Power factor correction

[aero]

I have an Oxford 180 welding transformer (wired for single phase). Recently I read somewhere that there used to be a 'condenser' (capacitor) available that would correct the power factor and thereby reduce the mains consumption of the set?

Any of you professional electricians/welders able to shed any more light on this?

 

[shenion]

Usually the caps are placed across a separate winding on the transformer. Could be directly on the mains though. Most i have seen are basically A/C motor run capacitors (oil filled).

It helps filter the spikes and will also add a capacitive load to balance the mostly inductive load of the welder. Will increase the idle current a bit though.

 

[aero]

I'd like to know exactly how to connect a cap up and what value/rating should be. I'm guessing the value of the capacitor must have some relationship to the transformer. And is there any benefit as I've never seen one fitted.

 

[zzr1200]

I can't see a capacitor reducing the power required to run the set at a given ampage/voltage, but it may help dampen the current spike on the mains supply as you short circuit the transformer while arcing up.

 

[shenion]

It will reduce the current required. Here's info from the DIalarc 250 vs 250P:

250  DC 50% load, 230V: 61A , 14.1 KVA, 7.9 KW
250P DC 50% load, 230V: 41.1A, 9.5 KVA, 7.7KW

As you can see, the current is reduced by about 1/3. Notice with power factor correction, the KVA vs the KW out is much closer.

 

[drivewasher]

It will reduce the current required. Here's info from the DIalarc 250 vs 250P:

250  DC 50% load, 230V: 61A , 14.1 KVA, 7.9 KW
250P DC 50% load, 230V: 41.1A, 9.5 KVA, 7.7KW

As you can see, the current is reduced by about 1/3. Notice with power factor correction, the KVA vs the KW out is much closer.

Power factor correction as far a I undestand is to make the current move along the sign wave to line it up with the voltage. Cos an inductive loads current lags behind the voltage IE an heating elements load can be calculated by its resistance but an inductive load cant

 

[aero]

Yeah resistive loads (heating elements) have a PF of 1.

Let me rephase the question. What size and volteage rating capacitor would I need to put on my Oxford to bring the PF nearer to 1?

And where should it be connected?

seems to be getting a few brians ticking over

 

[madkayaker]

As a transformer is a almost pure resistive load PF is not an issues. Amperage should be in line with voltage if it is out at an angle that gives you the PF. It is only a real issue on 3PH when you are billed for the angle of the PF. As for spikes on the primary side of the transformer there shouldn't be much interference and setting up surge arresters big enough for the current would be expensive.

 

[alistairlee]

I have an Oxford 180 welding transformer (wired for single phase). Recently I read somewhere that there used to be a 'condenser' (capacitor) available that would correct the power factor and thereby reduce the mains consumption of the set?

Any of you professional electricians/welders able to shed any more light on this?

Hi
Not sure why you want to correct the PF as I understood it the meter reads KWH which will be less than KVA if the PF is inductive or capacative, (ie the last thing you want is a PF =1) this applies to domestic supplies, industrial may have a penalty if PF is not =1.

This advice is worth what you paid for it.

Al

 

[madkayaker]

Hi
Not sure why you want to correct the PF as I understood it the meter reads KWH which will be less than KVA if the PF is inductive or capacative, (ie the last thing you want is a PF =1) this applies to domestic supplies, industrial may have a penalty if PF is not =1.

This advice is worth what you paid for it.

Al

That's all there is to it really as domestics and light commercial are billed KWH not KVA the power factor is irrelevant. Better saving you money and fitting more energy saving light bulbs

 

[aero]

This advice is worth what you paid for it.

Ok My thinking was going along the lines of this:

A transformer is a bloody great inductor as it consists of windings, which is what an inductor is (not a resistor).

This will cause a shift in phase between the current and the voltage.

To provide the same output and make up for the shift in phase you use more power to achieve the same output.



BUT

You say the way in which domestic electricity is metered means that you will not pay more?


(industrial users can cause severe phase shifts in the national grid which is why they are heavily monitored and billed for phase angle shifts. But getting the PF as near to 1 means they use less so it works in their favour too)

 

[alistairlee]

Ok My thinking was going along the lines of this:

A transformer is a bloody great inductor as it consists of windings, which is what an inductor is (not a resistor).

This will cause a shift in phase between the current and the voltage.

To provide the same output and make up for the shift in phase you use more power to achieve the same output.



BUT

You say the way in which domestic electricity is metered means that you will not pay more?


(industrial users can cause severe phase shifts in the national grid which is why they are heavily monitored and billed for phase angle shifts. But getting the PF as near to 1 means they use less so it works in their favour too)

Hi
A long time ago it was explained why a domestic meter would read less KWH used for a given Amps if the PF was leading or lagging, and that it could cost £0 if the PF is 0 (hypothetical situation obviously). -bugged me a bit when I found out that they were adding caps to the inside of PC powers supplies to correct PF

I don't know how input PF alters the amps required at the welding torch though (way out of my depth).

Regards
Al

 

[aero]

They add caps to correct the phase shift.....so you use less power.

I think this has been done to death now. I'll ring some electrical contarctors who specialise in PF correction and see what they say.

Thanks though.

 

[madkayaker]

Have a look on this website
http://energystar.custhelp.com/cgi-...std_adp.php?p_faqid=4941&p_created=1204908170

 

[aero]

OK I have the answer. I've been on to Technical arc services, the people who make the Portamig.

I asked about the power factor of the Portamig (as I have one as well) to get their interest. It seems the PF of the Portamig is 0.92, so not worth trying to correct the PF as its already very good.

I asked about my Oxford as a second question leading on from the first. The Oxford's PF is terrible, about 0.6! That means for every amp it comsumes its only doing 0.6 amp worth of work. Or to put it anotherway you need to use 1.4 amps to get 1 amps worth of work from it (almost 50% more).

They supply PF correction capacitors and are finding out which one I'll need. I'm waiting for a quote as well, the price might be prohibetly expensive, but I'll let you know.

This will be the same for any basic TRANSFORMER type arc welder, so by adding a capacitor you could save a fair amout of mains power and therefore keep your electricity bill down.

I read the link to the web site. Its talking about your average domestic device. A bloody great welding transformer is not your average bit of domestic equipment, despite what I keep telling my other half.

This is a specific case where the item in question has a very very poor power factor. If the capacitor were fitted to the welder it would only be working when you used the welder, so my thinking is this could be worthwhile in this specific instance. Depending on how much the capacitor costs.

I also have a 3 phase invertor, which warns strongly against PF correction devices so I'd have to ensure the welder wasn't being used at the same time.

 

[abrogard]

Very ancient thread I stumbled across while researching concerning my Transarc 200 transformer single phase stick welder.

My problem was wiring it up - it came without input lead. 10A or 15A or what?

I got informally told they were 15A machines. Meant running a 15A supply to my shed.

Looked into it a little (I'm no electrician) and learned a little and saw 10KVA peak load written on the back of the thing.

Figured from that I needed (140V supply here in Aus.) I needed something like a 40A supply. Because 10000/240 = 41 (about).

That's when I learned about Power Factor.

The thing is it simply doesn't draw 10KW. It has a PF of about 0.6. It draws about 25A peak. 10A home wiring can handle that at max for short times.

And I get billed, as a domestic user, for that 25A. I get billed on KW. I get billed for 6KW, not 10KVA.

And it wires up and was intended to be wired up as a 15A supply machine.

I thought I might chuck this in because I feel this is the truth of the matter and the thread left things in an inconclusive state.

Like I say I'm not an electrician. I believe the above to be the truth. If I'm wrong I'd like to know. No one should take my words as gospel and do electrical things based on them without professional advice.