**POWER FACTOR**

**Definition of Power Factor**

In a system of alternating voltages and currents, power factor is defined as the ** ‘cosine’** of the phase angle that exists between the voltage applied to a circuit and the current that flows through that circuit. Power factor is always lies between zero to plus or minus one. The relationship between the voltage and current of any specified load can also be expressed as the following ratio.

**Inductive Loads**

Inductive (or magnetic) devices are those which contain coils, where the current passing through the coil generates magnetic lines of ‘flux’ of a magnetic field. e.g. transformers, induction motors, lighting chokes or ballast’s, solenoids, induction furnaces and arc welders.As the current passes through the inductive loads, part of the current used for magnetization. In this case the current lags the voltage and the p.f. is lagging.

**Vector Relationship**

**compensation of Reactive Power**

**Equipment with Low Power Factor**

S.No | Equipment Device | Power Factor |

1 | Induction motors | 50% – 90% |

2 | Small Transformers | 30% – 95% |

3 | Fluorescent & high intensity
discharge lamp (with no p.f. comp.) | 40% – 80% |

4 | Induction heating equipment | 60% – 90% |

5 | Arc welders | 50% – 70% |

6 | Solenoids | 20% – 50% |

**Effects of Low Power Factor **

System Capacity Improvement

Low p.f. reduces the capacity of the electrical system since the system must carry the total current but only

the active current provides useful power.

For example, a transformer rated at 500 kVA can only provide 400 kW of power if the p.f. is 0.8. If the p.f. were improved to 0.9, 450 kW of power can be provided.

**Power Factor Improvement**

The power factor for lightning an heating loads supplied from 3-phase power supply ranges from 0.95 to unity but the power factor for the motor loads ranges from 0.5 to 0.9.Power factor for single phase motors may be low as 0.4(lagging) and for electric welding may be even 0.2 or 0.3(lagging).If the power factor of the supply or power station is raised to unity ,the current for the same amount of power to be supplied is reduced to minimum.

*Advantage of good power factor*

- line losses are reduced to minimum
- efficiency of plant is improved.
- voltage drop and voltage regulation is considerably reduced.
- load output of given plant is better utilized.

*Disadvantages of low power factor*

At the higher power factor ,the current for a given load supplied at a constant voltage will be lower and at low power factor current will be higher.

Let us consider, load as P is supplied at terminal voltage V and at power factor cosΦ by a 3-phase balanced system then load current is given by

If we keep p and V constant,the load current IL is inversely proportional to the power factor,cosΦi. lower the power factor,higher the current and vice-versa.When the current is higher then the low power factor affects the system and has some major disadvantages:

- Rating of generators and transformers are proportional to their output current.Therefore large generators and transformers are required to deliver same load but at a low power factor.
- At a low power factor,the cross section area of the bus-bar and the contact surface of switchgear should be enlarged for the same power to be delivered.
- At a low power factor,for the same amount of power to be transmitted ,the transmission line or the distributor or cable will have to carry more current.Therefore it requires more conductor material for transmission lines,distributors and cables.
- copper loss i.e (I^2R) are proportional to the square of the current and inversely proportional to the square of the power factor.hence results in poor efficiency as more copper losses incur at low power factor.
- causes the large voltage drop in generators,transformers,transmission lines and distributors results in poor regulation.Hence we need extra regulating devices to keep the voltage drop within permissible limits.
- Results in the high costs for alternators,switchgears,transformers,transmission lines ,distributors and cables.

**Causes of Low power factor**

Ac motor(except overexcited synchronous motors and certain type of commutators motors) and transformers operates at lagging power factor.The power factor decreases with the decrease in load.Arc lamps also operates at low power factor i. lagging power factor.Industrial heating furnance such as arc and induction furnances operate on very low lagging power factor.

**Methods of Power factor Improvement**

The following methods can be implied for the improvement of Power Factor.

1. By Use of Static Capacitor

a)Shunt Capacitor

b)Series Capacitor

2. By use of Synchronous Machine

a)Synchronous Motor

b)Synchronous Condenser

3. By use of induction motors with Phase advancer

4. Other measures

**I^2R Loss Reduction due to p.f. Correction**

**By Use of Synchronous Machine**

• The power factor of the synchronous machine can be made leading by simply changing the excitation.

• Hence these device can be employed for power factor correction.

• Various synchronous machine used are

– Synchronous motor

– Synchronous condenser

pic source:your electrical home

**Synchronous Motor**

• For constant speed application synchronous Motor having leading p.f. of about 0.8 can be used.

**Synchronous Condensers**

These are overexcited Synchronous Motor running at no load.power factor can be improved by using synchronous condensers like shunt capacitors connected across the supply.

*pic source:Electronics Hub*

In phasor diagram IL represents the current taken by the industrial load,lagging the voltage V by a large angle ΦL and phasor IM represents the current drawn by the synchronous condenser leading the voltage v by an angle ΦM.The resultant current I is the phasor sum of IL and IM. The angle Φ is much smaller than ΦL.Hence the power factor is improved from cosΦL to cosΦ by the use of synchronous motor.

*Advantages*

• Finer control can be obtained by variation of field excitation

• Helps in voltage regulation

• Improves system stability and reduction of the effect of sudden changes in load owing to its inertia.

*Drawbacks*

• Costlier

• High maintenance and operating cost

• Low efficiency then static capacitor

• Noisy

• Starting is cumbersome

**By use of induction motors with phase Advancers**

- Power factor of an induction motor falls due to exciting current drawn from the ac supply mains as exciting currents lags behind the voltage by π/2.
- Phase advancer is a simple AC exciter which is connected on the main shaft of the induction motor and operates for power factor improvement
- The poor p.f of an induction motor is due to the fact that it takes lagging current 90 degrees out of phase with the voltage.
- This exciting out of phase current if supplied from a different source helps to improves it pf.

**By use of static capacitor**

- power factor is improved by connecting the capacitors in parallel with the equipment operating at lagging power factor such as induction motors,fiuorescent tubes.
- current drawn by induction motors,fiuorescent tubes can be resolved into two components active components which is in phase with the supply voltage and quadrature component of constant magnitude.
- capacitors draw current leading the supply voltage by π/2 and neutralize the quadrature components of current drawn by the equipment across which these are connected.
- capacitor can be either connected in star or delta.

*pic sources:electrical4pocket.com*

The value of static capacitors for improvement of power factor cab be determined as follows:

Let the load current drawn by an inductive load be of I amperes lagging behind the applied voltage V by an angle Φ1 and pf be improved to cosΦ2(lagging) by connecting capacitor across the inductive load.

Active components of current drawn by load=I cosΦ1 represented by phasor OB

Reactive components of current drawn by load =I sinΦ1=I cosΦ1 *tanΦ1 represented by phasor AB

Let the current drawn by capacitor C be IC amperes leading the applied voltage by π/2 represented by phasor BD.

Lagging components of current drawn from supply mains represented by phasor BC=OB tanΦ2

=I cosΦ1tanΦ2

The current drawn by the capacitor shiuld neutralize the reactive power components.Thus

IC=current represented by phasor AB-current represented by phasor BC

=I cosΦ1 tanΦ1 -I cosΦ1 tanΦ2

=I cosΦ1 (tanΦ1-tanΦ2)

value of capacitance required is ,C=IC/2πfV where v is supply voltage and f is the supply frequency.

**By other Measures**

• Special types of machine such as synchronous induction motor, high pf motor like compensated induction motor,

Schrage motors etc..

## For more notes on Electrical Engineering:

- https://www.notesforengineering.com/pn-junction/
- http://abhinavbhattarai.com.np/2020/09/20/from-maxwells-equations-to-electrical-engineering/