3-Winding XFMR Rating

[mityeltu]

I have started studying for the PE and so I picked up a copy of Stevenson's Elements of Power System Analysis, 4th ed.

I was looking at the section on 3p, 3winding xfmrs and didn't follow why something WASN'T done.

Here's the problem: pg. 154 for those following along.

Primary: Y-15MVA, 66kV
Secondary: Y-10MVA, 13.2kV
Tertiary: D-5MVA, 2.3kV

Zps: 7% on 15MVA, 66kV base
Zpt: 9% on 15MVA, 66kV base
Zst: 8% on 10MVA, 13.2kV base

Now the example states to convert all the impedances to a common15MVA, 66kV base.

Zps and Zpt don't need to be modified (already on the correct base). So, the conversion to the base for Zst, according to the book, is Zst = 8% x 15/10 = 12%

Here's what I don't get. The impedance was measured on a different base for voltage and VA. So why is there not a term for the voltage?

That is, IMO, it should have read: Zst = 8%x(15/10)x(13.2/66)^2 = 0.48%

I'm willing to conceded that I'm mistaken, but I don't understadn why the term is not used. Can someone explain what I'm missing?

Thanks.

 

[dkarst]

I have started studying for the PE and so I picked up a copy of Stevenson's Elements of Power System Analysis, 4th ed.

I was looking at the section on 3p, 3winding xfmrs and didn't follow why something WASN'T done.

Here's the problem: pg. 154 for those following along.

Primary: Y-15MVA, 66kV
Secondary: Y-10MVA, 13.2kV
Tertiary: D-5MVA, 2.3kV

Zps: 7% on 15MVA, 66kV base
Zpt: 9% on 15MVA, 66kV base
Zst: 8% on 10MVA, 13.2kV base

Now the example states to convert all the impedances to a common15MVA, 66kV base.

Zps and Zpt don't need to be modified (already on the correct base). So, the conversion to the base for Zst, according to the book, is Zst = 8% x 15/10 = 12%

Here's what I don't get. The impedance was measured on a different base for voltage and VA. So why is there not a term for the voltage?

That is, IMO, it should have read: Zst = 8%x(15/10)x(13.2/66)^2 = 0.48%

I'm willing to conceded that I'm mistaken, but I don't understadn why the term is not used. Can someone explain what I'm missing?

Thanks.

I don't have my copy of Stevenson in front of me but I'll take a shot. The formula you are applying at the end is used for example when you are given a generator that has a different rating in kVA and kV than the system base kVA and base voltage at the point of implementation. The transfomer has a kVA rating but the kV rating is determined by turns ratio and connection. Be very careful as if you have 3 single phase transfomers connected as a 3 phase unit, the base voltage is NOT the turns ratio, but the line-line voltage ratio on each side. I have no idea of current test content but I would be surprised if proper conversion of this type was not covered on your upcoming PE exam.

Now getting back to your question... if I told you that 8% of rated voltage (8% *13.2kv) when applied to secondary winding created 1 pu current in a shorted tertiary winding per the definition of 8% Z, you would agree, correct? Remember 1 pu current is related to the 10 MVA tertiary base rating MVA. Now if I want to rate the tertiary at 15MVA, I would raise that applied secondary voltage to say 8% *(15/10) = 12% of the rated (12% * 13.2kV), now 1 pu current also flows but I've changed my rating to the 15MVA desired.

Not sure if this helps, I'll try to remember to look in Stevenson over weekend.

 

[mityeltu]

I don't understand that.

how can 8% of 13.2 kv (1056v) on the secondsry produce 1pu in the shorted tertiary, and so can 12% of 13.2kv (1584v)?

If the windings don't change, applying more potential, produce the same amount of current? Even if you say the bases chaged, the impedance didn't, and hence the current MUST change.

1pu is rated current, right? You can't arbitrarily change rated current can you?

 

[mityeltu]

Between symmetrical components and per unit, I think I have made a horrible mistake in getting into EE. 4 years after graduating I always still feel dumber after I ask a question than before I asked.

not that your answer doens't make sense to someone, it just doesn't make sense to me.

 

[dkarst]

I don't understand that.

how can 8% of 13.2 kv (1056v) on the secondsry produce 1pu in the shorted tertiary, and so can 12% of 13.2kv (1584v)?

If the windings don't change, applying more potential, produce the same amount of current? Even if you say the bases chaged, the impedance didn't, and hence the current MUST change.

1pu is rated current, right? You can't arbitrarily change rated current can you?

Between symmetrical components and per unit, I think I have made a horrible mistake in getting into EE. 4 years after graduating I always still feel dumber after I ask a question than before I asked.

not that your answer doens't make sense to someone, it just doesn't make sense to me.

Hang in there... symmetrical components for one thing are quite difficult to understand... but let's solve this first.

I tried to state clearly but did not emphasize enough... I think you are swapping machine ratings and base ratings. In this example, the tertiary winding is rated for 5MVA at 2.3kV. If you draw 10 MVA at 2.3kV from that winding you will likely let all the smoke out. But the base impedance from the secondary to the tiertary is specified on a 10MVA base. That means if I apply 8% of rated secondary voltage, the base current will flow... Not the rated secondary current.

In the same way, let's say I give you a generator with a base impedance of 3% on 13.2 kV 20 MVA rating. I could ask you to give me the base impedance on a 20kV, 100MVA base and you would use you equation you used earlier to give me the new base impedance. Note this doesn't say by magic you can now extract 100MVA out of the machine. Just the same, I could ask you to give me the base impedance at 8kV, 10 MVA and you could do it even though the machine capabilities haven't changed.

Now back to where I should have been clearer... as you said, 8% of 13.2 kv (1056v) on the secondary produces 1pu in the shorted tertiary... but this is 1pu on that 10MVA base, not the rated so the 1pu is on the 10MVA base (you can calculate the current in amps, I think roughly 2500 amps).

Now if I raise the applied voltage at secondary to 12% of 13.2kv (1584v), proportionally more current flows because as you said I didn't do anything to the windings or construction... so ... approx 3750 amps flows. Now on the new base of 15MVA, that is 1pu current.

 

[dkarst]

Hang in there... That means if I apply 8% of rated secondary voltage, the base current will flow... Not the rated secondary current.

meant to say the base current will flow in the tertiary ... Not the rated tertiary current.

 

[mityeltu]

I thought 1pu WAS rated current/voltage/kva. Is that not right?

If I have a 5MVA, 2.3kV xfmr, I say the base_MVA = 5, base_kV = 2.3.

If the voltage changes to 2kV isn't that 2/2.3 = 0.87 pu?

and then the base_I = (5M/3)/2.3k = 724.6, rated L-N current or 1 pu.

Is this not right?

I think I am only gettign more confused.

 

[dkarst]

I thought 1pu WAS rated current/voltage/kva. Is that not right?

If I have a 5MVA, 2.3kV xfmr, I say the base_MVA = 5, base_kV = 2.3.

If the voltage changes to 2kV isn't that 2/2.3 = 0.87 pu?

and then the base_I = (5M/3)/2.3k = 724.6, rated L-N current or 1 pu.

Is this not right?

I think I am only gettign more confused.

I think you may be missing a "sqrt" on the 3 in your base current equation depending if you are considering total 3 phase power but you are correct in your assumptions... because in a standard transfomer there is only ONE kVA rating for the transformer which applies to both primary and secondary. As the Stevenson section you referenced states, for a three winding transformer, you can have different kVA ratings.

I think (but have to admit now that I've reread the section) that given they have defined Zst = 8% on a 10MVA, 13.2kV base; in order to convert this to the 15MVA base desired you apply the per unit new = p.u given * (kVA.new/kVA.given) = 12%.

I would also suggest posting this over in engineerboards.com as a number of people are posting PE test/studying questions over there.

 

[mityeltu]

Ok, well, engineerboards.com wouldn't let me register for some reason. I'll work on that later. I appreciate the added link.

If I have a 3 winding xfmr, the turns ratio P-S and P-T are fixed as well as the rated MVA per pair of windings, right? So, can't I just treat this as 2 separate 2-winding xfmrs? One xfmr Y:Y 10MVA 66:13.2 kV and the other Y 5MVA 66:2.3 kV.

I think my problem is that the rated MVA for the P-T transformer is 5MVA/2.3kV, but the impedance is given on a 10MVA/13.2kV base.

Also, if I am given impedance in %Z, then is that really a pu impedance?

I calculate my Zbase P-T as Z = (2.3k/srt(3))^2/(5M/3) = 1.058 ohm

So, just for my own edification, if this %z were given on the rated 5MVA/2.3kV bases, then my impedance would be 8% = 0.08 pu = 0.08 * 1.058 = 0.08464 ohm. Is this right?

 

[mityeltu]

Alright, I think I might understand why there is no voltage term in the conversion.

The base for the tertiary winding is 5MVA, 2.3kV. The impedance though, is reported on a 10MVA, 13.2kV base. This, I guess, is stating the the impedance has been converted already from it's original 5MVA/2.3kV base to the new base. Since the 10MVA, 13.2kV base is the same as the secondary wonding, then converting it to the 15MVA, 66kV needs only include MVA rating because the kV bases for the primary and secondary are already related through the turns ratio of the respective coils.

Does this sound right, or is there another facet I'm missing?

 

[dkarst]

If I have a 3 winding xfmr, the turns ratio P-S and P-T are fixed as well as the rated MVA per pair of windings, right?

No, this is where a 3 winding transfomer is different, instead of one MVA rating per winding pair you have multiple. The rated MVA of the primary-secondary pair of windings is not the same.... in this case the primary is rated 15MVA and the secondary 10MVA. If you don't even think about % Z for now but load the secondary to 15MVA just because the primary can handle that load, the secondary will be overloaded. In this case and I think normally true, the secondary and tertiary windings at full load will require full load primary current (or full MVA)
.

If I have a 3 winding xfmr, the turns ratio P-S and P-T are fixed as well as the rated MVA per pair of windings, right? So, can't I just treat this as 2 separate 2-winding xfmrs? One xfmr Y:Y 10MVA 66:13.2 kV and the other Y 5MVA 66:2.3 kV?

I don't that works and think you will see how if you look at the very next example problem in Stevenson where they employ this 3-winding transformer in a real one-line. The other piece you have to be careful of is we have been trying to get Zst in the correct base. To use this in a one-line, you have to use the equivalent circuit shown in Fig. 6.23 on pg 154 with Zp, Zs, Zt etc. (not Zpt, Zst...)

You mentioned two equivalent xfmers but you also have the circuit from secondary to tertiary to worry about....so you have three terminals...

Once we get Zst on the correct base, you solve for Zp, Zs, and Zt which are j0.02, j0.05 and j0.07 pu. If you look at Fig 6.24, you can see that for example a fault at the secondary with tertiary open (with source on primary)for example sees an impedance of j0.02 and j0.05 in series or j0.07pu. This is the same as the original 7% on the 15MVA 66kV base given in the original problem statement. So from this standpoint it does look like your indepedent view but you should really use the circuit diagram in Fig. 6.24

The same is true for a fault on the tertiary winding with secondary open... j0.02 + j0.07 = j0.09 p.u.

When it comes to a fault at tertiary with source being secondary, it sees j0.12 total which is Zst but on the required 15MVA base.

I think my problem is that the rated MVA for the P-T transformer is 5MVA/2.3kV, but the impedance is given on a 10MVA/13.2kV base.

I think you meant to say S-T winding pair here but other than that you are correct.

 

[dkarst]

Alright, I think I might understand why there is no voltage term in the conversion.

The base for the tertiary winding is 5MVA, 2.3kV. The impedance though, is reported on a 10MVA, 13.2kV base. This, I guess, is stating the the impedance has been converted already from it's original 5MVA/2.3kV base to the new base. Since the 10MVA, 13.2kV base is the same as the secondary wonding, then converting it to the 15MVA, 66kV needs only include MVA rating because the kV bases for the primary and secondary are already related through the turns ratio of the respective coils.

Does this sound right, or is there another facet I'm missing?

I think you are understanding correctly and although it is just semantics, I would prefer the term The RATING for the tertiary winding is 5MVA, 2.3kV. Base voltages will change for example throughout a one-line, the the ratings of a transformer are only changed through additional cooling for example.

Remember once you have chosen a base kVA, the base voltages are given by turns ratio/connections of the transfomers... you can't just change the base voltage in the tertiary randomly.

 

[mityeltu]

I think that's got it. I believe we were actually speaking the same language, but different dialects.

The breakdown I think was because when I started dismantling the 3-winding xfmr into 2, I was really thinking of power in vs power out.

IOW, if I have a black box rated 15MVA with 2 xfmr's, one rated 10MVA and the other 5 MVA, then I have 15MVA in and 15MVA out.

This only helps with the impedance question if I then assume the 2 xfmr's just happen to be sharing a common primary coil, making the 3-winding xfmr again.

I really appreciate your help. I don't feel nearly as dumb as I did.

 

[dkarst]

I think that's got it. I believe we were actually speaking the same language, but different dialects.

The breakdown I think was because when I started dismantling the 3-winding xfmr into 2, I was really thinking of power in vs power out.

IOW, if I have a black box rated 15MVA with 2 xfmr's, one rated 10MVA and the other 5 MVA, then I have 15MVA in and 15MVA out.

This only helps with the impedance question if I then assume the 2 xfmr's just happen to be sharing a common primary coil, making the 3-winding xfmr again.

I really appreciate your help. I don't feel nearly as dumb as I did.

I hope you were able to follow the next example problem in Stevenson Fig 6.24 as I think it will be helpful.

Back a few posts ago, you seemed a bit down so I'd like to close this post out with some encouragement. No matter how knowledgeable the various people seem on this board or other places for that matter, I am certain even the most knowledgeable with admit they still "don't know everything about everything". As an example, you may be a code expert and a wizard on VFDs, but you fall short on protective relaying. The benefit of a forum like this is there is usually an expert on almost everything "on call".

I also want to wish you luck on the upcoming PE exam and also encoourage you to somehow get registered over on engineerboards.com as I think that will help you out.