Potential transformers

[rodinedoruelo]

Good day! Can I ask something. Aside from overloading what is usually the main cause while the PT destroyed? And how can you troubleshoot the fault.

And another one, usually what is the lifespan of motor bearings if it is running 24hours everyday? And what is the effective preventive maintenance to extend their lifespan?


I need your help!
Please Help Me!
Thank you very much!!

 

[7anoter4]

I think the insulation failure is the cause of most of transformer failures. See:

http://www.bplglobal.net/eng/knowledge-center/download.aspx?id=191

From Siemens Motor ABC:
"The lifetime of electrical machinery is mainly defined by the lifetime of the insulating system and the roller bearings.
The nominal (calculated) lifetime of a roller bearing also depends on the particular load and the speed. For standard motors, this lifetime is 40.000 operating hours under the assumption that there is a coupling out-drive. The larger the bearing, then the higher the churning work and grease consumption. However, consistent lubrication properties are a prerequisite for a high bearing lifetime. The lifetime of a bearing always depends on the load."

 

[LSpark]

Can't help with the bearings but the most common failure mechanism for PTs in my company is shorted HV turns, occassionally catastrophic but normally picked by that phase reading high.

 

[electricpete]

I've often wondered whether shorted turns would cause ct output current high or low.

One one hand, we could simply say the turms ratio decreases, so the current stepdown ratio decreases, so the secondary current increases for a given primary current.

On the other hand, the short allows higher circulating current in the shorted turns which will contribute to cancelling the primary amp turns and wont' shot up on the output. This logic leads us to believe the ct secondary current would decrease for a given primary current.

I've never measured the results, but my thinking is that the first explanation neglects the important effect of amp turns in shorted turns and reality should be closer to the second explanation (resulting in secondary current decrease). You have observed the opposite?

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[electricpete]

And another one, usually what is the lifespan of motor bearings if it is running 24hours everyday? And what is the effective preventive maintenance to extend their lifespan?

As a starting point, tou can calculate an ESTIMATED L10 life (beyond which 90% of bearings are expected to last) using load. Similar to
Life = (P/C)^3 * 1,000,000 revolutions

where P is actual load and C is load capacity. Load incorporates radial and axial loads (P = X*Fr+Y*Fa)

SKF publishes a variety of life adjustment factors which modify/refine the above equations and certainly lubrication plays a big role.

In the end it's always somewhat of a "guess" but at least you can take advantage of substantial work done to try to quantify the life statistically to improve your guess

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[waross]

I would suggest that a shorted turn on a 100:5 Amp CT would act as a 100 to 100 ratio. eg. 1:1
The remaining turns would act as a 95:5 ratio. This would be the condition for light loads. I would expect that loads in excess of about 25% or 30% would result in rapid failure from overheating.
Note: This example is for a specific ratio of 100:5. The higher the ratio, the less per unit load will be required to initiate total failure. Lower ratio CTs under light loading may not fail completely for some time.

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

 

[electricpete]

Life = (P/C)^3 * 1,000,000 revolutions

Correction: Actually should be Life = (C/P)^3 * 1,000,000 revolutions
I’d suggest to find some references and look at the equation, assumptions, correction factors for yourself.

=======

I don’t particularly see the logic to arrive at Bill’s conclusion for shorted CT secondary turns.

Here’s my attempt
Assume N secondary turns, one of which is shorted.
RPT = resistance per turn
Rc = external CT secondary circuit resistance (in addition to resistance per turn)
Rs = resistance of short (for example shorting jumper)
Ic = current in external CT secondary circuit
Is = curent circulating in the shorted turn
(reference direction for Is is arbitrarily assigned the same as reference direction for Ic)
* ASSUME - we can Neglect self-inductance of coils
* ASSUME - we can Neglect magnetizing branch
Ec = voltage induced from primary into secondary CT circuit (Ic loop)
Es = voltage induced from primary into shorted CT turn loop (Is loop)

KVL:
Ec = Ic*(N*RPT+Rc) + Is*RPT {note there is one turn that has Ic + Is flowing in it)
Es = Is*(1*RPT+Rs)

Also we know the induced voltage Ec is N times higher than induced voltage Es
Ec = N*Es

Substitute above expressions for Ec, Es:
Ic*(N*RPT+Rc) + Is*RPT = N*(Is*RPT+Is*Rs) Equation 1

Amp turn balance (where I1 = primary current)
N*Ic+Is=I1 Equation 1

Solve Eq1 and Eq2 for the unknowns Ic and Is

Ic = I1 * [RPT*(N-1)+N*Rs] / [Rc+(RPT+Rs)*N^2]

Is = I1 * [N*RPT+Rc] / [Rc+(RPT+Rs)*N^2]

Is can be on the order of I1 (very high) if Rc dominates the resistances in the circuit. Supports Bills idea the circuit may not last long in this configuraiton for CT under load.

Ic has the pleasing required result that it simplifies to Ic = I1/N if we substitute Rs = Infinity (no short)

Ic will always be less than ideal I1/N otherwise.

That is based on assumptions stated above and analysis provided above. Could be wrong on assumptions or analysis...

Seems like there was a thread on this maybe 5 years ago but I can't find it and I don't remember the result.


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[waross]

Hi Pete;
Try this;
A 100:5 window type CT has 20 turns on the secondary.
One turn becomes shorted. 19 turns are still active.
The secondary may now be considered as two windings with a jumper joining them. The current in the shorted turn will be the same as the primary current. the current in the active winding is now Primary current over the original ratio minus one.
The current in the active winding flows in the shorting "jumper".
The current in the shorted turn flows in the shorting "jumper".
Mr. Kirkoff assures us that there will be no transfer of current between the shorted turn and the active winding.



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

 

[stevenal]

"I would suggest that a shorted turn on a 100:5 Amp CT would act as a 100 to 100 ratio. eg. 1:1
The remaining turns would act as a 95:5 ratio."

If the shorted turn is acting as 1:1, then the remaining turns are carrying zero current. This is the only way Ampere's law will be satisfied. Iprimary*1 turn = Ishort* 1 turn+ Isecondary* 19 turns.

But since the 19 turns remain connected in a low impedance path, the current will split in a manner that is inversely proportional to the two path impedances and still satisfy Ampere's Law above.

I think this may be electricpete's point, but with a little less math.

 

[magoo2]

Why all the talk about CTs. The post is about a PT failure. Am I missing something?

 

[electricpete]

I made the first comment/question about CT's. ...didn't read closely.

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[waross]

Thanks stevenal.

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

 

[electricpete]

In the unlikely event anyone is interested, attached is the sketch I made to accompany my writeup posted above (17 Jul 13 10:45)

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[iop995]

Regarding this eq. "Ec = Ic*(N*RPT+Rc) + Is*RPT", I think Is*RPT term must not be included. Turn in short have no "external" voltage component, induced voltage Es will delivery an Is current on RPT resistance. For internal short-circuited turn, Rs must be zero.
Ec and Es induced voltage equations are valid for steady state only. In transient phases will need to add voltage components proportional with secondary winding inductance, so proportional with N^2.

 

[electricpete]

Hi - iop995! Thanks for taking an interest in my post.

Regarding this eq. "Ec = Ic*(N*RPT+Rc) + Is*RPT", I think Is*RPT term must not be included.

It looks correct to me. Referring to my figure....

I have assumed the short bypasses exactly one turn (I may not have stated that clearly).

So, the current that flows in that single shorted turn is Ic + Is, and therefore we have a voltage drop [(Ic+Is)*RPT * 1] in that single turn.

The current in remaining N-1 turns is Ic, which results in voltage drop [Ic*(N-1) * RPT] in those remaining N-1 turns.

Also of course to complete the loop we need voltage drop in exteral resistance [Ic*Rc].

Add those three voltage drops [ ] together:
[(Ic+Is)*RPT * 1] + [Ic*(N-1) * RPT] + [Ic*Rc]
= Ic *RPT * (1 + N-1) + Is * RPT + Ic * Rc
= Ic *RPT*N + Is * RPT + Ic * Rc
= Ic * ( N*RPT + Rc) + Is * RPT
That is equated to Ec for KVL. I don't see any problem there. Does it change your view?

For internal short-circuited turn, Rs must be zero

Rs is the resistance we assign to the fault. There is no loss of generality to include possible non-zero fault resistance in the model. If you want it to be a perfect short, set Rs=0. But I found it was a helpful double check on the solution as shown in my post above to verify expected results when assigning Rs =infinity (that would not have been possible if Rs was assumed zero before equations were solved)

Ec and Es induced voltage equations are valid for steady state only.

Agreed, that was my intent - sinusoidal steady state solution.

In transient phases will need to add voltage components proportional with secondary winding inductance, so proportional with N^2.

Agreed (even in steady state). I made the simplifying assumption that the secondary leakage inductance was neglible. It would be many orders of magnitude lower than magnetizing inductance. Exactly how that leakage inductance impedance typically compares to resistive impedances in the circuit I'm not sure offhand. (open to comments).

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[electricpete]

The two overlapping loop currents is a little tricky.

Above I justified (to my thinking) the KVL equation: Ec = Ic*(N*RPT+Rc) + Is*RPT
where Ec is voltage induced in loop defined by Ic

Now if I turn my attention to the other loop equation: Es = Is*(1*RPT+Rs) ... seems like I have treated this differently... maybe should be Es = Is*(1*RPT+Rs) + Ic*1*RPT

I'll think about it some more.

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[iop995]

If shorted turn is in one point, Ic pass through this single point not through entire turn (it can't being a closed loop and a single contact point); in shorted turn pass only Is developed by Es. See these two voltage source connencted in a single point, so can't have power / current exchange between them. It's something wrong in this?
So, this eq Ec = Ic*(N*RPT+Rc) + Is*RPT can't include last term.