POWER SYS. ANALYSIS+DESIGN
6th Edition
ISBN: 9780357700907
Author: Glover
Publisher: INTER CENG
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
error_outline
This textbook solution is under construction.
Students have asked these 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 MVAR
1.
FIGURE 52 shows the one-line diagram of a simple three-bus power system with
generation at bus I. The voltage at bus l is V1 = 1.0L0° per unit. The scheduled
loads on buses 2 and 3 are marked on the diagram. Line impedances are marked in
per unit on a 100 MVA base. For the purpose of hand calculations, line resistances
and line charging susceptances are neglected
a) Using Gauss-Seidel method and initial estimates of Va
0)-1.0+)0 and V o)-
(
1.0 +j0, determine V2 and V3. Perform two iterations
(b) If after several iterations the bus voltages converge to
V20.90-j0.10 pu
0.95-70.05 pu
determine the line flows and line losses and the slack bus real and reactive power.
2
400 MW
320 Mvar
Slack
0.0125
0.05
300 MW
270 Mvar
FIGURE 52
A network consisting of a set of generator and load buses is to be modeled with a DC power flow, for the sake of conducting a
contingency analysis. The initial flows calculated with the DC power flow give the following information: f°2-4 = - 65.3 MW and fº4-5 = 13.6
MW. The following values of LODF and PTDF factors are given: PTDF54,2-4 = -0.2609, LODF2-4,4-5 = -0.6087. Calculate the contingency flow
on line 2-4 due to outage of line 4-5.
Select one:
O a. -75.5MW
O b. None of these
O c. -68.85MW
O d. -73.58MW
O e. 75.5MW
O f. -61.75MW
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
- To convert a per-unit impedance from old to new base values, the equation to be used is Zp.u.new=Zp.u.old(VbaseoldVbasenew)2(SbasenewSbaseold)Zp.u.new=Zp.u.old(VbaseoldVbasenew)2(SbasenewSbaseold)Zp.u.new=Zp.u.old(VbaseoldVbasenew)2(SbasenewSbaseold)arrow_forward6.1. A power system network is shown in Figure 47. The generators at buses 1 and 2 are represented by their equivalent current sources with their reactances in per unit on a 100-MVA base. The lines are represented by n model where series reactances and shunt reactances are also expressed in per unit on a 100 MVA base. The loads at buses 3 and 4 are expressed in MW and Mvar. (a) Assuming a voltage magnitude of 1.0 per unit at buses 3 and 4, convert the loads to per unit impedances. Convert network impedances to admittances and obtain the bus admittance matrix by inspection. j0.25 50.25 -j4 j0.4 j0.1 j0.16 j0.2 -j4+3 4 S3 -j4 S4 FIGURE 47 One-line diagram for Problem 6.1. 100 MW +j25 Mvar 200 MW +j50 Mvararrow_forwardQ6/ The per-unit reactance for a given system are shown in figure below. (1 MVA) is being delivered to the receiving end bus of the system at unity power factor and unit voltage. A three phase short circuit occurs at F. Find the critical clearing angle and the critical clearing time? Take H = 3 p. u. second tc X=j0.3 j0.1 w Peralat j0.25 j0.25 V20 x 1arrow_forward
- Discuss the role of FACTS (Flexible Alternating Current Transmission Systems) devices in power system control and optimization.arrow_forward400Ω A k=0.9 600Ω B k=0.8 66Ω 600Ωarrow_forwardThe 6-bus power system network of an electric utility company is shown in the Figure below. The line and transformer data containing the subtransient series resistance and reactance in per unit, and one-half of the total capacitance in per unit susceptance on a 100-MVA base, is tabulated below. The prefault voltage profile of the power system as obtained from four iterations of the newton Raphson power flow method are provided below as well.arrow_forward
- The one-line diagram of a simple power system is shown in Figure below. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA base. The system data expressed in per unit on a common 100-MVA base is tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phase. G Stark Item Base MVA Voltage Rating X' x² 20 kV 20 kV 20/220 kV 20/220 kV 100 0.05 0.15 0.15 0.10 0.10 220 kV 0.125 0.125 0.30 0.15 0.25 025 0.7125 0.15 100 100 0.15 0.05 0.10 0.10 0.10 100 0.10 100 100 Lu La 220 kV 0.15 220 kV 0.35 100 A balanced three-phase fault at bus 3 through a fault impedance Zf= jo.I per unit. The magnitude of the fault current in amperes in phase b for this fault is: Select one: A. 345.3 B. 820.1 C. 312500 3888888 产产arrow_forwardFor the system in Figure 4 with given generation and load dispatch determine the voltages after two itterations of Gauss-Seidel method. Assume the initial voltage to be 1.01 at angle of 0◦ pu at bus 1, 1.015 at angle of 0◦ pu at bus 2, and 1.0 at angle of 0◦ pu at bus 3. All line impedances are in per unit on a common base, and charging is neglected. Take base power of 100 MVA.arrow_forwardThe figure below shows the one-line diagram of a four- bus power system. The voltages, the scheduled real power and reactive powers, and the reactances of transmission lines are marked at this one line diagram (The voltages and reactances are in PU referred to 100 MW base. The active power P2 in MW is the last three digits (from right) of your registration number (i.e for the student that has a registration number 202112396, P2 =396). [10] Starting from an estimated voltage at bus 2, bus 3, and bus 4 equals V2 (0) = 1.15<0°, V3 = 1.15 < 0°, V4 1.1< 0°. 1- Specify the type of each bus and known & unknown quantities at each bus. 2- Find the elements of the second row of the admittance matrix (i.e. [Y21 Y22 Y23 Y24]). 3- Using Gauss-Siedal fınd the voltage at bus 2 after the first iteration. 4- Using Newton-Raphson, calculate: |- The value of real power (P2), at bus 2 after the first iteration. Il- The second element in the first row of the Jacobian matrix after the first iteration. 2 P2…arrow_forward
- please solve for nodal stress method. (if it is possible to apply supernodes)arrow_forward6. For a three bus power system assume bus 1 is the swing with a per unit voltage of 1.020 , bus 2 is a PQ bus with a per unit load of 2.0 + j0:5, and bus 3 is a PV bus with 1.0 per unit generation and a 1.0 voltage setpoint. The per unit line impedances are j0.1 between buses 1 and 2, j0.4 between buses 1 and 3, and j0.2 between buses 2 and 3. Using a flat start, use the Newton-Raphson approach to determine the first iteration phasor voltages at buses 2 and 3.arrow_forwardProblem 6.10: A 2-MVA, 4160/480 V transformer has series impedance of R = 2% and X = 6%. Determine the (i) full load voltage regulation of the trans- former at 0.85 pf lagging, and (ii) secondary output voltage on unloading the transformer completely.arrow_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