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Phase to phase wiring question

  1. Dec 21, 2008 #1
    I am servicing a machine that uses a 3 phase power supply. Some of the components (motor, compressor, fan) use 3 phase power. Other components (solenoids) use single phase (phase to phase) tapped from L1 and L3. I understand regular single phase-120v and a nuetral. The nuetral creates the potential for the current to flow. How does the two hots, in this case (L1 and L3) create a potential without using a nuetral wire? If a soleniod is using two hots out of phase, how is that single phase and how does the current flow? How is there a potential created between the two phases?

    Thanks for any help.
     
  2. jcsd
  3. Dec 21, 2008 #2
    You derive a neutral by having a transformer with a 'star' (wye) secondary.


    http://ecmweb.com/mag/electric_understanding_basics_wye/


    The voltage between phases is sq rt 3 times the phase to neutral because the phases are 120 deg out with one another and the resulant sine is the vector sum.

    So the phase to phase voltage is 208 Volts.
     
  4. Dec 22, 2008 #3
    I know all that but how do you get current to flow between two phases? What is the potential?
     
  5. Dec 22, 2008 #4

    stewartcs

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    There is a potential difference since they are out of phase. Take a look here for a graphic example.

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/hsehld.html

    In this household type of circuit (i.e. 180 electrical degrees out of phase), when one leg is at a positive peak potential, the other leg is at a negative peak potential. Hence the peak-to-peak voltage is the vector sum of the two. Since both are connected in the same circuit current flows.

    If you are only using two legs of a three-phase system, the voltage (and current) will be 120 electrical degrees out of phase as stated above.

    CS
     
  6. Dec 22, 2008 #5
    Thanks, that clears it up a bit. But how do you get 240 from that? I can see how L1 is finding its potential when L2 goes negative. when the top of the wave at L1 is at peak value, is that 120? if so, how can the total voltage be 240? If the bottom of the wave at L2 is only a potential, it would seem to me that your only getting 120 because the bottom half cancles out the other 120

    . Thanks again for your help.
     
  7. Dec 22, 2008 #6

    russ_watters

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    If L1 is at 120, then L2 is at -120. 120 - (-120) = 240
     
  8. Dec 22, 2008 #7
    Ok that makes sense. Now what about this:

    When I measure L1 to ground I get 220 voltage when I measure L2 to ground I get 0 and when I measure L3 to ground I get 200. How can L2 to ground be 0 if its 3 phases? Im in Japan by the way. Thanks for any help.
     
  9. Dec 22, 2008 #8

    russ_watters

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    Sounds like you dropped a phase. The other two should be better balanced, too. Can you check at the panel to make sure you aren't missing something?
     
  10. Dec 22, 2008 #9
    No. The japanese verified it as well. L2 is a nuetral, grounded at the transformer in a delta configuration. I dont understand it, how L2 can be nuetral in a 3 phase system. Also,if I dropped a phase, the 3 phase motor and compressor would be single phasing and making troubles.

    Thanks for any help.
     
  11. Dec 23, 2008 #10

    stewartcs

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    Sounds like you have a corner-grounded delta connection. This will not affect the way a three-phase system works. The three phases are still 120 electrical degrees apart. Hence, you have 0 volts between ground and ground (i.e. ground and L2) and your equipment still works. Check the phase-to-phase voltage to confirm.

    Hope this helps.

    CS
     
  12. Dec 24, 2008 #11
    Sounds like Ground is being tapped off of L2.

    L2 is not at 0 electrical potential..... The dilema here is that you are mesuring the difference between the same leg....think about it.....L2 is zero in regards to what?....("Itself")
    In other words you are just measuring L2/GRND in contrast to L2/GRND.

    Corner-grounded delta connection configurations are typically for "ungrounded" systems.....
    Thats why sparks arn't flying.

    So L2 is a normal phase....its just being used as a ground or "return" path to compensate your ungrounded 3 phase system.

    Hope this helps.
     
  13. Dec 24, 2008 #12

    stewartcs

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    That's how corner-grounded delta connections work. The corner is grounded at one leg.

    L2 is at 0 electrical potential when he references it to ground (see post #7 above). In a corner-grounded delta connection 0 volts would be read at the corner ground with respect to ground since it is grounded. In other words there is no potential difference between a leg that is grounded and ground (actually there is a very very small difference but it is normally neglected).

    Corner-grounded delta connections can be used for both ungrounded an grounded systems. Since transformers electrically decouple a circuit, grounds from the rest of the system are not passed along. There just can't be two grounds in the transformer (i.e. no center tap ground for lighting circuits in addition to the corner ground - only one or the other).

    Grounds are not used as return paths for various reasons.

    CS
     
  14. Dec 24, 2008 #13
    Please explain what I convoluted.
     
  15. Dec 24, 2008 #14

    berkeman

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    I will send you a personal message (PM) about why your post was deleted. It violated PF forum guidelines. You have already received a PM from the initial deletion.
     
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