3V0 (broken delta) Resistor Sizing

[CaptElectron]

Hello All,

I have a 3V0 (broken delta) resistor sizing question…

We have a 4160V, 1388kVA generator connected to a step-up transformer using a short distribution line (approx 3800 feet). The generator has a HRG system on it.

The distribution line is at 4160V and is configured as an ungrounded delta.

We put a set of PTs on the distribution line to detect ground faults. The PTs are connected wye (grounded) on the primary and broken delta on the secondary. They are rated 4200/120V and 1000 VA each.

I’ve read that it may be a good idea to connect a resistor across the broken delta to diminish the effects of ferroresonance. Following the calculation methods of an application guide, I came up with a maximum voltage (under fault conditions) on the resistor of 206V. Wanting to limit the current so as to not over-duty the 1000 VA rating of the PTs (using the continuous current rating, as it may be a while before the fault is cleared) I calculate a max current of 8.33 Amps.

To achieve the 8.33 Amps, I calculate that I’d need a 24 Ohm resistor (minimum) with a power rating of 1,715 Watts.

My question is… why not increase the resistance significantly and limit the power requirements of the resistor?... for example, why not use a 100 KOhm resistor with a 1/2 Watt rating, which would be much cheaper and easier to find?

Thanks in advance.

 

[waross]

Under fault conditions I would expect the healthy phases to develop 119 volts. Two 119 volt windings in open delta = 119 volts.
1000VA/119V = 8.4A. 8.4A at 119V = 14.2ohms. Power = 1000W.
Why are you concerned with ferro-resonance? I understood that it was an issue with longer distribution lines.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[CaptElectron]

The guide I had said to use this equation:

Vresistor(max) = 3x(Vlg/VTR), where Vlg is Voltage (line to ground) of the primary and VTR is the turns ratio of the PTs. So in my case it would be 3 x (2402V/35) = 206V.

But either way you bring up a good point…

Do I need the resistor to begin with?

 

[dpc]

The issue is the ferro-resonance between the PTs themselves and the capacitance in the ungrounded system.

Blackburn suggests 125 ohms with a 350 watt rating (208 V) for a 4160 V system "derived from experience."

This assumes that the VT ratio is such that the line-neutral voltage corresponds to 69.3 volts on the PT secondary.

See "Protective Relaying - Principles and Applications" by J. Lewis Blackburn.

Cheers,

Dave

 

[waross]

Draw a sketch of the line to line voltages. This sketch should be an equilateral triangle. With a ground on the system, say "A" phase, the "A" corner of the triangle will be grounded. The PT primaries are connected phase to ground, so the ground fault will be connecting the wye point of the CTs to "A" phase. "A" phase PT will be shorted out by the ground fault and develop zero volts on its secondary. "B" and "C" phase PTs will be connected from "A" to "B" and "A" to "C" respectively. This is a line to line connection and the PTs will develop full voltage on their secondaries. These voltages will be in open delta so the resultant voltage will be equal to the developed voltages but at a different phase angle. "A" phase PT is not developing any voltage but it is still in series with the other PTs and contributes a small amount of reactance and resistance to the circuit.
Remember that the HRG resistor is limiting the ground fault current to a low level so we can have a ground fault on one phase for a time.
Remember also that with HRG the phase to ground voltage on the unfaulted phases rises to phase to phase voltage. Thus an HRG system must have all insulating components rated for line to line voltage rather than line to neutral voltage.
When connecting PTs in wye configuration it is good practice to evaluate the voltage they will see in the event of a ground fault. Your selection of 4200V PTs for a 4160V is appropriate.
I'll let some of the Gurus comment on the possibility of ferroresonance.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[slavag]

Hi.
Please see attached
thread238-210935
Best Regards.
Slava

 

[waross]

With all respect, Slava, I think that some of the Basler information is misleading.

VT Voltage Rating
It is important to note that, during a ground fault, two
of the VTs must withstand and reproduce full line-line
voltage. It would be an error to use a VT rated only for
line-ground voltage. For instance, in the wye-broken delta
system described above, the normal system voltage is
13.8kVLL and 7.97kVLG, and secondary voltage is 69.3
under normal operating conditions but rising to 120V
during a fault. The VT in this case needs to be rated for
13.8kV/120V.

It would make this a little clearer to add that this describes the individual PT voltages and not the broken delta voltage.
The normal broken delta voltage is zero and the broken delta voltage with a ground fault on the primary is 120V.
Some of the explanations are somewhat confusing in that it is not made clear whether the voltages discussed are PT voltages or broken delta voltages.
Some of the descriptions do not make it clear that we must consider the vector sum, not the simple sum of the PT voltages.
In normal operation, each PT develops 69 volts, the vector sum is zero volts. In a normal delta, the vector sum of the three phase voltages is zero. Breaking the delta does not change this.
With one phase grounded, the PTs on the healthy phases each develop 120 volts. They are in effect in open delta or the IEC V connection. The vector sum is 120 volts. This is the same as an open delta or V power transformer circuit. All three line to line voltages are equal. Now to this resultant 120 open delta voltage is added the PT on the grounded phase. The ground has effectively shunted this PT and it develops zero volts. It is in series with the other two PTs and adds some impedance to the measuring circuit but no voltage. As a result, 120 volts is seen across the break.
I think that multiplying the line to neutral voltage times 3 and then dividing by root 3 is not a valid operation and only accidentally gives the correct answer.
But we haven't answered the prime question.
Do we always consider the issue of ferro resonance or can we ignore it for small systems? Is ferro resonance more of an issue with metering transformers than with distribution transformers where it is relatively rare?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[slavag]

Hi Bill.
We must separate open delta and broken delta terms.
What is standard practic of VT/PT choise with broken delta connection:
(xxx/sqr3)/(110/sqr3)/(110/3). (standard secondary 100/110/115/120 V).
A last 110/3 is used for the broken delta connection.
Now, in case of ground fault:
healthy phase is (110/3)^sqr3 and vector sum of them is 110V.this vector sum is voltage on the broken delta.
Regards.
Slava

 

[waross]

Hi Slava. I take exception to the use of "3" in the equations.
The sum of the voltages is zero, not 3 times anything.
Under normal conditions the delta is balanced. With a wye or star primary connection, the voltage on each phase will be 1/1.73 of normal voltage because that is the normal phase to neutral voltage. Each phase will be equal.
A phase = 69 volts.
B phase = 69 volts.
C phase = 69 volts.
No times three. Just divide by root 3.
Total voltage isn't three times anything, it is the vector sum of the three phases and equals zero.
I have never seen this confusion applied to a power delta circuit, why confuse a metering delta circuit.
Yes, I have had occasion to use a broken delta on a large power transformer.
Consider a broken delta as a normal delta connection with a very high PU impedance.
Under a ground fault, the impressed primary voltages change and so also do the secondary voltages. The delta voltages no longer vector sum to zero. As a result very high circulating currents will result, limited by the PU impedance of the transformers. Breaking the delta limits these currents by raising the effective PU impedance of the transformer circuit.
Under a short circuit, two phases see full line voltage and develop 120 volts on the secondary. The third phase develops zero volts. The vector sum of two 120 volt phases is still 120 volts. No times three, and we have now also lost the root 3. The only thing times three is the transformer resistances and this is negligible on a broken delta circuit. (Note that the voltage drops across the resistances are out of phase and do not equal 3 times anything.)
We have a root 3 ratio between the line voltages and the neutral voltages, but no factor of three.
Yes, there is a difference between an open delta and a broken delta. An open delta takes two transformers and a broken delta takes three transformers.
BUT, with a ground fault on a broken delta circuit it is valid to consider the circuit to be an open delta circuit with an impedance across the open side equal to either the relay or relay resistor combination in series with the third PT impedance with the primary shorted.
Am I missing or misunderstanding some convention here?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

I think that I see the problem. The formula is for the sizing of resistors. That would depend on the VA rating of the PTs times three for the total power, except; Under ground fault conditions, only two transformers will be supplying voltage. One transformer will have the primary shorted by the ground fault and will not contribute.
How about this formula:
Each PT = 1000VA
1000VA/120V = 8.33 A.
We will limit the maximum current to 8.33A.
Maximum broken circuit voltage will be 120 volts.
120V/8.33A = 14.4 Ohms minimum resistance.
Determine the maximum allowable current for the smallest PT and limit the current to below that value.
Do not be confused by three phase power calculations, the power in an open delta circuit is single phase power.
Does this reconcile our friendly misunderstanding?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[scottf]

A few points:

You need to make sure you're using 2-bushing VTs rated to take the L-L voltage continuously....understanding you're connecting them line-to-ground.

Also, I think too much effort it used when sizing resistors for ferroresonance suppression using the broken-delta method. Best advice is to size the resistor such that it results in the thermal burden at rated voltage, i.e. for a 1000VA thermal burden and a 120 V rated secondary, use an 8 ohm resistor.

I don't know of many, if any, situations where this simply application of a resistor didn't work on a distribution line.

 

[davidbeach]

I would think that you could use a one-bushing VT, but it would certainly need to be rated for line-line voltage. If the non-bushing connection is to ground, it should never see a significant voltage regardless of faults on the system. No?

 

[slavag]

Hi.
Im agree with David, used one bushing VT.
Bill, please see attached sketch of broken delta voltage in the ground fault time. It's show why U0 is 3 times of phase voltages.
Best Regards.
Slava

 

[waross]

Hello Slava;
Thank you for your time and thank you for the education.
I see the source of my error.
Considering the two healthy phases only;
At first glance it would seem that a delta bank with one transformer at zero volts would be an open delta. However, when the primary transitions from normal to ground fault conditions the healthy phasers rotate, one clockwise and one counter clockwise.
The primary of the two transformers is now connected in open delta BUT the polarity of one of the transformers is reversed from a normal delta or open delta connection.
The result is that the secondary connection is not an open delta as it seems at first look. The secondary is now two phases of a wye or star connection. The vector sum is of course the same as it would be with two phases of a wye connection. Look at it as an open delta and then rotate one of the vectors 180 degrees.
Normal voltage is 69.3 volts. With a ground fault this rises by a factor of 1.73 to 120 volts. Two 120 volt windings in wye sum to 120 times 1.73 The result is the original voltage times 3, or 69.3 x 3 = 208 volts.
I have inherited so many grounded wye/delta power systems, open, closed, and one broken that I did not stop to realize the effect the phase shift was having on the metering transformers.
Again thank you for your patience and help, Slava.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[CaptElectron]

Great discussion everyone. Thanks for the help.

CE

 

[Kdanijel]

I agree, good and useful discussion.
I just want to give some points regarding ferroresonance damping.
Use of 100 KOhm resistor proposed by CaptElectron is not a solution, since it would not do much to damp ferroresonance.
Practice and some calculations show that decreasing resistance connected to broken delta increases chance of ferroresonance prevention.
We had experience with 35kV station (with 3 power lines) where normally used 16 Ohm resistor was not enough. After tests and calculations, conclusion was that 11 Ohm resistor is required to damp ferroresonance.
Value of effective damping resistor resistance depends on too many things: VT magnetizing characteristics, configuration of substation, possible switching situations... Very difficult to model and calculate and get reasonable result.
What resistance should be used? As small as VTs thermally allow, determined by conclusions of this discussion.

Danijel