Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN: 9781305632134
Author: J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 8, Problem 8.16MCQ
For a balanced-Y impedance load with per-phase impedance of
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Question 3. A three-phase impedance load balanced D a connected load in parallel with a balanced ? connected load
made of bonding. D each arm for the load ?D = 6 + 6? and ? each arm for the connected load
It is given as ?? = 2 + 2?. ? the connected load ?? = 1? is earthed through the neutral impedance.
To this impedance load, the symmetrical component voltages ?0 = 10∠60°, ?1 = 100∠0° and ?2 =
Unbalanced phase-to-earth voltages ??, ?? and ?? of 15∠200° volts were applied.
a) Draw the zero, positive and negative sequence circuits.
b) Find the complex powers imparted to each component circuit.
A-3-phase transmission line is shown in figure:
.AV:
ΔV,
AVb
AV
-AV₂-
=
-AV
I₁
a
Ib
Voltage drop across the transmission line is given
by the following equation :
I
Z₂ Zm Zm
Zm Zs Zm
Zm Zm Zs
][
Ia
Ib
Ic
Shunt capacitance of the line can be neglect. If
the line has positive sequence impedance of 15
and zero sequence impedance of 48 , then the
values of Z, and Zm will be
S
A balanced three phase voltage of 360-V line-to-neutral is applied to a balanced Y-connected load
with ungrounded neutral, as shown in figure-1. The three phase load consists of three mutually-
coupled reactances. Each phase has a series reactance of
Z, = j24 N, and the mutual coupling between phases Zm = j6 N.
(i)
Evaluate the line currents by mesh analysis without using symmetrical components.
Evaluate the line currents using symmetrical components.
(ii)
j6
j24
j6
V.
Figure-1
Chapter 8 Solutions
Power System Analysis and Design (MindTap Course List)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.Similar questions
- Q2. Figure Q2 shows the single-line diagram. The scheduled loads at buses 2 and 3 are as marked on the diagram. Line impedances are marked in per unit on 100 MVA base and the line charging susceptances are neglected. a) Using Gauss-Seidel Method, determine the phasor values of the voltage at load bus 2 and 3 according to second iteration results. b) Find slack bus real and reactive power according to second iteration results. c) Determine line flows and line losses according to second iteration results. d) Construct a power flow according to second iteration results. Slack Bus = 1.04.20° 0.025+j0.045 0.015+j0.035 0.012+j0,03 3 |2 134.8 MW 251.9 MW 42.5 MVAR 108.6 MVARarrow_forwardThe bus impedance matrix for a three-bus power system has per-unit elements as: Z11 = 0.5 Z12 = 0.4 Z13 = 0.8 Z21 = 1 Z22 = 0.5 Z23 = 0.6 Z31 = 0.5 Z32 = 1.2 Z33 = 1 Prefault voltage is 1 per unit and prefault current is neglected. A three-phase short circuit occurs at bus 3. Determine the subtransient fault current. Round your answer to 2 decimal places.arrow_forwardQuestion 3arrow_forward
- The Gauss-Seidel is an iterative method used to solve a set of nonlinear equations in the power flow. For the power network shown in Figure below, bus 1 is a slack/swing bus with V1 =1.0 L per unit and bus 2 is a load bus with P2 = 280 MW and Q2 = 60 MVar. The line impedance on a base of 100 MVA is Z = 0.02 +j0.04 per unit. Using Gauss-Seidel method, show how you determine V2 (first iteration only), and the power loss in line 1-2. Use an initial estimate of per unit but do not need to solve them numerically. After four iterations the voltage at bus 2 converges to per unit. Z12=0.02+j0.04 PLarrow_forwardFor the system shown in figure, voltages V2, V3 and angles 82, 83 are calculated using Newton-Raphson method. shunt line charging admittances are neglected. All the values are given in per unit on 100MVA base. Calculate the complex power flows S12 and S32 in actual units. 1) Z12 = 0.01+ j0.02 Z3 = 0.02+ j0.04 PL G QL Slack bus V2 = 0.96L -1.67° V3 = 0.884 - 5.48° Vi=1.0/0° (estimated time to answer this question: 13 minutes)arrow_forwardIn Figure 1, the 3-phase source has a line-to-line voltage of 440 volts. The details of the impedances are the following: Z₁ = 36-j15; Z₂ = 40 + /300; Z₂ = 45+j60 ; Z₂ = 75 + jo ; Z5 = 50+j50; Z6 = 20+ j48 2. Determine the following: (a) Line currents: labi, and I (c) Line currents: lar, Ibr, and ICT; (b) Line currents: Iaz, Ibz, and Iczi (d) PT, QT, ST, and pft for the system; (e) W, and W, readings for 2-wattmeter method at lines a and b. Balanced 3-Phase ab-bc-ca source C Figure 1 →laT -Ibr -KT 1 lal ²3 ↓ Ibi 1 Ici la2 lb2 le2ܝ 150 1²₂ ONDarrow_forward
- The 13.2 kV 50-Hz generator feeds a load over a line with (Zload) impedance Zline. Load is 800n + j500n and impedance of the line is 20 + j300 a) If the generator is directly connected to the load as given in (a), what will be the line losses? b) If the generator is connected to the load using step-up and step-down transformers as given in (b), what will be the line losses? c) How much is the increase in efficiency compared to (a) from (b)? Zline 2. (a) 1:10 10:1 Zline Tr Tr2 (b) Zload Zloadarrow_forwardIs it true??arrow_forwardK:17)arrow_forward
- 2.41 A A-Y connected bank of three identical 150-kVA, 2400-V:120-V, 60-Hz transformers is supplied at its high-voltage terminals through a feeder whose impedance is 6.4+j154 m2 per phase. The voltage at the sending end of the 2.12 Problems 119 feeder is held constant at 2400 V line to line. The results of a single-phase short-circuit test on one of the transformers with its low-voltage terminals short-circuited are VH = 131 V 1y = 62.5 A P= 1335 W %3D a. Calculate the series impedance of this three-phase transformer bank as referred to its high-voltage terminal. b. Determine the line-to-line voltage supplied to the feeder when the transformer bank delivers rated current at rated voltage to a balanced three-phase unity power factor load at its low-voltage terminal.arrow_forward2. The one-line diagram of a three-phase power system is as shown in the figure below. Impedances are marked in per unit on a 100-MVA, 400-kV base. The load at bus 2 is S2 = 15.93 MW-j33.4 Mvar, and at bus 3 is S3 = 77 MW +j14 Mvar. It is required to hold the voltage at bus 3 at 4006 0 kV. Working in per unit, determine the voltage at buses 2 and 1. V3 j0.5 pu V₂ S₂ j0.4 pu S3arrow_forwardIn a balanced three-phase Y-connected system with a positive-sequence source, the line-to-line voltages are 3 times the line-to-neutral voltages and lend by 30. (a) True (b) Falsearrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Power System Analysis and Design (MindTap Course ...Electrical EngineeringISBN:9781305632134Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. SarmaPublisher:Cengage Learning
Power System Analysis and Design (MindTap Course ...
Electrical Engineering
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:Cengage Learning
What is the Difference Between Single Phase and Three Phase???; Author: Electrician U;https://www.youtube.com/watch?v=FEydcr4wJw0;License: Standard Youtube License