Understanding CT magnetization curve 5

[GavinCSY]

I know when a CT goes into saturation it starts to produce errors and problems to produce sufficient secondary current. So we do a secondary injection to determine the knee point of the CT. Ok so now i have a graph showing the excitation voltage and the excitation current.

My question is how to relate this curve to when the CT is in operation? eg. i have a knee point at 2700V@4.25mA for a protection CT with a ratio of 10 000:1. So at what current in the primary would cause the CT to saturate?

How do i know when i look at a magnetization curve that the CT is the right one for the job?

Thanks

 

[slavag]

Hi.
Please see attached links.
thread238-182418
thread238-165896
For what you use this CT ( It's like to generator CT)?
What type of protection functionality?
You need check this curves with requirements of relay mnf.
Regards.
Slava

 

[slavag]

Hi.
Please see two option od Effective ALF
The effective accuracy limiting factor is calculated as follows:
n' =n x ( Pn + Pw)/(P + Pw)
n'= Vk / (Ins x [Rw + 2 x RL+ RB])

n rated accuracy limiting factor (ALF) E.g. 20 for 5P20
Pn rated VA's VA E.g. 30
Pw secondary winding losses in VA at rated current
P effective burden = (2 × RL + RB) × Ins^2
n' effective ALF (at effectively connected burden) or
Kssc according IEC 44-6
Vk Knee point voltage
Ins Rated secondary current
RL Resistance of secondary wiring (single length)
RB Total resistance of effective burden
Rw Resistance of CT secondary winding.
Good Luck
Slava

 

[GavinCSY]

Yes it's a CT used for generator protection. So to know if the CT is the right one i have to check with the protection relay manual

But what about the first question of finding the primary current which will cause it to saturate?

 

[slavag]

It's not so correct Q.
CT saturation is not function of only primary current.
It's function of secondary burden ( or voltages on econdary side of CT).
It's also function of DC offset in the fault current.
With low burden and no DC comp possible, I don't know, 70-100 times of CT rated current.
With high level of DC copm or high level of harmonics or with high secondary burden it's other value with same CT.
For you very important accuracy class of CT for 87G protection and high value of ALF. I recommend CT with min 5P30 acuraccy class and 30-60VA burden, secondary 1A.
Good Luck.
Slava

 

[GavinCSY]

I see, so that means if the fault current doesn't generate 2700V on the secondary side, the CT won't go into saturation, is that right?

 

[slavag]

No, NO.
Don't forget, CT it's shorted xfr, voltage on secondary side "0". If you open current circuit, you see HV on terminals.

 

[GavinCSY]

Oh yes, my bad.

Sorry but i still don't understand the point of a secondary injection. OK it's to find the knee point where the CT saturates and starts to give errors but the thing is it is found using a voltage injection to the secondary side when during normal operation no voltage is generated in the secondary side.

So the knee point is at 2700V@4.25mA which means what exactly?

Does it mean the CT is still accurate once the primary current exceeds say 3/4 rated?

Why protection CTs have high knee point and measuring low knee points?

OR the secondary injection test is just to prove the CT is in no way damaged during transport and still maintains the right characteristic?

 

[slavag]

Hi Gahvin.
On part of Q I have quik answers, on part I need more time and prefer wait good and simple explanation from others
(Scottf are our leader in these issues).
Protection core CT have high knee point for right operation
and accuracy in high level faults, 5P30: in 30 times of In your meas error up to 5%.
low knee points for meas cores for protect on current inputs
of meas devices.
That means you want CT saturation ASAP.
Targets of VA curves test:
1. Found Internal damages of CT
2. Eq. of all 6 CT's of generators for example .
Best Regards.
Slava

 

[jghrist]

2700 volts seems very high for a knee point.

 

[oldfieldguy]

jghrist--

Those high-ratio CT's in generator service typically have very high saturation voltages.

gahvin--

Here's a quick and dirty field test to see if you're headed for trouble. Using an AC current source, inject a current signal at your CT's terminals into your current circuit with ALL devices connected. You can use an amp or two. With the current flowing, measure the VOLTAGE at those terminals. This will give you a pretty accurate measure of your current circuit's actual burden. Multiply this figure times the expected fault current, and you'll see what that current circuit will look like when the CT is pushing fault current.

If that fault current produces a terminal voltage near the KNEE of your saturation curve, then you can expect all the hassles of operation under saturation to affect your relaying. If your calculated voltage is below the knee, you're safe.

In lieu of the calculations, you can inject actual secondary currents into the loop and measure voltage. The same criteria apply: Voltage at full expected fault current should be less than the knee of the CT saturation curve.

Incidentally, 4.25 mA is not saturated...

old field guy

 

[RalphChristie]

The magnetisation curve is the best method of determining a CTs performance. The amount of magnetizing current required to generate an open circuit voltage at the terminals of the unit, expressed in a graph. The knee point voltage is generally defined as the voltage at which a 10% increase in voltage will require a 50% increase in current.

A CT must be able to provide a high enough voltage to drive a load (burden) under certain conditions.

Instruments and meters are required to work accurately up to full load current, but above this it is better to saturate the CT to protect the instruments under fault conditions.
Protective gear are required to work under a wide range of currents from full load current up to fault current many times normal rating. It is then important to ensure that saturation is avoided wherever possible to ensure positive operation of relays.

What is the purpose of a magnetization (excitation) curve?

First of all, it shows you if there a any internal problems in a CT.
Secondly, especially on differential schemes, you can ensure that all the CTs have the same mag-curves. This will minimize spill currents in the operating coils and minimize nuisance operation under load or through fault conditions.
Thirdly, to ensure that a high enough voltage is provided under fault conditions to ensure relay operation.

during normal operation no voltage is generated in the secondary side

Although it might not be high, a voltage is generated during load flow conditions. the higher the current, the higher the voltage until it reach a point of saturation (this point will depend on the core material of the CT and the burden connected to it)

Regards
Ralph

And yes, 2700V seems a little high, although I have found voltages of 600V on a 1200/1 CT. So, on a 10000/1 CT it might be true, although I am not sure.





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[slavag]

Hi Ralph.
Special thanks from me.
Regards.
Slava

 

[scottf]

Ghavin-

It's been explained pretty well in this thread so far, but I'll just throw in my 2 cents worth.

You have asked how a secondary voltage on the CT tells you anything about the performance in terms of primary current.

Think about it this way, during an excitation test, you apply voltage to the secondary and plot that against how much current is drawn from the source. Since there is nothing on the primary side, the current measured is the excitation current.

During operation, imagine the CT as a constant current source. For your example, a 10,000:1A ratio. So when 10kA is on the primary, 1 A is on the secondary, but that is only part of the equation. The CT (constant current source) must drive that secondary current into some burden...with the burden consisting of the connected devices (relay) plus the lead impedance.

The CT's limit is how much voltage it can develop and still be a "constant current source". So...let's say your leads plus the relay is a 1 ohm impedance. With 10kA in the primary, that's 1A in the secondary and 1V across the secondary. A knee point voltage of 2700 V, tells you that with a 1 ohm burden connected to the CT, you can drive the primary up to 2700 times 10kA before the unit saturates.

Saturated is defined as the secondary current no longer behaving linearly to the primary current (within some tolerance, like 10% or 5%).

In the IEC world, protection accuracy class is expressed as some accuracy (like 5%) up to some accuracy limiting voltage (like 20) at some burden (like 30VA). If the secondary is rated at 1A, then a 30VA burden means a 30 ohm burden impedance.

In the IEEE world, protection performance is actually listed as a secondary voltage and you work backwards to the burden (i.e. C800).

 

[GavinCSY]

Guys thanks a lot for all the explanations here. Really appreciate it.

 

[GavinCSY]

old field guy,

Just to clarify your calculations,

Eg.

Burden = 1 A x 1.5 V
= 1.5 VA

Then multiply by fault current, lets say 10 times

= 1.5 x 100 000
= 150 000 A ?

Next is to find the terminal voltage which is?

= ???

A billion thanks, sorry if this seems like a really really silly question. Really new to this

 

[slavag]

Hi Gahvin.
if you have 1.5V on secondary with injection 1A:
you have only 15V in case of 10 times current.

This secondary test is very important in commissioning time, it show you that you current circuit wiring is OK.

 

[GavinCSY]

Ah ok i think i get it now,

So to summarize, CT saturation is no good because it causes errors and the secondary current drops low. So make a mag curve to find where this saturation takes place.

Measuring CT's you want it to saturate slightly above the rated current to ensure that the instruments are not damaged but for protection you want it to have a high knee point to ensure it still works through the fault current.

After deriving the knee point and the actual burden,eg. 2700V and 1.5VA, the CT primary can be driven up to 1800 (2700/1.5) times the nominal before it saturates but even before that everything else would be in flames or molten metal.

Making mag curves also ensures you have the same CT characteristics if used for differential protection, shows if the CT is damaged or not, determine that a measuring and a protection CT is not swapped

 

[oldfieldguy]

gahvin--

You got it right. The quick and easy view on CT saturation is that up until the saturation point, the current OUT of the secondary is essentially identical and proportional to the current going IN the primary. Your metering and protection equipment can use this current as equivalent to the real power.

If something is wrong and the terminal voltage at your CT exceeds the saturation voltage of the CT, then that "identical and proportional" business goes out the window. You get strange waveforms and magnitudes.

Older relays, especially electromechanicals, handled this poorly, reacting in barely comprehensible fashion, if at all. New microprocessor-based relays can see the characteristic of the waveform and determine that the CT is in saturation, and may alter their operation characteristics.

A quick and dirty rule of thumb is to establish the impedance of your current circuit, determine your maximum fault current, and then select a CT with a saturation "knee" at least twice the expected terminal voltage.

This works well for me.



old field guy

 

[LionelHutz]

I've always believed the curve shows how much secondary current is "lost" to develop the voltage. For example, when your the CT is developing 2700V into it's secondary loop it produces 4.25mA less current that the ideal or calculated. That's why, once you hit the knee of the curve, a lot of secondary current starts getting "lost" trying to produce the small increase in voltage. I hope someone will confirm this.

FYI, other CT considerations

A CT rated such as 10C800;
the 10 means it will have a 10% max ratio error at 20 times rated current.
The 800 is the max secondary voltage produced at 20 times rated current.
These is a relaying class CT rating.


A CT rated such as 0.3B0.2;
The 0.3 is the max ratio error when the CT is operated into a burden of 0.2VA at rated primary current.
This is a metering class CT rating.