Wye-Wye Transformer and grounding 3

[bspace123]

Hi All

I have been reading for hours and still confused about the grounding concepts for wye-wye transformers.

1. Say we have a wye-wye transfrormer and the primary side neutral is not connected to ground but the secondary side neutral is connected to ground, if we have a phase-ground fault on the secondary, will fault current flow from phase -> ground -> neutral of the secondary transformer? And will this fault current be seen as phase current on the primary side?

2. Say we now connect the primary side neutral to ground, will this change anything?

3. Is it common practice to bond the STAR points together if they are connected to ground for this type of transformer? If so, why?

4. If we have both the primary and secondary side neutrals connected to ground, but the primary side is via a neutral earthed resistor, if we have a ground fault on the primary side, will fault current still be limited by the resistor? or will the bonding mentioned in point 3 compromise this?

 

[RRaghunath]

1. If primary neutral is not grounded and if only source for secondary distribution system is from the same transformer primary, there cannot be any current in the secondary earth fault (except some small current due to capacitive coupling between primary neutral and earth).
2. If both primary and secondary neutral points are earthed, the earth fault current for fault on secondary will come from the primary and flows between corresponding primary phase and earth (through primary neutral). The zero sequence impedance for such an arrangement is nearly equal to the transformer rated %Z.
3. Star points will be individually earthed through dedicated earth electrodes. The earth mats will be interconnected below earth.
4. The transformer primary winding can contribute to fault in the primary power system if there is an embedded Delta in the transformer or if there is a source on the secondary side. Otherwise No.
The neutral earthing resistor on primary will affect the fault current magnitude on secondary side.
The fault current in primary power system will be decided by the source upstream and the NER on the transformer primary cannot affect (unless there is an embedded Delta winding in the transformer as mentioned earlier).

 

[VMMF]

@RRaghunath Great post. I would like to ask you

5. On delta(pri)-wye grounded(sec) systems if we have a phase-ground fault on the secondary, fault current will flow from phase -> ground -> neutral of the secondary transformer?
6. Could you please provide some links or name articles further explaining your answer # 1?
7. Is there a way to intuitively understand why if a fault occurs on the secondary ,in a delta(pri)-wye grounded(sec) transformer the fault current will flow, but in a wye ungrounded(pri)-wye grounded(sec) transformer the fault current will not flow?
8. If transformer is wye grounded-wye grounded. A fault on primary side will not be visible on secondary side?

 

[davidbeach]

The first rule of transformers is that for any given winding, if current can't flow on one side it can't flow on the other side. In the wye-wye with only one neutral grounded, current in one winding, from phase to neutral, typically associated with a ground fault, has no place to flow on the other side without the neutral grounded. Down one winding, but then nowhere to go. It can't go out the neutral as there isn't any connection, and it can't go back up other phases because the fault isn't on those phases on the faulted side.

I’ll see your silver lining and raise you two black clouds. - Protection Operations

 

[bspace123]

Makes perfect sense, thanks!

 

[waross]

Hello David.
I hesitate to suggest another alternative to yours.
We seldom disagree and when we do, I am usually wrong.
I understood that the primary impedance on a shorted transformer dropped.
With a connected wye point, the unbalanced current has a path to return to the source.
Now consider a floating primary wye point on a wye:wye transformer.
Compare this to three resistors in a three phase wye.
If the resistors are equal there is no need for a wye point connection.
Now replace one of the resistors with one of lower value.
For example, 100 Ohms, 100 Ohms and 10 Ohms.
The current through the 10 ohm resistor returns through the other two resistors.
The voltage rises through the 100 Ohms resistors until Ohms law is satisfied.
My understanding of a floating primary wye point was that in the event of a heavy load or a short circuit on one phase of the secondary, that the current would return through the other two phases with a resulting rise in voltage.
However the effect is non-linear.
The voltage will rise on the unfaulted phases until those cores saturate and the impedance drops.
A balance will be reached with the voltage across the unfaulted phases at the knee of the saturation curve and the remaining voltage driving current through the heavy load or fault.
Respectfully
Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[VMMF]

@davidbeach . Thank you for your answer. I'm not sure if you were answering my question #8, which is, if we have a wye grounded-wye grounded transformer, does a fault on primary side will trip relays on secondary side? I know secondary side faults are visible on primary side, but does primary side fault current trying to return to source affect secondary side which is not where the source is? Thank you

 

[bspace123]

I’m also still puzzled by the floating neutral on the primary leading to no earth fault current in the secondary.

If we have a Wye connected motor with no neutral connected and we increase voltage on one of the phases, the increased current will flow back through the other two phases? There’s no need for a neutral connection for this current to flow. But we are saying that does not happen for a Wye connected transformer with floating neutral?

 

[waross]

4. If we have both the primary and secondary side neutrals connected to ground, but the primary side is via a neutral earthed resistor, if we have a ground fault on the primary side, will fault current still be limited by the resistor? or will the bonding mentioned in point 3 compromise this?

Normally, a fault on the primary side will be an issue for the primary source protection.
There are two exceptions for a wye primary.
1. A grounded wye primary and a delta secondary or tertiary.
2. A three legged transformer core forming a phantom delta.
In both cases, the a fault on the primary will cause a back-feed of current into the fault from the healthy phases.
In the case of the grounded wye delta, the back-fed current will be limited by 3 times the transformer impedance. (1/3 the available short circuit current)

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[davidbeach]

Bill's resistor analogy is useful for how the voltages will put the neutrals and phases at unexpected voltages to ground during a fault on the grounded side, but I don't think it is complete for a transformer. After all, Ohm isn't the only guy with a law named after him, there's that Ampere guy as well. The resistors don't have to worry about amp*turn balance, it just isn't a thing for resistors. But it definitely is a thing for transformer windings.

Ignoring things like phantom tertiary effects and the like (build your mental image transformer from 3 single phase cans), there's no path for the ungrounded side fault current that meets Ampere's Law. What will happen is that the faulted phase on the grounded side will get pulled close to ground and the other phases will move away from ground. Anything connected phase-neutral on those will experience a significant overvoltage. The ungrounded side neutral (wye point more precisely) will also pull close to ground across the winding feeding the fault on the other side.

Sketch it out. Three single phase ideal transformers connected in wye-wye with one side grounded. Use 1:1 transformers to make the math easy. In that case each winding has equal and opposite currents, same magnitude but 180 degrees apart in the windings; each has the same voltage across the winding. All current must flow in loops, you can't have a current "stop" someplace but flow else where. You can have a source on the side with the ungrounded neutral that can supply any current at any angle, but all other nodes must follow Kirchhoff (another guy with a Law).

I’ll see your silver lining and raise you two black clouds. - Protection Operations

 

[waross]

As the voltage across the unfaulted windings rises the cores will eventually saturate.
With the cores saturated, the effective impedance will drop and the current will increase disproportionately.
The current at the actual fault will be less than if the primary wye point was grounded, but will still be significant.
The increased current in the saturated windings will balance the current in the faulted phase making Mr. Kirchhoff content.
Without knowing the saturation voltage of the transformers it is difficult to predict the actual voltages and fault currents, but a balance between the currents in all three phases will be reached when the unfaulted transformers are somewhere on the saturation knee.
As a guesstimate the fault current may be 1/4 to 1/2 of what the fault current would be with the primary wye point grounded.
(Too bad we can't have this discussion over coffee with a full napkin dispenser on the table. Yours, Bill)

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

Dear Mr bspace123
B. "... I ... confused about the grounding concepts for wye-wye transformers.
1. Say ...,,. the primary side neutral is not connected to ground but the secondary side neutral is connected to ground, if we have a phase-ground fault on the secondary, will fault current flow from phase -> ground -> neutral of the secondary transformer? And will this fault current be seen as phase current on the primary side?
2. Say ... connect the primary side neutral to ground, will this change anything?
3. Is it common practice to bond the STAR points together if they are connected to ground for this type of transformer? If so, why?
4. If ..... both the primary and secondary side neutrals connected to ground, but the primary side is via a neutral earthed resistor, if we have a ground fault on the primary side, will fault current still be limited by the resistor? or will the bonding mentioned in point 3 compromise this?..."
C. I tried to sketch it out by using 3 pcs of 1:1 YNyn 0 Group connection with primary connected to a 3-phase 120deg source L1, L2, L3 without N. See attached drawing (a). To see whether it make sense.
C1. See attached drawing (b)
a) There will be secondary current a1=af=an ; but a2=a3=0 ,
b) There will be primary current A1, A2 and A3; with A1>A2 also A1>A3; but A2=A3 ,
c) The primary AN = 0.
C2. See attached drawing (c)
Current distribution is as C1 above.
C3. Is it common practice to bond the STAR points together ....see (c). Reason being to maintain the same earth potential.
C4.1. " both the primary and secondary side neutrals connected to ground, but the primary side is via a neutral earthed resistor, "
The statement is incorrect.
See attached drawing (d) where the primary N is earthed through an earthing resistor R. The secondary n is directly earthed.
Current distribution is as C1 above.
Che Kuan Yau (Singapore)

 

[che12345]

Dear All,
I have problem to attach the dwg. Please help me. Thank you.
Che Kuan Yau (Singapore)

 

[che12345]

Dear All,
The attachment is in order.
Che Kuan Yau (Singapore)

 

[waross]

May I help?

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

This is a simplified explanation.
I will neglect magnetizing currents and quadrature effects to illustrate the effect of saturation on transformers in series with afloating neutral.
I will assume saturation occurs at a point rather than a knee.
Assumption.
120 Volt transformers in series across 240 Volts.
Transformers saturate at 120% (144 Volts)
Transformer rated current: 10 Amps
Load on the healthy transformer, T1: 1 Amp.
Available Short Circuit Current: 2000 Amps.
As the load on T2 increases, the effective impedance drops and the voltage across T2 increases.
Approaching saturation, the voltage across T1 will be approach 144 Volts. (120% of 120 Volts.)
The voltage across T2 will be 96 Volts. (240V - 96V)
The current will approach 14.4 Amps.
Once T1 is saturated, the impedance will decrease as the current increases.
With a short circuit on T2, the current will be limited by a voltage approaching 96 Volts. ASCC = (2000 Amps / 120%) = 1670 Amps.
The calculations are a little more difficult for three phase circuit.
The voltage will approach 1.73 times rather than 2 times normal voltage, the phase displacement between the healthy Phases will decrease from 120 degrees and approach 60 degrees.
However, for both single phase and three phase, the voltage across the healthy transformers will increase to the point of saturation, and once saturated the current will rise with the current demanded by the faulted phase.
The three phase fault current will be less than with a solidly connected neutral but will be significant.
I have ignored the effect of the X/R ratio and have ignored the error resulting from this oversight.
My intention is to demonstrated the effect of transformer saturation, not to accurately calculate the actual fault current.

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

Mr. Che.
Please try this.
When you want to quote a passage, highlight the passage and press "Cntrl C" to copy the passage.
Then click on the cartoon character with the text bubble. It is too the right of the Omega symbol in the line above.
You will see a pop-up box that gives you the option to enter the name of the person or authority quoted.
You may enter a name and press OK or enter or simply press OK or enter.
Now you will see the Quote code, with the cursor positioned to accept a quote.
Press "Cntrl V" and your quote will be entered into the quote code.
Press preview to view your result before Submitting your post.
This will make your replies much easier to read.

Example:
Dear Mr bspace123

B. "... I ... confused about the grounding concepts for wye-wye transformers.
1. Say ...,,. the primary side neutral is not connected to ground but the secondary side neutral is connected to ground, if we have a phase-ground fault on the secondary, will fault current flow from phase -> ground -> neutral of the secondary transformer? And will this fault current be seen as phase current on the primary side?
2. Say ... connect the primary side neutral to ground, will this change anything?
3. Is it common practice to bond the STAR points together if they are connected to ground for this type of transformer? If so, why?
4. If ..... both the primary and secondary side neutrals connected to ground, but the primary side is via a neutral earthed resistor, if we have a ground fault on the primary side, will fault current still be limited by the resistor? or will the bonding mentioned in point 3 compromise this?..."

C. I tried to sketch it out by using 3 pcs of 1:1 YNyn 0 Group connection with primary connected to a 3-phase 120deg source L1, L2, L3 without N. See attached drawing (a). To see whether it make sense.
....

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

Dear Mr waross
1. Thank you for your kind advice the procedure on when quoting a passage.
2. Looking forward to your learned advice on the current distribution per the sketch make sense at all. Thank you.
Che Kuan Yau (Singapore)

 

[waross]

You must consider saturation that will limit the voltage across the healthy transformers to about 115% to 120% of rated voltage.
The remaining voltage across the faulted transformer will drive the fault current.

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[VMMF]

Questions 5,6,7,and 8 that I made have been ignored so far. I think they were overlooked with so much activity in this thread. Could anyone please reply them in order? Thank you