Open Delta Claculations

Status
Not open for further replies.
B

bphgravity

Senior Member
Location
Florida
The recent topic discussing the use of 1.732 was very interesting and informative. I have a question about open delta systems.

It has been my understanding that the open delta is the utilization of two transformers instead of three to derive a 3? system. The capacity of an open delta is 57.7% of a typical closed delta arrangement. This is derived from from taking 1.732 and dividing it by 3.

Can anyone explain this formula and how it is determined? Thanks
 
Re: Open Delta Claculations

Assume that full load for the Delta secondary in Diagram A is a line current of 17.3 amps, which delivers 14,400 va.
That means that the full load current rating of each phase winding is 10 amps.

Now, when it is converted to open-delta, (Diagram B) the line current will have to be reduced to 10 amps to prevent overloading the remaining phases, A and C.

Ed

Trans13.gif
 
Re: Open Delta Claculations

Ed,

Explain to us why the currents in the two coils do not add to 17.3A at their junction? I think I know, but I have to wake up before I do any serious math.
 
Re: Open Delta Claculations

Ed,

Explain to us why the currents in the two coils do not add to 17.3A at their junction? I think I know, but I have to wake up before I do any serious math.
 
Re: Open Delta Claculations

Explain to us why the currents in the two coils do not add to 17.3A at their junction
Click to expand...
Do you mean at the junction of windings A and C in diagram B?

Now Rattus, you know I can't do that, mathmatically. :D

All I know is what the meters indicate. :p

Ed
 
Re: Open Delta Claculations

Ed,

I am not sure I can do it either, but I think it is a matter of the extra current which used to be supplied by the missing leg of the delta.

I do think your diagram is correct though. I will waste a few hours on it to satisfy myself
 
Re: Open Delta Claculations

Are you asking how the 17.3 A is derived?? By vectorial addition and or Pathagorem Therom ?? I think it can be solved by geometry and the addition of coordinates also!!

Charlie
 
Re: Open Delta Claculations

Charlie,

No, I am asking Ed why the 10A currents in the two legs of the tranformer do not add to 17.3A in the line. That is, he has 10A + 10A = 10A, and I am still scratching my head because it looks right. I think we have some unusual phase angles as a result of removing one xfrmr.
 
Re: Open Delta Claculations

I'm not sure what you mean by unusual phase angles, but consider this. The currents in the conductors are 120 degrees apart, and since they are all equal in magnitude they sum to zero, by definition of a balanced 3 phase system. The currents in the wires are also the 3 currents in the junction of the two transformers and the one common phase. Three equal currents differing in phase by 120 degrees.
Jim T
 
Re: Open Delta Claculations

Originally posted by rattus:
Charlie,

No, I am asking Ed why the 10A currents in the two legs of the tranformer do not add to 17.3A in the line. That is, he has 10A + 10A = 10A, and I am still scratching my head because it looks right. I think we have some unusual phase angles as a result of removing one xfrmr.
Click to expand...
I can't draw as ED does, but his diagram is essentially showing RMS values (what is generally measured). Assume the single-phase transformer has a max rating of 10A RMS through it.

If you draw the instantaneous currents through the system at any given point in time though, you will find Kirchoff's laws are intact.

Edit with applogies to Mr. Kirchoff

[ February 21, 2005, 04:19 PM: Message edited by: rbalex ]
 
Re: Open Delta Claculations

As Jim T and Bob have indicated, the (new) line currents of 10 amps are rms values. The instantaneous values must obey Kirchhoff's Law at any junction.

The currents in the circuit conductors at the instant indicated would be have to be-
:D

Ed

[ February 21, 2005, 04:26 PM: Message edited by: Ed MacLaren ]
 
Re: Open Delta Claculations

Ed's drawing of the closed delta (how does he do that???) illustrates two 10A currents meeting at point (AB) those current are not in phase and as rbalex indicated when you sum the positive and negative values of those two currents at a given point in time the answer is also a multiple of 1.73. or in this case 17.3 A.

If you look at the open delta the current drawn is only passing through one coil (not all three). It is a 10A coil and the line can only draw 10A with out damaging that phase coil. There is not a second path for the line current to enter point A and follow to point B actually in the diagram coil B is missing.

This open delta will still power a three phase load but the kVA rating of the transformer bank must be derated to 57.8% of the rating if the third transformer was present.

I don't know if this helps, but I hope so

Charlie P
 
Re: Open Delta Claculations

What?s happening here is a comparison of apples and pork chops: both are edible, and they go well together, but they are not the same thing. :D ). As far as the load is concerned, nothing happened. It was getting 10 amps on each line, and it still is getting 10 amps on each line. But the two remaining secondary coils will see the change. They were each supplying 5.77 amps, but now they are each supplying 10 amps.
 
Re: Open Delta Claculations

To all:

I could see that the remaining coils had to provide the current of the missing coil and also felt Ed's diagram was correct, but it threw me to see 10 + 10 = 10 even with vector addition.

I finally got my currents running the right direction and I see how it all works out.

Looking at Ed's diagram B, the 10A line current entering the junction of the two coils does indeed split into a 5.77A component in each coil.

Then the 5.77A current normally supplied by the missing coil adds vectorially to bring the phase/line currents up to 10A as Ed indicates.

Nuthin' to it!

[ February 22, 2005, 12:51 AM: Message edited by: rattus ]
 
Status
Not open for further replies.
Top