The question is that how kW is converted to kVA in complex form and how kV square is 2.4 for T1,T2,T3 can you plz explain it by seeing the solution

Example 2.3 For the system of Example 2.1, assume the voltage at N1 is 2400 volts and compute the secondary voltages on the three transformers using the diversity factors. The system of Example 2.1, including segment distances, is shown in Figure 2.12. Assume that the power factor of the loads is 0.9 lagging. The impedance of the lines are: z = 0.3 + j0.6 2/mile The ratings of the transformers are T1: 25 kVA, 2400-240 volts, Z = 1.8/40% T2: 37.5 kVA, 2400-240 volts, Z = 1.9/45% T3: 50 kVA, 2400-240 volts, Z = 2.0/50% From Example 2.1 the maximum diversified kW demands were computed. Using the 0.9 lagging power factor, the maximum diversified kW and kVA demands for the line segments and transformers are Segment N1-N2: P12 = 92.8 kW S12 = 92.8 + j45.0 kVA Segment N2-N3: P23 = 72.6 kW S23 = 72.6 + j35.2 kVA Segment N3-N4: P4 = 48.9 kW S34 = 48.9 + j23.7 kVA Transformer T1: P = 30.3 kW STI = 30.3 + j14.7 kVA Transformer T2: Pr2 = 35.5 kW S12 = 35.5 + j17.2 kVA Transformer T3: Pr3 = 48.9 kW S13 = 48.9 + j23.7 kVA The Nature of Loads 29 Convert transformer impedances to ohms referred to the high-voltage side kV - 1000 kVA 2.4 - 1000 T1: Zbase = = 230.4 2 25 ZTI = (0.018 /40) · 230.4 = 3.18 + j2.67 2 kV² - 1000 kVA 4² . 1000 37.5 T2: Zbase = = 153.6 2 ZT2 = (0.019/45) · 153.6 = 2.06 + j2.06 2 kV? . 1000 kVA 2.4 - 1000 T3: Zbase = = 115.2 Q %3D 50 Z73 = (0.02/50) 115.2 = 1.48 + j1.77 2
Transcribed Image Text:Example 2.3 For the system of Example 2.1, assume the voltage at N1 is 2400 volts and compute the secondary voltages on the three transformers using the diversity factors. The system of Example 2.1, including segment distances, is shown in Figure 2.12. Assume that the power factor of the loads is 0.9 lagging. The impedance of the lines are: z = 0.3 + j0.6 2/mile The ratings of the transformers are T1: 25 kVA, 2400-240 volts, Z = 1.8/40% T2: 37.5 kVA, 2400-240 volts, Z = 1.9/45% T3: 50 kVA, 2400-240 volts, Z = 2.0/50% From Example 2.1 the maximum diversified kW demands were computed. Using the 0.9 lagging power factor, the maximum diversified kW and kVA demands for the line segments and transformers are Segment N1-N2: P12 = 92.8 kW S12 = 92.8 + j45.0 kVA Segment N2-N3: P23 = 72.6 kW S23 = 72.6 + j35.2 kVA Segment N3-N4: P4 = 48.9 kW S34 = 48.9 + j23.7 kVA Transformer T1: P = 30.3 kW STI = 30.3 + j14.7 kVA Transformer T2: Pr2 = 35.5 kW S12 = 35.5 + j17.2 kVA Transformer T3: Pr3 = 48.9 kW S13 = 48.9 + j23.7 kVA The Nature of Loads 29 Convert transformer impedances to ohms referred to the high-voltage side kV - 1000 kVA 2.4 - 1000 T1: Zbase = = 230.4 2 25 ZTI = (0.018 /40) · 230.4 = 3.18 + j2.67 2 kV² - 1000 kVA 4² . 1000 37.5 T2: Zbase = = 153.6 2 ZT2 = (0.019/45) · 153.6 = 2.06 + j2.06 2 kV? . 1000 kVA 2.4 - 1000 T3: Zbase = = 115.2 Q %3D 50 Z73 = (0.02/50) 115.2 = 1.48 + j1.77 2
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