208v question

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dnem said:
That's what I was thinking



I wasn't thinking that. . But now that I've read it a few times, it's sinking in ..... kinda.

"Ztotal = 1.5/2 that of a full winding."
So Z is low because of the two paths from highleg to neutral. . Are you saying this is a disadvantage or a problem ? . Is "weak source impedance" a bad thing ? . It sounds like a good thing to me.

"For a resistive load, Ibn carries a phase angle of 90 degrees which is 90 degrees out of phase from Van and Vbn and 30 degrees out from Vba and Vbc."
So does this mean alot of wasted power because of the 90? out of phase ?



So besides the usual losses, you take an additional "hit" because of the phase angle differences of 2 series phases that are out of phase with each other ? . Impedance in the source transformer, impedance in the load motor, resistance thruout the entire circuit plus this additional loss ? . In addition to the usual losses, you would still need 115.4va in the secondary to deliver 100w at the load ?

Am I seeing the issue here ?
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I would say that the 120V legs are providing apparent power (VA) but no real power (W). This current still loads transformer A-C. The other phases would be providing power at angles of +/- 30 degrees.

In short it appears that even moderate use of the wild leg would make it difficult to balance the load as well as making poor use of the transformer ratings.

Furthermore, all this current would flow in the neutral--not good.
 
This needs some answering:
dnem said:
So Z is low because of the two paths from highleg to neutral. . Are you saying this is a disadvantage or a problem ? . Is "weak source impedance" a bad thing ? . It sounds like a good thing to me.
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No, Z is high because of the path, and that's a disadvantage. A source should have a 'strong' impedance, which means a low internal resistance. For a source, lower impedance means a greater ability to not be affected by the load.

A car 12v battery has a much lower impedance than an egress-light gel-cell. They're both 12v (nom.) but the larger battery is capable of supplying more current without the voltage dropping. Source impedance behaves as a resistor in series with the source.

If you're into audio, you may have heard the term "damping factor." This is the ability of an amplifier to not only start the speaker cone moving, but stop (or alter) it when the signal stops (or changes). It is directly based on the amp's impedance.

A speaker is a linear motor; supply voltage, it moves. But, when the signal stops, inertia keeps the cone vibrating for a few moments. A moving speaker is a permanent-magnet generator; a short across the terminals acts as a dynamic brake.

An amplifier with no input signal should be able to keep the output terminal at 0 volts, even if a voltage is attempting to be input. This ability is dependent on the amp's power supply and its output capacity; a higher damping factor means a low impedance.
 
coulter said:
There is only one issue I know of against using the wild leg for 1ph 208V feeders. Nothing I came up with, but it has been the subject of several threads. 1pole CB that fit normal 240/120D panels are not rated for 208V.

carl
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Good call, and after you find/purchase a single pole breaker rated for 208volts, you'd probabaly opt for a 2pole breaker and more like a 240 volt heater.
 
LarryFine said:
No, Z is high because of the path, and that's a disadvantage. A source should have a 'strong' impedance, which means a low internal resistance. For a source, lower impedance means a greater ability to not be affected by the load.
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So 'strong' impedance is low impedance ?

Whose idea was it to screw up those terms, the same guy that came up with flammable and inflammable ?
Maybe it was the same guy that came up with that brilliant rule:
I before E except when we want to put E before I.
 
IMHO saying something has 'strong impedance' to mean that it is a good voltage source is tremendously confusing.

You could have a 'stiff' or a 'strong' voltage source, which has _low_ impedance, or you could have a 'stiff' or 'strong' current source, which has _high_ impedance. But a 'strong impedance' seems to mean high impedance, exactly the opposite of what is required for a stiff voltage source.

-Jon
 
Challenge:

Challenge:

Does anyone dare post a phasor diagram of a wild leg system with no loads other than a resistive load between B and N? It should include the effects of resistance and leakage inductance.

Furthermore, the diagram need not be quantitative, only qualitative to provide a picture of what is happening. I have done one, but still am not sure if it is right.
 
Don't forget that a wild leg system derived from an open delta will behave slightly different than one from a closed delta. Circulating currents in a closed delta are one reason that a standard (single core) dry type 240/120 3PH 4W transformer is limited to 5% line to ground loading.
 
rattus said:
Does anyone dare post a phasor diagram of a wild leg system with no loads other than a resistive load between B and N?

I have done one, but still am not sure if it is right.
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Had anyone other than you asked the question, I would have suggested asking you, so . . . :confused:
 
jim dungar said:
Don't forget that a wild leg system derived from an open delta will behave slightly different than one from a closed delta. Circulating currents in a closed delta are one reason that a standard (single core) dry type 240/120 3PH 4W transformer is limited to 5% line to ground loading.
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What about line-to-line loads across the open side?
 
LarryFine said:
What about line-to-line loads across the open side?
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I don't know what you are asking about.

The loads do not actually close the delta so they do not cause circulating currents.

Open delta banks are sized different than closed deltas. Three single phase units are sized differently than a single core three-phase unit.

Open deltas do have more of a problem with voltage imbalance than closed deltas even without the addition effect cause by 120V loads.
 
jim dungar said:
I don't know what you are asking about.
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I think you caught it. To what degree can we treat the open side of an open delta as if it were not open, as far as 1-ph line-to-line loads are concerned.

It seems that 240v line-to-line loads across the two real secondaries (barring 120v line-to-neutral loads) help stabilize the voltage across the open side.

Added: I grasp that we can't pretend the missing secondary exists as far as its lack of contribution to supplying power goes. I'm asking more about the 240v across the open side being a stable 240 volts, within transformer capacity.
 
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LarryFine said:
It seems that 240v line-to-line loads across the two real secondaries (barring 120v line-to-neutral loads) help stabilize the voltage across the open side.
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I agree, the more balanced the loading the more balanced the transfromer (voltage) performance.
 
Why wouldn't the open delta loads suffer from high source impedance similar to the Vbn loads? Rattus would likely say the impedance is 2Z. I can't tell, I haven't eeen a model or the math yet.

carl
 
coulter said:
Why wouldn't the open delta loads suffer from high source impedance similar to the Vbn loads? Rattus would likely say the impedance is 2Z. I can't tell, I haven't eeen a model or the math yet.

carl
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I believe he might say 1.5Z.
 
That's okay. It's the Big R's comment - not mine. I don't even know if the VD are the same direction. Haven't seen a model or any math yet.

carl
 
coulter said:
That's okay. It's the Big R's comment - not mine. I don't even know if the VD are the same direction. Haven't seen a model or any math yet.

carl
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FWIW, I don't see a big voltage stability issue with using the high leg for line-neutral loads (See Jim's #17). I would think you would want the transformers designed to handle the load from the get-go. Of course, if you were doing that, why not use a different design?

As for 1-phase 240 volt loads across the open delta, I only remember the single-phase loads across one of the available coils. In that case:

The 3-phase coil supplies a load equal to 1/sqrt(3) of the balanced 3-phase load.

The 1-phase coil supplies a load equal to the resultant of the 1-phase load and the balanced 3-phase load. This supplied load can be up to the sum of the 1-phase load plus 1/sqrt(3) of the 3-phase load, depending on the power factor of the loads.
 
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