208v question

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bpk

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I have never tried this but have always wondered if you have a piece of equipment (like an AC unit) rated at 208 volts typically it would be fed with 2 hots, if you had a high leg delta would it work the same to feed it with a neutral and a the 208v high leg? I have always thought it wouldnt make a difference but never knew for sure, as long as there wasnt any integral parts that needed 120v.
 
Between two hots will always be 208.

It's if there's a neutral involved you need to make sure you're on the C and A phases.
 
He means if you use phase B and neutral. Would that operate a piece of equipment rated at 208v? For example...a water heater.
 
phase to phas 208
phase to neutral 208....

I don't see any problem with this. as connecting to a 208 single phase heater.:grin:
 
Yes, electrically speaking, it would work. Voltmeters can measure it, it's real voltage.

The issue (from what I hear) is that it performs poorly due to the relatively weak source impedance, meaning it doesn't maintain voltage under load well, and probably wreaks havoc on the primary system.

I haven't tried it myself, as far as a real load goes, but I'd be interested in others' experiences.
 
Some electronic equipment manufacturers use surge protection devices designed for 120 volts between incoming legs and the EGC. They carry no current unless there is an over-voltage fault. Applying 208 volts to ground on one leg could very well cause the protection device (MOV, gas tube, etc.) to fault, tripping the OCP. Additionally, high frequency noise filters have maximum ratings between each leg and ground. Applying this higher voltage could damage the capacitors and significantly increase the leakage current into the EGC. Listing agencies place strict limits on leakage currents.

If it is electronic equipment (motor drives, DC power supplies, battery chargers, etc.) either don't do it or inquire the manufacturer. Though it may well function such use probably violates listing and constitutes a hazard.
 
The delta system was really designed to give 240/120 service. It would most probably work but I wouldnt do it because it is just so wrong and such an odd application that I think you would have liability if someone less trained got hurt or burned something up on the odd voltage of the high leg to neutral.
I certainly wouldnt do it on my license working for annother contractor I would have to be convinced. I dont know of any rule that would prohibit it though.
I am a worry wart and often think about liability and or hurting someone all of the time I try very hard to do the right thing for all involved contractor customer and myself.
 
Problem is that this load current flows through all 3 xfrmrs and is not in phase with any of the phase voltages. Besides the extra impedance, your apparent power calculations would be in error, vis-a-vis the Oregon Fudge Factor!
 
I'm going to need some clarification on this weak source impedance, problems with apparent power calculations, and what's an "oregon fudge factor".

carl
 
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
 
coulter said:
I'm going to need some clarification on this weak source impedance, problems with apparent power calculations, and what's an "oregon fudge factor".

carl
Click to expand...

Since the wild leg voltage is obtained by windings in series, resistance and leakage reactance are increased.

The Oregon Fudge Factor refers to a trick question of sorts on the Oregon supervisor's test. It amounts to this:

For a 10A, 208V load on an open wye, what is the ratio between the apparent power delivered by the transformers to the apparent power delivered to the load?

Each transformer delivers 120V x 10A = 1200 VA while the apparent power computed at the load is 208V x 10A = 2080VA, therefore the ratio is,

OFF = 2400VA/2080VA = 1.154

You would have a similar situation if you loaded the wild leg.
 
rattus said:
Since the wild leg voltage is obtained by windings in series, resistance and leakage reactance are increased. ...
Click to expand...
Assuming this is the source of the "weak source impedance", given normal transformer nameplate data (Z is on the nameplate), can this weak source impedance be quantified - can you calculate it? A sample calculation with a diagram would be nice. I won't harass you on your arrow direction:) - I promise.

carl
 
r_merc said:
What is the maximum voltage on a 208 ph-n High leg vs 208 ph-ph
Click to expand...
You mean from the high leg to the neutral (208v) or anywhere on a high-leg system (240v)?

Or are you asking what's the difference between a Y's 208v and a Delta's 208v?
 
coulter said:
Assuming this is the source of the "weak source impedance", given normal transformer nameplate data (Z is on the nameplate), can this weak source impedance be quantified - can you calculate it? A sample calculation with a diagram would be nice. I won't harass you on your arrow direction:) - I promise.

carl
Click to expand...

Let Vbn = 208V @ 90

Think of windings A-N and B-A in series. Z is 1.5X that of winding B-A
Ditto for windings C-N and B-C. Then,

Ztotal = 1.5/2 that of a full winding. But the real culprit is the PF.

Vbn = Van + Vba = Vcn + Vbc =

= 120 @ 0 + 240 @ 120 = 120 @ 180 + 240 @ 60 = 208V @ 90

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.

(Unless I made a typo)
 
Theoretically there is no reason that you would not be able to use the 208V L-N. The 208V L-N is as stable as the 240V L-L that completes the delta.

Practically there are many more reasons not to than there are to.

1) Most panels used with 240/120V 3P4W systems do not have 1-pole breakers rated for 208V to ground.

2) The transformers have not been sized to supply this amount of unbalanced load.

The real answer is very actual installation dependent. In fact it is possible to completely overload a poorly sized grounded delta transformer bank by connecting enough 120V L-N loads to only one phase.
 
rattus said:
Let Vbn = 208V @ 90 ... Vbn = Van + Vba = Vcn + Vbc =

= 120 @ 0 + 240 @ 120 = 120 @ 180 + 240 @ 60 = 208V @ 90 ...
Click to expand...
Great derivation of how Vbn = 208V >90

carl

severely edited
 
jim dungar said:
Theoretically there is no reason that you would not be able to use the 208V L-N. The 208V L-N is as stable as the 240V L-L that completes the delta.

Practically there are many more reasons not to than there are to.

1) Most panels used with 240/120V 3P4W systems do not have 1-pole breakers rated for 208V to ground.

2) The transformers have not been sized to supply this amount of unbalanced load.

The real answer is very actual installation dependent. In fact it is possible to completely overload a poorly sized grounded delta transformer bank by connecting enough 120V L-N loads to only one phase.
Click to expand...

That's what I was thinking

rattus said:
Let Vbn = 208V @ 90

Think of windings A-N and B-A in series. Z is 1.5X that of winding B-A
Ditto for windings C-N and B-C. Then,

Ztotal = 1.5/2 that of a full winding. But the real culprit is the PF.

Vbn = Van + Vba = Vcn + Vbc =

= 120 @ 0 + 240 @ 120 = 120 @ 180 + 240 @ 60 = 208V @ 90

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.

(Unless I made a typo)
Click to expand...

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 ?

rattus said:
OFF = 2400VA/2080VA = 1.154

You would have a similar situation if you loaded the wild leg.
Click to expand...

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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