Can an iron core both generate and absorb vars?

[mbrooke]

Alright, say you had two transformers in parallel on the primary and secondary. All else is equal- except one has its taps set 2.5% higher. Current will circulate in the secondary.

My question is, why is the power factor lower with this circulating current? That would mean one transformer is generating vars while the other is absorbing them. But both of these are iron cores, not one core and one capacitor.

 

[paulengr]

1. Capacitors and Inductors draw current 90 degrees to the voltage. Pure vars. Power factor is zero. The math says the magnetic or electrostatic field builds and collapses which gets into the physics at the time domain but basic phasor math is that X sub L is 100% reactive.
2. No work is being done. So we know is is all vars, not kilowatts.

Not sure why this is confusing at all. V = L di/dt. Integrate and we get that current is the integral of voltage divided by the inductance. If V is a sine, a cosine pops out.

 

[mbrooke]

1. Capacitors and Inductors draw current 90 degrees to the voltage. Pure vars. Power factor is zero. The math says the magnetic or electrostatic field builds and collapses which gets into the physics at the time domain but basic phasor math is that X sub L is 100% reactive.
2. No work is being done. So we know is is all vars, not kilowatts.

Not sure why this is confusing at all. V = L di/dt. Integrate and we get that current is the integral of voltage divided by the inductance. If V is a sine, a cosine pops out.

I guess I should ask: why is the power factor less than 1 when transformers are not in the same tap or generators are not equally excited.

 

[GoldDigger]

I guess I should ask: why is the power factor less than 1 when transformers are not in the same tap or generators are not equally excited.

Can you give a source for this assertion? If the two transformers have the same %Z then the circulating currents will be essentially resistive when there is no load on the secondary.
If you are looking at the power factor on the input side, unloaded with matching taps, the power factor will be very low because the excitation current will be dominant. If you then mismatch the taps the PF will increase, although the additional resistive power will not be delivered to a load.
Adding load should increase the power factor of the input power, but the efficiency will remain low because of the circulating current.

 

[Hv&Lv]

All else is equal- except one has its taps set 2.5% higher. Current will circulate in the secondary.

Primary also. When it comes back on little iA, what happens to big IA?
Why would you want to even try this? The loading will be uneven, the no load losses will be ridiculous, and your setting someone up to get hurt.

 

[GoldDigger]

See http://www.facilitiesnet.com/whitepapers/pdfs/SchneiderElectric_111711.pdf for a rigorous analysis of all but the case proposed.

 

[GoldDigger]

Primary also. When it comes back on little iA, what happens to big IA?
Why would you want to even try this? The loading will be uneven, the no load losses will be ridiculous, and your setting someone up to get hurt.

The Schneider paper suggests one case for doing this is a situation where the two transformers are normally operating independently supplying different loads and hence on different taps, but a temporary paralleling is necessary for some operational reason.

 

[Hv&Lv]

Alright, say you had two transformers in parallel on the primary and secondary. All else is equal- except one has its taps set 2.5% higher. Current will circulate in the secondary.

My question is, why is the power factor lower with this circulating current? That would mean one transformer is generating vars while the other is absorbing them. But both of these are iron cores, not one core and one capacitor.

Wait. Answered too early in the morning...
Just really noticed 2.5%...
That’s not a big deal...

 

[winnie]

Terminology point. As paulengr says, both pure capacitors and pure inductors draw current 90 degrees out of phase with the supply voltage, and both can be described as consuming vars.

Implied but not explicitly stated is that capacitors draw current which leads the voltage by 90 degrees, and inductors draw current which lags behind the voltage by 90 degrees. This is why capacitive vars can cancel out inductive vars.

Because most loads are resistive or somewhat inductive in nature, practical grid loading consumes inductive vars. Because the utility has to add capacitive loads to cancel the inductive vars, we say that the capacitors are 'generating' vars, but really they are loads, consuming capacitive vars.

Back to the original question: if you have two transformers which as wired have slightly different inductance, and current is circulating between them, and that circulating current has a power factor, then clearly one transformer has to be supplying vars to the other. Similarly one transformer is supplying power to the circulating loop and one is absorbing power.

Think of it this way: you have a closed loop with two coils in series. Current is circulating in that entire loop, and current must be the same everywhere in that loop. The two coils are each generating an end that depends on the primary tapping and are otherwise similar. The phase relationship between the current and the emf produced by each coil will be almost inverted between the two coils.

Jon

 

[mbrooke]

Primary also. When it comes back on little iA, what happens to big IA?
Why would you want to even try this? The loading will be uneven, the no load losses will be ridiculous, and your setting someone up to get hurt.

Theoretical (knowledge) discussion.

 

[mbrooke]

Wait. Answered too early in the morning...
Just really noticed 2.5%...
That’s not a big deal...

Ok, try 5 taps away on one transformer boosting 5 taps on another bucking... or one on neutral and and one 10 taps heck 16 taps away.

 

[mbrooke]

Back to the original question: if you have two transformers which as wired have slightly different inductance, and current is circulating between them, and that circulating current has a power factor, then clearly one transformer has to be supplying vars to the other. Similarly one transformer is supplying power to the circulating loop and one is absorbing power.



Jon

Alright- I can picture on trafo supplying VARS, but how does the others, an iron core, absorb them? This is what I'm having trouble putting into words or picturing.

 

[Hv&Lv]

Ok, try 5 taps away on one transformer boosting 5 taps on another bucking... or one on neutral and and one 10 taps heck 16 taps away.

That’s basically bypassing a regulator that’s off neutral.
Look here, play around awhile

EveryCircuit: Animated interactive circuit simulator

Interactive real-time circuit simulation and animated visualization inspired students and engineers to design 2.6 million circuits online and in mobile app.

everycircuit.com

 

[winnie]

Alright- I can picture on trafo supplying VARS, but how does the others, an iron core, absorb them? This is what I'm having trouble putting into words or picturing.

There is a very small net voltage in the circuit, from the difference in tap settings. This small voltage causes current to circulate.

Assume that the circulating current lags the _net_ voltage, eg. that the entire circuit is slightly inductive.

Now look at the individual coils. The two transformers are in phase, so the net voltage is in phase with the higher voltage coil, and 180 degrees out of phase with the lower voltage coil (because in the circulating circuit that we are considering, you hit one of the transformer coils going in the backward direction).

Say the circulating current lags the net voltage by 30 degrees. This means that the circulating current lags the higher voltage by 30 degrees. But the lower voltage coil is 180 degrees out of phase, so the same circulating current has to _lead_ the lower voltage by 150 degrees.

-Jon

 

[mbrooke]

That’s basically bypassing a regulator that’s off neutral.
Look here, play around awhile

EveryCircuit: Animated interactive circuit simulator

Interactive real-time circuit simulation and animated visualization inspired students and engineers to design 2.6 million circuits online and in mobile app.

everycircuit.com

Ok- picture two transformers running in parallel not on the same tapes. Why does said circulating current has a power factor of less than one?

 

[mbrooke]

There is a very small net voltage in the circuit, from the difference in tap settings. This small voltage causes current to circulate.

Assume that the circulating current lags the _net_ voltage, eg. that the entire circuit is slightly inductive.

Now look at the individual coils. The two transformers are in phase, so the net voltage is in phase with the higher voltage coil, and 180 degrees out of phase with the lower voltage coil (because in the circulating circuit that we are considering, you hit one of the transformer coils going in the backward direction).

Say the circulating current lags the net voltage by 30 degrees. This means that the circulating current lags the higher voltage by 30 degrees. But the lower voltage coil is 180 degrees out of phase, so the same circulating current has to _lead_ the lower voltage by 150 degrees.

-Jon

So what happens to the current entering the transformer at the lower taps?

 

[winnie]

It just circulates. The current flow of all components in series in a simple circuit is always the same.

Both coils apply voltage to the circuit. One coil applies say 120V, the other coil applies 117V. The net 3V is what causes the circulating current.

One coil is driving the current, the other coil is 'resisting' the current, and the actual current flow is the net result of the two coils.

-Jon

 

[GoldDigger]

And the circulating current in the case of mismatched taps will be close to 100% resistive. So one transformer is sourcing kW and the other is sinking it, both in the form of I**2R heating and in return of power through the primary.

 

[mbrooke]

And the circulating current in the case of mismatched taps will be close to 100% resistive. So one transformer is sourcing kW and the other is sinking it, both in the form of I**2R heating and in return of power through the primary.

Your explanation make more sense. But I've always been told it has a power factor.

 

[winnie]

Your explanation make more sense. But I've always been told it has a power factor.

I don't know if the circulating current has a power factor or not, but if you draw out the vector diagram it is clear that the power factor is 'opposite' in the two coils. If the circulating current is 0° (purely resistive) at one coil, then it is 180° (pure resistive power return) in the other coil. If the circulating current is say 10° lagging in one coil (mostly resistive with a bit of inductive loading) then it will be 190° lagging in the other coil, which is the same as 170° _leading_, in other words mostly resistive power return with a smidge of 'capacitive' loading, even though there is no capacitor present.

You can see the same sort of funny power factors if you search for the 'Oregon fudge factor' in previous posts. This has to do with the apparent power factor seen when a single phase resistive load is connected line-line to a wye source. Presuming no other loads, one supply coil sees a 30 degree leading power factor, the other a 30 degree lagging power factor. But the load itself sees 0° power.

-Jon