Reactive Power Management in Power Distribution System
An in-depth discussion on how to manage reactive power in a utility network. This is an important basic concept to know in times of increasing EV and solar renewables penetration.

An in-depth discussion on how to manage reactive power in a utility network. This is an important basic concept to know in times of increasing EV and solar renewables penetration.
Most of the loads in a modern power distribution system are inductive in nature. Typical examples being, loaded overhead lines, motors, transformers, cables, drives, and fluorescent lights. These inductive loads consume both active and reactive power. Active power is required to meet real output requirements and inductive reactive power is required to maintain the magnetic field in the core. Phasor diagram of these parameters is shown in Fig 1.1(a) and 1.1(b). The phasor diagram is described below.
Cos Φ = Power factor,
kW = Active power,
kVAri = Reactive power
kVA = Apparent power = (kW2 + kVAri2)1/2
Given this, the power supply system should be capable of supplying both active and inductive reactive power. Presence of inductive reactive power results in the power factor being less than unity and lagging in nature, which in turn results in the following ill effects on a power distribution system.
Thus, we need to reduce the reactive power to reduce all of the above ill effects. The most effective method for this is increasing the power factor of a power distribution system. The methodology used to achieve a higher power factor under modern application conditions is REACTIVE POWER MANAGEMENT. It can also be said that “REACTIVE POWER MANAGEMENT” is a tool for improvement in the power factor of a distribution system. Hence, both of these are two faces of the same coin. The methodologies used for power factor improvement are mentioned below
The concept of power factor improvement is shown in Fig 1.2(a). Here it is shown that kVAR1 is reactive power fed by the supply-side before the installation of capacitors. (kVAR1 - kVAR2) is reactive power fed by the capacitor. kVAR2 is new resultant reactive power to be fed by the supply system after the installation of capacitors. It can be seen that kVAR2 < kVAR1. In Fig 1.2(b) also, it is shown that kVARc is reactive power fed by the capacitors.
Fig 2.1(a) shows a simplified distribution system with a power factor of 0.75 without using capacitors. Fig 2.1(b) shows the same distribution system with a power factor of 0.95 by using capacitors.
kVA = kW + kVAR
3.1 – Providing compensation at the main incomer to power distribution system bus bar is called central compensation. This is shown in Fig. 3.1. Here capacitor is installed at position No. 1.
3.2 – Providing capacitor compensation at
This is suitable for installations where there is a number of power distribution boards for various load feeders. This is shown in Fig. 3.2.
3.3 – Providing capacitor compensation at all positions as described below.
This is suitable for installations consisting of the main receiving station, substations, several load feeders, and a variety of loads. This is shown in Fig. 3.3.
The estimation of kVAr required for compensation to achieve desired power factor is generally done depending on the type of loads to be compensated. For this purpose, the tables and formulae given in this section may be used.
Table 4.1 gives the recommended ratings of power capacitors, which are to be used directly with 3 ph. AC induction motors as at position No. 3 of fig 3.3
Note :
For power distribution systems, the capacitors are to be installed at position No.1 of Fig. 3.1. Initial pf, final pf, and multiplying factors are shown in table 4.2. The operating load kW and its average pf can be obtained from the electricity bill. Else, it can also be calculated from the following formula.
Average pf = kWh/kVAh, operating load kW = kVA demand X average pf. The average pf is considered as initial pf and the final pf is the pf to be achieved by “REACTIVE POWER MANAGEMENT”. This can be calculated using multiplying factor given in table 4.2, This is shown in the example given below.
Example –
Calculate the required kVAr compensation for a 500 kW installation to improve the pf from 0.75 to 0.96.
kVAr = kW X multiplying factor from table 4.2 = 500 X 0.59 = 295 kVAr
Note –
The table 4.2 is based on following formulae.
Multiplying factor = (tan Φ1 – tan Φ2 )
kVAr required = kW (tan Φ1 – tan Φ2 ), where Φ1 = Cos-1(pf1), Φ2 = Cos-1(pf2)
pf1 and pf2 are initial and final power factors respectively.
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