View source for Definition of Power Factor

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== Definition of power factor ==

{| cellspacing="1" cellpadding="1" width="65%" border="1"
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| bgcolor="#0099cc" | The power factor is the ratio of kW to kVA. The closer the power factor approaches its maximum possible value of 1, the greater the benefit to consumer and supplier.<br>PF = P (kW) / S (kVA)<br>P = Active power<br>S = Apparent power
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The power factor of a load, which may be a single power-consuming item, or a number of items (for example an entire installation), is given by the ratio of P/S i.e. kW divided by kVA at any given moment.<br>The value of a power factor will range from 0 to 1.<br>If currents and voltages are perfectly sinusoidal signals, power factor equals cos <span class="texhtml">φ</span>.<br>A power factor close to unity means that the reactive energy is small compared with the active energy, while a low value of power factor indicates the opposite condition. 

== Power vector diagram ==

*Active power P (in kW)

&nbsp; - Single phase (1 phase and neutral): P = V I cos <span class="texhtml">φ</span> <br>&nbsp; - Single phase (phase to phase): P = U I cos <span class="texhtml">φ</span><br>&nbsp; - Three phase (3 wires or 3 wires + neutral): P = <math>\sqrt 3</math> U I cos <span class="texhtml">φ</span> 

*Reactive power Q (in kvar)

- Single phase (1 phase and neutral): P = V I sin <span class="texhtml">φ</span><br>- Single phase (phase to phase): Q = U I sin <span class="texhtml">φ</span><br>- Three phase (3 wires or 3 wires + neutral): P = <math>\sqrt 3</math> U I sin <span class="texhtml">φ</span> 

*Apparent power S (in kVA)

&nbsp; - Single phase (1 phase and neutral): S = V I<br>&nbsp; - Single phase (phase to phase): S = U I<br>&nbsp; - Three phase (3 wires or 3 wires + neutral): P = <math>\sqrt 3</math> U I <br>where:<br>V = Voltage between phase and neutral<br>U = Voltage between phases<br>I = Line current<br><span class="texhtml">φ</span> = Phase angle between vectors V and I. <br>&nbsp; - For balanced and near-balanced loads on 4-wire systems 

== Current and voltage vectors, and derivation of the power diagram ==

The power “vector” diagram is a useful artifice, derived directly from the true rotating vector diagram of currents and voltage, as follows:

The power-system voltages are taken as the reference quantities, and one phase only is considered on the assumption of balanced 3-phase loading.

The reference phase voltage (V) is co-incident with the horizontal axis, and the current (I) of that phase will, for practically all power-system loads, lag the voltage by an angle <span class="texhtml">φ</span>.

The component of I which is in phase with V is the “wattful” component of I and is equal to I cos <span class="texhtml">φ</span>, while VI cos <span class="texhtml">φ</span> equals the active power (in kW) in the circuit, if V is expressed in kV.

The component of I which lags 90 degrees behind V is the wattless component of I and is equal to I sin <span class="texhtml">φ</span>, while VI sin<span class="texhtml">φ</span> equals the reactive power (in kvar) in the circuit, if V is expressed in kV.

If the vector I is multiplied by V, expressed in kV, then VI equals the apparent power (in kVA) for the circuit.

The simple formula is obtained: S<sup>2</sup> = P<sup>2</sup> + Q<sup>2</sup>

The above kW, kvar and kVA values per phase, when multiplied by 3, can therefore conveniently represent the relationships of kVA, kW, kvar and power factor for a total 3-phase load, as shown in '''Figure L3'''.


[[Image:FigL03.jpg|none]]
'''''Fig. L3:''''' ''' Power diagram''' 


== An example of power calculations (see '''Fig. L4''') ==


{| style="width: 797px; height: 177px" cellspacing="1" cellpadding="1" width="797" border="1"
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| bgcolor="#0099cc" colspan="2" | '''Type of circuit''' 
| bgcolor="#0099cc" | '''Apparent power S (kVA)''' 
| bgcolor="#0099cc" | '''Active power P (kW)''' 
| bgcolor="#0099cc" | '''Reactive power Q (kvar)'''
|-
| colspan="2" | Single-phase (phase and neutral) &nbsp; 
| S = VI 
| P = VI cos <span class="texhtml">φ</span> 
| Q = VI sin <span class="texhtml">φ</span>
|-
| colspan="2" | Single-phase (phase to phase) &nbsp; 
| S = UI 
| P = UI cos φ 
| Q = UI sin <span class="texhtml">φ</span>
|-
| valign="top" rowspan="2" | Example&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
| 5 kW of load 
| valign="top" rowspan="2" | 10 kVA 
| valign="top" rowspan="2" | 5 kW 
| valign="top" rowspan="2" | 8.7 kvar
|-
| cos <span class="texhtml">φ</span> = 0.5
|-
| colspan="2" | Three phase 3-wires or 3-wires + neutral 
| S = <math>\sqrt 3</math> UI 
| P = <math>\sqrt 3</math> UI cos <span class="texhtml">φ</span> 
| Q = <math>\sqrt 3</math> UI sin <span class="texhtml">φ</span>
|-
| valign="top" rowspan="3" | Example 
| Motor Pn = 51 kW 
| valign="top" rowspan="3" | 65 kVA 
| valign="top" rowspan="3" | 56 kW 
| valign="top" rowspan="3" | 33 kvar
|-
| cos <span class="texhtml">φ</span>= 0.86
|-
| <span class="texhtml">ρ</span>= 0.91 (motor efficiency)
|}

'''''Fig. L4:''''' '' Example in the calculation of active and reactive power''

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