Explanation Concept used: The 4 × 4 bus admittance matric can be written as Y bus = [ Y 11 − Y 12 − Y 13 − Y 14 − Y 21 Y 22 − Y 23 − Y 24 − Y 31 − Y 32 Y 33 − Y 34 − Y 41 − Y 42 − Y 43 Y 44 ] Here, the admittance Y 11 , Y 22 , Y 33 and Y 44 are the self-admittance and, Y 12 , Y 21 , Y 13 , Y 14 , Y 21 , Y 23 Y 24 , Y 31 , Y 32 , , Y 34 , Y 41 , Y 42 ,and Y 43 are the mutual admittance. Write the expression for the self-admittance. Y = 1 Z .....(1) Here, Y is the admittance and Z is the impedance. Calculation: The circuit diagram is redrawn as shown in Figure 1 The Y 1 is the admittance of the impedance j 1 , Y 2 is the admittance of the impedance j 0.2 , Y 3 is the admittance of the impedance j 0.4 , Y 4 is the admittance of the impedance j 0.2 , Y 5 is the admittance of the impedance j 0.8 and Y 6 is the admittance of the impedance j 0.08 . Substitute Y 1 for Y and j 1 for Z in equation (1). Y 1 = 1 j 1 = − j Substitute Y 2 for Y and j 0.2 for Z in equation (1). Y 2 = 1 j 0.2 = − j 5 Substitute Y 3 for Y and j 0.4 for Z in equation (1). Y 3 = 1 j 0.4 = − j 2.5 Substitute Y 4 for Y and j 0.2 for Z in equation (1). Y 4 = 1 j 0.2 = − j 5 Substitute Y 5 for Y and j 0.8 for Z in equation (1). Y 5 = 1 j 0.8 = − j 1.25 Substitute Y 6 for Y and j 0.08 for Z in equation (1). Y 6 = 1 j 0.08 = − j 12.5 The admittance diagram for the system is redrawn as shown in Figure 2. The expression for admittance Y 11 is written as Y 11 = Y 1 + Y 2 + Y 3 Substitute − j for Y 1 , − j 5 for Y 2 and − j 2.5 for Y 3 in the above equation. Y 11 = − j − j 5 − j 2.5 = − j 8.5 The expression for admittance Y 12 is written as Y 12 = Y 3 Substitute − j 2.5 for Y 3 in the above equation. Y 12 = − j 2.5 The expression for admittance Y 13 is written as Y 13 = Y 2 Substitute − j 5 for Y 2 in the above equation. Y 13 = − j 5 The expression for admittance Y 14 is written as Y 14 = 0 The expression for admittance Y 21 is written as Y 21 = Y 3 Substitute − j 2.5 for Y 3 in the above equation. Y 21 = − j 2.5 The expression for admittance Y 22 is written as Y 22 = Y 3 + Y 4 + Y 5 Substitute − j 2.5 for Y 3 , − j 5 for Y 4 and − j 1.25 for Y 5 in the above equation. Y 22 = − j 2.5 − j 5 + − j 1.25 = − j 8.75 The expression for admittance Y 23 is written as Y 23 = Y 4 Substitute − j 5 for Y 4 in the above equation. Y 23 = − j 5 The expression for admittance Y 24 is written as Y 24 = 0 The expression for admittance Y 31 is written as Y 31 = Y 2 Substitute − j 5 for Y 2 in the above equation

MindTap Engineering, 1 term (6 months) Printed Access Card for Glover/Overbye/Sarma's Power System Analysis and Design, 6th

6th Edition

ISBN: 9781305636323

Author: Glover, J. Duncan, Overbye, Thomas, Sarma, Mulukutla S.

Publisher: Cengage Learning

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Question

Chapter 2, Problem 2.38P

To determine

To draw:

The admittance diagram for the system. Also, determine the 4×4 bus admittance matric Ybus.

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Chapter 2 Solutions

MindTap Engineering, 1 term (6 months) Printed Access Card for Glover/Overbye/Sarma's Power System Analysis and Design, 6th

Ch. 2 - The rms value of v(t)=Vmaxcos(t+) is given by a....Ch. 2 - If the rms phasor of a voltage is given by V=12060...Ch. 2 - If a phasor representation of a current is given...Ch. 2 - Prob. 2.4MCQ Ch. 2 - Prob. 2.5MCQ Ch. 2 - Prob. 2.6MCQ Ch. 2 - Prob. 2.7MCQ Ch. 2 - Prob. 2.8MCQ Ch. 2 - Prob. 2.9MCQ Ch. 2 - The average value of a double-frequency sinusoid,...

Ch. 2 - The power factor for an inductive circuit (R-L...Ch. 2 - The power factor for a capacitive circuit (R-C...Ch. 2 - Prob. 2.13MCQ Ch. 2 - The instantaneous power absorbed by the load in a...Ch. 2 - Prob. 2.15MCQ Ch. 2 - With generator conyention, where the current...Ch. 2 - Consider the load convention that is used for the...Ch. 2 - Prob. 2.18MCQ Ch. 2 - The admittance of the impedance j12 is given by...Ch. 2 - Consider Figure 2.9 of the text, Let the nodal...Ch. 2 - The three-phase source line-to-neutral voltages...Ch. 2 - In a balanced three-phase Y-connected system with...Ch. 2 - In a balanced system, the phasor sum of the...Ch. 2 - Consider a three-phase Y-connected source feeding...Ch. 2 - For a balanced- load supplied by a balanced...Ch. 2 - A balanced -load can be converted to an...Ch. 2 - When working with balanced three-phase circuits,...Ch. 2 - The total instantaneous power delivered by a...Ch. 2 - The total instantaneous power absorbed by a...Ch. 2 - Under balanced operating conditions, consider the...Ch. 2 - One advantage of balanced three-phase systems over...Ch. 2 - While the instantaneous electric power delivered...Ch. 2 - Given the complex numbers A1=630 and A2=4+j5, (a)...Ch. 2 - Convert the following instantaneous currents to...Ch. 2 - The instantaneous voltage across a circuit element...Ch. 2 - For the single-phase circuit shown in Figure...Ch. 2 - A 60Hz, single-phase source with V=27730 volts is...Ch. 2 - (a) Transform v(t)=75cos(377t15) to phasor form....Ch. 2 - Let a 100V sinusoidal source be connected to a...Ch. 2 - Consider the circuit shown in Figure 2.23 in time...Ch. 2 - For the circuit shown in Figure 2.24, compute the...Ch. 2 - For the circuit element of Problem 2.3, calculate...Ch. 2 - Prob. 2.11P Ch. 2 - The voltage v(t)=359.3cos(t)volts is applied to a...Ch. 2 - Prob. 2.13P Ch. 2 - A single-phase source is applied to a...Ch. 2 - Let a voltage source v(t)=4cos(t+60) be connected...Ch. 2 - A single-phase, 120V(rms),60Hz source supplies...Ch. 2 - Consider a load impedance of Z=jwL connected to a...Ch. 2 - Let a series RLC network be connected to a source...Ch. 2 - Consider a single-phase load with an applied...Ch. 2 - A circuit consists of two impedances, Z1=2030 and...Ch. 2 - An industrial plant consisting primarily of...Ch. 2 - The real power delivered by a source to two...Ch. 2 - A single-phase source has a terminal voltage...Ch. 2 - A source supplies power to the following three...Ch. 2 - Consider the series RLC circuit of Problem 2.7 and...Ch. 2 - A small manufacturing plant is located 2 km down a...Ch. 2 - An industrial load consisting of a bank of...Ch. 2 - Three loads are connected in parallel across a...Ch. 2 - Prob. 2.29P Ch. 2 - Figure 2.26 shows three loads connected in...Ch. 2 - Consider two interconnected voltage sources...Ch. 2 - Prob. 2.35P Ch. 2 - Prob. 2.36P Ch. 2 - Prob. 2.37P Ch. 2 - Prob. 2.38P Ch. 2 - Prob. 2.39P Ch. 2 - A balanced three-phase 240-V source supplies a...Ch. 2 - Prob. 2.41P Ch. 2 - A balanced -connected impedance load with (12+j9)...Ch. 2 - A three-phase line, which has an impedance of...Ch. 2 - Two balanced three-phase loads that are connected...Ch. 2 - Two balanced Y-connected loads, one drawing 10 kW...Ch. 2 - Three identical impedances Z=3030 are connected in...Ch. 2 - Two three-phase generators supply a three-phase...Ch. 2 - Prob. 2.48P Ch. 2 - Figure 2.33 gives the general -Y transformation....Ch. 2 - Consider the balanced three-phase system shown in...Ch. 2 - A three-phase line with an impedance of...Ch. 2 - A balanced three-phase load is connected to a...Ch. 2 - What is a microgrid?Ch. 2 - What are the benefits of microgrids?Ch. 2 - Prob. CCSQ Ch. 2 - Prob. DCSQ

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To draw: The admittance diagram for the system. Also, determine the 4 × 4 bus admittance matric Y bus .

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To draw: The admittance diagram for the system. Also, determine the 4×4 bus admittance matric Ybus.

Chapter 2 Solutions

To draw:

The admittance diagram for the system. Also, determine the 4×4 bus admittance matric Ybus.