mbrooke
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What is the power factor on the secondary of a short circuited POCO transformer? Say I jumped X1 to X0 on a single or 3 phase unit. What would my meter read PF wise?
Transformer Short Circuit Power Factor
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mbrooke
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gar
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mbrooke:
You have not clearly defined your question.
Define the two points in your system that are to define where the load starts of which you want to measure the power factor.
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mbrooke:
You have not clearly defined your question.
Define the two points in your system that are to define where the load starts of which you want to measure the power factor.
.
mbrooke
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gar
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mbrooke:
The power factor of that load is determined by the resistance and inductance of that short piece of wire shorting the transformer secondary. My guess is that at 60 Hz it is close to resistive, and thus PF is close to 1.
If you look at the input to the transformer with the secondary shorted, then inductive reactance may be moderately greater than the equivalent series resistance.
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mbrooke:
The power factor of that load is determined by the resistance and inductance of that short piece of wire shorting the transformer secondary. My guess is that at 60 Hz it is close to resistive, and thus PF is close to 1.
If you look at the input to the transformer with the secondary shorted, then inductive reactance may be moderately greater than the equivalent series resistance.
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mbrooke
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mbrooke
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gar
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On a Signal Transformer model A81-175-16, rectifier supply type of transformer, with the secondary shorted I get the input current lagging by about 1 mS at 60 Hz. This is not a lot of leakage inductance, and not a low power factor.
In a power supply you want the leakage inductance low so as to have good output voltage regulation. If the transformer was wound on a core with no air gap the relative leakage inductance would be even lower.
Power company transformers may be designed with a relatively higher leakage inductance than rectifier power supply transformers to reduce short circuit current.
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On a Signal Transformer model A81-175-16, rectifier supply type of transformer, with the secondary shorted I get the input current lagging by about 1 mS at 60 Hz. This is not a lot of leakage inductance, and not a low power factor.
In a power supply you want the leakage inductance low so as to have good output voltage regulation. If the transformer was wound on a core with no air gap the relative leakage inductance would be even lower.
Power company transformers may be designed with a relatively higher leakage inductance than rectifier power supply transformers to reduce short circuit current.
.
gar
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mbrooke:
In your post #8. Yes the transformer is an inductor.
But in your unclear post #1, as clarified in your post #5, you were talking about the power factor of the short circuit load on the transformer. That power factor has nothing to do with the transformer. It only has to do with the actual load.
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mbrooke:
In your post #8. Yes the transformer is an inductor.
But in your unclear post #1, as clarified in your post #5, you were talking about the power factor of the short circuit load on the transformer. That power factor has nothing to do with the transformer. It only has to do with the actual load.
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mbrooke
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I'd imagine the volts and amps at the load would drop to zero, hence why I thought the trafo's PF being only concerned with its own widnings was obvious. My mistake in explaining/describing it the way I did.
Anyway, do you know what the PF is typically?
I ask because it appears like the PF of the secondary changes short circuit calculations...
gar
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mbrooke:
By the way this thread has flowed I do not believe you have any electrical engineering degree, and have not really had any formal education in basic electrical circuit theory. My following comments are based on that assumption.
A typical AC power transformer operating at 60 Hz is fundamentally made up from two isolated coils of wire (meaning no significant conductive path from primary to secondary or to the magnetic core), and a magnetic core that electrically couples the two windings together by magnetic linkage. For a power transformer it is assumed that there is rather good magnetic coupling from primary to secondary.
Each winding has a series resistance, and s series inductance (leakage inductance). Leakage inductance is an inductive component in series with the resistance that results from incomplete coupling of the primary with the secondary.
There are various ways to create an equivalent circuit for a transformer. None exactly represent a real transformer, but most do an adequate job. A typical way is to have a series resistance, a series inductance, going to the input of an ideal transformer. Then there is a shunt inductance across the ideal transformer primary. This shunt inductance is typically ignored. In other words not in the equivalent circuit.
If you short the secondary of the ideal transformer the equivalent circuit becomes a series resistance with a small inductance. The turns ratio of the ideal transformer has no effect on the power factor of the system as viewed at the transformer input terminals.
If you view your circuit from the secondary, load side, of the transformer, then the short circuit current changes as you change the turns ratio, but the transformer input power factor does not change. And there is no significance to describing a power factor of the load, because it is assumed to be a short circuit.
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mbrooke:
By the way this thread has flowed I do not believe you have any electrical engineering degree, and have not really had any formal education in basic electrical circuit theory. My following comments are based on that assumption.
A typical AC power transformer operating at 60 Hz is fundamentally made up from two isolated coils of wire (meaning no significant conductive path from primary to secondary or to the magnetic core), and a magnetic core that electrically couples the two windings together by magnetic linkage. For a power transformer it is assumed that there is rather good magnetic coupling from primary to secondary.
Each winding has a series resistance, and s series inductance (leakage inductance). Leakage inductance is an inductive component in series with the resistance that results from incomplete coupling of the primary with the secondary.
There are various ways to create an equivalent circuit for a transformer. None exactly represent a real transformer, but most do an adequate job. A typical way is to have a series resistance, a series inductance, going to the input of an ideal transformer. Then there is a shunt inductance across the ideal transformer primary. This shunt inductance is typically ignored. In other words not in the equivalent circuit.
If you short the secondary of the ideal transformer the equivalent circuit becomes a series resistance with a small inductance. The turns ratio of the ideal transformer has no effect on the power factor of the system as viewed at the transformer input terminals.
If you view your circuit from the secondary, load side, of the transformer, then the short circuit current changes as you change the turns ratio, but the transformer input power factor does not change. And there is no significance to describing a power factor of the load, because it is assumed to be a short circuit.
.
gar
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I pulled out an old textbook and found the following information on a 10 kVA 2400 V primary transformer --- R = 10.4 ohms. and Xl = 14.6 ohms. So this is a lower power factor than my rectifier type transformer. I reflected the secondary impedance to the primary to determine what one sees at the primary.
Power factor does not change much as you change turns ratio. And you do not normally consider a different power factor for a secondary vs primary.
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I pulled out an old textbook and found the following information on a 10 kVA 2400 V primary transformer --- R = 10.4 ohms. and Xl = 14.6 ohms. So this is a lower power factor than my rectifier type transformer. I reflected the secondary impedance to the primary to determine what one sees at the primary.
Power factor does not change much as you change turns ratio. And you do not normally consider a different power factor for a secondary vs primary.
.
mbrooke
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This is really good! As I think it, this results in voltage dropping on the secondary terminals as load increases, especially above its kva and/or that external load is inductive. However, please do not confuse this last statement as having anything to do with my question. At least not yet.
How does this compare with putting both on the output and assuming an infinite source? Or would my simplification be just to far off?
Right, but as the short circuit is happening, won't the current flowing between X1 and X0 have a lagging component ie if voltage from X1 and X0 is measured (assume 1 volt because the jumper will not be an ideal zero ohms) on an oscilloscope and current is measured via an amp clamp and overlayed said current would lag the voltage? Kind of like this?
mbrooke
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In my guesstimations, POCO transformers have a very low power factor.
My concerns rests with the secondary, in that I want to turn (model) the pole pig or pad mount as a series inductor and series resistor fed via in infinite source.
Given your values- can I determine short circuit current on the secondary? I know it gets complicated- the fault current doubles when X1 to X0 is shorted vs X1 to X2 on a 120/240 transformer- but my concern rests for line to ground.
gar
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mbrooke:
I think it is going to be hard for you to get a good grasp on electrical circuit theory on this forum. It is extremely inefficient here. You need to see if you can take some related college courses at a good school.
I will try to answer some of your questions.
Theoretically one can assume an ideal transformer where input, and output are related precisely by turns ratio, and there is no power loss. Then to this theoretical transformer you can place a series impedance on the input, or on the output, or part of that impedance on both sides of the transformer. In addition to this there is some shunt impedance somewhere or ignored.
The simplest circuit is to put all the series impedance on the input side and ignore any shunt impedance (by ignore means an open circuit).
So to further simplify the circuit eliminate the ideal transformer. This results in a series circuit of a resistance, an inductor, and a load.
The load can be a simple resistor, and thus the load power factor is unity. But the power factor at the input to the transformer is not 1.
If the load is zero impedance, then it has an undefined power factor. And the power factor at the transformer input is the power factor of the series resistance and inductance. Had there been a resistive load on the transformer, then the input power factor to the transformer would have moved to some degree toward 1.
Now consider a piece of wire Pi*6 ft long. No matter how I shape that wire its DC resistance remains a constant. However, its inductance and therefore its inductive reactance is very dependent on how that wire is shaped. If the wire is shaped into a circle its inductance is high compared to that same wire shaped into a very tight hairpin loop. So the power factor of the shorting wire is very dependent on how the wire is looped. Low inductance wire wound resistors are really made of a hairpin loop instead of just a wire wound around a core all in one direction.
If we make the short on the secondary of a transformer a very low impedance compared to the transformer internal impedance, then we really don't care what its power factor is because it won't affect the transformer input power factor very much.
If a transformer has a non-unity turns ratio, and we short the secondary, then the secondary short circuit current as viewed at the primary side will be different than as viewed on the secondary side, but the transformer power factor remains unchanged. Theoretically we study this, but can not perform an actual experiment. We can measure secondary current zero crossing, and compare with the primary voltage zero crossing. From this we can calculate power factor. But the results are almost identical to --- if we measure the primary current zero crossing. We have no way to get at and measure the equivalent secondary voltage zero crossing.
This may or may not answer some of your questions.
Secondary short circuit current can be much higher than primary because current is related by turns ratio.
If one assumes that a transformer is working in a linear range, then there is no step change in characteristics based on passing full load rating.
.
mbrooke:
I think it is going to be hard for you to get a good grasp on electrical circuit theory on this forum. It is extremely inefficient here. You need to see if you can take some related college courses at a good school.
I will try to answer some of your questions.
Theoretically one can assume an ideal transformer where input, and output are related precisely by turns ratio, and there is no power loss. Then to this theoretical transformer you can place a series impedance on the input, or on the output, or part of that impedance on both sides of the transformer. In addition to this there is some shunt impedance somewhere or ignored.
The simplest circuit is to put all the series impedance on the input side and ignore any shunt impedance (by ignore means an open circuit).
So to further simplify the circuit eliminate the ideal transformer. This results in a series circuit of a resistance, an inductor, and a load.
The load can be a simple resistor, and thus the load power factor is unity. But the power factor at the input to the transformer is not 1.
If the load is zero impedance, then it has an undefined power factor. And the power factor at the transformer input is the power factor of the series resistance and inductance. Had there been a resistive load on the transformer, then the input power factor to the transformer would have moved to some degree toward 1.
Now consider a piece of wire Pi*6 ft long. No matter how I shape that wire its DC resistance remains a constant. However, its inductance and therefore its inductive reactance is very dependent on how that wire is shaped. If the wire is shaped into a circle its inductance is high compared to that same wire shaped into a very tight hairpin loop. So the power factor of the shorting wire is very dependent on how the wire is looped. Low inductance wire wound resistors are really made of a hairpin loop instead of just a wire wound around a core all in one direction.
If we make the short on the secondary of a transformer a very low impedance compared to the transformer internal impedance, then we really don't care what its power factor is because it won't affect the transformer input power factor very much.
If a transformer has a non-unity turns ratio, and we short the secondary, then the secondary short circuit current as viewed at the primary side will be different than as viewed on the secondary side, but the transformer power factor remains unchanged. Theoretically we study this, but can not perform an actual experiment. We can measure secondary current zero crossing, and compare with the primary voltage zero crossing. From this we can calculate power factor. But the results are almost identical to --- if we measure the primary current zero crossing. We have no way to get at and measure the equivalent secondary voltage zero crossing.
This may or may not answer some of your questions.
Secondary short circuit current can be much higher than primary because current is related by turns ratio.
If one assumes that a transformer is working in a linear range, then there is no step change in characteristics based on passing full load rating.
.
mbrooke
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I can tell the whole concept of Zs=Ze+(R1+R2) is making you rather flustered, because we both know colleges in North America don't teach that
Understood.
Understood.
Right, however, I do not care about the input. I only care about the secondary current and voltage.
The way I see it is if the secondary short circuit consists of X and R, that the fault current will be lower due to the trafo's X vs if the transformer was pure R based upon secondary fault current determined by Z where . Am I correct? I will make a picture.
Understood, and thank you!
gar
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mbrooke:
You said
"Right, however, I do not care about the input. I only care about the secondary current and voltage.
The way I see it is if the secondary short circuit consists of X and R, that the fault current will be lower due to the trafo's X vs if the transformer was pure R based upon secondary fault current determined by Z where . Am I correct? I will make a picture."
I believe you are interested in determining output short circuit current.
Obviously the output voltage is zero.
If the output is loaded with a short circuit, then we don't care much about what is its actual impedance, we just assume the short is zero impedance. Thus. the short's L/R is no importance.
The input current and voltage to the transformer will tell you what the secondary short circuit current is.
From a paper work calculation perspective you can only determine output short circuit by analysis of the transformer input. You can consider power factor at both the primary and secondary sides, it is the same, except you have no way to measure it on the secondary side.
An equivalent circuit for the shorted secondary transformer is a series resistance and inductance on the primary side of an ideal transformer. To get the secondary short circuit current by calculation you determine the input current, and then multiply that by the ideal transformer turns ratio.
If the input current for a shorted secondary is 10 A and the turns ratio is 10 to 1, then the calculated output current is 100 A.
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mbrooke:
You said
"Right, however, I do not care about the input. I only care about the secondary current and voltage.
The way I see it is if the secondary short circuit consists of X and R, that the fault current will be lower due to the trafo's X vs if the transformer was pure R based upon secondary fault current determined by Z where . Am I correct? I will make a picture."
I believe you are interested in determining output short circuit current.
Obviously the output voltage is zero.
If the output is loaded with a short circuit, then we don't care much about what is its actual impedance, we just assume the short is zero impedance. Thus. the short's L/R is no importance.
The input current and voltage to the transformer will tell you what the secondary short circuit current is.
From a paper work calculation perspective you can only determine output short circuit by analysis of the transformer input. You can consider power factor at both the primary and secondary sides, it is the same, except you have no way to measure it on the secondary side.
An equivalent circuit for the shorted secondary transformer is a series resistance and inductance on the primary side of an ideal transformer. To get the secondary short circuit current by calculation you determine the input current, and then multiply that by the ideal transformer turns ratio.
If the input current for a shorted secondary is 10 A and the turns ratio is 10 to 1, then the calculated output current is 100 A.
.
mbrooke
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Alright, so, here is what I have in mind:
The graphic doesn't include the transformer's current limiting properties, which would be realistic if this 12AWG wire was coming off a 5000 amp service where 300 amps would just be another motor starting.
However, a fault at the main switchboard would have to include the transformer's impedance?
The scenario (albeit somewhat unrealistic for the real world, however frankensteined for discussion) 300kva 277/480 volt Y bank compromised of 3 100kva 1.9%Z units:
1000 foot drop giving a ground fault loop like this:
There are two ways that I'm thinking about doing this- Option 1 and Option 2.
Option 2 involves finding the full load kva then dividing by the % Z to obtain a secondary short circuit current of 19,000 amps. Using ohms law we obtain 0.01458 ohms. This value is then converted into a resistor in front of an infinite source.
Using:
I get:
Z= 0.0047032164 + 0.006084
Z= 0.0107872164
Z= 0.10 ohms
or 2,770 amps of fault current
However, in the real world the trafo would be a combination of R and X. Would the X change this value? Since its adding a lagging current component?
The graphic doesn't include the transformer's current limiting properties, which would be realistic if this 12AWG wire was coming off a 5000 amp service where 300 amps would just be another motor starting.
However, a fault at the main switchboard would have to include the transformer's impedance?
The scenario (albeit somewhat unrealistic for the real world, however frankensteined for discussion) 300kva 277/480 volt Y bank compromised of 3 100kva 1.9%Z units:
1000 foot drop giving a ground fault loop like this:
There are two ways that I'm thinking about doing this- Option 1 and Option 2.
Option 2 involves finding the full load kva then dividing by the % Z to obtain a secondary short circuit current of 19,000 amps. Using ohms law we obtain 0.01458 ohms. This value is then converted into a resistor in front of an infinite source.
Using:
I get:
Z= 0.0047032164 + 0.006084
Z= 0.0107872164
Z= 0.10 ohms
or 2,770 amps of fault current
However, in the real world the trafo would be a combination of R and X. Would the X change this value? Since its adding a lagging current component?
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