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XFMR test question - wye-delta TTR

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1. Member
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Originally Posted by Kalbi_Rob
WI | DE

So, easy way to remember how to perform turns ratio is to start with the acronym WIDE where WI stands for Wye where current is equal between phase and line, and DE stands for Delta where E (electromotive force which also defines voltage) or voltage is equal between phase and line.
So, on the Wye side of the transformer you need to take the voltage (assume to be phase voltage if not otherwise noted) and divide by square root of 3 (or ~1.73). So the equation is
(7200V/(sqrt(3))/480V = 4156.9V/480V = 8.66 so the TTR = 8.66:1

Turns ratio is only taken on single phases of a transformer, so all values need to be reduced to individual line values to represent each single phase. This is also why you see a 12kV/480 delta/wye transformer with voltage values of 12kV - 480Y/277V.
If for whatever reason you needed to convert current values from phase to line on a Delta Xfmr, then you take phase current and divide by sqrt(3).
thank you. this is very helpful.

2. Originally Posted by ElectricalTestTech
It is not a simple division. You have to pay attention to the Wye and Delta. Any winding that is Wye, you have to use phase to ground voltage. 7200/1.732. Delta windings use phase to phase.

7200/1.732= 4160ish / 480 = ?

Sorry I’m doing the math in my head and on my cell phone.

Hope this helps
Is there any illustration that can show this math? Sorry i am a visual learner lol

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Originally Posted by Ronwilson1801
Is there any illustration that can show this math? Sorry i am a visual learner lol
See if this helps

4. Originally Posted by dbryce89
A wye-delta transformer has a primary line voltage of 7200 V and a secondary line voltage of 480 V. What is the turns ratio?

Not sure what I did wrong on this question. They are both line voltages so I thought it was a simple division question.
Wye VP = VL ÷ √3

Delta VL = VP

Wye VP = 7200 ÷ √3 = 4157
Delta VL = 480
4157/480 = 8.7:1

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Originally Posted by dbryce89
A wye-delta transformer has a primary line voltage of 7200 V and a secondary line voltage of 480 V. What is the turns ratio?

Not sure what I did wrong on this question. They are both line voltages so I thought it was a simple division question.
You did nothing wrong. I do not know why are assuming that you did anything wrong. Just FYI - I am not an administrator here or at NETA. Just a regular member. I am just baffled by the confusion here that should not have existed.

The gentlemen on this thread have made all sorts of assumptions and tried their explanations. With all due respect to their efforts, I am not sure how many of these gentlemen or NETA question makers use a standard textbook definition of Turns Ratio.

For Turns Ratio - You take identical parameters. Line parameters (LL) or Phase parameters (LN). If nothing is mentioned, automatic assumption is that the values are Line parameters NOT phase. These are the first concepts of power engineering.

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Originally Posted by ElectricalTestTech
It is not a simple division. You have to pay attention to the Wye and Delta. Any winding that is Wye, you have to use phase to ground voltage. 7200/1.732. Delta windings use phase to phase.

7200/1.732= 4160ish / 480 = ?

Sorry I’m doing the math in my head and on my cell phone.

Hope this helps
"Delta windings use phase to phase"

The fundamental is: In delta windings, V Line to Line or V subscript LL = V Line to Neutral or V subscript LN. So no conversion is required on Delta side.

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The way I taught my guys that seemed to stick with them was to visualize all transformer ratios as single phase devices.

First off, it's important in this example to note that it's a WYE primary, which is not common and is an easy way to make a mistake.

With a line voltage 7.2kV, we divide by 1.732 to find a phase voltage ~4.16kV.
In the DELTA secondary, VL and VP are equal at 480V.

At this point you can ignore the fact that these are part of a three phase system. You have three individual transformers with primary voltage of 4,160 and secondary at 480.

4,160/480 = 8.67

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