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Thread: ? 630.11 & .12 Electric Welders

  1. #1
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    ? 630.11 & .12 Electric Welders

    Trying to calculate for Industrial Arc Welder
    It is a Miller DIALARC HF AC/DC +TIG
    The manual is available here
    It will be installed 40 ft. from the breaker panel, 230V
    What size conductors and OCP is required?
    Please show your work so that I may do the calculations next time.
    It may be necessary to do calculations for 30% and 100% Duty Cycle.
    It is NOT a Motor Generator!
    It does NOT have power factor correction.
    Thanks.



    ambidextrious wirenutter
    All quoted sections are from NEC 2008 unless otherwise stated

  2. #2
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    110509-0738 EDT

    Denis:

    I can not answer your question, but I have part of the approach for you.

    The welder has a switching power supply at its input and it is quite tolerant of input voltage variations. 200 V @ 105 A is its lowest for full power output.

    You have a source of 230 V and under load you can expect the source to drop a noticeable amount, assume 10 V. Your total loop wire length is 80 ft so assume 100 ft. Assume the maximum drop in this 100 ft is 20 V. This would mean the lowest voltage at the welder would be 200 V. Another assumption is that the source voltage will never be below 230 V. Round off the current to 100 A.

    From these assumptions the maximum wire resistance would be R = 20/100 = 0.2 ohms. Using 20 C resistance values for copper wire and 100 ft this means you could use #13 wire which is 2 ohms per 1000 ft. Obviously this is too small for other reasons, meaning temperature rise in the wire.

    What this calculation tells you is you do not have a voltage drop problem with a wire that will be adequate for the load current.

    Here is were I stop because I would have to go the library to look at the NEC tables.

    However, I would design based on the 105 A and a continuous load, but I expect your actual application won't be this severe.

    .
    Last edited by gar; 05-09-11 at 09:00 AM.

  3. #3
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    Quote Originally Posted by gar View Post
    110509-0738 EDT

    Denis:

    I can not answer your question, but I have part of the approach for you.

    The welder has a switching power supply at its input and it is quite tolerant of input voltage variations. 200 V @ 105 A is its lowest for full power output.
    interresting approach

    it does not have a switching power supply
    this machine is from the 70's
    the voltage range is provided by changing connections on the internal transformer
    ambidextrious wirenutter
    All quoted sections are from NEC 2008 unless otherwise stated

  4. #4
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    110509-0953 EDT

    Denis:

    Even with a transformer input the electronics on the output side of the transformer will produce the same effect in terms of input voltage and current but over a more limited voltage range. So as voltage goes down input current will increase for a constant group of welder settings.

    In any event your primary concern will be based on the wire and its current rating.

    #6 is 0.4 ohms/1000 ft at 20 C, and thus your approximate voltage drop would be 0.04*100 = 4 V with this size wire.

    #4 about 2.5 V.
    #2 about 1.6 V.

    .

  5. #5
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    you determine conductor size by using table 630.11(A). Model HF is the unit without power factor correction(which is a bit unusal because this was offered at a nominal add on $ when sold new and most dealers only stocked them HFP) anyway startting with 90 amps you use the 40% duty cycle multiplier because this is the rated output duty cycle. .63 X 90 is 56.7. This is not of course considered a continous load so you stick with that as your wire ampacity. Then depending on what wiring method you choose you get you wire size from 310.16. 6 thhn cu in pipe or 4 cu nm. hope this helps. Not sure where the other person replying was going with voltage drop ????????

  6. #6
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    630.12

    Stew, thanks, your explanation helps to make sense the method

    now let's work on 630.12
    ambidextrious wirenutter
    All quoted sections are from NEC 2008 unless otherwise stated

  7. #7
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    This is another case in the code where conductors can be protected by overcurrent devices that exceed the rating of the conductor. In your case the welder is rated at 90 amps and can be protected at not more than 200%. That being said my proceedure on transformer welders is to protect at or near namplate amps due to the fact that the machines are rarley ever asked to weld at thier maximum output. I would use no more than a 100 amp or even a 90 amp breaker on this especially for a machine that is to be used in a home workshop. Just the way I would do it.

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