## How to improve the power factor?

It’s quite simple. **By installing capacitors or capacitor banks.** Improving the power factor of an electrical installation consists of giving it the means to “produce” a certain proportion of the reactive energy it consumes itself.

**The capacitor is most frequently used, given:**

- Its non-consumption of active energy
- Its purchase cost
- Its ease of use
- Its service life (approximately 10 years)
- Its low maintenance (static device)

### Power diagram

Power factor is the ratio of working power to apparent power. It measures **how effectively electrical power is being used**.

To determine power factor (PF), divide working power (kW) by apparent power (kVA). In a linear or sinusoidal system, the result is also referred to as the cosine Î¸.

**PF = kW / kVA = cosine Î¸ kVA**

For example, if you had a boring mill that was operating at **100 kW** and the apparent power consumed was **125 kVA**, you would divide 100 by 125 and come up with a **power factor of 0.80**.

**(kW) 100 / (kVA) 125 = (PF ) 0.80**

**Where:**

**P**Â – Active power**S1 and S2**Â – apparent powers

(before and after compensation)**Qc**Â – capacitor reactive power**Q1**Â – reactive power without capacitor Q2: reactive power with capacitor

**Equations:**

- Q2 = Q1 – Qc
- Qc = Q1 – Q2
- Qc = PÃ—tg Ï†1 – PÃ—tgÏ†2
**Qc = PÃ—(tg Ï†1 – tg Ï†2)**

Where **Ï†1** is phase shift without capacitor andÂ **Ï†2** is phase shift with capacitor

The capacitor is a receiver composed of two conductive parts (electrodes) separated by an insulator.Â When this receiver is subjected to a sinusoidal voltage, the current and therefore its power (capacitive reactive) is leading the voltage by 90Â°.

Conversely, for all other receivers (motors, transformers, etc.) the current and therefore **its power (reactive inductive) is lagging the voltage by 90Â°**.

**gives a resulting current or power below the value which existed before the capacitors were installed**.

In simple terms, it is said that inductive receivers (motors, transformers, etc.) consume reactive energy whereas capacitors (capacitive receivers) produce reactive energy.

### How to calculate the power of capacitors

Based on electricity bills to calculate the capacitor banks to be installed, use the following method:

- Select the month in which the bill is highest (kVArh to be billed)
- Assess the number of hours the installation operates each month
- Calculate the capacitor power Qc to be installed

**Qc**= kVArh to be billed (monthly) / No. of hours’ operation (monthly)

**Example for the subscriber //**

- Highest reactive energy bill: December Number of kVArh to be billed:
**70,000** - Monthly operating times:
**High-load + peak times = 350 hours**

**Qc (bank to be installed)** = 70,000 / 350 = **200 kVAr**

#### Based on measurements taken on the HV/LV transformer secondary: PkW-cosFI

##### Example //

An establishment supplied from an **800 KVA HV/LV subscriber station** wanting to change the power factor of its installation to:

- CosÏ† = 0.928 (tgÏ† = 0.4) at the primary
- I.e. CosÏ† = 0.955 (tgÏ† = 0.31) at the secondary, with the following readings:
**Voltage:**400 V 3-phase 50 HZ**PkW**= 475**Cos (secondary)**= 0.75 (i.e. tg Ã¸ = 0.88)

Qc (bank to be installed) =Â PkW x (tgÏ† measured – tgÏ† to be obtained)

Qc = 475 x (0.88 – 0.31) = 270 kVAr

#### Calculation for future installations:

##### Example

1000 kva transformer, Q capacitor = **250 kVAr**

Note: This type of ratio corresponds to the following operating conditions:

- 1000 kVA transformer
- Actual transformer load =
**75%** - CosÏ† of the load =
**0.80**} k =**0.421** - CosÏ† to be obtained =
**0.95**} – see table below

**Qc** = 1000 x 75% x 0.80 x 0.421 = **250 kVAr**

### Capacitor power calculation table

#### Conversion table

Based on the power of a receiver in kW, this table can be used to calculate the power of the capacitors to change from an initial power factor to a required power factor. It also gives the equivalence between cos Ã¸ and tg Ã¸.

**Example:** 200 kW motor – cosÏ† = 0.75 – required cosÏ† = 0.93 – Qc = 200 x 0.487 = **98 kVAr**

**Reference //** Reactive energy compensation and power quality monitoring by Legrand

A very useful article

Thanks

Eng sirelkhatim

very good

nice