Best Book for Capacitor Bank.pdf

CAPACITOR BANK & POWER FACTOR MANAGEMENT

CAPACITOR B BANK & POWER F FACTOR M MANAGEMENT

Electrical Chargeman Category B0

ILSAS/MFF/2007 ILSAS/MFF/2007

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Contents Title 1.0

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Page No. FUNDAMENTALS OF BASIC ELECTRICITY

Definition of electrical power Apparent Power (VA) Real Power or True Power (Watts) Reactive Power (Var) Power Factor Relationship between Real, Reactive, and Apparent power power Equipment Causing Poor Power Factor Calculation for power factor

6 6 6 6 7 7 7 13 14

2.0

EFFECT OF LOW POWER FACTOR

20

3.0

POWER FACTOR CORRECTION

21

3.1 3.2 3.3 3.4

Capacitive Power Factor correction Function of PFC Capacitors How capacitors work Categories of Power Factor Correction

21 23 24 25

4.0

CAPACITOR BANK SIZING

30

4.1 4.2

30 33

4.3

Selecting kVAR for 3-Phase Motors Power Factor Correction Capacitors on Reduced Voltage Motors and Multi-Speed Motors Selecting kVAR for Bulk Correction

5.0

MV CAPACITOR BANK DESIGN

38

5.1 5.2 5.3 5.4

Function of a capacitor bank Capacitor unit & Capacitor Bank configuration Capacitor construction Basic design of power capacitor

38 38 40 69


34

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Contents Title 1.0

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Page No. FUNDAMENTALS OF BASIC ELECTRICITY

Definition of electrical power Apparent Power (VA) Real Power or True Power (Watts) Reactive Power (Var) Power Factor Relationship between Real, Reactive, and Apparent power power Equipment Causing Poor Power Factor Calculation for power factor

6 6 6 6 7 7 7 13 14

2.0


20

3.0


21

3.1 3.2 3.3 3.4

Capacitive Power Factor correction Function of PFC Capacitors How capacitors work Categories of Power Factor Correction

21 23 24 25

4.0


30

4.1 4.2

30 33

4.3

Selecting kVAR for 3-Phase Motors Power Factor Correction Capacitors on Reduced Voltage Motors and Multi-Speed Motors Selecting kVAR for Bulk Correction

5.0

MV CAPACITOR BANK DESIGN

38

5.1 5.2 5.3 5.4

Function of a capacitor bank Capacitor unit & Capacitor Bank configuration Capacitor construction Basic design of power capacitor

38 38 40 69


34

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6.0

PROTECTION OF Y-Y CAPACITORS

73

6.1 6.2 6.3 6.4 6.5 6.6

Capacitor fusing Relaying protection The concept of unbalance current flow Basics on relay unbalance & Unbalance CT Typical settings for a 11 kV 3-5 MVar Y-Y Capacitor Banks Lightning protection

73 74 81 82 83 84

7.0

PROTECTION FOR DELTA CAPACITORS

85

7.1 7.2

85 87

7.3

Fusing arrangement for Delta Capacitors Comparison between fusing arrangement for Delta & Y capacitors Relaying protection

8.0

MAINTENANCE PROCEDURES FOR A Y-Y CAPACITOR BANK

8.1 8.2 8.3

Safety Regulations Maintenance Methodology for Y-Y Capacitor Banks Maintenance program

9.0

MAINTENANCE PROCEDURES FOR A DELTA CAPACITOR BANK

9.1 9.2 9.3

Safety Regulations Maintenance Methodology for Delta Capacitor Banks Maintenance program

10.0

COMMISSIONING

10.1 10.2 10.3 10.4 10.5

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

90

91 91 95

108 108 111

121 121 129 132 134

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


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ILSAS/MFF/2007

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ILSAS/MFF/2007

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1.0

FUNDAMENTALS OF BASIC ELECTRICITY

1.1

Definition of electrical power

Power is the ability of a system to perform work. Water performs work when it turns the turbine blades of a hydroelectric plant. Electricity performs work when it heats up a heating element or turns a motor. It takes power to store energy, like in capacitive or inductive devices, while these devices then release some energy, or power, at a later time. 1.2

Apparent Power (VA)

Apparent power is defined as the power that is "apparently" absorbed by a system. That is, the product of current times voltage tells us a device appears to be using a certain amount of power. Generators are rated in volt-amperes (VA), or thousand-volt-amperes (kVA). 1.3

Real Power or True Power (Watts)

The real amount of power a device is using, or results in actual work performed, is called the "real power". The real power tells us how much actual work can be performed, or how many horsepower our motor is delivering. For a resistive and/or DC circuit, the apparent power and the real power are the same, but for a capacitive or inductive circuit, the real power is heavily dependent on the amount that the current or voltage is delayed. Real power is presented in Watts. There is mathematically no difference between watts and volt-amperes, except that we use one term for apparent power, and one for real power, but they are both units of power. We use the power factor to go from apparent power to real power. Real power = P=VI R = VI cos θ

(1)

IR = I cos θ cos θ = power factor

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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 θ

(2)

Q refers to the maximum value of the instantaneous power absorbed by the reactive component of the load. The instantaneous reactive power is alternately positive and negative, and it expresses the reversible flow of energy to and from the reactive component of the load. 1.5

Power Factor

When we use any capacitive or inductive device on an AC circuit, the current or voltage 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 a capacitor the voltage lags behind the current. For an inductive load, the current lags the voltage, and the power factor is said to be lagging. For capacitive loads, the current leads the voltage, and the power factor is said to be leading. For a purely capacitive or inductive circuit with zero resistance, the angle of lead/lag is 90°. Adding resistance to the circuit will decrease the leading/lagging angle. The term "Power factor", is the cosine of the phase angle. For a purely inductive circuit, the lag angle is 90°, and the power factor is zero [cosine (90 0)=0]. A common power factor for electric motors is 0.8, which gives us a lagging angle of 36° (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 zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. As mentioned above, this “phantom 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 being used, 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 reactive power and true power is called apparent power , and it is the product of a circuit's voltage and current, without reference to phase angle. Apparent power is measured 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, usually resistances (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're dealing 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. For instance, if I'm calculating true power from current and resistance, I must use the polar magnitude for current, and not merely the “real” or “imaginary” portion of the current. If I'm calculating apparent power from voltage and impedance, both of these formerly complex quantities must be reduced to their polar magnitudes for the scalar arithmetic. There are several power equations relating the three types of power to resistance, 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 and reactive power. There are three equations available for the calculation of apparent power, P=IE being useful only for that purpose. (E is equivalent to Voltage). 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 in Figure below and a resistive/reactive load in Figure below.

1.6.1 Resistive load only:

120 Volt 50 Hz

Figure 1:

ILSAS/MFF/2007

True power, reactive power, and apparent power for a purely resistive load.

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1.6.2 Reactive load only:

120 Volt 50 Hz

Figure 2:

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True power, reactive power, and apparent power for a purely reactive load.

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1.6.3 Resistive/reactive load:

120 Volt 50 Hz

Figure 3:

True power, reactive power, and apparent power for a resistive/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 (amount of 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:

1.7

•

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

•

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

•

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

•

These three types of power are trigonometrically related to one another. In a 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.

Equipment Causing Poor Power Factor

A great deal of equipment utilized by todays modern industry causes poor plant power 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 machines and weaving machines. 60% power factor and below: Single stroke presses, automated machine tools, finish grinders, welders. When the above equipment functions within a plant, savings can be achieved by utilizing 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 work output and more particularly is a good indicator of the effect of the load current on the efficiency of the supply system. All current flow will cause losses in the supply and distribution system. A load with a power factor of 1.0 results in the most efficient loading of the supply and a load with a power factor of 0.5 will result in much higher losses in the supply system. A poor power factor can be the result of either a significant phase difference between the voltage and current at the load terminals, or it can be due to a high harmonic content or distorted/discontinuous current waveform. Poor load current phase angle is generally the result of an inductive load such as an induction motor, power transformer, lighting ballasts, welder or induction furnace. 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 φ

Power Factor (PF) =

kW = KVA

|kVA| = √ [(kW)2 + ( kVAr)2]

(6)

True Power Apparent Power

(7)

(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 amount of absorbed/returned power. It also happens to be the same angle as that of the circuit's impedance in polar form. When expressed as a fraction, this ratio between true power and apparent power is called the power factor for this circuit. Because true power and apparent power form the adjacent and hypotenuse sides of a right triangle, respectively, the power factor ratio is also equal to the cosine 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 less quantity. For the purely resistive circuit, the power factor is 1 (perfect), because the reactive power equals zero. Here, the power triangle would look like a horizontal line, because the opposite (reactive power) side would have zero length. For the purely inductive circuit, the power factor is zero, because true power equals zero. Here, the power triangle would look like a vertical line, because the adjacent (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 purely capacitive circuit would again be a vertical line (pointing down instead of up as it was for the purely inductive circuit). Power factor can be an important aspect to consider in an AC circuit, because any power factor less than 1 means that the circuit's wiring has to carry more current than what would be necessary with zero reactance in the circuit to deliver the same amount of (true) power to the resistive load. If our last example circuit had been purely resistive, we would have been able to deliver a full 169.256 watts to the load with the same 1.410 amps of current, rather than the mere 119.365 watts that it is presently dissipating with that same current quantity. The poor power factor makes for an inefficient power delivery system. ILSAS/MFF/2007

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Poor power factor can be corrected, paradoxically, by adding another load to the circuit drawing an equal and opposite amount of reactive power, to cancel out the effects of the load's inductive reactance. Inductive reactance can only be canceled 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 opposing reactances in parallel is to bring the circuit's total impedance equal to its total resistance (to make the impedance phase angle equal, or at least closer, to zero). 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 the source (of known voltage), we'll use the power formula, which starts from voltage and reactance:

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Let's use a rounded capacitor value of 22 µF and see what happens to our circuit:

120 Volt 50 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. The main current has been decreased from 1.41 amps to 994.7 milliamps, while the power dissipated at the load resistor remains unchanged at 119.365 watts. The power 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 correction efforts had been perfectly on-target, we would have arrived at an impedance angle of exactly zero, or purely resistive. If we had added too large of a capacitor in 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 low power factor just as well as too much inductance. You must be careful not to over-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 power system voltages and the occasional voltage spike from lightning strikes, for continuous 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 is leading . 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 the load's reactance. If the load's reactance is inductive in nature (which it almost always will be), parallel capacitance is what is needed to correct the 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


"Power Factor" is an electrical term used to rate the degree of the synchronization of power supply current with the power supply voltage. It is important that we clearly understand the meaning of "Power Factor" and its effect on 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 power factor rating as a criterion in choosing among competing motors, especially when a large motor is involved.

The power factors in industrial plants are usually lagging due to the inductive nature 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 current must be transmitted than is actually used to perform useful work. This increased cost 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 electrical transmission system which means that the industrial power user must spend more on transmission lines and transformers to get a given amount of useful power through his plant.

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3.0


A poor power factor due to an inductive load can be improved by the addition of power factor correction (PFC) capacitor, but, a poor power factor due to a distorted current waveform requires change in equipment design or expensive harmonic filters to gain an appreciable improvement. Many inverters are quoted as having a power factor of better than 0.95 when in reality, the true power factor is between 0.5 and 0.75. The figure of 0.95 is based on the cosine of the angle between the voltage and current but does not take into account that the current waveform is discontinuous and therefore contributes to increased losses on the supply 3.1

Capacitive Power Factor correction

Capacitive Power Factor correction is applied to circuits, which include induction motors as a means of reducing the inductive component of the current and thereby reduce the losses in the supply. There should be no effect on the operation of the motor itself. An induction motor draws current from the supply that is made up of resistive components 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: ILSAS/MFF/2007

Magnetizing current Page 21 of 136


The current due to the leakage reactance is dependant on the total current drawn by the motor, but the magnetizing current is independent of the load on the motor. The magnetizing current will typically be between 20% and 60% of the rated 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 to operate. The magnetizing current does not actually contribute to the actual work output of the motor. It is the catalyst that allows the motor to work properly. The magnetizing current and the leakage reactance can be considered passenger components of current that will not affect the power drawn by the motor, but will contribute 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 of 0.75 The resistive component of the current is 75 Amps and this is what the KWh meter measures. The higher current will result in an increase in the distribution losses of (100x 100)/(75x75) = 1.777 or a 78% increase in the supply losses. In the interest of reducing the losses in the distribution system, power factor correction is added to neutralize a portion of the magnetizing current of the motor. 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 to reduce wasted energy in the distribution system, the consumer will be encouraged to apply power factor correction. Power factor correction is achieved by the addition of capacitors in parallel with the connected motor circuits and can be applied at the starter, or applied at the switchboard or distribution panel. The resulting capacitive current is leading current and is used to cancel the lagging inductive current flowing from the supply


Power Factor Correction (1) Page 22 of 136


3.2

Function of Power Factor Correction (PFC) Capacitors

PFC equipment provides the means of reducing the reactive power being supplied 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 by Capacitor Bank

b. With Power Factor Correction

Figure 7:

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Power Factor Correction (2)

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

Figure 8:

3.3

Power Factor Correction for Induction motors

How capacitors work

Induction motors, transformers and many other electrical loads require magnetizing current (kVAR) as well as actual power (kW). By representing these components of apparent power (kVA) as the sides of a right triangle, we can determine the apparent power from the right triangle rule: kVA 2 = kW2 + kVAR2. To reduce the kVA required for any given load, you must shorten the line that represents the kVAR. This is precisely what capacitors do. By supplying kVAR right at the load, the capacitors relieve the utility of the burden of carrying the extra kVAR. This makes the utility transmission/distribution system more efficient, reducing cost for the utility and their customers. The ratio of actual power to apparent power is usually expressed in percentage and is called 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 "Bulk Correction". 3.4.1 Bulk Correction The Power factor of the total current supplied to the distribution board is monitored by a controller which then switches capacitor banks In a fashion to maintain 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") as possible. 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 to the magnetizing current of induction motors, it is easy to correct each individual motor by connecting the correction capacitors to the motor starters. With static correction, it is important that the capacitive current is less than the inductive magnetizing current of the induction motor.

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In many installations employing static power factor correction, the correction capacitors are connected directly in parallel with the motor windings. When the motor is off-line, the capacitors are also off-line. When the motor is connected to the supply, the capacitors are also connected providing correction at all times that the motor is connected to the supply. This removes the requirement for any expensive power factor monitoring and control equipment. In this situation, the capacitors remain connected to the motor terminals as the motor slows down. An induction motor, while connected to the supply, is driven by 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, a magnetic field associated with the rotor. As the motor decelerates, it generates voltage out its terminals at a frequency which is related to it's speed. The capacitors connected across the motor terminals, form a resonant circuit with the motor inductance. If the motor is critically corrected, (corrected to a power factor of 1.0) the inductive reactance equals the capacitive reactance at the line frequency and therefore the resonant frequency is equal to the line frequency. If the motor is over corrected, the resonant frequency will be below the line frequency. If the frequency of the voltage generated by the decelerating motor passes through the resonant frequency of the corrected motor, there will be high currents and voltages around the motor/capacitor circuit. This can result in sever damage to the capacitors and motor. It is imperative that motors are never over corrected or critically corrected when static correction is employed. Static power factor correction should provide capacitive current equal to 80% of the 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 the rated current of the motor while smaller low speed motors can have a magnetizing current as high as 60% of the rated full load current of the motor. It is not practical to use a "Standard table" for the correction of induction motors giving optimum correction on all motors. Tables result in under correction on most motors but can result in over correction in some cases. Where the open shaft current cannot be measured, and the magnetizing current is not quoted, an approximate level for the maximum correction that can be applied can be calculated from the half load characteristics of the motor. It is dangerous to base correction on the full load characteristics of the motor as in 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 disconnected conditions. 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 the motor and one for the capacitors. Where one contactor is employed, it should be up sized for the capacitive load. The use of a second contactor eliminates the problems 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 a variable speed drive or inverter. The connection of capacitors to the output of an inverter can cause serious damage to the inverter and the capacitors due to the high frequency switched voltage on the output of the inverters. The current drawn from the inverter has a poor power factor, particularly at low load, but the motor current is isolated from the supply by the inverter. The phase angle of the current drawn by the inverter from the supply is close to zero resulting in very low inductive current irrespective of what the motor is doing. The inverter does not however, operate with a good power factor. Many inverter manufacturers quote a cos Ø of better than 0.95 and this is generally true, however the current is non sinusoidal and the resultant harmonics ILSAS/MFF/2007

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cause a power factor (KW/KVA) of closer to 0.7 depending on the input design of the inverter. Inverters with input reactors and DC bus reactors will exhibit a higher true power factor than those without. The connection of capacitors close to the input of the inverter can also result in damage 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 of the capacitors. It is recommended that capacitors should be at least 75 Meters away from inverter inputs to elevate the impedance between the inverter and capacitors and reduce the potential damage caused. Switching capacitors, Automatic bank correction etc, will cause voltage transients and these transients can damage the input circuits of inverters. The energy is proportional to the amount of capacitance being switched. It is better to switch lots 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 of a solid state soft starter. When a solid state soft starter is used, the capacitors must be controlled by a separate contactor, and switched in when the soft starter output voltage has reached line voltage. Many soft starters provide a "top of ramp" or "bypass contactor control" which can be used to control the power factor correction capacitors. The connection of capacitors close to the input of the soft starter can also result in damage to the soft starter if an isolation contactor is not used. The capacitors tend to cause transients to be amplified, resulting in higher voltage impulses applied to the SCRs of the Soft Starter, and the energy behind the impulses is much greater due to the energy storage of the capacitors. It is recommended that capacitors should be at least 50 Meters away from Soft starters to elevate the impedance between the inverter and capacitors and reduce the potential damage caused. Switching capacitors, Automatic bank correction etc, will cause voltage transients and these transients can damage the SCRs of Soft Starters if they are in the Off state without an input contactor. The energy is proportional to the amount of capacitance being switched. It is better to switch lots of small amounts of capacitance than few large amounts.

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Figure 11:

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Static correction for soft starter schemes

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4.0


The most common method for improving power factor is to add capacitors banks to the system. Capacitors are attractive because they're economical and easy to maintain. Not only that, they have no moving parts, unlike some other devices used for the same purpose. When you add a capacitor bank to your system, the capacitor supplies the reactive power needed by the load. If you size and select the capacitor bank to compensate to a unity power factor, it can supply all the reactive power needed by the load, and no reactive power is demanded from the utility. If you design the capacitor bank to improve the power factor to a quantity less than 1.0, the reactive power supplied by the bank will be its rated kVARs (or MVARs), while the 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 power factor 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.

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(9)

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Table 1

O r i g i n a l P o w e r F a c t o r i n %

50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

80 0.982 0.937 0.893 0.85 0.809 0.769 0.73 0.692 0.655 0.618 0.584 0.549 0.515 0.483 0.45 0.419 0.388 0.358 0.329 0.299 0.27 0.242 0.213 0.186 0.159 0.132 0.105 0.079 0.053 0.026 0

Power Factor Improvement Table 81 1.008 0.962 0.919 0.876 0.835 0.795 0.756 0.718 0.681 0.644 0.61 0.575 0.541 0.509 0.476 0.445 0.414 0.384 0.355 0.325 0.296 0.268 0.239 0.212 0.185 0.158 0.131 0.105 0.079 0.052 0.026 0

82 1.034 0.989 0.945 0.902 0.861 0.821 0.782 0.744 0.707 0.67 0.636 0.601 0.567 0.535 0.502 0.471 0.44 0.41 0.381 0.351 0.322 0.294 0.265 0.238 0.211 0.184 0.157 0.131 0.105 0.078 0.052 0.026 0

ILSAS/MFF/2007

83 1.06 1.015 0.971 0.928 0.887 0.847 0.808 0.77 0.733 0.696 0.662 0.627 0.593 0.561 0.528 0.497 0.466 0.436 0.407 0.377 0.348 0.32 0.291 0.264 0.237 0.21 0.183 0.157 0.131 0.104 0.078 0.052 0 .026 0

84 1.086 1.041 0.997 0.954 0.913 0.873 0.834 0.796 0.759 0.722 0.688 0.653 0.619 0.587 0.554 0.523 0.492 0.462 0.433 0.403 0.374 0.346 0.317 0.29 0.263 0.236 0.209 0.183 0.157 0.13 0.104 0.078 0 .052 0.026 0

85 1.112 1.067 1.023 0.98 0.939 0.899 0.86 0.822 0.785 0.748 0.714 0.679 0.645 0.613 0.58 0.549 0.518 0.488 0.459 0.429 0.4 0.372 0.343 0.316 0.289 0.262 0.235 0.209 0.183 0.156 0.13 0.104 0 .078 0.052 0.026 0

86 1.139 1.094 1.05 1.007 0.966 0.926 0.887 0.849 0.812 0.775 0.741 0.706 0.672 0.66 0.607 0.576 0.545 0.515 0.486 0.456 0.427 0.399 0.37 0.343 0.316 0.289 0.262 0.236 0.21 0.183 0.157 0.13 0 .105 0.079 0.053 0.027 0

87 1.165 1.12 1.076 1.033 0.992 0.952 0.913 0.875 0.838 0.801 0.767 0.732 0.698 0.666 0.633 0.602 0.571 0.541 0.512 0.482 0.453 0.425 0.396 0.369 0.342 0.315 0.288 0.262 0.236 0.209 0.183 0.157 0 .131 0.105 0.079 0.053 0.026

Desired Pow er Factor in % 88 89 90 91 1.192 1.22 1.248 1.276 1.147 1.175 1.23 1.231 1.103 1.131 1.159 1.187 1.06 1.088 1.116 1.144 1.019 1.047 1.075 1.103 0.979 1.007 1.033 1.063 0.94 0.968 0.996 1.024 0.902 0.93 0.958 0.986 0.865 0.893 0.921 0.949 0.828 0.856 0.884 0.912 0.794 0.822 0.85 0.878 0.759 0.787 0.815 0.843 0.725 0.753 0.781 0.809 0.693 0.721 0.749 0.777 0.66 0.688 0.716 0.744 0.629 0.657 0.685 0.713 0.598 0.626 0.554 0.682 0.568 0.596 0.624 0.652 0.539 0.567 0.595 0.623 0.509 0.537 0.565 0.593 0.48 0.508 0.536 0.564 0.452 0.48 0.508 0.536 0.423 0.451 0.479 0.507 0.396 0.424 0.452 0.48 0.369 0.397 0.425 0.453 0.342 0.37 0.398 0.426 0.315 0.343 0.371 0.399 0.289 0.317 0.345 0.373 0.263 0.291 0.319 0.347 0.236 0.264 0.292 0.32 0.21 0.238 0.266 0.294 0.184 0.212 0.24 0.268 0 .158 0 .186 0 .214 0 .242 0.132 0.16 0.188 0.216 0.106 0.134 0.162 0.19 0 .08 0.108 0.136 0.164 0.053 0.081 0.109 0.137 0.027 0.055 0.082 0.111 0 0.028 0.056 0.084 0.028 0.056 0.028

92 1.306 1.261 1.217 1.174 1.133 1.09 1.051 1.013 0.976 0.939 0.905 0.87 0.836 0.804 0.771 0.74 0.709 0.679 0.65 0.62 0.591 0.563 0.534 0.507 0.48 0.453 0.426 0.4 0.374 0.347 0.321 0.295 0 .269 0.243 0.217 0.191 0.167 0.141 0.114 0.086 0.058 0.03

93 1.337 1.292 1.248 1.295 1.164 1.124 1.085 1.047 1.01 0.973 0.939 0.904 0.87 0.838 0.805 0.774 0.743 0.713 0.684 0.654 0.625 0.597 0.568 0.541 0.514 0.487 0.46 0.434 0.408 0.381 0.355 0.329 0 .303 0.277 0.251 0.225 0.198 0.172 0.145 0.117 0.089 0.061 0.031 0

94 1.369 1.324 1.28 1.237 1.196 1.156 1.117 1.079 1.042 1.005 0.971 0.936 0.902 0.87 0.837 0.806 0.775 0.746 0.716 0.686 0.657 0.629 0.6 0.573 0.546 0.519 0.492 0.466 0.44 0.413 0.387 0.361 0 .335 0.309 0.283 0.257 0.23 0.204 0.177 0.149 0.121 0.093 0.063 0.032

95 1.403 1.358 1.314 1.271 1.23 1.19 1.151 1.113 1.076 1.039 1.005 0.97 0.936 0.904 0.871 0.84 0.809 0.779 0.75 0.72 0.691 0.663 0.634 0.607 0.58 0.553 0.526 0.5 0.474 0.447 0.421 0.395 0 .369 0.343 0.317 0.291 0.265 0.238 0.211 0.183 0.155 0.127 0.097 0.066 0.034

96 1.442 1.395 1.351 1.308 1.267 1.228 1.189 1.151 1.114 1.077 1.043 1.008 0.974 0.942 0.909 0.878 0.847 0.817 0.788 0.758 0.729 0.701 0.672 0.645 0.618 0.591 0.564 0.538 0.512 0.485 0.459 0.433 0 .407 0.381 0.355 0.329 0.301 0.275 0.248 0.22 0.192 0.164 0.134 0.103 0.071 0.037

97 1.481 1.436 1.392 1.349 1.308 1.268 1.229 1.191 1.154 1.117 1.083 1.014 0.982 0.949 0.918 0.887 0.857 0.828 0.798 0.769 0.741 0.712 0.685 0.658 0.631 0.604 0.578 0.552 0.525 0.499 0.473 0.447 0 .421 0.395 0.369 0.343 0.317 0.29 0.262 0.234 0.206 0.176 0.143 0.113 0.079 0.042 0 0

98 1.529 1.484 1.44 1.397 1.356 1.316 1.277 1.239 1.202 1.165 1.131 1.096 1.062 1.03 0.997 0.966 0.935 0.905 0.876 0.84 0.811 0.783 0.754 0.727 0.7 0.673 0.652 0.62 0.594 0.567 0.541 0.515 0 .489 0.463 0.437 0.417 0.39 0.364 0.237 0.309 0.281 0.253 0.223 0.192 0.16 0.126 0.089 0.047 0

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99 1.59 1.544 1.5 1.457 1.416 1.377 1.338 1.3 1.263 1.226 1.192 1.157 1.123 1.091 1.058 1.027 0.996 0.966 0.937 0.907 0.878 0.85 0.821 0.794 0.767 0.74 0.713 0.687 0.661 0.634 0.608 0.582 0 .556 0.53 0.504 0.478 0.451 0.425 0.398 0.37 0.342 0.314 0.284 0.253 0.221 0.187 0.15 0.108 0.061 0

100 1.732 1.687 1.643 1.6 1.559 1.519 1.48 1.442 1.405 1.368 1.334 1.299 1.265 1.233 1.2 1.169 1.138 1.108 1.079 1.049 1.02 0.992 0.963 0.936 0.909 0.882 0.855 0.829 0.803 0.776 0.75 0.724 0 .698 0.672 0.645 0 .62 0.593 0.567 0.54 0.512 0.484 0.456 0.426 0.395 0.363 0.328 0.292 0.251 0.203 0.142


EXAMPLE: A small machine tool plant used an average of 100 kW with an existing power factor 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 in this case gives 0.421 as the factor needed to complete the formula referenced above: 0.421 x 100 kW = 42 Kvar The customer may now choose the capacitor catalog number by kVAR and voltage from the complete ratings listed in this catalog. If kW or Present Power Factor are not known you can calculate from the following formulas to get the three 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 factor for calculation purposes.

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4.2

Power Factor Correction Capacitors on Reduced Voltage Motors and Multi-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 sure that your circuit exactly matches the circuit shown here before applying capacitors. Failure to do so may result in damage to the motor. The main contacts, illustrated in the diagrams below as M1, M2, M3, reference the contacts that must be closed to start or run the motor. Capacitors should be connected on the motor side of the main contacts.


Capacitor schemes for multi-speed motors Page 33 of 136


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 power factor of a total system for Bulk Correction you must have three pieces of information: • • •

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:

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

Exercise No.4


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4.3.2 Direct calculation Capacitor size (kVAr) = kW x (Tangent φ1 – Tangent φ2)

(13)

φ1 - Angle for existing power factor φ2 – 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 is charged power factor penalty. What is the size of capacitor bank to improve power factor from 0.65 to 0.90? Calculation

Existing power factor = Cos φ1 = 0.65 Maximum load = 2000 kW Desired power factor = Cos φ2 = 0.90 Angle for existing power factor = φ1 = Cos-1(0.65) = 49.45 0 Angle for desired power factor = φ2 = Cos-1(0.90) = 25.84 0

Capacitor size (kVAr)

= kW x (Tangent φ1 – Tangent φ2) = 2000 x (Tan 49.45 0 – Tan 25.84 0) = 1,369 kVAr

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Exercise No.5 (Direct calculation)


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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 value of Var, which is proportional to the square of the voltage applied. The reactance of a capacitor bank varies inversely with the frequency: Var ∝ V2

(14)

Xc = 1/(2πfC)

(15)

So for high frequencies, they provide low impedance. The leading current drawn by 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 voltage to 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 and split wye-connected capacitor bank. The following key points can be made in regard to bank connection under normal and abnormal system conditions. Bus Bar Connections: From looking at Figure 14, it should be evident that the ungrounded-wye connection is much simpler in design than either of the two delta connected banks. The crossover connection that connects phase "A" to phase "C" to close the delta is complicated at the medium voltage level due to clearance requirements.

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Figure 14:

Connection and fusing arrangements for ungrounded-wye and delta connected capacitor banks

Fusing: Figure 14 also shows common fusing practices for each of the bank arrangements. The Figure shows that the delta connected bank can be protected by placing the fuses inside or outside of the delta. Two fuses per single phase capacitor 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 are sized in the same way as the fuses for the ungrounded-wye connected capacitor bank. Capacitor: Except for voltage rating, the capacitors in both ungrounded-wye and delta-connected banks are the same and will have the same kvar rating. They consists of a double bushing design, meaning both terminals are fully insulated from their case (ground). On delta connected banks, the capacitors have a lineto-line voltage rating, and on a wye-connected banks, they have a line-to-neutral voltage rating. Fault Conditions: A capacitor typically fails in two ways: 1) A bushing to case fault occurs. 2) The internal sections fail (commonly known as a dielectric fault), which basically shorts the capacitor terminals. Whether the capacitors are delta or wye-connected, a bushing fault will have the same impact on the power system.

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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 to a phase-to-phase bolted fault. This fault will cause a major voltage sag on the facilities 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 can impose mechanical and thermal stress on components in the fault path. On a wye-ungrounded capacitor bank, internal section faults subject the power system to a fault current that is three times the banks rating (until the capacitor fuse blows). Therefore, the voltage sag, mechanical and thermal stressing associated with 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

Figure 15:

ILSAS/MFF/2007

Capacitor Unit

Capacitor element

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 and series arrangements to give the desired voltage rating and total kVar needed for the 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. The selection will depend on the location in the system where they are connected and the kVar capability of the bank. The types of bank most often applied are the outdoor rack and the pole mount design. This approach is suitable for large, MV capacitor banks of any designated kVar rating and voltage. The voltage range starts from 11 kV and goes up as high as the application requires. The kVar Capacity stars from 300 kVar and can be designed as large as required. These designs will include racks for mounting and individual unit fuses. The second type of equipment available is the enclosed or house construction. The housing is built to enclose the capacitor, switching device, and the automatic sensing and control equipment. These are used when where a relatively small bank of capacitors is needed to improve the voltage profile and capacity on a distribution feeder. The second variety of housed equipment can includes larger rated banks, up to 6,000 kVar or 6 Mvar. The capacitor units are individually fused. Suitable Vacuum switches or Contactors can be utilized and operated by automatic sensing and control. 5.3.3.2

Capacitor connection

There are a number of ways in which a capacitor bank may be connected, with the choice being dependent on:1. 2. 3. 4.

The voltage level of the system The kVar capacity of the Capacitor Bank The system grounding The desired relay protection

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

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Two criteria are applied to determine the minimum allowable number of paralleled capacitor units in each phase. These criteria are as follows:

• The loss of one capacitor unit in a phase should not produce a voltage across the 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 ensure clearing in 300s or less. It should be pointed out that the 300s time span is a maximum 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 ungrounded wye, with paralleled units to make up the total kvar. It is recommended that a minimum of 4 paralleled units to be applied to limit the over voltage on the remaining units when one is removed from the circuit. 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 an industrial or commercial power systems the capacitors are not grounded for a variety of reasons. Industrial systems are often resistance grounded. A grounded wye 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.

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5.3.3.4

Y-Y Capacitor Bank design


Typical Medium Voltage Y-Y Capacitor design Page 44 of 136


5.3.3.4.1

Y-Y Connection CT

REACTOR

REACTOR

REACTOR

BUSBAR 11 KV

red

40 uH REACTOR

yellow

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

333 kVar

40 uH REACTOR

Blue

40 uH REACTOR

BUSBAR 11 KV

UNBALANCE RELAY


Typical Medium Voltage Y-Y Capacitor single line diagram Page 45 of 136


CT

REACTOR

REACTOR

REACTOR

2 MVar

3 MVar

BUSBAR 11 KV

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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red

40 uH REACTOR

yellow

833 kVar

833 kVar

833 kVar

833 kVar

833 kVar

833 kVar

40 uH REACTOR

Blue

40 uH REACTOR

BUSBAR 11 KV

UNBALANCE RELAY

Figure 20:


Description of a typical 5 MVar 11 kV Capacitor Bank 1. Rating per capacitor unit = 833 kVar, 6.35 kV (V LN) 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 uH REACTOR

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

225 kVar

40 uH REACTOR

225 kVar

40 uH REACTOR

225 kVar

BUSBAR 11 KV

UNBALANCE RELAY

Figure 21:


Description of a typical 5.4 MVar 11 kV Capacitor Bank 1. Rating per capacitor unit = 225 kVar, 6.35 kV (V LN) 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 V k 1 1

333 kVar

333 kVar

V k 1 1

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

Example 2:

Configuration of 0.333 MVar, 33 kV Capacitor Bank Red Phase 333 kVar r a V k 1 1 1

333 kVar

333 kVar V k 3 3

Yellow Phase Blue Phase

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/ ωC, ω = 2x3.14x50 The equivalent reactance (X T) = 1/ ωC1 + 1/ ωC2 +1/ ωC3 = 1/ ωCT

∴ 1/ CT =1/ C1 + 1/ C2 +1/ C3 Equivalent kVar per phase = (Voltage LN)2 x ω x CT Three phase voltage rating = √3x(3x6.35 kV) = 33 kV Sample calculation:

1/CT = 1/(26.3uF) + 1/(26.3uF) + 1/(26.3uF) =0.114 CT = 8.77 uF Total Q per phase = 6.35kV 2 x ω x CT = 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/XC1 + 1/XC2 + 1/XC3 Formula: Capacitive reactance (X C) = 1/ ωC, ω = 2x3.14x50 The equivalent reactance (1/X T) = ωC1 + ωC2 +ωC3 = ωCT

∴ CT =C1 + C2 + C3 Equivalent kVar per phase = (Voltage LN)2 x ω x CT 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 ω x CT = 999 kVar Or simply adding up all the capacitor values in parallel = 333 kVar+333kVar+333 kVar = 999 kVar ILSAS/MFF/2007

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MEDIUM VOLTAGE CAPACITOR & POWER FACTOR MANAGEMENT

Exercise No.6

Y-Y connection for 5 MVAR 11 kV

NOTA:

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Single line diagram for 5 MVAR 11 kV

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Exercise No.7

Y-Y connection for 4 MVAR, 11 kV

Capacitor Unit 333 kVar 6.35 kV


Exercise No.7



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Exercise No.8




Exercise No.8



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Exercise No.9




Exercise No.9



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Exercise No.10

Current for 5 MVAR, 33 kV



Exercise No.10




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Exercise No.11




Exercise No.11




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5.3.4 Calculation of Capacitive Current (I kvar ) Power factor = cos φ Ic = kvar /( √3 x VLL)

3-phase

(16)

Ic = kvar /(V LN)

1-phase

(17)

From equation (13) kVar for PF Correction

kVar = kW x (tan φ1-tan φ2)



3-phase

(16)

Ic = kvar /(V LN)

1-phase

(17)

From equation (13) kVar for PF Correction

Figure 21:

kVar = kW x (tan φ1-tan φ2)

Capacitive current for a 5 Mvar Capacitor Bank 11 kV

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5.3.5 Capacitor current during failure of capacitor cans

Figure 22:

Calculation for 1 capacitor can failure



5.3.5.1

Technical equations

a. Ic = kVar /( √3 x VLL) 2

(18)

b. Var = V ωC

(19)

c. V=IxZ

(20)

d. I = V/Z

(21)

e. X= 1/ [ ωC]

(22)

f. Var= V2ωC

(23)

g. V2= Var / ωC

(24)

h. (IX)2= Var x X

(25)

i. I2= Var/X

(26)

j. I= √ (Var/X)

(27)



5.3.6 Relationship between kVar, current & impedance

Example 5 MVar, 11 kV

Figure 23:


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Figure 24

Connection diagram in uF for 5 MVar 11 kV

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Figure 25

Connection diagram in ohm for 5 MVar 11 kV

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Figure 26

Calculation of Amp based on Total Impedance

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Exercise No.12

Calculate the current if one capacitor can is damaged.


Calculate the current if one capacitor can is damaged.

Exercise No.12

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5.3.7 Estimation of neutral current

I A

=

[IR2 + IY2 + IB2]

(28)

IB

=

[ (IR x IY ) + ( I Yx IB ) + ( IB x IR) ]

(29)

I neutral

=

√ [I A – IB]

(30)


5.3.7 Estimation of neutral current

I A

=

[IR2 + IY2 + IB2]

(28)

IB

=

[ (IR x IY ) + ( I Yx IB ) + ( IB x IR) ]

(29)

I neutral

=

ILSAS/MFF/2007

√ [I A – IB]

(30)

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5.4

Design of Power Capacitor Banks < 4.16 kV

5.4.1 Delta connection

VCB

Fuses

Discharge resistors

Figure 30:

Delta connection Capacitors

Capacitor: Except for voltage rating, the capacitors in both ungrounded-wye and delta-connected banks are the same and will have the same kvar rating. They consists of a double bushing design, meaning both terminals are fully insulated from their case (ground). On delta connected banks, the capacitors have a lineto-line voltage rating, and on a wye-connected banks, they have a line-to-neutral voltage rating. 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 to a phase-to-phase bolted fault. This fault will cause a major voltage sag on the facilities 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 can impose mechanical and thermal stress on components in the fault path.

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On a wye-ungrounded capacitor bank, internal section faults subject the power system to a fault current that is three times the banks rating (until the capacitor fuse blows). Therefore, the voltage sag, mechanical and thermal stressing associated 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 ω x Volt 2 /1000

ω = 2xπ x f

(31)

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 Y-B B-R

375 uF 375 uF 375 uF

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kVar (3-phase) = 2/3 x (375uF + 375uF + 375 uF) x ω x Volt 2 /1000 = 2/3 x (375uF + 375uF + 375 uF) x 2x3.14x50 x 415 2/1000 = 40.6 kVar

Exercise No.13

Calculation of kVar for Delta Capacitor


3.3 kV

uF

Figure 32:

Capacitance measurement for Delta Capacitor using Capacitance meter


146.22 uF 146.22 uF 146.22 uF

kVar (3-phase) =

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3-phase

(32)

Example:

Capacitor kVar (3-phase delta)

= 50 kVar

Voltage rating

= 415 Volt

Ic = kvar /( √3 x VLL) = 50 kVar / ( √3 x 415 Volt ) = 69.56 Amp

Exercise No.14

Calculation of kVar for Delta 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 will consist 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 in the 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 continuity 2. To prevent damage to adjacent capacitors and equipment or injury to personnel 3. To provide visual indication of a failed unit 4. To limit the energy into a faulted unit to help prevent case rupture The requirements for proper fuse selection are as followsa. The rated voltage of the fuse should not be less than rated voltage of the capacitor with which it is used. Since capacitors are designed to operate continuously 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 fuses for large bank rating and for banks connected to buses with high short circuit capacity. c. The fuse should have a time-current clearing characteristic that lies below the time-current case rupture probability characteristics of the capacitor units to be applied. d. The selected fuse should have sufficient rating to carry 165 % of rated capacitor currents. This margin allows for temporary over voltage, harmonic currents, switching surges, and manufacturing tolerance in the capacitor itself. e. The fuse must clear the minimum over current resulting due to a failed unit within 300 seconds maximum and as stated earlier 30 seconds or less is desirable. 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 is more difficult since a failed unit causes about 3 times normal fault current to flow through the fuse protecting the faulted phase. In some cases, it becomes necessary to reduce the rating of the fuse to minimum current of 150 % of normal. This is permissible because ungrounded banks do not need allowance for zero sequence currents. f. The fuse must be capable of withstanding the energy contributed to a unit by other 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 the individual or group fusing, additional relay protection for the whole bank is required. The 2 basic type of relay protection used with medium voltage capacitors are; 1) over-current relaying of major equipment faults and 2) relaying to identify loss of units within a bank. Over-current relaying is needed for removal of the capacitor bank in the event of a fault between the switching device and the bank itself. The relay should be chosen so that the highest magnitude of inrush current associated with the capacitor switching will not trip the circuit breaker immediately as the bank is energized. Also, if a capacitor unit fault occurs the relay should delay operation until the fuse clears. Relays with an inverse time current characteristics are usually used and settings are selected to override these 2 conditions. However, the relay may also have a very inverse or extremely inverse time current characteristics if coordination with other system protective devices required it. Instantaneous over current relays are not often utilized since they are likely to trip unnecessarily when the bank is energized. Relaying for the detection of the loss of capacitor units and guarding against excessive operating voltages should be used as protective measure supplement to the periodic visual inspection of individual unit fuses. This is because there will be a compromise between a relaying method of adequate sensitivity which is also immune to undesirable operation due to harmonics, system voltage unbalances, external faults and unbalance due to the varying capacitance of the individual capacitor units.

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It is important to note that except for short periods of time, the voltage across the capacitors should not exceed 110 % of rated voltage. On large banks, it is recommended that some type of protection be provided which will relay the bank off 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 a foreign object.

6.2.1 Common scheme for ungrounded bank Figure 33 is an effective arrangement for ungrounded bank. The potential transformer (PT) will detect a change in voltage across the phase if one or more units 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.

PT Secondary PT Primary

UnGrounded Neutral Voltage Relay

Figure 33

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Unbalance Phase Voltage

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PT

Voltage relay

Figure 34:

Voltage Imbalance between neutrals

CT

Current relay

Figure 35:

ILSAS/MFF/2007

Current Imbalance between neutrals

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Potential transformers have an advantage over capacitor potential devices in that they furnish a discharge path for the capacitors, and will discharge the capacitors in a much softer time than the discharge resistors built into the capacitors. Use of the potential transformers as a discharge path will be effective only if the transformers have the thermal capability to pass the current associated with the stored energy in the bank. The kVar rating of the bank will determine the energy storage capability and this should be matched against the energy absorption capability of as transformer as shown in Figure 33. The internal resistors built into each individual capacitor unit will also discharge the bank but this requires about five minutes for high voltage units and one minute for low voltage units. For automatically controlled banks this may be too long 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 normal operating sequences of any switching controls should be reviewed to insure that the bank will have a sufficient time to discharge to a safe level before being re energized. The schemes shown in Figure 34 & 35 are known as double-wye schemes and they have considerable merit. They offer protection similar to that of Figure 34 and are less expensive since they require only a single PT or CT. For smaller banks, it may be expensive to split the banks into 2 wye groups, or it may be impossible 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 be achieved at less cost than Figure 33, making the double wye protection more attractive, particularly on industrial systems where there is no advantage to grounding 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 of Figure 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. As noted earlier, the ground arrangement is not often used by industrials for several reasons, one of them, being possible interference from ground relaying. Use of a third harmonic filter would permit setting the relay at a fairly low value of pickup.

Voltage relay

Voltage relay

PT

Resistor

Figure 36

Neutral Neutral Voltage

Figure 37

Neutral Neutral Current

No over voltage protection scheme will provide positive protection against over voltage all cases. The basic theory of all schemes is the detection of current or voltage unbalance. There is some inherent unbalance in capacitor banks, and the relaying must be set above the maximum inherent unbalance which can occur if false trip-outs are to be avoided. The inherent unbalance is due to mainly to harmonic currents and voltages, and varying capacitance of the capacitors. The tolerance of a power capacitor is minus zero to + 15 % with the average being about 4 %. When individual units are placed in racks and stacked for high voltage banks, the capacitance of the racks in each phase leg may vary. This causes normal condition unbalance and makes it more difficult to detect small unbalance due to removal of faulted units if a parallel group. If it is considered necessary to balance the phases closely, it is possible by special arrangement at the factory. The problem of phase unbalance tends to diminish as the bank rating increases. In cases where a large number of capacitor units, say 20 or more, are used in a parallel group, it may not be possible to detect the loss of one unit, since the relays 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 not raise the voltage on the remaining units above the allowable 10 % over voltage. In most cases, the relaying can be set to operate because dangerous conditions exist, without serious danger of false operation. On large installation, it is good practice to use 2 relays. One will sound an alarm when 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 procedure has the advantage of keeping the bank in service when possible while indicating that capacitors have failed yet still protect the capacitors from serious overvoltages. The summary of the characteristics of the several relay arrangements is shown in Table 2.

Table 2:

Characteristics of Protective Relaying Methods

Type of protective relaying

Unbalance phase voltage Voltage unbalance between neutral Current unbalance between neutrals Neutral voltage Neutral current


Figure

Sensitive to Switching Transient

33

Yes

Sensitive to 3rd harmonic voltage & currents No

Sensitive to system voltage unbalance

Number of CT required

Number of PT required

No

-

3

34

Yes

No

No

-

1

35

Yes

No

No

1

-

36

Yes

Yes

Yes

-

1

37

Yes

Yes

Yes

1

-

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REACTOR

REACTOR

REACTOR

11 kV Fuses Unbalance CT 2 MVar

3 MVar

CT Description of relays: A- Overcurrent & earthfault relays B-Unbalance relay A

B Relays

Figure 38

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Ammeter

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 the reactive 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. The overall capacitance for the particular phase the capacitor unit is connected will no longer equal to the other two phases. Different capacitance will give different values of reactive current flows. A neutral current will then be introduced in the neutral conductor. If the neutral current exceed the prescribed limits i.e. 6-9 Amp, then the voltage across 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 the neutral 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 current flow and for tripping purposes. The monitoring of the neutral current is done in two stages. The first stage, setting of 60 % of the CT rating, is aimed to be as an alarm to inform the user of the possible deterioration process. The second stage, setting of 90 % of the CT rating, is aimed for tripping purposes. If the amount of neutral current is ≥ 90 %, the circuit breaker will trip off the 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 capacitor bank.

• A healthy Capacitor bank that’s 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 experience failure. The circuit breaker for the feeder will not be activated if there is no existence 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 bank exceed 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 phase have 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 then detected at the neutral conductor by the unbalance CT. The first stage of the unbalance relay will then be activated.

Table 2 Capacitor

Example on relay settings kV

Rated

CT for

Bank (MVar) 5 4 3

11 11 11

Amp 262 210 157

relay 400 400 400

Capacitor

kV

Rated

CT for

11 11 11

Amp 262 210 157

relay 300 300 300

Bank (MVar) 5 4 3

ILSAS/MFF/2007

O/C

/5 /5 /5

/5 /5 /5

E/F

TMS

75% 75% 50%

20% 20% 20%

0.10 0.10 0.10

O/C

E/F

TMS

100% 75% 75%

20% 20% 20%

0.10 0.10 0.10

Un1st balance stage CT 10 / 5 60% 10 / 5 60% 10 / 5 60% Un1st balance stage CT 10 / 5 60% 10 / 5 60% 10 / 5 60%

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2nd stage 90% 90% 90% 2nd stage 90% 90% 90%


6.6

Lightning protection

In common with other apparatus connected to a power system, capacitors should be protected by surge arresters. Here the primary function is protection of the capacitor bank and not the system itself. The choice between surge arrester rated for grounded neutral service or ungrounded neutral service should be in accordance with established industry practices. The rating of surge arresters (line-to-ground) is determined by the system grounding, and is not related to the grounding or lack thereof, of the capacitors themselves. Occasionally, the suggestion is made that surge arresters are unnecessary for capacitor banks which are Y-Connected with grounded neutral. It is true that capacitors 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 by the size of the capacitor bank and the amount of energy to be expected in a given 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 system components by shunting the surge current to ground.

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Three phase capacitors use fuses in the line because they are connected delta Internally. Normally branch fuses are used for single-phase capacitors connected delta. However, on the smaller banks mentioned above, the single phase capacitors 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 of choice when the neutral of the capacitor bank is not grounded. When the bank has higher kvar ratings and units are placed in parallel, the in line fusing becomes large, and may not coordinate with the tank rupture curve of the capacitor and the upstream co-ordination may not be possible. For example, consider both fusing methods for a 450 kVAR, 4160 volts delta connected 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 = 93.68 amps ... a 100-amp fuse is required.

• In branch fusing 150 kVAR/4.16 KV = 36 amps x 1.5 = 54 amps ... a 60-amp fuse can be used.

There are other potential problems in fusing a delta-connected bank with "in branch" fusing. It is a normal practice utilized in metal enclosed banks to install two bushing capacitors connected phase to phase with the capacitor tank grounded to the frame. In some cases, the user only applies one fuse per phase. This could be dangerous. When a capacitor starts to fail and the fuse operates the capacitor is still in the circuit via the second bushing. The failure within the capacitor is fed thru this connection and eventually the major insulation of the can will fail and the capacitor tank will rupture. The other method is to use two fuses, i.e. one per bushing. This gives the user a false sense of security. In this case both fuses would have to operate before the failed capacitor can be effectively removed from the system. Normally only one of the fuses operates, which will be the one nearest the faulted packs. The other bushing remains connected to the system via the good fuse. The result is still an eventual major insulation failure if the bank is not removed from service.

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The burning between packs could possibly continue due to the second bushing still being energized via the second fuse. During this condition a low energy fault could be developed. The current limiting fuse still in the circuit will be getting warm while the capacitor could be boiling. Eventually the major insulation will be breached grounding the faulted capacitor through the tank to frame, and there will be a race between the capacitor and the fuse to see if the fuse will clear before the capacitor ruptures. 7.2 Comparison between fusing arrangement for Delta & Y capacitors

Complete protection must be provided for a capacitor installation. This will consist 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 in the system an 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 and delta connected capacitor banks

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Fusing: Figure 40 also shows common fusing practices for each of the bank arrangements. The Figure shows that the delta connected bank can be protected by placing the fuses inside or outside of the delta. Two fuses per single phase capacitor 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 are sized in the same way as the fuses for the ungrounded-wye connected capacitor bank. Fault Conditions: A capacitor typically fails in two ways: 1) A bushing to case fault occurs. 2) The internal sections fail (commonly known as a dielectric fault), which basically shorts the capacitor terminals. Whether the capacitors are delta or wye-connected, a bushing fault will have the same impact on the power system.

Internal section faults, or dielectric faults appear differently to a power system. On a delta connected bank, an intern al section fault subjects the power system to a phase-to-phase bolted fault. This fault will cause a major voltage sag on the facilities 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 can impose mechanical and thermal stress on components in the fault path. On a wye-ungrounded capacitor bank, internal section faults subject the power system to a fault current that is three times the bank s rating (until the capacitor fuse blows). Therefore, the voltage sag, mechanical and thermal stressing associated with the fault, and case rupture concerns are reduced .

7.2.1

Capacitor fusing

The major purposes of capacitor fusing are:

• To maintain service continuity • To prevent damage to adjacent capacitors and equipment or injury to • •

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

The requirements for proper fuse selection are as follows-

• The rated voltage of the fuse should not be less than rated voltage of the capacitor with which it is used. Since capacitors are designed to operate continuously at 110 % rated voltage, the fuse should also have a voltage rating, which has at least 110 % of the capacitor unit rating.

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• 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 fuses for large bank rating and for banks connected to buses with high short circuit capacity.

• The fuse should have a time-current clearing characteristic that lies below the time-current case rupture probability characteristics of the capacitor units to be applied.

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

• The fuse must clear the minimum over current resulting due to a failed unit within 300 seconds maximum and as stated earlier 30 seco nds or less is desirable. On grounded wye and delta-connected banks, obtaining this

• clearing time is not a problem. For ungrounded wye connected banks, it is more difficult since a failed unit causes about 3 times normal fault current to flow through the fuse protecting the faulted phase. In some cases, it becomes necessary to reduce the rating of the fuse to minimum current of 150 % of normal. This is permissible because ungrounded banks do not need allowance for zero sequence currents.

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

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7.3

Relaying protection

In addition to the individual capacitor unit protection through the use of the individual or group fusing, additional relay protection for the whole bank is required. The type of relay protection used with Delta capacitors are;

• Over-current relaying of major equipment faults and Over-current relaying is needed for removal of the capacitor bank in the event of a fault between the switching device and the bank itself. The relay should be chosen so that the highest magnitude of inrush current associate d with the capacitor switching will not trip the circuit breaker immediately as the bank is energized. Also, if a capacitor unit fault occurs the relay should delay operation until the fuse clears. Relays with an inverse time current characteristics are usually used and settings are selected to override these 2 conditions. However, the relay may also have a very inverse or extremely inverse time current characteristics if coordination with other system protective devices required it. Instantaneous over current relays are not often utilized since they are likely to trip unnecessarily when the bank is energized.

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8.0

MAINTENANCE PROCEDURES FOR A Y-Y CAPACITOR BANK

8.1

Safety Regulations

When working with capacitor banks, the following safety regulations, in particular, must be observed:

• Do not touch a capacitor bank until it has been completely discharged, short circuited and grounded. Short circuit the capacitor units individually also, as if there is an internal fault, such as a damaged discharge resistor, there may still be a voltage even though the capacitor bank has been discharged.

• Normally capacitors have an internal discharge resistor, which reduces the voltage of the unit to: 50 V within 5 minutes - AS2897 - 75V within 10 minutes – IEC 871.

• Avoid skin contact with the impregnation fluid in the event of leakage , and avoid breathing in fumes or gases from the impregnation fluid. In the event of skin contact, wash with soap and water. If fluid gets into your eyes, rinse with lukewarm water. 8.2

Maintenance Methodology for Y-Y Capacitor Banks

During normal operating conditions, capac itor bank needs no maintenance, but it should be checked regularly as described below to ensure trouble-free operation. To obtain trouble free operation a routine inspection recommended:

of the following is

a. Regular Inspection An annual inspection is sufficient unless the capacitor bank is exposed to abnormal conditions of any kind, such as excessive contamination. At each inspection, proceed as follows-

• Perform a visual inspection of pollution, damage to the finish, leaking capacitor units etc. The insulators and bushings should be wiped clean if necessary. Under heavy pollution condition, cleaning should be carried out more frequently.

• Check the setting values and operation of the protection relays. • The unbalance current should be measured before disconnection

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b. Check the Unbalance current in the Y-Y capacitor bank If the unbalance protection has tripped the capacitor bank, all capacitor bank units should be capacitance measured and fault units replaced. When replacing, the capacitance difference between the fault and replacement units should not differ by more than ± 1 character. The unbalance current is checked after reconnection and should be less than 20 % of the operating value for the protection. And also if the current readings differ from that obtained at the time of commissioning, there ma y be breakdown in one or more of the internal capacitor elements. The change in current due to breakdown in one element depends on the total number of elements in the capacitor unit and on the connection arrangement of the capacitor bank. However, the current deviation is seldom less than 15 % of the tripping value. When the capacitor bank is made up of capacitors with internal fuses, an unbalance current of up to 50 % of the tripping value of the protection system is acceptable. Failures of singled elements do not affect the service ability of the capacitor bank or significantly reduce its service life. If the unbalance current exceeds 50 % of the tripping value, a capacitance measurement of all capacitor units in the bank must be performed and faulty units replaced. The overall maintenance flowcharts are shown in Figure 41 & Figure 42 respectively. If a VCB Trip –When the VCB tripping, it may be due to –

• Unbalance neutral current varies more than 20 %. The action to take is to measure the capacitor units by Capacitor Bridge and to find out whether the capacitors are in good working conditions or not.

• Cable Faults • Test the main supply cable according to the present process for cable inspection

• Shorting Phase to phase shorting may be caused by conducting objects or conductors shorted phase to phase.

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FLOWCHART A ROUTINE MAINTENANCE

NO

START

DONE RECORD CURRENT (PHASE+ NEUTRAL)

(Neutral I) > 20% of Unbalance CT size? YES

START CAPACITOR BANK HAS TRIPPED OFF

DISCONNECT CAPACITOR

WITHDRAW THE VCB TRUCK AND LOCK MAIN BUSBAR SHUTTER WITH A NON STANDARD LOCK AND HANG SAFETY NOTICE

APPLY CIRCUIT MAIN EARTH AT VCB

OPEN CAPACITOR BANK AND DISCHARGE USING MV DISCHARGE ROD

INSPECTION & MAINTENANCE PROGRAM

11 kV CABLE

PROTECTION SCHEMES

CAPACITOR TOLERANCE

REACTOR TOLERANCE

WAIT FOR 10 MINUTES TO DISCHARGE

SAFETY INTERLOCK

REFER TO FLOWCHART B

CAPACITOR CAN BE COMMISSIONED YES MEET TECHNICAL REQUIREMENT ?

NO RECTIFICATION

Figure 41:

ILSAS/MFF/2007

Work process for maintenance of a Y-Y capacitor bank (1)

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FLOWCHART B

FROM FLOWCHART A (CAPACIT OR MAINTEN ANCE)

TO FLOWCHART A (CAPACI TOR CAN BE COMMIS SIONED)

11 kV CABLE

PROTECTION SCHEMES

CAPACITOR TOLERANCE

REACTOR TOLERANCE

TERMINATION

OCEF RELAY

CLEANING OF CONTACTS

CLEANING OF CONTACTS

INSULAT ION

UNBALANCE REL AY

CABLE ENTRY

UNBALANCE CT RATIO TEST

CURRENT INJECTION TEST CAPACIT ANC E TEST

MEET TECHNICAL REQUIREMENT ?

Figure 42:

S AFETY INT ERLOCK

R ECTIFICAT ION NO

Work process for main ten ance of a Y-Y capa citor bank (2)

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8.3

Maintenance program

8.3.1 Isolation procedure

• • • • •

Line voltage with the capacitor bank connected respectively disconnected. Disconnect and isolate the Cap-Bank to be checked. Wait a minimum of 10 minutes for the capacitor bank to be discharged Short Circuit and earth the Capacitor Bank Short Circuit the terminals of each capacitor unit parallel group


8.3

Maintenance program


• • • • •

Line voltage with the capacitor bank connected respectively disconnected. Disconnect and isolate the Cap-Bank to be checked. Wait a minimum of 10 minutes for the capacitor bank to be discharged Short Circuit and earth the Capacitor Bank Short Circuit the terminals of each capacitor unit parallel group

Figure 43:

ILSAS/MFF/2007

Discharge capacitor banks using Discharge Rod

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8.3.2

Exterior & interior maintenance program

At each inspection, proceed as follows: 1. Inspect the capacitor bank for dirt build up or leaking capacitor units. 2. Clean the insulators and bushings if necessary. Wash down the capac itor bank if it is dirty. 8.3.2.1 •

•

• • • •

Ensure enclosure is receiving adequate ventilation. Check to see if filters are clean and airflow is not restricted. Replace filters and remove obstacles as required. Ensure enclosure, enclosure doors, louvers, and rodent guards are adequate to prevent entry of liquids, insects, and rodents. Clean and correct as required. Inspect operation and adjustment of key interlock systems to determine if security features are working properly. Correct deficiencies. Examine enclosure for corrosion and paint adherence. Repaint scratched or marred exterior surfaces to closely match original finish, as required. Examine warning signs and placards. If not legible, replace. Clean all enclosure windows.

8.3.2.2 • • • • • •

Maintenance for the Enclosure Interior

Remove accumulated dust and dirt. Remove insects, nests, or animal material. Inspect for condensation. Clean and inspect strip heaters and check thermostats. Check proper operation of ventilation equipment (fans) and thermostats. Correct as required. Examine enclosure for corrosion and paint adherence. Repaint scratched or marred exterior surfaces to closely match original finish, as required. Examine warning signs and placards. If not legible, replace. Clean all enclosure windows.

8.3.2.3 • • • • •

Maintenance for the Enclosure Exterior

Bus Bar & Wiring

Inspect for loose bus bar connections and discoloration. Tighten as required. Inspect for proper phase to phase and phase to ground clearance. Remove excess surface oxides from aluminum connectors. Inspect control wire connections, tighten as required. Inspect wire insulation for cuts, breakdown, or burns. Replace as required

ILSAS/MFF/2007

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8.3.3 Maintenance on the capacitor units 8.3.3.1

To take capacitanc e measurement

If the capacitor bank is not equipped with unbalance protection, all capacitor units should be capacitance measured annually. Capacitance m easurement of the units in a capacitor bank with unbalance protection does not have to be includ ed in the regular inspection. However, it should always be performed if the protection indicates a fault or disconnects the capacitor bank. It is advisable to perform an annual inspection despite of the regular inspection. Measurement is easily performed using a Capacitance Bridge CB10.The internal connection in the capacitor bank does not need to be opened to perform the measurement. If the measured values deviate by more than 10 % from the value stated in the routine test report, the capacitor unit should be replaced. Normally, capacitance measurements do not have to a part of the regular inspection when the bank is equipped with unbalance protection. Capacitance measurement is only performed at fault indication or when the unbalance protection has tripped the bank. By using a capacitance bridge CB-10 the measurement can be performed without making any disconnection within the capacitor bank. 8.3.3.2

Inspection

A leaking capacitor must be normally replaced. It is advisable to refer to the Manufacturer of the product. • • • • •

Check for physical damage, leaks, bulges, or discoloration. Replace as required. Clean capacitor case, insulation bushings, and any connectors that are dirty or corroded. Check each capacitor for capacitive reactance by applying 120 volts to each phase and measuring corresponding current. Verify with specification. Confirm kVar, voltage, and BIL rating for each capacitor. Verify with specification. Verify internal discharge resistors are working properly. Replace cells as required.

ILSAS/MFF/2007

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8.3.3.3

Capacitance Measurement

Acceptable capacitance reading

• -5% to 10 % up to 3 MVAr • 0 % to 10 % from 3 MVAr to 30 MVAr • 0% to 5 % for capacitor > 30 MVAr CT

REACTOR

RE ACTOR

REACTOR

uF BUSBAR 11 KV CAPACITANCE BRIDGE

Figure 44:

Capacitance measurement using capacitance brid ge CB10 (1)

ILSAS/MFF/2007

Page 98 of 136

CAPACITOR BANK & POWER FACTOR MANAGEMENT CT

REACTOR

REACTOR

REACTOR

uF BUSBAR 11 KV CAPACIT ANCE BRIDGE

Figure 45:

Capacitance measurement using capacitance bridge CB10 (2)

ILSAS/MFF/2007

Page 99 of 136


Calculation of kVar using measurment values of uF = kVOLT2 x (2πf) x Capacitance (uF)

kVAR

333 kVar

6.35 kV 2x3.14x50

(33)

26.3 uF

Exercise No.15

Phase Red Yellow Blue

Nameplate uF Measured uF 26.3 26.1 27.4 25.8 26.5 23.0

% Diff

Please indicate the faulty capacitor can.

ILSAS/MFF/2007

Page 100 of 136

CAPAC ITOR BANK & POWER FACTOR MANAGE MENT

8.3.4 Reactor Test CT

REACTOR

REACTOR

REACTOR

X =VOLT/CURRENT VOLT METER

L = X / [2x3.14x 50) uH

CURRENT INJECTION TEST SET BUSBAR 11 KV

Figure 46:

Reactor test

ILSAS/MFF/2007

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Examples on test result

Description Volts, mV Current, Amp Z(Impedance) Inductance, uH

Red 63 5 0.0126 40.13

Note- This test requires a injection test set.

Z = Voltage/ Current = X L XL = ω x L

Yellow 63 5 0.0126 40.13

Blue 63 5 0.0126 40.13




Red 63 5 0.0126 40.13

Yellow 63 5 0.0126 40.13

Blue 63 5 0.0126 40.13



ω = 2 x 3.14 x 50 L = XL / (2x3.14x50)

ILSAS/MFF/2007

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Exercise No.16


Red 123 10

Please calculate the values of the impedances & inductances

Yellow 124 10

Blue 126 10


Exercise No.16


Red 123 10

Yellow 124 10


ILSAS/MFF/2007


8.3.5 Unbalance CT Ratio Test

Page 103 of 136

Blue 126 10


8.3.5 Unbalance CT Ratio Test

This test must be done using a current injection test set.

ILSAS/MFF/2007

Page 104 of 136


8.3.6 Insulators • • •

Check for cracks, chips, and signs of arc tracking. Replace as required. Clean insulators and barriers. Check all mounting hardware, tighten as required.

8.3.7 Fuses • • • • •

Check all capacitor fuses, control fuses, and PT fuses for blown fuses. Replace as required. Check all mounting hardware, tighten as required. Confirm proper fuse rating. Verify with specification. Check functioning of neutral unbalance sensor. Clean contact area of fuses and fuse holders.


8.3.6 Insulators • • •


8.3.7 Fuses • • • • •


ILSAS/MFF/2007

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8.3.8 Relay maintenance – OCEF & Unbalance

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

0.1


8.3.8 Relay maintenance – OCEF & Unbalance

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

0.1

Figure 47:

ILSAS/MFF/2007

OCEF & Unbalance relays

Page 106 of 136


8.3.9 Trip test for unbalance relay

NOTA:

ILSAS/MFF/2007

Page 107 of 136


9.0


9.1

Safety Regulations





9.0


9.1

Safety Regulations




• Avoid skin contact with the impregnation fluid in the event of leakage, and avoid breathing in fumes or gases from the impregnation fluid. In the event of skin contact, wash with soap and water. If fluid gets into your eyes, rinse with lukewarm water. 9.2

Maintenance Methodology for Delta Capacitor Banks

During normal operating conditions, capacitor bank needs no maintenance, but it should be checked regularly as described below to ensure trouble-free operation. To obtain trouble free operation a routine inspection of the following is recommended:

Regular Inspection

An annual inspection is sufficient unless the capacitor bank is exposed to abnormal conditions of any kind, such as excessive contamination. At each inspection, proceed as follows-

• Record the capacitor current • Perform a visual inspection of pollution, damage to the finish, leaking capacitor units etc. The insulators and bushings should be wiped clean if necessary. Under heavy pollution condition, cleaning should be carried out more frequently.

• Check the setting values and operation of the protection relays. ILSAS/MFF/2007

Page 108 of 136


If a VCB Trip – When the VCB tripping, it may be due to –

• Cable Faults • Test the main supply cable according to the existing process for cable inspection

• Shorting Phase to phase shorting may be caused by conducting objects or conductors shorted phase to phase.

Maintenance procedure

The overall maintenance flowcharts are shown in Figure 48.

ILSAS/MFF/2007

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ROUTINE MAINTENANCE

YES

START

CAPACITOR CURRENT = RATED CURRENT ?

RECORD CURRENT (PHASE+ NEUTRAL)

DONE

NO

START CAPACITOR BANK HAS TRIPPED OFF

DISCONNECT CAPACITOR

WITHDRAW THE VCB TRUCK AND LOCK MAIN BUSBAR SHUTTER WITH A NON STAND ARD LOCK AND HANG SAFETY NOTICE

APPLY CIRC UIT MAIN EARTH AT VCB

OPEN CAPACITOR BANK AND DISCHARGE USING MV DISCHARGE ROD

INSPECTION & MAINTENANCE PROGRAM

MV CABLE

PROTECTION SCHEMES

CAPACITOR TOLERANCE

REACTOR TOLERANCE

WAIT FOR 10 MINU TES TO DISCHARGE

SAFETY INTERLOCK

CAPACITOR CAN BE COMMISSIONED YES MEET TECHNICAL REQUIREMENT ?

NO RECTIFICATION

Figure 48:

ILSAS/MFF/2007

Work process for maintenance of a Delta capacitor bank

Page 110 of 136


9.3

Maintenance program


• • • • •

Line voltage with the capacitor bank connected respectively disconn ected. Disconn ect and isolate the Cap-Bank to be checked. Wait a minimum of 10 minutes for the capacitor bank to be discharged Short Circuit and earth the Capacitor Bank Short Circuit the terminals of each capacitor unit parallel group

Figure 49:

ILSAS/MFF/2007

Discharge capacitor banks using Discharge Rod

Page 111 of 136


9.3.2

Exterior & interior maintenance program

At each inspection, proceed as follows:

• Inspect the capacitor bank for dirt build up or leaking capacitor units. • Clean the insulators and bushings if necessary. Wash down the capacitor bank if it is dirty. 9.3.2.1 •

•

• • • •

Ensure enclosure is receiving adequate ventilation. Check to see if filters are clean and airflow is not restricted. Replace filters and remove obstacles as required. Ensure enclosure, enclosure doors, louvers, and rodent guards are adequate to prevent entry of liquids, insects, and rodents. Clean and correct as required. Inspect operation and adjustment of key interlock systems to determine if security features are working properly. Correct deficiencies. Examine enclosure for corrosion and paint adherence. Repaint scratched or marred exterior surfaces to closely match original finish, as required. Examine warning signs and placards. If not legible, replace. Clean all enclosure windows.

9.3.2.2 • • • • • •

Maintenance for the Enclosure Interior

Remove accumulated dust and dirt. Remove insects, nests, or animal material. Inspect for condensation. Clean and inspect strip heaters and check thermostats. Check proper operation of ventilation equipment (fans) and thermostats. Correct as required. Examine enclosure for corrosion and paint adherence. Repaint scratched or marred exterior surfaces to closely match original finish, as required. Examine warning signs and placards. If not legible, replace. Clean all enclosure windows.

9.3.2.3 • • • • •

Maintenance for the Enclosure Exterior

Bus Bar & Wiring

Inspect for loose bus bar connections and discoloration. Tighten as required. Inspect for proper phase to phase and phase to ground clearance. Remove excess surface oxides from aluminum connectors. Inspect control wire connections, tighten as required. Inspect wire insulation for cuts, breakdown, or burns. Replace as required

ILSAS/MFF/2007

Page 112 of 136


9.3.4 Maintenance on the capacitor units 9.3.4.1

To take capacitanc e measurement

If the capacitor bank is not equipped with unbalance protection, all capacitor units should be capacitance measured annually. Capacitance m easurement of the units in a capacitor bank with unbalance protection does not have to be includ ed in the regular inspection. However, it should always be performed if the protection indicates a fault or disconnects the capacitor bank. It is advisable to perform an annual inspection despite of the regular inspection. Measurement is easily performed using a Capacitance meter. The internal connection in the capacitor bank need to be opened to perform the measurement. If the measured values deviate by more than 10 % from the value stated in the routine test report, the capacitor unit should be replaced.

Figure 50:

ILSAS/MFF/2007

Capacitance measurement for a Delta Capacitor Bank

Page 113 of 136


Capacitance measurement is only performed at fau lt indication or when the value of the capacitor current does not equal to the rated capacitor current. Acceptable capacitance reading

• -5% to 10 % up to 3 MVAr • 0 % to 10 % from 3 MVAr to 30 MVAr • 0% to 5 % for capacitor > 30 MVAr Example: Voltage rating per Capacitor

3.3 kV

uF

Figure 51:

Capacitance measurement for Delta Capacitor us ing Capacitance meter


146.22 uF 146.22 uF 146.22 uF

kVar (3-phase) = 2/3 x (146.22 uF + 146.22 uF + 146.22 uF) x

ILSAS/MFF/2007

ω x Volt2 /1000 = 999 kVar

Page 114 of 136


Exercise No.17

P.F.C

The capacitor cans are rated at 3.3 kV line to line.

Cans Number 1 Number 2 Number 3

Nameplate uF R-Y Y-B B-R 146.22 146.22 146.22 146.22 146.22 135.60 146.22 146.22 146.22

kVar

% Diff

Please indicate the faulty capacitor can.

ILSAS/MFF/2007

Page 115 of 136


Calculation for Exercise 16.

ILSAS/MFF/2007

Page 116 of 136


9.3.4.2

Inspection on capacitor units

A leaking capacitor must be normally replaced. It is advisable to refer to the Manufacturer of the product. • • • •

Check for physical damage, leaks, bulges, or discoloration. Replace as required. Clean capacitor case, insulation bushings, and any connectors that are dirty or corroded. Confirm kVar, voltage, and BIL rating for each capacitor. Verify with specification. Verify internal discharge resistors are working properly. Replace cells as required.

9.3.4.3 • • •


9.3.4.4 • • • • •

Insulators

Fuses


ILSAS/MFF/2007

Page 117 of 136


9.3.4.5

Reactor Test

Volt Meter

Current Injection Test Set

Figure 52:

ILSAS/MFF/2007

Reactor test

Page 118 of 136




Red 63 5 0.0126 40.13

Yellow 63 5 0.0126 40.13



ω = 2 x 3.14 x 50 L = XL / (2x3.14x50)

ILSAS/MFF/2007

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Exercise No.18 Description Volts, mV Current, Amp Z(Impedance) Inductance, uH

Red 25 2

Yellow 25 2

Blue 25 2


Blue 63 5 0.0126 40.13


Exercise No.18 Description Volts, mV Current, Amp Z(Impedance) Inductance, uH

Red 25 2

Yellow 25 2

Blue 25 2


9.3.4.6

Relay maintenance – OCEF

The inspection of the OCEF relays must be done together with the maintenance of the other circuit breakers.

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

ILSAS/MFF/2007

Page 120 of 136

CAPACITOR CAPACITOR BANK & POWER FACTOR MANAGEMENT

10.0

COMMISSIONING PROCEDURES

10.1

Before energizing the capacitor bank, check:

• that the capacitors have been correctly installed. It is particularly important to ensure that the electrical connections have been retightened.

• that the settings of the protection circuits are correct and that the protective relays are operating properly.

• to ensure all foreign objects in the capacitors are removed. • to ensure all the cable entrance is properly covered. 10.2

Setting for Power Factor Regulator (Controller)

Some capacitor banks are controlled using a power factor regulator. A power factor regulator is a mini computer that regulates the switching in and out for the capacitor units to the supply networks by controlling the VCBs or contactors.

PFC controls the switching of the VCBs or contactors. P.F.C

Figure 53:

ILSAS/MFF/2007

Power Factor Regulator for a Delta Capacitor

Page 121 of 136


CT

REACTOR

REACTOR

REACTOR

PFC controls the switching of the VCBs or contactors.

Figure 54:

ILSAS/MFF/2007

PFC BUSBAR 11 KV

Power Factor Regulator for a Y-Y Capacitor

Page 122 of 136


The power factor regulator (controller) monitor the input current from only one phase and voltage inputs (phase to phase voltage) for the other two phases.

Figure 55:

Power Factor regulator (controller)

The power factor regulator must be programmed to ensure correct operation of capacitor bank. The general parameters required for the power factor regulator are:

• • • • •

Target power factor – 0.90, 0.95 or 1.0 C/K - To be calculated Sequence - either 1:1:1:1 or 1:2:2:2 Time delay - 60, 120, 180, 240 seconds Operating modes: Circular or linear mode

Target Power factor:

This is the final power factor value to be achieved by the capacitor bank.

C/K:

Ratio of the capacitor current for the 1 st step over the current transformer (CT) size.

Sequence:

The ratio of the 2 nd, 3rd, 4th etc steps a gainst the 1 st capacitor step.


Page 123 of 136


Time delay:

The delay time for the capacitor to switch in or switch out after receiving signal from the power factor regulator

Operating modes:

There are two modes – linear or circular modes.

10.2.1 Calculation of C/K ratio The C/K ratio is the ratio of the capacitor current for the first ste p and the current tr ansformer (CT) size. C/K = Q (Mvar) x 1000 = Capacitor current for the first step (1.732 x V LLx k) CT Size Capacitor current per step =

k

Q (Mvar) x 1000 (1.732 x V LL)

= CT Ratio

VLL = phase to phase voltage i.e. 11 kV


Page 124 of 136


Example:

Size of Capacitor Bank = 4 Mvar, Y-Y Voltage

=

Number of step =

11 kV 2

MVar value for each step = 2 Mvar (1 st step), 2 Mvar (2 nd Step) Size of CT =

300/5

Calculation: k

= CT Ratio = 300/5 = 60

VLL = phase to phase voltage i.e. 11 kV

Capacitor current per s tep =

2 MVar x 1000 = 104.9 Amp (1.732 x 11 kV)

C/K = Q (Mvar) x 1000 = Capacitor current for the first step (1.732 x V LLx k) CT Size = 104.9 / 60 = 1. 748


Page 125 of 136


Exercise No.19

Size of Capacitor Bank = 1.2 Mvar, Delta Voltage

=

Number of step =

3.3 kV 2

MVar value for each step = 600 kVar (1 st step), 600 kVar (2 nd Step) Size of CT =

400/5

Please calculate the value for C/K:

ILSAS/MFF/2007

Page 126 of 136


Sample C/K calculations for 11 kV Capacitor Banks

C/T ratio 50 /5 100 /5 150 /5 200 /5 300 /5 400 /5 500 /5 600 /5 800 /5 1000 /5 1600 /5 2000 /5 3000 /5

k 10 20 30 40 60 80 100 120 160 200 320 400 600

1000 5.249 2.624 1.750 1.312 0.875 0.656 0.525 0.437 0.328 0.262 0.164 0.131 0.087

ILSAS/MFF/2007

Capacitor step rating (kVar) 2000 3000 4000 10.498 15.746 20.995 5.249 7.873 10.498 3.499 5.249 6.998 2.624 3.937 5.249 1.750 2.624 3. 499 1.312 1.968 2. 624 1.050 1.575 2. 100 0.875 1.312 1. 750 0.656 0.984 1. 312 0.525 0.787 1. 050 0.328 0.492 0. 656 0.262 0.394 0. 525 0.175 0.262 0. 350

5000 26.244 13.122 8.748 6.561 4.374 3.280 2.624 2.187 1.640 1.312 0.820 0.656 0.437

Page 127 of 136


Sample C/K calculations for 3.3 kV Capacitor Banks

C/T ratio 50 /5 100 /5 150 /5 200 /5 300 /5 400 /5 500 /5 600 /5 800 /5

k 10 20 30 40 60 80 100 120 160

300 5.249 2.624 1.750 1.312 0.875 0.656 0.525 0.437 0 328

Capaci tor s tep rating (kVar) 600 900 1200 10.498 15.746 20.995 5.249 7.873 10.498 3.499 5.249 6.998 2.624 3.937 5.249 1.750 2.624 3.499 1.312 1.968 2.624 1.050 1.575 2.100 0.875 1.312 1.750 0 656 0 984 1 312

1500 26.244 13.122 8.748 6.561 4.374 3.280 2.624 2.187 1 640


Sample C/K calculations for 3.3 kV Capacitor Banks

C/T ratio 50 /5 100 /5 150 /5 200 /5 300 /5 400 /5 500 /5 600 /5 800 /5 1000 /5 1600 /5 2000 /5 3000 /5

k 10 20 30 40 60 80 100 120 160 200 320 400 600

300 5.249 2.624 1.750 1.312 0.875 0.656 0.525 0.437 0.328 0.262 0.164 0.131 0.087

Capaci tor s tep rating (kVar) 600 900 1200 10.498 15.746 20.995 5.249 7.873 10.498 3.499 5.249 6.998 2.624 3.937 5.249 1.750 2.624 3.499 1.312 1.968 2.624 1.050 1.575 2.100 0.875 1.312 1.750 0.656 0.984 1.312 0. 525 0.787 1.050 0.328 0.492 0.656 0.262 0.394 0.525 0.175 0.262 0.350

ILSAS/MFF/2007

1500 26.244 13.122 8.748 6.561 4.374 3.280 2.624 2.187 1.640 1.312 0.820 0.656 0.437

Page 128 of 136


10.3

Calculation of resonance frequency

A capacitor bank must not be exposed to resonance conditions. The capacitor bank will experience high voltage and this will lead to deterioration of the capacitor units. To prevent the res on ance co ndition, it is important to identify potential condition that might lead to such incidences. A capacitor bank must not be exposed to resonance condition between 5 th harmonic number to 13 th harmonic number. The calculation of resonance cond ition is done using Equation (34) h

=√ [kVAsc/ kVar] = √ [Xc/ Xsc] =√ [kVA-tx x 100/(kVar-cap x Z-tx %)]

(34)


10.3

Calculation of resonance frequency

A capacitor bank must not be exposed to resonance conditions. The capacitor bank will experience high voltage and this will lead to deterioration of the capacitor units. To prevent the res on ance co ndition, it is important to identify potential condition that might lead to such incidences. A capacitor bank must not be exposed to resonance condition between 5 th harmonic number to 13 th harmonic number. The calculation of resonance cond ition is done using Equation (34) h

=√ [kVAsc/ kVar] = √ [Xc/ Xsc] =√ [kVA-tx x 100/(kVar-cap x Z-tx %)]

(34)

h = calculated system harmonic kVAsc = Short circuit power of system kVar

= rating of the capacitors

kVA-tx = kVA rating of the step down Tx Z-tx

= Step-down Tx Impedance

If the resonance frequency is between 5 th to 13th harmonic number, the capacitor bank must be protected using detuned reactors with values of 7 %.

ILSAS/MFF/2007

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Example:

Capacitor size:

2 Mvar Y-Y, 2 step, 11 kV

Transformer details:

1.5 MVA, 5.25 %

Calculation for resonance condition for 1 st step: h

=√ [kVA-tx x 100/(kVar-cap x Z-tx %)] = √ [1500 x 100 /(1000 x 5.25)] = 5.35

Calculation for resonance condition for 1 st & 2nd step: h

=√ [kVA-tx x 100/(kVar-cap x Z-tx %)] = √ [1500 x 100 /(2000 x 5.25)] = 3.78

A serious resonance condition will happen when the 1 st capacitor step is switch in. Therefore a 7 % detuned reactor must be installed in series with the capacitor units. Installing a s eries reactor will prevent the resonance condition from happening and damaging the capacitors.

ILSAS/MFF/2007

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Exercise No.20

Capacitor size:

900 kVar Delta, 3 step, 3.3 kV


1.5 MVA, 5.25 %

Please calc ulate the resonance condition for 1 st, 2nd & 3rd capacitor steps.

At what step will the resonance condition happen?

What is the recommended measure to prevent the resonance conditions?

ILSAS/MFF/2007

Page 131 of 136


10.4

Calculation for the increase in line voltage

Commissioning a capacitor bank will slightly increase the line voltage. Its’ important to verify the voltage increase to ensure no potential problem for high voltage incidenc es. The equation to estimate the % increase in voltage is as follows: % Voltage rise =

Capacitor size (kVAr) x 100 Fault Level kVA(3-phase)

(35)

Equation (35) requires the value of th e fault level. The values can be obtained from the Operation & Maintenance Units in a plant.

However, an estimated value can be obtained using this methodology:

Assuming the maximum impedance of t he network is based on the distribution (incoming) transformer. For example: A TX 15 MVA, 33/11 kV Volt, % Z = 10.5 Ifault= MVA/ (VLL x %Z x√3)= 7.498 kA (3-phase) Sfault =Ifault x VLL x √3 = 142 MVA (3-phase)

ILSAS/MFF/2007

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Exercise No.21

Capacitor size:

900 kVar Delta, 3 step, 3.3 kV


1.0 MVA, 5.25 %

What is the % voltage increase?

Exercise No.22

Capacitor size:

4 MVar Y-Y, 2 step, 11 kV


1.5 MVA, 5.25 %

What is the % voltage increase?

ILSAS/MFF/2007

Page 133 of 136


10.5

Inspection after commissioning

Before commissioning the capacitor bank, please rec ord all the currents, voltages, power factors and harmonic levels at the incoming supply. Once the capacitor bank has b een commissioned, swi tch in all the steps one by one and record the same parameters for evaluation.

a. Volt Phase

Before

Step 1

2

3

4

commissioning R-Y Y-B B-R Step

b.Current

Phase

Before

1

2

3

4

2

3

4

commissioning Red Yellow Blue Step

c. Power Factor Phase

Before

1

commissioning Red Yellow Blue

ILSAS/MFF/2007

Page 134 of 136


d. THDV Phase

Before

Step 1

2

3

4

commissioning R-Y Y-B B-R 10.5.1 Checking operating data: Make the following measurements to ensure that th e capacitors are not subject to excessive stress: 1. Voltage before and after switching in the capacitor bank: The capacitor bank must not be subjected to a continuous voltage exceeding 110% of the nominal voltage, continuous meaning 12 hours in every 2 4 according to AS 2897-19 86 and IEC 871. 2. Capacitor cu rrent: The capacitor bank must not be subjected to a conti nuous current exceeding 130% of the rated current. 3. Ambient temperature: The capacitor bank must not be operated in a n ambient temperature outside the stated temperature class. 4. Unbalance current: (for banks with unbalance protection) The unbalance current should not exceed 10% of the tripping value of the protection system. Large capacitor banks are balanced at the factory to ensure that they do not exceed this value. Small capacitor banks are sometimes impossible to balance sufficiently to comply with this recommendation if the capacitances of the individual capacitor units differ too widely. If the individual capacitors are at different temperatures, for example because of solar radiation, the unbalance current immediately after switching in the bank may differ from the value after a few hours in service. The unbalance current should therefore be measured when the capacitors have reached their normal operating temperature. ILSAS/MFF/2007

Page 135 of 136

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