Bennett Prescott wrote on Wed, 12 October 2005 00:18 |
I would also be very appreciative if someone would explain to me what Kva is actually used for, and why they don't just spec the genny in, say, watts. |
RYAN LOUDMUSIC JENKINS wrote on Wed, 12 October 2005 18:19 |
Just watch out for the dimmer noise. |
Bennett Prescott wrote on Wed, 12 October 2005 20:54 | ||
People keep saying this to me. Am I the only one with a properly grounded rig that uses equipment sans pin-1 problems? I have yet to have lampies add noise to my signal, although I've certainly seen the problem on others rigs but haven't had the time to troubleshoot to the point where I could definitively blame some SCR dimmers. Perhaps it's just that my cable runs are usually neatly contained and I'm therefore unlikely to have, say, and XLR connector sitting on top of a dimmer rack. Saved by my small size, perhaps? |
Bob Healey wrote on Wed, 12 October 2005 21:38 |
I've found I generate the most noise when I have a large excess of feeder cable and no good place to put it. |
Phill wrote on Wed, 12 October 2005 |
The power factor is the ratio of the dissipated power (in phase sum) to total power (Volt x Amps ignoring phase). |
Rob Burgess wrote on Wed, 12 October 2005 22:37 | ||
You don't figure 8 the excess? -- Rob |
Langston Holland wrote on Thu, 13 October 2005 10:29 |
IOW does one have to use one of those fancy and expensive power meters to determine when the genny is about to fuss? |
Bennett Prescott wrote on Wed, 12 October 2005 18:06 |
I follow you until about halfway through there when you start talking about "stored" current and bouncing. Any chance I could get you to elaborate, and pretend I'm an idiot? |
Phillip Graham wrote on Mon, 17 October 2005 13:34 |
Hey Bennett, I'll give it another shot at explanation. First of all, it's stored energy, not current. Current is always pushed in the front and out the back, so to speak. Let's think about what electricity in a cable really consists of. First, and foremost, it consists of a propagating electric field set up by the rotation of the generator windings through the magnetic field (Wikipedia Faraday's Law of Induction). This field is represented by a net displacement of charges (i.e. electrons). The electric field propagates very rapidly (approx. 0.8 C depending on the dielectric constant of the cable) The second thing electricity consists of is the drift velocity of electrons in the cable, due to the net electric field causing a slow movement of the average position of an electron in the cable. The drift velocity is slow, only a few cm/second. Some of the energy the cable dissipates due to resistance contributes to causing the drift velocity. The net drift velocity for a pure sine signal, of course, is zero. Next we must consider the definition of work, because the work-energy theorem says that energy is the ability to do work. Work=Force (dot product) displacment, or in integral form, Now, the rubber meets the road for this situation in the dot product. Linear algebra tells that the dot product equals Force*displacement*cosine(theta) where theta is the angle between the Force and displacment vectors. (Please no one jump on me for not rigoursly defining the vector magnitudes) So, why does this vector math matter? Cosine of 0degrees is one, and cosine of 90degrees is zero. That means that the amount of work done is function of the angle between the direction of the force and the direction of the displacement. That is why no work is done to keep an object moving in a circle, because the centripetal force is normal to the displacment at every point. So, in your generator, you have some current that flows in phase (i.e. cos(theta)=0) with the voltage, and that in phase current is doing WORK. You also have an out of phase (i.e. cos(theta)=90) current component for power factors other than 1, and that component's energy cannot, by definition, do any work. So, for half of the AC cycle. your generator is creating an electric field that is trying to push electrons towards the stage. In spite of that, some of the electrons are not moving in sync with this electric field, but are doing their own thing on a phase lag. These electrons are not following your generators gentle reminders to head towards the stage, and therefore part of the electric field is "wasted" on them because they aren't doing any work for you at the stage end. Instead these electrons are content to bounce back and forth to the beat of their own drum, and therefore the electric field they produce does the same. Since it lags by exactly 90 degrees it's not doing any work. In a perfect system this energy would just be stored, but in a real system there are dissipations due to resisitive losses in the source and load. Those resistive losses (see my blurb about the drift velocity) result in joule heating of the components involved (generator windings, feeder, etc.) but this energy cannot be tapped to perform any work because of its phase relationship to the generators electric field. This turned out rather longer, and less clear, than I had hoped. Let me know if you need some clarification. |
JR [John Roberts |
wrote on Mon, 17 October 2005 15:00] Oh yeah... that clears it up... |
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While no doubt accurate, is that to the point? Perhaps a simpler answer is to ignore all this dot/scaler/mumbo_jumbo. |
Phillip Graham wrote on Mon, 17 October 2005 15:42 |
The dot product is fundmental to understanding the definition of work, unfortunately. I was looking to provide some insight into the physics of what is going on. It just didn't come out clear. I should probably provide a take home message for the parent post: 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. |
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 |
Langston Holland wrote on Tue, 18 October 2005 10:08 | ||
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? |
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. |
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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. |
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. |
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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. |
Langston Holland wrote on Tue, 18 October 2005 10:08 | ||
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? |
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. |
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I didn't say otherwise. |
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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. |
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. |
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. |
Ryan Lantzy wrote on Tue, 18 October 2005 18:23 | ||
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. |
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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 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. |
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. |
Ryan Lantzy wrote on Wed, 19 October 2005 08:58 | ||
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. |
JR [John Roberts |
wrote on Wed, 19 October 2005 10:25] 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? |
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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". |
Shane Ervin wrote on Wed, 19 October 2005 11:24 |
But to properly size a rented DG, isn't it clouding the picture to bring Hydro accounting into the analytical framework for selecting the nameplate rating of a standalone machine? There are better ways to arrive at Theory-Lite aimed at pragmatic application to Bennett's show-biz operations. |
Peter wrote on Wed, 19 October 2005 20:53 |
Absolutely! One of the big issues with things like amplifiers is related to the effect the DC power supplies has on the supply. The bridge rectifier / filter cap type supply tends only to draw current on the peak bits of the wave - to top up the filter caps so to speak --- it draws current only when the supply voltage is above the voltage stored in the filter cap (multiplied by the turns ratio of the TX etc.). This causes distortion of the supply, flattening of the tops of the waveform. Its one of the reasons for the push towards switch mode PSUs. There were millions and millions of computers and other similar equipment doing just that when they had those old power supplies. I believe it was becoming quite an issue for power authorities around the world. Using a single generator I believe its possible to see a power factor as bad as 0.5 –0.6 in some situations when the load is highly inductive and unlike a supply utility you will not be able to add any correction. (capacitor/ inductor banks/ rotating synchronous condensers/ controllable static var sources or var generators) I don’t know what to expect for PA equipment, but its generally combined with a lot of lighting equipment, which is probably the dominant part of the load with a power factor close to 1. As a guess if you use 0.8 for your cals and conservatively estimate your load I would not expect too many problems. This is not a bad link for everyone that explains this stuff - http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html Peter |