Power Factor

Correction and VAR

compensationWeek 8-9

Percentage of electricity that is being useful for doing work and is defined as the ratio

between the ‘active or actual power’ in kW or W, to the ‘apparent power’ expressed in volt-

ampere or kilo volt-ampere

What is power factor:

𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 =𝐴𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟

𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟

The apparent power also referred to as the total power supplied by the utility has two

components:

a- Productive component: powers the equipment and performs useful work and is

given in kW or W

b- Reactive power: generates magnetic field to produce flux necessary for the

operation of induction devices (AC motors, transformers, induction furnaces..etc.) and is

measured in kVar of Var. It produces no productive power

Industrial plants introduces inefficiencies into the electrical supply network by drawing

additional currents, called “inductive reactive currents”. Although these currents produce no

useful power, they increase the load on the supplier’s switch gear and distribution network

and on the consumer’s switchgear and cabling.

Typical un-improved power factor by industry

Industry Power Factor Industry Power Factor

Arc welding 35-60 Paint manufacturing 65-70

clothing 35-60 Metal working 65-70

Chemical 65-75 Plastic 75-80

Cement 80-85 Office building 80-90

Calculation of the necessary reactive power:

Once the power factor 𝑐𝑜𝑠𝜑1of the installation and the power factor to be obtained 𝑐𝑜𝑠𝜑2

are known, it is possible to calculate the reactive power of the capacitor bank necessary to

improve the power factor as:

𝑄𝑐 = 𝑃𝐿 × tan𝜑1 − tan𝜑2

Load PF= 𝑐𝑜𝑠𝜑1

Improved PF= 𝑐𝑜𝑠𝜑2

𝑃𝐿= OB= kW-load capacity

OA= ȁ𝑆 load capacity

OD= ȁ𝑆 reduced capacity

This is done using:

1. Calculations and Power factor triangle method

2. Tables

3. Curves

Using tables

𝑘𝑊 = 𝑘𝑉𝐴1 × 𝑐𝑜𝑠𝜑1, 𝑘𝑉𝐴𝑅1

= 𝑘𝑉𝐴1 × 𝑠𝑖𝑛𝜑1

∴ 𝑘𝑉𝐴𝑅1 = 𝑘𝑊 × 𝑡𝑎𝑛𝜑1

For correction to 𝑐𝑜𝑠𝜑2:

∴ 𝑘𝑉𝐴𝑅2 = 𝑘𝑊 × 𝑡𝑎𝑛𝜑2

Therefore, the required correction capacitors

∴ 𝑄𝑐 = 𝑘𝑉𝐴𝑅1 − 𝑘𝑉𝐴𝑅2

∴ 𝑄𝑐 = 𝑘𝑊 × 𝑡𝑎𝑛𝜑1 − 𝑡𝑎𝑛𝜑2

Finally,

𝑄𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 𝑄𝑐 ×𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟

Multiplying factor

found in table

Using curves:

In this case, choose the initial pf

value and select the current power

factor curve, the value of the

required capacitor is chosen from

the intersection of both values in

terms of the original kW load

Is it that simple?

Of course not!!!

Power factor correction in Distribution transformer:

Transformers are often in constant service. It is advisable that power factor

correction is carried out by keeping into account the transformer reactive power

so that an average power factor equal to 0.9 on the MV side is guaranteed.

Generally, the compensation power 𝑄𝑐 [kVar] in a transformer having a rated 𝑆𝑟

[kVA], shall not exceed the reactive power absorbed under minimum reference

load condition.

Apparent power kVA Transformer

7.2-23 kV 24 kV 36 kV

No load load No load load No load load

630 17 40.7 18.8 43.6 21.2 46

500 13.5 32.3 15.8 36.8 18 39

400 1.8 25.7 13.2 30 15.2 32

100 3.6 6.92 4.16 7.96 5.08 8.88

European standard

More than

double!

Power factor correction in Distribution transformer:

Deriving from the nameplate characteristics of the transformer the percentage no-load

current 𝐼𝑜%, the percentage short-circuit voltage 𝑢𝑘%, the iron losses 𝑃𝑓𝑒 and the copper

losses 𝑃𝑐𝑢 [kW], the required compensation power results to be about:

𝑄𝑐=𝐼𝑜%

100× 𝑆𝑟

2− 𝑃𝑓𝑒

2 + 𝐾𝑙2 ×

𝑢𝑘%

100× 𝑆𝑟

2− 𝑃𝑐𝑢

2

≈𝐼𝑜%

100× 𝑆𝑟 + 𝐾𝑙

2 ×𝑢𝑘%

100× 𝑆𝑟

Where 𝑘𝐿 is the load factor, defined as the ratio between the minimum reference load and

the rated power of the transformer

3 phase transformer with transformation ratio of 6.6kV/ 433V or 11 kV/433 V

Apparent power kVA

Transformer

Core losses

(W)

Copper

losses (W)

% magnetizing

current

1000 1770 11820 1.2

500 1030 6860 1.53

Power factor correction in distribution transformer:

𝑇𝑎𝑏𝑙𝑒 1: 𝑄𝑐 [kVar] to be connected to the secondary winding of an ABB transformer

Power factor correction in distribution transformer:

Egyptian code

USA recommendation practice: in case

of use of fixed capacitor banks placed

on the secondary transformer windings,

capacitors should be sized to 40-67%

from transformer apparent power rating

Tables are available for both

power transformers and

distribution transformers

Power factor correction in Three phase induction motor:

PF of 3-phase IM versus rated power output

in kW (or HP). Power factor improves with

higher motor power rating

Power factor versus speed performance. Better

power factor is achieved at higher speed

Power factor correction in Three phase induction motor:

As the rotor is loaded an increasing resistive

component is reflected from rotor to stator,

increasing the power factor. Power factor improves with higher

loading for the same power rating.

Power factor correction in Three phase induction motor:

Tables for individual power factor

correction for three phase induction

motors

Power factor correction capacitors can be

calculated using the following:

=𝐾𝑤 × 𝑎𝑐𝑡𝑢𝑎𝑙 % 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 × 𝑐𝑜𝑟𝑟. 𝑓𝑎𝑐𝑡𝑜𝑟

𝑒𝑓𝑓𝑒𝑐𝑖𝑒𝑛𝑐𝑦 𝑎𝑡 𝑎𝑐𝑡𝑢𝑎𝑙 𝑙𝑜𝑎𝑑

Correction factor: from tables

% loading: actual load / full load (usually 75%

from full load)

Efficiency at actual load: from chart

How much can I save by installing power capacitors? Power capacitors used for power factor correction provide many benefits:

Increase system capacity

Reduce utility power bills

Improve system operating characteristics (voltage gain)

Improve system operating characteristics (reduce line losses)

Increase system capacity:

PF improvement releases system capacity and

permits additional loads (motors, lighting

..etc.) to be added without overloading the

system. Capacitors reduce the current drawn

from the power supply, less current means

less load on transformers and feeder circuits,

leading to more investment in other devices

such as transformers.Switched capacitor panel

How much can I save by installing power capacitors?

Reduce utility power bills:

Electricity bill contains kW and kVAr of power plant. While reactive power

doesn’t register on kW demand or kW hour meters, the utility’s transmission

and distribution system must be large enough to provide the total power.

Utilities have various ways of passing along the expense of larger generators,

transformers, cables, switches..etc. Capacitors will save money no matter how

the utility bills on power. Utility charges according to the kW demand and add

a surcharge or adjustment for power factor. The adjustment may be a

multiplier applied to kW demand.

𝑛𝑒𝑤 𝑢𝑡𝑖𝑙𝑖𝑡𝑦 𝑏𝑖𝑙𝑙 =𝑘𝑊 − 𝑑𝑒𝑚𝑎𝑛𝑑 ∗ 0.9

𝑎𝑐𝑡𝑢𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟

If the power factor was 0.84, the utility

would require𝟎.𝟗

𝟎.𝟖𝟒= 𝟏. 𝟎𝟕 kW –normal

demand or 7% increase in utility billing

How much can I save by installing power capacitors?

Improve system operating characteristics (voltage gain):

Good PF provides “stiffer” voltage, typically a 1-2% voltage rise can be

expected when PF is brought to ±0.95. Excessive voltage drop make your motors

sluggish, and cause them to overheat.

Low voltage also interferes with lighting, the proper application of motor controls and

electrical and electronic instruments.

An estimation of the voltage rise from the improved power factor with the installation of

power capacitors can be made using the following:

% 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑖𝑠𝑒 =𝑘𝑉𝐴𝑟 𝑜𝑓 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟𝑠 ×% 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒

𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 − 𝑘𝑉𝐴

How much can I save by installing power capacitors?

Improve system operating characteristics (reduce line losses):

Improving PF at the load point shall relieve the system of transmitting reactive

current. Less current shall mean lower losses in the distribution system of the

facility since losses are proportional to the square of the current. Therefore, the

fewer kW-hr needed to be purchased from the utility

An estimation of the power losses can be made using the following:

% 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑝𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑠𝑒𝑠 = 1 −𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑃𝐹

𝑖𝑚𝑝𝑟𝑜𝑣𝑒𝑑 𝑃𝐹

2

× 100

Types of power factor compensation:

1. Individual correction

2. Group compensation

3. Centralized compensation

4. Combined compensation

5. Automatic compensation

Individual correction

Applied directly at the terminals of the load which demand reactive power.

Simple, capacitor and load can use the same protective devices against over current and

are connected and disconnected simultaneously.

Advisable in the case of large equipment with constant load and continuous operation

Used with motors and lamps

Individual correction

In case of direct connection (diagram 1 and 2), after the motor disconnection from the

supply, the motor will continue to rotate (residual KE) and self excite with the reactive

energy drawn from the capacitor and may turn into an asynchronous generator and

might damage with over voltage

With this type of correction the network on the supply side of the load works with a high

power factor, on the other hand, this solution is costly

Delta connected capacitors for 3rd harmonic cancellation

Group correction

Improving locally the power factor of groups of loads having similar functioning

characteristics by installing a dedicated capacitor bank.

Power factor is improved only by upstream the point where the capacitor bank is

located.

The capacitor requires its own switching device.

Centralized correction

System with permanently changing loads, several compensation units are usually

installed centrally in the main distribution switchboard.

Switched ON and OFF by a controller depending on the respective reactive power

demand.

Rated for the highest load, thus the total compensation power installed is lower than for

the other types of compensation. However, the load on the distribution system is not

reduced, and there must be enough room for all capacitor units in the main distribution

panel.

Combined PF Compensation

Derives from a compromise between the two solutions of individual and

centralized power factor correction and it exploits the advantages they offer.

In this case, distributed compensation is used for high power electrical

equipment and the centralized modality for the remaining part.

Combined solution is used in installations where large equipment only are

frequently used, in such circumstances their power factor is corrected

individually, whereas the PF of small equipment is corrected by the

centralized modality.

Automatic PF Compensation

Automatic correction is used in installations with variable absorption of reactive power.

This system monitors the installation and automatically switches different capacitor

banks.

This system required sensors, intelligent system comparing measured and desired

operation to connect/disconnect capacitors, and electric power boards with switches and

protective devices.

Power factor correction: Common Problems

1- Harmonic and Resonance:

• Capacitors connected to induction motors increase the chance of resonance

between the power factor capacitors and the motor’s inductive reactance.

• This issue makes the selection and filters design even more difficult.

• For these reason, NEMA standards (NEMA MG 1-1993 section 14.43.4)

have recommended not to use individual correction technique in presence of

large number of induction motors.

𝑓𝑟𝑒𝑠 =1

2𝜋 𝐿𝐶

Increasing C means that the square root increases so that the resonant frequency

reduces and could approach the 50 Hz. Non linear loads introduce currents of

harmonic ferq. 150-300 Hz so it even becomes nearer to the boundaries of

resonance

Power factor correction: Common Problems

2- Self Excitation:

• Capacitors are connected in parallel to motors

where both are fed/disconnected from the same

source together.

• Motors store energy in their rotating mass while

capacitors stores energy in their electric field.

• When motors are disconnected (with capacitors)

from the power source, the motor will continue to

rotate due to the energy stored in its inertia. At

this point energy is being exchanged between the

capacitor and the motor.

• Current will pass in the circuit between the motor

and capacitors and this current depends on the

voltage difference between both.

Capacitor and motor magnetization

curve for self excitation

Power factor correction: Common Problems

assume motor terminal voltage is 460 v and that the

capacitors used are 6 kVAr represented by the

straight line shown. The intersection point between

the capacitor load line and the magnetizing curve

shows that when the motor is disconnected both the

motor and capacitor will have the same voltage.

Assume another capacitor is used of 14.4 kVAr. At

460 volts, motor draws reactive current of 8 A while

capacitor draws 18 A.

When both are disconnected from source, the motor

terminal voltage will rise suddenly to 680 V. Leading

to that the capacitor tries to discharge this

“unbalanced” energy between capacitor and motor

leading to rise in motor terminal voltage.