Bad voltage imbalance.
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  1. #1
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    Default Bad voltage imbalance.

    I built a 10 hp ROTARY PHASE CONVERTER and tested it out with a old 7.5 Lincoln motor as an idler. The pas through legs were 247 volts and the generated leg was within 5%. I sold just the box only and the new owner hooked it up with a 7.5 hp as an idler. His pass thru legs are 255 volts and L1 to L3 is 295 volts. What the heck could be wrong? He is trying to run a binding machine and it trips the breaker on the machine. Two other of the same boxes tun other people’s equipment just fine. cfe5485d-ad4d-4b45-a1e7-daf6e4c76e62.jpgcfe5485d-ad4d-4b45-a1e7-daf6e4c76e62.jpg

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    Old motors are higher fla amps, higher core losses and lower power factor.

    If he has a high efficiency new motor it will need less capacitance.

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    Your motor could be wired star and his motor delta...Phil

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    I have read that putting capacitance between the pass thru legs will lower amperage draw as well as correcting power factor. Any thoughts?

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    The 255 line-to-line voltage seems very high.

    The manufactured leg may be very high based on the large amount of balancing capacitance present. If this
    were mine I'd probably remove them all to start, and re-measure.

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    2nd this approach.

    Start with just enough start cap to get things turning.

    Work from there under load conditions.

    The more motors in line , the better.

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    This is what I sent to the person having problems.

    Another comment from the forum "" Default
    “”””The 255 line-to-line voltage is very high.
    If your electric provider is supplying 255 volts that would be disastrous to others running 208 volt motors. Like 23% high. Almost 50 volts. I would ask them to tap down a bit.””””

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    Another tip is:

    After getting the service voltage in range........

    Measure leg to leg AMPS rather than voltage.

    Amps are much more meaningful as they do all the work ;-)

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    Very good point. Can you expand. It seems the amperage in other builds has always been much lower than on pass through legs?? How do you balance amperage with run capacitors. What does putting capacitors between L1 and L2 do?

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    Amps is what runs the motor. Voltage is just the "push" behind the amps that causes the current to flow. So if the current in all three "legs" is equal, the motor should be well-balanced and capable of full power.

    That said, if the 3 phase is the same voltage on all legs, and at correct 120 deg, then the amps will normally be correctly balanced, so there is a lot of sense in getting the voltage correct "just like powerco 3 phase".

    Yes, it is common for the manufactured leg amps to be low.

    To understand why, you have to remember what drives the current in the motor.

    It's voltage, of course, but WHAT voltage?

    The motor, when running, creates a voltage that opposes current flow, it is a "reverse" voltage that opposes the applied voltage ("back EMF"). It may be 85 to 95 % of the mains voltage, which leaves only a relatively small voltage left to actually "push current through". The voltage also changes with motor RPM in a somewhat complex way (because it is affected by "slip").

    What that means is that a small overall voltage change can make a lot of difference in current. If the back EMF is 90% of mains voltage, then a 5% voltage drop due to resistance in the idler motor could make a 50% drop in current on the generated leg, because that 5% is half the difference between back EMF and mains voltage.

    The generated leg starts out being lower than mains voltage because it is the same voltage as the "back EMF". So you start behind the eight ball. The "balance capacitors" are used to cancel some motor impedance and boost voltage a bit. However, they need to be split between the pass-through wires so that they do not create a phase change that can also affect current*.

    And, as explained, a small difference in voltage can make a big difference in current.

    Also, when using balance capacitors, a phase change can also make a fairly large difference in current.

    * the back EMF itself will not have any phase issues, it is set by the way the motor is mechanically constructed. But once capacitors are involved, they can interact with motor inductance to cause a change in phase.

    The way they work is similar to a resonant circuit, although the capacitors are not used to actually resonate the circuit, they just get "close" to a degree that raises the voltage just enough. However, the phase changes rapidly as you approach resonance, and that can cause issues with current even though the voltage looks correct. Nobody ever measures phase when setting up the RPC, but maybe they should.

    .

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    WOW That’s a lot of good knowledge. Thank you so much.

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    So I have read that placing capacitance between the pass through phases will reduce current. How the heck does that work? Does it correct power factor.

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    Quote Originally Posted by Don Gitzel View Post
    So I have read that placing capacitance between the pass through phases will reduce current. How the heck does that work? Does it correct power factor.

    yes, that's what it does.

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    Are you talking about the current on the generated phase while under load. Or current on the pass through phases under load. If so how does capacitance limit current. Seems to me the load would dictate current draw. Just asking.

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    Quote Originally Posted by Don Gitzel View Post
    Are you talking about the current on the generated phase while under load. Or current on the pass through phases under load. If so how does capacitance limit current. Seems to me the load would dictate current draw. Just asking.

    It actually corrects PF on either, depending on where put. If put on the pass-through, per your original question, it corrects the PF "looking into" the RPC. The "balance" capacitors correct the "outgoing" PF on the generated leg, if they are used, and that tends to raise the output voltage, which is why they are used.

    But... short story is that capacitors draw current that is in advance of voltage, and inductors are opposite, they draw current that lags behind the voltage. Those leading and lagging currents are called "reactive" currents. They are actual measurable currents, but do not contribute to producing any power.

    So, since they are opposite, one can counteract the other; the lagging current combined with the advance or "leading" current, adds up to be zero, which leaves only the "in phase" current.

    When "looking into" the RPC, the motor is pulling a fair bit of lagging current, plus some "power current" that is in-phase. The capacitors pull "leading" current. Looking from the mains input, the "average" of those two is zero, which leaves the actual "power" current only.

    The size of capacitor is sized to the inductance such that it mostly counteracts the inductance, leaving only the "resistive", or "in-phase" current, which is what produces power, and is required. Usually the correction is made only 85% or 90% so there is always a little of the "reactive" current left, to avoid over-correcting.

    So it is actually a genuine net reduction of total current.

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    Quote Originally Posted by Don Gitzel View Post
    Are you talking about the current on the generated phase while under load. Or current on the pass through phases under load. If so how does capacitance limit current. Seems to me the load would dictate current draw. Just asking.

    On my 7.5 HP RPC, I was told to put the power factor capacitor between L1 and L! (pass through) . It dropped the current considerably. I don't recall the value but it wasn't very large.

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    Quote Originally Posted by Newman109 View Post
    On my 7.5 HP RPC, I was told to put the power factor capacitor between L1 and L! (pass through) . It dropped the current considerably. I don't recall the value but it wasn't very large.

    The drop may not be large.

    When there is no load, 80% or more of the low current may be reactive current (useless current). But under load, the reactive current does not change much, however the "power" current increases, so in that condition, 60% to 80% of the current can be "power" current. You would be eliminating only part of the 20% or 40% that remains.

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    Wow again. That is so much useful info. Thank you.


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