**Home ⇒ Overview Courses ⇒ Electric drives ⇒ 3 phase motor ⇒ Three-phase asynchronous motor ⇒ Reactive Power Compensation**

## Compensation of three-phase asynchronous motors

**Why compensate?**

Because large inductive loads strain the power supply system, It is recommended that big induction motors should not be compensated.

Therefore, capacitors are added to improve the power factor PF or cos φ.

As a reference value motors above 5 kvar should be compensated. Calculate the reactive power consumption in rated operation for the motor shown below and decide whether this motor must be compensated or not.

Q_{l }= √3 U * I * sin φ *| auxiliary calculation: PF = cos φ = 0,85 => φ ≈ 31,7888 => sin φ ≈ 0,52678*

Q_{l }= √3 * 400V * 24A * 0,52678 = __8,763 kvar__ => The motor should be compensated.

In practice, you will not compensate all the reactive power that occurs at nominal load. The reason is: At low load (the extreme case would be no-load), lower reactive currents will flow and you would have overcompensated the motor, which is undesirable.

Either a target power factor is specified or the capacitor power can be taken from a table.

According to the specification the motor should be compensated at cos φ_{2} = 0.98. The power triangle with and without compensation you can take from the following sketch:

Note:

active power P in kW

apparent power S in kVA

reactive power in kvar

### Determine the required capacitive reactive power Q_{bc} to get the new active power factor:

**Q _{c ges }= P_{zu} (tan φ_{1} - tan φ_{2}) ** | φ

_{1}before compensation; φ

_{2}after compensation

In our case:

P = √3 U * I * cos φ = √3 * 400V * 24A * 0,85 = 14,133 kW

Before compensation: cos φ_{1} = 0,85 => φ_{1} ≈ 31,79°

After compensation: cos φ_{2} = 0,98 => φ_{2} ≈ 11,48°

=> Q_{c total }= 14,133 kW (tan 31,79° - tan11,48°) = __5,889 kvar__

### Determinate the capacity of each capacitor:

First of all, it is possible to connect the capacitors in star or delta:

The total reactive power of our motor is Qb_{c total} = 5.889 kvar. Whether in star or delta, 1/3 of the reactive power now takes a single capacitor:

Q_{bc} = 1/3 Q_{bc total} = 1/3 * 5.889 kvar = 1.963 kvar

To show how the capacitive reactive resistance is related to the reactive power, we make a "bridge" to the ohmic resistance:

P = __U ^{2}__ compared with Q

_{C}=

__U__… Eq. (1)

^{2}. R X

_{C}

and X_{C} = 1 / 2π f C … Eq. (2)

… Eq.(2) in …Eq.(1) results in: Q_{C} =__ U ^{2} __ = U

^{2}2π f C

. 1 / 2π f C

=> **C = Q_{C} = Q_{C}**

__…Eq.(3) | ω = 2π f__

**U**

^{2}2π f U^{2}ωFrom equation ...(3) it can be seen that to determine the capacitor size it is important to know whether they are connected in star or delta connection. Why? If the capacitors are connected in star, the capacitor voltage (phase voltage) is reduced by the factor √3 to 230 V (400 V line voltage), i.e. the capacitance of the capacitors is increased three times.

We check this statement:

Capacitors connected in delta:

C = __ Q _{C} __ =

__1963 var__= 3,905 * 10

^{-5}F ≈

__39 µF__

. U

^{2}2π f (400V)

^{2}2π50s

^{-1}

Capacitors connected in star:

C = __ Q _{C} __ =

__1963 var__= 1,181 * 10

^{-4}F ≈

__118 µF__

. U

^{2}2π f (230V)

^{2}2π50s

^{-1}