Date:

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

Unbalanced Fault Analysis: Line to Line Fault

Line to Line Fault

Line to line fault happens when two phases of the three phase line accidentally connected each other. In this case, the fault current will flow for both phases involved. From the given diagram, the fault occurs in phases B and C while phase A remain unfaulted.

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-A
  • Fault current at phase C = If-C. 
  • Here we can also see that voltages at phase B and phase C will be equal during fault, thus VB = VC.

Using sequence network matrix formula, we can plot the given values as follows. 

Symmetrical Components Matrix Formula in Line to Line Fault
From this matrix equation, we can get a value of,
  • Ia0 = 0
  • Ia1 = - Ia2
Which means, the zero sequence current in a Line to Line fault is not involved while the positive sequence current and negative sequence current is opposing each other. 

In the sames sense, the matrix formula can also gives us the value of the voltages, 

Symmetrical Components Matrix Formula of Line to Line Fault

Note that in this case, VB = VC. By manipulating equations we can get Va1 = Va2. 

From the result of two matrix equations involving current and voltage, we can develop a sequence network which describes as that the positive and negative sequence is in parallel with each other. 

Equivalent Sequence Network for Line to Line Fault

This figure satisfies the equation Va1 = Va2 and Ia1 = (-Ia2).

For sample calculation, see the following article Example: Line to Line Fault Calculation

Thursday, October 03, 2019

Example: Single Line-to-Ground Fault Calculation



The single line-to-ground fault on a transmission line happens when one conductor accidentally to the ground or comes in some cases in contact with the neutral conductor. 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: Single Line to Ground Fault

Indeed, it is very important to know the procedure of single line to ground fault calculation.

Related articles:

  • Double Line to Ground Fault
  • Line to line fault
  • Symmetrical three phase fault


Example: 
Assumed that the generator below is solidly grounded and neglect the fault impedance. Determine the phase currents and phase voltage at faulted location.

Generator

Solution:

Compute the base value:
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

Develop positive sequence network:

System Positive- Sequence Network Equivalent
where,
  • Ea = the generator EMF, which has a value of 1 (angle 0) per unit. 
  • Xd" = the positive sequence impedance of transformer (in subtransient state)
  • If-1 = positive sequence current. 
Develop negative sequence network:

System Negative- Sequence Network Equivalent
where, 
  • If-2 = negative sequence current
  • X2 = negative sequence reactance

Develop zero sequence network:

System Zero-Sequence Network Equivalent
where,
  • If-0 = zero sequence current
  • X2 = zero sequence reactance
By principle, the single line to ground fault will develop and equivalent network where all sequence networks are connected in series.

then, If-1 = 1 (angle 0)/ (Z0 + Z1 + Z2) = -j 2.22 pu or 2.22 (angle -90)
also it means that, If-2 = If-0 = - j 2.22 pu

Multiplying the base value

Ibase = 0.837 kA
then, If-1 = 2.22 (angle -90) * (0.837 kA) = 1.86 kA (angle -90) --> actual value
Since all sequence currents are equal, thus:
  • If-1 = 1.86 kA (angle -90) --> + S
  • If-2 = 1.86 kA (angle -90)  --> -S
  • If-0 = 1.86 kA (angle -90) --> 0S
Applying the formula of sequence to phase values matrix we can get: 
  • Fault current at A = 5.58 kA (angle -90 degrees)
  • Fault current at B = 0
  • Fault current at C = 0
The zero values of phases B and C confirmed that no fault current flows from it during fault condition.

Computing Voltage at the faulted point,



Looking at this diagram, the voltage values of the un-faulted phases (phase B and C) are the only one that has a value while phase A (the faulted phase), its voltage is equal to zero (neglecting impedance).

Thus,

  • Voltage at phase A = 0
  • Voltage at phase B = Vb
  • Voltage at phase C = Vc. 
Analyzing positive sequence network equivalent, 
  • Vf-1 = 1 (angle 0) - (If-1) * (Z1) = 1 - (-j 2.22) (j 0.25) = 0.445 (angle 0) (+ sequence voltage)
Analyzing negative sequence network equivalent, 
  • Vf-2 = 0 - (If2) * (Z2) = 0- (-j2.22) * (j0.1) = 0.222 (angle 180) ( neg. sequence voltage).
Analyzing zero sequence network equivalent, 
  • Vf-0 = 0 - (If-0)* (Z0) = 0- (-j 2.22) * (j0.1) = 0.222 (angle 180) (zero sequence voltage).
Applying the formula of sequence to phase values matrix we can get: 
  • Voltage at phase A = 0
  • Voltage at phase B = 0.667 (angle -120) pu
  • Voltage at phase C = 0.667 (angle 120) pu
Applying base value, 

Vbase = 13.8 kV/ 1.73

Therefore, the actual values of phase voltages are, 
  • Voltage at phase A = 0
  • Voltage at phase B = 5.31 kV (angle -120) 
  • Voltage at phase C = 5.31 kV (angle 120)

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