Apparent Power, KVA?
- Thread starter rattus
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Re: Apparent Power, KVA?
It?s more complicated than that. But here?s a general set of rules. It uses as an example, an electric generator. But the rules apply to other components also.
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Iron loss occurs because the windings are wound around an iron core. This helps to conduct the magnetic field from the primary windings to the secondary windings, and to concentrate its intensity. As the AC current flows back and forth, the iron is magnetized first one in direction, and then in the other. That is the reason for the loss of energy. But the magnetic field is created by the flow of current. Its strength is determined by the amount of amps and the number of times the wire is wound around the core. Voltage has nothing to do with this equation.
It?s more complicated than that. But here?s a general set of rules. It uses as an example, an electric generator. But the rules apply to other components also.
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- <font size="2" face="Verdana, Helvetica, sans-serif">The KVA rating is based on the ability of the generator windings and the surrounding components to dissipate the heat that will be created within the windings.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">The KW rating is based on the capabilities of the prime mover (e.g., the diesel engine, the gas turbine, the water turbine: whatever is providing the motive force to spin the generator).</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">The voltage rating is based on the ability of the insulation system to prevent leakage current from the windings to the outside world.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">The current rating is based on the cross-sectional area of the conductor used to construct the windings.</font>
Iron loss occurs because the windings are wound around an iron core. This helps to conduct the magnetic field from the primary windings to the secondary windings, and to concentrate its intensity. As the AC current flows back and forth, the iron is magnetized first one in direction, and then in the other. That is the reason for the loss of energy. But the magnetic field is created by the flow of current. Its strength is determined by the amount of amps and the number of times the wire is wound around the core. Voltage has nothing to do with this equation.
Re: Apparent Power, KVA?
Charlie, I agree with most of what you say, but I would argue that the hysteresis and eddy current losses in a xfrmr are determined by the primary voltage. More voltage, bigger loop, more current in the primary inductance, hence more iron loss. For that matter, the xfrmr equivalent circuit contains an inductor, Lp, and a resistor, Rc, in parallel. Lp is the primary inductance. Rc accounts for the core loss which is clearly a function of voltage.
If voltage did not enter into the formula, then we could merely use the current rating and dispense with the KVA rating.
[ December 12, 2004, 10:27 PM: Message edited by: rattus ]
Charlie, I agree with most of what you say, but I would argue that the hysteresis and eddy current losses in a xfrmr are determined by the primary voltage. More voltage, bigger loop, more current in the primary inductance, hence more iron loss. For that matter, the xfrmr equivalent circuit contains an inductor, Lp, and a resistor, Rc, in parallel. Lp is the primary inductance. Rc accounts for the core loss which is clearly a function of voltage.
If voltage did not enter into the formula, then we could merely use the current rating and dispense with the KVA rating.
[ December 12, 2004, 10:27 PM: Message edited by: rattus ]
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Re: Apparent Power, KVA?
Neither a voltage source nor a current source appears in the transformer model that you have described. Certainly, the higher the voltage, the higher the current, the higher the magnetic field, and therefore the higher the core losses. Just as certainly, you could conclude that voltage is what drives the core losses. But then, could you not say the same about copper losses? Is it not the same voltage that drives the same current, resulting in energy losses within the windings? Why would you describe the one as being a function of voltage and the other as being a function of current? Voltage drives both, but current is the more immediate cause of both.
Neither a voltage source nor a current source appears in the transformer model that you have described. Certainly, the higher the voltage, the higher the current, the higher the magnetic field, and therefore the higher the core losses. Just as certainly, you could conclude that voltage is what drives the core losses. But then, could you not say the same about copper losses? Is it not the same voltage that drives the same current, resulting in energy losses within the windings? Why would you describe the one as being a function of voltage and the other as being a function of current? Voltage drives both, but current is the more immediate cause of both.
Re: Apparent Power, KVA?
Charlie, from the model, it is clear that core loss increases as the square of the primary voltage, and copper loss increases as the square of the load current. Core loss is unaffected by the load current. Copper loss is affected by the primary voltage insofar as it increases load current, but we must still evaluate separate equations when computing the losses. You cannot do it with voltage or current alone.
You are right; there are no sources in the model. None are needed.
I am merely trying to understand why we use the KVA rating rather than just the current rating. I may have known at one time, but I am not nearly as smart now as I was at age 16.
Charlie, from the model, it is clear that core loss increases as the square of the primary voltage, and copper loss increases as the square of the load current. Core loss is unaffected by the load current. Copper loss is affected by the primary voltage insofar as it increases load current, but we must still evaluate separate equations when computing the losses. You cannot do it with voltage or current alone.
You are right; there are no sources in the model. None are needed.
I am merely trying to understand why we use the KVA rating rather than just the current rating. I may have known at one time, but I am not nearly as smart now as I was at age 16.
al hildenbrand
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Re: Apparent Power, KVA?
I'm confused by the transformer equivalent circuit model described above. Isn't the resistance in series with the inductance?
And more to the point of the opening question, the inductive impedance in the equivalent circuit predicts a phase difference between voltage and current which, in turn, makes the power in the circuit complex (in the vector sense). Most individuals that make use of the transformer (or motor) will only be aware of the real power available to them. The transformer or motor, however, will have to be constructed to withstand the stresses of the total complex power required to provide the real power component.
Rating a device that has a complex total impedance in KVA more accurately describes its capability.
I'm confused by the transformer equivalent circuit model described above. Isn't the resistance in series with the inductance?
And more to the point of the opening question, the inductive impedance in the equivalent circuit predicts a phase difference between voltage and current which, in turn, makes the power in the circuit complex (in the vector sense). Most individuals that make use of the transformer (or motor) will only be aware of the real power available to them. The transformer or motor, however, will have to be constructed to withstand the stresses of the total complex power required to provide the real power component.
Rating a device that has a complex total impedance in KVA more accurately describes its capability.
Re: Apparent Power, KVA?
Al, leakage inductance and resistance are in series with the windings of an IDEAL xfmr, but the primary inductance, Lp, and equivalent core loss resistance, Rc, connect across the primary of the IDEAL xfrmr.
This xfmr is usually replaced by the reflected impedances of the secondary.
So, an unloaded xfrmr still gets warm, right? That is core loss generated in Rc which I repeat is an equivalent, not actual resistance.
My text was first copyrighted in 1936 and printed in 1956, but I think little has changed in xfrmr theory since then.
I need to learn how to post schematics, or can only moderators do that?
I think though that I have answered my own question. The power companies like to think in power, that is in KW for which they get paid. By computing apparent power in KVA, they have an easy indication of what they are missing. That is to say that voltage, which is usually constant, is included as a matter of convenience.
Al, leakage inductance and resistance are in series with the windings of an IDEAL xfmr, but the primary inductance, Lp, and equivalent core loss resistance, Rc, connect across the primary of the IDEAL xfrmr.
This xfmr is usually replaced by the reflected impedances of the secondary.
So, an unloaded xfrmr still gets warm, right? That is core loss generated in Rc which I repeat is an equivalent, not actual resistance.
My text was first copyrighted in 1936 and printed in 1956, but I think little has changed in xfrmr theory since then.
I need to learn how to post schematics, or can only moderators do that?
I think though that I have answered my own question. The power companies like to think in power, that is in KW for which they get paid. By computing apparent power in KVA, they have an easy indication of what they are missing. That is to say that voltage, which is usually constant, is included as a matter of convenience.
al hildenbrand
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- Electrical Contractor, Electrical Consultant, Electrical Engineer
Re: Apparent Power, KVA?
So, the transformer exciting current is what your KVA rating question is about?
Is your "Rc" a complex impedance, or are you thinking of it as only real?
Seems to me that, for No-Load, the Lp and Rc are exactly the vector impedance Z sub e. When energized by a sinusoidal source, Z sub e lets a non sinusoidal current (the exciting current) flow that, in turn, produces a sinusoidal flux wave in the transformer core, and, in turn, produces a sinusoidal back emf in the primary.
[ December 13, 2004, 05:31 PM: Message edited by: al hildenbrand ]
OK, that's a more detailed equivalent circuit. And, for understanding the behavior of a transformer with an open secondary, it has some value, but quickly falls to unimportant levels under load.
So, the transformer exciting current is what your KVA rating question is about?
Is your "Rc" a complex impedance, or are you thinking of it as only real?
Seems to me that, for No-Load, the Lp and Rc are exactly the vector impedance Z sub e. When energized by a sinusoidal source, Z sub e lets a non sinusoidal current (the exciting current) flow that, in turn, produces a sinusoidal flux wave in the transformer core, and, in turn, produces a sinusoidal back emf in the primary.
[ December 13, 2004, 05:31 PM: Message edited by: al hildenbrand ]
Re: Apparent Power, KVA?
Al, you are making this too hard. The question was, "Why use the KVA rating instead of the current rating?" This has evolved into an interesting dialogue about those magical things we call transformers.
The "exact" equivalent circuit is as follows:
Imagine a "T" network with primary resistance and leakage reactance drawn horizontally.
Then draw Rc and Lp in parallel to the return line. Rc is a pure resistance, but it is an electrical analog for the core loss, not a real resistor.
Now draw the reflected secondary resistance and leakage reactance horizontally. Remember too tht the reflected impedances are multiplied by the square of the turns ratio.
Now draw the reflected load impedance vertically to the return line.
This equivalent circuit takes into account all the xfrmr impedances plus the load impedance.
None of these circuits is perfect, but this one should work well from no load to full load.
This equivalent is from Alternating Current Machines, Puchstein, Lloyd, and Conrad, Wiley, 1956.
Now let me reiterate some facts about xfmrs--to first order that is:
1. Primary voltage across Lp, the primary inductance, creates the exciting current which creates the magnetic flux which is NOT affected by load current.
2. The primary voltage applied across Rc determines the core loss.
3. Load currents through the primary and secondary resistances create copper loss.
[ December 13, 2004, 07:34 PM: Message edited by: rattus ]
Al, you are making this too hard. The question was, "Why use the KVA rating instead of the current rating?" This has evolved into an interesting dialogue about those magical things we call transformers.
The "exact" equivalent circuit is as follows:
Imagine a "T" network with primary resistance and leakage reactance drawn horizontally.
Then draw Rc and Lp in parallel to the return line. Rc is a pure resistance, but it is an electrical analog for the core loss, not a real resistor.
Now draw the reflected secondary resistance and leakage reactance horizontally. Remember too tht the reflected impedances are multiplied by the square of the turns ratio.
Now draw the reflected load impedance vertically to the return line.
This equivalent circuit takes into account all the xfrmr impedances plus the load impedance.
None of these circuits is perfect, but this one should work well from no load to full load.
This equivalent is from Alternating Current Machines, Puchstein, Lloyd, and Conrad, Wiley, 1956.
Now let me reiterate some facts about xfmrs--to first order that is:
1. Primary voltage across Lp, the primary inductance, creates the exciting current which creates the magnetic flux which is NOT affected by load current.
2. The primary voltage applied across Rc determines the core loss.
3. Load currents through the primary and secondary resistances create copper loss.
[ December 13, 2004, 07:34 PM: Message edited by: rattus ]
Re: Apparent Power, KVA?
Hello to everyone:
If we talk about KVA do we not also have to mention the PF after all if there is a purely resistive element the KVA is equal to the KW.
Afterall a power factor of unity there is no quadrature power involved, with transformers and motors this can not be achieved.
So if we define KVa we must consider a reactive power component. Otherwise it would be pointless to use KVA and KVa/Kw would be interchangable.
The average power delivered to any device which is purely reactive(i.e., contains no resistors)must be zero. This of course is a direct result of the 90 degree phase difference whichmust exist between current and voltage; Hence, Cos(theta - Phi)= cos +/- 90 = 0.
The average power delivered to any newwork composed entirely of ideal inductors and capacitors is zero; The instantaneous power is zero only at specific instants. Thus power flows into the network for a part of the cycle and out of the network during another portion of the cycle and no power is lost.
So if you are going to consider KVA then the whole power triangle must be considered.
Gary
Hello to everyone:
If we talk about KVA do we not also have to mention the PF after all if there is a purely resistive element the KVA is equal to the KW.
Afterall a power factor of unity there is no quadrature power involved, with transformers and motors this can not be achieved.
So if we define KVa we must consider a reactive power component. Otherwise it would be pointless to use KVA and KVa/Kw would be interchangable.
The average power delivered to any device which is purely reactive(i.e., contains no resistors)must be zero. This of course is a direct result of the 90 degree phase difference whichmust exist between current and voltage; Hence, Cos(theta - Phi)= cos +/- 90 = 0.
The average power delivered to any newwork composed entirely of ideal inductors and capacitors is zero; The instantaneous power is zero only at specific instants. Thus power flows into the network for a part of the cycle and out of the network during another portion of the cycle and no power is lost.
So if you are going to consider KVA then the whole power triangle must be considered.
Gary
Re: Apparent Power, KVA?
The loses that occur in real transformers have to be accounted for in any accurate model of transformer behavior. The major items to consider are:
1.) Copper (I^2*R) losses Copper loses are the resistive heating losses in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings.
2.) Eddy current losses, are resistive heating loses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer.
3.) Hysteresis losses. these are associated with the rearrangement of the magnetic domains in the core during each half-cycle. These are complex, nonlinear functionss of the voltage applied to the transformer.
and
4.) leakage flux, the fluxes that escape the primary and secondary cores and pass through only one of the transformer windings are leakage fluxes, these escaped fluxes produce a self-inductance that must be accounted for.
It must be remember the equivalent circuits used for a transformer are a mathematical model, but the impedance used for the primary and the model used for the secondary must be complex otherwise the model will be useless.
Gary
The loses that occur in real transformers have to be accounted for in any accurate model of transformer behavior. The major items to consider are:
1.) Copper (I^2*R) losses Copper loses are the resistive heating losses in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings.
2.) Eddy current losses, are resistive heating loses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer.
3.) Hysteresis losses. these are associated with the rearrangement of the magnetic domains in the core during each half-cycle. These are complex, nonlinear functionss of the voltage applied to the transformer.
and
4.) leakage flux, the fluxes that escape the primary and secondary cores and pass through only one of the transformer windings are leakage fluxes, these escaped fluxes produce a self-inductance that must be accounted for.
It must be remember the equivalent circuits used for a transformer are a mathematical model, but the impedance used for the primary and the model used for the secondary must be complex otherwise the model will be useless.
Gary
al hildenbrand
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Re: Apparent Power, KVA?
I wonder if we aren't looking at the leaves when more meaning comes from the whole tree.
I think the schematic of a transformer, a schematic that is an equivalent circuit of a single electrical impedance, i.e., the transformer, . . . that the schematic is a "leaf".
Looking at the current alone, one does not see the phase difference from the voltage.
Since power is the product of voltage and current, all the necessary information is included to express a meaningful "rating". In my opinion, that is why we rate transformers in KVA rather than current.
I wonder if we aren't looking at the leaves when more meaning comes from the whole tree.
I think the schematic of a transformer, a schematic that is an equivalent circuit of a single electrical impedance, i.e., the transformer, . . . that the schematic is a "leaf".
It occurs to me that the "tree" is that when a transformer is in operation, the voltage and the current are no longer "in sync" with each other. Whether the transformer is under load, has an open or shorted secondary, the "resistance" of the transformer is a "complex impedance" that forces the voltage and current out of sync with each other.
Looking at the current alone, one does not see the phase difference from the voltage.
Since power is the product of voltage and current, all the necessary information is included to express a meaningful "rating". In my opinion, that is why we rate transformers in KVA rather than current.
Re: Apparent Power, KVA?
Al, the "exact" equivalent circuit comprises 6 or more circuit elements in a "T" configuration and will predict the power losses, voltage drops, phase changes, etc. It can be simplified a bit to use only 4 circuit elements, but this is more of an approximation. I will post the diagram when I figure out how to do it.
Al, the "exact" equivalent circuit comprises 6 or more circuit elements in a "T" configuration and will predict the power losses, voltage drops, phase changes, etc. It can be simplified a bit to use only 4 circuit elements, but this is more of an approximation. I will post the diagram when I figure out how to do it.
al hildenbrand
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Re: Apparent Power, KVA?
Rattus,
I have a copy of the diagram here as well. But the elements of it are an engineering fiction. . .they are the leaves.
Rattus,
I have a copy of the diagram here as well. But the elements of it are an engineering fiction. . .they are the leaves.
Re: Apparent Power, KVA?
Al, why do you say the equivalent circuit is engineering fiction? Like most equivalent circuits, it is not perfect because the iron core transformer is not perfectly linear, and resistances change with temperature. Nowadays though, there must be computer models available that take the non-linearities into account. There are some fantastic programs and models available for electronic circuits.
If it is useless, I wasted a lot of time drawing vector diagrams, etc., back in engine school.
Al, why do you say the equivalent circuit is engineering fiction? Like most equivalent circuits, it is not perfect because the iron core transformer is not perfectly linear, and resistances change with temperature. Nowadays though, there must be computer models available that take the non-linearities into account. There are some fantastic programs and models available for electronic circuits.
If it is useless, I wasted a lot of time drawing vector diagrams, etc., back in engine school.
al hildenbrand
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Re: Apparent Power, KVA?
You did not hear from my words that the equivalent circuit is "useless", rather that it is a fiction, something that is made up that models the behaviour of the real thing within acceptable error ranges.
We also spent a lot of time in school with logarithms to turn multiplication and division into addition and subtraction. Do you still look up logs to avoid the more complex math of multiplication and division. . .?
You did not hear from my words that the equivalent circuit is "useless", rather that it is a fiction, something that is made up that models the behaviour of the real thing within acceptable error ranges.
We also spent a lot of time in school with logarithms to turn multiplication and division into addition and subtraction. Do you still look up logs to avoid the more complex math of multiplication and division. . .?
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Re: Apparent Power, KVA?
I can't speak for rattus, but I do not. Instead, I let my slide rule add the logarithms for me.
Re: Apparent Power, KVA?
Al, you can use the term "fiction" if you wish, but this circuit is based on actual physical properties. The equivalent core loss resistor is the only element derived from the "black box" approach.
I would be interested to hear if there are more sophisticated models today. That is, a model which accounts for the non-linearities, temperature, etc.
I don't see what logarithms have to do with it though. They are still very important in the audio and communications industries, although we don't need the tables anymore, thanks to the wondrous computers and calculators we now use.
Al, you can use the term "fiction" if you wish, but this circuit is based on actual physical properties. The equivalent core loss resistor is the only element derived from the "black box" approach.
I would be interested to hear if there are more sophisticated models today. That is, a model which accounts for the non-linearities, temperature, etc.
I don't see what logarithms have to do with it though. They are still very important in the audio and communications industries, although we don't need the tables anymore, thanks to the wondrous computers and calculators we now use.
Re: Apparent Power, KVA?
Rattus
Trig and log tables went out with the slide-rule, ie. obsolete,.
And as far as 'fiction' for the transformer circuit goes, well maybe its just these MUCH older guys see things differently than we do afterall things move much faster now than they did 45 years ago.(Just kidding)
Maybe 'fiction' was a bad choice of word, or maybe just a bad hairday..
Gary
Rattus
Trig and log tables went out with the slide-rule, ie. obsolete,.
And as far as 'fiction' for the transformer circuit goes, well maybe its just these MUCH older guys see things differently than we do afterall things move much faster now than they did 45 years ago.(Just kidding)
Maybe 'fiction' was a bad choice of word, or maybe just a bad hairday..
Gary
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