Date:

Tuesday, July 14, 2020

What are the Conditions in Selecting Current Transformer in Protective Relaying


Current Transformer

Current Transformer

The current transformer is one of the main essential components in protective relaying. Along with Potential Transformers (PT), this device is also the key element that will detect any abnormalities in the system. When it comes to protective relaying applications, the CT that will be used is those with large cores that allow the replication of the primary current during fault condition (high primary current). 

However, we should not rely on the physical sizes of selecting the right unit to use. We need to satisfy the essential requirements to prevent undesirable results. In order to achieve this goal, we need to comply with two conditions, namely: 

Condition 1: The current rating of the primary must be close to the full load current of the load or the system that it protects. For example, if the full load current is 387 Amperes, then it is better to choose a CT that has a conversion ratio of 400:5. 

Condition 2: Calculate the secondary voltage at the time of the fault to determine whether the CT will saturate during a fault condition. In this case, even if we satisfy condition 1, we need to select the next higher rating and do the same process. 

How to determine CT Saturation?

To determine whether a CT will saturate in a certain condition, we need to find the following values: 
  • Amount of fault current 
  • CT Secondary Resistance
  • Wire resistance that connects CT and the protection relay
  • The Burden resistance of the relay. 

Example: 

A protection relay is to be installed to a motor with full load current rating of 290. A CT with conversion ratio of 300:5 will be used and it has a total distance of 20 meters from the relay. The wire has a resistance of 1.5 ohms per kilometers. If the CT secondary resistance is 0.09 ohms and the relay burden resistance is 0.005 ohms. Determine if the CT ratio will saturate based on the curve given below. 

Note: Assume that the fault current in the system is 3.5 kA.



Solution:

In this case, we will enumerate the given values as, 
  • CT to be used is 300 : 5
  • Fault current = 3,500 Amperes
  • CT secondary resistance = 0.090 ohms
  • Wire distance to relay = 20 mtrs rated @ 1.5 ohms/ km
  • Relay burden resistance = 0.006 ohms

Calculate the wire resistance: 

Rwire = (2 x 20 m) x 1.5 ohms/ km 
                             1000 

Rwire = 0.06 ohms


Calculate the total burden

Total Burden = RCT (secondary) + Rwire + R relay

Total Burden = 0.090 + 0.06 + 0.006 = 0.156 ohms

Calculate the Secondary Voltage in the event of fault. 

V = (Total Burden * Fault Current)/ CT ratio

V = (0.156 * 3,500) / 60

V = 9.1 Volts

Check the CT Curve if the secondary voltage in the event of fault is below the KNEE POINT. 


The value of secondary voltage in the event of fault is 9.1 Volts which is below the knee point. Therefore, we can accept the 300:5 CT given in the problem. Otherwise, we will choose the next higher value and do the same process. 

Sunday, July 05, 2020

How TopSolid Helps in Making Models 3D Printing Ready?

source: https://drive.google.com/file/d/12DKIkCONwp9dVi16nYAhPaC-JHu0jPcC/view?usp=sharing

TopSolid is one of the most popular software solutions for 3D modeling and creating printable3D models. It's a program that's been around for years, widely used by professionals and amateur designers from all over the globe.

There's no doubt that TopSolid is a powerful software designed to help users create vivid and realistic 2D and 3D models. You can use it to create a sketch of the product you want to print, then perfect it is using advanced 3D tools, and finally export your model in a 3D file ready for printing. But, before we get to the details, let's see what TopSolid all is about.


TopSolid Background

TopSolid is a CAD/CAM program developed by a French company called Missler Software. The company was started in 1977, and it's still going strong ever since. The company has spent the last 20 years developing and perfecting products designed for the mechanical wood and sheet metal industries. As a result, TopSolid can cover everything needed to create a fully functional device. That includes creating a basic design, making molds, making 2D and 3D models, and so on. It's got a set of advanced features that will make any modeling job more comfortable than ever.


Advanced Features

We won't get into too many details because this software has many features to offer, but we will cover the ones that will make 3D modeling and printing seem effortless.

Integrated Project Data Management

While this feature isn't that useful for individual users, it saves a lot of time for larger companies. Namely, TopSolid comes with an integrated PDM platform that keeps a detailed library of all 3D models you've created. That means that you can share your work with the rest of your staff directly. All you need is to install TopSolid PDM on a dedicated computer on your network. Everyone connected to the network will then get access to the shared models, which can drastically improve mass production.

The PDM feature allows you to arrange data by time, date, and projects. You can create specific documents for each part needed to make your model work, create tutorials on how to assembly them, and so on. PDM will automatically track what's going on and provide you with a list of documents you can use to see how a project is going. Every change is then automatically applied to the documents in the vault, making sure that all project details are up to date. It even keeps a detailed history of all previous versions so that you can go back at any given time.


source: https://drive.google.com/file/d/1VLO8772kjBHdJPhs6PoYtWAx3ocsuGXv/view?usp=sharing


Fantastic UI

If you have any previous experience with CAD programs, you know how complicated they can be for beginners. Well, TopSolid has a different approach than most similar 3D printing programs. When you run it, you will arrive at the start page from where you can see and access recent projects and, of course, start new ones.

Beginners will love the tutorials section that holds an impressive 374-page user's guide with detailed explanations of all tools and features. You will also be able to access user forums and see what new features are added with every update.

Every time you open a document, you will be taken to a new tab surrounded by available tools. Every document will tell you details about when the last changes were made, and what documents are being worked on at the moment. The main menu below the document will show you a series of tabs packed with tool icons. There are dozens of options you can try to organize your projects, and the advanced tools allow you to make changes on the go. All of the information about your 3D projects will be visible at the same time, making it much easier to visualize the end product.

source: https://drive.google.com/file/d/1Gq-axgeSIqtmX7krthg6_zGHWuyhijVV/view?usp=sharing


Creating a Model From Scratch

TopSolid can create a 3D model from a 2D sketch quickly. The process is quite simple, and you don't need a lot of time to figure out how it works. You have to draw a rough 2D model, and contours will appear automatically. Once you're happy with your sketch, all you have to do is to extrude it and make adjustments to the newly created 3D model.

If your 3D model has a lot of different parts, you will have to create 2D sketches for each part separately. The sketching is done using a top view of the work area, but you can also create spatial sketches with some built-in tools.


Importing STL Files

If you already have a ready 3D model you want to print, you can add it to the software directly. You can make changes if you want, but you can also export the file in several different available formats designed for specific uses. That includes 3D printing as well as classic machine manufacturing techniques. That means that you can quickly select the type of file you need and export it directly to your 3D printer. The printer will then follow instructions taken from the file and print out the 3D model you wanted.


Import Any Type of File

Many people have great ideas and sketches, but they never go through with the realization because they don't know how to do it. That's where TopSolid has the most significant impact on beginners. You can create a doodle in any program, upload it to TopSolid, and get a 3D model preview with one click!

You will probably get a model that's not exactly as you've imagined it, but the available tools will allow you to make changes immediately. You won't need more than a few hours to master the process, and once you do that, you can print any 3D model in a matter of minutes.


Improve Your 3D Models with this Advanced Tool

TopSolid is a fantastic tool for 3D printing because it creates 3D models based on sketches instantly. Instead of having to create every part and model from nothing, you will start with more than half of the work done already. That's fantastic if you have large 3D printing projects, and once you see how it works in practice, you will love everything this software has to offer.


Author/ Contributor:

Natalie Hutchins

email: nathalie.hn77@gmail.com

Friday, February 14, 2020

The Role of Power System Operators

Power System Operator



The Role of Power System Operators 


The power system operators are mostly concerned with the reliability or security of the power grid. Operators control the actual flow of electric power from the power plant to the customers. They also work with neighboring control areas to manage the import and export of power. Matching the supply of power to the demand of consumers is one of the main focuses of these operators. 

Tasks and operations performed during normal conditions 

There are many tasks and operations performed during normal conditions. The following functions are the most fundamental operator role. 

Dispatch and Control -- They operate high voltage direct current converter equipment, tap changing transformers, phase shifting transformers, and circuit breakers as well as voltage control equipment. Dispatchers monitor equipment and record readings at a pilot board, which is a map of the transmission grid system showing the status of transmission circuits and connections with substations and industrial plants. There are many times when all you have to do is monitor the system. It is important that you be there, but the hours may seem long, especially when the system is running well and there is little maintenance activity being performed. 

Energy Demand Forecasting -- The most important task performed is the projection of the customer’s energy requirements on a near-term basis. This is normally a projection of hourly megawatts of power for the next 24 to 96 hours. This projection is based upon historical load patterns and weather projections. Depending on where located above the equator, your load patterns can very significantly between winter, summer, spring and fall due to time sunrise and sunset and temperature conditions. Other factors affecting load patterns include cloud cover, day of the week and holidays. 


Energy Demand

Market Authority -- Market and Interchange Transaction Operators’ focus is placed on the wholesale energy market. They also work with the rest of operations to mitigate problems in the event of a security violation.


European Electricity Spot Market

Monitor Load to Capacity Resources -- The next step is to enter the projected purchases and sales on an hourly total basis into the automatic generation control program. This automatic generation control program performs balancing between generation, customer load requirements and purchases and sales and provides its frequency bias obligation. 

Monitoring Frequency -- This automatic generation control program also maintains a constant frequency and accurate time. If a party's generation is unable to perform its balancing obligation, frequency will deviate from 60 hertz and create an inaccuracy in time. When a frequency deviation occurs, the automatic generation control program applies its frequency bias calculation to bring frequency back to normal. 

Energy Reconciliation -- Another task, energy accounting, is an after-the-fact reconciliation of scheduled transactions with other utilities and the actual deliveries including losses. During this reconciliation, differences between scheduled the actual deliveries are determined and accumulated inadvertent energy. In most cases, the inadvertent energy is corrected through a payback arrangement. 

Power Outage Coordination -- Another task, just as important, is the scheduling of equipment outages for maintenance and other reasons. This requires communication and coordination with the maintenance people requesting authorization to work on electrical equipment. In order to perform work on an electrical system, a permit or clearance is required. In order to provide a clearance or permit, studies must be done to determine the consequence of taking the equipment out of services during the requested time frame. 


Technician Repairing a Transmission Line during Power Outage

Many times this requires coordination with the regional reliability authority. These studies are performed on a first contingency basis. In other words, the system must remain stable and secure during the outage period with the worst contingency occurring. This analysis is done with modern computerized models of the electric transmission systems. The outage scheduling task is performed by the group responsible for transmission operations and requires coordination with the reliability coordinator. 


Commit Capacity Resources -- Prior to committing the generating units to meet these energy requirements, a predefined amount of operating reserves is added to the projected energy requirements. Then the generating units are committed to meet these requirements (including reserves) for the next 24 to 96 hours based on discussions with the generating stations to determine their availability. This step is referred to as Unit Commitment. 

Develop Transactions with Other Power Producers  -- Normally, after committing the units and determining whether there is access capacity available for sale or there is a need to purchase to cover a deficiency in generating capacity, an individual then goes to the market and sells the excess or purchases the deficiency. These transactions are reviewed by the interchange authority. 


Advanced Tools for Power System Operator 


SCADA -- Analog information is gathered from remote terminal units from substations throughout the system by the SCADA scanning program. This information is made available to you through real-time and study versions of diagrams and tabular displays. 


Power System SCADA

Information includes bus voltages, line flow, and transformer flow and transformer oil temperature. With the newest control systems you will have physical control of essentially all controllable apparatus and functions. SCADA means supervisory control and data acquisition. 

Power System Simulators -- Energy management computer systems have other tools available to assist in performing your duties. These tools include mathematical models of the transmission system such as power flow studies, contingency evaluation studies and state estimation programs. 


Power System Simulators


The state estimation program, which calculates unknown system values, checks actual metered values for relative accuracy and calculates unknown metered quantities and is the most useful tool today. In addition, to avoid major outages, operating guides based on detailed studies, are also available to the system operator and outage coordinator. These operating guides are reviewed and approved by the reliability authority prior to being made available for use by the system operator. 

System Controllers -- Control functions are available to you as the system operator. Using the one-line diagrams and tabular displays and these control functions, you have ability to turn breaker and automatic reclosure on or off, control first and second end reclosing, and turn transfer trip relaying on or off. 

Protection Relays in Power System

Protection Relays


Protection Relay Functions

Protective devices that are used to protect electrical systems & equipment usually have one or more functions. Protective relays detect fault conditions and initiate circuit breaker trips to de-energize faulted equipment or circuits before serious damage can occur.

Basic function includes:
  1. Alert operating personnel to abnormal or potentially dangerous conditions or to the fact that a trip circuit has been energized.
  2. Automatically interrupt current flow to equipment if a potentially dangerous fault occurs.
  3. Automatically energize stand-by equipment as necessary to maintain system operation.
Sample Protection Relay Scheme

Common Characteristics of Protection Relays
  1. Reliability - Must operate when they are supposed to.
  2. Speed - Must be able to respond to a fault and isolate the affected equipment before damage can occur.
  3. Simplicity - Important for economic reasons as well as for maintenance efficiency.
  4. Sensitivity - Able to detect a fault as soon as it occursSelectivity - Should isolate only the faulted area.

Coordination

The relay co-ordination refers to the tripping of protecting relay in a proper sequence or order in electrical power system. This is to avoid tripping of un-faulted branch in the system. Relay co-ordination is required to isolate the faulty part with minimized relay & circuit breaker operation.

The importance of coordination in Protection Relay Scheme

Zones of Protection

Protected zones are established to protect certain components such as:
  • Generator
  • Transformers
  • Buses
  • Motors

ANSI Device Numbers

Each relay in a protection scheme has a specific functions and responds to a certain type of fault encountered in the power system. 

ANSI device numbers. In the design of electrical power systems, the ANSI standard device numbers (ANSI /IEEE Standard C37. 2 Standard for Electrical Power System Device Function Numbers, Acronyms, and Contact Designations ) identifies the features of a protective device such as a relay or circuit breaker.

Sample ANSI Device Designation

Basic Protection  


Characteristic
Protection
ANSI Device Number
A
The differential relay operates whenever there is a difference between the currents going into and coming out of any of the three phases.
Generator Short Circuit Protection
87
B
The result of a breakdown of the insulation in one of the phase windings
Generator Ground Fault Protection
64
C
Can be caused by tripping of a field breaker or by short circuits in the field (rotor) windings
Generator Loss of Field Excitation
40
D
It can happen when there is not enough steam flow to the turbine to drive the generator
Generator Motoring Protection
32
E
The relay trips the feeder breaker if an overcurrent condition occurs
Bus Overcurrent Protection
51
F
If an undervoltage condition occurs, the relay trips the circuit breakers to the loads on the bus that could be damaged by the undervoltage condition.
Bus Undervoltage Protection
27
G
If a ground occurs, the relay senses it and operates an alarm to alert operating personnel
Bus Ground Detection and Protection
64
H
If a fault occurs in the transformer, current flow through the current transformers becomes unbalanced.
Transformer Differential Protection Relay
87



Electronic Protection Relays


Nowadays, most of the protection relays used are electronic type due to its greater precision and allow closer system coordination. Also, the accuracy of solid state relay is greater than the electromechanical relays. 

One of the great features of solid state relays is it keeps the history of the operation. And when the relays are networked together, it can be synchronized to a master clock and all significant events can be recorded. 

Communication


The electronic protection relay can be connected with the use of RS232 or RS 485 connections with a local computer. These can be used for relay configuration, monitoring and troubleshooting. Most especially, the relays now provide Ethernet capability for networking. Communication between relays in the power system makes it possible to exchange input and outputs thru a communication link, thus reducing the amount of hard wiring. 


Protection Relay Communication 

Friday, November 15, 2019

Example: Double Line to Ground Fault Calculation

Double Line to Ground Fault

Double line to ground fault when both phases on the three phase line are accidentally connected to the ground. In this case, fault current will flow from the line to the ground within the involved phases, say, Phase B and Phase C.

Example:

From the figure given below, assume that the generator is solidly grounded and neglect the fault impedance. Determine the phase currents and phase voltage when Line to Line fault occurs in the system.



The article Unbalanced Fault Analysis: Double Line to Ground Fault explained and derived that all sequence network are connected in parallel to satisfy the conditions in this type of fault.

Thus,

Equivalent Sequence Network of DLG Fault


By circuit analysis we can get the equivalent impedance as,

  • Zeq = Z1//Z2 + Z0 = j0.1//j0.1 + j0.25 = j0.3 pu
Let the value of Ea = 1 (angle 0). 

Therefore, the value of the +s current is, 
  • Ia1 = Ea / j0.3 pu = -j 3.33 pu

By current division, we can say that the respective values of -s and 0s current as,
  • Ia2 = Ia0 = - (-j 3.33/2) = j 1.67 pu
Summary of sequence currents, 
  • Ia1 = -j 3.33 pu
  • Ia2 = j 1.67 pu
  • Ia0 = j 1.67 pu
  • Fault Current at Phase A, IA = 0.
  • Fault Current at Phase B, IB = 5 (angle -30) pu
  • Fault Current at Phase C, IC = 5 (angle -150) pu
Apply base values, 
  • Choose: Sb = 20 MVA and kVb = 13.8 kV, then... 
  • Ea = 20 MVA/ 20 MVA = 1 (angle 0) per unit.
  • Ibase = 20 MVA / (1.73 x 13.8 kV)
  • Ibase = 0.837 kA

Therefore the actual Fault Current Values are, 
  • IA = 0
  • IB = 5 x 0.837 kA = 4.185 kA (angle -30) 
  • IC = 5 x 0.837 kA = 4.185 kA (angle -150) 
Voltage Values,

Based on the equivalent sequence network,

  • Va2 = Z2 x Ia2 = j 0.1 x (-j 1.67) = 0.167 (angle 0) pu
  • Therefore it follows that,  Va2 = Va0 = 0.167 (angle 0) pu
  • VA = 0.501 (angle 0) pu
  • VB = VC = 0
Apply voltage base, 
  • VA = 0.501 x (13.8 kV/1.73) = 3.996 kV
  • VB = VC = 0

Unbalanced Fault Analysis: Double Line to Ground Fault



Double line to ground fault when both phases on the three phase line are accidentally connected to the ground. In this case, fault current will flow from the line to the ground within the involved phases, say, Phase B and Phase C. 

Double Line to Ground Fault - Phase B and Phase C

From this scenario, the system parameters can be considered as follows: 

  • Fault current at phase A = 0 (since no fault current is flowing in phase A)
  • Fault current at phase B = If-B
  • Fault current at phase C = If-C. 
  • Here we can also see that voltages at phase B and phase C will be equal to zero (neglecting ground impedance). 


Using the symmetrical components equation matrix formula plus the values we get from the above conditions, we can plot the current equation as follows: 


Symmetrical Components Matrix Equation of Current for DLG Fault

From this matrix equation, we can get the following values: 

  • Ia0 + Ia1 + Ia2 = 0 (the sum of all sequence currents is equal to zero)
  • IA = 0 (since there is no fault current flowing in Phase A during the fault. 

In the same case, the symmetrical components for Voltages is: 

Symmetrical Components Matrix Equation of Voltage for DLG Fault
From the above matrix equation, we can get the following voltage values:

  • Va0 = Va1 = Va2 = Va/ 3
From the obtained voltage and current values, we can demonstrate it using the following sequence network,


Equivalent Sequence Network of DLG Fault

This figure satisfies the values obtained based on the given conditions.

For sample calculation, see: Example: Double Line to Ground Calculation

Saturday, October 26, 2019

How to Control Lighting Circuit in 2 or More Locations?



In building wiring, it is important to have an effective and flexible control of the electric lighting system. Irrespective to the type and purpose of the premises, the lighting controls in the building marks the beginning of the efficient use of energy.

For example in staircase, the efficient use of energy starts with the conventional control that enables a person to control the light whether he is in the upper floor or in the lower floor. A person can switch ON the light as soon as he climb the stairs and then he can also switch it OFF when he is already in the upper floor.

Staircase

The same thing happened in the long hallway, there must be a means where the occupant can switch ON and OFF the lights at any strategic point along the hallway.

Long Hallway

And so how can we make it possible?

This article covers only the conventional method, without the use of building automation system.

Controlling the light in two locations

Type of switch needed:

  • 2 units of 2-way switch (called 3-way switch in North America). 


Terminals of 2-way switch

As we can see the 2 way switch has 3 terminals namely L1, L2 and common. The connection of the circuit can be done using the following steps:
  1. From the source, connect the LINE/HOT conductor to the common terminal of the first switch. 
  2. From the LINE terminal of the light, use conductor to connect it to the COMMON terminal of second 2- way switch. 
  3. Connect L1 of the first switch to L1 of the second switch. 
  4. Connect L2 of the first switch to L2 of the second switch.

This can be best demonstrated in the diagram below, 

Light controlled in 2 locations 

Note: Always connect the switch in the LINE or HOT wire for safety. Read this article for more information >> How to conduct polarity testing?

Controlling the lighting circuit in 3 or more locations

Type of switch needed:
  • 2 units of 2-way switch (called 3-way switch in North America). 
  • 1 unit - Intermediate Switch (called 4-way switch in North America).
This method can be used efficiently in long hallways where a switching point is a must in a strategic point along the hallway. Here, the use of intermediate switch (4-way switch in North America) is needed.


Intermediate Switch


This type of switch has 4 terminals and to be installed in between 2-way switches. For example, if there is a need to control in 3 locations, the 2-way switches will be installed in each end while the intermediate switch will be installed in the middle.

If 4 locations is required then there will be 2 intermediate switch in the middle.

The connection of the circuit can be done using the following steps:
  1. From the source, connect the LINE/HOT conductor to the common of the first 2-way switch. 
  2. From the LINE/HOT terminal of the light, connect it to the common terminal of the second 2-way switch which is to be installed in the other end of the circuit. 
  3. Connect the intermediate switch in the middle. 
This can be best demonstrated in the diagram below, 

Light Controlled in 3 locations. 

Note that the 1- unit intermediate switch is just a minimum.  In case the requirement is controlling light in 4 locations, then 2 units intermediate switch are needed. 

In general, we need to understand that the Intermediate Switch is always installed in the middle no matter how many units are required. The 2-way switch on the other hand need to be installed in both ends of the circuit.

This can be further explained in the diagram below,

Light Controlled in 4 locations. 

building wiring, 2 way switch, 3 way switch, intermediate switch, switching control, domestic home wiring, how to control light in 2 locations, how to control lights in 3 locations, how to control light in 4 or more locations.

Saturday, October 05, 2019

Example: Line to Line Fault Calculation


The Line-to-Line fault on a transmission line happens when two current carrying conductors in a three phase system accidentally comes in contact with each other. For this reason, the protective devices of a power system needs to be accurately responsive to avoid severe damage in the system.

See Unbalanced Fault Analysis: Line to Line Fault

For this reason, it is very important to know the procedure of line to line fault calculation in an unbalanced system.

Related articles:



Example: 

From the figure given below, assume that the generator is solidly grounded and neglect the fault impedance. Determine the phase currents and phase voltage when Line to Line fault occurs in the system.



Solution:

From the article Unbalanced Fault Analysis: Line to Line Fault, we know that the positive and negative sequence network is connected in parallel and the zero sequence network is not involved in this type of fault.

From the article Unbalanced Fault Analysis: Line to Line Fault, we know that the equivalent positive sequence network is,

Positive Sequence Network
While the negative sequence network is, 

Negative Sequence Network
Connecting this networks in parallel and getting the Thevenin's equivalent looking at the faulted bus and the reference bus, we can get an equivalent impedance of j 0.25 // j 0.1. 

Thus, the equivalent sequence network can be simplified as one source of 1 (angle 0) per unit and an equivalent impedance of j 0.71 per unit. 

Thus, the positive sequence current will be, 

If-1 = 1 (angle 0)/ j 0.71 = -j 1.41 or 1.41 (angle -90) per unit

Since If-1 = (-If-2), we can directly conclude that the negative sequence current is, 

If-2 = 1.41 (angle 90) per unit --> See Unbalanced Fault Analysis: Line to Line Fault

Therefore we can summarize the sequence components as follows, 
  • If-1 = 1.41 (angle -90) per unit (positive sequence current)
  • If-2 = 1.41 (angle 90) per unit (negative sequence current)
  • If-0 = 0 (zero sequence network is not involved in Line to Line fault)
By using sequence to phase matrix formula, we can get the values of fault current as, 
  • Fault current at phase A = 0.
  • Fault current at phase B = 2.442 (angle 180) pu
  • Fault current at phase C = 2.442 (angle 0) pu
Consider base values, 

Choose: Sb = 20 MVA and kVb = 13.8 kV

then,
Ea = 20 MVA/ 20 MVA = 1 (angle 0) per unit.
Ibase = 20 MVA / (1.73 x 13.8 kV)
Ibase = 0.837 kA

Therefore the actual values of fault currents are, 
  • Fault current at phase A = 0.
  • Fault current at phase B = 2.04 kA (angle 180)
  • Fault current at phase C = 2.04 kA (angle 0) 
Voltage values,
Analyzing positive sequence network equivalent, 
Vf-1 = 1 (angle 0) - (If-1) * (Z1) = 1 - (-j 1.41) (j 0.25) = 0.6475 (angle 0) (+ sequence voltage)

since Vf-1 = Vf-2; therefore Vf-2 = 0.6475 (angle 0) (neg. sequence voltage) See Unbalanced Fault Analysis: Line to Line Fault

Applying the formula of sequence to phase values matrix we can get: 
  • Voltage at phase A = 1.295 (angle 0) pu
  • Voltage at phase B = 0.647 (angle 180) pu
  • Voltage at phase C = 0.647 (angle 180) pu
Applying base values, 

Vbase = 13.8 kV/ 1.73 

Therefore the actual voltage values are, 
  • Voltage at phase A = 10.31 kV (angle 0)
  • Voltage at phase B = 5.15 kV (angle 180)
  • Voltage at phase C = 5.15 kV(angle 180)

For more details See Unbalanced Fault Analysis: Line to Line Fault

power system analysis, fault calculation, unbalance fault, short circuit analysis

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