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A number of power quality issues including electrical harmonics, poor power factor, voltage instability and imbalance, impact on the efficiency of electrical equipment's.

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1)What is power factor ?

In Alternative current (AC) system have two quantity namely voltage and current, the power factor has cosine angle between the voltage and current.

Pf will be lagging & leading depends upon load connected.

if inductive load such as motor pf of the circuit lagging,where capacitive load such as capacitor

have leading pf.

**2)Lossless because low power factor(pf) ?**

~High energy uses and costs.

~ Failure of motors.

~Failure of electrical and electronics equipment.

~Malfunction of Fuses & Circuit Breakers.

~Unstable equipment operations.

~Overheating of transformers and switch-gears.

**3)How does Power Factor Correction work?**

**#see_video_how power factor improved**

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Power Factor is the ratio of Active Power(KW) to the Apparent Power(KVA) drawn by any load.

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Low power factor leads to increased transmission and distribution losses leads to increased electricity bills without increase in productivity, heating & hunting of DG with increased fuel consumption.

Mathematically

Pf =kw/KVA

Or

True power/apparent power.

Generally power factor in varies between 0 to 1.

it is best practice to maintain power factor unity(1).

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PFC systems increase the efficiency of power supply, delivering immediate cost savings on electricity.

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Real Power is the power that actually powers the equipment and performs useful, Productive work.

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Reactive Power is required by some equipment (e.g. transformers, motors and relays) to produce a magnetic field for operation; however it does not perform any real work.

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Apparent Power is the vector sum of Real and Reactive Power and corresponds to the total power required to produce the equivalent amount of real power for the load.

A capacitor generate leading power factor for compensation of inductive power factor,

In term of KVAR.

Where KVAR known reactive power,equal to

Voltage x current x sin Phi.

**4) how to calculate pf.?**

Q-1) a 10 KW induction motor operate in 220 volt 50 hz power supply, if motor 50 ampere current ,then calculate power factor of motor.

Q-2) if require maintain power factor 0.98,calculate VAR value of capacitor.

Answer-

Given in Q-1

Active power(KW)=10

apparent power =V x I x cosine

Cosine = KW/VI

10/(220x50/1000)

10/11

0.90

if require 0.98 pf.

KVAR =V x I x sin phi

= V x I x √[ 1-pf x pf]

11 x √[1-0.98 x0.98]

=11 x 0.4

=0.440 KVAR

Hence 440 VAR capacitor bank require to maintain 0.98 pf of induction motor, to efficient operation and save electricity bill.

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