# 4 example calculations of compensation for reactive power

## Reactive power

First, let’s say some words about

**basics of the reactive power in system**. Reactive current arises in every electrical system. 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.**

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.

### 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 cosÏ† = 0.75 to cosÏ† = 0.95**. Determine the required capacitive power. The power and current before compensation are:
The power and current after compensation are:

**The required capacitive power is:**

### Example 2 – Capacitive Power With k Factor

The capacitive power can be determined with the f

**actor 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 cosÏ† = 0.75 to cosÏ† = 0.95**, from the table 1 we find a**factor k = 0.55**:### 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 cosÏ† = 0.7 to cosÏ† = 0.95**. What is the required cable cross-section?
The current consumption before compensation is:

The current consumption after compensation is:

The effective resistance per unit length

**for 516 A**is:
According to Table 2 (see below) we must choose a cable with a cross-section of

**4 × 95 mm**. The effective resistance per unit length for^{2}**380 A**is:
Here, a cable cross-section of

**4 × 70 mm**is required. As this example illustrates, the improved power factor leads to lower costs because of the reduced cross-section.^{2}### 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.### Tables

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:**### 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 cosÏ† = 0.75 to cosÏ† = 0.95**. Determine the required capacitive power. The power and current before compensation are:
The power and current after compensation are:

**The required capacitive power is:**

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