## Dimensioning of the capacitor bank

For the dimensioning of the capacitor bank to be installed in order **to improve the power factor of a plant**, it is necessary to calculate correctly the power factor according to the consumption or to the load cycle of the plant.

This in order to avoid the intake of excess reactive energy, which is a condition normally forbidden by power supply authorities. To carry out distributed or group power factor correction, it is necessary **to calculate the cosφ of the single load or of the group of loads (factory areas)**.

**This can be carried out as follows:**

**Directly**– through direct measuring by means of a power factor meter**Indirectly**– through the reading of the active and reactive energy meters

**according to which the load is absorbing energy**. The reading of the instrument shall be carried out in different moments of the load cycle, so that an average power factor value can be obtained.

If the readings of the active and reactive energy absorbed by the load or by the whole of the loads constituting the factory areas during a work cycle are available, the average power factor can be calculated as follows:

where:

**E**and_{Pi}**E**are the values of active and reactive energy read at the beginning of the work cycle_{Qi}**E**and_{Pf}**E**are the values of active and reactive energy read at the end of the work cycle_{Qf}

## Calculation of the necessary reactive power

Once the power factor (cosφ1) of 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.

**Indicating by:**

**P**– the installed active power**φ**– the phase displacement angle before power factor correction_{1}**φ**– the phase displacement angle to be obtained with the power factor correction the power of the capacitor bank Q_{2}_{c}is:

Once the initial cosφ is known, Table 1 allows to calculate (in kvar per kW installed) **the power of the capacitor bank necessary to obtain a defined power factor**.

**can be delta connected or star connected**.

When selecting the connection modality, it is necessary to keep into account that with delta connection, each capacitance is subject to the supply line-to-line voltage, but, at the same level of generated reactive power, it has a value equal to 1/3 of the value it will have in case of star-connection:

In the low voltage field, where insulation problems are less important, the delta connection is usually preferred for the capacitor bank, since it allows a smaller sizing of the capacitances of each phase.

### Example calculation

In a plant with active power equal to **300 kW** at **400 V** and **cosφ= 0.75**, we want to increase the power factor up to **0.90**. In the table 1 above, at the intersection between the row “initial cosφ” 0.75 with the column “final cosφ” 0.9, **a value of 0.398 for the coefficient K is obtained**.

Therefore a capacitor bank is necessary with **power Q _{c}** equal to:

Q_{c} = K · P = 0.398 · 300 = **119.4 kvar**

The factor K can be determined also using the following nomograph:

As shown in the figure, tracing a line segment from the value of the initial cosφ to the value to be obtained, the intersection of the line with the middle graduated scale, gives the value of K which, **multiplied by the active power P of the load, defines the necessary reactive power Q _{c}**.

**Reference //** Power factor correction and harmonic filtering in electrical plants by ABB

Armando Medel Trujillo

Excelente información teórica y práctica ya que es un tema muy común en cualquier industria saludos y de favor si me pueden enviar la información en PDF gracias

Mina Hanaa

Very helpful informations, Thanks

Jamal

Greetings…Very good information …thanx alot

hari haran

Thank for your site. MCCB rating for 200 kVAR capacitor bank. Please guide me.

mohammadreza noorizadeh

Greeting,

thanks for this important article

Mallikarjuna Balimidi

Really it is a good site for electrical engineers. I don’t know the meaning of work cycle? Please explain about work cycle.

viswanatha reddy

i am explaining what i know, if any wrong please inform me

work cycle means

in a particular period (ie: 24 hrs or 1 week or month) working condition of the Machine/Equipment

Ex: in one day (ie: 24 hrs)

6am to 9 am – machine running on half load ( power consumption is less)

9 am to 6 pm- machine running on full load ( power consumption also more )

load may be vary from time to time during a period. that may be considered for power factor correction.

other wise if u maintain constant capacitor bank for power factor correction

it leads to Leading power factor.

ABDULRAHEEM ALLEHIANI

Greeting,

thanks for this important article,

i have question

the capacitor bank it can be used for power factor correction and voltage regulator (to maintain the drop voltage during start up the high capacity motor)?

Martin

Good day

Through this platform I have learnt really a lot

THANKS