Best Book for Capacitor Bank.pdf

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CAPACITOR BANK & POWER FACTOR MANAGEMENT

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Electrical Chargeman Category B0

ILSAS/MFF/2007 Page 1 of 136


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Contents

Title Page No.

1.0 FUNDAMENTALS OF BASIC ELECTRICITY 6

1.1 Definition of electrical power 61.2 Apparent Power (VA) 61.3 Real Power or True Power (Watts) 61.4 Reactive Power (Var) 71.5 Power Factor 71.6 Relationship between Real, Reactive, and 7

Apparent power1.7 Equipment Causing Poor Power Factor 13

1.8 Calculation for power factor 14

2.0 EFFECT OF LOW POWER FACTOR 20

3.0 POWER FACTOR CORRECTION 21

3.1 Capacitive Power Factor correction 213.2 Function of PFC Capacitors 233.3 How capacitors work 243.4 Categories of Power Factor Correction 25

4.0 CAPACITOR BANK SIZING 30

4.1 Selecting kVAR for 3-Phase Motors 304.2 Power Factor Correction Capacitors on Reduced Voltage 33

Motors and Multi-Speed Motors4.3 Selecting kVAR for Bulk Correction 34

5.0 MV CAPACITOR BANK DESIGN 38

5.1 Function of a capacitor bank 385.2 Capacitor unit & Capacitor Bank configuration 38

5.3 Capacitor construction 405.4 Basic design of power capacitor 69



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6.0 PROTECTION OF Y-Y CAPACITORS 73

6.1 Capacitor fusing 736.2 Relaying protection 74

6.3 The concept of unbalance current flow 816.4 Basics on relay unbalance & Unbalance CT 826.5 Typical settings for a 11 kV 3-5 MVar Y-Y Capacitor Banks 836.6 Lightning protection 84

7.0 PROTECTION FOR DELTA CAPACITORS 85

7.1 Fusing arrangement for Delta Capacitors 857.2 Comparison between fusing arrangement for Delta 87

& Y capacitors

7.3 Relaying protection 908.0 MAINTENANCE PROCEDURES FOR A Y-Y CAPACITOR BANK

8.1 Safety Regulations 918.2 Maintenance Methodology for Y-Y Capacitor Banks 918.3 Maintenance program 95

9.0 MAINTENANCE PROCEDURES FOR A DELTA CAPACITOR BANK

9.1 Safety Regulations 108 9.2 Maintenance Methodology for Delta Capacitor Banks 1089.3 Maintenance program 111

10.0 COMMISSIONING

10.1 Before energizing the bank, check: 12110.2 Setting for Power Factor Regulator (Controller) 12110.3 Calculation of resonance frequency 12910.4 Calculation for the increase in line voltage 13210.5 Inspection after commissioning 134

Sample for Inspection & Maintenance Forms are attached as Appendix A.

Sample for commissioning forms are attached as Appendix B.



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1.4 Reactive Power (Var)

The instantaneous power absorbed by the reactive part of the load is given by:

Reactive power =Q=VI X = VI sin (2) Q refers to the maximum value of the instantaneous power absorbed by thereactive component of the load. The instantaneous reactive power is alternatelypositive and negative, and it expresses the reversible flow of energy to and fromthe reactive component of the load.

1.5 Power Factor

When we use any capacitive or inductive device on an AC circuit, the current orvoltage flowing through the circuit will be slightly delayed, or out of phase. A

motor is an inductive element, and the current lags behind the voltage. In acapacitor the voltage lags behind the current.

For an inductive load, the current lags the voltage, and the power factor is said tobe lagging. For capacitive loads, the current leads the voltage, and the powerfactor is said to be leading.

For a purely capacitive or inductive circuit with zero resistance, the angle oflead/lag is 90. Adding resistance to the circuit will decrease the leading/laggingangle.

The term "Power factor", is the cosine of the phase angle. For a purely inductivecircuit, the lag angle is 90, and the power factor is zero [cosine (90 0)=0]. Acommon power factor for electric motors is 0.8, which gives us a lagging angle of36 (This is because there is some resistance inside the motor windings).

1.6 Relationship between Real, Reactive, and Apparent power

We know that reactive loads such as inductors and capacitors dissipate zeropower, yet the fact that they drop voltage and draw current gives the deceptiveimpression that they actually do dissipate power. As mentioned above, thisphantom power is called reactive power , and it is measured in a unit called Volt-

Amps-Reactive (VAR), rather than watts. The actual amount of power beingused, or dissipated, in a circuit is called true power , and it is measured in watts(symbolized by the capital letter P, as always). The combination of reactivepower and true power is called apparent power , and it is the product of a circuit'svoltage and current, without reference to phase angle. Apparent power ismeasured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S.



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As a rule, true power is a function of a circuit's dissipative elements, usuallyresistances (R). Reactive power is a function of a circuit's reactance (X).

Apparent power is a function of a circuit's total impedance (Z). Since we'redealing with scalar quantities for power calculation, any complex starting

quantities such as voltage, current, and impedance must be represented by their polar magnitudes , not by real or imaginary rectangular components. Forinstance, if I'm calculating true power from current and resistance, I must use thepolar magnitude for current, and not merely the real or imaginary portion ofthe current. If I'm calculating apparent power from voltage and impedance, bothof these formerly complex quantities must be reduced to their polar magnitudesfor the scalar arithmetic.

There are several power equations relating the three types of power toresistance, reactance, and impedance (all using scalar quantities):

(3)

(4)

(5)



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Please note that there are two equations each for the calculation of true andreactive power. There are three equations available for the calculation ofapparent power, P=IE being useful only for that purpose. (E is equivalent toVoltage). Examine the following circuits and see how these three types of power

interrelate for: a purely resistive load in Figure below, a purely reactive load inFigure below and a resistive/reactive load in Figure below.

1.6.1 Resistive load only:

120 Volt

50 Hz

Figure 1: True power, reactive power, and apparent power for a purelyresistive load.



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1.6.2 Reactive load only:

120 Volt50 Hz

Figure 2: True power, reactive power, and apparent power for a purelyreactive load.



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1.6.3 Resistive/reactive load:

120 Volt50 Hz

Figure 3: True power, reactive power, and apparent power for aresistive/reactive load.

These three types of power -- true, reactive, and apparent -- relate to one

another in trigonometric form. We call this the power triangle :



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Figure 4: Power Triangle

Using the laws of trigonometry, we can solve for the length of any side (amountof any type of power), given the lengths of the other two sides, or the length of

one side and an angle.



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REVIEW:

Power dissipated by a load is referred to as true power . True power issymbolized by the letter P and is measured in the unit of Watts (W).

Power merely absorbed and returned in load due to its reactive propertiesis referred to as reactive power . Reactive power is symbolized by the letterQ and is measured in the unit of Volt-Amps-Reactive (VAR).

Total power in an AC circuit, both dissipated and absorbed/returned isreferred to as apparent power . Apparent power is symbolized by the letterS and is measured in the unit of Volt-Amps (VA).

These three types of power are trigonometrically related to one another. Ina right triangle, P = adjacent length, Q = opposite length, and S =hypotenuse length. The opposite angle is equal to the circuit's impedance(Z) phase angle.

1.7 Equipment Causing Poor Power Factor

A great deal of equipment utilized by todays modern industry causes poor plantpower factor. One of the worst offenders is the lightly loaded induction motor.Examples of this type of equipment and their approximate power factors are:

80% power factor or better: Air conditioners, pumps, center less grinders,cold headers, up setters, fans or blowers.

60% to 80% power factor: Induction furnaces, standard stamping machinesand weaving machines.

60% power factor and below: Single stroke presses, automated machinetools, finish grinders, welders.

When the above equipment functions within a plant, savings can be achieved byutilizing industrial capacitors



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1.8 Calculation for power factor

Power factor is the ratio between the kW (Kilo-Watts) and the kVA (Kilo-Volt Amperes) drawn by an electrical load where the kW is the actual (true) load

power and the kVA is the apparent load power.It is a measure of how effectively the current is being converted into useful workoutput and more particularly is a good indicator of the effect of the load currenton the efficiency of the supply system.

All current flow will cause losses in the supply and distribution system. A loadwith a power factor of 1.0 results in the most efficient loading of the supply and aload with a power factor of 0.5 will result in much higher losses in the supplysystem.

A poor power factor can be the result of either a significant phase differencebetween the voltage and current at the load terminals, or it can be due to a highharmonic content or distorted/discontinuous current waveform.

Poor load current phase angle is generally the result of an inductive load such asan induction motor, power transformer, lighting ballasts, welder or inductionfurnace.

A distorted current waveform can be the result of a rectifier, variable speed drive,switched mode power supply, discharge lighting or other electronic load.

Power Factor (PF) = cos (6)

Power Factor (PF) = kW = True Power (7)KVA Apparent Power

|kVA| = [(kW)2 + ( kVAr) 2] (8)

As was mentioned before, the angle of this power triangle graphically indicates

the ratio between the amount of dissipated (or consumed ) power and the amountof absorbed/returned power. It also happens to be the same angle as that of thecircuit's impedance in polar form. When expressed as a fraction, this ratiobetween true power and apparent power is called the power factor for this circuit.Because true power and apparent power form the adjacent and hypotenusesides of a right triangle, respectively, the power factor ratio is also equal to thecosine of that phase angle.



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Using values from the last example circuit using Equation (7):

It should be noted that power factor, like all ratio measurements, is a unit lessquantity.

For the purely resistive circuit, the power factor is 1 (perfect), because thereactive power equals zero. Here, the power triangle would look like a horizontalline, because the opposite (reactive power) side would have zero length.

For the purely inductive circuit, the power factor is zero, because true powerequals zero. Here, the power triangle would look like a vertical line, because theadjacent (true power) side would have zero length.

The same could be said for a purely capacitive circuit. If there are no dissipative(resistive) components in the circuit, then the true power must be equal to zero,making any power in the circuit purely reactive. The power triangle for a purelycapacitive circuit would again be a vertical line (pointing down instead of up as itwas for the purely inductive circuit).

Power factor can be an important aspect to consider in an AC circuit, becauseany power factor less than 1 means that the circuit's wiring has to carry morecurrent than what would be necessary with zero reactance in the circuit to deliverthe same amount of (true) power to the resistive load. If our last example circuithad been purely resistive, we would have been able to deliver a full 169.256watts to the load with the same 1.410 amps of current, rather than the mere119.365 watts that it is presently dissipating with that same current quantity. Thepoor power factor makes for an inefficient power delivery system.



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Poor power factor can be corrected, paradoxically, by adding another load to thecircuit drawing an equal and opposite amount of reactive power, to cancel out theeffects of the load's inductive reactance. Inductive reactance can only becanceled by capacitive reactance, so we have to add a capacitor in parallel to our

example circuit as the additional load. The effect of these two opposingreactances in parallel is to bring the circuit's total impedance equal to its totalresistance (to make the impedance phase angle equal, or at least closer, tozero).

Since we know that the (uncorrected) reactive power is 119.998 VAR (inductive),we need to calculate the correct capacitor size to produce the same quantity of(capacitive) reactive power. Since this capacitor will be directly in parallel with thesource (of known voltage), we'll use the power formula, which starts from voltageand reactance:



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Let's use a rounded capacitor value of 22 F and see what happens to ourcircuit:

120 Volt50 Hz

Figure 4: Parallel capacitor corrects lagging power factor of inductive load.



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The power factor for the circuit, overall, has been substantially improved. Themain current has been decreased from 1.41 amps to 994.7 milliamps, while thepower dissipated at the load resistor remains unchanged at 119.365 watts. Thepower factor is much closer to being 1:

Since the impedance angle is still a positive number, we know that the circuit,overall, is still more inductive than it is capacitive. If our power factor correctionefforts had been perfectly on-target, we would have arrived at an impedanceangle of exactly zero, or purely resistive. If we had added too large of a capacitorin parallel, we would have ended up with an impedance angle that was negative,indicating that the circuit was more capacitive than inductive.

It should be noted that too much capacitance in an AC circuit will result in a lowpower factor just as well as too much inductance. You must be careful not toover-correct when adding capacitance to an AC circuit. You must also be very careful to use the proper capacitors for the job (rated adequately for powersystem voltages and the occasional voltage spike from lightning strikes, forcontinuous AC service, and capable of handling the expected levels of current).



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If a circuit is predominantly inductive, we say that its power factor is lagging (because the current wave for the circuit lags behind the applied voltage wave).Conversely, if a circuit is predominantly capacitive, we say that its power factor isleading . Thus, our example circuit started out with a power factor of 0.705

lagging, and was corrected to a power factor of 0.999 lagging.

REVIEW: Poor power factor in an AC circuit may be corrected, or re-established at

a value close to 1, by adding a parallel reactance opposite the effect of theload's reactance. If the load's reactance is inductive in nature (which italmost always will be), parallel capacitance is what is needed to correctthe poor power factor.

Exercise No.1

A transformer delivers maximum loading of 800kW & 700 kVar.

What is the value of the power factor for the transformer?

Exercise No.2

A 1000 kVA transformer has maximum loading of 800kW & power factor of 0.45.

What is the % loading of the transformer?



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2.0 EFFECT OF LOW POWER FACTOR

"Power Factor" is an electrical term used to rate the degree of thesynchronization of power supply current with the power supply voltage. It is

important that we clearly understand the meaning of "Power Factor" and its effecton the electrical supply system for the following reasons:

a low power factor can increase the cost of power to the user a low power factor can increase the cost of power transmission equipment to

the user a customer may request assistance in selecting equipment to correct a low

power factor over-correction of power factor by the addition of excessive capacitance is

sometimes dangerous to a motor and the driven equipment. (above 95%power factor)

a customer may, to some extent, use motor power factor rating as a powerfactor rating as a criterion in choosing among competing motors, especiallywhen a large motor is involved.

The power factors in industrial plants are usually lagging due to the inductivenature of induction motors, transformers, lighting, induction heating furnaces, etc.This lagging power factor has two costly disadvantages for the power user.

First, it increases the cost incurred by the power company because more currentmust be transmitted than is actually used to perform useful work. This increasedcost is passed on to the industrial customer by means of power factor

adjustments to the rate schedules.

Second, it reduces the load handling capability of the industrial plants electricaltransmission system which means that the industrial power user must spendmore on transmission lines and transformers to get a given amount of usefulpower through his plant.



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3.0 POWER FACTOR CORRECTION

A poor power factor due to an inductive load can be improved by the addition ofpower factor correction (PFC) capacitor, but, a poor power factor due to a

distorted current waveform requires change in equipment design or expensiveharmonic filters to gain an appreciable improvement.

Many inverters are quoted as having a power factor of better than 0.95 when inreality, the true power factor is between 0.5 and 0.75. The figure of 0.95 is basedon the cosine of the angle between the voltage and current but does not take intoaccount that the current waveform is discontinuous and therefore contributes toincreased losses on the supply

3.1 Capacitive Power Factor correction

Capacitive Power Factor correction is applied to circuits, which include inductionmotors as a means of reducing the inductive component of the current andthereby reduce the losses in the supply. There should be no effect on theoperation of the motor itself.

An induction motor draws current from the supply that is made up of resistivecomponents and inductive components. The resistive components are:-

1) Load current.2) Loss current.

and the inductive components are:3) Leakage reactance.4) Magnetizing current.

Figure 5: Magnetizing current



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The current due to the leakage reactance is dependant on the total current drawnby the motor, but the magnetizing current is independent of the load on themotor. The magnetizing current will typically be between 20% and 60% of therated full load current of the motor. The magnetizing current is the current that

establishes the flux in the iron and is very necessary if the motor is going tooperate. The magnetizing current does not actually contribute to the actual workoutput of the motor. It is the catalyst that allows the motor to work properly. Themagnetizing current and the leakage reactance can be considered passengercomponents of current that will not affect the power drawn by the motor, but willcontribute to the power dissipated in the supply and distribution system.

Take for example a motor with a current draw of 100 Amps and a power factor of0.75 The resistive component of the current is 75 Amps and this is what the KWhmeter measures.

The higher current will result in an increase in the distribution losses of (100x100)/(75x75) = 1.777 or a 78% increase in the supply losses.

In the interest of reducing the losses in the distribution system, power factorcorrection is added to neutralize a portion of the magnetizing current of themotor. Typically, the corrected power factor will be 0.92 - 0.95.

There are many ways that this is metered, but the net result is that in order toreduce wasted energy in the distribution system, the consumer will beencouraged to apply power factor correction.

Power factor correction is achieved by the addition of capacitors in parallel withthe connected motor circuits and can be applied at the starter, or applied at theswitchboard or distribution panel. The resulting capacitive current is leadingcurrent and is used to cancel the lagging inductive current flowing from thesupply

Figure 6: Power Factor Correction (1)



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3.2 Function of Power Factor Correction (PFC) Capacitors

PFC equipment provides the means of reducing the reactive power beingsupplied by the utility. Reducing the reactive power supplied by the utility results

in a cost reduction to electrical bills, since the kVA demand is also reduced.

a. No Power Factor Correction

kVar supplies byCapacitor Bank

b. With Power Factor Correction

Figure 7: Power Factor Correction (2)



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PFC capacitors are the main component in PFC equipment, with their size mostoften referred to in kVAr. Figure 8 illustrates how a PFC capacitor works wheninstalled on the line side of a motor.

Figure 8: Power Factor Correction for Induction motors

3.3 How capacitors work

Induction motors, transformers and many other electrical loads requiremagnetizing current (kVAR) as well as actual power (kW). By representing thesecomponents of apparent power (kVA) as the sides of a right triangle, we candetermine the apparent power from the right triangle rule: kVA 2 = kW 2 + kVAR 2.To reduce the kVA required for any given load, you must shorten the line thatrepresents the kVAR. This is precisely what capacitors do.

By supplying kVAR right at the load, the capacitors relieve the utility of theburden of carrying the extra kVAR. This makes the utility transmission/distributionsystem more efficient, reducing cost for the utility and their customers. The ratioof actual power to apparent power is usually expressed in percentage and iscalled power factor.



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3.4 Categories of Power Factor Correction

Capacitors connected at each starter and controlled by each starter is known as"Static Power Factor Correction" while capacitors connected at a distribution

board and controlled independently from the individual starters is known as "BulkCorrection".

3.4.1 Bulk Correction

The Power factor of the total current supplied to the distribution board ismonitored by a controller which then switches capacitor banks In a fashion tomaintain a power factor better than a preset limit. (Typically 0.95)

Ideally, the power factor should be as close to unity (Power factor of "1") aspossible. There is no problem with bulk correction operating at unity.

Figure 9: Bulk correction

3.4.2 Static Correction

As a large proportion of the inductive or lagging current on the supply is due tothe magnetizing current of induction motors, it is easy to correct each individualmotor by connecting the correction capacitors to the motor starters.

With static correction, it is important that the capacitive current is less than theinductive magnetizing current of the induction motor.



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In many installations employing static power factor correction, the correctioncapacitors are connected directly in parallel with the motor windings. When themotor is off-line, the capacitors are also off-line. When the motor is connected tothe supply, the capacitors are also connected providing correction at all times

that the motor is connected to the supply. This removes the requirement for anyexpensive power factor monitoring and control equipment.

In this situation, the capacitors remain connected to the motor terminals as themotor slows down. An induction motor, while connected to the supply, is drivenby a rotating magnetic field in the stator, which induces current into the rotor.

When the motor is disconnected from the supply, there is for a period of time, amagnetic field associated with the rotor. As the motor decelerates, it generatesvoltage out its terminals at a frequency which is related to it's speed.

The capacitors connected across the motor terminals, form a resonant circuitwith the motor inductance. If the motor is critically corrected, (corrected to apower factor of 1.0) the inductive reactance equals the capacitive reactance atthe line frequency and therefore the resonant frequency is equal to the linefrequency.

If the motor is over corrected, the resonant frequency will be below the linefrequency. If the frequency of the voltage generated by the decelerating motorpasses through the resonant frequency of the corrected motor, there will be highcurrents and voltages around the motor/capacitor circuit. This can result in severdamage to the capacitors and motor. It is imperative that motors are never overcorrected or critically corrected when static correction is employed.

Static power factor correction should provide capacitive current equal to 80% ofthe magnetizing current, which is essentially the open shaft current of the motor.

The magnetizing current for induction motors can vary considerably. Typically,magnetizing currents for large two pole machines can be as low as 20% of therated current of the motor while smaller low speed motors can have amagnetizing current as high as 60% of the rated full load current of the motor. Itis not practical to use a "Standard table" for the correction of induction motorsgiving optimum correction on all motors. Tables result in under correction onmost motors but can result in over correction in some cases. Where the openshaft current cannot be measured, and the magnetizing current is not quoted, anapproximate level for the maximum correction that can be applied can becalculated from the half load characteristics of the motor.

It is dangerous to base correction on the full load characteristics of the motor asin some cases, motors can exhibit a high leakage reactance and correction to



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0.95 at full load will result in over correction under no load, or disconnectedconditions.

Static correction is commonly applied by using one contactor to control both the

motor and the capacitors. It is better practice to use two contactors, one for themotor and one for the capacitors. Where one contactor is employed, it should beup sized for the capacitive load. The use of a second contactor eliminates theproblems of resonance between the motor and the capacitors.

Figure 10: Static correction

3.4.3 Precaution on use of Static Connection

3.4.3.1 Inverter.

Static Power factor correction must not be used when the motor is controlled by avariable speed drive or inverter. The connection of capacitors to the output of aninverter can cause serious damage to the inverter and the capacitors due to thehigh frequency switched voltage on the output of the inverters.The current drawn from the inverter has a poor power factor, particularly at lowload, but the motor current is isolated from the supply by the inverter. The phaseangle of the current drawn by the inverter from the supply is close to zeroresulting in very low inductive current irrespective of what the motor is doing. Theinverter does not however, operate with a good power factor.

Many inverter manufacturers quote a cos of better than 0.95 and this isgenerally true, however the current is non sinusoidal and the resultant harmonics



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cause a power factor (KW/KVA) of closer to 0.7 depending on the input design ofthe inverter. Inverters with input reactors and DC bus reactors will exhibit ahigher true power factor than those without.

The connection of capacitors close to the input of the inverter can also result indamage to the inverter. The capacitors tend to cause transients to be amplified,resulting in higher voltage impulses applied to the input circuits of the inverter,and the energy behind the impulses is much greater due to the energy storage ofthe capacitors. It is recommended that capacitors should be at least 75 Metersaway from inverter inputs to elevate the impedance between the inverter andcapacitors and reduce the potential damage caused.

Switching capacitors, Automatic bank correction etc, will cause voltage transientsand these transients can damage the input circuits of inverters. The energy isproportional to the amount of capacitance being switched. It is better to switchlots of small amounts of capacitance than few large amounts.

3.4.3.2 Solid State Soft Starter

Static Power Factor correction capacitors must not be connected to the output ofa solid state soft starter. When a solid state soft starter is used, the capacitorsmust be controlled by a separate contactor, and switched in when the soft starteroutput voltage has reached line voltage. Many soft starters provide a "top oframp" or "bypass contactor control" which can be used to control the powerfactor correction capacitors.

The connection of capacitors close to the input of the soft starter can also resultin damage to the soft starter if an isolation contactor is not used. The capacitorstend to cause transients to be amplified, resulting in higher voltage impulsesapplied to the SCRs of the Soft Starter, and the energy behind the impulses ismuch greater due to the energy storage of the capacitors. It is recommended thatcapacitors should be at least 50 Meters away from Soft starters to elevate theimpedance between the inverter and capacitors and reduce the potential damagecaused.

Switching capacitors, Automatic bank correction etc, will cause voltage transientsand these transients can damage the SCRs of Soft Starters if they are in the Offstate without an input contactor. The energy is proportional to the amount ofcapacitance being switched. It is better to switch lots of small amounts ofcapacitance than few large amounts.



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Figure 11: Static correction for soft starter schemes



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4.0 CAPACITOR BANK SIZING

The most common method for improving power factor is to add capacitors banksto the system. Capacitors are attractive because they're economical and easy to

maintain. Not only that, they have no moving parts, unlike some other devicesused for the same purpose.

When you add a capacitor bank to your system, the capacitor supplies thereactive power needed by the load. If you size and select the capacitor bank tocompensate to a unity power factor, it can supply all the reactive power neededby the load, and no reactive power is demanded from the utility. If you design thecapacitor bank to improve the power factor to a quantity less than 1.0, thereactive power supplied by the bank will be its rated kVARs (or MVARs), whilethe rest of the reactive power needed by the load will be supplied by the utility.

4.1 Selecting kVAR for 3-Phase Motors

To properly select the amount of kVAR required to correct the lagging powerfactor of a 3-phase motor you must have three pieces of information:

kW (kilowatts) Existing Power Factor in percent Desired Power Factor in percent

The formula to calculate the required kVAR is:

Factor from Table 1 x kW = kVAR of capacitors required. (9)



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Table 1 Power Factor Improvement Table

80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 950 0.982 1.008 1.034 1.06 1.086 1.112 1.139 1.165 1.192 1.22 1.248 1.276 1.306 1.337 1.369 1.403 1.442 1.481 151 0.937 0.962 0.989 1.015 1.041 1.067 1.094 1.12 1.147 1.175 1.23 1.231 1.261 1.292 1.324 1.358 1.395 1.436 152 0.893 0.919 0.945 0.971 0.997 1.023 1.05 1.076 1.103 1.131 1.159 1.187 1.217 1.248 1.28 1.314 1.351 1.392 53 0.85 0.876 0.902 0.928 0.954 0.98 1.007 1.033 1.06 1.088 1.116 1.144 1.174 1.295 1.237 1.271 1.308 1.349 154 0.809 0.835 0.861 0.887 0.913 0.939 0.966 0.992 1.019 1.047 1.075 1.103 1.133 1.164 1.196 1.23 1.267 1.308 55 0.769 0.795 0.821 0.847 0.873 0.899 0.926 0.952 0.979 1.007 1.033 1.063 1.09 1.124 1.156 1.19 1.228 1.268 156 0.73 0.756 0.782 0.808 0.834 0.86 0.887 0.913 0.94 0.968 0.996 1.024 1.051 1.085 1.117 1.151 1.189 1.229 157 0.692 0.718 0.744 0.77 0.796 0.822 0.849 0.875 0.902 0.93 0.958 0.986 1.013 1.047 1.079 1.113 1.151 1.191 158 0.655 0.681 0.707 0.733 0.759 0.785 0.812 0.838 0.865 0.893 0.921 0.949 0.976 1.01 1.042 1.076 1.114 1.154 59 0.618 0.644 0.67 0.696 0.722 0.748 0.775 0.801 0.828 0.856 0.884 0.912 0.939 0.973 1.005 1.039 1.077 1.117 60 0.584 0.61 0.636 0.662 0.688 0.714 0.741 0.767 0.794 0.822 0.85 0.878 0.905 0.939 0.971 1.005 1.043 1.083 161 0.549 0.575 0.601 0.627 0.653 0.679 0.706 0.732 0.759 0.787 0.815 0.843 0.87 0.904 0.936 0.97 1.008 1.014 162 0.515 0.541 0.567 0.593 0.619 0.645 0.672 0.698 0.725 0.753 0.781 0.809 0.836 0.87 0.902 0.936 0.974 0.982 63 0.483 0.509 0.535 0.561 0.587 0.613 0.66 0.666 0.693 0.721 0.749 0.777 0.804 0.838 0.87 0.904 0.942 0.949 64 0.45 0.476 0.502 0.528 0.554 0.58 0.607 0.633 0.66 0.688 0.716 0.744 0.771 0.805 0.837 0.871 0.909 0.918 065 0.419 0.445 0.471 0.497 0.523 0.549 0.576 0.602 0.629 0.657 0.685 0.713 0.74 0.774 0.806 0.84 0.878 0.887 066 0.388 0.414 0.44 0.466 0.492 0.518 0.545 0.571 0.598 0.626 0.554 0.682 0.709 0.743 0.775 0.809 0.847 0.857 67 0.358 0.384 0.41 0.436 0.462 0.488 0.515 0.541 0.568 0.596 0.624 0.652 0.679 0.713 0.746 0.779 0.817 0.828 68 0.329 0.355 0.381 0.407 0.433 0.459 0.486 0.512 0.539 0.567 0.595 0.623 0.65 0.684 0.716 0.75 0.788 0.798 069 0.299 0.325 0.351 0.377 0.403 0.429 0.456 0.482 0.509 0.537 0.565 0.593 0.62 0.654 0.686 0.72 0.758 0.769 70 0.27 0.296 0.322 0.348 0.374 0.4 0.427 0.453 0.48 0.508 0.536 0.564 0.591 0.625 0.657 0.691 0.729 0.741 071 0.242 0.268 0.294 0.32 0.346 0.372 0.399 0.425 0.452 0.48 0.508 0.536 0.563 0.597 0.629 0.663 0.701 0.712 072 0.213 0.239 0.265 0.291 0.317 0.343 0.37 0.396 0.423 0.451 0.479 0.507 0.534 0.568 0.6 0.634 0.672 0.685 073 0.186 0.212 0.238 0.264 0.29 0.316 0.343 0.369 0.396 0.424 0.452 0.48 0.507 0.541 0.573 0.607 0.645 0.658 074 0.159 0.185 0.211 0.237 0.263 0.289 0.316 0.342 0.369 0.397 0.425 0.453 0.48 0.514 0.546 0.58 0.618 0.631 75 0.132 0.158 0.184 0.21 0.236 0.262 0.289 0.315 0.342 0.37 0.398 0.426 0.453 0.487 0.519 0.553 0.591 0.604 076 0.105 0.131 0.157 0.183 0.209 0.235 0.262 0.288 0.315 0.343 0.371 0.399 0.426 0.46 0.492 0.526 0.564 0.578 77 0.079 0.105 0.131 0.157 0.183 0.209 0.236 0.262 0.289 0.317 0.345 0.373 0.4 0.434 0.466 0.5 0.538 0.552

78 0.053 0.079 0.105 0.131 0.157 0.183 0.21 0.236 0.263 0.291 0.319 0.347 0.374 0.408 0.44 0.474 0.512 0.525 079 0.026 0.052 0.078 0.104 0.13 0.156 0.183 0.209 0.236 0.264 0.292 0.32 0.347 0.381 0.413 0.447 0.485 0.499 080 0 0.026 0.052 0.078 0.104 0.13 0.157 0.183 0.21 0.238 0.266 0.294 0.321 0.355 0.387 0.421 0.459 0.473 081 0 0.026 0.052 0.078 0.104 0.13 0.157 0.184 0.212 0.24 0.268 0.295 0.329 0.361 0.395 0.433 0.447 0.582 0 0 .026 0.052 0.078 0.105 0.131 0.158 0.186 0.214 0.242 0.269 0.303 0.335 0.369 0.407 0.421 0.483 0 0.026 0.052 0.079 0.105 0.132 0.16 0.188 0.216 0.243 0.277 0.309 0.343 0.381 0.395 0.4684 0 0.026 0.053 0.079 0.106 0.134 0.162 0.19 0.217 0.251 0.283 0.317 0.355 0.369 0.43785 0 0.027 0.053 0 .08 0.108 0.136 0.164 0.191 0.225 0.257 0.291 0.329 0.343 0.417 86 0 0.026 0.053 0.081 0.109 0.137 0.167 0.198 0.23 0.265 0.301 0.317 0.39 087 0.027 0.055 0.082 0.111 0.141 0.172 0.204 0.238 0.275 0.29 0.364 0.88 0 0.028 0.056 0.084 0.114 0.145 0.177 0.211 0.248 0.262 0.237 0.389 0.028 0.056 0.086 0.117 0.149 0.183 0.22 0.234 0.309 0.390 0.028 0.058 0.089 0.121 0.155 0.192 0.206 0.281 0.342

91 0.03 0.061 0.093 0.127 0.164 0.176 0.253 0.31492 0.031 0.063 0.097 0.134 0.143 0.223 0.284 93 0 0.032 0.066 0.103 0.113 0.192 0.253 094 0.034 0.071 0.079 0.16 0.221 0.95 0.037 0.042 0.126 0.187 0.396 0 0.089 0.15 0.29297 0 0.047 0.108 0.25198 0 0.061 0.20399 0 0.142

Or i

ginal Power F

actor in

%

Desired Pow er Factor in %


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EXAMPLE:

A small machine tool plant used an average of 100 kW with an existing powerfactor of 80%. Their desired power factor is 95%. The kVAR of capacitors

necessary to raise the power factor to 95% is found by using Table 1 , which inthis case gives 0.421 as the factor needed to complete the formula referencedabove:

0.421 x 100 kW = 42 Kvar

The customer may now choose the capacitor catalog number by kVAR andvoltage from the complete ratings listed in this catalog. If kW or Present PowerFactor are not known you can calculate from the following formulas to get thethree basic pieces of information required to calculate kVAR:

PF = kW / kVA (7)

kVA = 1.732 x I x Volt / 1000 (10)

kW = 1.732 x I x Volt x PF / 1000 (11)

kW = HP x 0.746 / eff (12)

WHERE

I = full load current in amps Volt = voltage of motor PF = Present power factor as a decimal (80% = .80) HP = rated horsepower of motor eff = rated efficiency of motor as a decimal (83% = .83)

If Desired Power Factor is not provided, 95% is a good economical power factorfor calculation purposes.

http://www.ece.umr.edu/courses/f02/ee207/spectrum/pf/table01.htmlhttp://www.ece.umr.edu/courses/f02/ee207/spectrum/pf/table01.html


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4.2 Power Factor Correction Capacitors on Reduced Voltage Motors andMulti-Speed Motors

The following shows capacitor connections for typical starting circuits for reduced

voltage and multi-speed motors. Variations to these circuits do exist. Make surethat your circuit exactly matches the circuit shown here before applyingcapacitors. Failure to do so may result in damage to the motor. The maincontacts, illustrated in the diagrams below as M1, M2, M3, reference the contactsthat must be closed to start or run the motor. Capacitors should be connected onthe motor side of the main contacts.

Figure 12: Capacitor schemes for multi-speed motors



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4.3 Selecting kVAR for Bulk Correction

4.3.1 Power Factor Improvement Table

To properly select the amount of kVAR required to correct the lagging powerfactor of a total system for Bulk Correction you must have three pieces ofinformation:

kW (kilowatts) load profiles (Minimum 1 month) Existing Average Power Factor in percent for 1 month Desired Power Factor in percent

The formula to calculate the required kVAR is:

Factor from Table 1 x kW = kVAR of capacitors required.

Figure 13: Capacitor schemes for bulk correction



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Exercise No.3

One customer has maximum load of 3000 kW & average power factor at 0.74.

Due to having power factor less than 0.85 (Based on Tariff), the customer ischarged power factor penalty.

What is the size of capacitor bank to improve power factor from 0.74 to 0.95?

Exercise No.4

One customer has maximum load of 8500 kW & average power factor at 0.54.Due to having power factor less than 0.85 (Based on Tariff), the customer ischarged power factor penalty.




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4.3.2 Direct calculation

Capacitor size (kVAr) = kW x (Tangent 1 Tangent 2) (13)

1 - Angle for existing power factor

2 Angle for desired power factor

kW kW Load

EXAMPLE

One customer has maximum load of 2000 kW & average power factor at 0.65.Due to having power factor less than 0.85 (Based on Tariff), the customer ischarged power factor penalty.


Calculation

Existing power factor = Cos 1 = 0.65

Maximum load = 2000 kW

Desired power factor = Cos 2

= 0.90

Angle for existing power factor = 1 = Cos -1(0.65) = 49.45 0

Angle for desired power factor = 2 = Cos -1(0.90) = 25.84 0

Capacitor size (kVAr) = kW x (Tangent 1 Tangent 2)

= 2000 x (Tan 49.45 0 Tan 25.84 0)

= 1,369 kVAr



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Exercise No.5 (Direct calculation)

One customer has maximum load of 8500 kW & average power factor at 0.54.Due to having power factor less than 0.85 (Based on Tariff), the customer is

charged power factor penalty.What is the size of capacitor bank to improve power factor from 0.54 to 0.95?



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5.0 MV CAPACITOR DESIGN

5.1 Function of a capacitor bank

Power capacitors are inexpensive source of reactive power. They provide a valueof Var, which is proportional to the square of the voltage applied. The reactanceof a capacitor bank varies inversely with the frequency:

Var V2 (14)

Xc = 1/(2 fC) (15)

So for high frequencies, they provide low impedance. The leading current drawnby the capacitors gives a voltage rise through the inductive reactance of the

power system, which raises the operating voltage level. A fixed bank of capacitor will furnish a fixed amount of Var at a constant voltageto the load.

Power Capacitors can be designed to be operational for any voltage range.

5.2 Design of Power Capacitors

Capacitor Banks at 4.16 kV and below are normally connected in delta.

5.2.1 Difference between Delta and Wye Connection

Figure 14 shows the capacitor bank connections that are the topic of this section.The only other popular connection that is not shown is the grounded-wye andsplit wye-connected capacitor bank. The following key points can be made inregard to bank connection under normal and abnormal system conditions.

Bus Bar Connections: From looking at Figure 14, it should be evident that theungrounded-wye connection is much simpler in design than either of the twodelta connected banks. The crossover connection that connects phase "A" to

phase "C" to close the delta is complicated at the medium voltage level due toclearance requirements.



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Figure 14: Connection and fusing arrangements for ungrounded-wye anddelta connected capacitor banks

Fusing: Figure 14 also shows common fusing practices for each of the bankarrangements. The Figure shows that the delta connected bank can be protectedby placing the fuses inside or outside of the delta. Two fuses per single phasecapacitor are required when fusing inside of the delta, but their rating isdecreased to 57% of the outside fuse rating. The fuses outside of the delta aresized in the same way as the fuses for the ungrounded-wye connected capacitorbank.

Capacitor: Except for voltage rating, the capacitors in both ungrounded-wye anddelta-connected banks are the same and will have the same kvar rating. Theyconsists of a double bushing design, meaning both terminals are fully insulatedfrom their case (ground). On delta connected banks, the capacitors have a line-to-line voltage rating, and on a wye-connected banks, they have a line-to-neutralvoltage rating.

Fault Conditions: A capacitor typically fails in two ways: 1) A bushing to casefault occurs. 2) The internal sections fail (commonly known as a dielectric fault),which basically shorts the capacitor terminals. Whether the capacitors are deltaor wye-connected, a bushing fault will have the same impact on the powersystem.



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Internal section faults, or dielectric faults appear differently to a power system.On a delta connected bank, an internal section fault subjects the power system toa phase-to-phase bolted fault. This fault will cause a major voltage sag on thefacilities power system (until the capacitor fuse(s) blow) and may cause capacitor

case rupture if not properly protected.It also subjects the power system to high magnitude fault currents, which canimpose mechanical and thermal stress on components in the fault path. On awye-ungrounded capacitor bank, internal section faults subject the power systemto a fault current that is three times the banks rating (until the capacitor fuseblows). Therefore, the voltage sag, mechanical and thermal stressing associatedwith the fault, and case rupture concerns are reduced .

5.3 Design of Power Capacitor Banks > 4.16 kV

5.3.1 Capacitor unit & Capacitor Bank configuration

Capacitor bank Capacitor Unit Capacitor element

Figure 15: Power capacitor bank major components



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Figure 16: Components of a capacitor unit

5.3.2 Capacitor construction

Capacitors are manufactured in individual units that are combined in parallel andseries arrangements to give the desired voltage rating and total kVar needed forthe application.



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5.3.3 Basic design of power capacitor

5.3.3.1 MV Capacitor banks

There are two principal types of capacitor bank construction in MV range. Theselection will depend on the location in the system where they are connected andthe kVar capability of the bank.

The types of bank most often applied are the outdoor rack and the pole mountdesign.

This approach is suitable for large, MV capacitor banks of any designated kVarrating and voltage.

The voltage range starts from 11 kV and goes up as high as the applicationrequires. The kVar Capacity stars from 300 kVar and can be designed as largeas required. These designs will include racks for mounting and individual unitfuses.

The second type of equipment available is the enclosed or house construction.The housing is built to enclose the capacitor, switching device, and the automaticsensing and control equipment.

These are used when where a relatively small bank of capacitors is needed toimprove the voltage profile and capacity on a distribution feeder.

The second variety of housed equipment can includes larger rated banks, up to6,000 kVar or 6 Mvar. The capacitor units are individually fused. SuitableVacuum switches or Contactors can be utilized and operated by automaticsensing and control.

5.3.3.2 Capacitor connection

There are a number of ways in which a capacitor bank may be connected, withthe choice being dependent on:-

1. The voltage level of the system2. The kVar capacity of the Capacitor Bank3. The system grounding4. The desired relay protection

Once individual capacitor units are selected to meet the voltage requirements ofthe system, then the number of parallel unit are selected to meet the bank kVarrequirements.



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Two criteria are applied to determine the minimum allowable number ofparalleled capacitor units in each phase.

These criteria are as follows:

The loss of one capacitor unit in a phase should not produce a voltage acrossthe remaining units in that phase exceeding 110 % of rated voltage.

In the event of a failure of a unit, sufficient fault current should flow to ensureclearing in 300s or less. It should be pointed out that the 300s time span is amaximum and 30 s or less is a more desirable time span.

5.3.3.3 Y-Y Ungrounded Connection

Industrial and commercial capacitor banks are normally connected ungroundedwye, with paralleled units to make up the total kvar. It is recommended that aminimum of 4 paralleled units to be applied to limit the over voltage on theremaining units when one is removed from the circuit. If only one unit is neededto make the total kVar, the units in the other phases will not be overloaded if itfails.

In an industrial or commercial power systems the capacitors are not grounded fora variety of reasons. Industrial systems are often resistance grounded. Agrounded wye connection on the capacitor bank would provide a path for zerosequence currents and the possibility of a false operation of ground fault relays.

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



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5.3.3.4 Y-Y Capacitor Bank design

Figure 17: Typical Medium Voltage Y-Y Capacitor design



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5.3.3.4.1 Y-Y Connection

CT

REACTOR

REACTOR

REACTOR

BUSBAR 11 KV

40 uH REACTOR

333kVar

333kVar

333kVar

333kVar

333kVar

red

40 uH REACTOR

333kVar

333kVar

333kVar

333kVar

333kVar

yellow

40 uH REACTOR

333kVar

333kVar

333kVar

333kVar

333kVar

UNBALANCE RELAY

Blue

BUSBAR 11 KV

Figure 18: Typical Medium Voltage Y-Y Capacitor single line diagram



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CT

REACTOR

REACTOR

REACTOR

BUSBAR 11 KV

2 MVar 3 MVar

Figure 19: Y-Y connection for a 5 Mvar Capacitor Bank 11 kV (1)

Description of a typical 5 MVar 11 kV Capacitor Bank

1. Rating per capacitor unit = 333 kVar, 6.35 kV (V LN)

2. Number of capacitor units for a 5 MVar capacitor bank

= 15 (15 x 333 kVar = 5 MVar)

3. Rated reactive current (I kvar ) = 5 Mvar / ( 3 x 11 kV) = 262 Amp

4. Connection of capacitor bank = Ungrounded Y-Y (2Y)

5. Sizing per Y (First Y = 2 Mvar, Second Y = 3 Mvar)



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40 uH REACTOR

833kVar

833kVar

red

40 uH REACTOR

833kVar

833kVar

yellow

40 uH REACTOR

833kVar

833kVar

UNBALANCE RELAY

Blue

BUSBAR 11 KV


Description of a typical 5 MVar 11 kV Capacitor Bank


2. Number of capacitor units for a 5 MVar capacitor bank

= 6 (6 x 833 kVar = 5 MVar)

3. Rated reactive current (I kvar ) = 5 Mvar / ( 3 x 11 kV) = 262 Amp

4. Connection of capacitor bank = Ungrounded Y-Y (2Y)

5. Sizing per Y (First Y = 2.5 Mvar, Second Y = 2.5 Mvar)



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40 uHREACTOR

225kVar

225kVar

225kVar

225kVar

225kVar

40 uHREACTOR

225kVar

225kVar

225kVar

225kVar

225kVar

40 uHREACTOR

225kVar

225kVar

225kVar

225kVar

225kVar

UNBALANCE RELAY

BUSBAR 11 KV

225kVar

225kVar

225kVar

225kVar

225kVar

225kVar

225kVar

225kVar

225kVar


Description of a typical 5.4 MVar 11 kV Capacitor Bank


2. Number of capacitor units for a 5.4 MVar capacitor bank

= 24 (24 x 225 kVar = 5.4 MVar)

3. Rated reactive current (I kvar ) = 5.4 Mvar / ( 3 x 11 kV) = 283 Amp

Connection of capacitor bank = Ungrounded Y-Y (2Y)



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5.3.3.4.1.1 Series connection

Basic rating of MV capacitor unit: Q kVar, V LN (Voltage phase-neutral)

Example: 333 kVar, 6.35 kV, 26.3 uF

Configuration of 3-phase MV Capacitor: Ungrounded Star (Y)

Example 1: Configuration of 1.0 MVar, 11 kV Capacitor Bank

333 kVar

1 1 k V

1 1 k V

333 kVar 333 kVar

Three phase MVar rating = 0.33 MVar + 0.33 MVar + 0.33 MVar = 1.0 MVar

Phase to phase voltage rating= 3x6.35 kV = 11 kV



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Basic rating of MV capacitor unit: Q kVar, V LN (Voltage phase-neutral)

Example: 333 kVar, 6.35 kV, 26.3 uFConfiguration of 3-phase MV Capacitor: Ungrounded Star (Y)

Example 2: Configuration of 0.333 MVar, 33 kV Capacitor Bank

333 kVar

333 kVar

333 kVar

Red Phase

1 1 1 k V a r

Yellow Phase

Blue Phase

3 3 k V

Three phase MVar rating = 0.11 MVar + 0.11 MVar + 0.11 MVar = 0.33 MVar

Phase to phase voltage rating= 3x(3x6.35 kV) =33 kV



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5.3.3.4.1.2 Parallel connection

C1 C2 C3

a. Equivalent kVar (Q) per phase

Example:

Rating of each capacitor unit (C1, C2, C3): 333 kVar, 6.35 kV, 26.3 uF

Calculation:

The equivalent reactance (X T) = XC1 + XC2 + XC3

Formula: Capacitive reactance (X C) = 1/ C, = 2x3.14x50

The equivalent reactance (X T) = 1/ C1 + 1/ C2 +1/ C3 = 1/ CT

1/ C T =1/ C 1 + 1/ C 2 +1/ C 3

Equivalent kVar per phase = (Voltage LN)2 x x C T

Three phase voltage rating = 3x(3x6.35 kV) = 33 kV

Sample calculation:

1/C T = 1/(26.3uF) + 1/(26.3uF) + 1/(26.3uF) =0.114

CT = 8.77 uF

Total Q per phase = 6.35kV 2 x x C T = 111 kVar

Three phase Q = 3 x 111 kVar = 333 kVar = 0.333 Mvar



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C1 C2 C3

b. Equivalent kVar (Q) per phase

Example:

Rating of each capacitor unit (C1, C2, C3): 333 kVar, 6.35 kV, 26.3 uF

Calculation:

The equivalent reactance (1/X T) = 1/X C1 + 1/X C2 + 1/X C3

Formula: Capacitive reactance (X C) = 1/ C, = 2x3.14x50

The equivalent reactance (1/X T) = C1 + C2 +C3 = C T

C T =C 1 + C 2 + C 3

Equivalent kVar per phase = (Voltage LN)2 x x C T

Three phase voltage rating = 3x6.35 kV = 11 kV

Sample calculation:

CT = 26.3uF + 26.3uF+ 26.3uF = 78.9 uF

Total Q per phase = 6.35kV 2 x x C T = 999 kVar

Or simply adding up all the capacitor values in parallel

= 333 kVar+333kVar+333 kVar = 999 kVar



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MEDIUM VOLTAGE CAPACITOR & POWER FACTOR MANAGEMENT

Exercise No.6 Y-Y connection for 5 MVAR 11 kV

NOTA:

ILSAS/MFF/2007


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Single line diagram for 5 MVAR 11 kV

ILSAS/MFF/2007


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Exercise No.7 Y-Y connection for 4 MVAR, 11 kV

Capacitor Unit333 kVar6.35 kV

ILSAS/MFF/2007


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ILSAS/MFF/2007


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ILSAS/MFF/2007


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Exercise No.10 Current for 5 MVAR, 33 kV


ILSAS/MFF/2007


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Exercise No.11 Current for 3 MVAR, 11 kV


ILSAS/MFF/2007


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ILSAS/MFF/2007Page 60 of 136

5.3.4 Calculation of Capacitive Current (I kvar )

Power factor = cos

Ic = kvar /( 3 x V LL) 3-phase (16)

Ic = kvar /(V LN) 1-phase (17)

From equation (13)

kVar for PF Correction kVar = kW x (tan 1-tan 2)

Figure 21: Capacitive current for a 5 Mvar Capacitor Bank 11 kV


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ILSAS/MFF/2007Page 61 of 136

5.3.5 Capacitor current during failure of capacitor cans

Figure 22: Calculation for 1 capacitor can failure


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ILSAS/MFF/2007

Figure 24 Connection diagram in uF for 5 MVar 11 kV


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ILSAS/MFF/2007

Figure 25 Connection diagram in ohm for 5 MVar 11 kV


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ILSAS/MFF/2007

Figure 26 Calculation of Amp based on Total Impedance


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5.3.7 Estimation of neutral current

I A = [IR2 + IY2 + IB2] (28)

IB = [ (I R x IY ) + ( I Yx IB ) + ( I B x IR) ] (29)

I neutral = [I A I B] (30)


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On a wye-ungrounded capacitor bank, internal section faults subject the powersystem to a fault current that is three times the banks rating (until the capacitorfuse blows). Therefore, the voltage sag, mechanical and thermal stressingassociated with the fault, and case rupture concerns are reduced .

5.4.2 Calculation of Capacitor value (kVar) for Delta Capacitor

kVar (3-phase)

= 2/3 x (Total 3 phase uF) x x Volt 2 /1000 (31)

= 2x x f f = 50 Hz

Volt= Volt rating at Capacitor Can

Example:

Voltage rating per Capacitor 415 Volt

uF

Figure 31: Capacitance measurement for Delta Capacitor using Capacitance

meter

Capacitance values for phase-phase values

R-Y 375 uFY-B 375 uFB-R 375 uF


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5.4.3 Calculation of Capacitive Current (I kvar )

Power factor = cos

Ic = kvar /( 3 x V LL) 3-phase (32)

Example:

Capacitor kVar (3-phase delta) = 50 kVar

Voltage rating = 415 Volt

Ic = kvar /( 3 x V LL) = 50 kVar / ( 3 x 415 Volt ) = 69.56 Amp

Exercise No.14 Calculation of kVar for Delta Capacitor

Voltage rating per Capacitor 3.3 kV

Capacitor values (3-phase delta) 1 MVAR


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6.0 PROTECTION OF Y-Y CAPACITORS

Complete protection must be provided for a capacitor installation. This willconsist of protecting the individual units as well as the bank. Both fuses and

relays must be employed depending upon the rating of the bank, its location inthe system and other factors.

Fusing is used to remove a failed capacitor unit from the system.

Relays and breakers are applied for overall bank protection and for switching.

6.1 Capacitor fusing

The major purposes of capacitor fusing are:

1. To maintain service continuity2. To prevent damage to adjacent capacitors and equipment or injury topersonnel

3. To provide visual indication of a failed unit4. To limit the energy into a faulted unit to help prevent case rupture

The requirements for proper fuse selection are as follows-

a. The rated voltage of the fuse should not be less than rated voltage of thecapacitor with which it is used. Since capacitors are designed to operatecontinuously at 110 % rated voltage, the fuse should also have a voltage

rating, which has at least 110 % of the capacitor unit rating.b. The maximum interrupting rating of the fuse should be greater than the

available short circuit current, which can flow if a capacitor unit is shorted.This may the application of current limiting fuses in place of explosion fusesfor large bank rating and for banks connected to buses with high short circuitcapacity.

c. The fuse should have a time-current clearing characteristic that lies below thetime-current case rupture probability characteristics of the capacitor units tobe applied.

d. The selected fuse should have sufficient rating to carry 165 % of ratedcapacitor currents. This margin allows for temporary over voltage, harmoniccurrents, switching surges, and manufacturing tolerance in the capacitor itself.

e. The fuse must clear the minimum over current resulting due to a failed unitwithin 300 seconds maximum and as stated earlier 30 seconds or less isdesirable. On grounded wye and delta-connected banks, obtaining this


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clearing time is not a problem. For ungrounded wye connected banks, it ismore difficult since a failed unit causes about 3 times normal fault current toflow through the fuse protecting the faulted phase. In some cases, it becomesnecessary to reduce the rating of the fuse to minimum current of 150 % of

normal. This is permissible because ungrounded banks do not needallowance for zero sequence currents.

f. The fuse must be capable of withstanding the energy contributed to a unit byother capacitors in the same group as the faulted unit.

6.2 Relaying protection

In addition to the individual capacitor unit protection through the use of theindividual or group fusing, additional relay protection for the whole bank is

required. The 2 basic type of relay protection used with medium voltagecapacitors are;

1) over-current relaying of major equipment faults and2) relaying to identify loss of units within a bank.

Over-current relaying is needed for removal of the capacitor bank in the event ofa fault between the switching device and the bank itself.

The relay should be chosen so that the highest magnitude of inrush currentassociated with the capacitor switching will not trip the circuit breakerimmediately as the bank is energized. Also, if a capacitor unit fault occurs therelay should delay operation until the fuse clears.

Relays with an inverse time current characteristics are usually used and settingsare selected to override these 2 conditions. However, the relay may also have avery inverse or extremely inverse time current characteristics if coordination withother system protective devices required it.

Instantaneous over current relays are not often utilized since they are likely to tripunnecessarily when the bank is energized.

Relaying for the detection of the loss of capacitor units and guarding againstexcessive operating voltages should be used as protective measure supplementto the periodic visual inspection of individual unit fuses. This is because there willbe a compromise between a relaying method of adequate sensitivity which isalso immune to undesirable operation due to harmonics, system voltageunbalances, external faults and unbalance due to the varying capacitance of theindividual capacitor units.


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It is important to note that except for short periods of time, the voltage across thecapacitors should not exceed 110 % of rated voltage. On large banks, it isrecommended that some type of protection be provided which will relay the bankoff if the capacitors in a parallel group are subjected to over voltage, because of

the loss of a portion of the units in the group.This would also protect the bank when an entire parallel group was shorted by aforeign object.

6.2.1 Common scheme for ungrounded bank

Figure 33 is an effective arrangement for ungrounded bank. The potentialtransformer (PT) will detect a change in voltage across the phase if one or moreunits are removed. The secondary windings of the PT are connected in broken

delta and used in series with the coil of a voltage sensitivity relay.

UnGroundedNeutral

PT Primary

PT Secondary

Voltage Relay

Figure 33 Unbalance Phase Voltage


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PT

Voltagerelay

Figure 34: Voltage Imbalance between neutrals

Current

relay

CT

Figure 35: Current Imbalance between neutrals


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Potential transformers have an advantage over capacitor potential devices in thatthey furnish a discharge path for the capacitors, and will discharge the capacitorsin a much softer time than the discharge resistors built into the capacitors. Use ofthe potential transformers as a discharge path will be effective only if the

transformers have the thermal capability to pass the current associated with thestored energy in the bank. The kVar rating of the bank will determine the energystorage capability and this should be matched against the energy absorptioncapability of as transformer as shown in Figure 33.

The internal resistors built into each individual capacitor unit will also dischargethe bank but this requires about five minutes for high voltage units and oneminute for low voltage units. For automatically controlled banks this may be toolong a time.

If the bank is switched on while still charged from a previous energization higher

than normal transient currents may be experienced. The time delay and normaloperating sequences of any switching controls should be reviewed to insure thatthe bank will have a sufficient time to discharge to a safe level before being reenergized.

The schemes shown in Figure 34 & 35 are known as double-wye schemes andthey have considerable merit. They offer protection similar to that of Figure 34and are less expensive since they require only a single PT or CT. For smallerbanks, it may be expensive to split the banks into 2 wye groups, or it may beimpossible to do so and still keep the voltage on the remaining units below 110 %of rated in the event of the loss of an individual unit.

On larger banks it may be easier to connect the bank into 2 wye configurations.In such case, the protection provide by the relays is equivalent and can beachieved at less cost than Figure 33, making the double wye protection moreattractive, particularly on industrial systems where there is no advantage togrounding the neutral connection of the bank.


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6.2.2 Common scheme for single wye capacitor bank

When conditions are such that single wye banks must be used, the scheme ofFigure 36 & 37 can be used for over voltage protection. This provides a low cost

protective arrangement but it requires that the neutral point be grounded. Asnoted earlier, the ground arrangement is not often used by industrials for severalreasons, one of them, being possible interference from ground relaying. Use of athird harmonic filter would permit setting the relay at a fairly low value of pickup.

PT

Voltagerelay

Resistor

Voltagerelay

Figure 36 Neutral Voltage Figure 37 Neutral Current

No over voltage protection scheme will provide positive protection against overvoltage all cases. The basic theory of all schemes is the detection of current orvoltage unbalance. There is some inherent unbalance in capacitor banks, andthe relaying must be set above the maximum inherent unbalance which canoccur if false trip-outs are to be avoided. The inherent unbalance is due to mainlyto harmonic currents and voltages, and varying capacitance of the capacitors.

The tolerance of a power capacitor is minus zero to + 15 % with the averagebeing about 4 %. When individual units are placed in racks and stacked for highvoltage banks, the capacitance of the racks in each phase leg may vary. Thiscauses normal condition unbalance and makes it more difficult to detect smallunbalance due to removal of faulted units if a parallel group. If it is considerednecessary to balance the phases closely, it is possible by special arrangement atthe factory. The problem of phase unbalance tends to diminish as the bank ratingincreases.

In cases where a large number of capacitor units, say 20 or more, are used in aparallel group, it may not be possible to detect the loss of one unit, since therelays cannot be set to sensitive because of natural unbalances.


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This is not too important, since the loss of one unit in such a large group does notraise the voltage on the remaining units above the allowable 10 % over voltage.In most cases, the relaying can be set to operate because dangerous conditionsexist, without serious danger of false operation.

On large installation, it is good practice to use 2 relays. One will sound an alarmwhen one or more units have failed but dangerous voltages are not yet present.The second relay will trip if allowable over voltage is exceeded. Such procedurehas the advantage of keeping the bank in service when possible while indicatingthat capacitors have failed yet still protect the capacitors from serious over-voltages.

The summary of the characteristics of the several relay arrangements is shown inTable 2.

Table 2: Characteristics of Protective Relaying Methods

Type ofprotectiverelaying

Figure SensitivetoSwitchingTransient

Sensitiveto 3 rd harmonicvoltage &currents

Sensitiveto systemvoltageunbalance

Numberof CTrequired

Numberof PTrequired

Unbalancephasevoltage

33 Yes No No - 3

Voltageunbalancebetweenneutral

34 Yes No No - 1

Currentunbalancebetweenneutrals

35 Yes No No 1 -

Neutralvoltage

36 Yes Yes Yes - 1

Neutral

current

37 Yes Yes Yes 1 -


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CT

REACTOR

REACTOR

REACTOR

11 kV Fuses

2 MVar 3 MVar Unbalance CT

Ammeter

A B

Relays

Description of relays: A- Overcurrent & earthfault relaysB-Unbalance relay

Figure 38 Relay configurations for a 5 Mvar 11 kV capacitor


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6.3 The concept of unbalance current flow

If all the capacitor units in the Y-Y configuration are healthy, the value of thereactive currents at the phases are equal, thus there will be no current flow in the

neutral conductor.Once a capacitor unit deteriorates, its capacitance value will be reduced. Theoverall capacitance for the particular phase the capacitor unit is connected will nolonger equal to the other two phases.

Different capacitance will give different values of reactive current flows. A neutralcurrent will then be introduced in the neutral conductor.

If the neutral current exceed the prescribed limits i.e. 6-9 Amp, then the voltageacross the capacitors will exceed 110 % of the rated voltage.

This will lead to the damage of the other capacitor units.

The flow of the neutral current can be detected by installing a CT across theneutral conductor as shown in the diagram above. The size of the CT is 10/5.

The CT is then connected to a specific relay i.e. to be used to monitor the currentflow and for tripping purposes. The monitoring of the neutral current is done intwo stages.

The first stage, setting of 60 % of the CT rating, is aimed to be as an alarm toinform the user of the possible deterioration process.

The second stage, setting of 90 % of the CT rating, is aimed for trippingpurposes. If the amount of neutral current is 90 %, the circuit breaker will trip offthe circuit connected to the capacitor bank.


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6.4 Basics on relay unbalance & Unbalance CT

Description: -

The unbalance CT & relay monitor the neutral current inside the capacitorbank.

A healthy Capacitor bank thats operational will give neutral current = 0.

When one of the capacitor can fail, the neutral current 0

6.4.1 Types of protection schemes

a. Stage One

This scheme will be activated when one or more capacitor cans experiencefailure. The circuit breaker for the feeder will not be activated if there is noexistence of voltage in the capacitor bank more than 110 % of nominal.

b. Stage Two

This scheme will be activated when the voltage level inside the capacitor bankexceed 110 %.

6.4.2 Example on settings of unbalance relay & OCEF relay

Relays CT Ratio 1st stage 2nd stage Setting TMS

O/C, E/F 300/5 100% - 5 Amp 0.2

Unbalance 10/5 60% 90% 3 Amp 0.5


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6.5 Typical settings for a 11 kV 3-5 MVar Y-Y Capacitor Banks

In this example, the capacitance values of 4 capacitor units in the red phasehave deteriorated from 26.3 uF to 24.0 uF. The overall value of available reactive

power for the red phase is thus reduced from 1.7 Mvar to 1.5 Mvar.Due to this unbalance between the phases, a neutral current of 7.8 Amp is thendetected at the neutral conductor by the unbalance CT.

The first stage of the unbalance relay will then be activated.

Table 2 Example on relay settings

Capacitor

Bank (MVar)

kV Rated

Amp

CT for

relay

O/C E/F TMS Un-

balanceCT

1st

stage

2nd

stage5 11 262 400 / 5 75% 20% 0.10 10 / 5 60% 90%4 11 210 400 / 5 75% 20% 0.10 10 / 5 60% 90%3 11 157 400 / 5 50% 20% 0.10 10 / 5 60% 90%

Capacitor

Bank (MVar)

kV Rated

Amp

CT for

relay

O/C E/F TMS Un-balance

CT

1st

stage

2nd

stage5 11 262 300 / 5 100% 20% 0.10 10 / 5 60% 90%4 11 210 300 / 5 75% 20% 0.10 10 / 5 60% 90%3 11 157 300 / 5 75% 20% 0.10 10 / 5 60% 90%


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6.6 Lightning protection

In common with other apparatus connected to a power system, capacitors shouldbe protected by surge arresters. Here the primary function is protection of the

capacitor bank and not the system itself. The choice between surge arresterrated for grounded neutral service or ungrounded neutral service should be inaccordance with established industry practices.

The rating of surge arresters (line-to-ground) is determined by the systemgrounding, and is not related to the grounding or lack thereof, of the capacitorsthemselves.

Occasionally, the suggestion is made that surge arresters are unnecessary forcapacitor banks which are Y-Connected with grounded neutral. It is true thatcapacitors so connected do have some ability to slope off the crest front of an

incoming wave and to reduce its crest value. However, this ability is limited bythe size of the capacitor bank and the amount of energy to be expected in agiven surge or lighting stroke is indeterminate.

In the usual case for an industrial system with an ungrounded capacitor bank,properly applied surge arresters will help protect the capacitors and other systemcomponents by shunting the surge current to ground.


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Three phase capacitors use fuses in the line because they are connected deltaternally. Normally branch fuses are used for single-phase capacitors connected

hen the bank has higher kvar ratings and units are placed in parallel, the in line

or example, consider both fusing methods for a 450 kVAR, 4160 volts delta

93.68 amps

50 kVAR/4.16 KV = 36 amps x 1.5 = 54 amps

. a 60-amp fuse can be used.

here are other potential problems in fusing a delta-connected bank with "inranch" fusing. It is a normal practice utilized in metal enclosed banks to installo bushing capacitors connected phase to phase with the capacitor tank

rounded to the frame. In some cases, the user only applies one fuse per phase.his could be dangerous. When a capacitor starts to fail and the fuse operatese capacitor is still in the circuit via the second bushing. The failure within the

apacitor is fed thru this connection and eventually the major insulation of thean will fail and the capacitor tank will rupture.

he other method is to use two fuses, i.e. one per bushing. This gives the user alse sense of security. In this case both fuses would have to operate before theiled capacitor can be effectively removed from the system. Normally only one ofe fuses operates, which will be the one nearest the faulted packs. The other

ushing remains connected to the system via the good fuse. The result is still anventual major insulation failure if the bank is not removed from service.

Indelta. However, on the smaller banks mentioned above, the single phasecapacitors could be connected delta and fused outside the delta (In the line.) On

small banks that have only one capacitor per phase, this should be the method ofchoice when the neutral of the capacitor bank is not grounded.

Wfusing becomes large, and may not coordinate with the tank rupture curve of thecapacitor and the upstream co-ordination may not be possible.

Fconnected bank, using 150 kVAR per phase, will require the following fusing:

"In line fusing" Or, "group fusing"450 kVAR/(4.16 KV x 3) = 62.45 amps x 1.5 =

... a 100-amp fuse is required.

In branch fusing

1

..

TbtwgTthcc

Tfafathbe


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The burning between packs could possibly continue due to the second bushingtill being energized via the second fuse. During this condition a low energy fault

r Delta & Y capacitors

d other factors.

Fusing is used to remove a failed capacitor unit from the system.Relays and breakers are applied for overall bank protection and for switching.

Figure 40: Connection and fusing arrangements for ungrounded-wye anddelta connected capacitor banks

scould be developed. The current limiting fuse still in the circuit will be gettingwarm while the capacitor could be boiling. Eventually the major insulation will be

breached grounding the faulted capacitor through the tank to frame, and therewill be a race between the capacitor and the fuse to see if the fuse will clearbefore the capacitor ruptures.

7.2 Comparison between fusing arrangement fo

Complete protection must be provided for a capacitor installation. This willconsist of protecting the individual units as well as the bank. Both fuses andrelays must be employed depending upon the rating of the bank, its location inthe system an



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Fusing: Figure 40 also shows common fusing practices for each of the bankarrangements. The Figure shows that the delta connected bank can be protectedby placing the fuses inside or outside of the delta. Two fuses per single phasecapacitor are required when fusing inside of the delta, but their rating is

decreased to 57% of the outside fuse rating. The fuses outside of the delta aresized in the same way as the fuses for the ungrounded-wye connected capacitorank.

hich basically shorts the capacitor terminals. Whether the capacitors are delta

Internal section faults, or dielectric faults appear differently to a power system.

al section fault subjects the power system toa phase-to-phase bolted fault. This fault will cause a major voltage sag on themay cause capacitor

case rupture if not properly protected.

It also subjects the power system to high magnitude fault currents, which canp and thermal stress on components in the fault path. On a

wye-ungrounded capacitor bank, internal section faults subject the power systems rating (until the capacitor fuse

blows). Therefore, the voltage sag, mechanical and thermal stressing associatedconcerns are reduced .

To limit the energy into a faulted unit to help prevent case rupture

voltagerating, which has at least 110 % of the capacitor unit rating.

b Fault Conditions: A capacitor typically fails in two ways: 1) A bushing to casefault occur

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