The use of a breaker in a circuit is to protect the wire size used in the circuit from becoming overloaded. Using the wattage of the load does not help for breaker sizing because the breaker operates on amperage. Amperage can be found from wattage by using the following equation. I = W /E. Amps = Watts / Volts. As you can see the amperage can not be calculated because there is no voltage stated.
This formula will give you the amperage I = kw x 1000/1.73 x E x pf (pf = power factor). Take the amperage and multiply it by 125%. This will give you the breaker size that you need.
AWG #3 copper.
In standard construction this would require a 15 amp breaker and 14 gauge copper wire.
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.
3000 / 240 = Amps. You de-rate a breaker by 20 % for continuous load like an oven. You could get by with a 20 Amp breaker and 12 AWG wire. However, I would recommend 30 Amps and 10 AWG for an oven for the long run.
A breaker protects the wire size of the feeder that is connected to it. The amperage of the load must be found. Without a voltage stated the amperage from the wattage given can not be calculated. The equation for amperage when the kw is given is A = kW x 1000/1.73 x volts x pf. The pf constant to use is .9.
The formula you are looking for is I = W/E.
I=270000/380/1.732 I=410A USE: 500A CIRCUIT BREAKER
the given kw Divide by the your voltage
This formula will give you the amperage I = kw x 1000/1.73 x E x pf (pf = power factor). Take the amperage and multiply it by 125%. This will give you the breaker size that you need.
AWG #3 copper.
In standard construction this would require a 15 amp breaker and 14 gauge copper wire.
Calculate the ampereage at .8 pf and take a breaker of 150% of ampere nd set the over current protection at 125% of full load current.
Cant tell for sure. A good engine will have a small cc for its KW, while a crap one will require more cc. CC relates to the size of the engine, while kw refers to the power. Id say a 125 at a guess
7200
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.
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.