The comparison of the Heating Effect of Generator Operated at 0.8 and 1 Power Factor

Below is the comparison of heating between two synchronous generators rated 1000 KVA, 240 Amps, 2400 Volts, 900 RPM, and 0.8 pf, operated at 0.8 power factor and 1.0 power factor while maintaining an output KVA of 1000 KVA and output voltage of 2400 Volts. The assumption in this comparison is that the heat that is released by the generator is the real power losses in the windings of the generator. That is, Heat = Ploss = 3Ia^2(Ra) + If^2(Rf), where Ia is the phase current that flows in one phase of the generator armature winding, Ra is the per phase armature resistance, If is the field current and Rf is the resistance.

Case 1: 0.8 p.f.

0.8 is the rated power factor of the generator. At this value of power factor, the output power is 800 KW, and the phase angle between the voltage and the current at the terminal of the generator is displaced by 36.9 degrees. At terminal voltage of 2400 V (line to line), the magnitude of the phase current is,

Ia = 1000KVA/(2400*sqrt(3)), or 240.56 Amperes.

The per phase real power loss, of course, will be equal to the square of the magnitude of the current multiplied by the resistance of the armature winding. In equation form, power loss will be,

Paloss = (240.56)^2 x (Ra).

Thus, the total armature winding loss is,

PalossTOTAL = 3 x (240.56)^2 x (Ra).

Figure 1: Single phase model of a synchronous generatorCase 2: 1.0 power factorAt 1.0 power factor, the phase angle between the current and the voltage is unity, or 0 degrees, or in short, current and voltage are in phase with each other. Thus, the output KW is equal to the output KVA, or 1000 KW. But since the terminal voltage is maintained at 2400 Volts, as we had stated earlier, the magnitude of the per phase armature current is,

Ia = 1000KVA/(2400*sqrt(3)), or 240.56 Amperes.

Note that the value of Ia for 1.0 power factor is equal to Ia if the power factor is equal to 0.8. This is because the magnitude of the current does not depend on its phase angle difference with the voltage.
The same principle also applies for the calculation of the real power loss in the armature winding. Only the magnitude of the current is used for the computation. That is, for 1.0 power factor,

Paloss = (240.56)^2 x (Ra).

Then, the total armature winding loss is,

PalossTOTAL = 3 x (240.56)^2 x (Ra).

Conclusion for case 1 and 2

Theoretically, since the real power loss in the armature winding is equal for generators operated at 0.8 power factor and 1.0 power factor, the amount of heat dissipated in the armature winding for both cases should also be equal.

However, for a generator that is operated at 0.8 power factor at rated output voltage and KVA, the internally generated voltage, Eg, or the armature voltage is greater than the armature voltage of a generator that is operated at 1.0 power factor. Therefore, a generator that is operated at 0.8 power factor will require a higher field or excitation current which will result to additional heating due to field winding losses. (Note that armature winding and field or excitation winding are both compound in one casing. Excitation or field current is the output current of the DC generator that is coupled in the synchronous generator).

You can futher learn and compare the heating effect of a generator operated at higher and lower power factor using the Reactive Capability Curve

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