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[Phillip_Graham]

Ryan Lantzy wrote on Tue, 18 October 2005 13:00

Yeah, I agree with you Langston.  I think the resistive heating is there all the time.  No matter what the load.

Resistance is always there with real cables.  I am not saying any different.

Quote:

The measurable difference between kW and kVA is power factor.  Period.  Saying restitive heating is the cause is like saying that the heating loss of an engine is why you get worse fuel economy when driving hilly terrain vs. flat.

Resistive heating is NOT the cause of power factor.  The cause of power factor is the nature of the definition of Work when dealing with non-resistive components!  Capacitive elements store energy via the electric field, and inductive elements store energy via the magnetic field.

In a resistance free world, the fraction of the energy "trapped" by the phase relationship between voltage and current would be stored in the system, and oscillate back and forth between the source and the load.

However, in a real system there are dissipative losses, partially in the form of resistance heating.  There dissipative losses from the resistance of the cable due to the in phase voltage/current component (the component that is doing work for you), and there are also dissipative losses for the current that is 90deg. out of phase with the voltage, and therefore not doing work!  Electrons are still moving/bouncing around in the cable, even if it is not in a way that allows you to extract usuable work from the load of the system.

So, if you have a purely resistive load, and you get "X" amount of dissipative losses from resistive heating.  Now you have a load of power factor 0.8, with the same input power as the purely resistive load.  What is the amount of dissipative losses?  Still "X"!

Now, to get the same amount of work OUT of the 0.8 power factor system, you must apply 1/0.8=1.25x the input power, and your dissipative losses in the cable will be X/0.8=1.25*X.

I hope that clears up what I was trying to convey.

[Ryan Lantzy]

Phillip Graham wrote on Tue, 18 October 2005 13:31

Resistive heating is NOT the cause of power factor.  The cause of power factor is the nature of the definition of Work when dealing with non-resistive components!  Capacitive elements store energy via the electric field, and inductive elements store energy via the magnetic field.

I didn't say otherwise.

Quote:

In a resistance free world, the fraction of the energy "trapped" by the phase relationship between voltage and current would be stored in the system, and oscillate back and forth between the source and the load. However, in a real system there are dissipative losses, partially in the form of resistance heating.  There dissipative losses from the resistance of the cable due to the in phase voltage/current component (the component that is doing work for you), and there are also dissipative losses for the current that is 90deg. out of phase with the voltage, and therefore not doing work!  Electrons are still moving/bouncing around in the cable, even if it is not in a way that allows you to extract usuable work from the load of the system. So, if you have a purely resistive load, and you get "X" amount of dissipative losses from resistive heating.  Now you have a load of power factor 0.8, with the same input power as the purely resistive load.  What is the amount of dissipative losses?  Still "X"! Now, to get the same amount of work OUT of the 0.8 power factor system, you must apply 1/0.8=1.25x the input power, and your dissipative losses in the cable will be X/0.8=1.25*X. I hope that clears up what I was trying to convey.

I think you're saying the same thing everyone else was.  It just looked like you claimed that the resistive heating of cable was a component of the difference between kW and kVA.  You're clear now.  Crystal.

[Phillip_Graham]

Langston Holland wrote on Tue, 18 October 2005 10:08

Phill wrote on Mon, 17 Oct 2005 There is a portion of the energy transferred in an electrical system that has a phase relationship in such a way that is not capable of doing work for you, but it is still capable of causing resistive heating of the electrical conducting path of the system.  That is energy your generator must still provide that is not going towards making sound and light, and therefore requires a larger genny that with a purely resistive load Alright Mr. Wise guy, answer me this: Assume a genny or panel that can supply a maximum of 100 amps at 120 volts on each hot leg into a purely resistive load. This would be 100 x 120 = 12,000 watts per hot leg. If I measure the current, voltage, or wattage I will get readings of 100, 120, and 12,000 respectively. But, if I place a non-resistive sound and lighting load on the circuit instead, and draw the full available power from the power source, I will get readings of 100, 120, and something less (maybe much less) than 12,000 on the watt meter. Thus, the watt meter will tell me the power or work being done by the concert equipment and the voltage and current meters will tell me if I’m within the design limits of the genny or panel. True of False?

Hey Langston, this is true, as long as your wattmeter is measuring the true power, and not fudging in some way.

[Phillip_Graham]

[quote title=Ryan Lantzy wrote on Tue, 18 October 2005 16:04]

Phillip Graham wrote on Tue, 18 October 2005 13:31

Resistive heating is NOT the cause of power factor.  The cause of power factor is the nature of the definition of Work when dealing with non-resistive components!  Capacitive elements store energy via the electric field, and inductive elements store energy via the magnetic field.

Quote:

I didn't say otherwise.

Gotcha, thanks!

Quote:

I think you're saying the same thing everyone else was.  It just looked like you claimed that the resistive heating of cable was a component of the difference between kW and kVA.  You're clear now.  Crystal.

Figures I was trying to say the same thing.  Sorry about that.

While the difference between kW and kVA is NOT due to the resistive heating, I submit that resistive heating is the primary means for dissipating the energy that is not performing work, as represented by the power factor.

Sorry for the confusion.

[Phillip_Graham]

Bennett Prescott wrote on Mon, 17 October 2005 23:07

Thanks for keeping this going, JR and Phill. Just wanted to chime in to let you know I'm still here and following with much interest, if slowly and with much dictionary and internet checking. So far it makes enough sense to... make sense. Thanks for leaving out the calculus, although if you have to take it there I can follow as long as it's well explained.

Hey Bennett,

I think it's been pretty well covered in the subsequent posts.  I actually ended up thinking about it a lot more than I expected.

The only reason for bringing calculus into it was to show how work behaves over the period of a waveform.  I decided there was no point in that, as what really matters is seeing that the dot product is the key factor in defining what work is, and what it isn't.

[Ryan Lantzy]

Phillip Graham wrote on Tue, 18 October 2005 17:45

While the difference between kW and kVA is NOT due to the resistive heating, I submit that resistive heating is the primary means for dissipating the energy that is not performing work, as represented by the power factor. Sorry for the confusion.

Whoa, whoa, whoa.... I don't think so.

Resistive heating is not a component of the difference between kW and kVA and therefore NOT a component of power factor.

Guys, correct me if I'm wrong, but the power company doesn't charge you for resistive heating when they calculate your power factor.

[Phillip_Graham]

Ryan Lantzy wrote on Tue, 18 October 2005 18:23

Phillip Graham wrote on Tue, 18 October 2005 17:45 While the difference between kW and kVA is NOT due to the resistive heating, I submit that resistive heating is the primary means for dissipating the energy that is not performing work, as represented by the power factor. Sorry for the confusion. Whoa, whoa, whoa.... I don't think so. Resistive heating is not a component of the difference between kW and kVA and therefore NOT a component of power factor.

The resistance of the wire in your circuit is a part of the equation for the total reactance of the system, and therefore plays a role in determing the overall power factor.

If you have a generator connected to a purely reactive load through leads of zero resistance, it takes no power to turn the generator, because 100% of the power is returned to the generator.  But, if you add a resistance, then you will need to supply the power dissapated by that resistance.  This is because you no longer have a purely reactive load.

If you have a power distribution system, like Langston's example, the current in the system is I, and the voltage, V.  Together they represent the apparent power.

The resistive losses in the feeder is equal to I^2*R, so the resistive losses are a function of the amount of current flowing, regardless of the actual power transferred to the load.  This means that a system with unity power factor will have the same amount of dissipation as one with power factor less than one, AS LONG AS the resistive components of both systems are equal.

Quote:

Guys, correct me if I'm wrong, but the power company doesn't charge you for resistive heating when they calculate your power factor.

I^2*R losses are a BIG deal to the power company in transmission lines.  That's why transmission lines are at such high voltages, and why large industrial facilities are required to perform their own power factor correction.  For every extra amp the power company has to supply due to non-unity power factor, that is one less amp that earned them money.

What is CRITICAL to understand is that amp did not somehow magically dissapear into a black hole on a P<1 device, but was dissipated (via I^2*R) in the transmission line!  If the transmission line didn't dissipate that energy, the power would have been perfectly returned to the generator, and the generator would have therefore needed less power to turn, and everything would be cool.

However, since the transmission lines are not superconductors, they are dissipating that power.  When the power company is charging you for power factor, they are in effect charging you for the energy wasted to resistively heat their transmission line cables.  In that sense they are charging you directly for resistive heating.

This link has some detailed calculations for anyone that is interested:
http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html

[Ryan Lantzy]

Phillip Graham wrote on Tue, 18 October 2005 19:45

The resistance of the wire in your circuit is a part of the equation for the total reactance of the system, and therefore plays a role in determing the overall power factor. If you have a generator connected to a purely reactive load through leads of zero resistance, it takes no power to turn the generator, because 100% of the power is returned to the generator.  But, if you add a resistance, then you will need to supply the power dissapated by that resistance.  This is because you no longer have a purely reactive load.

While I agree that one must size a generator appropriately for the resistance of the wire, it does NOT contribute to difference in VA to W.

[John Roberts {JR}]

Ryan Lantzy wrote on Wed, 19 October 2005 07:58

While I agree that one must size a generator appropriately for the resistance of the wire, it does NOT contribute to difference in VA to W.

While to the limits of my understanding everything stated is accurate we seem to be missing how this impacts small generator systems. There has been nothing but silence to my query about what kind of reactive PFs could we expect from significant power consuming gear other than "lamps and amps" if any?

The one issue that I can speculate on is that most power amps and electronic gear draw current in two large bites per cycle at the mains voltage peaks. While still technically resistive (in phase with voltage) these concentrated current pulses will cause larger IR wiring caused voltage sag than would be predicted from the average current draw. Another possible parameter I have no idea about is the source impedance of the generator. Probably not as stiff as Con Edison but low enough to "git er done".

While I have no specific advice I expect there needs to be some power de-rating for this phenomenon. A PFC amplifier which spreads out these current spikes will do better from a lossy generator power distribution than a conventional amplifier.

JR

[Phillip_Graham]

Ryan Lantzy wrote on Wed, 19 October 2005 08:58

Phillip Graham wrote on Tue, 18 October 2005 19:45 The resistance of the wire in your circuit is a part of the equation for the total reactance of the system, and therefore plays a role in determing the overall power factor. If you have a generator connected to a purely reactive load through leads of zero resistance, it takes no power to turn the generator, because 100% of the power is returned to the generator.  But, if you add a resistance, then you will need to supply the power dissapated by that resistance.  This is because you no longer have a purely reactive load. While I agree that one must size a generator appropriately for the resistance of the wire, it does NOT contribute to difference in VA to W.

I am not saying the generator has anything to do do with the power factor.  I could write another long explanation, but it wouldn't say anything different that what I have said already.

Bottom line is that I^2*R losses are a function of the current, and are therefore a function of the apparent power, regardless of how much power is actually transferred to the load.  The closer the power factor is to 1, the more power is transferred to the load for a given level of I^2*R losses.

I'm just some guy on the internet that you've never met, so I understand your skepticism of me, but there are plenty of textbooks available at your local library that you can double-check on this topic.  I suggest looking into those if you have the time and interest.

[ProSoundWeb Community]