Phase converters and their three-phase voltages/currens

[tamtam402]

Let’s assume I use an ideal rotary phase converter to power a three-phase 10HP motor (100% efficient, PF = 1). Thus, I need S = 10 VA.

On a 120V single-phase line, I would get three-phase line-line voltages of 240V. Here’s my question. I was under the assumption that when you analyze motors, you find the single-phase power with the product of V(line-neutral) X I (one-phase current on the line). For the 10 VA situation described above, you’d want the sum of the three VI products I just described to be 10 VA. I was under the impression that a balanced system would have 3 phases, with each phase carrying (sorry if this isn’t the right word, english isn’t my primary language) 3.33 VA.


Question 1.

Here’s my problem: In the situation I described, two of the line-neutral are 120V, but the other line-neutral voltage is 208V! (read on phase converters if you do not understand why this is so) However, I know rotary phase converters are considered to be perfectly balanced when the currents of the three phases are the same. This is where my brain melts. Assuming two line-neutral voltages of 120V and a line-neutral voltage of 208V, with three perfectly-balanced currents on the lines, does that mean two of the lines carry a certain power, while the 3rd line (the 208V) carries more power?

Question 2.

I also have the following question: Assuming I have an ideal phase converter, I can easily find the single-phase line-neutral voltage from either the three-phase line-line voltage, or from the line-neutral voltages of any of the three lines (assuming I know which line is the 208V “wild leg”). However, how can I find the single-phase current (the one entering the phase converter) if I know the currents in the three phases? Will it simply be a sum of all three currents?

Thanks in advance!

 

[tamtam402]

Oh ya, I have one more question. I know real phase converters aren’t 100% efficient. Is the power lost in form of current or voltage? I would assume the power is lost in the current but I just want to make sure. Is it the same case for solid state phase converters?

 

[Windadct]

Hello Tam ... huh?

Single Phase would be 120-0-120 -- Totaling 240. ( Like standard household service here in the USA. The 240 is not part of a 3 phase arrangement - it is the 2 x 120 basically added, even as vectors.)

Single Phase the VA = V * I * PF in three phase the VA = Sqrt(3) * V (Line to Line) * I leg * PF.

10 HP = 7,546 W if PF = 1 then 7,546 VA... I so not know what "S" means or where you get 10VA.

Typically Phase Converters are rotary ( a motor and a generator) or Solid State ( like a UPS or motor drive).

They take A single phase and create 3 phase. 3 phase 120, is 120V Phase to Neutral and 208 Phase to Phase. ( You pretty much have to see the vectors for this to make sense - try this: http://www.askmehelpdesk.com/attach...-volt-120-208-volt-system-wiringdia3pwye1.jpg - the layout of the windings are effectively vectors.

So your question 1 makes no sense - in the phase converter the input single phase is 120-0-120 and output ( separate electrical circuit) is 3PH 120/208.

Question 2 : AN Ideal phase converter you have 3 x 120 Ph to Neutral and each phase to the other is 208.

Lastly - typically these are created so the output voltages are stable - the losses indicate then that the input current is more than the output current. Also - typically these devices are rated by OUTPUT - so a 7500W converter may consume more than 7500W - that will be defined on the specification sheet.

I think you are off track in your understanding of single phase vs three phase power.

 

[tamtam402]

Ya sorry I messed up the HP/W conversion, I was tired haha. Yes, an ideal converter would have the same line-neutral voltage on all it's legs, but that is not the case with rotary phase converters. Even digital converters have a high leg.

Google "Phase converter wild leg".

I know the 240V is single-phase. I said you'd get 240v line-line when using an ideal phase converter if your imput is taken from a 120V single-phase line...

My question still hold: Since the line-neutral voltages aren't the same even on a "perfect" rotary phase converter, am I right to assume the wild leg would contribute more power to the load (motor?). Usually one tries to balance the currents, but the line-neutral voltages cannot be balanced because of the wild leg.

 

[jim hardy]

but that is not the case with rotary phase converters. Even digital converters have a high leg.

You have not studied enough of them to make that statement.
It's not true and you have generalized from studying just one or two of them.
Time to get your thinking straight.

If you have three windings each with some voltage you can arrange them any way you want.
Some three phase systems indeed have a wild leg, some do not.
Rotary converters come in several flavors.
Simplest is just a three phase motor run single phase, the rotor's magnetic field will make a sloppy approximation of the third phase.
Best is a three phase generator on same shaft(or belted to) another motor which is perhaps single phase or different frequency. That generator can be earthed any way local electrical codes permit.

Google "IEEE Green Book", grounding practices.

 

[Windadct]

OK I think I am seeing the issue - in a 3 phase converter set up for universal household or farm use - they may make the output Delta and ground the middle of one of the legs making it a neutral. The "wild leg" - much to casual description - IMO. Is the phase opposite the ground point, this leg is only high relative to ground, even the term neutral here requires notation as the neutral applies to the 120/240 leg and the 208 V - but not for the 3PH 240. EG - your 3PH 10HP motor would see balanced 3Ph 240 - no high leg, as long as there is no other loads on the converter. Any other loads will cause a voltage imbalance - and degrade the performance of the motor.

This is center tapped delta arrangement. So you still have a 120/240 and 3PH 240.

If you are ONLY interested the 3 phase to be able to run 1 motor - I would set up a converter ( or motor drive) just to run that motor, and use the existing 120/240 for all other loads. KISS