# Open delta's



## MDShunk (Jan 7, 2007)

Draw it out and your qiestion might answer itself.


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## user4818 (Jan 15, 2009)

Draw a triangle and remove one of the legs. You still have 3 points to connect to. There's your open delta. Simple as that really. Oh, plus advanced calculus to explain why it works that way, but you don't need that.


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## RePhase277 (Feb 5, 2008)

You still get 3 phase because the ends of each coil are phase shifted in relation to each other, but the amount of available power is reduced to 57% of the full delta's capacity.


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## Dembones (Mar 24, 2007)

I understand how the open delta works, but I'm having trouble with why it doesn't work in some cases. Lets say we have a motor operating on 240V delta. A phase is lost, now we have a single phasing condition, which I understand to be the worst case scenario for a three phase motor. Assuming the overloads/fuses don't open from increased current, why doesn't the motor continue to operate normally as an open delta? For clarifiaction, let's say it's B phase that dropped, it is still tied to A&C just as an open delta connection would be right? Does it come down to conductor sizing for 57% greater current?


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## gcleary47 (Mar 23, 2008)

thanks for the responses. I think i'm just thinking into this one too much. Time to go to class


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## zod68 (Jan 27, 2009)

if a motor is a delta a phase b phase and c phase to opererate , can it operate without b phase ? consider t1 t2 n t3 . do not all three wires have to connected 9-6-3/ 8-5-2/ 7-4-1 . if you lose line too t2 what happens ?


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## RePhase277 (Feb 5, 2008)

Dembones said:


> I understand how the open delta works, but I'm having trouble with why it doesn't work in some cases. Lets say we have a motor operating on 240V delta. A phase is lost, now we have a single phasing condition, which I understand to be the worst case scenario for a three phase motor. Assuming the overloads/fuses don't open from increased current, why doesn't the motor continue to operate normally as an open delta? For clarifiaction, let's say it's B phase that dropped, it is still tied to A&C just as an open delta connection would be right? Does it come down to conductor sizing for 57% greater current?



Yes losing a phase is called single phasing and would occur if one pole of a three pole breaker went bad, or that conductor somehow became open. But this is not the same as removing one of the transformers from a delta bank. Think about it. In single phasing, you only have two good conductors going to the load. In an open delta, you still have all three conductors.


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## Dembones (Mar 24, 2007)

I know you're right, I'm not argueing theory, just wondering how the connections are physically made. Know of a good diagram? Can't find one online.


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## RePhase277 (Feb 5, 2008)

Dembones said:


> I know you're right, I'm not argueing theory, just wondering how the connections are physically made. Know of a good diagram? Can't find one online.



Like this?


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## Dembones (Mar 24, 2007)

All coming back to me now thanks.


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## kadams (Jan 29, 2009)

*480 v?*



InPhase277 said:


> You still get 3 phase because the ends of each coil are phase shifted in relation to each other, but the amount of available power is reduced to 57% of the full delta's capacity.


In a 240v secondary assume this is why there is still 240 v across the open transformer space something about a phase shift? Sometimes I look at the open delta and I wonder why there wouldn't be 480 v across the open leg? I know it only has 240v but it is a little counter-intuitive.


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## Richard Rowe (May 25, 2009)

We have a 480 3ph drop in one building that reads 504 to ground on two legs and 504 across those two legs. The third leg read 0 to ground and 504 to the other legs.... is this what you are talking about?


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## RePhase277 (Feb 5, 2008)

Richard Rowe said:


> We have a 480 3ph drop in one building that reads 504 to ground on two legs and 504 across those two legs. The third leg read 0 to ground and 504 to the other legs.... is this what you are talking about?


This sounds like a corner grounded delta to me. A different beast altogether than an open delta.


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## Richard Rowe (May 25, 2009)

I was thinking it is still called an open delta. I know you can get that an open delta from two trans, but not by droping a leg. Our other drops are 248 across and 121-121 and 214 to ground.


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## Buddha In Babylon (Mar 23, 2009)

intereting stuff. I am studying transformers right now, and i confess it's more than i expected. So much to learn...so little time...
I can only pray i live long enough to be a good electrician with a solid understanding of this kind of thing.:thumbsup:


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## RePhase277 (Feb 5, 2008)

Richard Rowe said:


> I was thinking it is still called an open delta. I know you can get that an open delta from two trans, but not by droping a leg. Our other drops are 248 across and 121-121 and 214 to ground.


You could have a corner grounded open delta, but that would be a rare find indeed. The other you describe is a delta high leg. You get 240 between phases, and 120 from two phases to the neutral. And 208 from one phase to the neutral. It is usually the B phase that is the "high leg". Also known as the "stinger" or "snake" or "wild leg", or several other names.


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## zgozvrm (Jun 17, 2009)

kadams said:


> In a 240v secondary assume this is why there is still 240 v across the open transformer space something about a phase shift? Sometimes I look at the open delta and I wonder why there wouldn't be 480 v across the open leg? I know it only has 240v but it is a little counter-intuitive.


See my post from another discussion:
http://www.electriciantalk.com/f30/explain-120-208v-7538/index3/#post133747


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## jimosborne (Feb 23, 2014)

*240 Volt Open Delta Transformer*



zgozvrm said:


> See my post from another discussion:
> http://www.electriciantalk.com/f30/explain-120-208v-7538/index3/#post133747


Each of the two phases of your open-delta has a magnitude of 240 Volts.
Intuitivedly, you would think the voltage across the open phase would be 240 + 240 = 480 Volts. There is, however, a phase difference of 120 degrees between the two phases. The resultant voltage between the open ends is the vectorial sum of the two phases:
(Ec)^2 = (Ea)^2 + (Eb)^2 + (2 * Ea * Eb * Cos(120 Deg)
(Ec)^2 = (240)^2 + (240)^2 + (2 * 240 * 240 * (-0.5))
(Ec)^2 = (57,600) + (57,600) + (-57,600)
(Ec)^2 = 57,600
Ec = SQRT(57,600)
Ec = 240

Hope that helps!


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## chicken steve (Mar 22, 2011)

The standard open delta & associated math>

Open Delta or V-V Connection of 3-Phase Transformer

But it gets better.....>:whistling2:










~CS~


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