We have a synchronous, brushless (not permenant magnet) generator with the following ratings:(adsbygoogle = window.adsbygoogle || []).push({});

3 phase brushless alternator

Model TFW-24

Volt 120/208

Frequency 60 Hz.

Phase 3

KW (P) 24

KVA 30

Power Factor (Cos) 0.8

Insulation class B

Protection Type IP21

Rat S1

Standard JB/T3320-1 2000

If a load with a power factor of 0.8 is applied and the nameplate full load current is drawn, The killowatt is 24 and the killovoltamp is 30. (corresponding exactly to nameplate ratings)

(Since KW = I * E *1.73 * PF and KVA = I * E * 1.73)

or in this case: 24Kw = 83 * 208 * 1.73 * 0.8 and

30KVA = 83 * 208 * 1.73

However if a resistive load with a power factor of 1.00 is applied and the nameplate full load current is drawn, The killowatt is 30 and the killovoltamps is also 30. (Killowatt rating is 125% of nameplate rating while current is at nameplate rating.)

Since KW = I * E *1.73 * PF and KVA = I * E * 1.73

or in this case: 30Kw = 83 * 208 * 1.73 * 1.00 and

30Kva = 83 * 208 * 1.73

Can the generator handle nameplate full load amps without overheating in both situations described above regardless of the fact that the nameplate killowatt rating is overreached when a resistive load is applied?

From what I understand a generator's power factor rating is how low the power factor the load can go at full load amps, and although it is uncommon that it is needed, load power factors can be lower but the load amperage will need to be lowered to prevent overheating of the generator stator. This leads me to believe that if the power factor of a load is 1.00 it can run at full load amperage, which as shown in the above calculations is 125% of rated killowatt. Please let me know if this is wrong.

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# Killowatt Versus Killovoltamp

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