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Transformer In rush current 1
- Thread starter Pgee630
- Start date
Hi Pete.
When the transformer saturates it becomes an air core reactor for further increases of applied voltage.
given the very high ratio between the air core inductive reactance and the iron core inductive reactance, I tens to disregard the inductive reactance and just use the resistance when a core is saturated.
I expect that the error will be in the same ballpark as rounding errors.
Residual magnetism:
Energizing at the zero crossing is very bad for DC offset.
Energizing at the zero crossing with opposite residual magnetism is worse.
It results in:
A larger window of opportunity for maximum DC offset.
The core will saturate at more degrees before or after the zero crossing.
A closure at the zero crossing will result in a larger DC offset and a greater inrush than in the case of a core with no residual magnetism.
Example;
A core with no residual magnetism is energized at the zero crossing.
The core saturates when the instantaneous voltage reaches 90% of the maximum voltage.
The current resulting from the last 10% of applied voltage is limited by resistance only. (mostly)
A core with residual magnetism is energized at the zero crossing.
The core saturates when the instantaneous voltage reaches 80% of the maximum voltage.
The current resulting from the last 20% of applied voltage is limited by resistance only. (mostly)
I look forward to your comments regarding this simplification, Pete.
Bill
--------------------
"Why not the best?"
Jimmy Carter
When the transformer saturates it becomes an air core reactor for further increases of applied voltage.
given the very high ratio between the air core inductive reactance and the iron core inductive reactance, I tens to disregard the inductive reactance and just use the resistance when a core is saturated.
I expect that the error will be in the same ballpark as rounding errors.
Residual magnetism:
Energizing at the zero crossing is very bad for DC offset.
Energizing at the zero crossing with opposite residual magnetism is worse.
It results in:
A larger window of opportunity for maximum DC offset.
The core will saturate at more degrees before or after the zero crossing.
A closure at the zero crossing will result in a larger DC offset and a greater inrush than in the case of a core with no residual magnetism.
Example;
A core with no residual magnetism is energized at the zero crossing.
The core saturates when the instantaneous voltage reaches 90% of the maximum voltage.
The current resulting from the last 10% of applied voltage is limited by resistance only. (mostly)
A core with residual magnetism is energized at the zero crossing.
The core saturates when the instantaneous voltage reaches 80% of the maximum voltage.
The current resulting from the last 20% of applied voltage is limited by resistance only. (mostly)
I look forward to your comments regarding this simplification, Pete.
Bill
--------------------
"Why not the best?"
Jimmy Carter
electricpete
Electrical
Hi Bill, hope things are going well with you up there.
I think the current is still limited by magnetic effects during inrush. Even when deep into saturation you have at least as much inductive reactance as an air core and I think it's typically still more than resistance. The reason I think that: if it were limited by resistance only then I think the current would be a heckuva lot more than 10 or 15 times rated current.
I talked about a model I use for inrush because it is a simple that I’ve relied on a lot before and it makes sense to me personally (everyone has their own favorite way of looking at these things). To explain it a little more, if you had a one turn coil, v = d/dt{Phi[i(t)]}. By the chain rule for differentiation, v = dPhi/di * di/dt. Then you see dPhi/di plays the role of L(i) and gives the interpretation that L(i) is the slope of the Phi vs i curve (simply L if that curve is a straight line through the origin). Using the approach that I have described so far, it was necessary to assume Phi vs i is a single-valued curve resulting in a single valued function L(i) and that gives a simple model which is good/easy for numerical simulation because the rate of change at time t can be computed from the variables at time t. But what’s missing from that is that it doesn’t have the ability to capture hysteresis (and therefore remnant flux). We can correct that by replacing L(i) with L(i,t) which now represents the slope of the Phi vs i as you trace out the hysteresis loop in the Phi vs i plane. In that way you can still visualize integrating to find i(t)=1/L(i,t) * Int{v(t)*dt} . Now if you still close at v(t)=0 but at a point on the hysteresis curve where i is zero but Phi is already positive (remnant flux), then Phi is already closer to saturation and you'll reach saturation in a shorter time after closing and therefore reach a higher peak current than you would closing in at the same angle if the initial flux was 0. That was a long way to go to say I agree with you, remnant flux will definitely give you a higher peak current at the worst case closing angle. In terms of quantitativly how much more, I don't know without thinking about it some more. I'll take your word for it on those numbers.
=====================================
(2B)+(2B)' ?
I think the current is still limited by magnetic effects during inrush. Even when deep into saturation you have at least as much inductive reactance as an air core and I think it's typically still more than resistance. The reason I think that: if it were limited by resistance only then I think the current would be a heckuva lot more than 10 or 15 times rated current.
I talked about a model I use for inrush because it is a simple that I’ve relied on a lot before and it makes sense to me personally (everyone has their own favorite way of looking at these things). To explain it a little more, if you had a one turn coil, v = d/dt{Phi[i(t)]}. By the chain rule for differentiation, v = dPhi/di * di/dt. Then you see dPhi/di plays the role of L(i) and gives the interpretation that L(i) is the slope of the Phi vs i curve (simply L if that curve is a straight line through the origin). Using the approach that I have described so far, it was necessary to assume Phi vs i is a single-valued curve resulting in a single valued function L(i) and that gives a simple model which is good/easy for numerical simulation because the rate of change at time t can be computed from the variables at time t. But what’s missing from that is that it doesn’t have the ability to capture hysteresis (and therefore remnant flux). We can correct that by replacing L(i) with L(i,t) which now represents the slope of the Phi vs i as you trace out the hysteresis loop in the Phi vs i plane. In that way you can still visualize integrating to find i(t)=1/L(i,t) * Int{v(t)*dt} . Now if you still close at v(t)=0 but at a point on the hysteresis curve where i is zero but Phi is already positive (remnant flux), then Phi is already closer to saturation and you'll reach saturation in a shorter time after closing and therefore reach a higher peak current than you would closing in at the same angle if the initial flux was 0. That was a long way to go to say I agree with you, remnant flux will definitely give you a higher peak current at the worst case closing angle. In terms of quantitativly how much more, I don't know without thinking about it some more. I'll take your word for it on those numbers.
=====================================
(2B)+(2B)' ?
H Pete.
We are doing great up here.
We are in one of the small areas of Alberta that has so few cases that we are not in lockdown.
Re saturation:
I may be wrong but I use a rough estimate for permeability of 4,000 to 1.
Consider a transformer with a working voltage of 120 Volts.
On energization the transformer goes into full saturation at 110 Volts.
The voltage continues to rise but the winding now acts as an air core inductance.
The 10 Volt difference between 110 Volts and 120 Volts is limited by the resistance and by 1/4000 of the normal inductive reactance.
The voltage at which the transformer saturates is influenced by the amount and polarity of any residual magnetism.
The effect of residual magnetism may range from saturation at a lower voltage to no saturation.
I expect that worst case inrush PU will be influenced by both the X/R ratio and the hysteresis of the transformer core.
I may be wrong.
It's been a long time and I don't have time this morning to review my old texts.
Am I making sense?
Bill
--------------------
"Why not the best?"
Jimmy Carter
We are doing great up here.
We are in one of the small areas of Alberta that has so few cases that we are not in lockdown.
Re saturation:
I may be wrong but I use a rough estimate for permeability of 4,000 to 1.
Consider a transformer with a working voltage of 120 Volts.
On energization the transformer goes into full saturation at 110 Volts.
The voltage continues to rise but the winding now acts as an air core inductance.
The 10 Volt difference between 110 Volts and 120 Volts is limited by the resistance and by 1/4000 of the normal inductive reactance.
The voltage at which the transformer saturates is influenced by the amount and polarity of any residual magnetism.
The effect of residual magnetism may range from saturation at a lower voltage to no saturation.
I expect that worst case inrush PU will be influenced by both the X/R ratio and the hysteresis of the transformer core.
I may be wrong.
It's been a long time and I don't have time this morning to review my old texts.
Am I making sense?
Bill
--------------------
"Why not the best?"
Jimmy Carter
electricpete
Electrical
I agree permeability on the order of thousands, so that seems to contradict what I said about air core inductance limiting the current. I didn't really mean to throw air core out there except to explain the inductance doesn't go away, but I hadn't thought about it in the way you said it. My thought process was that resistance doesn't seem enough to explain it... if that were the case than more efficient transformers would have higher inrush current... but our super efficient generator stepup transformer still has 10x rated peak current peaks when energized from the high side with low side open and back when I looked at that I didn't think the CT's were saturating. My general understanding is more efficient trasnformers and higher X/L systems result in longer duration of the inrush but not higher inrush peaks. I can't explain that with resistance.
But if magnetics were limiting the current as I suggested then why wouldn't current be higher? I don't have a good answer without digging into it closer. I might be wrong, you might be right.... it wouldn't be the first time.
Edit - If I could remember our measured dc winding resistance values that might shed some light one way or the other. I'll poke around to see if I can find some of those from our large power transformers.
=====================================
(2B)+(2B)' ?
But if magnetics were limiting the current as I suggested then why wouldn't current be higher? I don't have a good answer without digging into it closer. I might be wrong, you might be right.... it wouldn't be the first time.
Edit - If I could remember our measured dc winding resistance values that might shed some light one way or the other. I'll poke around to see if I can find some of those from our large power transformers.
=====================================
(2B)+(2B)' ?
Just a point on terminology:
- our transformers are generally unloaded via secondary circuit breaker [ although there are remote locations where a load break switch is used to perform this operation ]
- the trafo is then removed from potential by the slow-opening high-side disconnect switch, meaning [ specifically pointing out ] that only charging / magnetizing current is interrupted, NOT load current
I'm trying to preclude confusion and/or the use of murky descriptions.
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
electricpete
Electrical
Well I was back at work today so I went digging for some more info to shed light on whether or not the resistance controls the inrush (for an example generator stepup transformer, not necessarily representative of others)
Attached is some info from one of our Generator Stepup Transformers.
One document is the ratings
362.25kv/25kv wye/delta
850MVA
Rated HV current 1426
On-load losses 1629kw
If we compute an efficiency as 1-(1629kw/850mva) it is something like 99.8% efficient.
If we had no other info and tried to estimate conservatively high estimate of winding resistance, we might assume 100% of the losses come from the HV winding I^2*R (in fact there is also LV winding I^2*R, core losses, stray losses).
A HIGH estimate for HV winding phase to neutral resistance would be
R = Losses / (3*I^2) = 1,629,000W/(3*1426A^2) = 0.27 ohms
If resistance controls the peak current, what would the peak inrush current be?
I = sqrt2*(V/sqrt3) / R = sqrt2*209,145v/0.27 = 1,095,466A which is 476 times as high as the peak rated current sqrt2*1426A
In fact when measured directly during a test at room temperature the HV winding resistance phase to neutral was 0.11 ohms per second attachment(maybe a little higher at operating temperature). So if resistance controlled the current then the previous result would easily double to something like 1000 times rated current.
That does not compute for me for a number of reasons. For one I've never heard anyone mention a multiple anywhere near that high (I've heard numbers up to 40x rated). For another thing I’d think our waveforms would have shown obvious evidence of CT saturation but the waveforms I remember from inrush were somewhat smooth for most of the waveform (I wasn’t able to track down any waveforms… wish I would have been better at saving stuff). Also the transformer and incoming lines would heat quite a bit. For a third thing I’d wonder how those big long HV bushings could stand the forces (I realize they are well separated but they would also interact with tremendous induced circulting currents in the tank).
Another thing to consider aside from the high magnitude is the dc component which lasts for a long time (10 or 20 seconds = 600 or 1200 cycles). It seems to iondicate that inductance remains relevant in the circuit (you’d think resistive controlled circuit would decay a lot quicker).
With all that said I still don’t have a good explanation for how the magnetics controls the current when pushed so far into saturation. The best I can come up with is that it is well beyond the linear limit for saturation but not so far that it acts like an air core . Also my simple thinking involved single phase transformer, there may be complications when we think about the different legs of the core and how this interacts with the other phases that I’m not understanding. More questions than answers. I am happy to leave it an open question.
=====================================
(2B)+(2B)' ?
Attached is some info from one of our Generator Stepup Transformers.
One document is the ratings
362.25kv/25kv wye/delta
850MVA
Rated HV current 1426
On-load losses 1629kw
If we compute an efficiency as 1-(1629kw/850mva) it is something like 99.8% efficient.
If we had no other info and tried to estimate conservatively high estimate of winding resistance, we might assume 100% of the losses come from the HV winding I^2*R (in fact there is also LV winding I^2*R, core losses, stray losses).
A HIGH estimate for HV winding phase to neutral resistance would be
R = Losses / (3*I^2) = 1,629,000W/(3*1426A^2) = 0.27 ohms
If resistance controls the peak current, what would the peak inrush current be?
I = sqrt2*(V/sqrt3) / R = sqrt2*209,145v/0.27 = 1,095,466A which is 476 times as high as the peak rated current sqrt2*1426A
In fact when measured directly during a test at room temperature the HV winding resistance phase to neutral was 0.11 ohms per second attachment(maybe a little higher at operating temperature). So if resistance controlled the current then the previous result would easily double to something like 1000 times rated current.
That does not compute for me for a number of reasons. For one I've never heard anyone mention a multiple anywhere near that high (I've heard numbers up to 40x rated). For another thing I’d think our waveforms would have shown obvious evidence of CT saturation but the waveforms I remember from inrush were somewhat smooth for most of the waveform (I wasn’t able to track down any waveforms… wish I would have been better at saving stuff). Also the transformer and incoming lines would heat quite a bit. For a third thing I’d wonder how those big long HV bushings could stand the forces (I realize they are well separated but they would also interact with tremendous induced circulting currents in the tank).
Another thing to consider aside from the high magnitude is the dc component which lasts for a long time (10 or 20 seconds = 600 or 1200 cycles). It seems to iondicate that inductance remains relevant in the circuit (you’d think resistive controlled circuit would decay a lot quicker).
With all that said I still don’t have a good explanation for how the magnetics controls the current when pushed so far into saturation. The best I can come up with is that it is well beyond the linear limit for saturation but not so far that it acts like an air core . Also my simple thinking involved single phase transformer, there may be complications when we think about the different legs of the core and how this interacts with the other phases that I’m not understanding. More questions than answers. I am happy to leave it an open question.
=====================================
(2B)+(2B)' ?
Lets break this down.
For the sake of simplicity we will consider tje onset of saturation to be a definite point.
Assume that the transformer saturates at 90% of the applied voltage.
Up to saturation the current increase proportionally to the voltage and is limited by the impedance.
The impedance is composed of resistance and iron core inductive reactance.
Above the saturation point, the increase in current is limited by the new impedance.
The saturated impedance is composed of the resistance and the air core inductive reactance.
So we have unsaturated current based on 90% of applied voltage PLUS saturated current based on 10% of applied voltage.
This is a simplification for the sake of explanation.
There are two other factors.
1. Phase angles; The unsaturated current will not be at the same phase angle as the saturated current and so simple addition may not be used.
2. While the X/R ration before saturation may be 1/10, above saturation the X/R ratio may be 400/1. At this X/R ratio the inductive reactance may be ignored, as the circuit will be overwhelmingly resistive.
But, This applies only to the current developed due to the 10% voltage above saturation.
We have assumed that the transformer saturates at 90% of applied voltage. This percentage will vary from energization to energization depending on the residual magnetism and the point on wave at which the transformer is energized.
Another way of looking at it.
In an imaginary transformer, below saturation, each volt increase in applied voltage will cause a 1 Amp increase in current.
In the same transformer above saturation, each each volt increase in applied voltage will cause a 10 Amp increase in current.
If this transformer saturates at 95% voltage the current will be:
95% x 1 Amp = 95 Amps, plus 5% x 10 Amps = 50 Amps for a total of 145 Amps inrush.
If this transformer saturates at 90% voltage the current will be:
90% x 1 Amp = 90 Amps, plus 10% x 10 Amps = 100 Amps for a total of 190 Amps inrush.
Bill
--------------------
"Why not the best?"
Jimmy Carter
For the sake of simplicity we will consider tje onset of saturation to be a definite point.
Assume that the transformer saturates at 90% of the applied voltage.
Up to saturation the current increase proportionally to the voltage and is limited by the impedance.
The impedance is composed of resistance and iron core inductive reactance.
Above the saturation point, the increase in current is limited by the new impedance.
The saturated impedance is composed of the resistance and the air core inductive reactance.
So we have unsaturated current based on 90% of applied voltage PLUS saturated current based on 10% of applied voltage.
This is a simplification for the sake of explanation.
There are two other factors.
1. Phase angles; The unsaturated current will not be at the same phase angle as the saturated current and so simple addition may not be used.
2. While the X/R ration before saturation may be 1/10, above saturation the X/R ratio may be 400/1. At this X/R ratio the inductive reactance may be ignored, as the circuit will be overwhelmingly resistive.
But, This applies only to the current developed due to the 10% voltage above saturation.
We have assumed that the transformer saturates at 90% of applied voltage. This percentage will vary from energization to energization depending on the residual magnetism and the point on wave at which the transformer is energized.
Another way of looking at it.
In an imaginary transformer, below saturation, each volt increase in applied voltage will cause a 1 Amp increase in current.
In the same transformer above saturation, each each volt increase in applied voltage will cause a 10 Amp increase in current.
If this transformer saturates at 95% voltage the current will be:
95% x 1 Amp = 95 Amps, plus 5% x 10 Amps = 50 Amps for a total of 145 Amps inrush.
If this transformer saturates at 90% voltage the current will be:
90% x 1 Amp = 90 Amps, plus 10% x 10 Amps = 100 Amps for a total of 190 Amps inrush.
Bill
--------------------
"Why not the best?"
Jimmy Carter
A couple of comments on your last post.
1. GSUs are not generally energized DOL and may not be optimized to reduce the inrush.
2. That efficiency may suggest a lower than normal resistance.
That may not be the best transformer on which to base estimates.
I may be wrong.
Bill
--------------------
"Why not the best?"
Jimmy Carter
1. GSUs are not generally energized DOL and may not be optimized to reduce the inrush.
2. That efficiency may suggest a lower than normal resistance.
That may not be the best transformer on which to base estimates.
I may be wrong.
Bill
--------------------
"Why not the best?"
Jimmy Carter
electricpete
Electrical
Thanks Bill, lots of good points.
> In the same transformer above saturation, each each volt increase in applied voltage will cause a 10 Amp increase in current.
I agree. The interesting question that got my attention is what limits the current to 10, given your point about iron relative permeability being in the 1000's. I'm not 100% sure but maybe it's a gradual transition both in terms of the material B vs H properites and the spatial variation of flux density (some parts of the core can be deeper into saturation than others at a given point in time)...and as a result it still might retain some effective mu > mu0 during the worst case inrush.
> 2. While the X/R ration before saturation may be 1/10, above saturation the X/R ratio may be 400/1. At this X/R ratio the inductive reactance may be ignored, as the circuit will be overwhelmingly resistive.
Since you said the circuit becomes overwhelmingly resistive, I think you meant to say the X/R goes down (not up) as we go into saturation, and I'd agree the effective L decreases in saturation.
Interestingly, the fact that the inrush tends to last a long time might suggest the effective L/R (or X/R) remains somewhat high and the effective L (or X) doesn't decrease that dramatically (at least not dramatically enough to drive the dc component to zero quicker).
> GSUs are not generally energized DOL and may not be optimized to reduce the inrush.
I don't understand DOL in the context of transformers. At any rate we have a low-voltage generator circuit breaker that we sync our generator across, so our GSU's are always energized by closing a high-side circuit breaker with the low side circuit breaker open. That is the condition where the prolonged inrush occurs. It's my understanding for any transformer energizing under no-load is a worst case for inrush.
> this may not be the best transformer on which to base estimates.
I agree there are a wide variety of transformers, and certainly the winding resistance will play a bigger role in the smaller ones than it does on our GSU. I know that some of our HV switchyard circuit breakers have preinsertion resistors, we might have those on the HV breakers that energize the GSU but if so the resistors are only in the circuit for a very short time.
=====================================
(2B)+(2B)' ?
> In the same transformer above saturation, each each volt increase in applied voltage will cause a 10 Amp increase in current.
I agree. The interesting question that got my attention is what limits the current to 10, given your point about iron relative permeability being in the 1000's. I'm not 100% sure but maybe it's a gradual transition both in terms of the material B vs H properites and the spatial variation of flux density (some parts of the core can be deeper into saturation than others at a given point in time)...and as a result it still might retain some effective mu > mu0 during the worst case inrush.
> 2. While the X/R ration before saturation may be 1/10, above saturation the X/R ratio may be 400/1. At this X/R ratio the inductive reactance may be ignored, as the circuit will be overwhelmingly resistive.
Since you said the circuit becomes overwhelmingly resistive, I think you meant to say the X/R goes down (not up) as we go into saturation, and I'd agree the effective L decreases in saturation.
Interestingly, the fact that the inrush tends to last a long time might suggest the effective L/R (or X/R) remains somewhat high and the effective L (or X) doesn't decrease that dramatically (at least not dramatically enough to drive the dc component to zero quicker).
> GSUs are not generally energized DOL and may not be optimized to reduce the inrush.
I don't understand DOL in the context of transformers. At any rate we have a low-voltage generator circuit breaker that we sync our generator across, so our GSU's are always energized by closing a high-side circuit breaker with the low side circuit breaker open. That is the condition where the prolonged inrush occurs. It's my understanding for any transformer energizing under no-load is a worst case for inrush.
> this may not be the best transformer on which to base estimates.
I agree there are a wide variety of transformers, and certainly the winding resistance will play a bigger role in the smaller ones than it does on our GSU. I know that some of our HV switchyard circuit breakers have preinsertion resistors, we might have those on the HV breakers that energize the GSU but if so the resistors are only in the circuit for a very short time.
=====================================
(2B)+(2B)' ?
Inrush current (worst case) can be calculated and would be used in design spec for core bracing etc
There is no IEC60076 requirement to physically test it - there are many variables involved to get WC - it could be done if absolutely required, but unsure why anyone would want to
I.Eng MIMMM MIET MIPowerE AIOSH
There is no IEC60076 requirement to physically test it - there are many variables involved to get WC - it could be done if absolutely required, but unsure why anyone would want to
I.Eng MIMMM MIET MIPowerE AIOSH
sarawakmiri
Electrical
why specify transformer oa / fa / 55/ 65c
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