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240VAC single phase from three-phase?

Discussion in 'Home Power and Microgeneration' started by Ulysses, Mar 29, 2010.

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  1. Ulysses

    Ulysses Guest

    Hi. I've been googling and read a lot about three phase power but didn't
    find the answer to my specific question...

    I have a vacuum-heat press that operates on 240 VAC single phase. The plug
    is a 20 amp twistlock so I'm assuming it draws less than 20 amps at that
    voltage. The vacuum motor runs from 120 volts and is included in the
    circuitry. I can't seem to find the wiring diagram so I'm guessing that
    either one leg of the 240 supplies the pump and the other supplies the the
    heating elements OR one leg supplies the pump and one heating element and
    the other leg only one heating element (this seems unlikely to me).

    So, I have three-phase power in my store and the voltage across any two hot
    lines is about 209. BUT, the voltage between any hot line and ground or
    neutral is 120 volts. SO, since I'm not going to be running any 240 volt
    motors etc, and the pump is running from 120 volts anyway, can I simply
    connect the vacuum-heat press to two legs of the three-phase plus neutral
    and ground? As far as I can figure I will be supplying the pump and heating
    elements individually with 120 anyway. Unless for some reason the heating
    elements require 240 but that seems like it would be an unbalanced load with
    the pump drawing only 120 from one of the lines.

    The wiring diagram is somewhere and I'll consult it before doing anyway. If
    this won't work what are my options?
     
  2. Gordon

    Gordon Guest

    Most likely the heating elements are wired across the 240 volt legs
    and the vacuume pump goes from one leg to neutral. Sure it's
    imballanced,
    but that is not a problem.
    The reason that two legs of a 3-phase supply read 208 volts is because
    they are 120 degrees out of phase. A single phase 240 volt (residential)
    secondary. The two legs are 180 degrees apart.

    So on a 3 phase supply, one leg will rise up to 120volts, but the other
    leg will not be at -120 volts. It will be at about -88volts (if i did
    the head math correctly).

    Any way, hooking that machine up the 3 phase will cause the heating
    elements to run cooler. that may or may not be a problem.
     
  3. misc note: That "120 degrees out of phase" is only coincidentally
    the same number as 120 Volts.

    Misc note #2: in some areas of the country, such as NYC, you
    don't get that "single phase 240 volt", but rather, you're likely
    to get the 208 volt deal.

    The way the wiring works is that you have four wires from
    the street. You've got "hot legs" of "A", "B", and "C", each
    of which are 120 degrees apart. You also have a neutral.

    If you tap from any of the legs to ground (neutral) you'll
    get 120VAC.

    If you tap hot-to-hot, you'll get the equivalent of 208V.

    In a typical setting the first apartment has legs "A" and "B"
    (plus neutral and a safety ground). The next one has "B"
    and "C". The third has "A" and "C". Rinse lather, repeat.

    All three legs go to the roof to feed the elevator motor
    and the central AC chiller.
    Note that it's NOT a mere 14 percent difference (208 vs. 240),
    but is in the case of resistance heaters, about a 25 percent
    reduction. That's because with the voltage dropping,
    you get a corresponding reduction in amperage...

    So you've got roughly 75 percent of the heating capability.

    If you've got, say, a typical kitchen oven, it simply means
    it'll take longer to heat up. But oce it does, everything
    will be fine.

    But if you're trying to boil water on the top heater, the lower
    voltage will take about 33 percent longer...
     
  4. Ulysses

    Ulysses Guest

    I found the installation instructions (finally) but there is no wiring
    diagram. You seem to be sure about how the heating elements are wired so
    perhaps you are familiar with the type of press I'm talking about. It turns
    out the 208 volts is "acceptable" as per the instructions so that should not
    be a problem. I guess my question was not clear (or too many questions at
    once) but IF each heat element was connected to a different leg and neutral
    THEN each would be supplied with 120 volts and the 208 volts would be
    irrelevant. Is this correct? The only three-phase I ever worked on before
    was with small alternators and I always used all three legs and was only
    concerned with the total output. In any case I've worked with 240 volts so
    I'm sure I can do the wiring without any major mishaps. The vacuum press
    has controls for setting the desired temperature and shows the present
    temperature so it should work, just take a little longer to heat up as was
    mentioned. There is a card included with the instructions that tells what
    the voltage readings should be at an accessory outlet (120 volt) on the
    front of the machine so that should tell me if everything is wired
    correctly. I does, however, leave out one hot lead..... The hot leads
    that I indend to use are each connected to 20 amp breakers and nothing else
    is being used on those circuits so I think I'm good-to-go. Thanks everyone
    for the help :-D
    That's what I have.
    As of this moment I have no doubts, but if any arise I'll get the
    electrician.
     
  5. amdx

    amdx Guest

    Your math is ahh, hmm, depends how you think about it :)
    You almost need to think in peak and peak to peak terms when you
    are discussing 3 phase and measuring phase to phase.
    When one phase peaks it is 169.68 volts above ground, (1.414 x 120V)
    The other leg must be -124.43V, because 124.43 + 169.68 = 294.11 Vpp.
    And 294.11V x .707 = 208 V
    And yes you're correct 88V x 1.414 = 124.43Vp
    What I'm a bit puzzled by is I can't find a proper graph.
    The negative leg is never at the correct point to show what I have said.

    What should be 208 on this graph calculates out to 192Vrms (assumes 0.6 on
    the graph)
    http://upload.wikimedia.org/wikiped...ase-voltage.svg/575px-3-phase-voltage.svg.png

    Ands this one shows 240Vrms (assumes 70.7 on this graph) note labels change
    from page to page.
    http://www.automotive-res.com/EX/11-14-02/C0190-Figure1.gif

    I think both graphs are wrong, ie, poor representation of a sinewave.
    The line is in the wrong position at 120 degrees.
    Mike
     
  6. Ulysses

    Ulysses Guest

    Well, I didn't *think* I was presuming, maybe just hoping. If I understand
    it correctly then IF each heating element was powered by a different leg of
    the three-phase supply then there would be no loss of power being supplied
    to the elements whether they were connected to single phase or three-phase.
    I'm not trying to be difficult, I'm just trying to understand it completely.
    The total voltage of my three-phase power that reads 208 volts between two
    legs would be 360 volts, right (208*sq/rt of 3)? That would make each leg
    produce 120 volts between hot and ground, right? With single phase each leg
    would also be 120 volts between hot and ground. From what everyone has said
    it is common for such heating elements to be connected to 208 or 240 (or
    230) volts so I'm just trying to understand why. It seems to me it would be
    more versatile if the elements were powered by 120.

    The motor has a 120 volt type power cord that appears to have Hot, N, and G.
    I did not read the label on the motor to see if it could be wired for 240
    but I'll have to do that. The motor plugs into it's own reserved receptacle
    at the rear of the machine. I will test the voltage at that outlet once I
    get the thing connected but since the power cord has a plug exactly like a
    desk-top computer I'm *assuming* it's 120 volts. I mean, they wouldn't use
    the same type of power cord for 240 would they and be running two hots and
    ground through it?
    Yes, and thank you.
     
  7. amdx

    amdx Guest

    I was responding to Gordon where he said,
    "So on a 3 phase supply, one leg will rise up to 120volts, but
    the other leg will not be at -120 volts. It will be at about -88volts"
    This is true but only at one instant in time, and that would NOT be at
    a peak.
    A little study of rms vs peak vs peak to peak will show you where
    my numbers came from.
    I thought a graph showing the voltage over time would be a good way
    to show the voltages. I was disappointed that I didn't find an accurate
    graph. None have the proper voltage, when line A peaks positive, line B
    should be at -124V on the graph. Any graph I have found is either to high
    or to low. This just a poor drawing of a sinewave.
    Yes it is 208 (Vrms)
    It would seem to me they would be 180 degrees out of phase.
    Mike
     
  8. amdx

    amdx Guest

    Snip
    Snip 12 volt battery data.

    If you want to talk DC polarity and (phase) note one end of your two
    batteries is
    '+' of one and the '-' or opposite polarity (phase)

    I tried to post graphics from the page, they didn't post so please go
    to the URL and page down 3/4 page.
    Here is a website that describes the phase relationship of a typical
    120/240V system showing the 180 degree phase.
    http://www.allaboutcircuits.com/vol_2/chpt_10/1.html

    Also so see the second graph about half way down on this page, and
    description.
    http://www.tpub.com/content/construction/14027/css/14027_75.htm

    Mike
     
  9. Nope, you got it backwards... If they are 0 Degrees apart the meter
    would read Zero Volts, HOWEVER if they are 180 degrees apart the Meter
    would read 240 Vac. Draw it out, and you can see the Phase relationship.
     
  10. amdx

    amdx Guest

    Thanks Bruce,

    Things get dificult to explain when the depend on a reference point.

    This website shows a graph showing the phase of the two lines 1808 out of
    phase.
    See the second graph about half way down on this page, and
    description.
    http://www.tpub.com/content/construction/14027/css/14027_75.htm
    Mike
    PS. I still have interest in a graph that shows 3 phase ac where I can
    accurately extraopolate 208 V.
     
  11. Bob F

    Bob F Guest

    Really???
     
  12. Bob F

    Bob F Guest

    As I see it, if the voltage polarities were the same - 0 degrees apart, the
    neutral would carry the sum of the load currents. If they are 180 degrees apart,
    the neutral would carry 0 current.
     
  13. Yea, Mike's thinking is kind of "Out there" electrically speaking. You
    don't compare DC and AC when trying to explain AC and Power & Phase
    Relationships. As another commenter pointed out Vectors don't work in AC
    very well, because they lack the Time Element, and that time element is
    a critical part of AC evaluations. You need to know just where in the
    cycle you are talking about, when looking at instantaneous Phase and
    Power with the vector analysis.

    I still remember the quiz in EE 102 that dealt with all this... and I
    got 95%, but not without some heavy bookwork, the night before and lots
    of scrap paper.... back in my college days... many decades ago....
     
  14. amdx

    amdx Guest

    I think all our difference with mike boils down to the reference point and
    maybe
    some semantics. I kinda gave up on the argument.
    I did get some understanding on the 3 phase power problem I had.
    I was looking at the wrong point in the waveform, I was 30 degrees of peak.
    So using .707 x peak didn't work. Now that I have found the real peak and a
    nice graph it all came together.
    http://forums.mikeholt.com/showthread.php?t=121543
    Note: the dotted line is the voltage that results from adding the voltages
    from two of
    120 volt lines.
    Mike (amdx)
     
  15. amdx

    amdx Guest

    I'll give it my best shot.
    I'll define the system 240v, 2 wires, no neutral. let me know if that is a
    problem.
    We'll start with your scenario as below.We'll start with a sinewave at 0 degrees and 0 Volts. Both wires, since they
    are in phase.
    As the sinewave starts to rise and gets to 30 degrees the voltage is 147
    volts,
    at 60 degrees the voltage has reached 260 volts and then at 90 degrees, 294
    volts.
    Now lets get out the oscilloscope, one lead on each wire, at 30 degrees the
    voltage on
    Line 1 has reached 85 volts and the voltage on Line 2 has reached 85 volts
    also, because they're in phase
    So the oscilloscope measures the potential difference between L1 and L2 and
    gets 0 volts.
    This continues through the whole cycle, since the are in phase there is no
    potential difference.
    If they are in phase, how can there be any potential difference to do work.
    *If they are out of phase the real question is, at 30 degrees, does L1 go
    negative 42.5 volts and
    L2 go positive 42.5 volts. :)
    Mike (amdx)
    *Depends on the reference point.
     
  16. amdx

    amdx Guest

  17. Ulysses

    Ulysses Guest

    I probably shouldn't say this but it muddies it up for me. Without seeing
    the sine waves and seeing the phases I can't tell if they are in-phase or
    out-of-phase. And aside from all of that I thought you got 240 from the two
    120 volt coils if they were OUT of phase with each other, not IN phase.
    Perhaps I'm just not understanding what IN and OUT of phase means. I
    thought OUT of phase meant that when one sine wave is peaking at the top the
    other is at zero. Is this wrong? I have rewired generator heads many times
    in order to either get 240 volts OR 120 volts in parallel (to balance the
    load on the coils) and, in my mind (which can be a frightening place) the
    sine waves were going up and down at the same time so 120 was all I was
    going to be able to get. Kinda like connecting two batteries in parallel.
     
  18. amdx

    amdx Guest

    Hi Ulysses,
    Out of phase could be only 1 degree out of phase. What you described,

    The could be 90 degrees or 270 degrees out of phase, depends whether the
    zero line is headed pos or neg.
    180 degrees out of phase would be if L1 is a the positive peak
    and L2 is at the negative peak.
    I added some more info to support my assertion at 8:59 PM and
    9:37 pm , 04-02-10.
    See if it helps or if you disagree.
    Mike (amdx)
     
  19. amdx

    amdx Guest

    Hey guys I got this wrong, The peak of the 120 volt line is 169.68 volts,
    however,
    when you use two lines to get 208Vrms the 294Vpeak is 30 degrees from the
    120Vrms peak. (peaks at 169.68)
    At that point in the waveform one 120V line has reached +146.97 and the
    other
    is at -146.97 for a difference of 293.94Vpeak. This equals 208Vrms.
    Here is a very nice graph to show the relationship between the 120V 3 ph
    and the
    208 V line.
    http://i145.photobucket.com/albums/r...3waveforms.gif
    Mike (amdx)
     
  20. amdx

    amdx Guest

    That was a bad URL, this ones correct.
    http://i145.photobucket.com/albums/r204/Smart_S/3waveforms.gif
    Mike
     
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