The Solid State Relay (Static Relay) Overview (on photo: Basler Electric BE1-27 Solid State Protective Relay, Over/Under Voltage)

History of Relay

The static relay is the next generation relay after electromechanical type.The Solid Static relays was first introduced in 1960’s. The term ‘static’ implies that the relay has no moving mechanical parts in it.

Compared to the Electromechanical Relay, the Solid Static relay has longer life-span, decreased noise when operates and faster respond speed.

However, it is not as robust as the Electromechanical Relay.

Static relays were manufactured as semiconductor devices which incorporate transistors, ICs, capacitors, small microprocessors etc.

The static relays have been designed to replace almost all the functions which were being achieved earlier by electromechanical relays.

Measuring principles

The working principle of the Solid Static relays is similar to that of the Electromechanical Relay which means the Solid Static relays can perform tasks that the Electromechanical Relay can perform.

The Solid Static relays use analogue electronic devices instead of magnetic coils and mechanical components to create the relay characteristics. The measurement is carried out by static circuits consisting of comparators, level detectors, filter etc while in a conventional electromagnetic relay it is done by comparing operating torque (or force) with restraining torque (or force). The relaying quantity such as voltage/current is rectified and measured.

In a solid state relay, the incoming voltage and current waveforms are monitored by analog circuits, not recorded or digitized. The analog values are compared to settings made by the user via potentiometers in the relay, and in some case, taps on transformers.

In some solid state relays, a simple microprocessor does some of the relay logic, but the logic is fixed and simple.

For instance, in some time over current solid state relays, the incoming AC current is first converted into a small signal AC value, and then the AC is fed into a rectifier and filter that converts the AC to a DC value proportionate to the AC waveform. An op-amp and comparator is used to create a DC that rises when a trip point is reached. Then a relatively simple microprocessor does a slow speed A/D conversion of the DC signal, integrates the results to create the time-over current curve response, and trips when the integration rises above a set point.

Though this relay has a microprocessor, it lacks the attributes of a digital/numeric relay, and hence the term “microprocessor relay” is not a clear term.

Function of Relay

Early versions used discrete devices such as transistors and diodes in conjunction with resistors, capacitors, inductors, etc., but advances in electronics enabled the use of linear and digital integrated circuits in later versions for signal processing and implementation of logic functions.

While basic circuits may be common to a number of relays, the packaging was still essentially restricted to a single protection function per case, while complex functions required several cases of hardware suitably interconnected.

Basler Electric BE1-27 Solid State Protective Relay, Over/Under Voltage

User programming was restricted to the basic functions of adjustment of relay characteristic curves.

Therefore it can be viewed in simple terms as an analogue electronic replacement for electromechanical relays, with some additional flexibility in settings and some saving in space requirements.

In some cases, relay burden is reduced, making for reduced CT/VT output requirements. In a static relay there is no armature or other moving element and response is developed by electronic, magnetic or other components without mechanical motion.

The constraint in static relay is limited function/features.

In the last decade, some microprocessors were introduced in this relay to achieve the functions like:

1. Fuse failure features
2. Self check feature
3. Dead Pole detection and
4. Carrier aided protection features

Operation of Relay

The essential components of static relays are shown in figure below. The output of CT and PT are not suitable for static components so they are brought down to suitable level by auxiliary CT and PT. Then auxiliary CT output is given to rectifier.

Rectifier rectifies the relaying quantity i.e., the output from a CT or PT or a Transducer.

Solid state relay - Operation

The rectified output is supplied to a measuring unit comprising of comparators, level detectors, filters, logic circuits.

The output is actuated when the dynamic input (i.e., the relaying quantity) attains the threshold value. This output of the measuring unit is amplified by amplifier and fed to the output unit device, which is usually an electromagnetic one.

The output unit energizes the trip coil only when relay operates.

 Advantages of Solid State Relay

• Static Relay burden is less than Electromagnetic type of relays. Hence error is less.
• Low Weight
• Required Less Space which results in panel space saving.
• Arc less switching
• No acoustical noise.
• Multi-function integration.
• Fast response.
• Long life (High Reliability): more than 109 operations
• High Range of Setting compared to electromechanical Relay
• More Accurate compared to electromechanical Relay
• Low Electromagnetic Interference.
• Less power consumption.
• Shock and vibration resistant
• No contact bounce
• Microprocessor compatible.
• Isolation of Voltage

No moving parts: There are no moving parts to wear out or arcing contacts to deteriorate that are often the primary cause of failure with an Electro Mechanical Relay.

No mechanical contact bounce or arcing: A solid-state relay doesn’t depend on mechanical forces or moving contacts for its operation but performs electronically. Thus, timing is very accurate even for currents as low as the pickup value. There is no mechanical contact bounce or arcing, and reset times are extremely short.

Low input signal levels: Ideal for Telecommunication or microprocessor control industries. Solid state relays are fast becoming the better choice in many applications, especially throughout the telecommunication and microprocessor control industries.

Cost Issues: In the past, there has been a rather large gap between the price of an electromechanical relay and the price of a solid state relay. With continual advancement in manufacturing technology, this gap has been reduced dramatically making the advantages of solid state technology accessible to a growing number of design engineers.

References

• Handbook of Switchgear –Bhel
• Digital/Numerical Relays -T.S.M. Rao

Few Words About Digital Protection Relay (on photo: Siprotec microprocessor based protective relay)

History of Protective Relay

Around 1980s the digital relay entered the market. Compared to the Solid State Relay, the digital relay takes the advantages of the development of microprocessors and microcontrollers. Instead of using analog signals, the digital relay converts all measured analog quantities into digital signals.

In Digital Relay Microprocessors and micro controllers are used in replacement of analogue circuits used in static relays to implement relay functions. Digital protection relays introduced in 1980.

However, such technology will be completely superseded within the next five years by numerical relays.

By the mid-1990s the solid state and electromechanical relay had been mostly replaced by digital relay in new construction. In distribution applications, the replacement by the digital relay proceeded a bit more slowly.

While the great majority of feeder relays in new applications today are digital, the solid state relay still sees some use where simplicity of the application allows for simpler relays, and which allows one to avoid the complexity of digital relays.

Measuring principles

Compared to static relays, digital relays introduce Analogue to Digital Convertor (A/D conversion) of all measured analogue quantities and use a microprocessor to implement the protection algorithm.

The microprocessor may use some kind of counting technique, or use the Discrete Fourier Transform (DFT) to implement the algorithm.

Function of Relay

The functionality tends therefore to be limited and restricted largely to the protection function itself. Additional functionality compared to that provided by an electromechanical or static relay is usually available, typically taking the form of a wider range of settings, and greater accuracy.

A communications link to a remote computer may also be provided.

SEPAM relays in medium voltage switchgear

The limited power of the microprocessors used in digital relays restricts the number of samples of the waveform that can be measured per cycle. This, in turn, limits the speed of operation of the relay in certain applications. Therefore, a digital relay for a particular protection function may have a longer operation time than the static relay equivalent.

However, the extra time is not significant in terms of overall tripping time and possible effects of power system stability.

Operation of Relay

Digital relay consists of:

1. Analogue input subsystem,
2. Digital input subsystem,
3. Digital output subsystem,
4. A processor along with RAM (data scratch pad),
5. main memory (historical data file) and
6. Power supply

Operation diagram of digital relay

Digital relaying involves digital processing of one or more analog signals in three steps:

1. Conversion of analogue signal to digital form
2. Processing of digital form
3. Boolean decision to trip or not to trip.

 Advantages of Digital Relay

• High level of functionality integration.
• Additional monitoring functions.
• Functional flexibility.
• Capable of working under a wide range of temperatures.
• They can implement more complex function and are generally more accurate
• Self-checking and self-adaptability.
• Able to communicate with other digital equipment (pear to pear).
• Less sensitive to temperature, aging
• Economical because can be produced in volumes
• More Accurate.
• plane for distance relaying is possible
• Signal storage is possible

Limitations of Digital Relay

• Short lifetime due to the continuous development of new technologies.
• The devices become obsolete rapidly.
• Susceptibility to power system transients.
• As digital systems become increasingly more complex they require specially trained staff for Operation.
• Proper maintenance of the settings and monitoring data.

References

• Handbook of Switchgear –Bhel
• Digital/Numerical Relays -T.S.M. Rao

Flexibility and Reliability of Numerical Protection Relay (on photo: ABB's numerical relay type SPAD 330 C designed to be used as a fast interwinding short-circuit and interturn fault protection for two-winding power transformers and power plant generator-transformer units.

History of Relay

The first protection devices based on microprocessors were employed in 1985. The widespread acceptance of numerical technology by the customer and the experiences of the user helped in developing the second generation numerical relays in 1990.

Conventional electromechanical and static relays are hard wired relays. Their wiring is fixed, only their setting can be manually changed. Numeric relays are programmable relays. The characteristics and behaviour of the relay are can be programmed.

First generation numerical relays were mainly designed to meet the static relay protection characteristic, whereas modern numeric protection devices are capable of providing complete protection with added functions like control and monitoring.

Numerical protection devices offer several advantages in terms of protection, reliability, troubleshooting and fault information.

Typically, they use a specialized digital signal processor (DSP) as the computational hardware, together with the associated software tools.

Measuring Principles

The input analogue signals are converted into a digital representation and processed according to the appropriate mathematical algorithm. Processing is carried out using a specialized microprocessor that is optimized for signal processing applications, known as a digital signal processor or DSP for short. Digital processing of signals in real time requires a very high power microprocessor.

The measuring principles and techniques of conventional relays (electromechanical and  static) are fewer than those of the numerical technique, which can differ in many aspects like the type of protection algorithm used, sampling, signal processing, hardware selection, software discipline, etc.

These are microprocessor-based relays in contrast to other relays that are electromechanically controlled.

Function of Relay

Numerical Directional Overcurrent Relay (Relay operating time is determined by selecting definite time characteristics or one of the four inverse time characteristics i.e. 3s normal inverse, 1.3s normal inverse, very inverse and extremely inverse.)

Modern power system protection devices are built with integrated functions. Multifunction like protection, control, monitoring and measuring are available today in numeric power system protection devices.

Also, the communication capability of these devices facilitates remote control, monitoring and data transfer.

Traditionally, electromechanical and static protection relays offered single-function, single characteristics, whereas modern numeric protection offers multi-function and multiple characteristics. Numerical protection devices offer several advantages in terms of protection, reliability, and trouble shooting and fault information.

Numerical protection devices are available for generation, transmission and distribution systems.

Numerical relays are microprocessor based relays and having the features of recording of parameter used as disturbance recorder flexibility of setting & alarms & can be used one relay for all type of protections  of one equipment hence less area is required.

Numeric relays take the input analog quantities and convert them to numeric values.  All of the relaying functions are performed on these numeric values.

The following sections cover:

1. Relay hardware,
2. Relay software,
3. Multiple protection characteristics,
4. Adaptive protection characteristics,
5. Data storage,
6. Instrumentation feature,
7. Self-check feature,
8. Communication capability,
9. Additional functions,
10. Size and cost-effectiveness.

The disadvantages of a conventional electromechanical relay are overcome by using microcontroller for realizing the operation of the relays.

Microcontroller based relays perform very well and their cost is relatively low.

Operation of Relay

A current signal from CT is converted into proportional voltage signal using I to V converter.

The AC voltage proportional to load current is converted into DC using precision rectifier and is given to multiplexer (MUX) which accepts more than one input and gives one output.

Microprocessor sends command signal to the multiplexer to switch on desired channel to accept rectified voltage proportional to current in a desired circuit.

Microprocessor Based Numerical Relay

Microprocessor Relay - Operation diagram

Output of Multiplexer is fed to analog to digital converter (ADC) to obtain signal in digital form. Microprocessor then sends a signal ADC for start of conversion (SOC), examines whether the conversion is completed and on receipt of end of conversion (EOC) from ADC, receives the data in digital form.

The microprocessor then compares the data with pick-up value.

If the input is greater than pick-up value the microprocessor send a trip signal to circuit breaker of the desired circuit.

In case of instantaneous overcurrent relay there is no intentional time delay and circuit breaker trips instantly. In case of normal inverse, very inverse, extremely inverse and long inverse overcurrent relay the inverse current-time characteristics are stored in the memory of microprocessor in tabular form called as look-up table.

 Advantages of Numerical relays

Compact Size

Electromechanical Relay makes use of mechanical comparison devices, which cause the main reason for the bulky size of relays. It uses a flag system for the indication purpose whether the relay has been activated or not.

While numerical relay is in compact size and use indication on LCD for relay activation.

Flexibility

A variety of protection functions can be accomplished with suitable modifications in the software only either with the same hardware or with slight modifications in the hardware.

Reliability

A significant improvement in the relay reliability is obtained because the use of fewer components results in less interconnections and reduced component failures.

Multi Function Capability

Traditional electromechanical and static protection relays offers single-function and single characteristics. Range of operation of electromechanical relays is narrow as compared to numerical relay.

Different types of relay characteristics

It is possible to provide better matching of protection characteristics since these characteristics are stored in the memory of the microprocessor.

Digital communication capabilities

The microprocessor based relay furnishes easy interface with digital communication equipment. Fibre optical communication with substation LAN.

Modular frame

The relay hardware consists of standard modules resulting in ease of service.

Low burden

The microprocessor based relays have minimum burden on the instrument transformers.

Sensitivity

Greater sensitivity and high pickup ratio.

Speed

With static relays, tripping time of ½ cycle or even less can be obtained.

Fast Resetting

Resetting is less.

Data History

Availability of fault data and disturbance record. Helps analysis of faults by recording details of:

1. Nature of fault,
2. Magnitude of fault level,
3. Breaker problem,
4. C.T. saturation,
5. Duration of fault.

Auto Resetting and Self Diagnosis

Electromechanical relay do not have the ability to detect whether the normal condition has been attained once it is activated thus auto resetting is not possible and it has to be done by the operating personnel, while in numerical relay auto resetting is possible.

Limitations of Numerical Relay

Protection quality

Numerical relay offers more functionality, and greater precision. Unfortunately, that does not necessarily translate into better protection.

Faster Decisions

Numerical Relay can make faster decisions. However, in the real world, faster protection itself is of no value because circuit breakers are still required to interrupt at the direction of the protective equipment, and the ability to make circuit breakers interrupt faster is very limited.

Risk Of Hacking

Numerical Relay protection often relies on non-proprietary software, exposing the system to potential risk of hacking.

Interference

Numerical Relay protection sometimes has exposure to externally-sourced transient interference that would not affect conventional technology.

Failure Impact

Numerical Relay protection shares common functions. This means that there are common failure modes that can affect multiple elements of protection.

For example, failure of a power supply or an input signal processor may disable an entire protective device that provides many different protection functions.

But it remains something to be aware of.

Functions

A multifunction numeric relay can provide three phase, ground, and negative sequence directional or non-directional overcurrent protection with four shot recloser, forward or reverse power protection, breaker failure, over/under frequency, and over/under voltage protection, sync check, breaker monitoring and control.

It would take 10 – 11 single function Solid state or electromechanical relays at least 5 to 6 times the cost.

Additionally Numeric relays have communications capabilities, sequence-of-events recording, fault reporting, rate-of-change frequency, and metering functions, all in an integrated system.

Differences Between Earthed and Unearthed Cables

Introduction

In HT electrical distribution, the system can be earthed or unearthed.

The selection of unearthed or earthed cable depends on distribution system. If such system is earthed, then we have to use cable which is manufactured for earthed system. (which the specifies the manufacturer). If the system is unearthed then we need to use cable which is manufactured for unearthed system.

The unearthed system requires high insulation level compared to earthed system.

For earthed and unearthed XLPE cables, the IS 7098 part2 1985 does not give any difference in specification. The insulation level for cable for unearthed system has to be more.

Earthed System

Earlier the generators and transformers were of small capacities and hence the fault current was less. The star point was solidly grounded. This is called earthed system.

Where in unearthed system, (if system neutral is not grounded) phase to ground voltage can be equal to phase to phase voltage. In such case the insulation level of conductor to armor should be equal to insulation level of conductor to conductor.

In an earthed cable, the three phase of cable are earthed to a ground. Each of the phases of system is grounded to earth.

Example: 1.9/3.3 KV, 3.8/6.6 KV system

Unearthed System

Today generators of 500MVA capacities are used and therefore the fault level has increased. In case of an earth fault, heavy current flows into the fault and this lead to damage of generators and transformers. To reduce the fault current, the star point is connected to earth through a resistance. If an earth fault occurs on one phase, the voltage of the faulty phase with respect to earth appears across the resistance.

Therefore, the voltage of the other two healthy phases with respect to earth rises by 1.7 times.

In an unearth system, the phases are not grounded to earth .As a result of which there are chances of getting shock by personnel who are operating it.

Example: 6.6/6.6 KV, 3.3/3.3 KV system.

Unearthed cable has more insulation strength as compared to earthed cable. When fault occur phase to ground voltage is √3 time the normal phase to ground voltage. So if we used earthed cable in unearthed System, It may be chances of insulation puncture.

So unearthed cable are used. Such type of cable is used in 6.6 KV systems where resistance type earthing is used.

Thumb Rule

As a thumb rule we can say that 6.6KV unearthed cable is equal to 11k earthed cable i.e 6.6/6.6kv Unearthed cable can be used for 6.6/11kv earthed system.

Because each core of cable have the insulation level to withstand 6.6kv so between core to core insulation level will be 6.6kV+6.6kV = 11kV

For transmission of HT, earthed cable will be more economical due to low cost where as unearthed cables are not economical but insulation will be good.

In such cases, unearthed cables may be used so that the core insulation will have enough strength but current rating is de-rated to the value of earthed cables.

But it is always better to mention the type of system earthing in the cable specification when ordering the cables so that the cable manufacturer will take care of insulation strength and de rating.

Comparison of Protection Relay Types (photo by switchgearsupport.com)

Comparison Table

This comparison summarize characteristics of all protection relay types described in previously published technical articles:

1. Using Protective Relay For Fighting Against Faults
2. The Good Old Electromechanical Protective Relay
3. The Solid State Relay (Static Relay) Overview
4. Few Words About Digital Protection Relay
5. Flexibility and Reliability of Numerical Protection Relay

Relay’s Nomenclature as per ANSI

Relays for Transmission and Distribution Lines protection

References:
- Handbook of Switchgear – Bhel
- Digital/Numerical Relays - T.S.M. Rao

Types, applications and connections of Overcurrent relay (on photo: Transmission lines from Gillam to Churchill)

Types of protection

Protection schemes can be divided into two major groupings:

1. Unit schemes
2. Non-unit schemes

1. Unit Type Protection

Unit type schemes protect a specific area of the system, i.e., a transformer, transmission line, generator or bus bar.

Any deviation from this must indicate an abnormal current path. In these schemes, the effects of any disturbance or operating condition outside the area of interest are totally ignored and the protection must be designed to be stable above the maximum possible fault current that could flow through the protected area.

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2. Non unit type protection

The non-unit schemes, while also intended to protect specific areas, have no fixed boundaries. As well as protecting their own designated areas, the protective zones can overlap into other areas. While this can be very beneficial for backup purposes, there can be a tendency for too great an area to be isolated if a fault is detected by different non unit schemes.

The most simple of these schemes measures current and incorporates an inverse time characteristic into the protection operation to allow protection nearer to the fault to operate first.

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2.1 Overcurrent protection

This is the simplest of the ways to protect a line and therefore widely used.

It owes its application from the fact that in the event of fault the current would increase to a value several times greater than maximum load current. It has a limitation that it can be applied only to simple and non costly equipments.

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2.2 Earth fault protection

The general practice is to employ a set of two or three overcurrent relays and a separate overcurrent relay for single line to ground fault. Separate earth fault relay provided makes earth fault protection faster and more sensitive.

Earth fault current is always less than phase fault current in magnitude.

Therefore, relay connected for earth fault protection is different from those for phase to phase fault protection.

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Various types of Line Faults

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Overcurrent Relay Purpose and Ratings

A relay that operates or picks up when it’s current exceeds a predetermined value (setting value) is called Overcurrent Relay.

For feeder protection, there would be more than one overcurrent relay to protect different sections of the feeder. These overcurrent relays need to coordinate with each other such that the relay nearest fault operates first.

Use time, current and a combination of both time and current are three ways to discriminate adjacent overcurrent relays.

OverCurrent Relay gives protection against:

Overcurrent includes short-circuit protection, and short circuits can be:

1. Phase faults
2. Earth faults
3. Winding faults

Short-circuit currents are generally several times (5 to 20) full load current. Hence fast fault clearance is always desirable on short circuits.

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Primary requirement of Overcurrent protection

The protection should not operate for starting currents, permissible overcurrent, current surges. To achieve this, the time delay is provided (in case of inverse relays).

Overcurrent relay is a basic element of overcurrent protection.

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Purpose of overcurrent Protection

These are the most important purposes of overcurrent relay:

• Detect abnormal conditions
• Isolate faulty part of the system
• Speed Fast operation to minimize damage and danger
• Discrimination Isolate only the faulty section
• Dependability / reliability
• Security / stability
• Cost of protection / against cost of potential hazards

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Overcurrent Relay Ratings

In order for an overcurrent protective device to operate properly, overcurrent protective device ratings must be properly selected. These ratings include voltage, ampere and interrupting rating.

Current limiting can be considered as another overcurrent protective device rating, although not all overcurrent protective devices are required to have this characteristic

Voltage Rating: The voltage rating of the overcurrent protective device must be at least equal to or greater than the circuit voltage. The overcurrent protective device rating can be higher than the system voltage but never lower.

Ampere Rating: The ampere rating of a overcurrent protecting device normally should not exceed the current carrying capacity of the conductors As a general rule, the ampere rating of a overcurrent protecting device is  selected at 125% of the continuous load current.

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Difference between Overcurrent and Overload protection

Overcurrent protection protects against excessive currents or currents beyond the acceptable current ratings, which are resulting from short circuits, ground faults and overload conditions.

The overcurrent protection is a bigger concept So that the overload protection can be considered as a subset of overcurrent protection.

The overcurrent relay can be used as overload (thermal) protection when protects the resistive loads, etc., however, for motor loads, the overcurrent relay cannot serve as overload protection Overload relays usually have a longer time setting than the overcurrent relays.

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Types of Overcurrent Relay

These are the types of overcurrent relay:

1. Instantaneous Overcurrent (Define Current) Relay
2. Define Time Overcurrent Relay
3. Inverse Time Overcurrent Relay (IDMT Relay)

• Moderately Inverse
• Very Inverse Time
• Extremely Inverse

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1. Instantaneous Overcurrent relay (Define Current)

Definite current relay operate instantaneously when the current reaches a predetermined value.

Instantaneous Overcurrent Relay - Definite Current

• Operates in a definite time when current exceeds its Pick-up value.
• Its operation criterion is only current magnitude (without time delay).
• Operating time is constant.
• There is no intentional time delay.
• Coordination of definite-current relays is based on the fact that the fault current varies with the position of the fault because of the difference in the impedance between the fault and the source
• The relay located furthest from the source operate for a low current value
• The operating currents are progressively increased for the other relays when moving towards the source.
• It operates in 0.1s or less

Application: This type is applied to the outgoing feeders.

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2. Definite Time Overcurrent Relays

In this type, two conditions must be satisfied for operation (tripping), current must exceed the setting value and the fault must be continuous at least a time equal to time setting of the relay.

Definite time of overcurrent relay

Modern relays may contain more than one stage of protection each stage includes each own current and time setting.

1. For Operation of Definite Time Overcurrent Relay operating time is constant
2. Its operation is independent of the magnitude of current above the pick-up value.
3. It has pick-up and time dial settings, desired time delay can be set with the help of an intentional time delay mechanism.
4. Easy to coordinate.
5. Constant tripping time independent of in feed variation and fault location.

Drawback of Relay:

1. The continuity in the supply cannot be maintained at the load end in the event of fault.
2. Time lag is provided which is not desirable in on short circuits.
3. It is difficult to co-ordinate and requires changes with the addition of load.
4. It is not suitable for long distance transmission lines where rapid fault clearance is necessary for stability.
5. Relay have difficulties in distinguishing between Fault currents at one point or another when fault impedances between these points are small, thus poor discrimination.

Application:

Definite time overcurrent relay is used as:

1. Back up protection of distance relay of transmission line with time delay.
2. Back up protection to differential relay of power transformer with time delay.
3. Main protection to outgoing feeders and bus couplers with adjustable time delay setting.

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3. Inverse Time Overcurrent Relays (IDMT Relay)

In this type of relays, operating time is inversely changed with current. So, high current will operate overcurrent relay faster than lower ones. There are standard inverse, very inverse and extremely inverse types.

Discrimination by both ‘Time’ and ‘Current’. The relay operation time is inversely proportional to the fault current.

Inverse Definite Minimum Time (IDMT)

The operating time of an overcurrent relay can be moved up (made slower) by adjusting the ‘time dial setting’. The lowest time dial setting (fastest operating time) is generally 0.5 and the slowest is 10.

• Operates when current exceeds its pick-up value.
• Operating time depends on the magnitude of current.
• It gives inverse time current characteristics at lower values of fault current and definite time characteristics at higher values
• An inverse characteristic is obtained if the value of plug setting multiplier is below 10, for values between 10 and 20 characteristics tend towards definite time characteristics.
• Widely used for the protection of distribution lines.

Based on the inverseness it has three different types:

Inverse types

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3.1. Normal Inverse Time Overcurrent Relay

The accuracy of the operating time may range from 5 to 7.5% of the nominal operating time as specified in the relevant norms. The uncertainty of the operating time and the necessary operating time may require a grading margin of 0.4 to 0.5 seconds.

Normal inverse time Overcurrent Relay is relatively small change in time per unit of change of current.

Application:

Most frequently used in utility and industrial circuits. especially applicable where the fault magnitude is mainly dependent on the system generating capacity at the time of fault.

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3.2. Very Inverse Time Overcurrent Relay

• Gives more inverse characteristics than that of IDMT.
• Used where there is a reduction in fault current, as the distance from source increases.
• Particularly effective with ground faults because of their steep characteristics.
• Suitable if there is a substantial reduction of fault current as the fault distance from the power source increases.
• Very inverse overcurrent relays are particularly suitable if the short-circuit current drops rapidly with the distance from the substation.
• The grading margin may be reduced to a value in the range from 0.3 to 0.4 seconds when overcurrent relays with very inverse characteristics are used.
• Used when Fault Current is dependent on fault location.
• Used when Fault Current independent of normal changes in generating capacity.

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3.3. Extremely Inverse Time Overcurrent Relay

• It has more inverse characteristics than that of IDMT and very inverse overcurrent relay.
• Suitable for the protection of machines against overheating.
• The operating time of a time overcurrent relay with an extremely inverse time-current characteristic is approximately inversely proportional to the square of the current
• The use of extremely inverse overcurrent relays makes it possible to use a short time delay in spite of high switching-in currents.
• Used when Fault current is dependent on fault location
• Used when Fault current independent of normal changes in generating capacity.

Application:

• Suitable for protection of distribution feeders with peak currents on switching in (refrigerators, pumps, water heaters and so on).
• Particular suitable for grading and coordinates with fuses and re closes
• For the protection of alternators, transformers. Expensive cables, etc.

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3.4. Long Time Inverse Overcurrent Relay

The main application of long time overcurrent relays is as backup earth fault protection.

4. Directional Overcurrent Relays

When the power system is not radial (source on one side of the line), an overcurrent relay may not be able to provide adequate protection. This type of relay operates in on direction of current flow and blocks in the opposite direction.

Three conditions must be satisfied for its operation: current magnitude, time delay and directionality. The directionality of current flow can be identified using voltage as a reference of direction.

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Application of Overcurrent Relay

Motor Protection:

• Used against overloads and short-circuits in stator windings of motor.
• Inverse time and instantaneous overcurrent phase and ground
• Overcurrent relays used for motors above 1000 kW.

Transformer Protection:

• Used only when the cost of overcurrent relays are not justified.
• Extensively also at power-transformer locations for external-fault back-up protection.

Line Protection:

• On some sub transmission lines where the cost of distance relaying cannot be justified.
• primary ground-fault protection on most transmission lines where distance relays are used for phase faults.
• For ground back-up protection on most lines having pilot relaying for primary protection.

Distribution Protection:

Overcurrent relaying is very well suited to distribution system protection for the following reasons:

• It is basically simple and inexpensive.
• Very often the relays do not need to be directional and hence no PT supply is required.
• It is possible to use a set of two O/C relays for protection against inter-phase faults and a separate Overcurrent relay for ground faults.

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Connections Of Overcurrent Relay (part 2)

Continued from first part: Types and Applications Of Overcurrent Relay (part 1)

Connections of Overcurrent and Earth Fault Relays

1. 3 Nos O/C relay for overcurrent and earth fault protection

It’s used for:

• 3-phase faults the overcurrent relays in all the 3-phases act.
• Phase to phase faults the relays in only the affected phases operate.
• Single line to ground faults only the relay in the faulty phase gets the fault current and operates.

Even then with 3 overcurrent relays, the sensitivity desired and obtainable with earth leakage overcurrent relays cannot be obtained in as much as the high current setting will have to be necessarily adopted for the overcurrent relay to avoid operation under maximum load condition.

3 Nos O/C Relay for Over Current and Earth Fault Protection

Over current relays generally have 50% to 200% current setting while earth leakages over current relays have either 10% to 40% or 20% to 80% current settings.

A scheme without the neutral conductor will be unable to ensure reliable relay operation in the event of single phase to earth faults because the secondary current in this case (without star-point interconnection) completes its circuit through relay and C.T. windings which present large impedance.

This may lead to failure of protection and sharp decrease in reduction of secondary currents by CTs.

It is not sufficient if the neutral of the CTs and neutral of the relays are separately earthed. A conductor should be run as stated earlier.

2.  3 No O/C Relay+ 1 No E/F Relay for Overcurrent and Earth Fault Protection

The scheme of connection for 3 Nos Over current Relay 1 No Earth Fault Relay is shown in figure below.

3 No O/C Relay+ 1 No E/F Relay for Overcurrent and Earth Fault Protection

Under normal operating conditions the three phase fault conditions and current in the 3-phase are equal and symmetrically displaced by 12 Deg. Hence the sum of these three currents is zero. No current flow through the earth fault relay.

In case of phase to phase faults (say a short between R and Y phases) the current flows from R-phase up to the point of fault and return back through ‘Y’ phase. Thus only O/L relays in R and Y phases get the fault and operate.

Earthing of both will short circuit the E/L relay and make it inoperative for faults.

3.  2 No O/C Relay + 1 No E/F Relay for Over Current and Earth Fault Protection

The two over current relays in R and B phases will respond to phase faults. At least one relay will operate for fault involving two phase.

2 No O/C Relay + 1 No E/F Relay for Over Current and Earth Fault Protection

For fault involving ground reliance is placed on earth fault relay.

This is an economical version of 3-O/L and 1-E/L type of protection as one overcurrent relay is saved. With the protection scheme as shown in Figure complete protection against phase and ground fault is afforded.

Current Transformer Secondary Connections

For protection of various equipment of Extra High Voltage class, the star point on secondary’s of CT should be made as follows for ensuring correct directional sensitivity of the protection scheme.

Transmission Line , Bus Bar and Transformer:

• For Transmission Lines – Line side
• For Transformers – Transformer side
• For Bus bar – Bus side

Transmission Line , Bus Bar & Transformer scheme

Generator Protection:

• Generator Protection – Generator Side

Generator protection scheme

The above method has to be followed irrespective of polarity of CT’s on primary side.

For example, in line protection, if ‘P1’ is towards bus then ‘S2’s are to be shorted and if ‘P2’ is towards bus then ‘S1’s are to be shorted.

Standard Overcurrent and Earth Fault Protection

Defining Size and Location of Capacitor in Electrical System (1)

Content

• Type of Capacitor Bank as per Its Application:

1. Fixed type capacitor banks
2. Automatic type capacitor banks
3. Types of APFC – Automatic Power Factor Correction

1. Star-Solidly Grounded
2. Star-Ungrounded
3. Delta-connected Banks

1. Parallel Connection
2. Series Connection

Type of Capacitor Bank as per Its Application

1. Fixed type capacitor banks

The reactive power supplied by the fixed capacitor bank is constant irrespective of any variations in the power factor and the load of the receivers. These capacitor banks are switched on either manually (circuit breaker / switch) or semi automatically by a remote-controlled contactor.

These capacitors are applied at the terminals of inductive loads (mainly motors), at bus bars.

Disadvantages:

• Manual ON/OFF operation.
• Not meet the require kvar under varying loads.
• Penalty by electricity authority.
• Power factor also varies as a function of the load requirements so it is difficult to maintain a consistent power factor by use of Fixed Compensation i.e. fixed capacitors.
• Fixed Capacitor may provide leading power factor under light load conditions, Due to this result in overvoltages, saturation of transformers, mal-operation of diesel generating sets, penalties by electric supply authorities.

Application:

• Where the load factor is reasonably constant.
• Electrical installations with constant load operating 24 hours a day
• Reactive compensation of transformers.
• Individual compensation of motors.
• Where the kvar rating of the capacitors is less than, or equal to 15% of the supply transformer rating, a fixed value of compensation is appropriate.
• Size of Fixed Capacitor bank Qc ≤ 15% kVA transformer

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2. Automatic type capacitor banks

The reactive power supplied by the capacitor bank can be adjusted according to variations in the power factor and the load of the receivers.

The equipment is applied at points in an installation where the active-power or reactive power variations are relatively large, for example:

• At the bus bars of a main distribution switch-board,
• At the terminals of a heavily-loaded feeder cable.

Where the kvar rating of the capacitors is less than, or equal to 15% of the supply transformer rating, a fixed value of compensation is appropriate.

Above the 15% level, it is advisable to install an automatically-controlled bank of capacitors.

Control is usually provided by contactors. For compensation of highly fluctuating loads, fast and highly repetitive connection of capacitors is necessary, and static switches must be used.

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Types of APFC – Automatic Power Factor Correction

Automatic Power Factor correction equipment is divided into three major categories:

1. Standard = Capacitor + Fuse + Contactor + Controller
2. De tuned = Capacitor + De tuning Reactor + Fuse + Contactor + Controller
3. Filtered = Capacitor + Filter Reactor + Fuse + Contactor + Controller.

Advantages:

• Consistently high power factor under fluctuating loads.
• Prevention of leading power factor.
• Eliminate power factor penalty.
• Lower energy consumption by reducing losses.
• Continuously sense and monitor load.
• Automatically switch on/off relevant capacitors steps for consistent power factor.
• Ensures easy user interface.
• Automatically variation, without manual intervention, the compensation to suit the load requirements.

Application:

• Variable load electrical installations.
• Compensation of main LV distribution boards or major outgoing lines.
• Above the 15% level, it is advisable to install an automatically-controlled bank of capacitors.
• Size of Automatic Capacitor bank Qc > 15% kVA transformer.

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Type of Capacitor as per Construction

1. Standard duty Capacitor

Construction: Rectangular and Cylindrical (Resin filled / Resin coated-Dry)

Application:

1. Steady inductive load.
2. Non linear up to 10%.
3. For Agriculture duty.

2. Heavy-duty

Construction: Rectangular and Cylindrical (Resin filled / Resin coated-Dry/oil/gas)

Application:

1. Suitable for fluctuating load.
2. Non linear up to 20%.
3. Suitable for APFC Panel.
4. Harmonic filtering

3. LT Capacitor

Application:

• Suitable for fluctuating load.
• Non linear up to 20%.
• Suitable for APFC Panel & Harmonic filter application.

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Selecting Size of Capacitor Bank

The size of the inductive load is large enough to select the minimum size of capacitors that is practical.

For HT capacitors the minimum ratings that are practical are as follows:

Unit sizes lower than above is not practical and economical to manufacture.

When capacitors are connected directly across motors it must be ensured that the rated current of the capacitor bank should not exceed 90% of the no-load current of the motor to avoid self-excitation of the motor and also over compensation.

Crane motors or like, where the motors can be rotated by mechanical load and motors with electrical braking systems, should never be compensated by capacitors directly across motor terminals.

For direct compensation across transformers the capacitor rating should not exceed 90 % of the no-load KVA of the motor.

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Selection of Capacitor as per Non Liner Load

For power Factor correction it is need to first decide which type of capacitor is used.

Selection of Capacitor is depending upon many factor i.e. operating life, Number of Operation, Peak Inrush current withstand capacity.

• Calculation of Non liner Load, Example: Transformer Rating 1MVA,Non Liner Load 100KVA
• % of non Liner Load = (Non Liner Load/Transformer Capacity) x100 = (100/1000) x100=10%.
• According to Non Linear Load Select Capacitor as per Following Table.

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Configuration of Capacitor

Power factor correction capacitor banks can be configured in the following ways:

1. Delta connected Bank.
2. Star-Solidly Grounded Bank.
3. Star-Ungrounded Bank.

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1. Star-Solidly Grounded

• Initial cost of the bank may be lower since the neutral does not have to be insulated from ground.
• Capacitor switch recovery voltages are reduced
• High inrush currents may occur in the station ground system.
• The grounded-Star arrangement provides a low-impedance fault path which may require revision to the existing system ground protection scheme.
• Typically not applied to ungrounded systems. When applied to resistance-grounded systems, difficulty in coordination between capacitor fuses and upstream ground protection relays (consider coordination of 40 A fuses with a 400 A grounded system).
• Application: Typical for smaller installations (since auxiliary equipment is not required)

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2. Star-Ungrounded

Industrial and commercial capacitor banks are normally connected ungrounded Star, with paralleled units to make up the total kvar.

If only one unit is needed to make the total kvar, the units in the other phases will not be overloaded if it fails.

In industrial or commercial power systems the capacitors are not grounded for a variety of reasons. Industrial systems are often resistance grounded. A grounded Star connection on the capacitor bank would provide a path for zero sequence currents and the possibility of a false operation of ground fault relays.

Also, the protective relay scheme would be sensitive to system line-to-ground voltage Unbalance, which could also result in false relay tripping.

Application: In Industrial and Commercial.

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3. Delta-connected Banks

Delta-connected banks are generally used only at distributions voltages and are configured with a Single series group of capacitors rated at line-to-line voltage. With only one series group of units no overvoltage occurs across the remaining capacitor units from the isolation of a faulted capacitor unit.

Therefore, unbalance detection is not required for protection and they are not treated further in this paper.

Application: In Distribution System.

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Effect of series and Parallel Connection of capacitor

Parallel Connection

This is the most popular method of connection. The capacitor is connected in parallel to the unit. The voltage rating of the capacitor is usually the same as or a little higher than the system voltage.

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Series Connection

This method of connection is not much common. Even though the voltage regulation is much high in this method,

It has many disadvantages.

One is that because of the series connection, in a short circuit condition the capacitor should be able to withstand the high current. The other is that due to the series connection due to the inductivity of the line there can be a resonance occurring at a certain capacitive value.

This will lead to very low impedance and may cause very high currents to flow through the lines.

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Defining Size and Location of Capacitor in Electrical System (2)

Continued from part 1: Defining Size and Location of Capacitor in Electrical System (2)

Content

• Size of circuit breaker (CB), fuse and conductor of capacitor bank:
• A. Thermal and magnetic setting of a circuit breaker
• B. Fuse selection
• C. Size of conductor for capacitor connections
• Size of capacitor for transformer no-load compensation:
• Fixed compensation
• Sizing of capacitor for motor compensation:

1. If no-load current is known
2. If the no load current is not known

• Location 1 (the line side of the starter)
• Location 2 (between the overload relay and the starter)
• Location 3 (the motor side of the overload relay)

• A. Global compensation
• B. Compensation by sector
• C. Individual compensation

Size of CB, Fuse and Conductor of Capacitor Bank

A. Thermal and Magnetic setting of a Circuit breaker

1. Size of Circuit Breaker

1.3 to 1.5 x Capacitor Current (In) for Standard Duty/Heavy Duty/Energy Capacitors

• 1.31×In for Heavy Duty/Energy Capacitors with 5.6% Detuned Reactor (Tuning Factor 4.3)
• 1.19×In for Heavy Duty/Energy Capacitors with 7% Detuned Reactor (Tuning Factor 3.8)
• 1.12×In for Heavy Duty/Energy Capacitors with 14% Detuned Reactor (Tuning Factor 2.7)

2. Thermal Setting of Circuit Breaker

1.5x Capacitor Current (In) for Standard Duty/Heavy Duty/Energy Capacitors

3. Magnetic Setting of Circuit Breaker

5 to 10 x Capacitor Current (In) for Standard Duty/Heavy Duty/Energy Capacitors

Example: 150kvar,400v, 50Hz Capacitor

• Us = 400V, Qs = 150kvar, Un = 400V, Qn = 150kvar
• In = 150000/400√3 = 216A
• Circuit Breaker Rating = 216 x 1.5 = 324A
• Select a 400A Circuit Breaker.
• Circuit Breaker thermal setting = 216 x 1.5 = 324 Amp

Conclusion: Select a Circuit Breaker of 400A with Thermal Setting at 324A and Magnetic Setting (Short Circuit) at 324A

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B. Fuse Selection

The rating must be chosen to allow the thermal protection to be set to:

1.5 to 2.0 x Capacitor Current (In) for Standard Duty/Heavy Duty/Energy Capacitors.

• 1.35×In for Heavy Duty/Energy Capacitors with 5.7% Detuned Reactor (Tuning Factor 4.3)
• 1.2×In for Heavy Duty/Energy Capacitors with 7% Detuned Reactor (Tuning Factor 3.8)
• 1.15×In for Heavy Duty/Energy Capacitors with 14% Detuned Reactor(Tuning Factor 2.7)

For Star-solidly grounded systems:
Fuse > = 135% of rated capacitor current (includes overvoltage, capacitor tolerances, and harmonics).

For Star -ungrounded systems:
Fuse > = 125% of rated capacitor current (includes overvoltage, capacitor tolerances, and harmonics).

Example 1: 150kvar,400v, 50Hz Capacitor

• Us = 400V; Qs = 150kvar, Un = 400V; Qn = 150kvar.
• Capacitor Current =150×1000/400 =375 Amp

To determine line current, we must divide the 375 amps by √ 3

• In (Line Current) = 375/√3 = 216A
• HRC Fuse Rating = 216 x1.65 = 356A to
• HRC Fuse Rating = 216 x 2.0 = 432A so Select Fuse Size 400 Amp

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Problems with Fusing of Small Ungrounded Banks

Example: 12.47 kV, 1500 Kvar Capacitor bank made of three 3 No’s of 500 Kvar single-phase units.

• Nominal Capacitor Current = 1500/1.732×12.47 = 69.44 amp
• Size of Fuse = 1.5×69.44 = 104 Amp = 100 Amp Fuse

If a capacitor fails, we say that It may approximately take 3x line current. (3 x 69.44 A = 208.32 A).

It will take a 100 A fuse approximately 500 seconds to clear this fault (3 x 69.44 A = 208.32 A). The capacitor case will rupture long before the fuse clears the fault.

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C. Size of Conductor for Capacitor Connections

Size of capacitor circuit conductors should be at least 135% of the rated capacitor current in accordance with NEC Article 460.8 (2005 Edition).

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Size of capacitor for Transformer No-Load compensation

Fixed compensation

The transformer works on the principle of Mutual Induction. The transformer will consume reactive power for magnetizing purpose. Following size of capacitor bank is required to reduce reactive component (No Load Losses) of Transformer.

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Sizing of capacitor for motor compensation

The capacitor provides a local source of reactive current. With respect to inductive motor load, this reactive power is the magnetizing or “no load current“ which the motor requires to operate.

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1. If no-load current is known

The most accurate method of selecting a capacitor is to take the no load current of the motor, and multiply by 0.90 (90%).

Example:

Size a capacitor for a 100HP, 460V 3-phase motor which has a full load current of 124 amps and a no-load current of 37 amps.

Size of Capacitor = No load amps (37 Amp) X 90% = 33 Kvar

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2. If the no load current is not known

If the no-load current is unknown, a reasonable estimate for 3-phase motors is to take the full load amps and multiply by 30%. Then multiply it by 90% rating figure being used to avoid overcorrection and overvoltages.

Example:

Size a capacitor for a 75HP, 460V 3-phase motor which has a full load current of 92 amps and an unknown no-load current.

No-load current of Motor = Full load Current (92 Amp) x 30% = 28 Amp estimated no-load Current.

Size of Capacitor = No load amps (28 Amp) X 90% = 25 Kvar.

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Placement of Power Capacitor Bank for Motor

Capacitors installed for motor applications based on the number of motors to have power factor correction. If only a single motor or a small number of motors require power factor correction, the capacitor can be installed at each motor such that it is switched on and off with the motor.

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Required Precaution for selecting Capacitor for Motor:

The care should be taken in deciding the Kvar rating of the capacitor in relation to the magnetizing kVA of the machine.

If the rating is too high, It may damage to both motor and capacitor.

As a general rule the correct size of capacitor for individual correction of a motor should have a kvar rating not exceeding 85% of the normal No Load magnetizing KVA of the machine. If several motors connected to a single bus and require power factor correction, install the capacitor(s) at the bus.

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Where do not install Capacitor on Motor:

Do not install capacitors directly onto a motor circuit under the following conditions:

1. If solid-state starters are used.
2. If open-transition starting is used.
3. If the motor is subject to repetitive switching, jogging, inching, or plugging.
4. If a multi-speed motor is used.
5. If a reversing motor is used.
6. If a high-inertia load is connected to the motor.

Fixed power capacitor banks can be installed in a non-harmonic producing electrical system at the feeder, load or service entrance. Since power capacitor banks are reactive power generators, the most logical place to install them is directly at the load where the reactive power is consumed.

Three options exist for installing a power capacitor bank at the motor.

Installing a power capacitor bank at the motor

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Location 1 (The line side of the starter)

Install between the upstream circuit breaker and the contactor.

This location should be used for the motor loads with high inertia, where disconnecting the motor with the power capacitor bank can turn the motor into a self excited generator, motors that are jogged, plugged or reversed, motors that start frequently, multi-speed motors, starters that disconnect and reconnect capacitor units during cycling and starters with open transition.

Advantage

Larger, more cost effective capacitor banks can be installed as they supply kvar to several motors. This is recommended for jogging motors, multispeed motors and reversing applications.

Disadvantages

• Since capacitors are not switched with the motors, overcorrection can occur if all motors are not running.
• Since reactive current must be carried a greater distance, there are higher line losses and larger voltage drops.

Applications

• Large banks of fixed kVAR with fusing on each phase.
• Automatically switched banks

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Location 2 (Between the overload relay and the starter)

Install between the contactor and the overload relay.

• This location can be used in existing installations when the overload ratings surpass the National Electrical Code requirements.
• With this option the overload relay can be set for nameplate full load current of motor. Otherwise the same as Option 1.
• No extra switch or fuses required.
• Contactor serves as capacitor disconnect.
• Change overload relays to compensate for reduced motor current.
• Too much Kvar can damage motors.

Application:

Usually the best location for individual capacitors.

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Location 3 (The motor side of the overload relay)

Install directly at the single speed induction motor terminals (on the secondary of the overload relay).

• This location can be used in existing installations when no overload change is required and in new installations in which the overloads can be sized in accordance with reduced current draw.
• When correcting the power factor for an entire facility, fixed power capacitor banks are usually installed on feeder circuits or at the service entrance.
• Fixed power capacitor banks should only be used when the facility’s load is fairly constant. When a power capacitor bank is connected to a feeder or service entrance a circuit breaker or a fused disconnect switch must be provided.
• New motor installations in which overloads can be sized in accordance with reduced current draw
• Existing motors when no overload change is required.

Advantage

• Can be switched on or off with the motors, eliminating the need for separate switching devices or over current protection. Also, only energized when the motor is running.
• Since Kvar is located where it is required, line losses and voltage drops are minimized; while system capacity is maximized.

Disadvantages

• Installation costs are higher when a large number of individual motors need correction.
• Overload relay settings must be changed to account for lower motor current draw.

Application

Usually the best location for individual capacitors.

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Placement of capacitors in Distribution system

The location of low voltage capacitors in Distribution System effect on the mode of compensation, which may be global (one location for the entire installation), by sectors (section-by-section), at load level, or some combination of the last two.

In principle, the ideal compensation is applied at a point of consumption and at the level required at any instant.

Placement of capacitors in distribution system

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A. Global compensation

Principle

The capacitor bank is connected to the bus bars of the main LV distribution board to compensation of reactive energy of whole installation and it remains in service during the period of normal load.

Advantages

• Reduces the tariff penalties for excessive consumption of kvars.
• Reduces the apparent power kVA demand, on which standing charges are usually based
• Relieves Reactive energy of Transformer , which is then able to accept more load if necessary

Limitation

• Reactive current still flows in all conductors of cables leaving (i.e. downstream of) the main LV distribution board. For this reason, the sizing of these cables and power losses in them are not improved by the global mode of compensation.
• The losses in the cables (I²R) are not reduced.

Application

• Where a load is continuous and stable, global compensation can be applied
• No billing of reactive energy.
• This is the most economical solution, as all the power is concentrated at one point and the expansion coefficient makes it possible to optimize the capacitor banks
• Makes less demands on the transformer.

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B. Compensation by sector

Principle

Capacitor banks are connected to bus bars of each local distribution Panel.

Most part of the installation System can benefits from this arrangement, mostly the feeder cables from the main distribution Panel to each of the local distribution panel.

Advantages

• Reduces the tariff penalties for excessive consumption of kvar.
• Reduces the apparent power Kva demand, on which standing charges are usually based.
• The size of the cables supplying the local distribution boards may be reduced, or will have additional capacity for possible load increases.
• Losses in the same cables will be reduced.
• No billing of reactive energy.
• Makes less demands on the supply Feeders and reduces the heat losses in these Feeders.
• Incorporates the expansion of each sector.
• Makes less demands on the transformer.
• Remains economical

Limitations

• Reactive current still flows in all cables downstream of the local distribution Boards.
• For the above reason, the sizing of these cables, and the power losses in them, are not improved by compensation by sector
• Where large changes in loads occur, there is always a risk of overcompensation and consequent overvoltage problems.

Application

Compensation by sector is recommended when the installation is extensive, and where the load/time patterns differ from one part of the installation to another.

This configuration is convenient for a very widespread factory Area, with workshops having different load factors

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C. Individual compensation

Principle

• Capacitors are connected directly to the terminals of inductive circuit (Near to motors). Individual compensation should be considered when the power of the motor is significant with respect to the declared power requirement (kVA) of the installation.
• The kvar rating of the capacitor bank is in the order of 25% of the kW rating of the motor.
• Complementary compensation at the origin of the installation (transformer) may also be beneficial.
• Directly at the Load terminals Ex. Motors, a Steady load gives maximum benefit to Users.
• The capacitor bank is connected right at the inductive load terminals (especially large motors). This configuration is well adapted when the load power is significant compared to the subscribed power. This is the technical ideal configuration, as the reactive energy is produced exactly where it is needed, and adjusted to the demand.

Advantages

• Reduces the tariff penalties for excessive consumption of kvars
• Reduces the apparent power kVA demand
• Reduces the size of all cables as well as the cable losses.
• No billing of reactive energy
• From a technical point of view this is the ideal solution, as the reactive energy is produced at the point where it is consumed. Heat losses (RI²) are therefore reduced in all the lines.
• Makes less demands on the transformer.

Limitations

• Significant reactive currents no longer exist in the installation.
• Not recommended for Electronics Drives.
• Most costly solution due to the high number of installations.
• The fact that the expansion coefficient is not incorporated.

Application

Individual compensation should be considered when the power of motor is significant with respect to power of the installation.

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Common Capacitor Reactive Power Ratings

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How to Determine Correct Number of Earthing Electrodes (Strips, Plates and Pipes) - part 1

Introduction

Number of Earthing Electrode and Earthing Resistance depends on the resistivity of soil and time for fault current to pass through (1 sec or 3 sec). If we divide the area for earthing required by the area of one earth plate gives the number of earth pits required.

There is no general rule to calculate the exact number of earth pits and size of earthing strip, but discharging of leakage current is certainly dependent on the cross section area of the material so for any equipment the earth strip size is calculated on the current to be carried by that strip.

First the leakage current to be carried is calculated and then size of the strip is determined.

For most of the electrical equipment like transformer, diesel generator set etc., the general concept is to have 4 number of earth pits. 2 no’s for body earthing with 2 separate strips with the pits shorted and 2 nos for Neutral with 2 separate strips with the pits shorted.

For example for 100kVA transformer, the full load current is around 140A.

The strip connected should be capable to carry at least 70A (neutral current) which means a strip of GI 25x3mm should be enough to carry the current and for body a strip of 25×3 will do the needful. Normally we consider the strip size that is generally used as standards.

However a strip with lesser size which can carry a current of 35A can be used for body earthing. The reason for using 2 earth pits for each body and neutral and then shorting them is to serve as back up. If one strip gets corroded and cuts the continuity is broken and the other leakage current flows through the other run thery by completing the circuit.

Similarly for panels the no of pits should be 2 nos. The size can be decided on the main incomer circuit breaker.

Number of earth pits is decided by considering the total fault current to be dissipated to the ground in case of fault and the current that can be dissipated by each earth pit. Normally the density of current for GI strip can be roughly 200 amps per square cam. Based on the length and dia of the pipe used the number of earthing pits can be finalized.

1. Calculate numbers of pipe earthing

A. Earthing resistance and number of rods for isolated earth pit
(without buried earthing strip)

The earth resistance of single rod or pipe electrode is calculated as per BS 7430:

R=ρ/2×3.14xL (loge (8xL/d)-1)

Where:

ρ = Resistivity of soil (Ω meter),
L = Length of electrode (meter),
D = Diameter of electrode (meter)

Example:

Calculate isolated earthing rod resistance. The earthing rod is 4 meter long and having 12.2mm diameter, soil resistivity 500 Ω meter.

R=500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =156.19 Ω.

The earth resistance of single rod or pipe electrode is calculated as per IS 3040:

R=100xρ/2×3.14xL (loge(4xL/d))

Where:

ρ = Resistivity of soil (Ω meter),
L = Length of electrode (cm),
D = Diameter of electrode (cm)

Example:

Calculate number of CI earthing pipe of 100mm diameter, 3 meter length. System has fault current 50KA for 1 sec and soil resistivity is 72.44 Ω-Meters.

Current Density At The Surface of Earth Electrode (As per IS 3043):

• Max. allowable current density  I = 7.57×1000/(√ρxt) A/m2
• Max. allowable current density  = 7.57×1000/(√72.44X1) = 889.419 A/m2
• Surface area of one 100mm dia. 3 meter Pipe = 2 x 3.14 x r x L = 2 x 3.14 x 0.05 x3 = 0.942 m2
• Max. current dissipated by one Earthing Pipe = Current Density x Surface area of electrode
• Max. current dissipated by one earthing pipe = 889.419x 0.942 = 837.83 A say 838 Amps
• Number of earthing pipe required = Fault Current / Max.current dissipated by one earthing pipe.
• Number of earthing pipe required = 50000/838 = 59.66 Say 60 No’s.
• Total number of earthing pipe required = 60 No’s.
• Resistance of earthing pipe (isolated) R = 100xρ/2×3.14xLx(loge (4XL/d))
• Resistance of earthing pipe (isolated) R = 100×72.44 /2×3.14x300x(loge (4X300/10)) = 7.99 Ω/Pipe
• Overall resistance of 60 no of earthing pipe = 7.99/60 = 0.133 Ω.

B. Earthing resistance and number of rods for isolated earth pit
(with buried earthing strip)

Resistance of earth strip (R) As per IS 3043:

R=ρ/2×3.14xLx (loge (2xLxL/wt))

Example:

Calculate GI strip having width of 12mm , length of 2200 meter buried in ground at depth of 200mm, soil resistivity is 72.44 Ω-meter.

• Resistance of earth strip(Re) = 72.44/2×3.14x2200x(loge (2x2200x2200/.2x.012)) = 0.050 Ω
• From above calculation overall resistance of 60 no of earthing pipes (Rp) = 0.133 Ω.
  And it connected to bury earthing strip. Here net earthing resistance = (RpxRe)/(Rp+Re)
• Net eatrthing resistance = (0.133×0.05)/(0.133+0.05) = 0.036 Ω

C. Total earthing resistance and number of electrode for group
(parallel)

In cases where a single electrode is not sufficient to provide the desired earth resistance, more than one electrode shall be used. The separation of the electrodes shall be about 4 m. The combined resistance of parallel electrodes is a complex function of several factors, such as the number and configuration of electrode the array.

The total resistance of group of electrodes in different configurations as per BS 7430:

Ra=R (1+λa/n) where a=ρ/2X3.14xRxS

Where:

S = Distance between adjustment rod (meter),
λ = Factor given in table below,
n = Number of electrodes,
ρ = Resistivity of soil (Ω meter),
R = Resistance of single rod in isolation (Ω)

For electrodes equally spaced around a hollow square, e.g. around the perimeter of a building, the equations given above are used with a value of λ taken from following table.

For three rods placed in an equilateral triangle, or in an L formation, a value of λ = 1.66 may be assumed.

For Hollow square total number of electrodes (N) = (4n-1).

The rule of thumb is that rods in parallel should be spaced at least twice their length to utilize the full benefit of the additional rods. If the separation of the electrodes is much larger than their lengths and only a few electrodes are in parallel, then the resultant earth resistance can be calculated using the ordinary equation for resistances in parallel.

Typically, a 4 spike array may provide an improvement 2.5 to 3 times. An 8 spike array will typically give an improvement of maybe 5 to 6 times.

The  Resistance of Original Earthing Rod will be lowered by Total of 40% for Second Rod, 60% for third Rod,66% for forth rod.

Originally published at – Electrical Notes - Calculate Numbers of Plate/Pipe/Strip Earthings (Part-1)

Installation of vertical electrodes of earthing

Continued from first part:
How to Determine Correct Number of Earthing Electrodes (Strips, Plates and Pipes) – part 1

2. Calculate Number of Plate Earthing

The Earth Resistance of Single Plate electrode is calculated as per IS 3040:

R=ρ/A√(3.14/A)

Where:

ρ = Resistivity of Soil (Ω Meter),
A = Area of both side of Plate (m2),

Example: Calculate Number of CI Earthing Plate of 600×600 mm, System has Fault current 65KA for 1 Sec and Soil Resistivity is 100 Ω-Meters.

Current Density At The Surface of Earth Electrode (As per IS 3043):

• Max. Allowable Current Density  I = 7.57×1000/(√ρxt) A/m2
• Max. Allowable Current Density  = 7.57×1000/(√100X1)=757 A/m2
• Surface area of both side of single 600×600 mm Plate= 2 x lxw=2 x 0.06×0.06 = 0.72 m2
• Max. current dissipated by one Earthing Plate = Current Density x Surface area of electrode
• Max. current dissipated by one Earthing Plate =757×0.72= 545.04 Amps
• Resistance of Earthing Plate (Isolated)(R)=ρ/A√(3.14/A)
• Resistance of Earthing Plate (Isolated)(R)=100/0.72x√(3.14/.072)=290.14 Ω
• Number of Earthing Plate required =Fault Current / Max.current dissipated by one Earthing Pipe.
• Number of Earthing Plate required= 65000/545.04 =119 No’s.
• Total Number of Earthing Plate required = 119 No’s.
• Overall resistance of 119 No of Earthing Plate=290.14/119=2.438 Ω.

3. Calculating Resistance of Bared Earthing Strip

1) Calculation for earth resistance of buried Strip (As per IEEE)

The Earth Resistance of  Single Strip of Rod buried in ground is:

R=ρ/Px3.14xL (loge (2xLxL/Wxh)+Q)

Where:

ρ = Resistivity of Soil (Ω Meter),
h = Depth of Electrode (Meter),
w = Width of Strip or Diameter of Conductor (Meter)
L = Length of Strip or Conductor (Meter)
P and Q are Koefficients

2) Calculation for earth resistance of buried Strip (As per IS 3043)

The Earth Resistance of  Single Strip of Rod buried in ground is:

R=100xρ/2×3.14xL (loge (2xLxL/Wxt))

Where:

ρ = Resistivity of Soil (Ω Meter),
L = Length of Strip or Conductor (cm)
w = Width of Strip or Diameter of Conductor (cm)
t = Depth of burial (cm)

Example: Calculate Earthing Resistance of Earthing strip/wire of 36mm Diameter, 262 meter long buried at 500mm depth in ground, soil Resistivity is 65 Ω Meter.

Here:

R = Resistance of earth rod in W.
r = Resistivity of soil(Ω Meter) = 65 Ω Meter
l = length of the rod (cm) = 262m = 26200 cm
d = internal diameter of rod(cm) = 36mm = 3.6cm
h = Depth of the buried strip/rod (cm)= 500mm = 50cm

• Resistance of Earthing Strip/Conductor (R)=ρ/2×3.14xL (loge (2xLxL/Wt))
• Resistance of Earthing Strip/Conductor (R)=65/2×3.14x26200xln(2x26200x26200/3.6×50)
• Resistance of Earthing Strip/Conductor (R)== 1.7 Ω

4. Calculate Min. Cross Section area of Earthing Conductor

Cross Section Area of Earthing Conductor As per IS 3043

(A) =(If x√t) / K

Where:

t = Fault current Time (Second).
K = Material Constant.

Example: Calculate Cross Section Area of GI Earthing Conductor for System has 50KA Fault Current for 1 second. Corrosion will be 1.0 % Per Year and No of Year for Replacement is 20 Years.

Cross Section Area of Earthing Conductor (A) =(If x√t) / K

Here:

If = 50000 Amp
T = 1 Second
K = 80 (Material Constant, For GI=80, copper K=205, Aluminium K=126).

• Cross Section Area of Earthing Conductor (A) = (50000×1)/80
• Cross Section Area of GI Earthing Conductor (A) = 625 Sq.mm
• Allowance for Corrosion = 1.0 % Per Year & Number of Year before replacement say = 20 Years
• Total allowance = 20 x 1.0% = 20%
• Safety factor = 1.5
• Required Earthing Conductor size = Cross sectional area x Total allowance x Safety factor
• Required Earthing Conductor size = 1125 Sq.mm say 1200 Sq.mm
• Hence, Considered 1Nox12x100 mm GI Strip or 2Nox6 x 100 mm GI Strips

Calculating Resistance & Number of Earthing Rod

Reference: As per EHV Transmission Line Reference Book page: 290 and Electrical Transmission & Distribution Reference Book Westinghouse Electric Corporation, Section-I Page: 570-590.

Earthing Resistance of Single Rods:

R = ρx[ln (2L/a)-1]/(2×3.14xL)

Earthing Resistance of Parallel Rods:

R = ρx[ln (2L/A]/ (2×3.14xL)

Where:

L = length of rod in ground Meter,
a = radius of rod Meter
ρ = ground resistivity, ohm-Meter
A = √(axS)
S = Rod separation Meter

Earthing rod arrangements

Factor affects on Ground resistance

The NEC code requires a minimum ground electrode length of 2.5 meters (8.0 feet) to be in contact with the soil.  But, there are some factor that affect the ground resistance of a ground system:

• Length / Depth of the ground electrode: double the length, reduce ground resistance by up to 40%.
• Diameter of the ground electrode: double the diameter, lower ground resistance by only 10%.
• Number of ground electrodes: for increased effectiveness, space additional electrodes at least equal to the depth of the ground electrodes.
• Ground system design: single ground rod to ground plate.

The GI earthing conductor sizes for various equipment

Selection of Earthing System:

Size of Earthing Conductor

Ref IS 3043 and Handbook on BS 7671: The Lee Wiring Regulations by Trevor E. Marks.

Standard earthing strip/plate/pipe/wire weight

GI Earthing Strip:

GI Earthing Plate

GI Earthing Pipe

GI Earthing Wire

How to calculate voltage regulation of distribution line (on photo: Distribution Lines - Oaxaca, Mexico, 2013 via FlickR)

Content

1. Introduction to voltage regulation
2. Voltage Regulation for 11KV, 22KV, 33KV Overhead Line
3. Permissible Voltage Regulation (As per REC)
4. Voltage Regulation Values
5. Required Size of Capacitor
6. Optimum location of capacitors
7. Voltage Rise due to Capacitor installation
8. Calculate % Voltage Regulation of Distribution Line

Introduction to voltage regulation

Voltage (load) regulation is to maintain a fixed voltage under different load.Voltage regulation is limiting factor to decide the size of either conductor or type of insulation.

In circuit current need to be lower than this in order to keep the voltage drop within permissible values. The high voltage circuit should be carried as far as possible so that the secondary circuit have small voltage drop.

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Voltage Regulation for 11KV, 22KV, 33KV Overhead Line

% Voltage Regulation = (1.06 x P x L x PF) / (LDF x RC x DF)

Where:

P – Total Power in KVA
L –  Total Length of Line from Power Sending to Power Receiving in KM.
PF – Power Factor in p.u
RC – Regulation Constant (KVA-KM) per 1% drop.

RC = (KV x KV x 10) / ( RCosΦ + XSinΦ)

LDF – Load Distribution Factor.
LDF = 2 for uniformly distributed Load on Feeder.
LDF > 2 If Load is skewed toward the Power Transformer.
LDF = 1 To 2 If Load is skewed toward the Tail end of Feeder.

DF – Diversity Factor in p.u

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Permissible Voltage Regulation (As per REC)

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Voltage Regulation Values

The voltage variations in 33 kV and 11kV feeders should not exceed the following limits at the farthest end under peak load conditions and normal system operation regime.

• Above 33kV (-) 12.5% to (+) 10%.
• Up to 33kV (-) 9.0% to (+) 6.0%.
• Low voltage (-) 6.0% to (+) 6.0%

In case it is difficult to achieve the desired voltage especially in Rural areas, then 11/0.433 kV distribution transformers(in place of normal 11/0.4 kV DT’s) may be used in these areas.

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Required Size of Capacitor

Size of capacitor for improvement of the Power Factor from Cos ø1 to Cos ø2 is:

Required size of Capacitor (Kvar) = KVA1 (Sin ø1 – [Cos ø1 / Cos ø2] x Sin ø2)

Where KVA1 is Original KVA.

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Optimum location of capacitors

L = [1 – (KVARC / 2 KVARL) x (2n - 1)]

Where:

L – distance in per unit along the line from sub-station.
KVARC – Size of capacitor bank
KVARL – KVAR loading of line
n – relative position of capacitor bank along the feeder from sub-station if the total capacitance is to be divided into more than one Bank along the line. If all capacitance is put in one Bank than values of n=1.

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Voltage Rise due to Capacitor installation:

% Voltage Rise = (KVAR(Cap) x Lx X) / 10 x Vx2

Where:

KVAR (Cap) – Capacitor KVAR
X – Reactance per phase
L – Length of Line (mile)
V – Phase to phase voltage in kilovolts

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Calculate % Voltage Regulation of Distribution Line

Calculate Voltage drop and % Voltage Regulation at Trail end of following 11 KV Distribution system:

• System have ACSR DOG Conductor (AI 6/4.72, GI7/1.57)
• Current Capacity of ACSR Conductor = 205Amp,
• Resistance = 0.2792Ω and Reactance = 0 Ω,

Permissible limit of % Voltage Regulation at Trail end is 5%.

Method-1 (Distance Base)

Voltage Drop  = ( (√3x(RCosΦ+XSinΦ)x I ) / (No of Conductor/Phase x1000))x Length of Line

Voltage drop at Load A

• Load Current at Point A (I) = KW / 1.732xVoltxP.F
• Load Current at Point A (I) =1500 / 1.732x11000x0.8 = 98 Amp.
• Required No of conductor / Phase =98 / 205 =0.47 Amp =1 No
• Voltage Drop at Point A = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
• Voltage Drop at Point A =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x1500) = 57 Volt
• Receiving end Voltage at Point A = Sending end Volt-Voltage Drop= (1100-57) = 10943 Volt.
• % Voltage Regulation at Point A = ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
• % Voltage Regulation at Point A = ((11000-10943) / 10943 )x100 = 0.52%
• % Voltage Regulation at Point A =0.52 %

Voltage drop at Load B

• Load Current at Point B (I) = KW / 1.732xVoltxP.F
• Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.
• Distance from source= 1500+1800=3300 Meter.
• Voltage Drop at Point B = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
• Voltage Drop at Point B =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x3300) = 266 Volt
• Receiving end Voltage at Point B = Sending end Volt-Voltage Drop= (1100-266) = 10734 Volt.
• % Voltage Regulation at Point B= ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
• % Voltage Regulation at Point B= ((11000-10734) / 10734 )x100 = 2.48%
• % Voltage Regulation at Point B =2.48 %

Voltage drop at Load C

• Load Current at Point C (I) = KW / 1.732xVoltxP.F
• Load Current at Point C (I) =2000 / 1.732x11000x0.8 = 131 Amp
• Distance from source= 1500+1800+2000=5300 Meter.
• Voltage Drop at Point C = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
• Voltage Drop at Point C =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x5300) = 269 Volt
• Receiving end Voltage at Point C = Sending end Volt-Voltage Drop= (1100-269) = 10731 Volt.
• % Voltage Regulation at Point C= ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
• % Voltage Regulation at Point C= ((11000-10731) / 10731 )x100 = 2.51%
• % Voltage Regulation at Point C =2.51 %

Here Trail end Point % Voltage Regulation is 2.51% which is in permissible limit.

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Method-2 (Load Base)

% Voltage Regulation =(I x (RcosǾ+XsinǾ)x Length ) / No of Cond.per Phase xV (P-N))x100

Voltage drop at Load A

• Load Current at Point A (I) = KW / 1.732xVoltxP.F
• Load Current at Point A (I) =1500 / 1.732x11000x0.8 = 98 Amp.
• Distance from source= 1.500 Km.
• Required No of conductor / Phase =98 / 205 =0.47 Amp =1 No
• Voltage Drop at Point A = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
• Voltage Drop at Point A =((98x(0.272×0.8+0×0.6)x1.5) / 1×6351) = 0.52%
• % Voltage Regulation at Point A =0.52 %

Voltage drop at Load B

• Load Current at Point B (I) = KW / 1.732xVoltxP.F
• Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.
• Distance from source= 1500+1800=3.3Km.
• Required No of conductor / Phase =118 / 205 =0.57 Amp =1 No
• Voltage Drop at Point B = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
• Voltage Drop at Point B =((118x(0.272×0.8+0×0.6)x3.3)/1×6351) = 1.36%
• % Voltage Regulation at Point A =1.36 %

Voltage drop at Load C

• Load Current at Point C (I) = KW / 1.732xVoltxP.F
• Load Current at Point C (I) =2000 / 1.732x11000x0.8 = 131Amp.
• Distance from source= 1500+1800+2000=5.3Km.
• Required No of conductor / Phase =131/205 =0.64 Amp =1 No
• Voltage Drop at Point C = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
• Voltage Drop at Point C =((131x(0.272×0.8+0×0.6)x5.3)/1×6351) = 2.44%
• % Voltage Regulation at Point A =2.44 %

Here Trail end Point % Voltage Regulation is 2.44% which is in permissible limit.

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Total Losses in Power Distribution and Transmission Lines (on photo: Power sunset over Kampala City, Uganda)

Introduction

Power generated in power stations pass through large and complex networks like transformers, overhead lines, cables and other equipment and reaches at the end users. It is fact that the unit of electric energy generated by Power Station does not match with the units distributed to the consumers. Some percentage of the units is lost in the distribution network.

This difference in the generated and distributed units is known as Transmission and Distribution loss. Transmission and Distribution loss are the amounts that are not paid for by users.

T&D Losses = (Energy Input to feeder(Kwh) – Billed Energy to Consumer(Kwh)) / Energy Input kwh x 100

There are two types of Transmission and Distribution Losses:

1. Technical Losses
2. Non Technical Losses (Commercial Losses)

1. Technical Losses

The technical losses are due to energy dissipated in the conductors, equipment used for transmission line, transformer, subtransmission line and distribution line and magnetic losses in transformers.

Technical losses are normally 22.5%, and directly depend on the network characteristics and the mode of operation.

The major amount of losses in a power system is in primary and secondary distribution lines. While transmission and sub-transmission lines account for only about 30% of the total losses. Therefore the primary and secondary distribution systems must be properly planned to ensure within limits.

• The unexpected load increase was reflected in the increase of technical losses above the normal level
• Losses are inherent to the distribution of electricity and cannot be eliminated.

There are two Type of Technical Losses.

1. Permanent / Fixed Technical losses

• Fixed losses do not vary according to current. These losses take the form of heat and noise and occur as long as a transformer is energized
• Between 1/4 and 1/3 of technical losses on distribution networks are fixed losses. Fixed losses on a network can be influenced in the ways set out below
• Corona Losses
• Leakage Current Losses
• Dielectric Losses
• Open-circuit Losses
• Losses caused by continuous load of measuring elements
• Losses caused by continuous load of control elements

2. Variable Technical losses

Variable losses vary with the amount of electricity distributed and are, more precisely, proportional to the square of the current. Consequently, a 1% increase in current leads to an increase in losses of more than 1%.

• Between 2/3 and 3/4 of technical (or physical) losses on distribution networks are variable Losses.
• By increasing the cross sectional area of lines and cables for a given load, losses will fall. This leads to a direct trade-off between cost of losses and cost of capital expenditure. It has been suggested that optimal average utilization rate on a distribution network that considers the cost of losses in its design could be as low as 30 per cent.
• Joule losses in lines in each voltage level
• Impedance losses
• Losses caused by contact resistance.

Main Reasons for Technical Losses

1. Lengthy Distribution lines

In practically 11 KV and 415 volts lines, in rural areas are extended over long distances to feed loads scattered over large areas. Thus the primary and secondary distributions lines in rural areas are largely radial laid usually extend over long distances.

This results in high line resistance and therefore high I²R losses in the line.

• Haphazard growths of sub-transmission and distribution system in to new areas.
• Large scale rural electrification through long 11kV and LT lines.

2. Inadequate Size of Conductors of Distribution lines

The size of the conductors should be selected on the basis of KVA x KM capacity of standard conductor for a required voltage regulation, but rural loads are usually scattered and generally fed by radial feeders. The conductor size of these feeders should be adequate.

3. Installation of Distribution transformers away from load centers

Distribution Transformers are not located at Load center on the Secondary Distribution System.

In most of case Distribution Transformers are not located centrally with respect to consumers. Consequently, the farthest consumers obtain an extremity low voltage even though a good voltage levels maintained at the transformers secondary.

This again leads to higher line losses. (The reason for the line losses increasing as a result of decreased voltage at the consumers end therefore in order to reduce the voltage drop in the line to the farthest consumers, the distribution transformer should be located at the load center to keep voltage drop within permissible limits.)

4. Low Power Factor of Primary and secondary distribution system

In most LT distribution circuits normally the Power Factor ranges from 0.65 to 0.75. A low Power Factor contributes towards high distribution losses.

For a given load, if the Power Factor is low, the current drawn in high  And  the losses proportional to square of the current will be more. Thus, line losses owing to the poor PF can be reduced by improving the Power Factor.

This can be done by application of shunt capacitors.

• Shunt capacitors can be connected either in secondary side (11 KV side) of the 33/11 KV power transformers or at various point of Distribution Line.
• The optimum rating of capacitor banks for a distribution system is 2/3rd of the average KVAR requirement of that distribution system.
• The vantage point is at 2/3rd the length of the main distributor from the transformer.
• A more appropriate manner of improving this PF of the distribution system and thereby reduce the line losses is to connect capacitors across the terminals of the consumers having inductive loads.
• By connecting the capacitors across individual loads, the line loss is reduced from 4 to 9% depending upon the extent of PF improvement.

5. Bad Workmanship

Bad Workmanship contributes significantly role towards increasing distribution losses.

Joints are a source of power loss. Therefore the number of joints should be kept to a minimum. Proper jointing techniques should be used to ensure firm connections.

Replacement of deteriorated wires and services should also be made timely to avoid any cause of leaking and loss of power.

6. Feeder Phase Current and Load Balancing>

One of the easiest loss savings of the distribution system is balancing current along three-phase circuits.

Feeder phase balancing also tends to balance voltage drop among phases giving three-phase customers less voltage unbalance. Amperage magnitude at the substation doesn’t guarantee load is balanced throughout the feeder length.

Feeder phase unbalance may vary during the day and with different seasons. Feeders are usually considered “balanced” when phase current magnitudes are within 10.Similarly, balancing load among distribution feeders will also lower losses assuming similar conductor resistance. This may require installing additional switches between feeders to allow for appropriate load transfer.

Bifurcation of feeders according to Voltage regulation and Load.

7. Load Factor Effect on Losses

Power consumption of customer varies throughout the day and over seasons.

Residential customers generally draw their highest power demand in the evening hours. Same commercial customer load generally peak in the early afternoon. Because current level (hence, load) is the primary driver in distribution power losses, keeping power consumption more level throughout the day will lower peak power loss and overall energy losses.

Lower power and energy losses are reduced by raising the load factor, which, evens out feeder demand variation throughout the feeder.

The load factor has been increase by offering customers “time-of-use” rates. Companies use pricing power to influence consumers to shift electric-intensive activities during off-peak times (such as, electric water and space heating, air conditioning, irrigating, and pool filter pumping).

8. Transformer Sizing and Selection

Distribution transformers use copper conductor windings to induce a magnetic field into a grain-oriented silicon steel core. Therefore, transformers have both load losses and no-load core losses.

Transformer copper losses vary with load based on the resistive power loss equation (P loss = I²R). For some utilities, economic transformer loading means loading distribution transformers to capacity-or slightly above capacity for a short time-in an effort to minimize capital costs and still maintain long transformer life.

However, since peak generation is usually the most expensive, total cost of ownership (TCO) studies should take into account the cost of peak transformer losses. Increasing distribution transformer capacity during peak by one size will often result in lower total peak power dissipation-more so if it is overloaded.

Transformer no-load excitation loss (iron loss) occurs from a changing magnetic field in the transformer core whenever it is energized. Core loss varies slightly with voltage but is essentially considered constant. Fixed iron loss depends on transformer core design and steel lamination molecular structure. Improved manufacturing of steel cores and introducing amorphous metals (such as metallic glass) have reduced core losses.

9. Balancing 3 phase loads

Balancing 3-phase loads periodically throughout a network can reduce losses significantly. It can be done relatively easily on overhead networks and consequently offers considerable scope for cost effective loss reduction, given suitable incentives.

10. Switching off transformers

One method of reducing fixed losses is to switch off transformers in periods of low demand. If two transformers of a certain size are required at a substation during peak periods, only one might be required during times of low demand so that the other transformer might be switched off in order to reduce fixed losses.

This will produce some offsetting increase in variable losses and might affect security and quality of supply as well as the operational condition of the transformer itself. However, these trade-offs will not be explored and optimized unless the cost of losses are taken into account.

11. Other Reasons for Technical Losses

• Unequal load distribution among three phases in L.T system causing high neutral currents.
• leaking and loss of power
• Over loading of lines.
• Abnormal operating conditions at which  power and distribution transformers are operated
• Low voltages at consumer terminals causing higher drawl of currents by inductive loads.
• Poor quality of equipment used in agricultural pumping in rural areas, cooler air-conditioners and industrial loads in urban areas.

Electrical Thumb Rules You MUST Follow (Part 1)

Electrical Thumb Rules For:

• Cable Capacity
• Current Capacity of Equipment
• Earthing Resistance
• Minimum Bending Radius
• Insulation Resistance
• Lighting Arrestor
• Transformer
• Diesel Generator
• Current Transformer
• Quick Electrical Calculation

Cable Capacity

• For Cu Wire Current Capacity (Up to 30 Sq.mm) = 6X Size of Wire in Sq.mm
  Ex. For 2.5 Sq.mm = 6×2.5 = 15 Amp, For 1 Sq.mm = 6×1 = 6 Amp, For 1.5 Sq.mm = 6×1.5 = 9 Amp
• For Cable Current Capacity = 4X Size of Cable in Sq.mm, Ex. For 2.5 Sq.mm = 4×2.5 = 9 Amp.
• Nomenclature for cable Rating = Uo/U
• where Uo = Phase-Ground Voltage, U = Phase-Phase Voltage, Um = Highest Permissible Voltage

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Current Capacity of Equipment

• 1 Phase Motor draws Current = 7Amp per HP.
• 3 Phase Motor draws Current = 1.25Amp per HP.
• Full Load Current of 3 Phase Motor = HPx1.5
• Full Load Current of 1 Phase Motor = HPx6
• No Load Current of 3 Phase Motor = 30% of FLC
• KW Rating of Motor = HPx0.75
• Full Load Current of equipment = 1.39xKVA (for 3 Phase 415Volt)
• Full Load Current of equipment = 1.74xKw (for 3 Phase 415Volt)

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Earthing Resistance

• Earthing Resistance for Single Pit = 5Ω, Earthing Grid = 0.5Ω
• As per NEC 1985 Earthing Resistance should be < 5Ω.
• Voltage between Neutral and Earth <= 2 Volt
• Resistance between Neutral and Earth <= 1Ω
• Creepage Distance = 18 to 22mm/KV (Moderate Polluted Air) or
• Creepage Distance = 25 to 33mm/KV (Highly Polluted Air)

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Minimum Bending Radius

• Minimum Bending Radius for LT Power Cable = 12 x Dia of Cable.
• Minimum Bending Radius for HT Power Cable = 20 x Dia of Cable.
• Minimum Bending Radius for Control Cable = 10 x Dia of Cable.

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Insulation Resistance

• Insulation Resistance Value for Rotating Machine = (KV+1) MΩ.
• Insulation Resistance Value for Motor (IS 732) = ((20xVoltage (L-L)) / (1000+ (2xKW)).
• Insulation Resistance Value for Equipment (<1KV) = Minimum 1 MΩ.
• Insulation Resistance Value for Equipment (>1KV) = KV 1 MΩ per 1KV.
• Insulation Resistance Value for Panel = 2 x KV rating of the panel.
• Min Insulation Resistance Value (Domestic) = 50 MΩ / No of Points. (All Electrical Points with Electrical fitting & Plugs). Should be less than 0.5 MΩ
• Min Insulation Resistance Value (Commercial) = 100 MΩ / No of Points. (All Electrical Points without fitting & Plugs).Should be less than 0.5 MΩ.
• Test Voltage (A.C) for Meggering = (2X Name Plate Voltage) +1000
• Test Voltage (D.C) for Meggering = (2X Name Plate Voltage).
• Submersible Pump Take 0.4 KWH of extra Energy at 1 meter drop of Water.

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Lighting Arrestor

Arrestor have Two Rating:

1. MCOV=Max. Continuous Line to Ground Operating Voltage.
2. Duty Cycle Voltage. (Duty Cycle Voltage > MCOV).

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Transformer

• Current Rating of Transformer = KVA x 1.4
• Short Circuit Current of T.C /Generator = Current Rating / % Impedance
• No Load Current of Transformer =< 2% of Transformer Rated current
• Capacitor Current (Ic) = KVAR / 1.732xVolt (Phase-Phase)
• Typically the local utility provides transformers rated up to 500kVA For maximum connected load of 99kW,
• Typically the local utility provides transformers rated up to 1250kVA For maximum connected load of 150kW.
• The diversity they would apply to apartments is around 60%
• Maximum HT (11kV) connected load will be around 4.5MVA per circuit.
• 4No. earth pits per transformer (2No. for body and 2No. for neutral earthing),
• Clearances, approx.1000mm around TC allow for transformer movement for replacement.

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Diesel Generator

• Diesel Generator Set Produces = 3.87 Units (KWH) in 1 Litter of Diesel.
• Requirement Area of Diesel Generator = for 25KW to 48KW = 56 Sq.meter, 100KW = 65 Sq.meter.
• DG less than or equal to 1000kVA must be in a canopy.
• DG greater 1000kVA can either be in a canopy or skid mounted in an acoustically treated room
• DG noise levels to be less than 75dBA at 1 meter.
• DG fuel storage tanks should be a maximum of 990 Litter per unit. Storage tanks above this level will trigger more stringent explosion protection provision.

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Current Transformer

Nomenclature of CT:

• Ratio: input / output current ratio
• Burden (VA): total burden including pilot wires. (2.5, 5, 10, 15 and 30VA.)
• Class: Accuracy required for operation (Metering: 0.2, 0.5, 1 or 3, Protection: 5, 10, 15, 20, 30).
• Accuracy Limit Factor:
• Nomenclature of CT: Ratio, VA Burden, Accuracy Class, Accuracy Limit Factor.Example: 1600/5, 15VA 5P10  (Ratio: 1600/5, Burden: 15VA, Accuracy Class: 5P, ALF: 10)
• As per IEEE Metering CT: 0.3B0.1 rated Metering CT is accu­rate to 0.3 percent if the connected secondary burden if imped­ance does not exceed 0.1 ohms.
• As per IEEE Relaying (Protection) CT: 2.5C100 Relaying CT is accurate within 2.5 percent if the secondary burden is less than 1.0 ohm (100 volts/100A).

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Quick Electrical Calculation

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Electrical Thumb Rules You MUST Follow // PART 2 (on photo: Osprey Lunch Among The Electrical Wires! by Kathleen Jackson via Flickr)

Continued from first part: Electrical Thumb Rules You MUST Follow (Part 1)

Useful Electrical Equations

• For Sinusoidal Current: Form Factor = RMS Value/Average Value = 1.11
• For Sinusoidal Current: Peak Factor = Max Value/RMS Value = 1.414
• Average Value of Sinusoidal Current (I_av) = 0.637 x Im (Im = Max.Value)
• RMS Value of Sinusoidal Current (I_rms) = 0.707 x Im (Im = Max.Value)
• A.C Current = D.C Current/0.636.
• Phase Difference between Phase = 360/ No of Phase (1 Phase=230/1=360°, 2 Phase=360/2=180°)
• Short Circuit Level of Cable in KA (I_sc) =
  (0.094 x Cable Dia in Sq.mm) /√ Short Circuit Time (Sec)
• Max.Cross Section Area of Earthing Strip (mm2) = √(Fault Current x Fault Current x Operating Time of Disconnected Device ) / K
  K = Material Factor, K for Cu = 159, K for Al = 105, K for steel = 58 , K for GI = 80
• Most Economical Voltage at given Distance = 5.5 x √ ((km/1.6) + (kw/100))
• Cable Voltage Drop (%) =
  (1.732 x current x (RcosǾ+jsinǾ) x 1.732 x Length (km) x 100) / (Volt(L-L) x Cable Run.
• Spacing of Conductor in Transmission Line (mm) = 500 + 18 x (P – P Volt) + (2 x (Span in Length)/50).
• Protection radius of Lighnting Arrestor = √h x (2D-h) + (2D+L).
  Where h= height of L.A, D-distance of equipment (20, 40, 60 Meter), L=V x t (V=1m/ms, t=Discharge Time).
• Size of Lightning Arrestor = 1.5x Phase to Earth Voltage or 1.5 x (System Voltage/1.732).
• Maximum Voltage of the System = 1.1xRated Voltage (Ex. 66KV = 1.1 × 66 = 72.6KV)
• Load Factor = Average Power/Peak Power
• If Load Factor is 1 or 100% = This is best situation for System and Consumer both.
• If Load Factor is Low (0 or 25%) = you are paying maximum amount of KWH consumption. Load Factor may be increased by switching or use of your Electrical Application.
• Demand Factor = Maximum Demand / Total Connected Load (Demand Factor <1)
• Demand factor should be applied for Group Load
• Diversity Factor =
  Sum of Maximum Power Demand / Maximum Demand (Demand Factor >1)
  Diversity factor should be consider for individual Load
• Plant Factor (Plant Capacity) = Average Load / Capacity of Plant
• Fusing Factor = Minimum Fusing Current / Current Rating (Fusing Factor>1).
• Voltage Variation (1 to 1.5%) = ((Average Voltage – Min Voltage) x 100)/Average Voltage
  Ex: 462V, 463V, 455V, Voltage Variation= ((460 – 455)  x 100)/455 = 1.1%.
• Current Variation (10%) = ((Average Current – Min Current) x 100)/Average Current
  Ex: 30A,35A,30A, Current Variation = ((35-31.7) x 100)/31.7 = 10.4%
• Fault Level at TC Secondary
  = TC (VA) x 100 / Transformer Secondary (V) x Impedance (%)
• Motor Full Load Current = Kw /1.732 x KV x P.F x Efficiency

Electrical Thumb Rules You MUST Follow (photo by megstewart via Flickr)

Continued from second part: Electrical Thumb Rules You MUST Follow (Part 2)

Size of Capacitor for Power Factor Correction

Earthing Resistance value

Voltage Limit (As per CPWD & KEB)

Voltage Variation

Insulation Class

Standard Voltage Limit

Standard Electrical Connection (As per GERC)

Standard Meter Room Size (As per GERC)

Electrical Load as per Sq.ft Area (As per DHBVN)

Contracted Load in case of High-rise Building

Sizing The DOL Motor Starter Parts (Contactor, Fuse, Circuit Breaker and Thermal Overload Relay)

Calculate size of each part of DOL motor starter for the system voltage 415V, 5HP three phase house hold application induction motor, code A, motor efficiency 80%, motor RPM 750, power factor 0.8 and overload relay of starter is put before motor.

Basic Calculation of Motor Torque and Current

• Motor Rated Torque (Full Load Torque) = 5252xHPxRPM
• Motor Rated Torque (Full Load Torque) = 5252x5x750 = 35 lb-ft.
• Motor Rated Torque (Full Load Torque) = 9500xKWxRPM
• Motor Rated Torque (Full Load Torque) = 9500x(5×0.746)x750 = 47 Nm
• If Motor Capacity is less than 30 KW than Motor Starting Torque is 3xMotor Full Load Current or 2X Motor Full Load Current.
• Motor Starting Torque = 3xMotor Full Load Current.
• Motor Starting Torque = 3×47 = 142Nm.
• Motor Lock Rotor Current = 1000xHPx figure from below Chart/1.732×415

Locked Rotor Current

• As per above chart Minimum Locked Rotor Current  = 1000x5x1/1.732×415 = 7 Amp
• Maximum Locked Rotor Current = 1000x5x3.14/1.732×415 = 22 Amp.
• Motor Full Load Current (Line) = KWx1000/1.732×415
• Motor Full Load Current (Line) = (5×0.746)x1000/1.732×415 = 6 Amp.
• Motor Full Load Current (Phase) = Motor Full Load Current (Line)/1.732
• Motor Full Load Current (Phase) = 6/1.732 =4Amp
• Motor Starting Current = 6 to 7xFull Load Current.
• Motor Starting Current (Line) = 7×6 = 45 Amp

1. Size of Fuse

Fuse as per NEC 430-52

• Maximum Size of Time Delay Fuse = 300% x Full Load Line Current.
• Maximum Size of Time Delay Fuse = 300%x6 = 19 Amp.
• Maximum Size of Non Time Delay Fuse = 1.75% x Full Load Line Current.
• Maximum Size of Non Time Delay Fuse = 1.75%6 = 11 Amp.

2. Size of Circuit Breaker

Circuit Breaker as per NEC 430-52

• Maximum Size of Instantaneous Trip Circuit Breaker = 800% x Full Load Line Current.
• Maximum Size of Instantaneous Trip Circuit Breaker = 800%x6 = 52 Amp.
• Maximum Size of Inverse Trip Circuit Breaker = 250% x Full Load Line Current.
• Maximum Size of Inverse Trip Circuit Breaker = 250%x6 = 16 Amp.

3. Thermal Overload Relay

Thermal Overload Relay (Phase):

• Min. Thermal Overload  Relay setting = 70%x Full Load Current(Phase)
• Min. Thermal Overload Relay setting = 70%x4 = 3 Amp
• Max. Thermal Overload  Relay setting = 120%x Full Load Current(Phase)
• Max. Thermal Overload Relay setting = 120%x4 = 4 Amp

Thermal Overload Relay (Phase):

• Thermal Overload Relay setting = 100% x Full Load Current (Line).
• Thermal Overload Relay setting = 100%x6 = 6 Amp

4. Size and Type of Contactor

As per above chart:

• Type of Contactor = AC7b
• Size of Main Contactor = 100%X Full Load Current (Line).
• Size of Main Contactor = 100%x6 = 6 Amp.
• Making/Breaking Capacity of Contactor = Value above Chart x Full Load Current (Line).
• Making/Breaking Capacity of Contactor = 8×6 = 52 Amp.

Electrical Thumb Rules You MUST Follow - part 4 (on photo: distribution substation bus on a clear night in February by Eli Nelson) at Flickr

Continued from: Electrical Thumb Rules You MUST Follow (Part 3)

Eight rules to follow:

1. Substation Capacity and Short Circuit Current Capacity (As per GERC)
2. Substation Capacity and Short Circuit Current Capacity (As per Central Electricity Authority)
3. Minimum Ground Clearance and Fault Clearing Time
4. Busbar Ampere Rating
5. Busbar Spacing
6. Sound Level of Diesel Generator (ANSI 89.2 and NEMA 51.20)
7. IR Value of Transformer
8. Standard Size of MCB, MCCB, ELCB, RCCB, SFU and Fuse

1. Substation Capacity and Short Circuit Current Capacity

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2. Substation Capacity and Short Circuit Current Capacity

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3. Minimum Ground Clearance and Fault Clearing Time

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4. Busbar Ampere Rating

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5. Busbar Spacing

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6. Sound Level of Diesel Generator (ANSI 89.2 and NEMA 51.20)

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7. IR Value of Transformer

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8. Standard Size of MCB, MCCB, ELCB, RCCB, SFU and Fuse

Electrical Thumb Rules You MUST Follow, part 5 (photo by engineeringourfreedom.blogspot.com)

Continued from: Electrical Thumb Rules You MUST Follow (Part 4)

Eight rules to follow:

1. Standard Size of Transformer (IEEE/ANSI 57.120)
2. Standard Size of Motor (HP)
3. Approximate RPM of Motor
4. Motor Line Voltage
5. Motor Starting Current
6. Motor Starter
7. Impedance of Transformer (As per IS 2026)
8. Standard Size of Transformer

1. Standard Size of Transformer (IEEE/ANSI 57.120)

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2. Standard Size of Motor (HP)

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3. Approximate RPM of Motor

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4. Motor Line Voltage

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5. Motor Starting Current

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6. Motor Starter

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7. Impedance of Transformer (As per IS 2026)

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8. Standard Size of Transformer

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Electrical Thumb Rules You MUST Follow, Part 6 (On photo: View in the Transformer Room in the underground area during White House renovation on March 26, 1952 - via trumanlibrary.org)

Continued from: Electrical Thumb Rules You MUST Follow (Part 5)

12 electrical thumb rules to follow:

1. Transformer earthing wire / Strip size
2. Motor earthing wire / Strip size
3. Panel earthing wire / Strip size
4. Electrical equipment earthing
5. Earthing wire (As per BS 7671)
6. Area for transformer room: (As per NBC-2005)
7. Span of transmission line (Central electricity authority)
8. Max. lock rotor amp for 1-phase 230V motor (NEMA)
9. Three phase motor code (NEMA)
10. Service factor of motor
11. Type of contactor
12. Contactor coil

1. Transformer earthing wire / Strip size

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2. Motor earthing wire / Strip size

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3. Panel earthing wire / Strip size

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4. Electrical equipment earthing

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5. Earthing wire (As per BS 7671)

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6. Area for transformer room: (As per NBC-2005)

- The capacitor bank should be automatically switched type for substation of 5MVA and higher.
- Transformer up to 25 KVA can be mounted direct on pole.
- Transformer from 25 KVA to 250KVA can be mounted either on “H” frame of plinth.
- Transformer above 250 KVA can be mounted plinth only.
- Transformer above 100 MVA shall be protected by drop out fuse or circuit breaker.

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7. Span of transmission line (Central electricity authority)

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8. Max. lock rotor amp for 1-phase 230V motor (NEMA)

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9. Three phase motor code (NEMA)

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10. Service factor of motor

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11. Type of contactor

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12. Contactor coil

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