Compensation of three-phase induction motors

Why compensate?

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

Motor plate 3 Phase Induction motor

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 capacítor power can be taken from a table.

According to the specification the motor should be compensated at cos φ₂ = 0.98. The power triangle with and without compensation you can take from the following sketch:

Power vector diagram

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 φ₁  -  tan φ₂)   |  φ₁  before compensation;   φ₂  after compensation

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

Before compensation: cos φ₁ = 0,85  =>  φ₁ ≈ 31,79°
After compensation: cos φ₂ = 0,98  => φ₂ ≈ 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 capacítors in star or delta:

Capacítors for compensation

The total reactive power of our motor is Q_{c total} = 5.889 kvar. Whether in star or delta, 1/3 of the reactive power now takes a single capacítor:
Q_c = 1/3 Q_{c total} = 1/3 * 5.889 kvar = 1.963 kvar
To show how the capacítive reactive resístance is related to the reactive power, we make a "bridge" to the ohmic resístance:

P = U²                    compared with       Q_C  =  U²                      … Eq. (1)
.      R                                                               X_C

and   X_C  =  1 / 2π f C           … Eq. (2)

… Eq.(2) in …Eq.(1) results in:  Q_C =     U²         =   U² 2π f C
.                                                             1 / 2π f C

=>       C =     Q_C      =    Q_C                 …Eq.(3)      | ω = 2π f
                   U² 2π f       U² ω

From equation ...(3) it can be seen that to determine the capacítor size it is important to know whether they are connected in star or delta connection. Why? If the capacítors are connected in star, the capacítor voltage (phase voltage) is reduced by the factor √3 to 230 V (400 V line voltage), i.e. the capacítance of the capacítors is increased three times.
We check this statement:
Capacítors connected in delta:

C =     Q_C      =     1963 var             = 3,905 * 10^-5 F  ≈  39 µF
.      U² 2π f      (400V)² 2π50s^-1

Capacítors connected in star:

C =     Q_C       =     1963 var              = 1,181 * 10^-4 F  ≈  118 µF
.      U² 2π f        (230V)² 2π50s^-1