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 9, Problem 9.56P
(a)
To determine
The bus impedance matrix for each of the three sequence networks.
(b)
To determine
The fault current, the current out of phase C of machine 2 and the line-to-ground voltages at the terminals of machine 2during the fault.
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Consider the system shown in the single-line diagram of Figure (3). All reactances are shown in per
unit to the same base. Assume that the voltage at both sources is 1 p.u.
a Find the fault current due to a bolted- three-phase short circuit at bus 3
b- Find the fault current supplied by each generator and the voltage at each of the buses 1 and 2
under fault conditions
0.06 p.u.
0.2 p.u.
0.04 p.u.
0.25 p.u.
0.2 p.u.
0.2 p.. 0.2 p.u.
0.06 p.u.
0.25 p.u.
Figure (3) Single-line diagram
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0.06 p.u.
0.25 p.u.
b) A fault occurs at bus 3 of the network shown in Figure Q4. Pre-fault nodal
voltages throughout the network are of 1 p.u. and the impedance of the electric
arc is neglected. Sequence impedance parameters of the generator,
transmission lines, transformer and load are given in Figure Q4.
V₁ = 120° p.u.
V₂ = 120° p.u. V₂ = 1/0° p.u.
V₂= 120° p.u.
jXj0.1 p.u.
JX2) 0.1 p.u.
jX0j0.15 p.u.
jXn-j0.2 p.u.
1 JX(2)-j0.2 p.u. 2
jX)=j0.25 p.u.
JX20-10.15 p.u.
jXa(z)-j0.2 p.u. 4
jX2(0)=j0.2 p.u.
jXT(1) j0.1 p.u.
jXT(2)=j0.15 p.u.
jXT(0)=j0.1 p.u.
Figure Q4. Circuit for problem 4b).
=
jXj0.1 p.u.
j0.1 p.u.
-
JX(2)
JXL(0) 10.1 p.u.
=
(i) Assuming a balanced excitation, draw the positive, negative and zero
sequence Thévenin equivalent circuits as seen from bus 3.
(ii) Determine the positive sequence fault current for the case when a three-
phase-to-ground fault occurs at bus 3 of the network.
(iii) Determine the short-circuit fault current for the case when a one-phase-
to-ground fault occurs at bus…
b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 + j0 p.u, and the impedance of the electric arc is neglected (Zf = 0 + j0 p.u.). The positive, negative and zero
sequence impedance parameters of the generator, transmission lines and transformer are given in Figure Q3.
(i) Determine the positive sequence fault current for the case when a three-phase-to-ground fault occurs at bus 4 of the network
(ii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus 4. Recall that phasors can be expressed in terms of their symmetrical components as shown in the picture attached. where F stands for any three-phase quantity (e.g., current, voltage)
(iii) Determine the short-circuit fault current for the case when a phase-to-phase fault occurs at bus 4
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Chapter 9 Solutions
Power System Analysis and Design (MindTap Course List)
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- Consider the oneline diagram of a simple power system shown in Figure 9.20. System data in per-unit on a 100-MVA base are given as follows: The neutral of each generator is grounded through a current-limiting reactor of 0.08333 per unit on a 100-MVA base. All transformer neutrals are solidly grounded. The generators are operating no-load at their rated voltages and rated frequency with their ENIFs in phase. Determine the fault current for a balanced three-phase fault at bus 3 through a fault impedance ZF=0.1 per unit on a 100-MVA base. Neglect -Y phase shifts.arrow_forwardb) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) = j0.1X p.u. V₁ = 120° p.u. V₂ = 120° p.u. (i) (ii) 0 jX(1) = j0.2 p.u. 1 jx(2) j0.2 p.u. 2 jX1(0) = j0.25 p.u. jXT(1) jXT(2) 종 3 j0.1X p.u. JX3(1) j0.1Y p.u. j0.1X p.u. JX3(2) j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0 = x = 1, jX2(1) j0.2Y p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V3 = 120° p.u. Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: y = 7 = = jXa(r) = j0.13 p.u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from…arrow_forwardThe one line diagram of a simple three bus power system is shown in figure Each generator is represented by an emf behind the subtraction reactance. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. Athree phase fault occurs at bus 3 through a fault impedance of Zf=j0.19 per unit. a-using thevenin’s theorem obtain the impedance to the point of fault and the fault current in per unit. b-determine the bus voltage and line currents during faultarrow_forward
- b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. V₁ = 120° p.u. V₂ = 120° p.u. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) j0.1X p.u. - 0 jX(1) = j0.2 p.u. 1JX(2) = 0.2 p.u. 2 jX1(0) = j0.25 p.u. jX2(1) j0.2 p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V₂ = 120° p.u. jXT(1) j0.1X p.u. jXT(2) j0.1X p.u. JX3(1) j0.1Y p.u. JX3(2)=j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0- = 3 = Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(n) = j0.13 p. u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. 4 (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 4. (ii)…arrow_forwardQ2. The single-line diagram of a simple three-bus power system is shown in Figure-2. Each generator is represented by an emf behind the sub-transient reactance. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. A three-phase fault occurs at bus 3 through a fault impedance of Zf = j0.19 per unit. (i) Using Th'evenin's theorem, obtain the impedance to the point of fault and the fault current in (ii) Determine the bus voltages per unit. ) j0.05 j0.075 j0.75 2 j0.30 j0.45 Figure-2: Single line diagram of the power system network for Q2 3arrow_forwardThe one-line diagram of a simple power system is shown in Figure 9.20. Each generator is represented by an an emf behind the transient reactance. All impedances are expressed in per unit on a common MVA base. All resistance and shunt capacitances are neglected. The generation are operating on no load at their rated voltage with their emfs in phase. A three-phase fault occurs at bus 1 through a fault impedance of Z_f = j0.08 per unit. Using Thevenin's theorem obtain the impedance to the point of fault and the fault current in per unit. Determine the bus voltages and line currents during fault.arrow_forward
- Problem 5 Consider the system shown in the single-line diagram of Figure (3). All reactances are shown in per unit to the same base. Assume that the voltage at both sources is 1 p.u. a- Find the fault current due to a bolted- three-phase short circuit at bus 3. b- Find the fault current supplied by each generator and the voltage at each of the buses I and 2 under fault conditions. 0.04 p.u. 0.2 p.u. 0.06 p.u. 0.2 p.u. 0.25 p.u. G, 0.2 p.u. 0.2 p.u. 0.06 p.u. 0.06 р.и. 3 0.25 p.u. 0.25 p.u. G, Figure (3) Single-line diagram for Problem 5 elearrow_forwardFigure shows a sample power system network and Zpus Matrix elements. For a solid three phase fault take place at bus 3. Determine a) Fault current (b) V1 , V21 and Var (c) Post fault current in lines 1-2 and 1-3. The line from bus 1-2 impedance =j1.2p.u The line from bus 2-3 impedance =j0.16p.u The line from bus 3-1 impedance =j1.37p.u The Zbus matrix element values are Zbus Matrix Z13 = 0.4; Zbus Matrix Z23 =1.01; Zbus Matrix Z33 =1.55; Fault Post fault voltage at bus 1 (V1) in p.u Post fault voltage at bus 2 (V21 ) in p.u Post fault voltage at bus 3 (V31 ) in p.u Post fault current( I12) in line between 1-2 in p.u Post fault current (I13) in line between 1-3 in p.uarrow_forwardEXAMPLE 9.1 Power-system sequence networks and their Thévenin equivalents A single-line diagram of the power system considered in Example 7.3 is shown in Figure 9.3, where negative- and zero-sequence reactances are also given. The neutrals of the generator and A-Y transformers are solidly grounded. The motor neutral is grounded through a reactance X, = 0.05 per unit on the motor base. (a) Draw the per-unit zero-, positive-, and negative- sequence networks on a 100-MVA, 13.8-kV base in the zone of the generator. (b) Reduce the sequence networks to their Thévenin equivalents, as viewed from bus 2. Prefault voltage is VF = 1.05/0° per unit. Prefault load current and A-Y transformer phase shift are neglected. FIGURE 9.3 T, T2 2. M Line Single-line diagram for Example 9.1 X, = X2 - 20 n 1 X, = 60 N %3! 100 MVA 13.8 kV 100 MVA 100 MVA X" = 0.15 13.8-kVA/138-kVY 138-kV Y/13.8-kV A 100 MVA X, = 0.17 X = 0.10 per unit X = 0.10 per unit %3D 13.8 kV X" = 0.20 X2 = 0.21 Xo = 0.10 = 0.05 per unit…arrow_forward
- Q.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' isarrow_forwardQ2 Figure Q2 shows a single line diagram of a power system and the associated data of this system are given in Table Q2. The pre-fault load current and A-Y transformer phase shift are neglected. (a) (b) If a Single Line-to-Ground (S-L-G) fault occurs at Bus 5 and the pre-fault voltage is 1.0 pu, calculate the subtransient fault current in Ampere. (c) (d) (e) Using base of 100 MVA and 11 kV at generator G₁, construct the positive sequence, negative sequence and zero sequence networks with their corresponding component values indicated. G₁ Recalculate (b) if the neutral on HV side of T3 is solidly grounded. Repeat part (b) with Line-to-line (L-L) fault. What will happen to L-L fault current in (d) if the neutral on the HV side of T3 is solidly grounded? Bus 1 T₁ Bus 4 Line 1 Line 2 Figure Q2 Bus 5 T2 T3 Bus 2 Bus 3 G₂ G3arrow_forward3.On a double line-to-ground faultImmersive reader A) The short-circuit currents in the faulted phases are equal. B) The voltages at the fault point on the faulted phases are zero. C) The phase not involved in the fault has a zero voltage at the fault point. D) There are no circulating currents through the neutrals to ground. E) N. A.arrow_forward
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