3 phase to single phase supply

[mike_kilroy]

Just purchase. Refu.com
Use std 415 to 1000v input xfmr and buy std 1000v mining drive with optional output filter if needed,

Or

Buy std 415v drive, optional output filter if needed, and buy std 415 to 1000v 3ph transformer. No design work, just off the shelf components

 

[Ingenieur]

Just purchase. Refu.com
Use std 415 to 1000v input xfmr and buy std 1000v mining drive with optional output filter if needed,

Or

Buy std 415v drive, optional output filter if needed, and buy std 415 to 1000v 3ph transformer. No design work, just off the shelf components

Brilliant minds think alike :lol:

post 6

sometimes putting the xfmr on the output causes issues due to harmonics and waveform quality
but if you get the right xfmr may not be an issue

 

[mike_kilroy]

Brilliant minds think alike :lol:

post 6

sometimes putting the xfmr on the output causes issues due to harmonics and waveform quality
but if you get the right xfmr may not be an issue

Yep; the OUTPUT transformer cannot be just picked off the shelf for the rated current and voltages, but picked properly (meaning oversized to handle the PWM heat). This transformer itself will add L to the output and may not require any C to clean the voltage enough for OP use. Not sure what his use is.

Example how we use output xfmrs: http://spindel.com/index.php?option=com_content&view=article&id=30&Itemid=146

 

[Ingenieur]

Yep; the OUTPUT transformer cannot be just picked off the shelf for the rated current and voltages, but picked properly (meaning oversized to handle the PWM heat). This transformer itself will add L to the output and may not require any C to clean the voltage enough for OP use. Not sure what his use is.

Example how we use output xfmrs: http://spindel.com/index.php?option=com_content&view=article&id=30&Itemid=146

thanks
good info

 

[don_resqcapt19]

...
My point is that the current flow though the 'angled' coils will be in phase with the _single phase_ output voltage. The same current has to pass through all coils in series. ...

-Jon

i am just not seeing that.

We are adding two voltage vectors that are 120° to create a third voltage vector that is the 3rd leg of the triangle. This third vector is at the same phase angle as the first transformer coil that provides the other half of the 240 volts. With a 240 volt load there is equal current in all of the coils of the transformer, both primary and secondary.

(note this applies to a zig-zag and am not sure it applies to a double delta)

 

[winnie]

i am just not seeing that.

We are adding two voltage vectors that are 120° to create a third voltage vector that is the 3rd leg of the triangle. This third vector is at the same phase angle as the first transformer coil that provides the other half of the 240 volts. With a 240 volt load there is equal current in all of the coils of the transformer, both primary and secondary.

To clarify: I absolutely agree with what you describe above. I am not arguing the point about the voltages; I am instead pointing out that the _current_ follows the current laws, not the voltage laws. The rule for voltages is that in a closed circuit the sum of all voltages must add up to zero. The rule for currents is that in a closed circuit the same current must flow everywhere.

Consider a simpler situation, an open delta secondary, with terminals A B and C. B is the terminal common to both legs, A and C are the terminals across the 'open jaw'. The voltage across the open jaw is equal (except for impedance differences) to the voltage that would have been created by the missing winding. This is simply restating the vector math you gave.

Now place a resistor across the open jaw, between terminal A and C. Make no connection to terminal B.

Because this is a series circuit, the exact same current must flow through the two legs of the transformer. Because this is a resistor, the current flow will be in phase with the voltage placed across the resistor, or in phase with the A to C voltage. This current current flow in the coils must be the same as the current flow through the resistor, and must be in phase with the A to C voltage. The load resistor has unity power factor, but the two coils each see either a leading or lagging 60 degree current flow.

If this open delta is a transformer secondary, then the primary currents will similarly have strange power factors.

If this were an open delta to open delta transformer, with a resistive load across the open jaw, then virtually no current would flow into the B primary terminal because no current is flowing out of the B secondary terminal.

-Jon

 

[don_resqcapt19]

To clarify: I absolutely agree with what you describe above. I am not arguing the point about the voltages; I am instead pointing out that the _current_ follows the current laws, not the voltage laws. The rule for voltages is that in a closed circuit the sum of all voltages must add up to zero. The rule for currents is that in a closed circuit the same current must flow everywhere.

...

-Jon

i am still not seeing it...I see the current as also being identical in all 6 coils of the transformer. The current vectors also add at 120° and will be the same in the two coils connected that way with a resultant vector that is an extension of the current vector from the other secondary coil.

 

[winnie]

i am still not seeing it...I see the current as also being identical in all 6 coils of the transformer. The current vectors also add at 120° and will be the same in the two coils connected that way with a resultant vector that is an extension of the current vector from the other secondary coil.

The difficulty is that you are applying voltage math to current flow.

For _voltage_ all of the voltages in a closed loop must add up to zero. You do this with vector addition. In a closed loop you can have different phase angles for the voltage across each circuit element. In fact you _must_ have different phase angles, because in a simple 2 element circuit the load voltage must oppose the source voltage.

For _current_ all of the currents flowing into a node (terminal) must add up to zero. Again you do this by vector addition. This is the equivalent of saying that in any series circuit the exact same current must flow through each element.

Another way of saying this: you use vector math to add up voltages in series, you use vector math to add up currents in _parallel_.

-Jon

 

[GoldDigger]

Note: Highly revised, ignore the first form of this post.

i am still not seeing it...I see the current as also being identical in all 6 coils of the transformer. The current vectors also add at 120° and will be the same in the two coils connected that way with a resultant vector that is an extension of the current vector from the other secondary coil.

Not if you only load one of the two line-to-line phases.

In a parallel circuit two out-of-phase currents can add to produce a net current of an intermediate phase (in this case lying along the same line as the third current vector). But in a series circuit both currents will have the same phase, but they will both be out of phase with their own coil voltage vector.

The current in the two coils in series will then be in parallel with the current from the third coil.

The commonly shown zig zag and double delta each allow a high output from the three phases of a generator, but does not allow full rated three phase kVA of the generator to be delivered to a single phase load. So, using either zig-zag or double delta a 75kVA generator will produce 50kVA into a single phase load. The prime mover mechanical load will be 50kW but the current in some windings will be that corresponding to a 75kVA load or some windings will deliver their current at a low power factor.

The circuit I am referring to uses three identical transformers each with to identical and isolated secondaries (although the two secondaries on one of the three will end up with a common terminal).
It cannot, even if there are no transformer losses, derive 75kVA single phase from a three phase 75kVA generator. Only 50kVA, since some of the component currents have a low power factor.

Note that as long as the generator provides two identical and completely isolated output windings for each phase it can be connected in the Leyton configuration. It will provide single phase 120/240 from a set of six 120V windings and should have a uniform power drain on the prime mover over the entire cycle.

The Leyton connection allows you to have the same effect as the generator but starting from a normal three phase source/service. It requires larger than normal single phase transformers to handle the extra current, but will draw only as much power from the primary as is consumed by the load.
In the generator case a 50kW generator with three winding pairs each rated at 25kVA can deliver 50kW output with only 50kW mechanical load on the prime mover.


For an accurate and complete description, see:
http://wpedia.goo.ne.jp/enwiki/User:Hugh_Leyton/sandbox

 

[Phil Corso]

Gentlepeople...

It is the geometric shape of a Genny's magnetic-structure that makes the double-D workable, not so, with a Xfmr!

Also, if the Genny is contemplated, bear in mind it must be over-sized by 33 to 50%! Hence, a 800 to 900 kVA machine is needed!

Phil