The formulas you are looking for are, single phase kW = I x E x pf/1000. For three phase kW = I x E x 1.73 x pf/1000.
Since the breaker that is installed on the generator set will be sized to the output of the 30 kW generator, the load will stay connected until the thermal trip of the breaker trips the load off line. This will be in the matter of seconds before it happens. To handle a 75 kW load and depending if it is an inductive or a resistive load you will need at least a 80 to 100 kW generator.
At 746 watts per horsepower (electric) a 35 kw generator would require about 470 horsepower. Round that up, and allowing for control loop margin and mechanical losses, I would guess that a 35 kw generator should need a 500 or 600 horsepower engine.
First you will need a three phase generator. Mathematically there are 746 Watts per horsepower, but I like to use 1000 Watts for ease of mental calculation. This would mean you would need a 30 kW generator. If using 746 Watts per HP, you would need 22380 Watts, or 23 kW. Make sure this 23 kW is the normal load rating of the generator, not the surge rating! 30 kW would provide more of a safety cushion.
the unit of generators power is KVA becoze the kva is the power that contain the active power (KW) and the reactive power mean that the name plate of any generator must contain the rated kva of it (like the transformer P (KW) = P (kva) * cos fi P (KW) = V I cos fi for single phase P (KVA) = V I when cos fi closed to 1 this will increase the useful power that exit from the generator or transformer with my pleasure
Use the following equation, kW = Amps x Volts x 1.73 x pf/1000.
5.274 KW
Depends upon the Kw rating of the generator. The higher the Kw rating the more fuel will be used.
Since the breaker that is installed on the generator set will be sized to the output of the 30 kW generator, the load will stay connected until the thermal trip of the breaker trips the load off line. This will be in the matter of seconds before it happens. To handle a 75 kW load and depending if it is an inductive or a resistive load you will need at least a 80 to 100 kW generator.
It depends on the total connected load (KW) of the house. If the total connected load is about 20 KW the alternator generator should be designed to meet peak 20 kw load
A 5 kW generator would turn it over but if the full 30 hp of mechanical power is needed, that would require about 30 kW of electric power from the generator.
480V.
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At 746 watts per horsepower (electric) a 35 kw generator would require about 470 horsepower. Round that up, and allowing for control loop margin and mechanical losses, I would guess that a 35 kw generator should need a 500 or 600 horsepower engine.
Frequency is the speed that the generator revolves, not the sizing of the generator.
About 8hp (5 kW).
For equations or formulas use 746 watts = 1 horsepower.
It does not matter, when testing a generator with a resistive load bank, if you load it to kVA or KW. For a resitive load, i.e. non-reactive load, the power factor is one, so kVA and kW are the same.