1

POWER SAVINGS AND APPLICATION GUIDE For power factor correction and harmonic mitigation techniques

Normally, the term Power Factor (PF) is misunderstood. This often results in consumers paying penalties due to poor Power Factor. In laymen’s terms, electric power is made up of two components, active power (measured in kilo Watts / KW) that produces useful work, and reactive power (measured in Kilo Volt Ampere Reactive / KVAR) necessary to generate magnetic fields for operation of inductive electrical equipment. Although reactive power does not perform useful work, customers are billed for it when it is supplied by their utility provider. Total power is measured in Kilo Volt Ampere / KVA, which is the algebraic sum of KW and KVAR. In an electrical system with substantial inductive loads such as induction motors, customers can lower their electrical utility bills by supplying their own reactive energy from Capacitors. This has several advantages, including improvement of cable, transformer and motor losses, voltage regulation, and capacity relief of current carrying distribution equipment. This is depicted in figures 1 and 2.

Figure 1 Induction motor without power factor correction and utility supplying both reactive and active power

The ratio of useful power (measured in kilo Watts, KW) to total power (measured in kilo Volt Ampere, KVA) is known as Power Factor. Thus, the objective of installing a capacitor bank is to increase such a factor by supplying the reactive power when installed at or near inductive electrical equipment.

( )θcos==KVAKW

PF

2

Figure 2 Capacitor bank supplying the reactive power required by the loads Figure 2 is a low voltage capacitor bank. A common reason for poor power factor is operating motors at less than rated capacity. PF drops rapidly as illustrated in Figure 3 and outlined discussed within IEEE Std 141-1993.

Figure 3 PF behavior at different load profiles with and without capacitor banks

3

From figure 3 it can be seen that the power factor of a motor at full load is usually between 80 % and 90 % depending upon the motor type and speed. In general, induction motors do no operate at full load. Despite the fact that PF varies from no load to full load conditions, the reactive power consumed by the motor does not vary much. The table below summarizes typical examples of electrical loads which have poor power factor.

Load Typical Power Factor HVAC systems, pumps, cold headers, fans, blowers 80% ~ 90% Induction furnaces, weaving machines, spot welders 60% ~ 80% Automated machine tools, presses, industrial welders Less than 60%

SAVING MONEY Usually, utilities include a penalty charge in their electrical rates for low power factor. This can be alleviated when power capacitors are installed to supply the necessary reactive power (KVAR). By supplying KVAR with PSS Capacitor banks at the load, the capacitor relieves the utility of the burden of supplying the reactive current as much as 20 percent without a corresponding increase in the size of transformers, conductors, etc. This makes the utility transmission/distribution system more efficient, reducing cost for the utility as well as their customers. The utility will relieve the customers of the power factor penalty when they maintain a proper power factor level. BENEFITS OBTAINED FROM POWER FACTOR CORRECTION Improving the PF of a system reduces the current flowing through it which provides many advantages. Figure 4 shows the current and power triangles.

Figure 4 Current and power triangles

This illustrates that as the PF is corrected, the inductive or magnetizing current is reduced, in turn minimizing power losses from the point of connection upstream to utility source.

4

The main benefits from PF correction are:

a) Avoid utility penalties b) Reduce transformer, cabling, and motor losses c) Reduce total plant KVA for the same KW working power d) Lower utility electrical billing e) Improved voltage regulation f) Minimize electrical distribution system investment costs

Avoid utility penalties As PF correction reduces the magnetizing current through the power circuit, the utility company can deliver electrical energy to other customers more efficiently. Reduce transformers, cables, and motor losses Since power losses are given by the Joule effect (I2 R), the per cent reduction of power losses (% RL) can be expressed as follows:

2

arg100100%

∗−=

PFettPFinitial

RL

Where,

=RL% Reduction in Losses in per cent =PFinitial Initial Power Factor before the application of capacitors

=PFfinal Target Power Factor, that is, the Power Factor value it is desired to achieve For instance, if a poor PF of 57 % is improved to 98 %, the per cent reduction in power losses will be:

%6698.057.0

100100%2

=

∗−=RL

Reduce total plant KVA for the same KW working power Improving PF frees up actual KVA for additional loads that in turn allows system transformers operate more efficiently.

5

The freed up KVA can be calculated as follows:

−=

PFdesiredPFactualKWKVA

11

For example, if an 800KW plant with a poor power factor of 50 % has a total power of 1600 KVA, the freed KVA obtained when power factor is improved to 80 % will be:

KVAKVA 6008.0

15.0

1800 =

−=

KVAKVAKVA 10006001600 =−∴ will now be required.

Now, if the power factor was improved to unity the freed KVA would be:

KVAKVA 8000.1

15.0

1800 =

−=

Therefore, it is obvious that the better the PF, the more KVA capacity is released to serve the system. Increase KW working power for the same KVA demand Improving PF will release system capacity permitting the addition of more loads without overloading the source transformers. By using the next formula, it is possible to verify what was stated above.

PFKVAKW ∗= For example, a system with a poor PF of 50 % and a 500 KVA load profile operates at 250 KW capacity. Upon improving the Power Factor to 90 %, the available real power improves to 450 KW.

Improved voltage Regulation

As PF correction helps to reduce line voltage drops, motors, conveyors and other equipment will have more efficient performance.

6

The voltage rise when capacitors are installed at the transformer is given by:

T

T

KVARZKVAR

VR%

%∗

=

Where VR is the Voltage Rise in per cent and ZT is the transformer impedance % from nameplate. In general, the lower the rated voltage of a motor, the higher the full load current, the lower the full load speed and the lower the starting torque. Less investment In new installations, the reduction in size of distribution equipment (transformers, cables, switchgears, etc) will imply less waste of money. The table below shows this idea.

KW KVAR KVA PF Conclusion 200 0 200 100 IDEAL 200 67.24 211 95 GOOD 200 150 250 80 POOR 200 300 360 55 BAD

Useful Power Factor Correction formulas:

10003 IV

KVA∗∗

= Cf

X C ∗∗∗=

π2106

.%746.0

EffHP

KW∗

= ( )

10002 2KVCf

KVAR∗∗∗∗

=π

Legend V = Line to line voltage in Volts or kilo Volts (KV) I = Line current in Amperes (Amps) HP = Horse Power C = Capacitance in micro Farads (µF) XC = Capacitive Reactance in Ohms (Ω) f = Fundamental Frequency in Hertz (60 Hz)

7

Note: Sometimes equipment supplying reactive energy (KVAR) are operated either at a frequency lower than the fundamental or at a voltage lower than the rated voltage. In that case, the following formula could be useful:

∗

∗=

R

A

R

ARE f

fVV

KVARKVAR2

Where:

KVAREffectiveKVARE = KVARRatedKVARR =

VoltageAppliedVA = VoltageRatedVR =

FrequencyAppliedf A = FrequencyRatedf R =

For example, if a 100KVAR, 240V rated capacitor bank is used in a 220V system, it will supply:

KVARKVARE 03.84240220

1002

≈

∗=

If the same equipment (designed to operate at 60 Hz) is used at a lower frequency (50 Hz), it will supply:

KVARKVARE 03.706050

240220

1002

≈

∗

∗=

CALCULATING SIZE OF CAPACITOR BANK When determining the right size of a capacitor bank all the potential benefits have to be analyzed in each situation. For example, if there are penalty clauses imposed, it is recommended that the PF be raised to at least 95%. If more loads are intended to be installed by using existing transformers, the degree of correction will depend upon the amount of new loads required. Table A on following pages is usually used as standard practice to calculate the required KVAR to be installed. It gives multipliers for KW to get the capacitor KVAR needed to raise from initial to target corrected PF. All the values contained in the table are obtained from the following formula:

( )( ) ( )( )( )CPFARCPPFARCKWKVARrequired costancostan −∗=

Where: PPF = Present Power Factor CPF = Corrected Power Factor.

8

Example (a) The installed load of a power plant working at a PF of 55 % demands 150 KW on average. The PF is desired to be raised at 95 %. Determine the required KVAR to perform this action. Step1: Locate the original (actual) PF in Table A (rows). Step2: Locate the desired PF (columns) Step3: The point at which both values intercept is the correction factor (multiplier) that will be used for further calculations. In this case it corresponds to 1.518. Step4: The KVAR necessary to raise the PF to 95 % is obtained by multiplying the 150 KW by the factor found in Table A. Therefore, KVARKW 5.178189.1150 == . Use 180 KVAR, this may be higher due to capacitor value tolerances inherent to capacitor cells. Example (b) When a plant has a load of 650 KW and a total power requirement of 930 KVA, what is the size of capacitor required to improve the actual PF to 98 %? Step1: Calculate actual PF from the available data.

%707.0~6989.0930650

====KVAKW

PFactual

Step2: Obtain the factor from the table A below (0.817) Step3: Calculate the required KVAR in the same way

05.531650817.0 =∗=CKVAR . Say, 535 CKVAR Example (c) A group of inductive loads of 113 KW is connected to a 240V system. What is the size of capacitor is required to correct the actual PF to 100% when the line current is 367 Amps? Step 1: Calculate the actual PF.

55.152367240.033 =∗∗=∗∗= IkVKVA

%7474.0~7407.055.152

113====

KVAKW

PFactual

Step 2: Obtain the factor from the table A below (0.909) Step 3: Calculate the required KVAR in the same way. CC KVARKVAR 105~7.102113909.0 =∗=

9

TABLE A

Another common way of calculating the required capacitor KVAR rating is by means of the following tables according to IEEE Std 141-1986.

10

11

POWER BILL SAVINGS When electrical systems have poor PF, many of the utilities will impose a penalty due to increased generation and transmission costs needed to provide and transport real and reactive power. Here is a typical example:

Demand charge per month $/KW First 50 KW 6.42

Next 100 KW 5.10 Next 150KW 4.43 Excess KW 3.78

If the utility applies a PF penalty below 0.95, and the assumed method of charging the consumer is:

actual

demand

PFKW

demandBilling95.0∗

=

12

Actual load = 537 KW Actual Power Requirement = 780 KVA Actual PF = 0.6884 Actual Excess = 120 KW

Utility demand charges at actual PF = KWdemandBilling 06.7416884.0

95.0537=

∗=

Therefore, the charges will be: 50 KW x 6.42 $ / KW = $321 100 KW x 5.10 $ / KW = $510 150 KW x 4.43 $ / KW = $664.5 621.06 KW x 3.78 $ / KW = $2347.60 Total = $3843.1 / month After improving the actual PF to the minimum accepted value (0.95):

KWdemandBilling 53795.0

95.0537=

∗=

50 KW x 6.42 $USD / KW = $321 100 KW x 5.10 $USD / KW = $510 150 KW x 4.43 $USD / KW = $664.5 417 KW x 3.78 $USD / KW = $1576.26 Total = $3071.76 / month Total savings per month = $3843.1 - $3071.76 = $771.34 Total annual savings = $771.34 X 12 = $9256.08

PAYBACK ANALYSIS In order to correct the PF from 0.6884 to 0.95, from table 1 we see that 371.604 KVAR is required. Or approximately 375 KVAR. From PSS list price it is recommended to use: LAC 0375 480 05S N1 Prescribed equipment will perform automatic correction using 5 stages of 75 KVAR. List price: $14,262 Payback: $14,262 / $9256.08 = 1.54 years That is, one year and a half payback (based on list price of the capacitor bank)

13

CH A

0 50 100 150 200 250-150

-100

-50

0

50

100

150

CH A

0 50 100 150 200 250-150

-100

-50

0

50

100

150

UNDERSTANDING HARMONICS Power transmission and distribution systems are designed for operation with sinusoidal voltage and current waveforms at a constant frequency. However, when non-linear loads (such as thyristor-based drives, converters and arc furnaces) are connected to the system, excessive harmonic currents are generated, causing both current and voltage distortion.

Harmonics and Standard Capacitors don’t mix Harmonics are a natural by-product of non-linear loads such as drives (AC and DC), motor soft starters, welders, UPS systems, robotics and other electronic loads. Harmonics are higher than 60 Hz currents and voltages which are produced by these devices in response to the manner in which they draw current (see figure 5). The amount of harmonic current rises as more and more non-linear loads are added. Harmonics can cause serious problems to arise when standard fixed or automatic power factor correction capacitors are added to the network.

Ideal Sine Wave

Distorted Sine Wave

Figure 5 There are two main concerns to consider when applying power factor correction capacitors. The first, capacitors are a natural low impedance path for harmonic currents and will therefore absorb these energies. The increased capacitor current results in higher element temperature which negatively impacts the life of the capacitor. Also because capacitors reduce the network impedance, capacitor can actually increase the level of harmonic current on the network. It’s important to remember that while capacitors do not produce harmonic currents, they can magnify their effects. Furthermore, harmonic voltages present on the network, create voltage stresses on the capacitor dielectric which can cause premature dielectric breakdown and capacitor failure.

14

The second, a potentially more serious concern, is network resonance. When capacitors are added to the network, they set up a parallel resonance circuit between the capacitors and the system inductance’s (see left figure below). Harmonic current components that are close to the parallel resonance point are magnified (see right figure below). The magnified currents can cause serious problems such an excessive voltage distortion, nuisance fuse operation, overvoltage trip of drives and insulation breakdown within motors, transformers and conductors.

FIGURE 6 Harmonic filtering is the best way to eliminate this distortion from the power system, while at the same time producing reactive power. Harmonic filters consist of capacitors connected in series with a reactor. The capacitors produce reactive power at the filter’s fundamental frequency, and the circuit is designed to achieve the required power factor correction. The inductance of the reactor is chosen so that the filter forms a very low impedance series resonant circuit at the harmonic frequency. This ensures that a high proportion of power system harmonic currents enter the filter.

15

FIGURE 7 A typical harmonic filter consists of three series resonant circuits tuned to the most common harmonics (5

th, 7

th and 11

th harmonics). These filters are housed in steel cubicles. Each consists of a contactor,

reactor and capacitor. The unit is generally connected to the fuse feeders on the main distribution board. Harmonic filters can be operated in the same way as automatic capacitor banks: they are controlled by means of reactive power controller according to reactive power load curves. PSS Harmonic filters are custom designed for each application using standard components. This ensures that the best possible power factor correction and filtering characteristics are achieved for a reasonable investment.