Understanding delta VT ratios
Understanding delta VT ratios
(OP)
Hi
I have come across a 11kV VT which has one 33kV primary winding, one three-phase secondary winding and one delta winding. Ratios are given as [11kV/SQRT(3)]:[110V/SQRT(3)]: [110V/3].
I am comfortable with the ratios of the three-phase windings being [11kV/SQRT(3)]:[110V/SQRT(3)] which boils down to 100:1.
But I cannot understand where the 110V/3 comes from for the delta winding. The way I see it the open-delta voltage is the sum of the 3 phase voltages, VA + VB + VC = 3V0 = 0 with a healthy system. With a fully blown EF (say on A phase), VA is missing and so output is VA. This would be 63.5V and the primary side on the A-phase is 33kV/SQRT(3) which leads to 100:1 again? So why 110V/3?
Thanks.
I have come across a 11kV VT which has one 33kV primary winding, one three-phase secondary winding and one delta winding. Ratios are given as [11kV/SQRT(3)]:[110V/SQRT(3)]: [110V/3].
I am comfortable with the ratios of the three-phase windings being [11kV/SQRT(3)]:[110V/SQRT(3)] which boils down to 100:1.
But I cannot understand where the 110V/3 comes from for the delta winding. The way I see it the open-delta voltage is the sum of the 3 phase voltages, VA + VB + VC = 3V0 = 0 with a healthy system. With a fully blown EF (say on A phase), VA is missing and so output is VA. This would be 63.5V and the primary side on the A-phase is 33kV/SQRT(3) which leads to 100:1 again? So why 110V/3?
Thanks.






RE: Understanding delta VT ratios
RE: Understanding delta VT ratios
RE: Understanding delta VT ratios
Bill
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"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
Thanks for that. If that be the case that the delta windings are 110V phase-neutral, then the voltage ratio is then [11kV/SQRT(3)]/110 = 173.2, correct?
Is that the value then I would enter for the VN ratio of the delta connected VT in the relay? I have set the phase VTR = 100.
Secondly, this means that the open delta voltage is 110V for an EF?
Thanks.
RE: Understanding delta VT ratios
Regards
Marmite
RE: Understanding delta VT ratios
A closed delta will stabilize a floating neutral on the primary.
If the delta winding is rated at 110 Volts then it will be three 110 Volt windings connected in delta. (Open or closed).
Bill
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"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
Note that under IEC, these windings are only intended to be connected in a broken-delta configuration for residual voltage sensing. The winding rated for standard metering/protection (a-n winding for example) is only tested with the residual voltage winding (da-dn) with 0 burden up to 100% Vrated. The assumption is that the residual voltage winding will only see an effective burden during transient conditions and thus not impact the steady-state performance of the other winding(s).
If the da-dn winding does not have a listed accuracy class on the nameplate, then it is only intended to be used for ferroresonance damping purposes.
RE: Understanding delta VT ratios
I have a temporary unearthed system whilst generator gets synchronised to grid. An EF will then result in a full neutral shift and so the output voltage will be 110/3 = 36.67V.
Marmite - I am curious to know how you got 110V?
RE: Understanding delta VT ratios
"I am curious to know how you got 110V?"
The voltage on the broken delta is the total zero sequence voltage (3*V0) that appear during an earth fault.
All three windings will contribute to this voltage and since all three voltages are equal in magnitude and have the same angle
they will add to each other;
(110/3)+(110/3)+(110/3)V=110V.
-BAK
RE: Understanding delta VT ratios
If what you are saying is true, then it implies that all three primary phase voltages are equal in phase and magnitude as well. But how can this be for under an A-phase EF condition the A-phase voltage is 0?
RE: Understanding delta VT ratios
-BAK
RE: Understanding delta VT ratios
This is an international forum. What is called a broken delta in some parts of the world is called an open delta in other parts of the world.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
I'm not sure that is completely accurate.
I believe the attached picture accurately describes the naming convention universally. If someone would refer to the broken-delta configuration as "open-delta", then what do they call what is shown as open-delta?
RE: Understanding delta VT ratios
Well, the output of the broken delta VT (or open delta) in the phase domain, is the sum of the three phase voltages, VA + VB + VC. If VA is missing then VN = VB + VC = 3V0. Now we are given that the rated voltage per phase is 110/3. This is a phase value which equals the positive sequence voltage under healthy conditions. For EF's the phase and sequence quantities will be different.
I realise that you are using 3V0 = V0A + V0B + V0C. At the end of the day though this must lead to the same result as a phase domain analysis. In this example a phase domain analysis is not that complicated as there is no fault current and the healthy voltages are not distorted.
However, purely from a phase domain perspective how can the output voltage be 110V? I still believe it is 110/3. If I'm wrong I'd very much like to know how.
Thanks.
RE: Understanding delta VT ratios
By the way, I call it open corner delta, to me it is more clear on the concept.
Well, missing one healthy phase under unfaulted balanced conditions is different than single phase fault conditions.
If just sum two phase VT secondary vectors, you are correct, you will get Vb+Vc=-Va. Be noticed I mention individual VT's who has no magnetic flux interaction between them.
In the open corner delta, if there is a L-G fault, whether the system is grounded or not, there is negative and zero sequence involved, it is not same as missing a healthy phase. So, positive and negative sequence sum out zero, but zero sequence tripped. What is the magnitude of 3V0, depends, I don't know.
RE: Understanding delta VT ratios
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
I'll check if I have some hand made calculations that explains this, or maybe someone else has an example that they would like to share...
-BAK
RE: Understanding delta VT ratios
Commonly, for ground detection on an 11 kV circuit, the primary will be connected in wye. The applied voltage will be 6350 Volts.
The voltage of the delta winding will be 63.5 Volts.
Under ground fault conditions the applied voltage will rise to 11 kV on two phases and the secondary voltage will rise to 110 Volts on two phases. The voltage on the secondary of the faulted phase will drop to zero.
The directed resultant of two equal voltages displaced by 120 degrees will be an equal voltage displaced by 240 degrees from the reference voltage. 110 Volts.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
RE: Understanding delta VT ratios
The two voltages will have an angle of 60° between them - 110V∠30° + 110V∠-30° = 190.5V.
This is why the VT rating is typically 110/3V rather than 110/√3V on the the delta winding - it gives a nice 110V during a fault (63.5∠30° + 63.5V∠-30° = 110V), which is a more typical level for voltage relays.
RE: Understanding delta VT ratios
With the loss of one phase, you have two windings displaced 120 degrees and developing 63.5 Volts each, summing to 63.5 volts.
With one phase grounded, that phase is shorted out and develops zero Volts. The other two phases now develop 110 Volts each, summing to 110 Volts.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
Refer to the below diagram of the primary voltages. The blue line is pre-fault system, the green line is the post-fault system (fault is A-phase to earth).
Before the fault, the angle between VB1-N and VC1-N is clearly 120°, and the magnitude (not drawn) is 6.35 kV.
When the fault occurs, VA moves to N, so VB and VC also shift to compensate, which is why VB-N and VC-N magnitudes are now 11 kV. It should be visually apparent that the angle between VB2-N and VC2-N (both shown as dashed red lines) is a lot more acute than previously - if you do the trig, it is now 60°. This same change in angle occurs in the secondary as well, hence why we are adding two voltages that are 60° apart, not 120°, giving an "extra" factor of √3, hence why the veritas's VT has a ratio of 6350:36.7 not 6350:63.5 on the broken delta-connected winding.
RE: Understanding delta VT ratios
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
RE: Understanding delta VT ratios
3.10.3
open-delta connection
the winding connection in which the phase windings of a three-phase transformer, or the windings for the
same rated voltage of single-phase transformers associated in a three-phase bank, are connected in series
without closing one corner of the delta
Whereas the IEEE 100
The Authoritative Dictionary of
IEEE Standards Terms
open-delta connection (power and distribution transformers)
A connection similar to a delta-delta connection utilizing
three single-phase transformer, but with one single-phase
transformer removed.
Regards
Marmite
RE: Understanding delta VT ratios
However, purely from a phase domain perspective how can the output voltage be 110V? I still believe it is 110/3. If I'm wrong I'd very much like to know how.
The key thing here is whether the system is grounded or not (ungrounded, hiZ-grounded).
I think the OP have got a really good answer from "mgprt" for an ungrounded network. If the network is effectively (solid) grounded things are different.
I'll attach my scribbles since it consider both scenarios and also include some calculations. Someone might enjoy them...
-BAK
RE: Understanding delta VT ratios
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
RE: Understanding delta VT ratios
The vector diagram posted by mgtrp actually nicely illustrates the "neutral inversion" that occurs. When the EF occurs on an impedance system the potential of the neutral point actually goes negative w.r.t. ground.
I think this an excellent thread and a good refresher if nothing else. I would like to thank all respondents.
Regards.
RE: Understanding delta VT ratios
Glad to help.
-BAK
RE: Understanding delta VT ratios
As mgtrp pointed out, The connection of the PTs is C1-A1-B1. As a fault on A1 brings A1 towards N, the phase angles of the PTs approach C1-N-B1
Most of my delta experience has been with power transformers. Full delta, 3 transformers; Broken delta, 3 transformers, and open delta, two transformers. I'll skip 3 pages of descriptions of the problems and just say that I am not a fan of delta secondaries for power transformers.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding delta VT ratios
There is one small thing - on page one of your calcs, should the Vb and Vc line to ground voltages not be 11kV and not 11kV/sqrt(3)?
Need to think more about waross' comment.
RE: Understanding delta VT ratios
Yes, you're right. Actually that part should indicate the pre-fault conditions. I'll correct that and post a new verion later today. I also need to correct the calculations on page two; I'm missing a "3" in the middle part. It doesn't affect the outcome though.
-SB
RE: Understanding delta VT ratios
-SB
RE: Understanding delta VT ratios
I thus ask myself is your pre-fault vector diagram still applicable? i.e is the virtual star point at ground potential. I would say yes in that the system has a reference point via the earth of the primary neutral winding. There is no fault and so your pre-fault phasor diagram looks okay to me.
Comments?
waross - regarding your last post, you are referring to the VT voltage vectors. I was referring to the primary system neutral - i.e. star point of a transformer winding. Let's suppose the 11kV winding supplying our network in question is a star winding and let's suppose it is earthed via a NER = R ohms. With an EF current flows from the ground through R and to the neutral point N. From there back to the fault via the faulted phase. Current flowing through R necessarily implies a volts drop across R and since current flows from the Ground to N via R, means that N must be at a lower potential than Ground - thus the term "neutral inversion".
The VT will not see this as it's neutral is solidly earthed. So whatever current flows in the VT neutral, no neutral inversion here. All the VT will effectively see is O volts on the faulted phase and sqrt(3) * Vnom on the healthy phases.
Or so I think??
RE: Understanding delta VT ratios
I'm not 100% sure that I follow your first question fully, but if your source that's providing the grounding trips off on the 11 kV side, then you now have an ungrounded system. If it trips off on the HV side, then it will continue to provide a ground reference.
For your second question, you are correct that there is some voltage between the neutral point and ground, but not correct that it is "negative" - this is an AC system, so the voltage between the two is constantly changing. All of the vector diagrams are constantly rotating.
RE: Understanding delta VT ratios
Yes, the 11kV network will be ungrounded once the source with the earth point gets disconnected. My 11kV delta winding in essence is now connected to an unearthed network.
Regarding the 2nd point - I have to disagree here. Please do not get confused between instantaneous and rms values. I am referring to rms quantities whe I refer to current flowing in the neutral. When we say a source delivers a current of say 100A to a load then we represent this as an rms current flowing from source to load. With phasors we actually have the length of the phasor being the magnitude of the rms quantity and the direction is relative to a reference which is usually the A-phase voltage. Yes, phasors rotate but the relative angles between the phasor quantities remain the same (assuming all in synchronism).
RE: Understanding delta VT ratios
The phasor diagram(s) can be thougth of as a representation of the VT windngs, nothing virtual about that.
Regarding the other issue of phasor voltage being negative; I would say no, it's not negative. IF it were negative the phasor arrow of phase "a" would point towards the neutral point. This would mean that the vector sum of the three phase voltages is non zero. This can't be, since we don't have any other voltage drop in the (ideal) system.
...and also another update of my document.
Regards,