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.
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| 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):
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| 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:
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| 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).
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| 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 current (B), whereas KVA (or apparent power) is equal to peak voltage times peak amperes, therefore, it can be seen that KW equals KVA only at unity power factor. | |
Using the nameplate information, The KVA for this particular three-phase generator would be:
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| Illustration 5 | g01059174 |
As previously stated, KW equals KVA times power factor. Therefore, the KW output of this generator equals 125 x 0.8 or 100 KW. One can readily see that, at unity or 1.0 power factor, KVA would equal KW. Further investigation of power factor would reveal it to be, "The ratio for expressing what part of the apparent power (KVA) flowing in an AC circuit is true power (KW) ",or "P.F. = KW/KVA".
Power factor can be either leading or lagging depending on the load. Illustration 4. A leading power factor can be caused by an excessive amount of capacitors connected to the load, a lightly loaded synchronous motor, or an induction motor being driven by its load. Lagging power factor loads are comprised mostly of induction motors. Unity Power factor loads are composed of electronic devices and resistance loads such as lights and heaters.
Standard industrial generators are rated 0.8 P.F. lagging since the average industrial load includes motors and other equipment, which operate at this power factor. The maximum value for power factor is 1.0 and the minimum is 0.0. The most common value range is from 1.0 to 0.8 (lagging). Leading power factor information is also included here although it will seldom be encountered in practice. The letters in the column of the table labeled "GRAPH LINE" furnish a cross reference between the tabulated data and the information in the graph.
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| Illustration 6 | g01059176 |
The Chart is showing how power factor affects engine and generator loading. The Engine is rated 149 horsepower at 1200 rpm. The three-phase generator is rated 240 volts, 300 amperes and 0.8 power factor (lagging) at 1200 rpm. | |
The table and its companion graph indicate how power factor affects engine and generator loading. Using the aforementioned engine, generator data and performing the necessary calculations with the previously described equations, various loaded conditions (including the rated 0.8 lagging power factor) have been simulated and plotted. Now with the aid of this graph, it can be seen why the simple solution could be given to the original power factor question. It can now be seen that:
- At any power factor in excess of rated (in this case anything over 0.8), the output of the electric set is limited by engine horsepower.
- At any power factor less than rated, the output of the electric set is limited by generator amperage.
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| Illustration 7 | g01059179 |