THE PHYSICS OF... POWER

[FionaZuppa]

Reactive power is not consumed and does not require any change in the average power produced by the prime mover at the generator. There will, however, be I²R losses associated with the reactive current in all of the wiring and transformers along the way. That will require real power to be generated. And the ampacity of the generator and delivery wiring will have to be higher than if the generator served only resistive loads. For that reason, POCO will be unhappy about your inductor.

unless the generator and transmission lines are super conducting, the amps from the imaginary components (reactance) are converted (consumed power) by the ckt, just not on the load side, thus the generator has to do work for those amps conversion. no?

look at this another way, if poco has to supply a load 240ac(rms)60Hz 100A and the dc ohms of the load was 240ohms then the load only consumes 240watts of energy (if and only if the reactance components are strictly passive), but the generator needs to do 24kW of work. where exactly does 23,760watt of energy get consumed, it has to get consumed somewhere.

 

[Carultch]

unless the generator and transmission lines are super conducting, the amps from the imaginary components (reactance) are converted (consumed power) by the ckt, just not on the load side, thus the generator has to do work for those amps conversion. no?

look at this another way, if poco has to supply a load 240ac(rms)60Hz 100A and the dc ohms of the load was 240ohms then the load only consumes 240watts of energy (if and only if the reactance components are strictly passive), but the generator needs to do 24kW of work. where exactly does 23,760watt of energy get consumed, it has to get consumed somewhere.

That's a very unrealistic scenario, because there will always be an unintentional resistive component in any practical transmission line.


For a purely theoretical situation that has exclusively reactive loads, once power is initially injected, the circuit will "coast" after the prime mover is removed. For a capacitor, this means that it will retain a state of charge in an open circuit. For an inductor, this means that it will retain a steady DC current if you short the circuit. If you have the two together, they will have an oscillation where the capacitor charges/discharges/charges opposite, and the inductor exhibits the electrical "inertia" that coasts the charges through during the intermediate state.


If the prime mover continues to supply power, you'll get a resonance among the reactive components.

 

[GoldDigger]

... but the generator needs to do 24kW of work. where exactly does 23,760watt of energy get consumed, it has to get consumed somewhere.

And that, dear sir, is where you are making your mistake.
The generator is only doing 240W of work.
The current going to the inductive load is retarding the rotation of the generator for 1/4 cycle and then aiding the rotation of the generator for 1/4 cycle. The result is that there is a micro variation in the rotational speed, but prime mover does not have to do any work to keep the shaft rotating at a constant speed.

Consider a rotating weight which is off center on a horizontal shaft. For 1/2 the rotation it takes torque to turn the shaft to raise the weight. For the next 1/2 rotation it would take braking force to keep the shaft from increasing speed.
But once you have the weight rotating fast enough you do not have to put any energy in to keep it rotating, except to make up frictional losses.
There will, however, be some rather large lateral forces on the shaft bearings as the unbalanced load spins, so the whole assembly has to be more strongly built that it would be to support a balanced spinning weight.
This analogy to the POCO generator is not perfect, but it is. IMHO, useful.

 

[FionaZuppa]

That's a very unrealistic scenario, because there will always be an unintentional resistive component in any practical transmission line.

isnt that what i eluded to? wasnt the example i gave very real? the load has LCR=Z but R is very low, and perhaps C=0 (for argument sake), just a large coil not doing much with the mag field. from looking at the components of the load only, it only dissipates 240watts, by sure as crap the generator is doing ~100x more work than that. did i miss something?

 

[FionaZuppa]

take a LCR problem

load (LCR vectors) = Z = 2.4ohms, @ 240vac(rms) = 100A
L_reactance = 2.377ohm
C_reactance = 0.001ohm
reactance = 2.376ohm
R = 0.33856ohm

I²R = 3385.6 W

I²Z = 24kW = VA

does the generator power(do work) for R or Z ??

 

[GoldDigger]

take a LCR problem

load (LCR vectors) = Z = 2.4ohms, @ 240vac(rms) = 100A
L_reactance = 2.377ohm
C_reactance = 0.001ohm
reactance = 2.376ohm
R = 0.33856ohm

I²R = 3385.6 W

I²Z = 24kW = VA

does the generator power(do work) for R or Z ??

It does work only for R. Simple as that.

The current through R will depend on overall Z though. And the current into the entire LCR circuit from an external source will be subject to additional resistive losses along the way, which will have to be supplied by the generator.

 

[winnie]

but here's a interesting question. if the reactive components load-side are not responsible for power consumption then the generator does not need to do work, but if no work is done by the generator for those components then where exactly is that reactive power being "absorbed" (aka consumed)? i have a good idea where that power is wasted, but wasting power means the generator is doing work for the lost power associated with the reactive components of the load. would a poco just hate a load that was a giant inductor super-cooled to zero ohms DC ??

In common parlance, devices with leading power factor _supply_ VARs, and devices with lagging power factor _consume_ VARs. This is only because of the historical accident that many customers have devices which put lagging loads on the grid, and the power company has to provide mechanisms to supply VARs to these loads.

As has been beaten to death, reactive 'power' represents energy being shuttled back and forth, not actual power that is (on average) consumed.

None the less, many devices require reactive power to function, for example induction motors. The energy _temporarily_ stored in the motors magnetic field is required for function. If you have an induction generator connected to a resistive load, you won't get any power delivered to the load unless you have something supplying VARs _to_ the generator.

If you placed a pure inductive load on the grid, and had commercial billing, then the POCO would figure out a way to charge you

If you placed a pure capacitive load on the grid, then you might even be able to get paid for supplying VARs.

-Jon

 

[GoldDigger]

If you have an induction generator connected to a resistive load, you won't get any power delivered to the load unless you have something supplying VARs _to_ the generator.

Not true. An ideal induction generator is perfectly capable of driving either capacitive or inductive loads. But with capacitive loads or excessively large inductive loads there are practical problems with voltage regulation and various instabilities of practical equipment.

POCOs do not have special non-induction generators just to create VARs.

 

[winnie]

My understanding was that the large generators on the grid were synchronous machines, and that the field would be adjusted to provide a leading power factor to supply VARs.

Additionally, I know that 'synchronous condensers' were at one time used to supply VARs to the grid. I don't know if this is still a common practice. A synchronous condenser is a synchronous motor with no output shaft (bearing loads and losses only) with the field adjusted to supply VARs.

-Jon

 

[jtinge]

My understanding was that the large generators on the grid were synchronous machines, and that the field would be adjusted to provide a leading power factor to supply VARs.

Additionally, I know that 'synchronous condensers' were at one time used to supply VARs to the grid. I don't know if this is still a common practice. A synchronous condenser is a synchronous motor with no output shaft (bearing loads and losses only) with the field adjusted to supply VARs.

-Jon

We still use synchronous condensers on some of our large wind tunnel drive systems, granted they are systems that are over 30 years old. Our newer drive systems utilize solid state adjustable speed drives which allows for a large speed range and fan load; however, we specify pf only at rated load. The harmonic filters are sized to realize the required pf at rated load.

 

[mike_kilroy]

That is a very good explanation. Thanks.

Yep.

Sent from my SM-G900V using Tapatalk

 

[mike_kilroy]

This whole thread make clear why there are engineers and technicians, and why neither should try to do the other's job.

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[FionaZuppa]

This whole thread make clear why there are engineers and technicians, and why neither should try to do the other's job.

Sent from my SM-G900V using Tapatalk

how can you tell, who do you see as the engineers, who are the technicians, and who are the physics folks?

 

[FionaZuppa]

lets take it further, who can explain this
if VA = J/s
so if the PF is 80% then where the heck does that other 20% of (VA=J/s) go over period delta t? imaginary power as they say? but yet the "j" component (imaginary) still has very real coulomb/sec associated with it. doesnt it take a force to build up voltage and create amps? hmmmm.

 

[Carultch]

lets take it further, who can explain this
if VA = J/s
so if the PF is 80% then where the heck does that other 20% of (VA=J/s) go over period delta t? imaginary power as they say? but yet the "j" component (imaginary) still has very real coulomb/sec associated with it. doesnt it take a force to build up voltage and create amps? hmmmm.

Volt-Amperes does not directly equal Joules per second.

Algebraically, it is true that you can manipulate the definitions of Volts and Amperes to get Joules per second. And it is true that if you multiply instantaneous values of Volts and Amperes together, you do get the instantaneous value of Joules per second.

However, when you multiply "time average" values of Volts together with the "time average" Amperes, you do not necessarily get a value for the long term Joules per second. The reason I put "time average" in quotations, is that it really is a root-mean-square value, as opposed to a direct mean value of the function.

What happens when the Volt-Amperes does not equal the Watts, is that the voltage and current are not synchronized. If you multiply the instantaneous Volts & Amperes together, and then accumulate (integrate) it over time, you get the Joules as you would expect. Divide it by the time period, and you get the average Watts.

This is an incandescent bulb filmed in slow motion, with a 50 Hz AC power source. You can see that the power is not steady, but instead comes in cycles. So the Wattage rating on the bulb, is really the average Watts over a cycle.
https://www.youtube.com/watch?v=eUprJS9sXYU

 

[Carultch]

Continued

With the pure resistive load, the RMS Volts * RMS Amps = Average Watts.

With any other load that has a reactive component to it, RMS Volts * RMS Amps is greater than Average Watts, due to the fact that the Volts and Amps are not synchronized. Volts attempt to apply power to the circuit, but the circuit's reactance causes a lag for the current to catch up. Or current may attempt to flow, but builds up a voltage as it comes to a capacitor as a blockage. This is energy being stored in the inductors' magnetic fields, or the capacitors electrostatic fields.

Considering a power factor of 80%, for a 100 VA load. Only 80 Watts are consumed, but it seems like there may have been another 20 Watts. There really wasn't another 20 Watts. It is just that we naively multiplied together a value of Volts and a value of Amps, that didn't occur simultaneously. Only simultaneous Volts and Amps actually carry power through the circuit. The non-simultaneous Volts and Amps are idle power flowing back and forth within the reactive energy storage components of the circuit.

 

[dionysius]

The mind's eye.....retrain HOW you think with visual cues......

The mind's eye.....retrain HOW you think with visual cues......

With the pure resistive load, the RMS Volts * RMS Amps = Average Watts.

With any other load that has a reactive component to it, RMS Volts * RMS Amps is greater than Average Watts, due to the fact that the Volts and Amps are not synchronized. Volts attempt to apply power to the circuit, but the circuit's reactance causes a lag for the current to catch up. Or current may attempt to flow, but builds up a voltage as it comes to a capacitor as a blockage. This is energy being stored in the inductors' magnetic fields, or the capacitors electrostatic fields.

Considering a power factor of 80%, for a 100 VA load. Only 80 Watts are consumed, but it seems like there may have been another 20 Watts. There really wasn't another 20 Watts. It is just that we naively multiplied together a value of Volts and a value of Amps, that didn't occur simultaneously. Only simultaneous Volts and Amps actually carry power through the circuit. The non-simultaneous Volts and Amps are idle power flowing back and forth within the reactive energy storage components of the circuit.

This is a very good description, Carultech, I see that you get it very well. Others need to read what you have submitted.

I have a very special way in my mind's eye that lets me visualize the situation very well. I can clearly lay down a sine wave for the voltage. Then below it I visualize a sine wave of same freq, different amplitude.

Now in my mind's eye I can "slide" the current wave back and forth relative to the fixed voltage wave.

This strategem has served me well from the first time I encountered the issue years ago.

Having read through the thread I can see that not every one has this "mind's eye" capability.

Having read Nikola's work I see that he had even more capability to "see" this way than I have. He was forced by shortage of resources in his young years to develop that great skill. He had such confidence in his capacity that he had the balls to direct the designs for the Niagara Falls gen plant without prototypes. It is this "shifting" of sine waves that introduces the complex number plane. If people change the way they visualize the issue this thread could be very short.

 

[mike_kilroy]

With the pure resistive load, the RMS Volts * RMS Amps = Average Watts.

With any other load that has a reactive component to it, RMS Volts * RMS Amps is greater than Average Watts, due to the fact that the Volts and Amps are not synchronized. Volts attempt to apply power to the circuit, but the circuit's reactance causes a lag for the current to catch up. Or current may attempt to flow, but builds up a voltage as it comes to a capacitor as a blockage. This is energy being stored in the inductors' magnetic fields, or the capacitors electrostatic fields.

Considering a power factor of 80%, for a 100 VA load. Only 80 Watts are consumed, but it seems like there may have been another 20 Watts. There really wasn't another 20 Watts. It is just that we naively multiplied together a value of Volts and a value of Amps, that didn't occur simultaneously. Only simultaneous Volts and Amps actually carry power through the circuit. The non-simultaneous Volts and Amps are idle power flowing back and forth within the reactive energy storage components of the circuit.

Nice description with exception of this incorrect line... That reactive power IS still going thru the 'circuit.' That is why it makes I^2R heat in the wires. If it were not 'power,' it could not make that heat. Drop this highlighted line and the rest is a good description.

 

[mike_kilroy]

lets take it further, who can explain this
if VA = J/s
so if the PF is 80% then where the heck does that other 20% of (VA=J/s) go over period delta t? imaginary power as they say? but yet the "j" component (imaginary) still has very real coulomb/sec associated with it. doesnt it take a force to build up voltage and create amps? hmmmm.

Just for kicks, I highlighted the semi-reduntant/incorrect(?) statement above... That force is voltage. Volts is the forcing function that makes the current appear. So the line should say "doesnt it take a force, called voltage, to build up and create amps?" ​Then answer can be YEP!

 

[mike_kilroy]

...
Now in my mind's eye I can "slide" the current wave back and forth relative to the fixed voltage wave.

Nice way to look at it. I prefer the current triangle tho since I work with vector drives all the time... in my mind's eye that reactive current (or power for those who prefer power triangle) just gets sent back to the POCO supply each half cycle, then like a mirror bounces it right back to me the next half cycle. Makes for a cool analogy I can 'see' that makes it crystal clear to me it is imaginary and not being supplied by the big boys at POCO.
Then, if I stick a vfd in series between POCO and my motor, this bouncing back and forth just goes on between my motor, its cables, and the dc bus capacitors in the vfd - POCO doesn't get any of it. I keep it all local.