Primary Current of Welding Machine

Is the input single- or three-phase? If three, the input current would be ~66A.

I think that your arithmetic is correct. The manufacturer has cited a kva rating based on the no-load output - 75 volts.

Supply and equipment are both single phase.

So, should I use the 114A to choose my conductor? Then, I have to use atleast #1AWG wire at my input?

Any advice on when can I utilize the 40% duty cycle as per nameplate?

Your welding device can accidentally by start, when used on the 300 Amp setting and when striking, use the rated amperage.

This will happen almost never.

Once you are welding, voltage is lower than the 75 volts and 300 Amps is for heavy welding or cutting. And on the other part the 40% duty cycle is not permitting you to work under these conditions for more than 40% of the time.

The other 60% of the time,you can drink beers while letting your device cool down or do other jobs, like preparing your materials to weld or... imagine what else.

Your duty cycle will be probably hopefully higher when you weld at lower outputs (weak amps) on your plate.

... happen almost never.

but i still have to rate my conductor and ocpd based on 114 Amperes, right?

or it's too much of an overkill?

The NEC does not give latitude on sizing conductors for what you THINK it might draw, you are required to size conductors based solely on the nameplate data. So yes, you must size your conductors based on 114A. The only leeway given for duty cycle is I believe that because that is not a continuous current rating, you may not be required to size the conductors at 125% of the nameplate rating. So you can use 114A instead of 142.5A. Continuous, as defined by the NEC is 3 hours or more.

Here in the spending zone is it easy to tell you to put a 100 amp breaker, because almost everyone has 120 Amps in its distributor panel.

Unless you are a professional welder, which I doubt, because you would have chosen for a 3 phase device in Europe when you need that 300 amps.

You probably will not have the opportunity to install a 100 amp breaker, because your main panel breaker is less and expanding to 100 or 120 Amps will need a different connection and a different tariff. Letting you weld on one phase with 300 Amps is looking for trouble with the neighbors, unless you are in a industrial zone.

It probably won't stop with a breaker and a dedicated line only.

If I were you I would first try with a breaker that is less than your main panel and see where it gets you.

If that breaker trips too often while welding, or you take a smaller size rod or make the extra expense.

So you decide, I have a strong feeling that you just need 50 or 60 amps to weld. But to get your 300 Amps for a long time, you might be easier with more.

dvmdsc-

Yes. In addition to your answer, which was very good, welders (the person vs. the machine, which is called a "welding machine") are not normally electricians or electrical engineers. Very seldom, in reality, is the wiring installed for the maximum current or the primary of the machine. Most welders will use wiring and a circuit breaker based on the highest secondary current they will be working with. This is done for economic reasons. Their reasoning is "why should I wire for a primary current that I won't ever use". Also if a smaller welding machine was used on that circuit they would probably just plug the machine in and weld. If the breaker tripped at normal use currents then they would change the primary circuit wires. Stingy economic reasons.

Very seldom would a secondary current of 300 amps be used in other than an industrial assembly situation. For two common type rods, 6010 would require a 1/4" diameter rod and 7018 would require a 7/32" rod. 1/8" is the most common rod size. Depending on the material and the configuration this could cause significant heat distortion.

I am not advocating that the OP operate with smaller wire than required. This is presented to show how such a big welding machine and the wiring for it may be a waste of money for the OP.

I have an over 30 year old Airco 150amp AC/DC/TIG welder and have never run into a situation where it was not sufficient current. If larger rods were needed I would simply increase the number of passes. It also means that I can use smaller and less expensive wiring for that circuit. Another advantage is I can construct extension cords for the primary and not have to use longer and bigger ground and electrode cables.

Good Luck, Old Salt

Is this your first welder?

What did the operator's manual say?

Duty cycle - Wikipedia, the free encyclopedia

Continuing on from that being I do a lot of welding I can say for sure that for about 95% of the time very few things need much over 150 amps on a stick welder so to be honest it's highly unlikely that the OP will ever even get up over a 100 amp input draw.

When I do the occasional 1/2" thick high strength steel cutting edge seam weld I am running 7/32" 7014 rods which at most are still running under 200 amps actual welding current and that's a rather large hot weld!

Similar with wire feed. To do that weld on my MIG system I would be running about 28 volts constant voltage and at most maybe seeing 200 - 250 amps peak welding amps with .045" wire which makes fore a very hot weld just the same!

My point is in realistic hobby welding applications its rather rare to ever need to run a 300 amp machine at it's full capacity. There are just too few times were it's justified and even less where having the line side power set up to handle that sort of load for more than a moment is highly unlikely.

These requirements are exactly the same in the Canadian code.

Very interesting blog, learnt a lot, thanks all.

I am surprised that such a high current is available on a single phase "at home" anywhere.....

Surely this is on an industrial site? E.G.Business premises?

Just being nosy......

Do you not have a manual with the machine? If not, you should be able to look up a manual on line. Manual shuld give you exactly what you need to know as far as what input amperage you need and even cable size required(which depends, of course, on how long the input cable is.

I am not familiar with the Yamato brand, but every other machine brand I have hooked up always had the required information in the manual. No need to do work that has already been done.

I don't have a manual, and I've tried looking online but could not find one.

Your welding unit according to its nameplate is rated at 13 KW.. Therefore the power transformed or reflected back to the primary will also be the same.. You can simply divide that same reflected power by the primary input voltage of 220 V, which would equate to about 60 Amperes! Then you can use this value to find the proper AWG size cable and corresponding protective device!

Good luck!

No, I think using the KW rating, one should use the power factor, which would result to a value the same as using the KVA rating. pls advice.

ianeill -

Stated KW rating reflects the real value of the power available from the machine, the electrical capacity that gets converted into real work.

KVA on the other hand reflects the apparent value due to the transformer together with all of its built-in losses. So the KVA rating only applies to the transformer alone, the transformer's electrical capacity!

Power factor is the ratio of the two differing powers. You only use this value if you want to correct and improve the welder's efficiency!..that is maybe due to its non-linear load?

My understanding was, you are only interested to find the proper size of the conductor that can handle the real work that can be delivered by the device?

vsar,

I am interested in the Input current, so I could determine the Input conductor and the protective device.

My understanding is that, to be able to deliver the real power, one needs the supply of apparent power, especially so in this kind of machine.

Another concern, will my supply be able to handle the load? Say, I have a 100A supply, with your computation of 60A, I am well within the limit. But, using the KVA rating which will result to about 115A, then my supply won't be able to handle the load. Please advice.

I would have thought that all the really good advice here must have nailed it for you by now.

Remember, even if the wiring is (theoretically) not strong enough to use the full power of the welder, as long as its properly fused/CB for that size of wiring, in a worst case situation you will drop the breaker/rupture the fuse.

No harm done......

Having a CB for a higher current than allowed is naturally dangerous.....surely that is completely clear?

The transformer rated capacity is termed as KVA. Within this figure is its ability to convert and deliver usable (real) energy, and all the losses or wasted energies spent in the conversion.

The (transformer + load) losses are not fix and not linear so to speak. As you raise the power setting, the greater the losses will be since lesser useful power is made available by the device. More energies are wasted in the conversion as more heat is generated since the transformer starts to saturate! The higher your power setting, the lower its welding efficiency of the machine becomes!

To safely use a machine that was not efficiently designed, it is not advisable to push it to its limit. You will just be wasting more energy by feeding it more amperage since most will only be lost during conversion at higher settings!

This is also where duty cycle comes in and why its important! Cooling down help improve its efficiency!

janeil18-

The power that a transformer type welding machines consumes, which it seems you have, is proportional to the power used in the secondary including the efficiency plus the basic power to operate the unit while no current is flowing.

Simplified, if the secondary voltage is 50 volts and the current is 100 amps, 5,000 watts are being consumed (P=I*E). Take this 5,000 watts and divide it by the input voltage 240volts and an efficiency estimated at 80 %_{( estimated)} and you have the input current, 21 amp, without the current to operate the machine without any secondary current flowing. This is for the fan and any other internal devices that draw power. With that at a hypothetical 1 amp, the total primary current would be 22 amps (21 amps power for the primary/secondary and 1 amp for fixed no load requirements.

At a secondary maximum for your machine it would be approx. 300 amps_s * 60 volts_s (estimated) = 18,000 watts, 18,000 watts/ (240 volts*80% efficiency) & 1 amp to run = 18,000/ (240*.8) & 1 amp or 95 amps maximum primary current. This is a basic example. All calculations are dependent upon the actual capabilities, output voltage at the set current, efficiency and basic no load requirements.

All these calculations are basic. There are other minor things that also affect the primary current and voltage. Also on the secondary any resistance of the connectors, wire, stinger resistances and ground clamp will also have some effect.

By substituting the parameters for your welding machine you can calculate the power requirements.

Good Luck, Old Salt