Trouble finding straight 240v breaker vs 120/240 breaker

[tortuga]

There is no real derating necessary for using the 'open' phase.

Typical formula for sizing the unequal transformers are:
Power pot = .58T
Lighting pot = .58T + S

Where T is the total three phase load and S is the total single phase load.
Note how the transformer sizing takes into account the open delta versus closed delta derate.

Thanks yeah was just going all theoretical / hypothetical thinking of Kirchhoff's Laws for current and asking:
If you put a 4500W (18.75A) single phase L-L 240V load
on the open phases of setup described in post #58 what would currents be in the two transformer windings?

 

[wwhitney]

If you put a 4500W (18.75A) single phase L-L 240V load
on the open phases of setup described in post #58 what would currents be in the two transformer windings?

The same as the current through the load, as the circuit consists of a single loop, so current everywhere is the same.

If you want to assume an idealized transformer, then as your water heater is a resistive load (with R_load =240V/18.75A =12.8 ohms) , that current is 18.75A in phase with the open coil voltage.

But you specified the %Z, so maybe you want a more realistic transformer model? Then you'd need to know the full coil impedance Z (as a vector or complex number). Which means one more real number in addition to %Z, e.g. the X/R ratio.

Once you have the impedance Z, then I believe you model each coil as an idealized 240V source (of the appropriate phase) in series with the impedance Z. If that's correct (I'm a bit new at transformers), as the circuit is a simple loop, you can lump all the impedances together (2 coils and the load), and the two idealized 240V sources add together the same as previously. In other words, you just have the 240V open coil voltage across an impedance of 2Z + R_load.

I have no idea what typical values are for Z, so I can't calculate a reasonable example.

Cheers, Wayne

 

[Jraef]

And it likely cost same or more than the three pole anyway.

Plus the two pole might not stocked where the three pole likely is. If it is Square D and you are waiting for factory order on this, it might take months to get it as well.

That’s been my experience: 3pole being less than the non-slash rated, likely because of volume of sales in comparison. It just tends to bug people when you “waste” one of their precious pole spaces.

 

[tortuga]

The same as the current through the load, as the circuit consists of a single loop, so current everywhere is the same.

If you want to assume an idealized transformer, then as your water heater is a resistive load (with R_load =240V/18.75A =12.8 ohms) , that current is 18.75A in phase with the open coil voltage.

But you specified the %Z, so maybe you want a more realistic transformer model? Then you'd need to know the full coil impedance Z (as a vector or complex number). Which means one more real number in addition to %Z, e.g. the X/R ratio.

Once you have the impedance Z, then I believe you model each coil as an idealized 240V source (of the appropriate phase) in series with the impedance Z. If that's correct (I'm a bit new at transformers), as the circuit is a simple loop, you can lump all the impedances together (2 coils and the load), and the two idealized 240V sources add together the same as previously. In other words, you just have the 240V open coil voltage across an impedance of 2Z + R_load.

I have no idea what typical values are for Z, so I can't calculate a reasonable example.

Cheers, Wayne

Thanks Wayne I did not have a specific transformer in mind. I was just contemplating potential side effects of connecting a single phase load that way, like each transformer is now carrying 18.75A
So how much of the 4500 W load would each transformer carry?
Here is a better diagram:

 

[wwhitney]

So how much of the 4500 W load would each transformer carry?

Sounds like you're happy with the idealized case (I'm not sure I got the transformer model correct for considering the transformer impedances).

Then in terms of VA (transformer capacity), each transformer is carrying the full 4500VA, since it is carrying the full 18.75A current. So it's inefficient in transformer capacity compared to a closed delta.

As far as W (real power), by symmetry each transformer is providing half the real power. Or, the current through each transformer is in phase with the voltage AC, which is out of phase with the coil voltage by 60 degrees. So the power factor for each transformer is cos(60) = 0.5.

Cheers, Wayne

 

[tortuga]

Ok so on a load calculation single phase (240V) loads connected across B-C (in the diagram above) get added to the 25kVA and the 75kVA.
If instead you just added a single phase 240V load across A-B that would only add to the 25kva's load.
Typically with these I find the "power pot" will have spare capacity when a lighting pot is heavily loaded.

 

[jim dungar]

Ok so on a load calculation single phase (240V) loads connected across B-C (in the diagram above) get added to the 25kVA and the 75kVA.
If instead you just added a single phase 240V load across A-B that would only add to the 25kva's load.
Typically with these I find the "power pot" will have spare capacity when a lighting pot is heavily loaded.

Think of the 75kVA transformer as being 25kVA 3-phase 240v and 50kVA single-phase 120/240V for maximum loading. If you want more 120/240 1-phase you simply need to reduce your 3-phase by the same amount. The worst loading condition is un-balanced 120V (which is also caused by using a 208V load from B-N).

There are slightly different loading/sizing issues if the transformer is a closed delta configuration, which is why you often see these sold with a 5% maximum rating for the neutral.

 

[tortuga]

Think of the 75kVA transformer as being 25kVA 3-phase 240v and 50kVA single-phase 120/240V for maximum loading. If you want more 120/240 1-phase you simply need to reduce your 3-phase by the same amount.

Right that is a good way to think about it.
If you have B-C connected single phase 240 load as I described in post #64 that loads up both though.

 

[jim dungar]

Right that is a good way to think about it.
If you have B-C connected single phase 240 load as I described in post #64 that loads up both though.

Not really, the BC loads are considered as part of the 25KVA 3 phase component of the AC 75kVA winding.

 

[synchro]

So how much of the 4500 W load would each transformer carry?
Here is a better diagram:
View attachment 2567623

As Wayne mentioned, if you put a single phase load across BC, all of that load current will flow through the AB winding.
Also as mentioned, that BC load current is at 60° from the AB voltage.

Now if you also place a resistive load across AB, the two load currents will be at a 60° from each other. And so if you have a 4.5 kW load across AB and a 4.5 kW load across BC, instead of doubling the AB winding current it will be √3 ≈ 1.732 times instead. But to be conservative, you could just add the AB and BC load currents to get the current drawn from terminal B of the AB winding in the open delta.

If you want to get into more detailed calculations:
Let I_B be the load current at terminal B of the open delta, and I_AB and I_BC be the currents through resistive loads across AB and BC respectively.
Also let I_C be the load current at terminal C of the open delta, and I_AC and I_BC be the currents through resistive loads across AC and BC respectively.

I_B² = I_AB² + I_BC² + I_AB x I_BC
I_C² = I_AC² + I_BC² + I_AC x I_BC

If you want to fully load the AB winding to its rated current I_Bmax when there's already a load current across the open jaw of I_BC, then allowable load current across AB would be:
I_AB = 0.5 x [ √ ( 4 I_Bmax² - 3 I_BC²) - I_BC ]

Similary, the allowable load current placed across AC when the AC winding is rated at I_Cmax and with a load current across the open jaw of I_BC is:
I_AC = 0.5 x [ √ ( 4 I_Cmax² - 3 I_BC²) - I_BC ]

Since the L-L voltages are the same, you can substitute kVA for all currents in these equations.
Here's some examples of the maximum single phase L-L loading on both the 25 kVA power pot and 75 kVA lighing pot, with numbers in kVA:

BC load	AB load	AC load	Total load
0.0	25.0	75.0	100.0
12.5	16.3	68.0	96.8
14.4	14.4	66.7	95.5
25.0	0.0	59.3	84.3

 

[kwired]

I don't really think that is an issue, I put 240 single phase loads across the "open " legs all the time.

Any three phase load you connect is kind of doing the same thing. Bottom line is you can't easily balance loading on open delta systems as it is not a balanced system to begin with. Some them have a higher capacity lighting pot than the stinger pot as well so you don't really balance them, you match the load to what the source can supply for the most part.

 

[brother]

Back to the OP, what kind and brand of panel is it? If it is a bolt-on style panel board then yes you can use a 277/480 breaker, for example with Siemans it would be a BQD. If it's a stab on, I'm pretty sure everyone makes a straight rated breaker. For Siemens it has an R suffix, i.e. Q230R (I recently priced them they are about a hundred bucks). For SQ D it's an H suffix, i.e. QO230H. Not sure what the part number is off hand for Joseph Eaton. All three poles are straight rated so you can also use a three-pole. (Some of this has already been mentioned, just summarizing).

Sorry, been away for bit. urgent matters had to deal with. We found our 240 rated breaker. And yes it is as someone pointed out, the phase to ground voltage is what is needed to protect the breaker. I have been told in the past when people used the wrong breaker on these high delta systems that they have been known to blow up.

 

[kwired]

Sorry, been away for bit. urgent matters had to deal with. We found our 240 rated breaker. And yes it is as someone pointed out, the phase to ground voltage is what is needed to protect the breaker. I have been told in the past when people used the wrong breaker on these high delta systems that they have been known to blow up.

I honestly don't believe that is too likely on very many installations. But at same time they still not designed for the application, maybe could result in failure but not necessarily that spectacular of a failure, more like contact failure that you can't even see but doesn't close properly when you reset it either.

Many open delta systems with a smaller stinger pot possibly don't even have that high of an available fault current to physically blow up the breaker. Open delta is likely the most common application out there with more than 120 volts to ground on a 240 or less system. Full delta with high leg is common here on farms or light industrial sometimes, but those usually have three phase motors as the majority of the load on them and would be using three pole breakers with somewhat limited 120 volt loads mixed in. Corner ground or ungrounded delta would be the other applications though those just don't seem to be too common and when they are they are likely separately derived on site for a specific load(s) and not used for general premises distribution.

 

[Birken Vogt]

Another note here: if the poco supplies open delta, it might be nice to put a few straight 240 loads across the wing pot if it saves spaces or whatever. But avoid the phantom pot. Can anyone think of a way to determine which set of phases are directly connected to the wing pot? Perhaps by applying a test load and seeing how the supply holds up? Just an idea I have had for years, probably not practical for anything.

 

[kwired]

Another note here: if the poco supplies open delta, it might be nice to put a few straight 240 loads across the wing pot if it saves spaces or whatever. But avoid the phantom pot. Can anyone think of a way to determine which set of phases are directly connected to the wing pot? Perhaps by applying a test load and seeing how the supply holds up? Just an idea I have had for years, probably not practical for anything.

Test load from a to b and then b to c might work, then compare voltage drop both ways?

In particular I would think if you paid attention to B to neutral volts that if you were connected across the open side you likely see some voltage drop but if connected across the closed side you won't see any voltage drop.

Add: Well maybe drop that is in the high leg conductor but shouldn't be any drop across B to N if you are only connected across the wing pot.