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# Power Factor System

Power Factor System is the ratio between the KW and the KVA drawn by an electrical load where the KW is the actual 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 causes losses both in the supply and distribution system. A load with a power factor of 1.0 results in the most efficient loading of the supply. A load with a power factor of, say, 0.8, results in much higher losses in the supply system and a higher bill for the consumer. A comparatively small improvement in power factor can bring about a significant reduction in losses since losses are proportional to the square of the current.
• When the power factor is less than one the ‘missing’ power is known as reactive power which unfortunately is necessary to provide a magnetizing field required by motors and other inductive loads to perform their desired functions. Reactive power can also be interpreted as wattles, magnetizing or wasted power and it represents an extra burden on the electricity supply system and on the consumer’s bill.
• A poor power factor is usually the result of a significant phase difference between the voltage and current at the load terminals, or it can be due to a high harmonic content or a distorted current waveform.
• A poor power factor is generally the result of an inductive load such as an induction motor, a power transformer, and ballast in a luminary, a welding set or an induction furnace. A distorted current waveform can be the result of a rectifier, an inverter, a variable speed drive, a switched mode power supply, discharge lighting or other electronic loads.
• A poor power factor due to inductive loads can be improved by the addition of power factor correction  equipment, but a poor power factor due to a distorted current waveform requires a change in equipment    Design or the addition of harmonic filters.
• Some 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.
• An inductive load requires a magnetic field to operate and in creating such a magnetic field causes the current to be out of phase with the voltage (the current lags the voltage). Power factor correction is the process of compensating for the lagging current by creating a leading current by connecting capacitors to the supply.
• P.F (Cos Ǿ)= K.W / KVA  Or
• P.F (Cos Ǿ)=  True Power / Apparent Power.
• KW is Working Power (also called Actual Power or Active Power or Real Power).
• It is the power that actually powers the equipment and performs useful work.
• KVAR is Reactive Power.
• It is the power that magnetic equipment (transformer, motor and relay)needs to produce the magnetizing flux.
• KVA is Apparent Power.
• It is the “vectorial summation” of KVAR and KW.

## Displacement Power Factor Correction

An induction motor draws current from the supply that is made up of resistive components and inductive components. The resistive components are:
2)  Loss current.
And the inductive components are:
3)  Leakage reactance.
4)  Magnetizing current.

• The current due to the leakage reactance is dependent 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 (100 x 100) /(75 x 75) = 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
• 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.

## Why Should I Improve My Power Factor?

• You want to improve your power factor for several different reasons.  Some of the benefits of improving your power factor include:

1) Lower utility fees by:

(a). Reducing peak KW billing demand:

• Inductive loads, which require reactive power, caused your low power factor.  This increase in required reactive power (KVAR) causes an increase in required apparent power (KVA), which is what the utility is supplying. So, a facility’s low power factor causes the utility to have to increase its generation and transmission capacity in order to handle this extra demand.
• By lowering your power factor, you use less KVAR.  This results in less KW, which equates to a dollar savings from the utility.

(b). Eliminating the power factor penalty:

• Utilities usually charge customers an additional fee when their power factor is less than 0.95.  (In fact, some utilities are not obligated to deliver electricity to their customer at any time the customer’s power factor falls below 0.85.)  Thus, you can avoid this additional fee by increasing your power factor.

2) Increased system capacity and reduced system losses in your electrical system

• By adding capacitors (KVAR generators) to the system, the power factor is improved and the KW capacity of the system is increased.
• For example, a 1,000 KVA transformer with an 80% power factor provides 800 KW (600 KVAR) of power to the main bus.
• By increasing the power factor to 90%, more KW can be supplied for the same amount of KVA.
• 1000 KVA =            (900 KW)2  +  ( ?  KVAR)2
• KVAR = 436
• The KW capacity of the system increases to 900 KW and the utility supplies only 436 KVAR.
• Uncorrected power factor causes power system losses in your distribution system.  By improving your power factor, these losses can be reduced.  With the current rise in the cost of energy, increased facility efficiency is very desirable.  And with lower system losses, you are also able to add additional load to your system.

3) Increased voltage level in your electrical system and cooler, more efficient motors

• As mentioned above, uncorrected power factor causes power system losses in your distribution system.  As power losses increase, you may experience voltage drops.  Excessive voltage drops can cause overheating and premature failure of motors and other inductive equipment. So, by raising your power factor, you will minimize these voltage drops along feeder cables and avoid related problems.  Your motors will run cooler and be more efficient, with a slight increase in capacity and starting.