Definition of reactive power
For most electrical loads like motors, the current I is lagging behind the voltage V by an angle φ.
If currents and voltages are perfectly sinusoidal signals, a vector diagram can be used for representation.
In this vector diagram, the current vector can be split into two components: one in phase with the voltage vector (component Ia), one in quadrature (lagging by 90 degrees) with the voltage vector (component Ir). See Fig. L1.
Ia is called the active component of the current.
Ir is called the reactive component of the current.
The previous diagram drawn up for currents also applies to powers, by multiplying each current by the common voltage V. See Fig. L2.
We thus define:
- Apparent power: S = V x I (kVA)
- Active power: P = V x Ia (kW)
- Reactive power: Q = V x Ir (kvar)
In this diagram, we can see that:
- Power Factor: P/S = cos φ
This formula is applicable for sinusoidal voltage and current. This is why the Power Factor is then designated as "Displacement Power Factor".
- Q/S = sinφ
- Q/P = tanφ
A simple formula is obtained, linking apparent, active and reactive power:
A power factor close to unity means that the apparent power S is minimal. This means that the electrical equipment rating is minimal for the transmission of a given active power P to the load. The reactive power is then small compared with the active power.
A low value of power factor indicates the opposite condition.
Useful formulae (for balanced and near-balanced loads on 4-wire systems):
- Active power P (in kW)
- Single phase (1 phase and neutral): P = V.I.cos φ
- Single phase (phase to phase): P = U.I.cos φ
- Three phase (3 wires or 3 wires + neutral): P = √3.U.I.cos φ
- Reactive power Q (in kvar)
- Single phase (1 phase and neutral): Q = V.I.sin φ
- Single phase (phase to phase): Q = U.I.sin φ
- Three phase (3 wires or 3 wires + neutral): Q = √3.U.I.sin φ
- Apparent power S (in kVA)
- Single phase (1 phase and neutral): S = V.I
- Single phase (phase to phase): S = U.I
- Three phase (3 wires or 3 wires + neutral): S = √3.U.I
V = Voltage between phase and neutral
U = Voltage between phases
I = Line current
φ = Phase angle between vectors V and I.
An example of power calculations (see Fig. L3)
|Type of circuit||Apparent power S (kVA)||Active power P (kW)||Reactive power Q (kvar)|
|Single-phase (phase and neutral)||S = VI||P = VI cos φ||Q = VI sin φ|
|Single-phase (phase to phase)||S = UI||P = UI cos φ||Q = UI sin φ|
|Example: 5 kW of load cos φ = 0.5||10 kVA||5 kW||8.7 kvar|
|Three phase 3-wires or 3-wires + neutral||S = UI||P = UI cos φ||Q = UI sin φ|
|Example||Motor Pn = 51 kW||65 kVA||56 kW||33 kvar|
|cos φ= 0.86|
|ρ= 0.91 (motor efficiency)|
The calculations for the three-phase example above are as follows:
Pn = delivered shaft power = 51 kW
P = active power consumed
S = apparent power
So that, on referring to Figure L16 or using a pocket calculator, the value of tan φ corresponding to a cos φ of 0.86 is found to be 0.59
Q = P tan φ = 56 x 0.59 = 33 kvar (see Figure L4).