Generator vs Inverter Generator (Genset)

[chala]

Originally Posted by Captain Bill

I have a 5 KW diesel Onan genset on my boat capable of putting out about 40 amps at 110V single phase.

So 40A * 110V = 4400 VA or 4.4 kVA and at unity (1) power factor is 4.4 kW
at a power factor of 0.8, 4.4 kVA * 0.8 = 3.5 kW

Originally Posted by Captain Bill

I would like to put a dive compressor on my boat that requires 2.2 KW 110V single phase.

Ignoring the PF of the compressor motor, (could be 0.8 to 0.9, the current in A of the motor could be more useful) 2200 W / 110 V = 20 A. An unloaded electrical motor without any type of soft starter may require between 8 to 10 times the rated current to start so 20 A * 10 = 200 A.
Any generator should come with a set of specifications for example a Markon Saw Fuji generator will state that for a BL 105 a motor starting current can be no more than twice the full load current of the generator and for an SL 105 not to exceed 1.5 time.
40A * 2 = 80A. 200A is much in excess of the generators Manufacturer recommendations.
In regard to engine power the generator set output may be limited by the horsepower rating of the engine. In some instance the manufacturer of a genset may limit the power of the engine in such a way that the generator cannot be over loaded (burned out). In any case Markon recommend 2 hp (736W metric, 746W imp * 2) off engine power for each 1000W of electrical load. Depending how an engine is constructed, the size of the flywheel, the torque available at given rpm and the type of governor installed will give that engine a different ability of responding to an overload.
Most generators are suitable for supplying continuously most types of load with a total load current not exceeding the current quoted on the nameplate.
In regard to Power Factor, a generator may not perform properly if the load pf is not in accordance with the manufacturer recommendation. For instance a BL 105 is only suitable for a load near unity (1), a SL 105 is suitable for a power factor between 1 and 0.8. If not stated on a nameplate it is possible to calculate the power factor of an appliance. For example a compact light show 11W 110V 200mA the pf will be 0.5
11/ (110*0.2) = 11W/ 22VA = 0.5, an appliance not recommended for the above generators. In any case a load with a fractional pf will use less engine power than a load at unity and a capacitive load should be avoided.
Without automatic load shedding it is difficult to load appropriately a genset when the load contain a motor with a large starting current so running a diesel like it should be run is difficult to achieve.