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# Physical interpretation of the Reactive Power

Is the reactive power an effect of energy oscillation?

#1

09-17-2013 12:10 AM

Top #2

sinusoidally diverted component of true power is reactive power.

09-17-2013 02:36 AM

Top #3

Reactive power is not used by the load. It is actually due to low power factor of the load. The reactive power keeps on moving between load and source and is only responsible for the losses.

09-17-2013 05:10 AM

Top #4

Physical interpretation of Reactive Power (Q)? For motors and generators, Q is the power needed to produce Magnetic Flux (B). Remember Fleming's Right Hand Rule (for Generator) and Fleming's Left Hand Rule (for motor), both rules explain about FBI: F for Force or rotation, B for Magnetic Flux and I for current.

Also remember Fleming's James Bond. Haha....

Also, Q can also be considered losses. wallahu a'lam

Also remember Fleming's James Bond. Haha....

Also, Q can also be considered losses. wallahu a'lam

09-17-2013 08:06 AM

Top #5

It a mathematical tool for sinusoidal circuits. It is defined as the mathematical difference between Apparent Power and Real Power.

09-17-2013 10:45 AM

Top #6

You cannot supply active power to a resistance without establishing the flux in the inductance associated with the circuit. In positive half cycle there is additional power drawn from the source. However this energy is stored, not consumed. In negative half of the cycle power is returned to the source. Average power over a cycle is zero. Thus it is indeed oscillating power flow. Average over cycle is zero. There are however heated arguments in literature about this concept since it is can be mathematically extended to even non-linear diode circuits where reverse power flow cannot take place..! Back to square one..!

09-17-2013 01:24 PM

Top #7

Thanks to all for your ans.

The question is because we have an unbalanced system under study and i found that power factor calculation does not obeys the well known geometrical relationship called power triangle nor the arithmetical relationship also.

However if I consider that reactive power is only an effect of the phase shift between the voltage and current the calculated power factor is almost the power factor measured.

So the well know arithmetical and geometrical equations do not describe energy flow between the supply and the load under unbalanced conditions.....this is my point of view.

From here my original question about the physical interpretation of reactive power...in other words: is reactive power a physical entity or is just an effect?

Another curious thing is that load under study are heating resistances and theory says that resistor have a unity power factor...however such resistors demands reactive power from the supply source, so cannot be considered as loads with unity power factor under unbalanced conditions but again if i consider the reactive power as an effect I can calculate the power factor almost matching the measured values.....!to calculate a power factor for a resistor!!, ...yes i know it sounds estrange....

In my opinion reactice power is just an effect and not a physical entity but i want to know other opinions and points of view.

Thank you all.

The question is because we have an unbalanced system under study and i found that power factor calculation does not obeys the well known geometrical relationship called power triangle nor the arithmetical relationship also.

However if I consider that reactive power is only an effect of the phase shift between the voltage and current the calculated power factor is almost the power factor measured.

So the well know arithmetical and geometrical equations do not describe energy flow between the supply and the load under unbalanced conditions.....this is my point of view.

From here my original question about the physical interpretation of reactive power...in other words: is reactive power a physical entity or is just an effect?

Another curious thing is that load under study are heating resistances and theory says that resistor have a unity power factor...however such resistors demands reactive power from the supply source, so cannot be considered as loads with unity power factor under unbalanced conditions but again if i consider the reactive power as an effect I can calculate the power factor almost matching the measured values.....!to calculate a power factor for a resistor!!, ...yes i know it sounds estrange....

In my opinion reactice power is just an effect and not a physical entity but i want to know other opinions and points of view.

Thank you all.

09-17-2013 04:11 PM

Top #8

Unbalance circuits will have positive, negative and possibly, zero sequence components of voltages and currents. They will have their respective contributions to power. Power factor is defined generally for positive sequence. The 3 components may have different reactances if transmission lines, machines and transformers are in the circuit. Hence pf for the three is in general different. Your question 'is reactive power a physical entity or is just an effect?' is still an open one!

09-17-2013 07:09 PM

Top #9

no resistor is said to be pure resistor . all components will be a combination of RLC. but as we all know according to the effects of different components we call them R or L or C.

now coming to reactive power :- the sinusoidal wave is a development of Simple harmonic motion. can we think the centrifugal/ centripetal force as reactive energy and the actual movement as real power. any takers?

now coming to reactive power :- the sinusoidal wave is a development of Simple harmonic motion. can we think the centrifugal/ centripetal force as reactive energy and the actual movement as real power. any takers?

09-17-2013 09:47 PM

Top #10

My opinion is that reactive power only fiction which is made by engineers with purpose to be explained need for improvement of value of power factor... Reactive power is power which is a consequence of power factor which value is lower than 1. In cases when we are talking about consumers which is value of power factor lower than 1 (electrical machines), we need to improve value of power factor by using synchronous generator or capacitor bank.

So, we don't have need to use reactive power anywhere... We can't use reactive power for anything which isn't same case when we are talking about active power...

But, by other side, we have a lot of reactive power losses in power system. For example, we have reactive power losses in transmission lines and distribution lines as a consequence of supplying of mentioned lines by voltage. In these situations, transmission lines and distribution lines are behaving like capacitors in relation to the earth. Other example, we have reactive power losses in electrical machines which are a consequence of dissipation of magnetic flux and magnetising of ferromagnetic core.

Also, if we have a very high voltage drops in our power system, we know that we need to improve value of power factor at the places which are the most critical. We can improve value of power factor at the places which are the most critical by using synchronous generator or capacitor bank. On this way, we will reduce reactive power losses at the places where we need to install equipment for improving of value of power factor, we will reduce voltage drops in our transmission lines and distribution lines and we will unload our transmission lines and distribution lines what is the most important thing in aspect of warming of mentioned lines and occupancy of capacity (size) of mentioned lines...

So, my opinion is the significance of reactive power very important and very complicated for understanding, because reactive power depends from a lot of things and conditions...

So, we don't have need to use reactive power anywhere... We can't use reactive power for anything which isn't same case when we are talking about active power...

But, by other side, we have a lot of reactive power losses in power system. For example, we have reactive power losses in transmission lines and distribution lines as a consequence of supplying of mentioned lines by voltage. In these situations, transmission lines and distribution lines are behaving like capacitors in relation to the earth. Other example, we have reactive power losses in electrical machines which are a consequence of dissipation of magnetic flux and magnetising of ferromagnetic core.

Also, if we have a very high voltage drops in our power system, we know that we need to improve value of power factor at the places which are the most critical. We can improve value of power factor at the places which are the most critical by using synchronous generator or capacitor bank. On this way, we will reduce reactive power losses at the places where we need to install equipment for improving of value of power factor, we will reduce voltage drops in our transmission lines and distribution lines and we will unload our transmission lines and distribution lines what is the most important thing in aspect of warming of mentioned lines and occupancy of capacity (size) of mentioned lines...

So, my opinion is the significance of reactive power very important and very complicated for understanding, because reactive power depends from a lot of things and conditions...

09-18-2013 12:06 AM

Top #11

For a "physical" interpretation, reactive current(power/KVA flow), in my opinion is best looked at from the perspective of a generator connected directly to an infinite bus (in LV generators this is the norm).

The generator when connected to the system, "see's/feels" the parallel impedance combination of all other generators (circa 3 ohms each) with respect to ground - which basically parallel to equate to a zero impedance in terms of restriction to any current flow out of our generator.

Post initial synchronization, the system voltage prevents currents from flowing into or out of the generator due to pressure (voltage) balance of our generator matching that of the system voltage.

If you (as the generator operator), try to lift the generator voltage, the result will only be heaps of current output flowing into the system – but with no actual extra power generated!

This is due to the fact that to achieve the extra generator voltage setpoint you desired, the generator must send out enough current into the system impedance to create the back emf required to achieve the new desired generator terminal voltage setpoint.

But because the system impedance to ground is very low (as it actually is) – then despite the extra current sent out in that fruitless attempt, the generator is near impotent to make any substantial effect on raising the “system” voltage – “fruitless” current sent out.

In a DC sense you can equate this to a small DC generator trying to lift the voltage of a load system that has a zener diode installed across that system load.

Back to the AC world, ….that current sent out in the fruitless attempt to lift system voltage must flow through the parallel low impedance of the other connected generators (each of those working against you – lowering their own generator excitation, hell bent on keeping their own same old voltage setpoints), thwarting our futile attempt to achieve a raise in the system voltage.

All those generators, although collectively of low impedance, compose virtually no resistance, compared to their inductive reactance. Hence all our little generators current flow - in its futile attempt to lift system volts - is virtually purely inductive.

So we have heaps of current flowing out in our attempt to lift generator volts, but because the current is 90 degrees lagging the voltage, the only power imposed on the generator prime mover is that due to the resistance of the generator windings (circa 1% of the full load current rating – hence basically un-noticeable).

Hence the physical interpretation of VAR’s, is actually simply a look at the voltage balance perspective of an electricity network. It’s the collective attempt of many parallel-connected generators to influence the system voltage – either trying to raise the voltage at a particular node (positive VAR’s) or trying to reduce the voltage at a particular node (negative VAR’s flowing back through our generator due to our attempt to lower our generator setpoint – which “lets current in”).

What a ramble… sorry about that.

The generator when connected to the system, "see's/feels" the parallel impedance combination of all other generators (circa 3 ohms each) with respect to ground - which basically parallel to equate to a zero impedance in terms of restriction to any current flow out of our generator.

Post initial synchronization, the system voltage prevents currents from flowing into or out of the generator due to pressure (voltage) balance of our generator matching that of the system voltage.

If you (as the generator operator), try to lift the generator voltage, the result will only be heaps of current output flowing into the system – but with no actual extra power generated!

This is due to the fact that to achieve the extra generator voltage setpoint you desired, the generator must send out enough current into the system impedance to create the back emf required to achieve the new desired generator terminal voltage setpoint.

But because the system impedance to ground is very low (as it actually is) – then despite the extra current sent out in that fruitless attempt, the generator is near impotent to make any substantial effect on raising the “system” voltage – “fruitless” current sent out.

In a DC sense you can equate this to a small DC generator trying to lift the voltage of a load system that has a zener diode installed across that system load.

Back to the AC world, ….that current sent out in the fruitless attempt to lift system voltage must flow through the parallel low impedance of the other connected generators (each of those working against you – lowering their own generator excitation, hell bent on keeping their own same old voltage setpoints), thwarting our futile attempt to achieve a raise in the system voltage.

All those generators, although collectively of low impedance, compose virtually no resistance, compared to their inductive reactance. Hence all our little generators current flow - in its futile attempt to lift system volts - is virtually purely inductive.

So we have heaps of current flowing out in our attempt to lift generator volts, but because the current is 90 degrees lagging the voltage, the only power imposed on the generator prime mover is that due to the resistance of the generator windings (circa 1% of the full load current rating – hence basically un-noticeable).

Hence the physical interpretation of VAR’s, is actually simply a look at the voltage balance perspective of an electricity network. It’s the collective attempt of many parallel-connected generators to influence the system voltage – either trying to raise the voltage at a particular node (positive VAR’s) or trying to reduce the voltage at a particular node (negative VAR’s flowing back through our generator due to our attempt to lower our generator setpoint – which “lets current in”).

What a ramble… sorry about that.

09-18-2013 02:33 AM

Top #12

Mr. Mocevic

I'm afraid you did not get some very fundamental aspects of the operation of AC systems.

The proper operation of an AC system requires magnetization/energization of the apparatus, machines, transformers, lines, etc.

If you do not properly energize the equipment, no active power conversion, MW conversion, (mechanical power being converted to electrical power in generators, electrical power being converted to mechanical power in motors, etc.) can happen. Similarly, you cannot transfer active power from one winding of a transformer to another winding, without first building up the magnetic flux at the core of said transformer.

In other words: no reactive power? Then you will also not have any active power in an AC system!

Some equipment (mostly synchronous machines) have their own excitation systems, so they have control over its own magnetization and, more important, can build up such magnetization without the help of a "system".

Induction motors, on the other hand, require the "system" to supply its magnetization. Therefore, every induction motor/generator requires an external source for magnetization and, as a result, will operate with a power factor different than 1. Always "sucking" reactive power from the "system". Similar argument can be made for transformers and even transmission lines (just look up "inrush currents" for instance). Transmission lines are a bit more complicated than that, they have inductances and capacitances and that is FUN!

We call "reactive power" the power that flows back and forth, with an average of zero (therefore, no energy associated with) as a result of charging and discharging electrical and magnetic fields in all these devices. This charge and discharge happen every cycle of the AC system, it is in the nature of the AC fields.

Thus, reactive power is NOT "losses". It is just necessary, it is the "grease" that allows the operation of an AC system. Losses are related to current magnitudes on resistors. There are no losses for these same currents flowing through capacitors or inductors.

Losses are strictly R.I^2

Therefore, any current will cause losses. Therefore, even at perfect unity power factor you will have losses! Because you will have a current, corresponding to the active power. Yes, very poor power factor will result in an increase in the current magnitude, increasing losses. But reactive power is NOT losses.

Finally, even if you don't know what reactive power is, you know exactly how to "measure" it: if you do not have enough reactive power available (to sustain the necessary magnetization of the AC apparatus) voltages will drop. If you have too much reactive power around, voltages will rise. In other words, as long as we are capable of controlling voltages (mostly via control of excitation systems in synchronous generators), we will have reactive power flows "just right".

I'm afraid you did not get some very fundamental aspects of the operation of AC systems.

The proper operation of an AC system requires magnetization/energization of the apparatus, machines, transformers, lines, etc.

If you do not properly energize the equipment, no active power conversion, MW conversion, (mechanical power being converted to electrical power in generators, electrical power being converted to mechanical power in motors, etc.) can happen. Similarly, you cannot transfer active power from one winding of a transformer to another winding, without first building up the magnetic flux at the core of said transformer.

In other words: no reactive power? Then you will also not have any active power in an AC system!

Some equipment (mostly synchronous machines) have their own excitation systems, so they have control over its own magnetization and, more important, can build up such magnetization without the help of a "system".

Induction motors, on the other hand, require the "system" to supply its magnetization. Therefore, every induction motor/generator requires an external source for magnetization and, as a result, will operate with a power factor different than 1. Always "sucking" reactive power from the "system". Similar argument can be made for transformers and even transmission lines (just look up "inrush currents" for instance). Transmission lines are a bit more complicated than that, they have inductances and capacitances and that is FUN!

We call "reactive power" the power that flows back and forth, with an average of zero (therefore, no energy associated with) as a result of charging and discharging electrical and magnetic fields in all these devices. This charge and discharge happen every cycle of the AC system, it is in the nature of the AC fields.

Thus, reactive power is NOT "losses". It is just necessary, it is the "grease" that allows the operation of an AC system. Losses are related to current magnitudes on resistors. There are no losses for these same currents flowing through capacitors or inductors.

Losses are strictly R.I^2

Therefore, any current will cause losses. Therefore, even at perfect unity power factor you will have losses! Because you will have a current, corresponding to the active power. Yes, very poor power factor will result in an increase in the current magnitude, increasing losses. But reactive power is NOT losses.

Finally, even if you don't know what reactive power is, you know exactly how to "measure" it: if you do not have enough reactive power available (to sustain the necessary magnetization of the AC apparatus) voltages will drop. If you have too much reactive power around, voltages will rise. In other words, as long as we are capable of controlling voltages (mostly via control of excitation systems in synchronous generators), we will have reactive power flows "just right".

09-18-2013 05:01 AM

Top #13

Mr:Lima:- Agreed mostly with your explanation, but there will be hysteresis and eddy current losses for inductive circuit. similarly capacitance or dielectric losses in cap circuit. But what about the question raised by Mr.Enrique?

09-18-2013 07:23 AM

Top #14

Mr. Nambiar

You are correct about hysteresis and eddy currents, no question about it. For the sake of completeness, hysteresis would also lead to harmonics, complicating things even further. And, when considering unbalanced three-phase systems and/or the presence of harmonics, the conventional tools for power system analysis might not be applicable.

The losses due to hysteresis are limited by using better materials in transformer core. Eddy current losses are limited by using laminated construction. These losses are a relatively small portion of the total losses in a power system. Most of the losses are Joule losses (currents and resistances).

Mr. Enrique's question could use some clarification, because "energy" in his question might be misinterpreted. Sure, reactive power is associated with charging and discharging (actually, charging with the opposite polarity) magnetic and electric fields. So, electromagnetic energy "flows" back-and-forth in these fields.

But they do so twice (one positive, one negative) on every cycle of the AC system, so the average energy is zero.

So, I have to answer Mr. Enrique's question with a "yes and no". Yes, there is an energy "exchange" between magnetic and electric fields. But no, that is not an oscillation in energy (kWh), not something that you could measure, for instance, in the torques on a mechanical shaft (that is purely kW, active power).

You are correct about hysteresis and eddy currents, no question about it. For the sake of completeness, hysteresis would also lead to harmonics, complicating things even further. And, when considering unbalanced three-phase systems and/or the presence of harmonics, the conventional tools for power system analysis might not be applicable.

The losses due to hysteresis are limited by using better materials in transformer core. Eddy current losses are limited by using laminated construction. These losses are a relatively small portion of the total losses in a power system. Most of the losses are Joule losses (currents and resistances).

Mr. Enrique's question could use some clarification, because "energy" in his question might be misinterpreted. Sure, reactive power is associated with charging and discharging (actually, charging with the opposite polarity) magnetic and electric fields. So, electromagnetic energy "flows" back-and-forth in these fields.

But they do so twice (one positive, one negative) on every cycle of the AC system, so the average energy is zero.

So, I have to answer Mr. Enrique's question with a "yes and no". Yes, there is an energy "exchange" between magnetic and electric fields. But no, that is not an oscillation in energy (kWh), not something that you could measure, for instance, in the torques on a mechanical shaft (that is purely kW, active power).

09-18-2013 09:55 AM

Top #15

Hi, sir Leonardo,

I need to say that I agree with your opinion that "Reactive power is needed for magnetising ferromagnetic core of electrical machines" and that is the opinion which I have already exposed in my previous comments at different topics in various discussions...

But also, I need to say that I disagree with your opinion that "Losses are strictly R.I^2"...

Losses which you mentioned in your previous comment (R.I^2) are power losses which are a consequence of inside resistance of copper or aluminum and they called active power losses or Joules' losses (they depend from square of current which flows through copper or aluminum)...

But, in case of electrical machines, we also have dissipation of magnetic flux and this phenomenon can be modeled as inductance through which flows current, so we have also losses which can express as X*I^2. Also, we can model phenomenon known as magnetising of ferromagnetic core, like inductance through which flows magnetising current, so we have losses which can express as Xm*I^2. These kinds of power losses are reactive power losses.

In case of transmission lines and distribution lines, we also have dissipation of magnetic flux and this phenomenon can be modeled as inductance through which flows current, so we have also losses which can express as X*I^2. Also, transmission lines and distribution lines, when they are supplied by voltage, they behave as capacitors in relation to earth, so we have losses which can express as B*U^2 (B - susceptance). These kinds of power losses are reactive power losses.

So, I wanted to explain that reactive power unusable power which means that we can't use it for anyone (useful) work (except for magnetising of ferromagnetic core of electrical machines) which isn't case with active power...

I need to say that I agree with your opinion that "Reactive power is needed for magnetising ferromagnetic core of electrical machines" and that is the opinion which I have already exposed in my previous comments at different topics in various discussions...

But also, I need to say that I disagree with your opinion that "Losses are strictly R.I^2"...

Losses which you mentioned in your previous comment (R.I^2) are power losses which are a consequence of inside resistance of copper or aluminum and they called active power losses or Joules' losses (they depend from square of current which flows through copper or aluminum)...

But, in case of electrical machines, we also have dissipation of magnetic flux and this phenomenon can be modeled as inductance through which flows current, so we have also losses which can express as X*I^2. Also, we can model phenomenon known as magnetising of ferromagnetic core, like inductance through which flows magnetising current, so we have losses which can express as Xm*I^2. These kinds of power losses are reactive power losses.

In case of transmission lines and distribution lines, we also have dissipation of magnetic flux and this phenomenon can be modeled as inductance through which flows current, so we have also losses which can express as X*I^2. Also, transmission lines and distribution lines, when they are supplied by voltage, they behave as capacitors in relation to earth, so we have losses which can express as B*U^2 (B - susceptance). These kinds of power losses are reactive power losses.

So, I wanted to explain that reactive power unusable power which means that we can't use it for anyone (useful) work (except for magnetising of ferromagnetic core of electrical machines) which isn't case with active power...

09-18-2013 12:50 PM

Top #16

Mr. Mocevic

I believe our misunderstanding is related to definitions.

We need to better define terms, here. What exactly are you calling "losses"? Is that a real energy loss, measured in Joules or kWh? If we limit the meaning of "losses" to Joules or kWh, then such losses cannot be represented as X.I^2.

Basically, I really think it is misleading to talk about reactive power losses (although it is done so frequently).

For instance, let's take a transformer for an example. You test a transformer and you get short-circuit and open-circuit impedances and losses. Then, the most basic model for such transformer is a T circuit, with series RL impedances (R and X) for each winding, with a central magnetizing shunt branch usually represented by parallel-connected R and L (G and B for their conductance and susceptance, respectively).

Furthermore, we often "lump" the series impedance in just one "side" of the transformer.

Then, in the most simplistic way, we calculate the series resistance to match the losses from the short-circuit test (assuming voltages are really low across the magnetizing branch, so the current across the shunt-connected resistance is low, thus the losses introduced by the shunt resistance can be neglected in these results). The reactance of the series impedance (leakage reactance X) is calculated from the measured impedance (test result) and the calculated value for the series resistance.

Conversely, the magnetizing branch (the parallel RL branch) is calculated from the open-circuit test results. Once again, G is calculated to match the losses measured in that test (assuming very low currents, therefore the R.I^2 losses can be neglected). And B is then calculated from the measured impedance for the open-circuit test and the calculated value of G.

Long story short: the test results show all losses in the transformer, hysteresis, eddy currents, copper, etc. The equivalent model "lumps" these losses in a resistance.

Why? Because there are no losses (in Joules) in a capacitance or an inductance, in a RLC circuit (circuit theory, ideal capacitors/reactors. In real life, capacitors and inductors have resistances and losses that can be measured as Joules, or we can detect an increase in temperature when in service).

What is X.I^2? That is a "sink" of reactive power, if this is an inductance, but it is a "source" of reactive power if this is a capacitance. You might call these reactive power losses, but then you might end up with negative losses, if you have more capacitance than inductance in a given example (for instance, the pi-model of a long transmission line in open circuit or light load).

Again, you have properly identified these terms as reactive power losses. So far we are in agreement. The problem is that thinking about reactive power as "losses" might give the impression we are talking Joules. We are not.

Finally, as I said before, if you don't have enough reactive power to "supply" all these X.I^2 "losses" voltages will drop. If you have excess "negative losses" from capacitors (cables, lines, etc.) voltages will rise.

My point: if you control voltages properly, you don't have to worry about reactive power. If you struggle to control voltages, you don't have enough reactive "sources" to sustain the necessary energization/magnetization of the AC system.

I believe our misunderstanding is related to definitions.

We need to better define terms, here. What exactly are you calling "losses"? Is that a real energy loss, measured in Joules or kWh? If we limit the meaning of "losses" to Joules or kWh, then such losses cannot be represented as X.I^2.

Basically, I really think it is misleading to talk about reactive power losses (although it is done so frequently).

For instance, let's take a transformer for an example. You test a transformer and you get short-circuit and open-circuit impedances and losses. Then, the most basic model for such transformer is a T circuit, with series RL impedances (R and X) for each winding, with a central magnetizing shunt branch usually represented by parallel-connected R and L (G and B for their conductance and susceptance, respectively).

Furthermore, we often "lump" the series impedance in just one "side" of the transformer.

Then, in the most simplistic way, we calculate the series resistance to match the losses from the short-circuit test (assuming voltages are really low across the magnetizing branch, so the current across the shunt-connected resistance is low, thus the losses introduced by the shunt resistance can be neglected in these results). The reactance of the series impedance (leakage reactance X) is calculated from the measured impedance (test result) and the calculated value for the series resistance.

Conversely, the magnetizing branch (the parallel RL branch) is calculated from the open-circuit test results. Once again, G is calculated to match the losses measured in that test (assuming very low currents, therefore the R.I^2 losses can be neglected). And B is then calculated from the measured impedance for the open-circuit test and the calculated value of G.

Long story short: the test results show all losses in the transformer, hysteresis, eddy currents, copper, etc. The equivalent model "lumps" these losses in a resistance.

Why? Because there are no losses (in Joules) in a capacitance or an inductance, in a RLC circuit (circuit theory, ideal capacitors/reactors. In real life, capacitors and inductors have resistances and losses that can be measured as Joules, or we can detect an increase in temperature when in service).

What is X.I^2? That is a "sink" of reactive power, if this is an inductance, but it is a "source" of reactive power if this is a capacitance. You might call these reactive power losses, but then you might end up with negative losses, if you have more capacitance than inductance in a given example (for instance, the pi-model of a long transmission line in open circuit or light load).

Again, you have properly identified these terms as reactive power losses. So far we are in agreement. The problem is that thinking about reactive power as "losses" might give the impression we are talking Joules. We are not.

Finally, as I said before, if you don't have enough reactive power to "supply" all these X.I^2 "losses" voltages will drop. If you have excess "negative losses" from capacitors (cables, lines, etc.) voltages will rise.

My point: if you control voltages properly, you don't have to worry about reactive power. If you struggle to control voltages, you don't have enough reactive "sources" to sustain the necessary energization/magnetization of the AC system.