Post by: Bennett Prescott on October 12, 2005, 12:16:43 AM
This show, however, there's lighting, and the lighting guy wants to know what kind of cans to bring and how many dimmers he can run. I told him I could only give him a rough estimate, and to plan on having xKw available, where x is about what I think I'm safe pulling off the generator divided by three (one third for me, one third for him, one third for the blowers on some space heaters).
Rough calculations tell me that I should have about 140 amps available from this generator when it's running in single phase mode, 250v. Does that sound right? The same calculations (with an appropriation for the extra phase) tell me I should have 100 amps available in three phase mode, 208v.
So the question is, am I right? Do I really have all sorts of juice? Also, is the spec on this generator continuous or peak? If peak (or in any case) how much should I de-rate the generator in order to get a realistic continuous (i.e. resistive) power capacity so I can tell the lighting guy what he can get away with? Assume I'll be connected over 4/0, so minimal voltage drop. My system probably averages 20a, seeing as it's an audio load and there's no backline or huge amp racks.
Attached is the Excel spreadsheet I roughed this out on.
Post by: Bennett Prescott on October 12, 2005, 12:18:41 AM
Post by: Mat Goebel on October 12, 2005, 12:24:00 AM
I think VA is just the term more widely used by electricians and the like, for one reason or another.
Update: I should have looked at the spreadsheet first. It seems that you're spot on.
Post by: Rob Timmerman on October 12, 2005, 12:36:37 AM
kW = kVA * pf
Without going into too much detail, power factor is the ratio between the amount of power that is actually dissipated and the apparent amount of power dissipated. A purely resistive load has a power factor of 1. Because of how electronic dimmers and switching power supplies work, the power factor of your load will be significantly less than 1.
A power factor of between .65 and .85 is probably appropriate for your application. Personally, I'd err to the low side (.65) to be safe. So this means that you have about 24kW available, total.
Post by: Jason Ellis on October 12, 2005, 12:55:31 AM
My office has a 25KW split-phase which is good for 100A or so of 240V (200A of 120V). Even with A/C on we barely break 25% of capacity (I like having a little extra room
)Just for fun I'll estimate lighting cans...lets say Mr. Squint has 10 20A dimmer packs he's running 4 lamps apiece on (no movers and shakers) that would give him 40 lamps for the show...of course I may be way off in my math as well...don't the lights have a pretty good PF? Not that you two should be pushing it to the limit of the genny, but you should still have a good amount of genny headroom with the above layout...
Post by: Ken Freeman on October 12, 2005, 12:56:14 AM
...The other piece of this to keep in mind is that a generator will have a lot easier time going from 40% to 60% percent of capacity than from 5% to 25%. If your lighting tech is fond of blackouts and fast changes, you may have a tough day. Also beware that many dimmers kick a lot of noise back on to the A/C line which your audio system may well amplify thereby creating a nice buzz every time the lights fade up or down. I suggest a second plant for the lighting guys unless they promise not to make any big changes until you make sure it will not mess with the audio system.
Have a good show!
Ken
Post by: Philip Roberts on October 12, 2005, 01:28:27 AM
As Rob and Ken said is exactly correct W=VA*pf,
Wikipedia has a pretty good article on Power Factor.
Lights are almost the perfect example of resistive loads, however I know dimmer mess this up royally, though I don't know in which direction. My understanding is the DC power supplies in audio gear also pull the power factor away from 1.
We saw an example of this with some of QSC's big amps in the PL series if I remember right the PL x.x-PFC had what ever additional circuitry what was needed to get the power factor back up closer to 1, I believe that is was EU regs that required it for items with such a large current draw.
Hopefully there's more light that mud here.
[edit: trying to make the link work right]
Post by: Jim Brooks on October 12, 2005, 09:52:03 AM
I run a 13,000 watt sound system and a 20,000 watt light system on a 25KW (single phase) Wisperwatt generator on a regular basis.
Two 20 amp circuits for the amp rack, one 20 amp circuit for the remaining audio.
All other circuits are for lights.
I kick the crap out of the generator but it works.
I do have a voltage regulator on all of the sensitive equipment just in case I have a brown out.
If the light system was at full for any long length of time,
I'd be tripping breakers.
I know I'm pushing the envelope.
Just the way it it.
Jim Brooks
casperband.com
Post by: Bob Healey on October 12, 2005, 11:32:42 AM
I'm going to need somewhere between 35A and 65A for the load based on past experience/breaker ratings on the gear, though I've never worked with a generator before.
Post by: Bennett Prescott on October 12, 2005, 12:24:53 PM
In any case, it looks like if my equipment draws 20a at 240 and Bob's draws 40a or so (that lets him have a 10Kw rig, and he's only asking for 6), we should still have plenty of genny "headroom" left over... more than 50%, in fact, which allows for some slack in that I have no idea what those heaters are drawing, but it can't be enormous since they can't be purely resistive (well, in a world without crazy people, but I assume that's not the case here).
[edit: posted updated Excel spreadsheet, although there's something wrong with my three-phase amperes calculation]
Post by: Phillip_Graham on October 12, 2005, 12:31:51 PM
| 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. |
Power = voltage x current
KW is the same as KVA for purely resistive loads.
KVA is designed to be a reminder of the fact that you really need to do the phasor sum of the voltage and current, so break out your trig.
The component of the power where the voltage and current are in phase is effectively disspated in the load, performing work over time (power=energy/time or work/time)
The out of phase component is effectively "stored" in that it bounces back and forth to the load with each alternating cyle.
This return of energy is what is difficult for generators to handle, and the reason for power factor correction.
The power factor is the ratio of the dissipated power (in phase sum) to total power (Volt x Amps ignoring phase).
Post by: Bob Healey on October 12, 2005, 12:37:10 PM
A/C power is a complex number of the form r + c*j. A purely resistive load is a real number. Power factor is the real power over the absolute value of the actual power. Watts divided by power factor gives you KVa which is the actual power a generator can put out.
Its been a few years since I took a circuits course, but here are the notes I am working off of:
http://www.ecse.rpi.edu/Courses/S03/ECSE-2010/lecture_materi als/Class%20slides%20(pdf)/class25S03.pdf
Edited to fix URL
Post by: Bennett Prescott on October 12, 2005, 06:06:05 PM
Post by: RYAN LOUDMUSIC JENKINS on October 12, 2005, 06:19:01 PM
I couldn't find the Power Factor for the GE generators but FYI the MQ Power generators that I spec for all of my events have a Power Factor of .8 The 25KVA gives 20kw output prime output and 22kw standby output. I can't tell you exacty what "Standby" refers to but I go by the prime output rating. I use them single phase 120/240 volts. I have a friend that owns his own and he runs them as 3 phase. In most instances I just use the 50 amp twistlock connectors and it works perfectly for me.
Hope that helps. If I am not mistaken your rig is about 11,000 Watts F.O.H. and 5000 watts Monitors. you could run you whole rig and backline off of one 50 amp connector and give the other to the light guy if two are available. I assume your rig probably doesn't use any more than about 60 amps at 120 Volts? you'll be just fine. Just watch out for the dimmer noise.
Post by: Tim Duffin on October 12, 2005, 07:58:35 PM
hope this helps
Tim
Post by: Bennett Prescott on October 12, 2005, 08:54:53 PM
| RYAN LOUDMUSIC JENKINS wrote on Wed, 12 October 2005 18:19 |
Just watch out for the dimmer noise. |
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?
Post by: Bob Healey on October 12, 2005, 09:38:23 PM
Post by: Rob Burgess on October 12, 2005, 10:36:19 PM
| 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? |
Perhaps
Some dimmers add gack to the AC, some not so much. Some times I'm able to share power with the squints and other times It's been so bad that I had to rewire using wall outlets. This is only a problem on medium sized gigs: small gigs I can run off the wall easily and large gigs have a seperate feed. It's the medium sized gigs that I worry about.--
Rob
Post by: Rob Burgess on October 12, 2005, 10:37:19 PM
| 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. |
You don't figure 8 the excess?
--
Rob
Post by: Jim Cutshall on October 12, 2005, 10:51:28 PM
I've run audio and lighting off the same genset many times for fairly large festivals - the key is to make sure the gen is large enough to handle the surges of the lights "flashing". It will also cause the Hz to flux so if you have any old organs (B3) it will go out of tune from the Hz not being correct but I've never had any noise from a properly setup rig.
FWIW,
Jim
Post by: RYAN LOUDMUSIC JENKINS on October 13, 2005, 12:28:12 AM
Another thing that helps keep dimmer noise out of the audio is to have them full on or full off, no actual "Dimming." That what I do and I never hear any buzz from them. Keep in mind that I only do very basic lighting myself and refer out anything beyond what I can do. That may be what the light guys have been doing at your events and it really helps. Also like you said about neat, organized cables, that helps big time too. If everything crosses at 90' angles whenever possible thats a big help.
Post by: Langston Holland on October 13, 2005, 10:29:04 AM
| 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). |
So... Please fix anything broken in the following. :)
1. Assuming the use of Bennett's 36kVA genny in single phase mode, this means there should be 150 amps at 120v available on each of the two hot legs into a purely resistive load.
2. Real world concert equipment places a very non-resistive load on the power source, a product of reactive components and distortion, which reduces the useable amount of power available to run the stuff you plug into the distro.
Question:
Can current and voltage meters be used to monitor the system such that a reading approaching 150 amps (while maintaining a reasonable RMS voltage) on each leg means that the actual limit of the genny is being approached? Or will the current reading be reduced by the mysterious power factor present in the circuit, thus giving one a false sense of security about the "headroom" available to power additional loads? IOW does one have to use one of those fancy and expensive power meters to determine when the genny is about to fuss?
Post by: RYAN LOUDMUSIC JENKINS on October 13, 2005, 11:50:17 AM
Here is a website that sells generators and has almost all of the generator specs you will ever need.
http://www.portable-electric-power-generators.com/
I refer to this site all the time amd it saves me a lot of time.
Post by: Bob Healey on October 13, 2005, 01:00:54 PM
| Rob Burgess wrote on Wed, 12 October 2005 22:37 | ||
You don't figure 8 the excess? -- Rob |
50 feet of cam x 5 in 10 sq ft of space. Figure-eighted, but doesn't do much.
Post by: Lee Patzius on October 13, 2005, 01:53:54 PM
| 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? |
Langston,
I just finished installing a bunch of Hitachi 3 phase variable frequency drive inverters, with Simpson true rms ammeter instruments, along with current toroids on a skid mounted process unit here at the plant, and spend another couple of days doing power studies.
I noticed quite a few things that threw me off at first... But I won't go into all the testing procedures, nor the inverter's V/f constant torque output and operation etc., but the testing was very interesting.
We had four ammeters to compare by, a Simpson true RMS, an analog standard Amprobe Clamp around, a freshly calibrated NIST Traceable Amprobe digital True RMS AC/DC clamp around meter, and an analog shunt type ammeter in series.
Even with variable freq drives and horrible motor harmonics (which I reduced with inductive line reactors installed) the Amps was Amps with the exception of one meter. We took the motors to full mechanical load, at various speeds and as always, produced accurate ammeter results.
EXCEPT for the analog shunt type ammeter. It tracked nicely with the others, until the motor was driven beyond it's full load amperage rating. In overcurrent conditions, the analog meter read HIGHER than the true RMS types. The true rms types locked in at 13.xx amps while the shunt type went up to 18 amps.
Anyway, what I wanted to point out was, the generator has to produce the full kVA regardless of the power factor. According to our studies, a clamp around Amprobe will read true amps rms, I supposed in reference to the true Watts, so it may be affected by kVARs. A shunt type analog ammeter did read higher in overload conditions for us though.
Post by: Marty McCann on October 13, 2005, 03:41:26 PM
Recently I had occasion to use a 5.5 KVA after Katrina. I had 7 evacuee's from New Orleans staying with me in a 1 Bathroom household. Along with the pet's, Katrina raised my total household to 9 people, 5 dogs, 2 fat guinea pigs, and one squawking parrot. Since my water comes from a deep well, and we had no electricity for 8 or 9 days, after the third day of using the above ground pool water to flush the commode, I went to Lowe's and bought a generator rated at 5,550 Watts.
I had no master breaker to disconnect, so I removed the Power companies meter and wired into the 240 VAC buss directly. When I powered up the gen set and turned on a minimum of lights and such, I was drawing about 4.5 Amps on one leg and 5.2 on another. I quickly found out how women can not be trusted to be frugal when it comes to power allocation. After they attempted to plug in curling irons, hair dryers, etc., I had to be the hard nosed EE, and insist on one light per room, refrigerator, and a television. I would turn on the well pump and it would draw 9.5 amps of current per leg (240 VAC). I would pump enough water to pressurize the storage tank, which was then good for about three flushes of the commode. (I have a rental house that gets water from the same well). When I would heat up the hot water, I would draw about 14 amps per leg. But I quickly realized (from the dimming lights) that even though I was below the max draw of the generator, that I was loosing electrical pressure in the form of voltage drop. Since I have a collection of multimeter's and Clamp Amp-meters, I began a routine of power consumption maintainence. The problem is, if you allow too much of a voltage drop, you can actually burn out your electrical motors. With the Hot water heater elements at max draw and my well pump running, the voltage dropped below 98 VAC, and current draw was 17.5 Amp on one leg and 18.2 Amps on the other. 98 VAC is not good for a deep well pump more than 360 Feet below surface. Think about this if it was 98 Volts at the panel, and 100 Ft to the well and another 360+ feet to the pump motot, what could my voltage have been at the pump?
After a few days, since my friends from New Orleans brought very few clothes, we had to turn on the washing machine, which would draw nearly 11 Amps more on a single leg. Even though this was considerably below the max generator capability, the generator would blow the breaker driving the washing machine in 12 - 14 minutes before a complete washing cycle ended. It was then that I learned to use my above ground pool's pump motor to balance the two legs from the generator, when I did this I could run the washing machine as long as I balanced the line with the pump motor.
At one point kinda for grins, I turned on the clothes dryer and I was impressed when the generator produced 23.4 Amps per leg (however only at about 94 Volts), after about 6 - 8 minutes the breakers would blow. With very little current draw, the gen set produced about 124 Volts.
So I have come to the conclusion that those of us that use generators for audio purposes would still do well to try to at least maintain a reasonably balanced current draw between the two legs of a single phase 240 VAC system. I would think that this would also apply to a three phase system.
I wish I would have kept a more accurate log of some the the things that I experienced with this emergency generator. I did learn that I need to redistribute some of my circuits, as I am drawing more current thru my Red leg than my black when I have a lot of things running at the same time. I have four different electrical panel boxes in my house to to various additions that were made over a 26 year period.
I also learned not to complain about the 5 or 6 dollars a day that I pay to the electric company for power. The generator cost me $780.00, but the electricity was 35 - 40 dollars a day, and it was only $2.29 a gal at the time.
marty
Post by: RYAN LOUDMUSIC JENKINS on October 13, 2005, 04:03:12 PM
I am sure I speak for all of us here that we are proud of you for steping up to the plate and providing a roof for the extra people during the aftermath of the huricane. Spending $800 on a gas generator was a good investment for your circumstances.
Post by: Michael E. on October 13, 2005, 06:37:33 PM
Post by: Tim McCulloch on October 14, 2005, 12:04:16 PM
Tim Mc
Post by: Don Lanier on October 16, 2005, 06:02:20 PM
Great little tool for calculating these mysterious numbers to you non sparky types
Don Lanier
http://www.pearlproaudio.com
Post by: Don Lanier on October 16, 2005, 06:19:35 PM
For my money Ill stay with the diesel. The ability to adjust frequency and voltage especially to adjust output is a big advantage over the gasoline brethren...Check my website for info on the MQ or Cummins lines.
http://www.pearlproaudio.com
Don
Post by: Bennett Prescott on October 16, 2005, 10:03:07 PM
Power-wise, I handed the genny/tent rental folks tails, said "I need 20 amps at 240" and half an hour later had the other end of my feeder energised (at 150v, but that was quickly rectified). Power was rock solid throughout the gig, even with 12,000w of "suprise" house lighting. I copied down the chart on the side of the genny that says how much you can expect to actually draw off the thing, which turns out to be about 27Kw.
Post by: RYAN LOUDMUSIC JENKINS on October 17, 2005, 10:44:27 AM
Post by: Phillip_Graham on October 17, 2005, 02:34:28 PM
| 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? |
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.
Post by: John Roberts {JR} on October 17, 2005, 03:00:39 PM
| 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. |
Oh yeah... that clears it up...
While no doubt accurate, is that to the point? Perhaps a simpler answer is to ignore all this dot/scaler/mumbo_jumbo. Audio amplifiers, even those not power factor corrected, will technically look resistive since the power supply rectifier diodes only conduct at the top of the mains voltage waveform so the current while not sinusoidal will clearly be in phase with the voltage waveform. Rather than worrying about reactive components bouncing energy back and forth, the primary issue as I see it, is perhaps some amount of de-rating for the fact the current is all concentrated at the peak of the waveform rather than spread across the entire waveform like a true PFC amp (or perfect resistance).
I regret I do not have practical advice to offer on a precise de-rating factor but having a safety margin is usually a good idea anyhow.
JR
Post by: Phillip_Graham on October 17, 2005, 04:42:31 PM
| JR [John Roberts |
wrote on Mon, 17 October 2005 15:00] Oh yeah... that clears it up... |
Yeah, it reads awful, though it's really quite elegant if you understand the vectors. The original version of the post had calculus in it!
| Quote: |
While no doubt accurate, is that to the point? Perhaps a simpler answer is to ignore all this dot/scaler/mumbo_jumbo. |
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.
Post by: John Roberts {JR} on October 17, 2005, 05:21:09 PM
| 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. |
Now we're getting somewhere. To complete this thought what kind of loads is he likely to encounter in his PA gig that require such de-rating for? While I expect amps and lamps to look resistive, I'm no expert on this subject. The loudspeakers are clearly reactive but won't have an impact beyond the power amp output stage.
JR
Post by: Bennett Prescott on October 17, 2005, 11:07:19 PM
Post by: Langston Holland on October 18, 2005, 10:08:37 AM
| 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?
Post by: Ryan Lantzy on October 18, 2005, 01:00:03 PM
| 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? |
Yeah, I agree with you Langston. I think the resistive heating is there all the time. No matter what the load.
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.
Post by: Phillip_Graham on October 18, 2005, 01:31:57 PM
| 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.
Post by: Ryan Lantzy on October 18, 2005, 04:04:28 PM
| 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.
Post by: Phillip_Graham on October 18, 2005, 05:05:47 PM
| 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? |
Hey Langston, this is true, as long as your wattmeter is measuring the true power, and not fudging in some way.
Post by: Phillip_Graham on October 18, 2005, 05:45:25 PM
| 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.
Post by: Phillip_Graham on October 18, 2005, 05:56:29 PM
| 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.
Post by: Ryan Lantzy on October 18, 2005, 06:23:59 PM
| 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.
Post by: Phillip_Graham on October 18, 2005, 07:45:21 PM
| 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. |
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
Post by: Ryan Lantzy on October 19, 2005, 08:58:03 AM
| 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.
Post by: John Roberts {JR} on October 19, 2005, 10:25:14 AM
| 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
Post by: Phillip_Graham on October 19, 2005, 10:38:06 AM
| 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. |
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.
Post by: Shane Ervin on October 19, 2005, 11:24:33 AM
Hi Phill,
Great topic I've been watching unfold. Kudos to Phill as usual for kind and thoughtful explanations. Kudos to Ryan for applying healthy scrutiny.
As an electrical engineer with a background in power distribution and who paid his way through school in the engine room of oil tankers (that have diesel gennies), I thought I'd try to mediate (wrong word?).
I'm very much a close enough for rock'n'roll kind of guy, though, and that usually makes me want to skim over lower level details and retain just the broad strokes necessary for good, safe operation of gear so the beer can continue to be sold and audience members get laid after the show.
Power Factor, as a concept so named, can serve the very purpose of simplifying things for guys like me who just want to have an easy, yet solid engineering basis for safe and effective operation of distribution systems and the rotary machines feeding them (and loading them). As such, and in this light, the good advice might be to learn the material in the earlier chapters of the EE textbook prior to delving (publicly at least) into the subtler details of subsequent chapters.
There's no harm in bringing calculus or dot product into the explanations on a forum like this, especially when it's Phill doing the talking and he simultaneously offers "theory lite" in follow-on sentences as he does for broader reader appeal (and accessibility I guess, too).
Equally there's no harm in responding with questions aimed at steering the thread towards basic application of theory-lite for the rock'n'roll purposes we're here for.
So all is well.
OK, then. We're talking real power, apparent power, reactive power, power company pricing policies, genny rentals, cable sizes, resistive losses, mechanical torque applied to the prime mover's crankshaft to keep the thing spinning at the desired speed for 60 Hz or so, and uh... what was the original poster's question again...? ( Hi Bennett! )
Ryan, Phill's just asking to you to consider that the prime mover only needs to overcome rotation friction and supply sufficient torque for real power to remain at the desired RPM. He describes how purely reactive loads can be fed from a DG up to the nameplate rating with surprisingly little mechanical torque, since the only real power loading down the DG comes from I2R losses in the connecting cable in such a case.
He's also correct in describing the rationale power companies use in establishing the terms and conditions of industrial-grade service to factories with lots of squirrel cage motors. Power factor most certainly is a concern for precisely the reasons he cites (and also explains why synchronous capacitors are used local to the inductively reactive load of a typical industrial plant).
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.
Post by: Phillip_Graham on October 19, 2005, 01:41:53 PM
| 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? |
Yeah JR, I have no idea of the practical power factor numbers here. Langston?
| Quote: |
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". |
I think you are exactly right on the real issues here. Wish Gene Pink was around, he might know something on this practically.
Lee Patzius, do you know anything about this?
So yeah, what JR said in terms of the relevant issues for small PD systems.
Post by: Phillip_Graham on October 19, 2005, 01:49:01 PM
| 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. |
Hey Shane, I totally agree with you. I guess I went down the power factor trail because it is something I understand and could talk about in a half-way intelligent manner.
I think JR has brought up the salient points for the live sound setting. I don't know the answers. I would be very interested to hear your experiences on this from working with those ship generators.
This thread is now officially passed to those with practical experience on de-rating portable gennies
Post by: Iain_Macdonald on October 19, 2005, 05:37:21 PM
I thought 0.85 was the standard figure used.
For a USA viewpoint http://www.abrconsulting.com/Conversions/elec-con.htm
Iain
Post by: Peter Morris on October 19, 2005, 09:53:03 PM
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
Post by: John Roberts {JR} on October 19, 2005, 11:46:19 PM
| 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 |
The only thing I would expand upon is that most amplifiers with switch mode PS use a standard FW bridge in their front end so they too will present concentrated current pulses. In fact they will be slightly worse because their charging current is not limited by a transformer in series. Regarding consumer gear this is indeed an issue for marginal utility infrastructure and Europe is far ahead of the US in mandating PF consumer gear. While there are only a handful of PFC audio power amps in production, every serious amp manufacturer developed the technology to avoid getting locked out of the European marketplace. The effectivity date for the rule kept getting postponed and I am no longer in a loop that cares so I don't know what the current status is but expect the benefit to European power distribution systems to full (professional power amp) implementation would have been minimal while impact on small Euro power amp companies would have been significant.
JR