Hi Bob,

I stand by my post, but Iâ€™ll take a shot at clarifying.Â Â And concede that I did make an approximationâ€¦

while i agree he is limited by engine power, the head is not maxed out.

at unity power factor (for the sake of discussion) theoretically all the kva generated go to making useful work, measured in kwatts

Agreed.

if however the powerfactor of the load drops to .7 there is a significant amount of power kvars? that is doing no useful work?

this power is circulating between the alternator and the load, creating nothing but heat

Ummmâ€¦ Sort of.Â Hereâ€™s the area where precise language is really beneficial.Â The thing is that, yes there is extra current flowing (circulating as you say) but it doesnâ€™t equate to heat in the magnitude you might be thinking.Â Some heat, to be sure, but Iâ€™ll leave it to you as to whether its significantâ€¦

Hereâ€™s an example that may help put it into perspective.

Letâ€™s say he needs 2.6 kW, but he has a 0.7 PF, as you say.Â

At 120V, this means his line current is 31A.Â (If he had aÂ perfect 1.0Â PF, his line current would be 22A)

So, clearly, the line current is higher with the low PF.Â But how big of a problem is this?Â

The 31A is still within the rating of his 5kW ST head, so no problem there.

Now, If youâ€™ll allow me to pull a number out of my ass, Iâ€™d like to estimate the resistance in the ST head stator windings and wiring to the load.Â And Iâ€™m going to estimate it at 0.3 ohms.Â (Actually this is not a total rectal extraction â€“ My 12 kW head is 0.1 ohms if wired for 120V, which I doubled to 0.2 ohms for this case, plus I added 0.1 ohms for an assumed 100 feet of 10AWG copper wire to the load)

Thus, in the 0.7PF case, the total power lost as heat in the ST head winding, and copper wire due to I2R loss is 31^2 * 0.3 =Â 290 Watts

Now, in the 1.0PF case, we have 22^2 * 0.3 =Â 150 Watts

So, yes, some power is being lost due to the poor power factor. (290-150 = 140 extra Watts lost)

This 140 Watts of loss, compared to the 2600W being delivered represents about 5% of the engineâ€™s output power.

Significant?Â You decide.

BTW, your first post said this

at .7 pf you might only be able to cover about 1.8kwatt of loading

Which is way out of whack, based on the analysis I show above.Â And it indicated to me a misunderstanding whereÂ you multiplied 2600W by the 0.7 PF.Â This is irrelevant, and not the correct approach to this problem.