10 hp inverter

[synchro]

Wow! This is interestin
My motors like higher voltages because they are from Europe. 230V is fine.
I oversized the transformer already.
Should I go this way?

It's up to you. I'd be interested in hearing what others have to say about it.

 

[synchro]

I've had no dealings with this firm but it looks interesting. It has a 4.4% impedance vs. the nominal 6% of the ACME DTGA0202S 20 kVA unit.

https://www.ebay.com/itm/372062395803

 

[synchro]

I just thought of a problem that could arise if using a wye-delta configuration for generating the 3rd phase. There would be zero sequence flux in the transformer core, and that could pose a problem especially for 3-legged cores because there's not a good path for zero sequence flux within the core structure. It's likely that most of the commonly available delta-wye's for the sizes of interest have three legged cores.
You could build a wye-delta from individual single phase transformers, but the two transformer version would be simpler if you can get suitable 120V/120V units. Three transformers would allow you to have one of the windings on the transfomer be a different voltage than 120V, such as 240V, 277V, 480V, etc.

 

[kwired]

Yes, you can use two 120V/120V single phase transformers like gar mentioned in post #6 to synthesize the third phase. I showed a connection of such transformers in posts 8 and 10 of this thread:
https://forums.mikeholt.com/threads/1-ph-to-3-ph-converter.2546600/#post-2546705

Like I said above you can also use a delta-wye. The voltage rating of the delta side is not important because it would not be externally connected.
In the drawing below windings 1, 2, and 3 of the wye are each coupled magnetically to windings 1, 2, and 3 of the delta, respectively.
You would connect your L1, L2, N to terminals A, B, and N of of the wye shown below. Windings 1 and 2 of the wye will therefore provide power to windings 1 and 2 of the delta (equivalently forming an "open delta" which then provides power to winding 3 of the delta). Delta winding 3 then powers winding 3 of the wye to produce the desired third phase C.

View attachment 2557786

Taps on the delta could be selected to achieve fine adjustments of the voltage on phase C.

Or just use two 120 x 120/240 transformers and connect primary A to N and B to N and the load to the open delta secondary. If no 120 volt loads being supplied can corner ground it and just have 240 three wire delta for output.

You need to keep in mind if both transformers are 1000VA you only get 2000 VA output capacity and not 3000VA like you would get from a full delta made of 1000 VA units.

 

[synchro]

Or just use two 120 x 120/240 transformers and connect primary A to N and B to N and the load to the open delta secondary. If no 120 volt loads being supplied can corner ground it and just have 240 three wire delta for output.

You need to keep in mind if both transformers are 1000VA you only get 2000 VA output capacity and not 3000VA like you would get from a full delta made of 1000 VA units.

On an open delta with a 3-phase load, the current through the windings will be at a 30° angle to the voltage across them and therefore have a power factor of cos(30°) ≈ 0.866. Hence two 1000 VA transformers will provide a 0.866 x 2 x 1000 VA ≈ 1730 VA output capability.

 

[synchro]

I just thought of a problem that could arise if using a wye-delta configuration for generating the 3rd phase. There could be zero sequence flux in the transformer core, and that could pose a problem especially for 3-legged cores because there's not a good path for zero sequence flux within the core structure. It's likely that most of the commonly available delta-wye's for the sizes of interest have three legged cores.

After further consideration, I think that zero sequence flux should not be an issue if there's a delta winding present on the transformer. The delta winding will cause the 3rd phase output from X3 to be such that the zero-sequence voltage on the three wye terminals is very small. A circulating current in the delta winding would be produced that cancels out zero sequence flux.
And so I think a delta-wye transformer is still in the game as a way to create a 3rd phase.

 

[gar]

210913-0910 EDT

In our area primary distribution is either wye or floating delta ( probably grounded at the substation ). At my home it is the three wire delta, and at my son's shop it is four wire wye. However, at the shop only three wires are the supply, L1, L2,and N. Extremely common in our area. The pole has two transformers with the secondaries connected as an open delta with a wild leg. This is probably a lower cost approach. If more three power is required at some later time, then they just add a third transformer.

Relative to the original post --- is a 3 phase neutral required by the load?

.

 

[MD Automation]

And so I think a delta-wye transformer is still in the game as a way to create a 3rd phase.

I have a little time this week and enough equipment in the shop to experiment with creating the 3rd leg with a Delta-Wye transformer (using 2 of the 3 legs of our 3 phase 208Y/120 POCO supply). I can post up results if any here are interested.

 

[synchro]

I have a little time this week and enough equipment in the shop to experiment with creating the 3rd leg with a Delta-Wye transformer (using 2 of the 3 legs of our 3 phase 208Y/120 POCO supply). I can post up results if any here are interested.

One thing to consider is that I believe methods using transformers to create a 3rd phase from an "open wye" L1, L2, N will typically draw about 3 times the current on the neutral as is drawn from the 3rd phase.
This is how I see it: The currents drawn from lines L1 and L2 by the transformer(s) will each be equal to the load current IL drawn from the transformer output L3. These two currents will add in phase on the neutral and result in 2 x IL through the neutral. Then because L3 is an output instead of an input to the transformer(s), the polarity of the load current IL drawn from L3 will cause it to add to the neutral current and result in a total of 3 x IL.

A simplified explanation for this is the following: When the 3rd phase L3 is at its peak voltage, the two phases L1 and L2 that provide power to the transformer(s) are at cos(60°) = 0.5 of their peak voltage (and of opposite polarity to L1 and L2). Therefore to provide the kW drawn by load current IL on L3 there must be a total of 2 x IL on the transformer inputs L1 and L2.
It should be noted that although the voltages across the transformer windings will have different phase angles as required to achieve the phase angle of the L3-N voltage, the currents through the windings will be in-phase with each other (or at 180° which is just a polarity reversal).

I believe the above to be the case. I may consider running a Spice simulation to verify this if I get around to it.
So if you experiment I would try a smaller 3-phase load and monitor the currents through the different conductors as well as the voltages across them.

 

[synchro]

A simplified explanation for this is the following: When the 3rd phase L3 is at its peak voltage, the two phases L1 and L2 that provide power to the transformer(s) are at cos(60°) = 0.5 of their peak voltage (and of opposite polarity to L1 and L2).

The should have been: "A simplified explanation for this is the following: When the 3rd phase L3 is at its peak voltage, the two phases L1 and L2 that provide power to the transformer(s) are at cos(60°) = 0.5 of their peak voltage (and of opposite polarity to L3 )."

 

[ptonsparky]

I have a little time this week and enough equipment in the shop to experiment with creating the 3rd leg with a Delta-Wye transformer (using 2 of the 3 legs of our 3 phase 208Y/120 POCO supply). I can post up results if any here are interested.

Yes, post. Thank you.

 

[kwired]

One thing to consider is that I believe methods using transformers to create a 3rd phase from an "open wye" L1, L2, N will typically draw about 3 times the current on the neutral as is drawn from the 3rd phase.
This is how I see it: The currents drawn from lines L1 and L2 by the transformer(s) will each be equal to the load current IL drawn from the transformer output L3. These two currents will add in phase on the neutral and result in 2 x IL through the neutral. Then because L3 is an output instead of an input to the transformer(s), the polarity of the load current IL drawn from L3 will cause it to add to the neutral current and result in a total of 3 x IL.

A simplified explanation for this is the following: When the 3rd phase L3 is at its peak voltage, the two phases L1 and L2 that provide power to the transformer(s) are at cos(60°) = 0.5 of their peak voltage (and of opposite polarity to L1 and L2). Therefore to provide the kW drawn by load current IL on L3 there must be a total of 2 x IL on the transformer inputs L1 and L2.
It should be noted that although the voltages across the transformer windings will have different phase angles as required to achieve the phase angle of the L3-N voltage, the currents through the windings will be in-phase with each other (or at 180° which is just a polarity reversal).

I believe the above to be the case. I may consider running a Spice simulation to verify this if I get around to it.
So if you experiment I would try a smaller 3-phase load and monitor the currents through the different conductors as well as the voltages across them.

But looking at just the primary circuit how do you increase current on the neutral without increasing current on one or both the other lines? This circuit is independent of the secondary and the two are only associated by magnetic coupling.

 

[synchro]

But looking at just the primary circuit how do you increase current on the neutral without increasing current on one or both the other lines? This circuit is independent of the secondary and the two are only associated by magnetic coupling.

If we are talking about using a delta-wye transformer in a wye-delta configuration, then the delta secondary with nothing connected to it makes any current within the secondary the same on all of its windings (i.e., the only currents present are circulating currents). Therefore the three L-N currents on the wye will be equal to each other. The currents through L1-N and L2-N of the wye come directly from the 3-wire feeder. The current through L3-N is returning from the 3-phase load that's also connected to L1 and L2 of the feeder, and therefore the L3-N current is part of the total current drawn from L1 and L2 of the feeder.

In a sense the wye-delta is acting in the same way as a grounding transformer by forcing equal currents with the same phase on each of the L-N windings in the wye. But in this case L3 is connected to the load and not to the source of power, and so the load determines the amount of current in all of the L-N windings.

I think another way to look at this is to consider how a T transformer configuration could create a 3rd phase from L1, L2, and N of an open-wye. The main winding would be across L1 and L2, and its center tap connected to a teaser winding which provides the created phase L3 at its other end. The neutral would be connected to a tap on the teaser that's 1/3 of the way up toward L3. And so the teaser would have 60V between its tap and the main winding's center tap, and 120V between its tap and L3.
In this case the teaser acts as an autotransformer boosting the 60V supplied between the main winding center tap and the neutral conductor up to 60V + 120V = 180V across the entire teaser winding.

I'm not suggesting to use a T configuration but I'm mentioning it because it shows more clearly how the neutral of an open wye would have a significant current when a third phase is produced from L1, L2, and N. I think it ultimately boils down to the fact that in a 120V/208V open wye, the component of the L1-N, L2-N voltages that's aligned with (i.e., in-phase with or 180° from) the created third phase L3-N is only 60V (because the 30°, 120°, 30° vector triangle formed by L1, L2, N is quite "squashed"). Therefore a 3:1 boost on this 60V component is needed to create the third phase L3, resulting in a total neutral current of three times the L3 load current.

 

[synchro]

With a wye-delta configuration to create a third phase, using individual single phase transformers to create the wye-delta might be more cost effective that a 3-phase transformer if the neutral current is the limiting factor and not the total KVA.

 

[MD Automation]

Long story short (tldr) – the experiment to create the 3rd leg from L1 and L2 of a 3 phase supply using a Delta-Wye transformer worked. Though it’s certainly not a perfect waveform as if it were delivered from your local friendly POCO.

As I understood it, the theory discussed here is to use L1/L2 AND the Neutral, in order to energize 2 windings on the Wye side. These windings would, in turn, (sorry, bad pun) magnetize the corresponding 2 windings on the Delta side which would then energize the 3rd Delta winding, since those are all connected in series. Energizing the 3rd Delta winding would then magnetize the remaining coil back over on the Wye side, thus creating the L3 leg.

I used 2 random legs of our shop 208Y/120 VAC supply to create a 3 wire feed to a generic 30 kVA 480D – 208Y step down transformer. Nothing connected to the HV Delta side. Connected shop supply L1 to X1, L2 to X2 and the supply Neutral to XO. Then hooked up a small ¾ HP gear motor to L1/L2/L3, anticipating the created L3 leg would work with the shop L1/L2 legs to power the motor. And it did.

A quick detour here about the neutral. Clearly the connection to XO is crucial, without it the L1-L2 supply is simply a single phase sine wave. It’s the Neutral reference that allows L1 and L2 to have (120 degree) phase angles established between them. But this is where I ran into a little bit of hesitancy. I was always taught to never connect the XO when backfeeding a Delta-Wye transformer like this. And this is, more or less, what we are doing here. There was a long thread a month or two ago about if this really was something to be avoided or not. I did not chime into that discussion, since I felt that smarter brains than mine were hashing it out. However, I always thought that if the balance between the supply phases and supply neutral was not perfectly matched by the “balance” of the 3 Wye transformer windings and their XO (and it never is) then you can create large neutral currents as they try and fight to locate a common mid-point. When I was first wiring this experiment up, I backfed all the POCO legs to X1/X2/X3 and measured the difference between my supply Neutral and the floating XO (normally the new SDS neutral when the Wye side is the secondary). It was around 4 or 5 volts. So no perfect balance there, not terrible, but not tiny. Then, with all 3 supply legs still hooked up, I connected the supply neutral to the XO and, when powered up and idling, it carries ~16 amps (measured with an old Fluke 322) presumably because of this imbalance between the POCO legs and the transformer windings. The 3 supply phases showed around 7 amps for idle current.

I then I removed the X3 leg and taped it off for this experiment. With the transformer powered on and “idling” (no motor load) that supply neutral carries ~18 amps. Just thought to mention this XO neutral current, as it made me somewhat nervous.

Here is a snapshot of the setup…




I used an old Hioki Power analyzer for displaying voltage and current waveform(s). Note - the current data pictured is only for the 3/4 HP motor load – not the motor load and transformer magnetizing current. Also note that I looped the motor leads around the clamps 2x, to get double the current reading. So where you see 3.2 amps motor draw, it’s really 1.6. The pictured current waveforms are still too small to be of any real use.

You can see here the display of motor voltage (measured to gnd) and current, using this experiment to derive the L3 leg…









The blue line is the L3 phase created by this setup. To my eye it has some distortion, especially around the top and bottom - this is also clearly measured in the THD in the last picture above. It’s slightly more peaked than the smooth(er) sine waves on L1 and L2. And the sides connecting the min and max peaks are straighter than the L1 and L2 POCO legs. It really looks like it is morphing into a sawtooth wave, maybe the 2 magnetic "couplings" that created it (from Y->D, then back again from D->Y) added some small distortion shaping? Also, obviously, there is some voltage imbalance here and associated current imbalance on the motor windings. Nothing you’d want to run 24/7 on a larger loaded motor with. Presumably, I could try and compensate for some of this imbalance by adjusting taps. Since this is a Delta-Wye Stepdown, the taps are on the Delta side – so I’d have to scratch my head for a bit before knowing which tap to try first and in which direction to move it. And then I'd simply guess anyway

Other current data of interest, measured with my Fluke (so not artificially doubled)…

Supply current on L1 (Black) 10.7A idle 12.3A motor running
Supply current on L2 (Red) 9.1A idle 11.8A motor running
Supply current on XO (White) 18.3A idle 21.4A motor running

At any rate, this is what I found out today. You can clearly create the 3rd leg using this setup. Not elegant, but it worked. If you needed the missing leg only for dumb motors, you’d be better off using a VFD I’d think (single phase or de-rated 3 phase unit). And I did not like connecting the XO up – though it was clearly necessary for this experiment. I feel like you’d be creating unnecessary current flow (and transformer heating) trying to drag/align the X=0 Y=0 vector “origin” of the transformer on top of the (presumably stiffer) POCO zero point. But maybe that's up for debate?

It also might be the case that my motor load was too small (compared to the size of the xfmr) to really identify / validate any of the mathematical predictions that were made above (like Synchro’s). But I simply wanted to start easy and small so as to minimize any risk.

 

[synchro]

Thanks MD Automation for doing all that testing with the delta-wye transformer to create a 3rd phase. It was very thorough and informative. It does look like the waveform on X3 has at least some third and fifth harmonic content to make it look somewhat like a rounded sawtooth.

With the amount of idling current that you're seeing, I think the concern I expressed in my post #23 of this thread may be raising its head. A common 3-legged transformer core works fine as long as all of the magnetic flux goes between one leg and another. But if there is zero sequence flux in which all three legs of the core get magnetized in the same direction together and at the same point in time, then there's no good path for this flux to return between the two ends of the legs (the top and bottom in your picture). A 5-legged core has a better return path through the outside legs but it's not perfect. Creating a wye-delta from individual single phase transformers should avoid this particular issue altogether.

 

[MD Automation]

Happy to help – I thought the idea would make an interesting experiment and easy enough to do here because I had all the right parts lying around. Always fun to take the theory and test it in the real world to see if the “airplane” really does fly.

It would be nice to do the same test with 3 separate xfmrs to test your flux return problem. I used to have a few identical little SquareD 1 kVA 480 x 240 single phase control transformers lying around, but they are gone now. Sad.

Anyway – here is another screenshot of some harmonic distortion for that generated L3 leg, all listed out by components. As you suspected there are some substantial 3rd and 5th harmonics in that leg (and a 9th too). Compared to the same list for L2, the 3rd harmonic is ~2x as big and the 5th is ~3.5x larger. And I should correct myself when I said it’s starting to resemble a sawtooth – I should have said triangle waveform.

 

[kwired]

If we are talking about using a delta-wye transformer in a wye-delta configuration, then the delta secondary with nothing connected to it makes any current within the secondary the same on all of its windings (i.e., the only currents present are circulating currents). Therefore the three L-N currents on the wye will be equal to each other. The currents through L1-N and L2-N of the wye come directly from the 3-wire feeder. The current through L3-N is returning from the 3-phase load that's also connected to L1 and L2 of the feeder, and therefore the L3-N current is part of the total current drawn from L1 and L2 of the feeder.

In a sense the wye-delta is acting in the same way as a grounding transformer by forcing equal currents with the same phase on each of the L-N windings in the wye. But in this case L3 is connected to the load and not to the source of power, and so the load determines the amount of current in all of the L-N windings.

I think another way to look at this is to consider how a T transformer configuration could create a 3rd phase from L1, L2, and N of an open-wye. The main winding would be across L1 and L2, and its center tap connected to a teaser winding which provides the created phase L3 at its other end. The neutral would be connected to a tap on the teaser that's 1/3 of the way up toward L3. And so the teaser would have 60V between its tap and the main winding's center tap, and 120V between its tap and L3.
In this case the teaser acts as an autotransformer boosting the 60V supplied between the main winding center tap and the neutral conductor up to 60V + 120V = 180V across the entire teaser winding.

I'm not suggesting to use a T configuration but I'm mentioning it because it shows more clearly how the neutral of an open wye would have a significant current when a third phase is produced from L1, L2, and N. I think it ultimately boils down to the fact that in a 120V/208V open wye, the component of the L1-N, L2-N voltages that's aligned with (i.e., in-phase with or 180° from) the created third phase L3-N is only 60V (because the 30°, 120°, 30° vector triangle formed by L1, L2, N is quite "squashed"). Therefore a 3:1 boost on this 60V component is needed to create the third phase L3, resulting in a total neutral current of three times the L3 load current.

Guess I had two single phase units on my mind as that is what I am used to seeing for open delta sources, but you may be right with a single core unit.

 

[darekelec]

This is a substantial research and discovery. I can find a transformer for 200$ whereas rotary phase converter (RPC) costs 2,5k$ .
so far I told customer a price for RPC and he replied ‘too expensive’

Will this aeropoane ‘fly’?

 

[ptonsparky]

Not sure which part(s) is too expensive, but maybe he should have invested in something that would have worked in the first place.