Q: in single line diagram of fig (2) below the three bus system. The sequencedata for transmission lines and two generators are given in table below. Ifsingle line to ground fault occurs at F, calculate the fault current. If the faultimpedance is j0.1p.u what will be the fault current? Repeat the calculations forL-L fault occurs.DescriptionGenerator - AGenerator- BTransmission Line 1Transmission Line 2Transmission Line 3Zeroj0.03j0.02j0.14jo.10j0.12Sequence Data in puPositivej0.25j0.20j0.08j0.06j0.06Negativej0.15jo.12j0.08j0.06jo.06T.L.2GBT.L.LT.L.3Fig.(2)

The first step is to find out the Sequence impedance at the fault location point Diagram for each…

Description Generator - A Generator -B Transmission Line 1 Transmission Line 2 Transmission Line 3 Zero j0.03 jo.02 j0.14 jo.10 j0.12 Sequence Data in pu Positive j0.25 j0.20 jo.08 jo.06 jo.06 Negative j0.15 jo.12 jo.08 j0.06 jo.06 GA T.L.2 GB T.L.L T.L.3 Fie O)

Description Generator - A Generator -B Transmission Line 1 Transmission Line 2 Transmission Line 3 Zero j0.03 jo.02 j0.14 jo.10 j0.12 Sequence Data in pu Positive j0.25 j0.20 jo.08 jo.06 jo.06 Negative j0.15 jo.12 jo.08 j0.06 jo.06 GA T.L.2 GB T.L.L T.L.3 Fie O)

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

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

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

Chapter9: Unsymmetrical Faults

Section: Chapter Questions

Problem 9.35P

See similar textbooks

Related questions

Q: Suppose two generators supplying a load. Generator-1 has a no-load frequency of 61.5 Hz and a slope…

A: The resulting power frequency diagram is shown below

Q: For a power station the yearly load duration curve is a straight line from 50,000 to 4000 kw. To…

A: Capacity of Generator 1 & 2 = 20,000 kW Capacity of Generator 3 = 7000 kW Pmax = 50000 kW Pmin =…

Q: A power system consists of 4 buses as shown in Figure Q1. Generators are connected at buses 1 and 3…

A: In a power system, Bus Admittance Matrix represents the nodal admittances of the various buses. With…

Q: A generator operating at 50 Hz delivers 1 p.u. power to an infinite bus through a transmission…

A: Given- frequency f=50 Hz, 1 p.u. power, Maximum power = 0.5 p.u. To Find- The value of δmax.

Q: A 13.2 kV, 60 Hz, 3-phase synchronous generator is connected in star and its neutral point is…

A: Given, The supply voltage (E)= 13.2 kV(rms) Frequency (f)= 50 Hz Per phase inductance (L)= 0.02 H…

Q: Why does an alternator's voltage drop sharply when it is loaded down with a lagging load?

A: With legging load the voltage drop across reactances increases because of negative angle of current

Q: A three phase star connected, 1500 kVA, 13 kV alternator has an armature resistance of 0.9 ohm and a…

A: Given : Power = 1500 Kva Voltage =13 kV Ra= 0.9 ohm Xa= 8 ohm

Concept explainers

Short Transmission Line

A short transmission line is a transmission line that has a length less than 80 kilometers, an operating voltage level of less than 20 kV, and zero capacitance effect.

Power Flow Analysis

Power flow analysis is a topic in power engineering. It is the flow of electric power in a system. The power flow analysis is preliminary used for the various components of Alternating Current (AC) power, such as the voltage, current, real power, reactive power, and voltage angles under given load conditions and is often known as a load flow study or load flow analysis.

Complex Form

A power system is defined as the connection or network of the various components that convert the non-electrical energy into the electric form and supply the electric form of energy from the source to the load. The power system is an important parameter in power engineering and the electrical engineering profession. The powers in the power system are primarily categorized into two types- active power and reactive power.

Question

Q: in single line diagram of fig (2) below the three bus system. The sequence
data for transmission lines and two generators are given in table below. If
single line to ground fault occurs at F, calculate the fault current. If the fault
impedance is j0.1p.u what will be the fault current? Repeat the calculations for
L-L fault occurs.
Description
Generator - A
Generator- B
Transmission Line 1
Transmission Line 2
Transmission Line 3
Zero
j0.03
j0.02
j0.14
jo.10
j0.12
Sequence Data in pu
Positive
j0.25
j0.20
j0.08
j0.06
j0.06
Negative
j0.15
jo.12
j0.08
j0.06
jo.06
T.L.2
GB
T.L.L
T.L.3
Fig.(2)

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1 Solution Part 1

VIEW

Step 2 Solution part 2

VIEW

Step 3 Solution part 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

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

Please solve Q1(B) ONLY. Q1(A) is for reference.
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. 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…
Please solve Q1(C) ONLY. Q1(A) is for reference.
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 ele
Figure 2 shows the circuit of a simple power system. The ratings of the generators, transformers and the motors are as shown in the figure. The reactance of lines 1 and 2 are 40 ohms and 50 ohms respectively. Assume that system is unloaded, and the generators are generating their rated voltage. Select the generator G₁ ratings as base values. Choose the fault location in bus 1 And give the below answers For a symmetrical three phase fault on the bus (a).Determine the Thevenin's equivalent circuit which can be used to calculate the fault current. (b) Calculate the fault current contribution from the generator connected to the bus on which you have selected for fault calculation
Q.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' is r
b) A fault occurs at bus 3 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. jx2(1) = j0.1X p.u. jx2(2) = j0.1X p.u. jX2(0)=j0.2Y p.u. = V₁ 120° p.u. V₂ = 120° p.u. V₁ = 120° p.u. jx)=j0.1X p.u. jX2)= j0.1X p.u. jxo)=j0.1Y p.u. jx(1) = j0.25 p.u. 2 X (2) = 0.25 p.u. 3 jx1(0)=j0.3 p.u. jXT(I)=j0.1Y p.u. jX3(1)=j0.1X p.u. = jXT(2) j0.1Y p.u. jx13(2)=j0.1X p.u. jXT(0) = j0.1Y p.u. jX3(0) = j0.05 p.u. 0 0- Figure Q3. Circuit for problem 3b). (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…
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)…
b) A fault occurs at bus 2 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. JX20 /0.1X p.u. jXa2) 0.1X p.u. JX20 j0.2Y p.u. V,= 120° p.u. V, 120° p.u. V, 120° p.u. jX4-70.2X p.u. jX2 j0.2X p.u. jX o 0.2Y p.u. jXncay J0.25 p.u. jXna J0.25 p.u. 3 jXno0.3 p.u. jXTu) /0.2Y p.u. jXra j0.2Y p.u. - j0.2Y p.u. Xp-10.1X p.u. jXa j0.1X p.u. jXp0)- j0.05 p.u. 0 Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXac1) = j0.22 p.u., jXac2) = j0.22 p.u., and jXaco) = j0.23 p. u. X-2 Y=8 (iv) Determine the short-circuit fault current for the case when a phase-to- phase fault occurs at bus 2.
Consider the single-line diagram of a power system shown below. The transient reactance of cach part of the system is as shown in table and expressed in p.u. on a common 100 MVA base. A three-phase fault with impedance of 0.16 p. occurs at bus (5). Determine short- circuit capacity (fault level) and short circuit current? Assume the fault voltage at bus (5) equal 112 KV
Q2 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 G3
The one line diagram of a simple 3-bus power system shown below. All impedances are expresses in per unit on a common base. A three phase fault occurs at bus 3 through a fault impedance Z =j 0.19 pu, calculate the per unit fault current. A. 1.5 p.u. B.2.5 p.u. G₁ j 0.05 j0.75 j 0.3 3 C. 5.0 p.u. D. 3.0 p.u. G₂ j 0.075 j0.45 2