I have a welding transformer that has 6 taps on it. The transformer it rated at 75 kva at 50% duty cycle. The max secondary voltage is 8.81 and the max short circuit amps are listed at 23,000. Here is my question if I am running on Tap #3 (which is 6.57 volts) what is my current. I am a mech guy but my electronics math is lacking. Do I have enough information here or no. What I come up with is an estimated 17,152 (max current x sec volt of tap / max sec volt). I don't have a ton of faith in that calculation, if anyone can elighten me I would appreciate it.
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Resistance Welding Transformer 1
- Thread starter aamoroso
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Skogsgurra
Electrical
aamoroso,
"I don't have a ton of faith in that calculation" I liked that.
I do not think that the calculation per se is dubious, but the inputs probably are. Those numbers are sales numbers and are usually on the high side.
The figures usually assume that your mains voltage stays constant even if heavily loaded (a welding tranformer takes all it can from the mains - and then some). It is usually also measured as a "short circuit current" which means that cables, tongs and the actual plates that you weld are excluded from the calculation. A simple test will show if your mains is tough enough; connect a lamp and see if it "browns out" during the weld. If it does, you really have to do something about the mains.
It is not uncommon to find half of the total voltage drop in the mains and the transformer and the other half in cables, tongs and welding material. If there is a rectifier, you usually have one third in mains and transformer, one third in the rectifier and one third in cables etc. So anything between your calculated current and half of it is possible. I would guess that you get between 60 and 80 percent of what you calculate. But I cannot guarantee it. Measure it.
There are very few 20 kA current transducers around, so it could be a problem to measure that current. Look for rogowski coils. They usually do a good job at these current levels, but be prepared that some of them cannot measure DC (If your current source is DC, which I do not know).
"I don't have a ton of faith in that calculation" I liked that.
I do not think that the calculation per se is dubious, but the inputs probably are. Those numbers are sales numbers and are usually on the high side.
The figures usually assume that your mains voltage stays constant even if heavily loaded (a welding tranformer takes all it can from the mains - and then some). It is usually also measured as a "short circuit current" which means that cables, tongs and the actual plates that you weld are excluded from the calculation. A simple test will show if your mains is tough enough; connect a lamp and see if it "browns out" during the weld. If it does, you really have to do something about the mains.
It is not uncommon to find half of the total voltage drop in the mains and the transformer and the other half in cables, tongs and welding material. If there is a rectifier, you usually have one third in mains and transformer, one third in the rectifier and one third in cables etc. So anything between your calculated current and half of it is possible. I would guess that you get between 60 and 80 percent of what you calculate. But I cannot guarantee it. Measure it.
There are very few 20 kA current transducers around, so it could be a problem to measure that current. Look for rogowski coils. They usually do a good job at these current levels, but be prepared that some of them cannot measure DC (If your current source is DC, which I do not know).
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Thanks skogsgurra, I think your information is actually alot more helpful than anything I have been able to get elsewhere. I am just trying to put together some rounded numbers to use as general start up parameters which will be adjusted based on performance but to have an almost scientific approach to it I needed to get in the ball park. It only took me over a week of research to get there but its a nice ball park better than we were a week ago.
Thanks again.
Thanks again.
I agree with skogs. I could be way off base, but 23kA is likely into a bolted short circuit—that’s the way it is with circuit-breaker test sets. The reactance (less so resistance) will ‘eat up’ the secondary voltage very easily. Secondary-side leads should be kept as close together [ideally twisted] as possible. The area of the secondary “loop” should contain as small an open area as feasible.
It may be economical to use a commercial current shunt for AC or DC measurement. For short duty cycles, it may be worth using a much lower-rated shunt—so you measure 200mV instead of 50. []
It may be economical to use a commercial current shunt for AC or DC measurement. For short duty cycles, it may be worth using a much lower-rated shunt—so you measure 200mV instead of 50. []
Skogsgurra
Electrical
Yes, it is possible to measure current quite easily by using a shunt (or a copper bar, if you keep track of temperature). But there is a trap to fall into. You can easily get between three and ten times more curent reading than you expect.
The reason is that a shunt for, say 50 kA at 100 mV, has a resistance that is equal to 2 micro-ohms. It will probably be one or two inches long, which translates into something like 50 nano-henries inductance, which is equal to 15,7 micro-henries. So most of your signal will be an inductive signal that is a lot higher than the resistive part, which the shunt was calibrated for.
You can reduce this effect by building the shunt with concentric voltage signal leads (a copper tube with twisted leads attached to its inside walls), but I prefer the rogowski. Especially if you are dealing with AC. But the shunt is OK for DC where the inductive effect is less disturbing.
The reason is that a shunt for, say 50 kA at 100 mV, has a resistance that is equal to 2 micro-ohms. It will probably be one or two inches long, which translates into something like 50 nano-henries inductance, which is equal to 15,7 micro-henries. So most of your signal will be an inductive signal that is a lot higher than the resistive part, which the shunt was calibrated for.
You can reduce this effect by building the shunt with concentric voltage signal leads (a copper tube with twisted leads attached to its inside walls), but I prefer the rogowski. Especially if you are dealing with AC. But the shunt is OK for DC where the inductive effect is less disturbing.
electricuwe
Electrical
aamoroso,
the answer to your question depends mainly on the transformer configuration.
Are the taps on the primary or on the secondary ?
the answer to your question depends mainly on the transformer configuration.
Are the taps on the primary or on the secondary ?
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