There are 3 types of electrical:
1) Real Power, measured in Watts (W).
This is the power drawn by a resistive load, e.g. a heater element in a kettle, and has a power factor of 1. (unity power factor, cos F=1, 1.0pf or pf=1)
2) Reactive Power, measured in Volt Amperes reactive (VAr’s).
This is the power drawn by a reactive load (a load with a winding around a core), e.g. an electro-magnet, and has a power factor of 0. (zero power factor,
cos F =0, 0pf or pf=0)
3) Apparent Power, measured in Volt Amperes (VA).
Many loads have a combination of resistive and reactive elements. (in fact it is not possible to produce a purely inductive load, since the wire used to
form the windings has a resistance). This combination of elements means that both real power (W) and reactive power (VAr) are drawn together.
The proportion of Real Power to Reactive Power is defined as the power factor. [Nearly all resistive load (e.g. Universal motor used in hand tools) then
power factor 0.95 to 1.0, nearly all inductive load then power factor ~ 0.3] The vast majority of single-phase loads have power factors approaching
Therefore, single-phase generator power ratings are taken at power factor =1, and are consequently in Watts (W) or kilo Watts (kW), where 1 kW = 1000 W.
Three-phase loads tend to have lower power factors, approaching 0.8, therefore, three phase generator power ratings are taken at power factor =0.8 and
are in VA or kVA.
There is obviously a relationship between real power, reactive power, apparent power and power factor…
i) Apparent Power (VA) = Ö [(real power (W))2 + (reactive power (VAr))2] And
ii) Power factor = Real Power (W) Apparent Power (VA)
Therefore…Apparent Power (VA) x Power factor = Real power (W)
If the Power factor =1, then all the Power is real, and Apparent Power (VA) = Real Power (W) (W = VA @ 1.0 pf)
For a single-phase generator, the rating should be at 1.0 pf, in which case Watts = Volt Amperes. But, for a three-phase generator the rating is at 0.8pf.
This is where confusion can arise!
Example. A three-phase generator has a continuous rating of 5 kVA at 0.8 pf. Now, at this rated load, the Real power (kW) will be …Real Power (kW)
= Apparent Power (kVA) x Power factorReal Power = 5 x 0.8 = 4kW
This means that a generator producing 5kVA at 0.8pf is actually producing 4 kW of Real power, but it is also producing some reactive power.
From i)… 5000 VA = Ö [( 4000 W)2 + ( Reactive Power)2] Reactive Power = 3000 VAr’s
It is this combination of 4kW of real power and 3kVAr’s of reactive power that has defined the limit for the generator rating. If the same generator was
loaded with a resistive load only, then it may be capable of more than 4kW, however, there is no formula that can be used to find this limit from the
0.8pf rating. It can only be found through testing of each machine. Similarly, a single-phase generator rated at 4kW, cannot be expected to produce
5kVA at 0.8pf !!!!