Electronic – Power factor correction consequences

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.

I think that trying to use RMS values of voltage and current for this is not going to work. Imagine shifting the current waveform 4ms later; neither the RMS voltage nor the RMS current would change at all, but the power drawn would change by an order of magnitude.

The instantaneous power drawn by your circuit is V * I. In any small time dt, the energy consumed will be V * I * dt. The energy consumed in 1s, the power drawn, will be the integral of V * I * dt from T=0 to T=1s. You could compute this directly from the sample values in your excel spreadsheet. At each time sample, multiply the instantaneous voltage by the instantaneous current, and that gives the instantaneous power drawn. Multiply that by the sample interval, and that is the energy consumed in that sample interval. Add all those up over an AC cycle, and multiply by the number of cycles per second and that is the energy drawn per second, otherwise known as the power.

Looking at the scope traces, the current drawn by the circuit is usually 0. Once per AC half-cycle, the current increases to about 90mA very quickly, then drops linearly to 0 over about 820us. It's a 60Hz circuit so it does this every 8.3ms. When the circuit is drawing current, the voltage is more-or-less constant at 170V. That's an average current of 45mA over the 820us at 170V = 7.65 W, but it only takes this power 1/10 of the total time, so the final power consumption is 0.76 W.

In my experience, the probability of wiring up a current probe backwards is exactly 0.5!