Voltage regulation wrt resistive and inductive loads...

Hi, In real world situation you will have hard time finding purely resistive load. Most loads are inductive affecting power factor. IMO, your generator maybe border line under powered for the welder.

thats actually a real interesting question...

with a pure reactive (inductive or capacitive) load, the current flow in the windings will cause a voltage drop but if i'm not mistaken since there is no (or very little) actual power flow, there will not be a load on the engine so the engine speed will not be a factor.

with a pure resistive load, the current through the windings will cause a voltage drop AND there will be a load on the engine that will try to slow it down. It is the job of the speed governor on the engine to keep the speed constant. If the speed drops the voltage and frequency will drop due to the engine speed drop.

So for a given amps, you may get more of a drop with a resistive load depending on how tight the governor speed control the engine.

Mark

All the generator cares is how much real power it takes to spin. It's not measuring LC, just measuring the final outcome. 1-2 % really surprises me.

Greg

Depending on what's loading the transformer, a transformer load isn't necessarily substantially inductive.

That's the egghead theory. Unfortunately, my knowledge of welders starts with being able to stick weld, then has this long, dark, gap, then gets to the circuits theory that I know as an electrical engineer. So I couldn't tell you just what the power factor of a welder is (PF = how "resistive" it is). I can tell you that it probably varies with the type of welder, what its technology is, and probably by whether it's cheap Chinese or quality late-model 'merican.

I'd like to see some measurements. We don't know much about the load. If it's a transformer and a stick welder, is there reason to believe that the load voltage/current presented by the arc isn't relatively in phase? Would be interesting to see the V-I curve of a plasma under welding conditions.

Just get an old electric range top. You can switch the loads on and off as necessary.

John

Infinate burner controls don't present a steady load.

In purely technical terms,you generator should be rated on Kva not Kw.

If you have a resistive load, Kva and Kw are the same. A purely resistive load would be incandescent lights or electric heaters.

However, once you get into motor/transformers, the current and the voltage are no longer synchronised. (power factor) In practice it means that the load is drawing more current than you might think. So you can only connect a smaller load than you might think looking at the Kw rating. To know the load the generator will meet accurately you must know the Kva rating of everything. (ie both generator and load)

Welding transformers are particularly suspect, they have a "Poor power factor".

It's not a good idea to load a generator up to capacity anyway,especially in warm weather.

Not true. If power factor is bad, ( ie inductive) the generator can be overloaded at below it's rated capacity in Kw.

No, it cares about the current being supplied. Heating is done by the current, not the real power generated. Generators are rated in KVA, not watts.

PURELY resistive? Where in the heck would you find a PURELY RESISTIVE load?

Awl --

Oh, Attenuators. I didn't know that. I guess that part of the circuit is purely resistive or resistive/inductive.

(I just looked up "purely resistive")

n 12, 5:10=A0pm, "Existential Angst" wrote:> Awl --

An electric heating element, like a range element, toaster, heater without a fan, water heater, light bulb etc are examples. They all have some theoretical small inductance, capacitance, too, but it's so tiny it can be ignored. The voltage and current through those devices is in phase.

An electric heating element, like a range element, toaster, heater without a fan, water heater, light bulb etc are examples. They all have some theoretical small inductance, capacitance, too, but it's so tiny it can be ignored. The voltage and current through those devices is in phase. ====================================================

Turns out I have a not-so-bad testing solution. Altho an electric range top would be great, I have a bunch of 240 V electric baseboard heaters, 10 A at 240 V. So just five of them will give me a 50 A load at 240 V, about what the unit is rated at, and I will be able to plot V vs. I, ito regulation. I also have a bunch of regular 120 V space heaters, but the 240 V stuff makes wiring easier. Won't get to this 'til about Wed/Thurs, tho. Still have to deliver my CO report.... now that I've recovered..... LOL

Ill repeat my reply here for the sake of those that filter google groups

thats actually a real interesting question...

with a pure reactive (inductive or capacitive) load, the current flow in the windings will cause a voltage drop but if i'm not mistaken since there is no (or very little) actual power flow, there will not be a load on the engine so the engine speed will not be a factor.

with a pure resistive load, the current through the windings will cause a voltage drop AND there will be a load on the engine that will try to slow it down. It is the job of the speed governor on the engine to keep the speed constant. If the speed drops the voltage and frequency will drop due to the engine speed drop.

So for a given amps, you may get more of a drop with a resistive load depending on how tight the governor speed control the engine.

Mark

I'm sort of guessing that myself, but I hope it's not so. With out-of-phase voltage current, I'm ALREADY worse than 12% regulation, and if this hunch is correct, pure resistive would make it even worser!

We'll see what happens. I'm amp-probing the current, metering the voltage, so I can repeat the welder load, do the pure resistive load, and compare both V vs. I plots.

Hold yer brefs until Thurs/Fri.... LOL

A heater?

A large block of carbon crystal with plates on the outside. Compression of the plates changes the resistance.

Used them in lab experiments. 200 amp switcher supplies.

Mart> >> Awl --

Watts generated (in any conductor =3D Amps squared X resistance.

Poor power factor increases amps which means more heat has to be dissipated in the generator. (in the whole circuit in fact)