Phase loss Y-y transformer 1

[Electrical1948]

Hi all

For a Y-y transformer ungrounded neutral, what will be the secondary and the primary voltages if phase A in primary is open?

 

[waross]

With C phase missing;
Secondary A to B voltage. 87% of normal voltage.
Secondary A to C and B to C voltage, 43% of normal line to line voltage with the impedance of the dead C phase winding in series.

This assumes that a Y-y with ungrounded neutral* implies balanced loads.
With unbalanced loads, it depends.

*ungrounded neutral. This term is commonly used to describe a neutral that is not connected to the source neutral. I have used that common usage in this description.
However, if the term is used to describe a neutral that is connected to the source neutral, but there is no connection from either the source neutral to ground nor a local ground at the transformer, you will have normal voltages on two phases to neutral, normal voltage from A to B, and zero volts to neutral on the missing phase.
We don't see a lot of ungrounded Y-y connections in distribution.

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

 

[che12345]

Electrical1948 (Electrical)(OP)15 Jul 24 12:09
" ..For a Y-y transformer ungrounded neutral, what will be the secondary and the primary voltages if phase A in primary is open? "
I have the following results for your kind scrutiny. I stand to be corrected.
1. Assumption: the traffo is YNyn 0, with turn ratio 1:1. Therefore, Pri: VAB = VBC = VCA, = Sec: Vab = Vbc = Vca; taken as 1 pu. Where VAN =VBN = VCN = Van = Vbn = Vcn = 0.577 pu.
2. That is, vector VAB = VNA-VNB , vector VBC = VNB-VNC.... Where vector VAB and vector VBC... are 120 electrical degree apart.
3. Now consider that pri phase A is open, i.e. = VNA = 0 pu.
3.1 Pri: VAB = VNB = 0.577 pu.; VBC = 1 pu ; VCA = VCN = 0.577 pu.
3.2 Sec: Vab = Vnb = 0.577 pu.; Vbc = 1 pu ; Vca = Vcn = 0.577 pu.
Che Kuan Yau (Singapore)

 

[Electrical1948]

@Waross

I have made a circuit diagram with phase loss on A which provides the following conditions for Y-side;

UA-n = 0 V
UB-n = 0.5*UBC
UC-n = -0.5*UBC

Shouldn't the voltages on the y-side be the same i.e.;

Ua-n = 0 V
Ub-n = 0.5*Ubc
Uc-n = -0.5*Ubc

 

[waross]

Yes.
But remember that those voltages are based on the assumption of balanced loads on the secondary.
In the event of unbalanced loads;
Ub-n and Uc-n will sum to Ub-c.
The division will be in the inverse ratio of the load impedances.
With badly unbalanced loads, simple arithmetic predicts that the voltage on the lightly loaded phase will approach Ub-c.
But the second thing to remember is that transformer saturation may limit the voltage rise on the lighter loaded phase.

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

 

[che12345]

@ Mr Electrical1948
1. Thank you for your learned advice, which arrived at the same values as my post dated 18th instant.
2. Please advise what is the significance by indicating UC-n = -0.5*UBC and Uc-n = -0.5*Ubc ; both with - sign. The voltage value indicated is the magnitude read on a volt meter. An AC volt meter would NOT read - or negative refection.
2.1 However, if the intention is to indicate it as a vector, it should indicate the magnitude and the angle.
3. "...But remember that those voltages are based on the assumption of balanced loads on the secondary..." is irrelevant and out of scope.
Che Kuan Yau (Singapore)

 

[waross]

UC-n

Most of us consider that use to be as a punctuation mark.
We read UC-n as;
"UC to n."

Ub-n = 0.5*Ubc
Uc-n = -0.5*Ubc

While possibly not up to your standards, most of us understand that convention to indicate opposite polarity to the preceding valie.

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

 

[che12345]

"...most of us understand that convention to indicate opposite polarity to the preceding valie".
1. I have no issue with e.g. Ua-n = 0 V ; Ub-n = 0.5*Ubc ; Uc-n = -0.5*Ubc , to mean
Ua-n=Van, Ub-n = Vbn, Uc-n = Vcn.
2. I can not understand the intention of e.g. -0.5*Ubc, the - sign.
3. Let us hear the clarification from learned Mr Electrical1948, who may have his reasons; as the negative (-) sign does not appear on the other vectors.
Che Kuan Yau (Singapore)

 

[Electrical1948]

Hi

As Waross pointed out, it is due to polarity. See image below.

 

[che12345]

@ Mr Electrical1948 (Electrical)(OP)19 Jul 24 05:30
"...As Waross pointed out, it is due to polarity. See image below".

1. Thank you for your learned clarification. Case closed.
2. I am taught of the convention e.g. VAN = 0.577 / 0; VBN = 0.577 /-120; VCN = 0.577 /-240; VAB = 1 /30 ...etc. Where 0.577 is the magnitude and followed by the angle . In this case VAN with /0 is taken as the reference.
Che Kuan Yau Singapore)

 

[Electrical1948]

Hi

The angles will not be like a balanced three phase system, they will we 180 degrees phase shifted.

If you look at the circuit provided the voltages across winding B (ZB) and winding C (ZC) will be as stated earlier. Consider that the winding impedances are equal (1 pu), then you get by voltage division; UB-n = UBC*(ZB/(ZB+ZC)) = 0.5*UBC. UC-n = -0.5*UBC = 0.5*UBC<180 deg due to opposite polarity.

 

[waross]

Mr Che;
I respect your knowledge and experience. It is evident from your posts that you have many years of experience and much technical knowledge.
With the greatest respect, may I point out that this is an international forum, and that conventions change from time to time.
Example #1
I have seem electricity flow three ways.
I suspect that you may be old enough to have seen the same.
When I was in high school and trade school, electricity flowed from Positive to Negative. Everyone knew that.
Then, by the time I was teaching at trade school, electron theory had become popular and electrons and electricity flowed from Negative to Positive.
I had to teach my students to use their left hand for the right hand rule.
Considering the flow of electrons to be the flow of electricity caused a lot of confusion with existing text books.
This was resolved by declaring that a flow of electrons from Negative to Positive caused a flow of electricity from Positive to Negative.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

How true. Thanks 1503-44
Example #2
kW, kVAh
Since the dawn of generator time, Manufactures rated their machines in KW and KVA.
Meters were KWHr and KVARHr meters.
When the convention was changed, no-one changed the millions of nameplates already in use.
It was also difficult to effect change with the older electrical persons who spent their time in the field with little or no contact with academia.
3# A pet peeve, Celsius;
Who remembers a n old temperature scale based on 100 divisions and called, appropriately, centigrade. That seemed to me to be very metric if nor SI.
That was changed to Celsius, but

Celsius used 0° for the boiling point of water and 100° for the melting point of snow.

Yes, I understand and accept the recognition given to Anders Celsius for his many notable and original accomplishments in several branches of science.
For years whenever I heard Celsius I thought, without malice, You got that backwards, SI.

The point is, not "What is right and wrong?" but more, "What is current?"

Again, respectfully I have proposal that may avoid future controversy and at the same time add to the general knowledge.
If you wish to quote me using conventions or labels not to your liking, simply add a discrete correction, without comment.
eg:

The generator was rated at 500 KW

Simply add

The generator was rated at 500 KW (kW)

You may comment thus; "Mr waross, in regards to the 500 kW generator (KW)."
I pray that you find this helpful and do not take offence.
These are international and also multi-generational fora.
Respectfully.





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