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Reactive power

First, let’s say some words about basics of the reactive power in system. Reactive current arises in every electrical system.

4 example calculations of compensation for reactive power
4 example calculations of compensation for reactive power (photo credit: mavinelectric.files.wordpress.com)

Not only large loads, but smaller loads as well require reactive power. Generators and motors produce reactive power, which causes unnecessary burdens to and power losses in the lines.

Figure 1 shows the block diagram for the network loading.

Equivalent circuit diagram of a network with different loading: a) Equivalent circuit; b) Phasor diagram
Figure 1 – Equivalent circuit diagram of a network with different loading: a) Equivalent circuit; b) Phasor diagram

Reactive power is necessary to generate magnetic fields, e.g. in motors, transformers and generators. This power oscillates between the source and the load and represents an additional loading.

Power supply companies and the consumers of this electrical energy are interested in reducing these disadvantages as well as possible. On the other hand, non-linear loads and phase-controlled inverters cause harmonics, which lead to voltage changes and a decrease in the power factor. In order to reduce these harmonics, series resonant (filter) circuits are used.

Now, let’s take few examples to calculate the following:

  1. Determination of Capacitive Power
  2. Capacitive Power With k Factor
  3. Determination of Cable Cross-Section
  4. Calculation of the c/k Value

Example 1 – Determination of Capacitive Power

A load has an effective power of P = 50 kW at 400 V and the power factor is to be compensated from cosφ = 0.75 to cosφ = 0.95. Determine the required capacitive power. The power and current before compensation are:

S1 formulae

The power and current after compensation are:

S2 formulae

The required capacitive power  is:

Qc formulae

Go back to calculations ↑

Example 2 – Capacitive Power With k Factor

The capacitive power can be determined with the factor k for a given effective power. The k factor is read from a table 1 – Multipliers to determine capacitor kilovars required for power factor correction (see below) and multiplied by the effective power. The result is the required capacitive power.

For an increase in the power factor from cosφ = 0.75 to cosφ = 0.95, from the table 1 we find a factor k = 0.55:

Required capacitive power Qc

Go back to calculations ↑

Example 3 – Determination of Cable Cross-Section

A three-phase power of 250 kW, with Un = 400 V, at 50 Hz is to be transmitted over a cable 80 m in length. The voltage drop must not exceed 4% =16 V. The power factor is to be increased from cosφ = 0.7 to cosφ = 0.95. What is the required cable cross-section?

P formulae

The current consumption before compensation is:

Current consumption before compensation

The current consumption after compensation is:

Current consumption after compensation

The effective resistance per unit length for 516 A is:

Effective resistance per unit length

According to Table 2 (see below) we must choose a cable with a cross-section of 4 × 95 mm2. The effective resistance per unit length for 380 A is:

Effective resistance per unit length

Here, a cable cross-section of 4 × 70 mm2 is required. As this example illustrates, the improved power factor leads to lower costs because of the reduced cross-section.

Go back to calculations ↑

Example 4 – Calculation of the c/k Value

Given a 150 condenser battery, i.e. 5 stages of 30 each, a supply voltage of 400 V, and an instrument transformer with a k of 500 A/5 A, how large is the c/k value? The ratio c/k is given by.

Ratio c/k formulae

Go back to calculations ↑

Tables

Table 1 – Multipliers to Determine Capacitor Kilovars Required for Power Factor Correction

Multipliers to Determine Capacitor Kilovars Required for Power Factor Correction
Table 1 – Multipliers to Determine Capacitor Kilovars Required for Power Factor Correction

Table 2 – Resistance per unit length for (Cu) cable with plastic insulation

Resistance per unit length for (Cu) cable with plastic insulation
Table 2 – Resistance per unit length for (Cu) cable with plastic insulation

Go back to calculations ↑

References

  1. Analysis and design of low voltage power systems by Ismail Kasikci (Purchase hardcover from Amazon)
  2. Power factor correction – A guide for the plant engineer by EATON

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About Author

author-pic

Edvard Csanyi

Edvard - Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry fascilities. Professional in AutoCAD programming. Present on

6 Comments


  1. Tomáš
    Feb 17, 2017

    Hey, I do not think that the pic b is correct. There are strange angles at Ur.
    cos phi = 1 means 0 deg between current and voltage so it is not capacitave nor reactive load but pure resistive.


  2. Abass
    Jan 24, 2017

    Many thanks Edvard.


  3. Josè Severino Queiroz
    Jan 23, 2017

    This is a good and practical method to get better the performance of the system mainly the losses and voltage regulations. Thank you..


  4. Aleksandar Kovac
    Jan 23, 2017

    very good.
    thanks


  5. Amine
    Jan 20, 2017

    very nice thank you


  6. gamal
    Jan 20, 2017

    Thank for usefull report

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