Xfrmr inrush and Available short circuit current

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sceepe

sceepe

Senior Member
Today I got an article emailed via Mr. Holts newsletter. This was a good article titled :Part 2: ?High Inrush Currents for Dry Type Transformers? Unfortunatly thinking about this has eatten up most of my morning. The article made a statement that:
"Even if the inrush value is known from the manufacturer, the actual value may be less. The available fault current and long conductor lengths feeding the transformer will reduce this maximum value."

My question is why does available fault current reduce the inrush current seen at the upstream breaker?
 
Mathematically speaking, the short circuit rating takes into account the impedance inherent in the transformer; the higher it is the lower the kAIC rating. Also, the longer the wires feeding the transformer, the lower the kAIC. So your available current is dependent on those factors, among others.

Someone with more technical knowledge can truly explain this... I hope what I said was accurate and helpful!

Keri
:grin: :grin:
 
It's a series circuit, so the current is the same everywhere.

You have two impedences: the transformer input impedence, and the upstream circuit impedence.

The available fault current is a function of the source impedence. The less available fault current, the higher the source impedence.

At normal steady state operation, the source impedence is small enough that it can be ignored. But for inrush calculations, it may be higher than the initial transformer impedence.

These two impedences are in series. So when you first apply a voltage, there is a higher total impedence than the transformer alone would be. That's what reduces the current.

Or did I totally misunderstand what you are asking?
 
steve66 said:
It's a series circuit, so the current is the same everywhere.

You have two impedences: the transformer input impedence, and the upstream circuit impedence.

The available fault current is a function of the source impedence. The less available fault current, the higher the source impedence.

At normal steady state operation, the source impedence is small enough that it can be ignored. But for inrush calculations, it may be higher than the initial transformer impedence.

These two impedences are in series. So when you first apply a voltage, there is a higher total impedence than the transformer alone would be. That's what reduces the current.

Or did I totally misunderstand what you are asking?
Click to expand...
Well, that's how I read it too.
Supply impedance will have some influence on transformer inrush, fault currents and harmonic voltage distortion from non-linear loads.
Maybe not much, but in some cases it can be significant.
For harmonic voltage distortion calculations, I generally include it even if only by estimation.
 
sceepe said:
Today I got an article emailed via Mr. Holts newsletter. This was a good article titled :Part 2: ?High Inrush Currents for Dry Type Transformers? Unfortunatly thinking about this has eatten up most of my morning. The article made a statement that:
"Even if the inrush value is known from the manufacturer, the actual value may be less. The available fault current and long conductor lengths feeding the transformer will reduce this maximum value."

My question is why does available fault current reduce the inrush current seen at the upstream breaker?
Click to expand...

I think a key phrase that is being overlooked is the phrase "may be less". I could give an example of where it is less, and I could give an example where it is not less.
 
SIMROX said:
Circuit breaker curve (D Curve) for high inrush applkications prevent breaker from tripping.
Click to expand...

Trip curves are manufacturer specific. There is no such thing as a universal "D curve".
 
sceepe said:
Today I got an article emailed via Mr. Holts newsletter. This was a good article titled :Part 2: ?High Inrush Currents for Dry Type Transformers? Unfortunatly thinking about this has eatten up most of my morning. The article made a statement that:
"Even if the inrush value is known from the manufacturer, the actual value may be less. The available fault current and long conductor lengths feeding the transformer will reduce this maximum value."

My question is why does available fault current reduce the inrush current seen at the upstream breaker?
Click to expand...

I also believe that the magnitude of the transformer inrush depends on where on the voltage cycle you close your switching device (breaker contactor etc...) I'm not sure where on the cycle leads to a higher current magnitude whether it is at a peak or a zero crossing.
 
Magnitude of Transformer Inrush Current

Magnitude of Transformer Inrush Current

Yes, the magnitude of inrush current also depends on "where you are" on the voltage waveform when you de-energize the xfmr.

1. If you de-energize the transformer when the voltage waveform is at its peak then the current through the core will be at a zero-crossing. This is due to the fact that the current lags the voltage by 90 degrees.

2. When the current is at a zero-crossing there will be no remnant flux in the core and it will not be magnetized with any particular bias.

3. If you energize the transformer when the conditions of 1, 2 are satisfied the inrush current will be minimized. However if the transformer is de-energized when the voltage waveform is not at peak, then there will be current flowing in the core, and there will be a remnant flux at the instant of de-energization, which would be in a particular direction based upon whether the voltage was at positive or negative half-cycle.

4. If the transformer is energized with the core magnetized, and biased in the opposite direction relative to the applied voltage's induced current, then you will have a very large inrush current. It has to overcome the magnetized core and magnetize it in the opposite direction.

If anyone sees holes in this let me know. I am reaching back and there are cobwebs...

Best regards,

Matt
 
mpross said:
Yes, the magnitude of inrush current also depends on "where you are" on the voltage waveform when you de-energize the xfmr.

1. If you de-energize the transformer when the voltage waveform is at its peak then the current through the core will be at a zero-crossing. This is due to the fact that the current lags the voltage by 90 degrees.
Click to expand...

Does the current always lag the voltage by exactly 90deg or does it depend on the power factor of the transformer.

I konw that with motors this vaule is not always exactly 90 deg but varies from motor to motor and determines the different power factors of motors.
 
mpross said:
Yes, the magnitude of inrush current also depends on "where you are" on the voltage waveform when you de-energize the xfmr.

1. If you de-energize the transformer when the voltage waveform is at its peak then the current through the core will be at a zero-crossing. This is due to the fact that the current lags the voltage by 90 degrees.

2. When the current is at a zero-crossing there will be no remnant flux in the core and it will not be magnetized with any particular bias.

3. If you energize the transformer when the conditions of 1, 2 are satisfied the inrush current will be minimized. However if the transformer is de-energized when the voltage waveform is not at peak, then there will be current flowing in the core, and there will be a remnant flux at the instant of de-energization, which would be in a particular direction based upon whether the voltage was at positive or negative half-cycle.

4. If the transformer is energized with the core magnetized, and biased in the opposite direction relative to the applied voltage's induced current, then you will have a very large inrush current. It has to overcome the magnetized core and magnetize it in the opposite direction.

If anyone sees holes in this let me know. I am reaching back and there are cobwebs...

Best regards,

Matt
Click to expand...

Another thing that I thought of about this explanation was that this same thing does not happen in motors simply due to the fact that there is not core in a motor to magnetize with remnant flux.

How does where on the voltage waveform you are when energizing factor into inrush currents
 
Transformer Inrush Current

Transformer Inrush Current

Hello mull982,

For your question- "How does where on the voltage waveform you are when energizing factor into inrush currents?".

It has to do with the voltage drop across an inductor: V = L*(di/dt) = (time rate of change of the current) times a proportionality constant (L). Don't know if you are calculus saavy, but the (di/dt) part is the derivative of a cosine function (which represents the current at a given point in time, and at a certain frequency), and causes a 90 degree phase shift in the current...

so to make that muddy explanation clear- when the voltage is at its peak the current will be out of phase by 90 degrees, or at its zero- crossing.

Back to your question- so it has to do with when you de-energize the unit, which is hard to do. If you can do this with the voltage at its peak the current will be at zero then you are minimizing the amount of current in the core, thus not leaving it magnetized in any particular direction. So when you go to energize at any given magnitude of voltage, you are minimizing inrush current to a certain degree.

If there is any remnant flux in the core, and you could by chance, or some other control scheme energize at close to the same value of voltage when the transfo was de-energized this would be interesting to see. Would it almost take the inrush current to zero?

Thanks,

Matt
 
mull982 said:
Does the current always lag the voltage by exactly 90deg or does it depend on the power factor of the transformer.

I konw that with motors this vaule is not always exactly 90 deg but varies from motor to motor and determines the different power factors of motors.
Click to expand...
Input magnetising current does pretty much lag by 90 deg on transformers (and similar for motors).
When you load the transformer, there will also be reflected load current and the power factor of that component depends on the load characteristics of course.

It is a similar story for induction motors. What you generally get on the nameplate is PF at rated load and speed. Run it off load (uncoupled) and what you get is mostly magnetising current plus a little active current to overcome windage and friction losses resulting in a very low PF.
 
mpross said:
Hello mull982,

For your question- "How does where on the voltage waveform you are when energizing factor into inrush currents?".

It has to do with the voltage drop across an inductor: V = L*(di/dt) = (time rate of change of the current) times a proportionality constant (L). Don't know if you are calculus saavy, but the (di/dt) part is the derivative of a cosine function (which represents the current at a given point in time, and at a certain frequency), and causes a 90 degree phase shift in the current...

so to make that muddy explanation clear- when the voltage is at its peak the current will be out of phase by 90 degrees, or at its zero- crossing.

Back to your question- so it has to do with when you de-energize the unit, which is hard to do. If you can do this with the voltage at its peak the current will be at zero then you are minimizing the amount of current in the core, thus not leaving it magnetized in any particular direction. So when you go to energize at any given magnitude of voltage, you are minimizing inrush current to a certain degree.

If there is any remnant flux in the core, and you could by chance, or some other control scheme energize at close to the same value of voltage when the transfo was de-energized this would be interesting to see. Would it almost take the inrush current to zero?

Thanks,

Matt
Click to expand...

Matt

Thanks for the explanation. Yes I am calculus savy and do understand the concept of di/dt. I also understand the concept of the phase shift causing the current to lag the voltage by 90 deg. With the current lagging the voltage by 90deg that would imply a power factor of 0. Do all transformers have a power factor of 0 causing this exact 90deg phase shift, or does it vary such as power factors in motors which can be 85%, 90% etc therefore leading to different phase shifts.

My question about re-energizing the transfrormer was more geared towards ignoring where on the waveform we de-energized and strictly looking at the effects of where we energize on the waveform. For example if we de-energized with voltage at peak and therefore had zero remnant flux in the core and then stricly focus on the inrush current as a result of where on the voltage waveform we are during energizing what effects does this have. Will there be more current at the voltage peak or zero corssing when just focusing on this case.

Thanks for the help!
 
Just to add a bit to this:

In addition to the remnant magnetization issue that Matt mentioned, where you close the circuit on startup will also alter inrush. The V=Ldi/dt (Voltage across an inductor is equal to the inductance times the rate change in current over time) that causes the 90 degree phase shift in the steady state (after everything has settled down) is also active from the very beginning.

A slightly different way to think about that di/dt term is to instead think of the 'volt-second' product. If you apply X volts for T seconds to an inductor, then the magnetic flux and its associated current flow will change by a given amount. The key here is that the volt-second product does not determine how much flux the transformer is carrying; it determines the _change_ in the amount of flux that the transformer is carrying.

When you apply AC voltage to a transformer, than the maximum change in magnetic flux is cause by a whole half cycle. In the steady state case, when the _positive_ voltage half cycle starts, the magnet flux is maximum in the negative direction. By the end of the positive half cycle, the magnetic flux is maximum in the positive direction; thus you have a change from -max to +max. Note that this is the 90 degree phase shift that Matt mentions.

Now imagine that you have a perfect transformer core, with _zero_ remnant magnetization. Close the circuit on this core just at the point where the positive half cycle starts. Since the magnetic flux is starting at zero, and a given half cycle results in a specific _change_ in magnetic flux, then by the end of this first half cycle, the magnetic flux needs to be _twice_ the normal steady state maximum.

A real transformer won't reach this double flux state, but instead will saturate the core. In trying to get to this double flux state, quite a bit of current will flow, since a saturated core takes more current to get a change in flux level. If you want to minimize the inrush current, then you not only want to open the circuit at the peak voltage point, you also want to close the circuit at the peak voltage point.

-Jon
 
winnie said:
Now imagine that you have a perfect transformer core, with _zero_ remnant magnetization. Close the circuit on this core just at the point where the positive half cycle starts. Since the magnetic flux is starting at zero, and a given half cycle results in a specific _change_ in magnetic flux, then by the end of this first half cycle, the magnetic flux needs to be _twice_ the normal steady state maximum.

A real transformer won't reach this double flux state, but instead will saturate the core. In trying to get to this double flux state, quite a bit of current will flow, since a saturated core takes more current to get a change in flux level. If you want to minimize the inrush current, then you not only want to open the circuit at the peak voltage point, you also want to close the circuit at the peak voltage point.

-Jon
Click to expand...

winnie I understand everything that you are saying except the above two paragraphs. Why does the magnetic flux need to be twice the the normal steady state during the first half cycle. When the half cycle starts the flux will be negative and when the half cycle ends the flux will be positive. Does the flux not change direction when the current crosses zero or does the flux add during this half cycle. I understand what you are saying about the core going into saturation but do not see why the flux doubles as you mentioned.
 
081018-1503 EST

mull982:

If you integrate the voltage from 0 to 180 deg, 1/2 half cycle, then under steady state conditions that takes you from maximum negative flux to maximum positive flux, or vice versa.

If you start at zero residual flux and 0 degrees on the voltage waveform, then you approximately reach the normal maximum flux level when the voltage gets to the 90 deg point. Continuing from 90 to 180 deg you drive the core more into saturation.

That is why if you stopped flux change at +90 deg on the voltage waveform, approximately 0 flux, you would want to start flux change at +90 deg.

.
 
gar said:
081018-1503 EST

mull982:

If you integrate the voltage from 0 to 180 deg, 1/2 half cycle, then under steady state conditions that takes you from maximum negative flux to maximum positive flux, or vice versa.

If you start at zero residual flux and 0 degrees on the voltage waveform, then you approximately reach the normal maximum flux level when the voltage gets to the 90 deg point. Continuing from 90 to 180 deg you drive the core more into saturation.

That is why if you stopped flux change at +90 deg on the voltage waveform, approximately 0 flux, you would want to start flux change at +90 deg.

.
Click to expand...

Ok so if you energize or start the flux change at +90 deg what would the flux do for the half cycle after that. It would start at zero and then go to a positive maximum before returning to 0 again in the half cycle before it would continue on to a negative flux maximum.

Is the flux going from zero to positive back to zero again not as severe as going from a positive to negative or vise versa?

When we talk about the flux changing I'm assuming we are also taking about the current magnitude and there for explaining the current change (2x) etc as well.
 
081018-2006 EST

mull982:

If you look at a hysteresis curve and assume excitation was terminated at a high flux level. With normal design this is partially into saturation. The residual flux level remains at this high level.

If voltage is applied at 0 deg in the correct direction, then with time this flux is reduced. When the voltage reaches the 90 deg point the flux is about 0. Continuing the voltage for the next 1/4 cycle will take flux to the maximum flux level in the opposite direction.

It is hard to talk about L di/dt in a magnetic core application because L is not constant as you move around the hysteresis curve.

.
 
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