3 phase welder amps question

I am confused. The nameplate on my hobart cybertig says that at 230V, it consumes 44 amps.
http://igor.chudov.com/projects/Welding/00-Hobart-CyberTig-Welder/dscf0003.jpg
Absent the power factor number, that means roughly 220*44*sqrt(3) = 16 kW power consumption.
Yet, it is a 28V, 200A welder, which means approximately 6 kW power output.
How can I reconcile these numbers? How can it take in 16 kW and yet only produce 6 kW? That seems awfully wasteful.
I tried to estimate what it would take, assuming more reasonable power efficiency.
It if produces 6 kW and wastes 2 kW as heat, then it would need 8 kW. Estimating amps, 8 kW/220/sqrt(3) = 21 amp.
This is not an idle question for me. I have 3 phase cabling that is capable of conducting 21 amps (combo of a 12 and 10 gauge cables). For 44 amps, I would need to spend big bucks to buy 4-6 flexible cable.
Also, I can easily support 8 kW power consumption after adding an extra idler to my RPC. 16 kW is out of my realm of possibility due to branch circuit capacity and common sense.
Some clarification would be appreciated.
i
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I always thought the rated input current was the peak current (ie when you initially strike an arc), with the rated output being the continuos current?
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wrote:

That makes sense... So, what would realistically happen if less power was available to the welder, enough for continuous welding but not enough for, say, rated starting power usage?
i
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IFF that is true, then you'd proly just have more trouble starting yer arc, it would seem to me. Power factors always confused me, but the bottom line can be solved w/ a relatively inexpensive AmpProbe, or clone. Clamp it on, see what it sez. Def'ly a tool you should own, Ig, considering your interests.
Ito of your wire size, it all depends on the *length of the run*. For a short run, and given that most welding is done in spurts (given 20% duty cycles), #12 wire is proly good for 50 amps, esp. if 'exposed" (ie, not crammed into pipe w/ 30 other wires). Purists here will squawk like hell, but take this in context of the caveats I mentioned. #12 wire can *easily* handle 30-40 Amps continuous, mebbe getting a little warm. Not recommending it, just saying you can get away with it. So your 10/12 ga combos are proly more than adequate, if you run is not much over 60 feet.
You can tell the effect of length on voltage drop w/ a simple bulb hooked up to one leg, or w/ a voltmeter. I think a voltage drop of 10% is tolerable , if electronics are not too persnickety. Ito of your rpc, the voltmeter will also be a good guide.
The Ampprobe is interesting as you watch it, relative to the types of arc you strike. You can also put the ampprobe on the welding cable itself, to see how the arc affects things.
If you are able to control things mechanically (arc stability, length, etc), you can "verify" the nameplate for yourself, using sed amp probe/voltmeter, ie, comparing the input/output kVA ratios you actually measure. ---------------------------- Mr. P.V.'d formerly Droll Troll

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On Sun, 25 Sep 2005 22:02:10 -0400, Proctologically Violated

Makes sense. Note also that I have HF arc starting feature.

Agreed 100%.

That's interesting. My current 3 phase cabling includes 40 feet of 10/4 SJOOW, then in-line 3 phase receptacle and a plug, and then another about 40 feet of 12/4 SJOOW. All sizes approximate. Note that the voltage is approximately 240V, which allows some wiggle room for some voltage drop along the line.
According to my AWG table, voltage drop at 40 amps along the 10 gauge cable would be 3.2 volt, and then in the 12 gauge cable it would be 5 volt, of 8.2 volt total. That would be a drop from 240V to 231.8 volts. This calculation does not consider any heating effects.

My run is about 60-80 feet (just eyeballing). It is nice to have long cabling and be able to drag out my welding machine out to the yard.

that would be a 220V bulb...

Agreed. For the arc, I can just hook it up to my oscilloscope, I think.

I pretty much have to experiment on this one. Thanks PV.
i
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On Mon, 26 Sep 2005 11:06:13 GMT, Ignoramus12004

While sitting on the toilet, I made some rough calculations. Looks like for a voltage drop of 4 volts per leg, at 30A consumption, I would look at power loss or 4*30 = 120 amps per leg, or 360 watts on three legs.
So, the cabling would act as a 360 watt heater. The energy output per foot of 80 foot cable would amount to 4.5 watts per foot. It would, at worst, make it barely noticeably warm.
i
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On Sun, 25 Sep 2005 22:02:10 -0400, "Proctologically Violated"

The one HF is currently selling for $11 is pretty decent. I use it everyday..or nearly so.
Gunner
"Pax Americana is a philosophy. Hardly an empire. Making sure other people play nice and dont kill each other (and us) off in job lots is hardly empire building, particularly when you give them self determination under "play nice" rules.
Think of it as having your older brother knock the shit out of you for torturing the cat." Gunner
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I have a large-ish clamp on ammeter, military surplus, I will check if it is good for smaller amperages. I recall that mine goes to hundreds of amps.
i

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On Mon, 26 Sep 2005 17:56:47 GMT, Ignoramus12004

Just checked it, it measures AC amps from 0A up to 1000 A.
i

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Ignoramus18928 wrote:

>

It's really just connected between two phases - essentially single phase welder.
That's normal (unless it is inverter) for AC output. DC output can use all 3 phases (3-phase rectifier bridge with 6 diodes), but AC can only use single phase. For cheaper welders, even the DC uses just single phase (transformer input from two phases) to save on amount of diodes - 2 vs 6 or 4 vs 6, depending on xformer topology.
To verify my hypothesis, measure phase currents of input 3-phase with clamp-on ampere meter.. It should have current on only 2 phase wires.. Of course, opening the case and looking at connections should help as well. Propably ground, neutral and 2 phases connected, or even without neutral (used for electronics of welder, if present).
Kristian Ukkonen.
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wrote:

The schematic for this welder shows a delta shaped, 3 phase transformer and rectifier bridge. The actual transformer is in the shape of letter E.

My welder does not have AC output, unless I am missing something very big.

Exactly my case.

Not in my instance.

Unfortunately, that is not the case.
i
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It is extremely rare to find AC output on a 3-phase, transformer based, welding power supply.
AC is almost exclusively found on either 3-phase inverter based machines or Single-phase transformer machines.
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That makes sense. So, Ernie, what are your thoughts on this welder's amp requirements? Just how big should the phase converter be to drive it. Again, 200 amp tig. I posted a picture of the nameplate in my original post.
I did run it at low settings with a 10 HP idler, but I am not sure just what it takes to run it at full power.
i
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Ig, If the output (arc) is not AC, then you proly can't put an amprobe on the welding cable itself, unless they've come out with DC-measuring amprobes, while I been asleep. If they have, I sure would like one! ---------------------------- Mr. P.V.'d formerly Droll Troll
wrote:

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On Mon, 26 Sep 2005 16:07:54 -0400, Proctologically Violated

I have amp meter for the output, right on the cybertig. Measuring output is not really an issue, except that I cannot look at anything but the arc while arc is produced, for obvious reasons. I would hook up the clamp on ammeter on the input (AC) side.
Besides having an ammeter for output,
http://igor.chudov.com/projects/Welding/00-Hobart-CyberTig-Welder/dscf0004.jpg
my welder has settable amperage where I would, say, set the arc to 78 amps and it would actually be 78 amps. It is programmable. So, I know that amps would be what I set them to be. (unless I am missing something)
i
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On Mon, 26 Sep 2005 16:07:54 -0400, "Proctologically Violated"

Yup, I've had one for some time. Different principle, works the same. I think it uses the Hall effect. I used it to calibrate instrumentation on big DC drives for extruders before we went to AC drives.
Pete Keillor

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I would imagine it's a bit pricey-er than the AC version. ---------------------------- Mr. P.V.'d formerly Droll Troll
wrote:

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On Mon, 26 Sep 2005 21:24:20 -0400, "Proctologically Violated"

I don't recall. I bought it for work. Fluke stuff ain't cheap.
Pete

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It might use a Hall effect device, but more likely a transductor. A couple of square loop cores acting as current transformers driven with a AC current. The AC current is proportional to the DC current.
Dan Pete Keillor wrote:

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I think you are right. The Hall-effect devices tend to be used for small currents typically in the low amps and milliamps range. billh

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