Reactive power loss

[WastefulMiser]

What is reactive power loss?

On edit: I guess I should state not the formula, but what causes the loss.

 

[mpross]

Reactive Power Loss

Reactive Power Loss

WastefulMiser,

Well, in a nutshell, and to state in mathematical terms...

Impedance Z = R + jX.

We know that real power loss is I^2 * R, so reactive power loss is I^2 * X.

Ciao,
Matt

 

[mivey]

Losses are I^2 * R. The R is from the conductor resistance. Part of the "I" is used to provide real power to the load. Part of the "I" is used to transport reactive power back and forth and is the extra "I" needed for the reactive power.

In delivering reactive power and receiving reactive power, this extra "I" running back and forth heats up the wire and creates real losses. That is reactive losses.

 

[LarryFine]

It's also why the closer to the offending load you can get the corrective equipment, the less of the system has to be sized for this wasted current.

 

[mivey]

It's also why the closer to the offending load you can get the corrective equipment, the less of the system has to be sized for this wasted current.

I am constantly amazed at how much stuff gets put in without that consideration. Installation time is the cheapest time handle it.

 

[WastefulMiser]

Losses are I^2 * R. The R is from the conductor resistance. Part of the "I" is used to provide real power to the load. Part of the "I" is used to transport reactive power back and forth and is the extra "I" needed for the reactive power.

In delivering reactive power and receiving reactive power, this extra "I" running back and forth heats up the wire and creates real losses. That is reactive losses.

According to the definition of Inductive reactance: X_L is the opposition to current flow offered by an inductor without the dissipation of energy.

Now if a purely inductive circuit (0 p.f.) could be created what is reactive power loss?

R = 0

X_L = 2*pi*f*L

I = V / X_L

Q_loss = X_L * I^2

 

[mivey]

According to the definition of Inductive reactance: X_L is the opposition to current flow offered by an inductor without the dissipation of energy.

Now if a purely inductive circuit (0 p.f.) could be created what is reactive power loss?

R = 0

X_L = 2*pi*f*L

I = V / X_L

Q_loss = X_L * I^2

#1:

I guess I should state not the formula, but what causes the loss.

#2:
If there is no dissipation of energy, why would you think you lost power?

 

[topgone]

According to the definition of Inductive reactance: X_L is the opposition to current flow offered by an inductor without the dissipation of energy.

Now if a purely inductive circuit (0 p.f.) could be created what is reactive power loss?

R = 0

X_L = 2*pi*f*L

I = V / X_L

Q_loss = X_L * I^2

I guess you don't incur losses that way! It is the real power that will be lost. IMO, what the phrase means is that reactive power loss is the amount of power lost along the line supplying power to a load as a result of the increase in current I due to the delivery of power at a more reactive load. I^2R loss will be larger as the I in the equation gets bigger for the same amount of kW load. Greater I results from a load that is more reactive (PF low) compared to the load being just plain resistive. On the load side, if it were physically possible to create an ideal conductor, that inductive load will not incur a power loss, but the lines supplying power to that load will - hence a "reactive power loss".
Hope this helps.

 

[mivey]

I guess you don't incur losses that way! It is the real power that will be lost. IMO, what the phrase means is that reactive power loss is the amount of power lost along the line supplying power to a load as a result of the increase in current I due to the delivery of power at a more reactive load. I^2R loss will be larger as the I in the equation gets bigger for the same amount of kW load. Greater I results from a load that is more reactive (PF low) compared to the load being just plain resistive. On the load side, if it were physically possible to create an ideal conductor, that inductive load will not incur a power loss, but the lines supplying power to that load will - hence a "reactive power loss".
Hope this helps.

In addition to what we both have agreed on, there are reactive losses that are not strictly from a "conductor", like the re-alignment of molecules in a transformer core every time the magnetic field changes. Even so, it is still dissipated as heat and is the result of reactive load.

 

[WastefulMiser]

#1:


#2:
If there is no dissipation of energy, why would you think you lost power?

Either it was being stored in the moon or it was all a lie.

Thanks all. I can die a happy person now.

 

[mivey]

Either it was being stored in the moon

Not too far from the truth. For a conductor with no "R" component but only an "X" component, the energy is stored in the wire for 1/2 of the cycle, then returned back on the other 1/2 cycle.

 

[Hameedulla-Ekhlas]

What is reactive power loss?

On edit: I guess I should state not the formula, but what causes the loss.

The diagram above shows a generator that provides electrical energy to a motor. If the motor is unloaded and is located a short distance from the genrator, the energy from the generator causes magnetic fields to form around the motor coils.
Since the coils of the motor are reactive, the current causes a magnetic field to expand around the coils during one portion of the alternation.
During another portion of the alternation, the magnetic field collapses, induces a voltage at the coil, and causes current from motor to flow back to the generator.

The rate which a reactive component stores energy in its magnetic field and then returns it to the source, is know as reactive power. Since it does not represent an energy loss, because there is no heat dissipation, it is sometimes called Wattless power.

 

[mivey]

Since it does not represent an energy loss, because there is no heat dissipation, it is sometimes called Wattless power.

No direct loss anyway. As was stated, the losses show up in the resistive pathway as well as in the jostling of molecules when the field is created/collapsed in the coil.

 

[rcwilson]

Reactive Power can get "lost".

We get complaints from generating plant clients because their generator is putting out 0.90 power factor 20 MVAr's, but the high voltage meters only show 0.95 power factor 13.5 MVAR's at the same 41.3 MW. They want us to recalibrate the meters or do something to find the "lost" 6.5 VARs.

We show them the math for the kVAR losses in the step-up transformer which are a lot more % wise than the kW losses. It usually takes a while before they understand.

 

[Besoeker]

Losses are I^2 * R. The R is from the conductor resistance. Part of the "I" is used to provide real power to the load. Part of the "I" is used to transport reactive power back and forth and is the extra "I" needed for the reactive power.

In delivering reactive power and receiving reactive power, this extra "I" running back and forth heats up the wire and creates real losses. That is reactive losses.

The current running back and forth and heating up the wire would be I^2*R as you correctly stated, the R being from the conductor resistance. That would thus be resistive losses.

That the current is higher as a result of the circuit being lower than unity power factor doesn't make those I^2*R being anything other than resistive.

 

[mull982]

I have wondered this exact question myself recently.

So I understand that for current to a device such as a motor, transformer etc.. there will be a real and reactive portion to the current. Are the kW losses in the cables only due to the real portion of the current or is it the total combination of real plus reactive current that is used in the I^2R calculation for kw loss?

So in other words the kW loss is Itotal^R and not just Ireal^R? Is this correct

If I have only a resistor and inductor in a series circuit will this circuit have both a real and reactive current portion to it? If so will it be the combination of the real and reactive currents of the total current that goes across the resistor to calculate I^2R losses across the resistor, and not just the real portion of the current?

 

[steve66]

Even so, it is still dissipated as heat and is the result of reactive load.

Umm.. I'm pretty sure if it causes heat it is going to appear as a resistive load. So for an inductor that stores 100 VA in reactive power, and disipates 1 watt in real power (due to core losses), the real power would raise the PF slightly to above 0 (maybe .01 or something like that), and the phase angle between the voltage and current would be something slightly less than 90 degrees (maybe something like 89 degrees - as a wild guess without doing the trig.)

Anyhow, I'm pretty sure (but not 100% positive) that the core losses would appear as a resistive load.

Agreed??

 

[Besoeker]

Are the kW losses in the cables only due to the real portion of the current or is it the total combination of real plus reactive current that is used in the I^2R calculation for kw loss?

The I^2R losses in the conductor depend on just the magnitude of the current and the resistance of the conductor.
The magnitude of the current is current. It's what you would measure with an ammeter. Splitting it into real and reactive components is a useful tool for circuit calculations. The conductor sees it just as current.

 

[Besoeker]

Anyhow, I'm pretty sure (but not 100% positive) that the core losses would appear as a resistive load.

Agreed??

Yes. Denoted as Rm in the Steinmetz induction motor equivalent circuit.
I think it's similar with transformers.

Edit.
I happen to have the J&P Transformer Book on the shelf above my computer.
I'll scan in a diagram that shows the equivalent circuit for a transformer.

 

[philly]

The I^2R losses in the conductor depend on just the magnitude of the current and the resistance of the conductor.
The magnitude of the current is current. It's what you would measure with an ammeter. Splitting it into real and reactive components is a useful tool for circuit calculations. The conductor sees it just as current.

That makes sense. Thanks!

Will a series circuit with just a resistor and inductor have both a real and reactive current portion?