CT shorting blocks

[sgtgt1977]

Can someone help me find where it might say in the code book that CT shorting blocks are required? The company I work for use's them on a PowerScout 24 meter the CT's are millivolt CT's.The manufacture says the shorting blocks are not necessary due to the millivolt CT. I just want to know the requirements and make sure it is safe.

 

[iceworm]

Can someone help me find where it might say in the code book that CT shorting blocks are required? The company I work for use's them on a PowerScout 24 meter the CT's are millivolt CT's.The manufacture says the shorting blocks are not necessary due to the millivolt CT. I just want to know the requirements and make sure it is safe.

It is a design decision - not regulatory.

ice

 

[rbalex]

It is a design decision - not regulatory.
ice

Not quite:

110.23 Current Transformers.
Unused current transformers associated with potentially energized circuits shall be short-circuited.

The NEC makes no distinction with power/metering/instrumentation CTs although it doesn't mandate shorting blocks as the shorting method.

 

[iceworm]

Not quite:

The NEC makes no distinction with power/metering/instrumentation CTs although it doesn't mandate shorting blocks as the shorting method.

I do not agree to your not agreeing.hmy:
Specifically, where the OP is using the term "CT" he is meaning a generic reference to any device that puts out a signal roughly portionally to the current which includes high impedance devices such as a Rogowski coil as well as a very low impedance device such as a "Current Transformer". And where 110.23 refers to a "Curent Transformer", they infact mean a low impedance device such as a "Current Transformer"

That was susposed to be funny, but you can't exactly see the body language. I'm goofing with you.

I don't think 110.23 applying to Rogowski coils.

ice

 

[liquidtite]

What are shorting blocks

 

[GoldDigger]

They are devices which connect together (short) the CT output leads when no current meter is attached.
The idea is that if the CT is not designed to saturate at a low current, then the voltage across the open secondary leads could be dangerously high.

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[GoldDigger]

Not quite:


The NEC makes no distinction with power/metering/instrumentation CTs although it doesn't mandate shorting blocks as the shorting method.

And in this context, "unused" means not having a current meter connected at the moment, not abandoned.

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[Nom Deplume]

I was told that if an unused CT was not shorted, and around a conductor with current flowing, that the CT would burn up.



True?

False?

 

[iceworm]

I was told that if an unused CT was not shorted, and around a conductor with current flowing, that the CT would burn up. ...

Generally speaking yes. But the discussion is about current transformers, not high impedance rogowsli coils. A CT is a transformer wound to put out a constant current as opposed to a constant voltage. CT are rated in Ampsout:ampsin like 500:5 So, for example, with 500A flowing throught the CT, there will be 5A flowing in the secondary. If the secondary is shorted, R = zero, V = zero. If the secondary load is 10 ohms, the voltage is 10 ohms x 5A = 50V. If the secondary load is 100 ohms, secondary voltage is 500V. If secondary is open circuited, (infinite load resistance) the sec voltage goes up until the coil shorts - most of the time they don't even make any extraneous energy emmission (heat, light, smoke) they just die.

ice

 

[GoldDigger]

Once again, the voltage will only increase rapidly up to the point that the core saturates. Beyond that it will depend on the direct magnetic coupling between the windings.
In any case your scenario would end up predicting an infinite voltage across all inductors, since they have no secondary at all.
If you had perfect coupling between the windings the voltage across the open secondary would be at most the product of the turns ratio and the drop across the primary because of the inductance.
The credible failure mechanism for the CT would be heating of the core from hysteresis losses.

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[iceworm]

Once again, the voltage will only increase rapidly up to the point that the core saturates. Beyond that it will depend on the direct magnetic coupling between the windings.
In any case your scenario would end up predicting an infinite voltage across all inductors, since they have no secondary at all.
If you had perfect coupling between the windings the voltage across the open secondary would be at most the product of the turns ratio and the drop across the primary because of the inductance.
The credible failure mechanism for the CT would be heating of the core from hysteresis losses.

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Yes you are mostly right. However I still like the concept for introducing people to why is is not a good idea to open circuit a Current Transformer. Yes, if one draws out the model of a CT, lists the equations accounting for saturation, your model shows the voltage does not go to infinity. That is absolutely true. Still, I like my model of "Summation (NI) = zero" for first order effects. Be that as it may, you are absolutely correct.

So tell us, how high might the voltage get? Don't just creep up the burden slowly, increasing the load resistance while the CT heats up, rather cut the wire with a suitable meter across the cut ends. Did I just fry my fluke 87?

Different Subject:
The physics of a multi-turn inductor are different from a one-turn primary CT. Normally one does not drive inductors with an infinite compliance current source. I'll leave that for a later discussion.

Change subjects again:
As for the creditable failure modes, I'm certain you know better than I. I have never seen any papers on CT failure modes. And i have never personnaly done an autopsy.

ice

 

[SG-1]

An open CT in a medium voltage structure can result in catastrophic failure. Such as a bushing type CT used in switchgear.
The CT circuit opens.
The voltage output exceeds the insulation level.
The windings begin to arc. You can hear this.
The arcing produces explosive gas.
Gas density reaches critical mass.
BOOM !!

The Fluke 87 is fried before the BOOM. Some of these CTs have a 800V knee voltage.

I would suspect that if the CT type is able to produce high voltages, the limiting factor is the winding insulation.

 

[GoldDigger]

Fair enough. The CTs used in MV and HV service are very different beasts than CTs used on systems of 600V or lower.

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[SG-1]

Fair enough. The CTs used in MV and HV service are very different beasts than CTs used on systems of 600V or lower.

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With these CTs you get more bang for the buck.

 

[iceworm]

Fair enough. The CTs used in MV and HV service are very different beasts than CTs used on systems of 600V or lower. ...

Lately (last ten years or so) I've been seeing rogowski coils on MV gear. As all know, they are way different beasts. Open circuit makes no difference. The information in the OP suggest he is dealing with rogowski coils.

ice

 

[iceworm]

With these CTs you get more bang for the buck.

Is that a pun as in they explode better?

ice

 

[GoldDigger]

Rogowski coil ?=? millivolt CT?

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[iceworm]

Rogowski coil ?=? millivolt CT?

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Disclaimer: I am not knowledgable on the physics of a Rogowski coil.
Most of the literature calls them "Transducers" - a "CT" being an iron core with a coil wrapped on it.

Rogowski coils are a helix, the ends pulled together around the conductor to be measureed. The output is high impedance. The ones I have seen are 400A to 1200A rated with the output at 150mV at rated current. They are not very linear like a CT is.

ice

 

[GoldDigger]

Key features of a Rogowski coil, AFAIK, are:
1. No core means still linear at high current levels.
2. No need to thread the conductor or use a split core
3. Output voltage is proportional to dI/dT, not I, so you need high impedance meter and the waveform is 90 degrees out of phase with I for a sine wave. And proportional to frequency. Both of those are usually dealt with by including an integrator circuit in the package.

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[iceworm]

Rogowski coil ?=? millivolt CT?

Key features of a Rogowski coil, AFAIK, are:
1. No core means still linear at high current levels.
2. No need to thread the conductor or use a split core
3. Output voltage is proportional to dI/dT, not I, so you need high impedance meter and the waveform is 90 degrees out of phase with I for a sine wave. And proportional to frequency. Both of those are usually dealt with by including an integrator circuit in the package.

You seem to be up on the physics. Tell me again - Why is it you asked?

ice