Electronic – Unstable diesel generator output voltages causing irregular behaviour in uninterupted power supply (UPS) units

Reactive power would put no extra load on a generator shaft if everything were perfect. However, real generators have real losses, with some of those proportional to the square of the current. The reactive load causes more current in the wires than there would be with a purely resistive load of the same real power. The extra current causes additional real power to be lost.

So the answer is that the engine will see a somewhat higher load and therefore use a bit more fuel. This is because of more inefficiencies and losses in the system, not be reactive power itself makes the generator harder to turn.

Added:

I should have mentioned this before, but somehow it slipped thru my mind at the time.

A reactive load on a perfect generator does not require more shaft power averaged over a cycle, but it does add "bumps" to the torque. One attribute of a 3 phase AC generator is that the torque is constant over a cycle with a resistive load. However, with a reactive load parts of the cycle will require more power and other parts less. The average power is still the same, but constant pushing forwards and backwards relative to the average torque can cause undesirable mechanical stresses and vibrations.

You can think of this a bit like moving two magnets past each other. Let's say they are oriented to repell. At a distance there is little force. You have to apply force to move them close together, meaning you put energy into the system. The magnets push in the direction of motion as they move away, thereby giving you back the energy you put in earlier. The net energy spent is 0, but there was definitely energy flow back and forth. There is always some loss as energy is moved around or converted back and forth in real systems.

Again, the reactive power itself doesn't cause the problem, but real power is lost because energy can't be moved around and converted with perfect efficiency. This real power loss has to be made up with more real power input. In addition, the extra mechanical forces can decrease the life of the generator and engine driving it.

On a generator, you have a prime mover (say, an engine) connected to the actual generator, which consists of either rotating coils of wire within a magnetic field, or rotating magnets surrounded by coils of wire. The number of poles (magnetic poles) and the rotational speed determine the output frequency: Freq = Engine_RPM * Number_Of_Poles / 120.

Typically, a United States portable generator runs at 3600 RPM, with 2 poles, for a design frequency of 60Hz. Larger portable generators run at 1800 RPM with 4 poles here.

That is how frequency is determined. The number of turns and the magnetic structure determine how many volts are produced at the design frequency, voltage and frequency aren't related in any fashion except for design. Again, in the States, most portable generators are wound to have a 240VAC single phase output, which is center tapped and delivered as two 120VAC hots with one neutral, but virtually any voltage can be delivered.

The current output of a generator is determined by its load, as long as the load doesn't exceed the maximum capacity of the generator's prime mover (engine) plus the conversion losses of the actual generator. Prime mover power is often rated in horsepower (US) or kilowatts (everywhere else). With no losses, a 10 horsepower engine could deliver 7457 watts (actually VA for non-resistive loads) continuously, or 62.1 amps at 120VAC continuously. Try to take more, and the engine will slow down (reducing both the frequency and the voltage, which will also drop the current) until you reach a point that the engine actually stalls.

You get fluctuation of frequency and voltage as the load changes because the engine cannot respond immediately to the actual load change. There are regulators controlling the engine throttle that attempt to keep the engine at a fixed (design) speed, but it takes time for the engine to respond to new commands as it has to deal with varying fuel/air mixtures and combustion which aren't instantaneous.

As a clarification to other discussions here:

For a purely resistive load, halving the voltage would halve the current, and result in one quarter the power consumed. You can't say that just cutting the voltage in half cuts the power consumed in half. With some devices, that may be true, but it entirely depends on the load.