V-BELT 7L5001 - Caterpillar

POWER FACTOR, Voltage, efficiency, KVA, Phase, amperage and KW. All these terms seem to have a specific mystery about them. They are terms whose definitions a generally misunderstood and often misquoted. This lack of complete understanding is reflected in such questions as, "If an AC generator is rated 240 volts and 300 amperes at 0.8 power factor, why can't the generator produce 240 volts and 300 amperes a 1.0 power factor?" Very simply, the answer to this question is – the generator is capable of this output, but it is possible the engine will not be set to deliver enough horsepower to permit the generator to carry the load at rated speed. To help clarify these terms and to explain further the simple solution to this problem, let us explore a typical engine generator setup.Taking, for example, a naturally-aspirated D342 Electric Set, we will investigate the characteristics of both the engine and generator. In starting, we will make a few assumptions and explain a few terms. First, we will assume the generator to be 90% efficient, which is about average for generator efficiency. This efficiency indicates that if 100 horsepower were applied to the generator shaft, only 90 horsepower could be extracted from the generator as electrical energy. The 10% loss within the generator is due basically to bearing friction, windage, and heat losses. We can further define one horsepower as being equal to 0.746 KW of electrical power and that KW (kilowatts or 1000 watts) is equal to KVA (kilovolt-amperes or 1000 volt-amperes) time the power factor.The naturally-aspirated D342 Electric Set is rated 100 KW at 1200 rpm. For an engine to deliver 100 KW from a 90% efficient generator, the following horsepower is necessary.
Illustration 1 g01059155
Now we have established that to produce 100 KW from a 90% efficient generator, 149 horsepower will be required. This does not include any capability for overload.The D342 N.A. Electric Set Engine, set for 1236 rpm high idle and 1200 rpm full load, has a factory rack setting of 0.180. This produces a parallel operation rating of 170 horsepower without fan. The engine is then capable of driving a 90% efficient generator with the following rating (again without including any capability for overload):
Illustration 2 g01059161
Actually, this engine is capable of powering a generator rated at 114 KW. Standard generator ratings do not include a 114 KW rating, so a 100 KW generator will be used. This also gives the unit a capability for small overload.The nameplate data for this sample generator are:
Illustration 3 g01059169
Considering these items one at a time; first, KVA is equal to the rated voltage times the amperage times 1.732 (or required for three-phase "8730;3" -- a constant required for three-phase generator) divided by 1000 (because KVA means 1000 VA).
Illustration 4 g01059171
In the illustration, note that for either leading or lagging power factor loads at time (T), voltage is at its peak (A) and current is less than peak (B). KW (or true power) is then the result of voltage (A) times