Capacitor Chargeman

CAPACITOR BANK & POWER FACTOR MANAGEMENT

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Electrical Chargeman Category B0

ILSAS/MFF/2007 Page 1 of 136

CAPACITOR BANK & POWER FACTOR MANAGEMENT Contents Title Page No. 1.0 FUNDAMENTALS OF BASIC ELECTRICITY 6 1.1 Definition of electrical power 6 1.2 Apparent Power (VA) 6 1.3 Real Power or True Power (Watts) 6 1.4 Reactive Power (Var) 7 1.5 Power Factor 7 1.6 Relationship between Real, Reactive, and 7

Apparent power 1.7 Equipment Causing Poor Power Factor 13 1.8 Calculation for power factor 14 2.0 EFFECT OF LOW POWER FACTOR 20 3.0 POWER FACTOR CORRECTION 21 3.1 Capacitive Power Factor correction 21 3.2 Function of PFC Capacitors 23 3.3 How capacitors work 24 3.4 Categories of Power Factor Correction 25 4.0 CAPACITOR BANK SIZING 30 4.1 Selecting kVAR for 3-Phase Motors 30 4.2 Power Factor Correction Capacitors on Reduced Voltage 33

Motors and Multi-Speed Motors 4.3 Selecting kVAR for Bulk Correction 34 5.0 MV CAPACITOR BANK DESIGN 38 5.1 Function of a capacitor bank 38 5.2 Capacitor unit & Capacitor Bank configuration 38 5.3 Capacitor construction 40 5.4 Basic design of power capacitor 69


CAPACITOR BANK & POWER FACTOR MANAGEMENT 6.0 PROTECTION OF Y-Y CAPACITORS 73 6.1 Capacitor fusing 73 6.2 Relaying protection 74 6.3 The concept of unbalance current flow 81 6.4 Basics on relay unbalance & Unbalance CT 82 6.5 Typical settings for a 11 kV 3-5 MVar Y-Y Capacitor Banks 83 6.6 Lightning protection 84 7.0 PROTECTION FOR DELTA CAPACITORS 85 7.1 Fusing arrangement for Delta Capacitors 85 7.2 Comparison between fusing arrangement for Delta 87

& Y capacitors 7.3 Relaying protection 90 8.0 MAINTENANCE PROCEDURES FOR A Y-Y CAPACITOR BANK 8.1 Safety Regulations 91 8.2 Maintenance Methodology for Y-Y Capacitor Banks 91 8.3 Maintenance program 95 9.0 MAINTENANCE PROCEDURES FOR A DELTA CAPACITOR BANK 9.1 Safety Regulations 108 9.2 Maintenance Methodology for Delta Capacitor Banks 108 9.3 Maintenance program 111 10.0 COMMISSIONING 10.1 Before energizing the bank, check: 121 10.2 Setting for Power Factor Regulator (Controller) 121 10.3 Calculation of resonance frequency 129 10.4 Calculation for the increase in line voltage 132 10.5 Inspection after commissioning 134 Sample for Inspection & Maintenance Forms are attached as Appendix A. Sample for commissioning forms are attached as Appendix B.






CAPACITOR BANK & POWER FACTOR MANAGEMENT 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=VIR = VI cos θ (1) IR = I cos θ cos θ = power factor


CAPACITOR BANK & POWER FACTOR MANAGEMENT 1.4 Reactive Power (Var) The instantaneous power absorbed by the reactive part of the load is given by: Reactive power =Q=VIX = 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 (900)=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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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)


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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: True power, reactive power, and apparent power for a purely resistive load.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 1.6.2 Reactive load only:

120 Volt 50 Hz

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


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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:



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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT REVIEW:

• 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.

1.7 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


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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 (6) Power Factor (PF) = kW = True Power (7)

KVA Apparent Power |kVA| = √ [(kW)2 + ( kVAr)2] (8) As was mentioned before, the angle of this “power triangle” graphically indicates the ratio between the amount of dissipated (or consumed) power and the 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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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:


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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).


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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?


CAPACITOR BANK & POWER FACTOR MANAGEMENT 2.0 EFFECT OF LOW POWER FACTOR "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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 3.0 POWER FACTOR CORRECTION

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: Magnetizing current


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.

Figure 6: Power Factor Correction (1)


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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: Power Factor Correction (2)


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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: Power Factor Correction for Induction motors

3.3 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: kVA2 = 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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.



Figure 11: Static correction for soft starter schemes


CAPACITOR BANK & POWER FACTOR MANAGEMENT 4.0 CAPACITOR BANK SIZING 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. (9)




Table 1 Power Factor Improvement Table

80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 10050 0.982 1.008 1.034 1.06 1.086 1.112 1.139 1.165 1.192 1.22 1.248 1.276 1.306 1.337 1.369 1.403 1.442 1.481 1.529 1.59 1.73251 0.937 0.962 0.989 1.015 1.041 1.067 1.094 1.12 1.147 1.175 1.23 1.231 1.261 1.292 1.324 1.358 1.395 1.436 1.484 1.544 1.68752 0.893 0.919 0.945 0.971 0.997 1.023 1.05 1.076 1.103 1.131 1.159 1.187 1.217 1.248 1.28 1.314 1.351 1.392 1.44 1.5 1.64353 0.85 0.876 0.902 0.928 0.954 0.98 1.007 1.033 1.06 1.088 1.116 1.144 1.174 1.295 1.237 1.271 1.308 1.349 1.397 1.457 1.654 0.809 0.835 0.861 0.887 0.913 0.939 0.966 0.992 1.019 1.047 1.075 1.103 1.133 1.164 1.196 1.23 1.267 1.308 1.356 1.416 1.55955 0.769 0.795 0.821 0.847 0.873 0.899 0.926 0.952 0.979 1.007 1.033 1.063 1.09 1.124 1.156 1.19 1.228 1.268 1.316 1.377 1.51956 0.73 0.756 0.782 0.808 0.834 0.86 0.887 0.913 0.94 0.968 0.996 1.024 1.051 1.085 1.117 1.151 1.189 1.229 1.277 1.338 1.4857 0.692 0.718 0.744 0.77 0.796 0.822 0.849 0.875 0.902 0.93 0.958 0.986 1.013 1.047 1.079 1.113 1.151 1.191 1.239 1.3 1.44258 0.655 0.681 0.707 0.733 0.759 0.785 0.812 0.838 0.865 0.893 0.921 0.949 0.976 1.01 1.042 1.076 1.114 1.154 1.202 1.263 1.40559 0.618 0.644 0.67 0.696 0.722 0.748 0.775 0.801 0.828 0.856 0.884 0.912 0.939 0.973 1.005 1.039 1.077 1.117 1.165 1.226 1.36860 0.584 0.61 0.636 0.662 0.688 0.714 0.741 0.767 0.794 0.822 0.85 0.878 0.905 0.939 0.971 1.005 1.043 1.083 1.131 1.192 1.33461 0.549 0.575 0.601 0.627 0.653 0.679 0.706 0.732 0.759 0.787 0.815 0.843 0.87 0.904 0.936 0.97 1.008 1.014 1.096 1.157 1.29962 0.515 0.541 0.567 0.593 0.619 0.645 0.672 0.698 0.725 0.753 0.781 0.809 0.836 0.87 0.902 0.936 0.974 0.982 1.062 1.123 1.26563 0.483 0.509 0.535 0.561 0.587 0.613 0.66 0.666 0.693 0.721 0.749 0.777 0.804 0.838 0.87 0.904 0.942 0.949 1.03 1.091 1.23364 0.45 0.476 0.502 0.528 0.554 0.58 0.607 0.633 0.66 0.688 0.716 0.744 0.771 0.805 0.837 0.871 0.909 0.918 0.997 1.058 1.265 0.419 0.445 0.471 0.497 0.523 0.549 0.576 0.602 0.629 0.657 0.685 0.713 0.74 0.774 0.806 0.84 0.878 0.887 0.966 1.027 1.16966 0.388 0.414 0.44 0.466 0.492 0.518 0.545 0.571 0.598 0.626 0.554 0.682 0.709 0.743 0.775 0.809 0.847 0.857 0.935 0.996 1.13867 0.358 0.384 0.41 0.436 0.462 0.488 0.515 0.541 0.568 0.596 0.624 0.652 0.679 0.713 0.746 0.779 0.817 0.828 0.905 0.966 1.10868 0.329 0.355 0.381 0.407 0.433 0.459 0.486 0.512 0.539 0.567 0.595 0.623 0.65 0.684 0.716 0.75 0.788 0.798 0.876 0.937 1.07969 0.299 0.325 0.351 0.377 0.403 0.429 0.456 0.482 0.509 0.537 0.565 0.593 0.62 0.654 0.686 0.72 0.758 0.769 0.84 0.907 1.04970 0.27 0.296 0.322 0.348 0.374 0.4 0.427 0.453 0.48 0.508 0.536 0.564 0.591 0.625 0.657 0.691 0.729 0.741 0.811 0.878 1.0271 0.242 0.268 0.294 0.32 0.346 0.372 0.399 0.425 0.452 0.48 0.508 0.536 0.563 0.597 0.629 0.663 0.701 0.712 0.783 0.85 0.99272 0.213 0.239 0.265 0.291 0.317 0.343 0.37 0.396 0.423 0.451 0.479 0.507 0.534 0.568 0.6 0.634 0.672 0.685 0.754 0.821 0.96373 0.186 0.212 0.238 0.264 0.29 0.316 0.343 0.369 0.396 0.424 0.452 0.48 0.507 0.541 0.573 0.607 0.645 0.658 0.727 0.794 0.93674 0.159 0.185 0.211 0.237 0.263 0.289 0.316 0.342 0.369 0.397 0.425 0.453 0.48 0.514 0.546 0.58 0.618 0.631 0.7 0.767 0.90975 0.132 0.158 0.184 0.21 0.236 0.262 0.289 0.315 0.342 0.37 0.398 0.426 0.453 0.487 0.519 0.553 0.591 0.604 0.673 0.74 0.88276 0.105 0.131 0.157 0.183 0.209 0.235 0.262 0.288 0.315 0.343 0.371 0.399 0.426 0.46 0.492 0.526 0.564 0.578 0.652 0.713 0.85577 0.079 0.105 0.131 0.157 0.183 0.209 0.236 0.262 0.289 0.317 0.345 0.373 0.4 0.434 0.466 0.5 0.538 0.552 0.62 0.687 0.82978 0.053 0.079 0.105 0.131 0.157 0.183 0.21 0.236 0.263 0.291 0.319 0.347 0.374 0.408 0.44 0.474 0.512 0.525 0.594 0.661 0.80379 0.026 0.052 0.078 0.104 0.13 0.156 0.183 0.209 0.236 0.264 0.292 0.32 0.347 0.381 0.413 0.447 0.485 0.499 0.567 0.634 0.77680 0 0.026 0.052 0.078 0.104 0.13 0.157 0.183 0.21 0.238 0.266 0.294 0.321 0.355 0.387 0.421 0.459 0.473 0.541 0.608 0.7581 0 0.026 0.052 0.078 0.104 0.13 0.157 0.184 0.212 0.24 0.268 0.295 0.329 0.361 0.395 0.433 0.447 0.515 0.582 0.72482 0 0.026 0.052 0.078 0.105 0.131 0.158 0.186 0.214 0.242 0.269 0.303 0.335 0.369 0.407 0.421 0.489 0.556 0.69883 0 0.026 0.052 0.079 0.105 0.132 0.16 0.188 0.216 0.243 0.277 0.309 0.343 0.381 0.395 0.463 0.53 0.67284 0 0.026 0.053 0.079 0.106 0.134 0.162 0.19 0.217 0.251 0.283 0.317 0.355 0.369 0.437 0.504 0.64585 0 0.027 0.053 0.08 0.108 0.136 0.164 0.191 0.225 0.257 0.291 0.329 0.343 0.417 0.478 0.6286 0 0.026 0.053 0.081 0.109 0.137 0.167 0.198 0.23 0.265 0.301 0.317 0.39 0.451 0.59387 0.027 0.055 0.082 0.111 0.141 0.172 0.204 0.238 0.275 0.29 0.364 0.425 0.56788 0 0.028 0.056 0.084 0.114 0.145 0.177 0.211 0.248 0.262 0.237 0.398 0.5489 0.028 0.056 0.086 0.117 0.149 0.183 0.22 0.234 0.309 0.37 0.51290 0.028 0.058 0.089 0.121 0.155 0.192 0.206 0.281 0.342 0.48491 0.03 0.061 0.093 0.127 0.164 0.176 0.253 0.314 0.45692 0.031 0.063 0.097 0.134 0.143 0.223 0.284 0.42693 0 0.032 0.066 0.103 0.113 0.192 0.253 0.39594 0.034 0.071 0.079 0.16 0.221 0.36395 0.037 0.042 0.126 0.187 0.32896 0 0.089 0.15 0.29297 0 0.047 0.108 0.25198 0 0.061 0.20399 0 0.142

Original Power Factor in

%

Desired Pow er Factor in %

CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.


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

CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.

Figure 12: Capacitor schemes for multi-speed motors


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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: Capacitor schemes for bulk correction


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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 One customer has maximum load of 8500 kW & average power factor at 0.54. Due to having power factor less than 0.85 (Based on Tariff), the customer is charged power factor penalty. What is the size of capacitor bank to improve power factor from 0.54 to 0.95?


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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) = 49.450 Angle for desired power factor = φ2 = Cos-1(0.90) = 25.840 Capacitor size (kVAr) = kW x (Tangent φ1 – Tangent φ2) = 2000 x (Tan 49.450 – Tan 25.840) = 1,369 kVAr


CAPACITOR BANK & POWER FACTOR MANAGEMENT Exercise No.5 (Direct calculation) One customer has maximum load of 8500 kW & average power factor at 0.54. Due to having power factor less than 0.85 (Based on Tariff), the customer is charged power factor penalty. What is the size of capacitor bank to improve power factor from 0.54 to 0.95?


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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 line-to-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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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 Capacitor Unit Capacitor element

Figure 15: Power capacitor bank major components



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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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. The voltage level of the system 2. The kVar capacity of the Capacitor Bank 3. The system grounding 4. 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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.


CAPACITOR BANK & POWER FACTOR MANAGEMENT 5.3.3.4 Y-Y Capacitor Bank design

Figure 17: Typical Medium Voltage Y-Y Capacitor design


CAPACITOR BANK & POWER FACTOR MANAGEMENT 5.3.3.4.1 Y-Y Connection

CT

REACTOR

REACTOR

REACTOR

BUSBAR 11 KV

40 uH REACTOR

333kVar

333kVar

333kVar

333kVar

333kVar

red

40 uH REACTOR

333kVar

333kVar

333kVar

333kVar

333kVar

yellow

40 uH REACTOR

333kVar

333kVar

333kVar

333kVar

333kVar

UNBALANCE RELAY

Blue

BUSBAR 11 KV

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



CT

REACTOR

REACTOR

REACTOR

BUSBAR 11 KV

2 MVar 3 MVar

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

Description of a typical 5 MVar 11 kV Capacitor Bank 1. Rating per capacitor unit = 333 kVar, 6.35 kV (VLN) 2. Number of capacitor units for a 5 MVar capacitor bank

= 15 (15 x 333 kVar = 5 MVar) 3. Rated reactive current (Ikvar) = 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)



40 uH REACTOR

833kVar

833kVar

red

40 uH REACTOR

833kVar

833kVar

yellow

40 uH REACTOR

833kVar

833kVar

UNBALANCE RELAY

Blue

BUSBAR 11 KV

Figure 20: Y-Y connection for a 5 Mvar Capacitor Bank 11 kV (2) Description of a typical 5 MVar 11 kV Capacitor Bank 1. Rating per capacitor unit = 833 kVar, 6.35 kV (VLN) 2. Number of capacitor units for a 5 MVar capacitor bank

= 6 (6 x 833 kVar = 5 MVar) 3. Rated reactive current (Ikvar) = 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)



40 uHREACTOR

225kVar

225kVar

225kVar

225kVar

225kVar

40 uHREACTOR

225kVar

225kVar

225kVar

225kVar

225kVar

40 uHREACTOR

225kVar

225kVar

225kVar

225kVar

225kVar

UNBALANCE RELAY

BUSBAR 11 KV

225kVar

225kVar

225kVar

225kVar

225kVar

225kVar

225kVar

225kVar

225kVar

Figure 21: Y-Y connection for a 5 Mvar Capacitor Bank 11 kV (3) Description of a typical 5.4 MVar 11 kV Capacitor Bank 1. Rating per capacitor unit = 225 kVar, 6.35 kV (VLN) 2. Number of capacitor units for a 5.4 MVar capacitor bank

= 24 (24 x 225 kVar = 5.4 MVar) 3. Rated reactive current (Ikvar) = 5.4 Mvar / (√3 x 11 kV) = 283 Amp Connection of capacitor bank = Ungrounded Y-Y (2Y)


CAPACITOR BANK & POWER FACTOR MANAGEMENT 5.3.3.4.1.1 Series connection Basic rating of MV capacitor unit: Q kVar, VLN (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

11 k

V11

kV

333 kVar333 kVar

Three phase MVar rating = 0.33 MVar + 0.33 MVar + 0.33 MVar = 1.0 MVar Phase to phase voltage rating= √3x6.35 kV = 11 kV


CAPACITOR BANK & POWER FACTOR MANAGEMENT Basic rating of MV capacitor unit: Q kVar, VLN (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

333 kVar

333 kVar

333 kVar

Red Phase

111 k

Var

Yellow Phase

Blue Phase

33 k

V

Three phase MVar rating = 0.11 MVar + 0.11 MVar + 0.11 MVar = 0.33 MVar Phase to phase voltage rating= √3x(3x6.35 kV) =33 kV


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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 (XT) = XC1 + XC2 + XC3 Formula: Capacitive reactance (XC) = 1/ ωC, ω = 2x3.14x50 The equivalent reactance (XT) = 1/ ωC1 + 1/ ωC2 +1/ ωC3 = 1/ ωCT ∴ 1/ CT =1/ C1 + 1/ C2 +1/ C3 Equivalent kVar per phase = (VoltageLN)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 = 6.35kV2 x ω x CT = 111 kVar Three phase Q = 3 x 111 kVar = 333 kVar = 0.333 Mvar



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/XT) = 1/XC1 + 1/XC2 + 1/XC3 Formula: Capacitive reactance (XC) = 1/ ωC, ω = 2x3.14x50 The equivalent reactance (1/XT) = ωC1 + ωC2 +ωC3 = ωCT ∴ CT =C1 + C2 + C3 Equivalent kVar per phase = (VoltageLN)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 = 6.35kV2 x ω x CT = 999 kVar Or simply adding up all the capacitor values in parallel = 333 kVar+333kVar+333 kVar = 999 kVar


MEDIUM VOLTAGE CAPACITOR & POWER FACTOR MANAGEMENT

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



Single line diagram for 5 MVAR 11 kV



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

Capacitor Unit333 kVar 6.35 kV

ILSAS/MFF/2007

Page 55 of 136



Capacitor Uni333 kVar 6.35 kV

ILSAS/MFF/2007

t

Page 56 of 136




ILSAS/MFF/2007

Page 57 of 136


Exercise No.10 Current for 5 MVAR, 33 kV


ILSAS/MFF/2007

Page 58 of 136


Exercise No.11 Current for 3 MVAR, 11 kV

Capacitor Uni333 kVar 6.35 kV

ILSAS/MFF/2007

t

Page 59 of 136



5.3.4 Calculation of Capacitive Current (Ikvar)

Power factor = cos φ

Ic = kvar /(√3 x VLL) 3-phase (16) Ic = kvar /(VLN) 1-phase (17) From equation (13) kVar for PF Correction kVar = kW x (tan φ1-tan φ2)

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



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) (18) b. Var = V

2ωC (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: Single line diagram for 5 MVAR 11 kV




Figure 24 Connection diagram in uF for 5 MVar 11 kV



Figure 25 Connection diagram in ohm for 5 MVar 11 kV



Figure 26 Calculation of Amp based on Total Impedance


Exercise No.12 Calculate the current if one capacitor can is damaged.




5.3.7 Estimation of neutral current IA = [IR2 + IY2 + IB2] (28) IB = [ (IR x IY ) + ( IYx IB ) + ( IB x IR) ] (29) I neutral = √ [IA – IB] (30)



5.4 Design of Power Capacitor Banks < 4.16 kV 5.4.1 Delta connection

Discharge resistors

Fuses

VCB

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 line-to-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.



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 Volt2 /1000 (31) ω = 2xπ x f f = 50 Hz Volt= Volt rating at Capacitor Can Example: Voltage rating per Capacitor 415 Volt

uF

Figure 31: Capacitance measurement for Delta Capacitor using Capacitance meter

Capacitance values for phase-phase values R-Y 375 uF Y-B 375 uF B-R 375 uF



kVar (3-phase) = 2/3 x (375uF + 375uF + 375 uF) x ω x Volt2 /1000 = 2/3 x (375uF + 375uF + 375 uF) x 2x3.14x50 x 4152/1000 = 40.6 kVar Exercise No.13 Calculation of kVar for Delta Capacitor Voltage rating per Capacitor 3.3 kV

uF

Figure 32: Capacitance measurement for Delta Capacitor using Capacitance meter

Capacitance values for phase-phase values R-Y 146.22 uF Y-B 146.22 uF B-R 146.22 uF kVar (3-phase) =



5.4.3 Calculation of Capacitive Current (Ikvar)

Power factor = cos φ

Ic = kvar /(√3 x VLL) 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 Voltage rating per Capacitor 3.3 kV Capacitor values (3-phase delta) 1 MVAR



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 follows- a. 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



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.



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.

UnGroundedNeutral

PT Primary

PT Secondary

Voltage Relay

Figure 33 Unbalance Phase Voltage



PT

Voltagerelay

Figure 34: Voltage Imbalance between neutrals

Currentrelay

CT

Figure 35: Current Imbalance between neutrals



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.



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.

PT

Voltagerelay

Resistor

Voltagerelay

Figure 36 Neutral Voltage Figure 37 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.



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 over-voltages. 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

Figure Sensitive to Switching Transient

Sensitive to 3rd harmonic voltage & currents

Sensitive to system voltage unbalance

Number of CT required

Number of PT required

Unbalance phase voltage

33 Yes No No - 3

Voltage unbalance between neutral

34 Yes No No - 1

Current unbalance between neutrals

35 Yes No No 1 -

Neutral voltage

36 Yes Yes Yes - 1

Neutral current

37 Yes Yes Yes 1 -



CT

REACTOR

REACTOR

REACTOR

11 kV Fuses

2 MVar 3 MVarUnbalance CT

Ammeter

A B

Relays

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

Figure 38 Relay configurations for a 5 Mvar 11 kV capacitor



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.



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

Unbalance 10/5 60% 90% 3 Amp 0.5



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 Example on relay settings

Capacitor

Bank (MVar)

kV Rated

Amp

CT for

relay

O/C E/F TMS Un-balance

CT

1st

stage

2nd

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

Capacitor

Bank (MVar)

kV Rated

Amp

CT for

relay

O/C E/F TMS Un-balance

CT

1st

stage

2nd

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



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.



Three phase capacitors use fuses in the line because they are connected delta

ternally. Normally branch fuses are used for single-phase capacitors connected

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

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

93.68 amps

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

. a 60-amp fuse can be used.

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

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

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

he other method is to use two fuses, i.e. one per bushing. This gives the user a lse sense of security. In this case both fuses would have to operate before the iled capacitor can be effectively removed from the system. Normally only one of e fuses operates, which will be the one nearest the faulted packs. The other ushing remains connected to the system via the good fuse. The result is still an ventual major insulation failure if the bank is not removed from service.

Indelta. 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. Wfusing becomes large, and may not coordinate with the tank rupture curve of the capacitor and the upstream co-ordination may not be possible. Fconnected bank, using 150 kVAR per phase, will require the following fusing:

• "In line fusing" Or, "group fusing"

450 kVAR/(4.16 KV x √3) = 62.45 amps x 1.5 =

... a 100-amp fuse is required.

• In branch fusing

1

..

TbtwgTthcc

Tfafathbe

CAPACITOR BANK & POWER FACTOR MANAGEMENT The burning between packs could possibly continue due to the second bushing till being energized via the second fuse. During this condition a low energy fault

r Delta & Y capacitors

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

Figure 40: Connection and fusing arrangements for ungrounded-wye and

delta connected capacitor banks

scould 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 fo

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




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

ank.

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

Internal section faults, or dielectric faults appear differently to a power system. al section fault subjects the power system to

a phase-to-phase bolted fault. This fault will cause a major voltage sag on the may cause capacitor

case rupture if not properly protected.

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

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

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

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

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

b 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), wor wye-connected, a bushing fault will have the same impact on the power system.

On a delta connected bank, an intern

facilities power system (until the capacitor fuse(s) blow) and

im ose mechanical

to a fault current that is three times the bank

with the fault, and case rupture

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

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



• 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

The fuse must clear the minimum over current resulting due to a failed unit 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.

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.

•

within 300 seconds maximum and as stated earlier 30 seco

•

•



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

d with the capacitor switching will not trip the circuit breaker mediately as the bank is energized. Also, if a capacitor unit fault occurs the

current characteristics if coordination with ther system protective devices required it.

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 associateimrelay 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 o Instantaneous over current relays are not often utilized since they are likely to trip unnecessarily when the bank is energized.



8.0 MAINTENANCE PROCEDURES FOR A Y-Y CAPACITOR BANK

Safety Regulations 8.1 When working with capacitor banks, the following safety regulations, in particular,

ust be observed:

• grounded. Short circuit the capacitor units individually also, as if

there is an internal fault, such as a damaged discharge resistor, there may

•

voltage of the unit to: 50 V within 5 minutes - AS2897 - 75V within 10 minutes

• , and

avoid breathing in fumes or gases from the impregnation fluid. In the event of

8.2 Du itor bank needs no maintenance, but it hould be checked regularly as described below to ensure trouble-free operation.

To of the following is commended:

. 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

m

Do not touch a capacitor bank until it has been completely discharged, short circuited and

still be a voltage even though the capacitor bank has been discharged.

Normally capacitors have an internal discharge resistor, which reduces the

– IEC 871.

Avoid skin contact with the impregnation fluid in the event of leakage

skin contact, wash with soap and water. If fluid gets into your eyes, rinse with lukewarm water.

Maintenance Methodology for Y-Y Capacitor Banks

ring normal operating conditions, capacs

obtain trouble free operation a routine inspectionre a

•

• •



b. Check the Unbalance current in the Y-Y capacitor bank

should not iffer by more than ± 1 character.

he unbalance current is checked after reconnection and should be less than 20

y be breakdown in one r more of the internal capacitor elements. The change in current due to

hen the capacitor bank is made up of capacitors with internal fuses, an

the unbalance current exceeds 50 % of the tripping value, a capacitance

he overall maintenance flowcharts are shown in Figure 41 & Figure 42 spectively.

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.

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 unitsd T% of the operating value for the protection. And also if the current readings differ from that obtained at the time of commissioning, there maobreakdown 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. Wunbalance 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. Ifmeasurement of all capacitor units in the bank must be performed and faulty units replaced. Tre If •

• •

•


NOTICE

MAINTENANCE

(PHASE+ NEUTRAL) Unbalance CT size?

NO

YES

MEET TECHNICALYES

RECTIFICATION

FLOWCHART A

CAPACITOR CANBE COMMISSIONED

ROUTINE

RECORD CURRENTSTART

(Neutral I) > 20% ofDONE

DISCONNECTCAPACITOR

CAPACITORBANK HAS

TRIPPED OFF

START

APPLY CIRCUIT MAINEARTH AT VCB

WAIT FOR 10 MINUTESTO DISCHARGE

WITHDRAW THE VCB TRUCK ANDLOCK MAIN BUSBAR SHUTTER WITHA NON STANDARD LOCK AND HANG

SAFETY

OPEN CAPACITOR BANK ANDDISCHARGE USING MV

DISCHARGE ROD

INSPECTION & MAINTENANCEPROGRAM

11 kV CABLE PROTECTIONSCHEMES

CAPACITORTOLERANCE

REACTORTOLERANCE

SAFETYINTERLOCK

REFER TOFLOWCHART B

REQUIREMENT ?

NO

Figure 41: Work process for maintenance of a Y-Y capacitor bank (1)


ITOR BANK & POWER FACTOR MANAGEMENT

Page 94 of 136

igure 42: ten e of capa nk (2)

TOR CANESIONED)

citor ba

AFETYERLOCK

(CAPA

COM

anc

ORNCE)

REACLER

CIN

CLEANINGO

CONTACTS

ECTI

a Y-Y

ORNCE

NTON

I

ION

Work process for main

(

CAPACTE

CLEANOF

CONTA

CAPACTOLERA

N

F

CABLE

TERMINATION

ION

FROM FLOWCHART A

MAIN A

IT EST

TTO A

IN

CT

URREJECTITEST

CTIS ME

ITON11 kV S

NT

AY

F

MEET TECHN LREQUIREMENT ?

R FICAT

O

TO FLOWCHART ACIB

MIS

INSULAT

CABLE ENTRY

FLOWCHART BCAPAC

TEN

ANC

G

S

RCE

IT

UNBALANCE CTRATIO TEST

PROTECHE

OCEF RELAY

UNBALANCEREL

ICA

ONS

CAPAC

ILSAS/MFF/2007



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

Figure 43: Discharge capacitor banks using Discharge Rod



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

bank if it is dirty.

8.3.2.1 Maintenance for the Enclosure Exterior

• 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 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

itor



8.3.3 Maintenance on the capacitor units

e measurement

If th csho d easurement of the uni in ed in t r

owever, 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. Capac nce 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.

• Cle r are dirty or corroded.

• Check each capacitor for capacitive reactance by applying 120 volts to each phase and measuring corresponding current. Verify with specification.

• onfirm kVar, voltage, and BIL rating for each capacitor. Verify with pecification.

• erify internal discharge resistors are working properly. Replace cells as quired.

8.3.3.1 To take capacitanc

e apacitor bank is not equipped with unbalance protection, all capacitor units ul be capacitance measured annually. Capacitance m

ts a capacitor bank with unbalance protection does not have to be includhe egular inspection.

H

ita

an capacito case, insulation bushings, and any connectors that

CsVre



0 % to 10 % from 3 MVAr to 30 MVAr

• 0% to 5 % for capacitor > 30 MVAr

ge CB10 (1)

8.3.3.3 Capacitance Measurement Acceptable capacitance reading

• -5% to 10 % up to 3 MVAr

•

ACTOR

BUSBAR 11 KV

CAPACITANCE BRIDGE

CT

REACTOR

RE

REACTOR

uF

Figure 44: Capacitance measurement using capacitance brid



CT

REACTOR

BUSBAR 11 KV

ANCE BRIDGE

REACTOR

REACTOR

uF

CAPACIT

Figure 45: Capacitance measurement using capacitance bridge CB10 (2)

CAPACITOR BANK & POWER FACTOR MANAGEMENT Calculation of kVar using measurment values of uF

citance (uF) (33)

333 kVar 6.35 kV 2x3.14x50 26.3 uF Exercise No.15

Please indicate the faulty capacitor can.

kVAR = kVOLT2 x (2πf) x Capa

Phase Nameplate uF Measured uF % DiffRed 26.3 26.1

Yellow 27.4 25.8Blue 26.5 23.0


ITOR BANK & POWER FACTOR MANAGE

MENT

Page 101 of 136

8.3.4 Reactor Test

CAPAC

ILSAS/MFF/2007

Figure 46: Reactor test

CT

REACTOR

REACTOR

REACTOR

BUSBAR 11 KV

CURRENTINJECTIONTEST SET

VOLTMETER

X =VOLT/CURRENT

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


Page 102 of 136

Examples on test result

Note- This test requires a injection test set. Z = Voltage/ Current = XL

XL = ω x L ω = 2 x 3.14 x 50

L = XL / (2x3.14x50)

Description Red Yellow B eVolts, mV 63 63 6Current, Amp 5 5Z(Impedance) 0.0126 0.0126 0.Inductance, uH 40.13 40.13 40 3

lu3

50126.1

CAPAC

ILSAS/MFF/2007


Exercise No.16

Please calculate the values of the impedances & inductances

Description Red Yellow BlueVolts, mV 123 124 126Current, Amp 10 10 10Z(Impedance)Inductance, uH



8.3.5 Unbalance CT Ratio Test

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



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.

•


CAPA

ILSAS/MFF/2007

CITOR BANK & POWER FACTOR MANAGEMENT

Page 106 of 136

8.3.8 Relay maintenance – OCEF & Unbalance

Figure 47: OCEF & Unbalance relays

Relays CT Ratio 1st stage 2nd stage Setting TMSO/C, E/F 300/5 100% - 5 Amp 0.2

Unbalance 10/5 60% 90% 3 Amp 0.1


8.3.9 Trip test for

unbalance relay

NOTA:


CAPACITOR BANK & POWER FACTOR MANAGEMENT 9.0 MAINTENANCE PROCEDURES FOR A DELTA CAPACITOR BANK

9.1 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.

.2 Maintenance Methodology for Delta Capacitor Banks

uring normal operating conditions, capacitor bank needs no maintenance, but it ould be checked regularly as described below to ensure trouble-free operation.

o obtain trouble free operation a routine inspection of the following is commended:

egular Inspection

n annual inspection is sufficient unless the capacitor bank is exposed to bnormal conditions of any kind, such as excessive contamination.

t 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.

•

•

9 Dsh Tre R Aa A

• •


CAPA

ILSAS/MFF/2007

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

•

Maintenance procedure The overall maintenance flowcharts are shown in Figure 48.


Page 109 of 136

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.


EARTH AT VCB TO DISCHARGE LOCK AND HANG NOTICE

MAINTENANCE

RECORD CURRENT(PHASE+ NEUTRAL) CURRENT = RATED

CURRENT ?

YES

NO

PROGRAM

MEET TECHNICALYES

RECTIFICATION

CAPACITOR CANBE COMMISSIONED

ROUTINE

STARTCAPACITOR

DONE

DISCONNECTCAPACITOR

CAPACITORBANK HAS

TRIPPED OFF

START

APPLY CIRCUIT MAIN WAIT FOR 10 MINUTESWITHDRAW THE VCB TRUCK AND

LOCK MAIN BUSBAR SHUTTER WITHA NON STANDARD

SAFETY

OPEN CAPACITOR BANK ANDDISCHARGE USING MV

DISCHARGE ROD

INSPECTION & MAINTENANCE

MV CABLE PROTECTIONSCHEMES

CAPACITORTOLERANCE

REACTORTOLERANCE

SAFETYINTERLOCK

REQUIREMENT ?

NO

Figure 48: Work process for maintenance of a Delta capacitor bank




9.3 Maintenance program 9.3.1 Isolation procedure

ected. ect and isolate the Cap-Bank to be checked.

• ait a minimum of 10 minutes for the capacitor bank to be discharged Circuit and earth the Capacitor Bank

• Short Circuit the terminals of each capacitor unit parallel group

Figure 49: Discharge capacitor banks using Discharge Rod

• Line voltage with the capacitor bank connected respectively disconn• Disconn

W• Short


.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 Maintenance for the Enclosure Exterior

• 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 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

9

• • •




9.3.4 Maintenance on the capacitor units

e measurement

If th csho d easurement of the uni in ed in t r

owever, 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: Capacitance measurement for a Delta Capacitor Bank

9.3.4.1 To take capacitanc

e apacitor bank is not equipped with unbalance protection, all capacitor units ul be capacitance measured annually. Capacitance m

ts a capacitor bank with unbalance protection does not have to be includhe egular inspection.

H



itan lt indication or when the value

f the capacitor current does not equal to the rated capacitor current.

cceptable capacitance reading

• 0 % to 10 % from 3 MVAr to 30 MVAr

Voltage rating per Capacitor 3.3 kV

ing Capacitance

nce values for phase-phase values

2/3 x (146.22 uF + 146.22 uF + 146.22 uF) x ω x Volt2 /1000 = 999 kVar

Capac ce measurement is only performed at fauo A

• -5% to 10 % up to 3 MVAr

• 0% to 5 % for capacitor > 30 MVAr

Example:

uF

Figure 51: Capacitance measurement for Delta Capacitor usmeter

Capacita R-Y 146.22 uF Y-B 146.22 uF B-R 146.22 uF kVar (3-phase)

=



Exercise No.17

.CP.F

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

lease indicate the faulty capacitor can.

Cans R-Y Y-B B-R kVar % DiffNumber 1 146.22 146.22 146.22Number 2 146.22 146.22 135.60Number 3 146.22 146.22 146.22

Nameplate uF

P



Calculation for Exercise 16.



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 Insulators

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

9.3.4.4 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.



9.3.4.5 Reactor Test

Figure 52: Reactor test

CurrentInjectionTest Set

Volt Meter


Page 119 of 136

Examples on test result

Note- This test requires a injection test set. Z = Voltage/ Current = XL

XL = ω x L ω = 2 x 3.14 x 50

L = XL / (2x3.14x50)

Description Red YellowVolts, mV 6Current, Amp 5Z(Impedance) 0.0 0 6 0 1 6Inductance, uH 40 3

B6

.040.

lue3

213

65

01.

3

21

.40

351.1263

CAPAC

ILSAS/MFF/2007

CAPACITOR BANK & POWER FACTOR MANAGEMENT Exercise No.18

Please calculate the values of the impedances & inductances 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.

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

ample for commissioning forms are attached as Appendix B.

Description Red Yellow BlueVolts, mV 25 25 25Current, Amp 2 2 2Z(Impedance)Inductance, uH

S S


CAPACITOR BANK & POWER FACTOR MANAGEMENT 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.

Figure 53: Power Factor Regulator for a Delta Capacitor

P.F.C

PFC controls the switching of or contactors.

the VCBs


CAPA

ILSAS/MFF/2007


Page 122 of 136

Figure 54: Power Factor Regulator for a Y-Y Capacitor

PFC PFC controls the switching of the VCBs or contactors.

CT

REACTOR

REACTOR

REACTOR

BUSBAR 11 KV

CAPACITOR BANK & POWER FACTOR MANAGEMENT The power factor regulator (controller) monitor the input current from only one hase and voltage inputs (phase to phase voltage) for the other two phases.

Figure 55: Power Factor regulator (controller)

he 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

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

/K: Ratio of the capacitor current for the 1st step over the

gainst the 1st capacitor step.

p

The power factor regulator must be programmed to ensure correct operation of capacitor bank. T

T

C

current transformer (CT) size. Sequence: The ratio of the 2nd, 3rd, 4th etc steps a




Time delay: The delay time for the capacitor to switch in or switch

out after receiving signal from the power factor

10.

p and the current ansformer (CT) size.

/K = Q (Mvar) x 1000

regulator Operating modes: There are two modes – linear or circular modes.

2.1 Calculation of C/K ratio The C/K ratio is the ratio of the capacitor current for the first stetr C = Capacitor current for the first step

(1.732 x VLLx k) CT Size

apacitor current per step = Q (Mvar) x 1000

C

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

(1.732 x VLL) k = CT Ratio



Example: Size of Capacitor Bank = 4 Mvar, Y-Y Voltage = 11 kV Number of step = 2 MVar value for each step = 2 Mvar (1st step), 2 Mvar (2nd Step) Size of CT = 300/5 Calculation: k = CT Ratio = 300/5 = 60 VLL = phase to phase voltage i.e. 11 kV Capac tep = 2 MVar x 1000itor current per s = 104.9 Amp (1.732 x 11 kV) C/K = Q (Mvar) x 1000 = Capacitor current for the first step

(1.732 x VLLx k) CT Size

= 10 74

4.9 / 60 = 1. 8



Exercise No.19 Size of Capacitor Bank = 1.2 Mvar, Delta

oltage = 3.3 kV

umber of step = 2

Var value for each step = 600 kVar (1st step), 600 kVar (2nd Step)

ize of CT = 400/5

lease calculate the value for C/K:

V N M S P


Page 127 of 136

Sample C/K calculations for 11 kV Capacitor Banks

k 1000 200 00 0 500050 /5 10 5.249 10.4 4 6.244

100 /5 20 2.624 5.24 7 13.122150 /5 30 1.750 3.49 49 8.748200 /5 40 1.312 2.62 93 6.300 /5 60 0.875 1.75 2 499 4.400 /5 80 0.656 1.312 68 624 3.2500 /5 100 0.525 1.050 75 100 2.6600 /5 120 0.437 0.875 1 12 750 2.1800 /5 160 0.328 0.656 0.984 312 1.6

1000 /5 200 0.262 0.525 0.787 050 1.31600 /5 320 0.164 0.328 0.492 656 0.82000 /5 400 0.131 0.262 0.394 525 0.63000 /5 600 0.087 0.175 0.262 350 0.4

C/T ratioCap it s r g

5613748024874012205637

24

20.10.6.5.3.2.2.1.1.1.0.0.0.

(kV00995498998249

ar)in

63

74

at3

15753.211

t0.7.8.2

.6

.9

.5

.3

epor0989940

ac

CAPAC

ILSAS/MFF/2007


Page 128 of 136

Sample C/K calculations for 3.3 kV Capacitor Banks

k 300 6 0 1 0 150050 /5 10 5.249 0 8 1 746 20 9 2 244

100 /5 20 2.624 . 7 73 0 1 122150 /5 30 1.750 . 5 49 6. 98 8 48200 /5 40 1.312 . 24 3 37 . 49 6 61300 /5 60 0.875 . 2.624 3.499 4 74400 /5 80 0.656 . 1.968 2. 24 3.280500 /5 100 0.525 . 1.575 2. 00 2.624600 /5 120 0.437 . 1.312 1. 50 2.187800 /5 160 0.328 . 0.984 1. 2 1.640

1000 /5 200 0.262 0. 0.787 1.050 1.3121600 /5 320 0.164 0.328 0.492 0.656 0.8202000 /5 400 0.131 0.262 0.394 0.525 0.6563000 /5 600 0.087 0.175 0.262 0.350 0.437

C/T ratioC p tor tep r ti k a

6.3..7.5.3

20.9.492

61731

V

598

r)

1

5

ng (0a

95..8.2.9

s0.424673086

a094999

5012507556

525

ci

153211100

a

CAPAC

ILSAS/MFF/2007



10.3 Calculation of resonance frequency

ank will experience high voltage and this will lead to deterioration of the

o ance ndition, it is important to identify potential condition at might lead to such incidences. A capacitor bank must not be exposed to

number.

of re onanc ition is done using Equation (34)

=√ [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 5th to 13th harmonic number, the capacitor bank must be protected using detuned reactors with values of 7 %.

A capacitor bank must not be exposed to resonance conditions. The capacitor bcapacitor units. To prevent the res n cothresonance condition between 5th harmonic number to 13th harmonic The calculation s e cond h

CAPA

ILSAS/MFF/2007

Example: Capac Transformer details: Calculation for resonance condition for 1

in. Therefore a 7 % detuned reactor must be units. Installing a shappening and damaging


Page 130 of 136

itor size: 2 Mvar Y-Y, 2 step, 11 kV

1.5 MVA, 5.25 %

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 1st & 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 1st capacitor step is switch

installed in series with the capacitor eries reactor will prevent the resonance condition from

the capacitors.

CAPA

ILSAS/MFF/2007

Exercise No.20 Capac Transformer details: Please calc At what step will the


Page 131 of 136

itor size: 900 kVar Delta, 3 step, 3.3 kV

1.5 MVA, 5.25 %

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

resonance condition happen?

What is the recommended measure to prevent the resonance conditions?



10.4 Calculation for the increase in line voltage

es.

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 The equation to estimate the % increase in voltage is as follows: % Voltage rise = Capacitor size (kVAr) x 100 (35)

Fault Level kVA(3-phase)

e fault level. The values can be obtained m the Operation & Maintenance Units in a plant.

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

he network is based on the distribution ncoming) transformer.

or example:

TX 15 MVA, 33/11 kV Volt, % Z = 10.5

= MVA/ (VLL x %Z x√3)= 7.498 kA (3-phase)

Equation (35) requires the value of thfro H Assuming the maximum impedance of t(i F A Ifault Sfault=Ifault x VLL x √3 = 142 MVA (3-phase)



Exercise No.21

kV

apacitor size: 4 MVar Y-Y, 2 step, 11 kV

ransformer details: 1.5 MVA, 5.25 %

hat is the % voltage increase?

Capacitor size: 900 kVar Delta, 3 step, 3.3 Transformer details: 1.0 MVA, 5.25 % What is the % voltage increase? Exercise No.22 C T W



10.5 Inspection after commissioning

ord all the currents, oltages, power factors and harmonic levels at the incoming supply. Once the

een c tch in all the steps one by one and cord the same parameters for evaluation.

Before commissioning the capacitor bank, please recvcapacitor bank has b ommissioned, swire

a. Volt StepPhase Before

commissioning

1 2 3 4

R-YY-BB-R

b.Current Step

Phase Before

commissioning

1 2 3 4

RedYellowBlue

c. Power Factor

Step

Phase Before

commissioning

1 2 3 4

RedYellowBlue



d. THDV Step

commissioning

Y-BB-R

Phase Before 1 2 3 4

R-Y

10.5.1 Checking operating data:

e capacitors are not subject excessive stress:

he capacitor bank must not be subjected to a continuous voltage exceeding 4

86 and IEC 871.

rrent:

nuous current exceeding 30% of the rated current.

. Ambient temperature:

n ambient temperature outside the tated temperature class.

. Unbalance current: (for banks with unbalance protection)

he unbalance current should not exceed 10% of the tripping value of the rotection system. Large capacitor banks are balanced at the factory to ensure at they do not exceed this value. Small capacitor banks are sometimes possible to balance sufficiently to comply with this recommendation if the

apacitances of the individual capacitor units differ too widely.

the individual capacitors are at different temperatures, for example because of olar radiation, the unbalance current immediately after switching in the bank ay differ from the value after a few hours in service.

he unbalance current should therefore be measured when the capacitors have ached their normal operating temperature.

Make the following measurements to ensure that thto 1. Voltage before and after switching in the capacitor bank: T110% of the nominal voltage, continuous meaning 12 hours in every 2according to AS 2897-19 2. Capacitor cu The capacitor bank must not be subjected to a conti1 3

he capacitor bank must not be operated in aTs 4 Tpthimc Ifsm Tre



10.5.2 Special precautions:

ded to allow operation uring temporary peaks. These margins should not be used under normal

refore be measured under all the load situations at occur, in order to ensure that the capacitor bank is not overloaded. Pay

pecial attention to periods of low load. A capacitor bank always causes a small crease in voltage. At low loads this can result in problems, since the voltage ay already be high even without the capacitor bank.

hen the load is low there is also a risk of resonance between capacitors and ansformers, and this may give rise to considerable over voltages.

The voltage and current margins stated above are intendservice conditions. Voltages and currents should thethsinm Wtr

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Capacitor Chargeman