Electrical distribution systems come in all sizes and carry-on different processes for our convenience. But there is one thing common in all inductive load appliances and machines. You have to maintain a constant magnetic field for the productive performance of the machine.

Yet, there is always a phase difference between voltage and current that wastes the electrical energy, and devices become unable to perform at their full potential.

A capacitor bank corrects the phase difference and enables a consistent magnetic field. So, electrical distribution systems can perform at their peak.

In such cases, you must install the capacitor with the required capacity. Otherwise, the phase difference will not be corrected entirely, and there will be a waste of power.

Here, you’ll learn everything about capacitor bank calculations. So, you can install the right capacitor bank in your electrical distribution systems.

## What is KVAR?

There are two types of power in any electrical distribution system: real power (KVA) and wasted power (KVAR).

Real power is the power that your electrical distribution system uses to deliver what you want from your device without any waste of energy. Electrical heaters, fluorescent bulbs, and similar devices use real power.

Reactive power or wasted power (KVAR) is generated when you supply electricity to an electrical distribution system, and it is used to create magnetic circuits. Its best example is an electrical motor that uses electrical energy to create a constant magnetic field to provide us with a rotatory motion that we can use for our purposes.

In the above example, the energy used by the motor does not produce any productive work, but is wasted in maintaining a magnetic field.

If the reactive power or wasted power is greater than the real power, your electrical distribution system will not be an efficient mechanism. Thus, it will consume more energy and produce little work.

That’s why we have to reduce reactive power by using capacitor banks, so our electrical distribution systems can produce the best work value for our operations.

## Difference Between KVA and KVAR

A system’s entire amount of power in use is indicated by the unit kVA, which measures perceived power. kW Equals kVA in an entirely efficient system. Electrical systems are never 100% efficient. Thus, not all of the perceived power is actually being put to good use.

Reactive power, or the power that hasn’t been transformed into kVA, is measured in kVAR. Reactive power, to put it simply, is the energy produced by a generator’s reactive parts but not consumed.

kW powers motion, light, heat, and sound when energy from a generator is used (real power). Reactive power, which is reliant on the power factor, is the energy that produces the magnetic fields that power rotating machinery.

## What is the Relation Between kVA, kW and kVAR?

A power factor between 0 and 1 represents electrical efficiency; the closer the power factor is to 1, the more effectively the kVA is transformed into usable kW. The power factor of generators is 0.8.

Active power (kW) and reactive power (kVA) combine to make apparent power (kVAR).

[kW + kVAR = kVA] Consequently, the Power factor formula is as follows:

100 x Active Power (kW) / Appearance Power (kVA)

## How to Calculate KVAR?

Reactive power is known as KVAR (Kilo Volt Ampere Reactive or Reactance). It is the force needed to magnetize flux that magnetic devices like transformers, motors, and relays create.

Understanding kW and kVA is necessary in order to comprehend kVAR.

The power that really drives the machinery and turns it into work is measured in kW. Working power, actual power, active power, or real power are additional names for it.

The amount of power required by magnetic devices like transformers, motors, and relays in order to produce the magnetizing flux is known as kVAR. It also goes by the name of reactive power. Use the formula listed below to calculate kVAR.

KVAR and KW are vectorially summed to form kVA. Alternating current or seeming power are other names for it (AC). It is calculated by dividing the root-mean-square voltage by the root-mean-square current (rms).

(Reactive Power)2 = (Apparent Power)2 – (True Power)2

kVAR(Reactive Power) = √ (Apparent Power)2 – (True Power)2

## Capacitor Bank Calculations or KVAR Calculations

### Capacitor Value Calculation in KVAR

Example 1

The power factor (P.F.) for a 3 Phase, 5 kW induction motor is 0.75 lagging. What size capacitor, measured in kVAR, is necessary to raise the power factor to 0.90?

Solution 1

Motor Input = 5kW

The multiplier in the table is 0.398, increasing PF from 0.75 to 0.90.

Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90

= 5kW x 0.398

= 1.99 kVAR

Now we have to divide this value by 3 to find the rating of each capacitor in a three-phase connection.

= 1.99kVAR / 3

= 0.663 kVAR

Solution 2

Motor input = P = 5 kW

Original P.F = Cosθ1 = 0.75

Final P.F = Cosθ2 = 0.90

θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819

θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843

Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = P (Tan θ1 – Tan θ2)

= 5kW (0.8819 – 0.4843)

= 1.99 kVAR

Now calculate the rating of each capacitor in a three-phase connection by dividing by 3.

1.99 kVAR / 3 = 0.663 kVAR

Example 2

At a power factor of 0.65, an alternator is supplying a load of 650 kW. What capacitance value, measured in kVAR, is necessary to increase the P.F (Power Factor) to unity (1)? Another question is how much more kW the alternator can produce for the same kVA demand as P.F. improves.

Solution 1 (Table Method)

Supplying kW = 650 kW

From Table 1, Multiplier to improve PF from 0.65 to unity (1) is 1.169

Required Capacitor kVAR to improve P.F from 0.65 to unity (1).

Required Capacitor kVAR = kW x Table 1 Multiplier of 0.65 and 1.0

= 650kW x 1.169

= 759.85 kVAR

We know that P.F = Cosθ = kW/kVA . . .or

kVA = kW / Cosθ

= 650/0.65 = 1000 kVA

When Power Factor is raised to unity (1)

No of kW = kVA x Cosθ

= 1000 x 1 = 1000kW

Hence increased Power supplied by Alternator.

1000kW – 650kW = 350kW

Solution 2 (Classical Calculation)

Supplying kW = 650 kW

Original P.F = Cosθ1 = 0.65

Final P.F = Cosθ2 = 1

θ1 = Cos-1 = (0.65) = 49°.45; Tan θ1 = Tan (41°.24) = 1.169

θ2 = Cos-1 = (1) = 0°; Tan θ2 = Tan (0°) = 0

Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = P (Tan θ1 – Tan θ2)

= 650kW (1.169– 0)

= 759.85 kVAR

### Calculation of Capacitor Value in Microfarad and KVAR

Example 3

A single-phase motor running at 500 volts, 60 cycles per second, can use 50 amps at trailing P.F 0.86. The capacitor bank must be connected across the motor in order to raise the power factor to 0.94. Calculate the necessary capacitor’s kVAR and -Farad capacity.

Solution 1 (Table Method)

Motor Input = P = V x I x Cosθ

= 500V x 50A x 0.86

= 21.5kW

From Table, Multiplier to improve PF from 0.86 to 0.94 is 0.230

Required Capacitor kVAR to improve P.F from 0.86 to 0.94

Required Capacitor kVAR = kW x Table Multiplier of 0.86 and 0.94

= 21.5kW x 0.230

= 4.9 kVAR

Solution 2 (Classical Calculation)

Motor Input = P = V x I x Cosθ

= 500V x 50A x 0.86

= 21.5kW

Actual or existing P.F = Cosθ1 = 0.86

Required or target P.F = Cosθ2 = 0.94

θ1 = Cos-1 = (0.86) = 30.68°; Tan θ1 = Tan (30.68°) = 0.593

θ2 = Cos-1 = (0.95) = 19.94°; Tan θ2 = Tan (19.94°) = 0.363

Required Capacitor kVAR to improve P.F from 0.86 to 0.95

Required Capacitor kVAR = P in kW (Tan θ1 – Tan θ2)

= 21.5kW (0.593 – 0.363)

= 4.954 kVAR

(2) To determine the amount of capacitance needed in farads to raise P.F. from 0.86 to 0.97 (Two Methods)

Solution 1 (Table Method)

Since we already know the required capacitance in kVAR, we can easily convert it to farads using this straightforward calculation.

C = kVAR / (2π x f x V2) in Farad

C = kVAR x 109 / (2π x f x V2) in Microfarad

= (4.954 kVAR) / (2 x π x 60Hz x 5002V)

= 52.56 μF

Solution 2

kVAR = 4.954 … (i)

We know that;

IC = V / XC

Whereas XC = 1 / 2π x f x C

IC = V / (1 / 2π x f x C)

IC = V x 2π x f x C

= (500V) x 2π x (60Hz) x C

IC = 188495.5 x C

And,

kVAR = (V x IC) / 1000 … [kVAR = ( V x I) / 1000 ]

= 500V x 188495.5 x C

IC = 94247750 x C … (ii)

Equating Equation (i) & (ii), we get,

94247750 x C = 4.954 kVAR x C

C = 4.954 kVAR / 94247750

C = 78.2 μF

## Conclusion

Now, you know the complete calculation methods to find the value of the capacitor required for your electrical distribution system in KVAR and microfarads. So, use the methods above for your benefit and install the capacitor of the right value in your devices.

Do you have a question in your mind? If so, make sure to fill out the form below!