Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN: 9781305632134
Author: J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher: Cengage Learning
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Chapter 6, Problem 6.37P
To determine
The value of
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The symmetrical components the value of(1+a^2+a^2)
Q1
A three bus system has the following bus power and voltage specifications:
Bus
Real
Reactive
Generated Generated
Voltage
Bus type
Power
Power
Real
Reactive
(p.u.)
Demand
Demand
Power
Power
(p.u.)
(p.u.)
(p.u.)
(р.и.)
0.4
0.25
unspecified unspecified
1.05 + j0.0
Slack
1.5
0.9
unspecified
P-Q
3
1
unspecified
1.02
Р-V
(magnitude)
Angle unspecified
The Ybus matrix is shown as follows:
[542-68° 22Z116°
Y = 22/116°
312108°
582-63° 33Z107° p.u.
bus
312108°
33/107° 672-67°
(a)Determine the voltages at bus 2 and bus 3 after first iterations using Gauss Seidel. Use flat
start voltage.
(b) If the final value of voltage at buses 2 and 3 are 12-5.13° p.u. and 1.02Z-8.32° p.u.
respectively, calculate the following;
(i) Real and reactive power generated at bus 1
(ii) Power losses in branch (2-3)
1.
2.
Draw the power system network and form bus admittance matrix (Ybus) matrix for the following power systems.
(ii)
Bus code
Impedance
Line charging admittance
1-2
4+j5
J0.05
1-3
6+j9
J0.02
2-3
13+j8
J0.03
2-4
6+j10
J0.05
3-4
12+j13
J0.06
Chapter 6 Solutions
Power System Analysis and Design (MindTap Course List)
Ch. 6 - For a set of linear algebraic equations in matrix...Ch. 6 - For an NN square matrix A, in (N1) steps, the...Ch. 6 - Prob. 6.9MCQCh. 6 - Prob. 6.11MCQCh. 6 - Using Gauss elimination, solve the following...Ch. 6 - Prob. 6.9PCh. 6 - Determine the bus admittance matrix (Ybus) for the...Ch. 6 - Prob. 6.34PCh. 6 - Prob. 6.37PCh. 6 - Prob. 6.38P
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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
- Determine the bus admittance matrix (Ybus) for the three-phase power system shown in Figure 6.23 with input data given in Table 6.11 and partial results in Table 6.12. Assume a three-phase 100 MVA per unit base. TABLE 6.11 Bus input data for Problem 6.20 TABLE 6.12 Partially Completed Bus Admittance Matrix (Ybus) for Problem 6.30arrow_forwardThe units of admittance, conductance, and susceptance are siemens. (a) True (b) Falsearrow_forwardIf the power system has 'n' number of generators, then the size of the B- coefficient matrix is (nxn). Select one: O True O Falsearrow_forward
- 2. For the 3-bus network shown in figure, the impedances indicated are in per unit. a) Draw pu admittance diagram and obtain the bus admittance matrix Ybus for the network. b) Find the source voltages Eai and Ecz so that buses 1 and 2 have the voltages V = 120°, V2 = 1.05490° X= 0.20 pu X = 0.20 pu X= 0.36 pu Xo = 0.36 pu Xo 0.15 pu X p= 0.36 pu X- 0.36 pu X - 0.3 pu Load %3Darrow_forwardPlease helparrow_forward3. The incremental cost function for generation units in a power system presented below. Write matrix equation for first iteration in second order gradient method. (IC) = 4.7 +0,007 P (IC)₂ = 4.2+0.004 P₁₂ (IC)₂ = 5.1 +0.012 P (IC) = 3.8+0.015₁ Consider unit 2 as P₁ = 300 MW dependent unit and initial = 200 MW P=250 MW guess as P₁ =100 MWarrow_forward
- Find the bus admittance matrix for the following power system network. Z-0.3+j0.5 Zı=0.6+j4 Z-0.4+j0.3 Slack Z13=6.5+j7 Z3=8.5+j9 PQ bus 3 Z-0.5+j0.4arrow_forwardThe order of the matrix increases during the formulation of Z bus , when a new branch is added between the existing bus and new bus True Falsearrow_forwardq11arrow_forward
- The bus admittance matrix for the power system shown in Figure 1 is given by 3-j9 −2+j6 -1+j3 0 Ybus - -2+j6 3.666-j11 -0.666+j2 −1+j3 -1+j3 -0.666+j2 3.666-j11 per unit −2+j6 0 -1+j3 −2+j6 3-j9 With the complex powers on load buses 2, 3, and 4 as shown in Figure 1. 2.1 Determine the value for V₂ that is produced by the first and second iterations of the Gauss Seidel procedure. Choose the initial guess V₂(0) = V3(0) = V4(0) = 1.0L0° per unit. (12) 2 V₁ = 1.04/0° P₁ + jQ₁ −1+j0.5 Figure 1 V₂ 0.5-10.2 0.3j0.1arrow_forwardI need answer within 20 minutes please please with my best wishesarrow_forwardi need the answer quicklyarrow_forward
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