# open delta transformer



## Rei (Mar 15, 2010)

I didn't post the complete solution, but my question is why phase voltage is 254V? This is delta which has Vline=Vphase. Therefore, Ia=37.8&lt;-36.9. Anyone see it differently?


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## tbob (Mar 15, 2010)

A-Phase as references..

Convert from Delta to Y with Van as a reference so angle is 0 degree.

Vab=sprt3 * Van Angle 0

Van= Vab/ (sprt3)

I think that's why Van=254 angle 0 degree.

We have a lot of expert in here will have some comment.

I don't know if it correct or not. But I am going to take my test this April and I hope someone will solve this problem for us.

Thanks.


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## DK PE (Mar 15, 2010)

This is a few more steps but may be easier to follow. Since they said assume Van = reference it is automatically 440/sqrt3 @ 0 = [email protected] 0.

Now assuming abc sequence Vab = [email protected] 30 and Vac = [email protected] This comes from just drawing the phasors tip to tail or look it up.

Some quick calculations tell you Z = 11.6 @ 36.9 since they said lagging and gave you each load is 16.7kVA @ 440V.

Now Iab = Vab/Z = 37.9 @ -6.9

and Iac = Vac/Z = 37.9 @ -67

Ia = Iab + Iac = 65.5 @ -36.9 if I did my math correct.

Hope this helps... I think they were looking to size the transfomers but I didn't see that part of solution.


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## DK PE (Mar 16, 2010)

Upon further reflection this morning, the solution given in the book is just working the problem on a *per-phase basis*. Don't worry about the open delta, it supplies a 3 phase *balanced load*. If you then divide that total load by 3 and use the line to neutral voltage which is 254, you get the current.


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## Rei (Mar 16, 2010)

DK PE said:


> Upon further reflection this morning, the solution given in the book is just working the problem on a *per-phase basis*. Don't worry about the open delta, it supplies a 3 phase *balanced load*. If you then divide that total load by 3 and use the line to neutral voltage which is 254, you get the current.


but in the delta configuration, line to neutral and line to line voltages are the same


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## Flyer_PE (Mar 16, 2010)

^In delta configuration, there is no line-neutral voltage. The reduction by the sqrt(3) actually applies to the current rather than the voltage. Mathematically, however, there is no real difference between a delta and a wye connected transformer so long as you are dealing with voltages and currents _external_ to the transformer. That enables you to treat either as if it is wye-connected for the purposes of determining line currents and how they affect voltage drops etc.


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## Bluekayak (Mar 16, 2010)

Flyer_PE said:


> ^In delta configuration, there is no line-neutral voltage. The reduction by the sqrt(3) actually applies to the current rather than the voltage. Mathematically, however, there is no real difference between a delta and a wye connected transformer so long as you are dealing with voltages and currents _external_ to the transformer. That enables you to treat either as if it is wye-connected for the purposes of determining line currents and how they affect voltage drops etc.


Flyer brings up an excellent point! So, in general mathematical terms, how does the load connected to this delta configured transformer differ from the 500kVA load placed between phases B and C on the 3-phase, 4-wire system given in NCEES #110?

DK_PE's approach is very elegant, however isn't the resultant impedance (i.e., Z=V^2/[email protected](0.8)[email protected]) the impedance of the load, not the delta phase impedance? The equivalent wye impedance would be Zy=Zd/3.

I apologize if I'm confusing the issue.


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## Rei (Mar 16, 2010)

I still don't get it. Yes, we are talking about the current which has the 1.73 different, but it's still 37.88A per phase, not 65.6A


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## Flyer_PE (Mar 16, 2010)

Try drawing the circuit a little differently:


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## DK PE (Mar 16, 2010)

My thinking was aligned with Flyer's but I believe I see where I may help with your "where did sqrt(3) go?"

Starting at left side of paper draw three horizontal lines. These are coming from your open delta which you know generate a standard set of three phase voltages. Draw those phasors with Van = 440/sqrt3 @ 0. Yes, I know there isn't a neutral "wire" but it does help to visualize the Vab etc. voltages and angles. The fact that these voltages were generated by an open delta is irrelevant. My opinion is you should have a reference sheet or page in a text that "falls open" to a figure like this.

Now extend the lines over and connect up your load in delta. I think you agree that each of the currents in EACH load is ~ 38A and the line current to phase current relationship in a delta load is sqrt 3 so the line current is 38* sqrt (3) = 65A.

Unless you are asked for the current the answer can be obtained more easily than all this work. For an open delta connection, the total kVA load possible is (1/sqrt(3)) x the two transformers ratings. See Wildi text for example.

Using Flyer's numbers 0.866 * 2 * 29kVA = ~ 50kVA load

BE CAREFUL. some references also use the 58% of the sum of the three transformers. KNOW which is which, under test pressure you don't want to be sorting it out.


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## DK PE (Mar 16, 2010)

Bluekayak said:


> Flyer brings up an excellent point! So, in general mathematical terms, how does the load connected to this delta configured transformer differ from the 500kVA load placed between phases B and C on the 3-phase, 4-wire system given in NCEES #110?
> DK_PE's approach is very elegant, however isn't the resultant impedance (i.e., Z=V^2/[email protected](0.8)[email protected]) the impedance of the load, not the delta phase impedance? The equivalent wye impedance would be Zy=Zd/3.
> 
> I apologize if I'm confusing the issue.


I don't have NCEES # 110 but the first difference is the load is balanced in this problem which enables you to handle it per-phase if you want. I would guess the best way to solve #110 is careful use of phasors but since I don't have problem just guessing a bit.

I meant the LOAD impedance is 11.6 ohms although didn't state it explicitly. The given diagram can be confusing when you are trying to sort out the transformer windings from the load. That is why in my last post I suggested breaking the problem into two parts and draw the load in delta on right side of page.

I don't see any advantage in this case to perform a delta - wye conversion


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## Rei (Mar 16, 2010)

> I don't see any advantage in this case to perform a delta - wye conversion


I agree and the answer is different when you make or don't make the conversion and so it's hard to tell on the exam whether you should or should not make the conversion.


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## HornTootinEE (Mar 17, 2010)

In a 3-ph balanced load system:

|S| = SQRT(3)xVL-LxIL

Why??

Y System:

VLL (V Line-Line) = SQRT(3)xVp (V phase or voltage across the load)

IL (Line current) = Ip (phase current or current actually flowing through the load)

Using substitution from phase values to line values and multiplying times SQRT(3)/SQRT(3):

|S| = 3VpIp = 3(VLL/SQRT(3))(IL) =&gt; *|S| = SQRT(3)xVLLxIL*

Now Delta:

VLL = Vp

IL = SQRT(3)xIp

math works out same as above.

NOTE: ONLY WORKS ON A BALANCED LOAD SYSTEM. And those systems with exactly balanced loads are very rare if not very very rare.

there is always a neutral point, even in a delta or open delta system. The question is whether or not it's physically there (in a Y it's where the phases meet, and is often grounded) or if it's only mathematically present as in a delta system. You can always go back and work this single phase by finding the voltage across one phase of the load and the current flowing through the load in a single phase.


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## Rei (Mar 18, 2010)

> math works out same as above.


for which part?


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## HornTootinEE (Mar 18, 2010)

The math to derive the formula |S|=SQRT(3)xVLLxIL works out the same way for a Wye or Delta system. I only showed it once. The difference is where the SQRT(3) comes from. In a Wye system it works out from the relationship between phase-phase and phase-neutral voltages, in a Delta the SQRT(3) comes from the relationship between phase current and line currents.


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## Rei (Mar 18, 2010)

djohnson.ee said:


> The math to derive the formula |S|=SQRT(3)xVLLxIL works out the same way for a Wye or Delta system. I only showed it once. The difference is where the SQRT(3) comes from. In a Wye system it works out from the relationship between phase-phase and phase-neutral voltages, in a Delta the SQRT(3) comes from the relationship between phase current and line currents.


But we are looking for the kVA rating of the two individual single phase transformers and so S = IV with no sqrt(3) factor. Therefore, it does matter whether we use line or phase value correctly.


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## HornTootinEE (Mar 19, 2010)

I was just showing you a quick way to find your line current external to the transformer. I come up with 65.6 A when using |S|=SQRT(3)xVLLxIL, which is exactly what the NCEES came up with for a line current... Then the transformer voltage being 440V, 65.6 A x 440V = =*28.9 kVA* rating per transformer. Is that the correct answer for the problem?


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## HornTootinEE (Mar 19, 2010)

One question for all: Would the NCEES ask for 28.9 kVA, 29 kVA, whatever or would they ask for the actual transformer rating? I don't think I've ever seen a transformer rated 29 kVA. It would be 30 kVA probably.

Just curious if give you mathematical answer choices or if they give you real-world answer choices.


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## Flyer_PE (Mar 19, 2010)

djohnson.ee said:


> I was just showing you a quick way to find your line current external to the transformer. I come up with 65.6 A when using |S|=SQRT(3)xVLLxIL, which is exactly what the NCEES came up with for a line current... Then the transformer voltage being 440V, 65.6 A x 440V = =*28.9 kVA* rating per transformer. Is that the correct answer for the problem?


It's the same answer I got above.



djohnson.ee said:


> One question for all: Would the NCEES ask for 28.9 kVA, 29 kVA, whatever or would they ask for the actual transformer rating? I don't think I've ever seen a transformer rated 29 kVA. It would be 30 kVA probably.
> Just curious if give you mathematical answer choices or if they give you real-world answer choices.


The short answer is that they could do either. I've seen both in sample questions. You have to read the questions carefully. They could ask for the minimum acceptable size and have an answer just below what is required. If you go with your first instinct and pick the closest answer (giving you a slightly under-sized transformer) rather than the next available size up...........


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## HornTootinEE (Mar 19, 2010)

Flyer_PE said:


> djohnson.ee said:
> 
> 
> > I was just showing you a quick way to find your line current external to the transformer. I come up with 65.6 A when using |S|=SQRT(3)xVLLxIL, which is exactly what the NCEES came up with for a line current... Then the transformer voltage being 440V, 65.6 A x 440V = =*28.9 kVA* rating per transformer. Is that the correct answer for the problem?
> ...


So in response to that, do they want you to think like a student or any engineer? I could as an engineer choose to use a 25 kVA knowing that it will be slightly overloaded, but that it could be well within what I find acceptable for use and it is a very common size. As a student I would say "Definitely it has to be a minimum of 28.9 kVA because the math says so" I suppose it goes back to "Read the question carefully!"


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## Flyer_PE (Mar 19, 2010)

djohnson.ee said:


> So in response to that, do they want you to think like a student or any engineer? I could as an engineer choose to use a 25 kVA knowing that it will be slightly overloaded, but that it could be well within what I find acceptable for use and it is a very common size. As a student I would say "Definitely it has to be a minimum of 28.9 kVA because the math says so" I suppose it goes back to "Read the question carefully!"


Here's my thought on this in the exam:

It's a multiple choice test. You don't get to write any kind of justification about choosing to live with a slightly overloaded transformer. If given a choice between a 25 and a 30 for this application, the 30 would get my vote. The only way it wouldn't is if the question got into the economics of how much the price difference is vs. service life etc. However, going there would result in a question that couldn't be answered in six minutes or less.


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## Kahrlo (Sep 6, 2010)

The easiest way to do this is to remember that in an open delta connection (2 transformer), that the maximum power that you can have is only 87% of the total kVA of the transformer. Therefore, the minimum power that each transformer must have is

50kVA/2/0.87= 28.735 kVA for each transformer


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## NoobEE (Mar 10, 2011)

The easiest way to do this is by just multiplying the Rated Power by sqrtof3 divided by 3 ..

[ 50 KVA x sqrtof 3 ] / 3 = 28.86 KVA ...

do you still want me to explain why I did it this way?


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## NoobEE (Mar 10, 2011)

Kahrlo said:


> The easiest way to do this is to remember that in an open delta connection (2 transformer), that the maximum power that you can have is only 87% of the total kVA of the transformer. Therefore, the minimum power that each transformer must have is
> 50kVA/2/0.87= 28.735 kVA for each transformer


..

kindly double check on your equation Sir.. I apologize but it doesn't equate to 28.735 based from your equation


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