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Author Topic: Specifying Generators in Kva  (Read 13818 times)

Don Lanier

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Re: Specifying Generators in Kva
« Reply #30 on: October 16, 2005, 06:19:35 pm »

I have to chime in here and say that the gasoline gennys are good for small non reactive loads such as a small amplified PA, Small meaning a powered head or such, Small lighting as commonly used for commercial TV shoots or Photography, While there have been great improvements in the Honda E Line they just dont cut the mustard for PRO audio/lighting use. The smaller diesel 20 Kw gensets are really economical and much more stable for small county fairs etc. The diesel genset is by far your best bet for stable and clean voltage/frequency, they also will run all nite and half the next day before needing fuel. Refueling a hot gas generator is a 50/50 shot at a fire or worse. MQ Power and Cummins both have silenced,electronically governed gensets in trailer enclosures that have either lug or cam outputs ready to go, and you can pull these smaller units with a pickup truck.

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.
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Bennett Prescott

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Re: Specifying Generators in Kva
« Reply #31 on: October 16, 2005, 10:03:07 pm »

Just to update y'all, the show went absolutely beautifully. I've got these gigs from now until eternity, and the band even tipped me afterwards ($50!) saying "Everyone else mangles our mix".

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.
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RYAN LOUDMUSIC JENKINS

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Re: Specifying Generators in Kva
« Reply #32 on: October 17, 2005, 10:44:27 am »

Congratulations, I told you that you would be just fine with that generator.
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Phillip_Graham

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Re: electricity, energy, work, etc.
« Reply #33 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.
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John Roberts {JR}

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Re: electricity, energy, work, etc.
« Reply #34 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... Rolling Eyes

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
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Phillip_Graham

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Re: electricity, energy, work, etc.
« Reply #35 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... Rolling Eyes



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!  Shocked

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.
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John Roberts {JR}

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Re: electricity, energy, work, etc.
« Reply #36 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

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Bennett Prescott

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Re: electricity, energy, work, etc.
« Reply #37 on: October 17, 2005, 11:07:19 pm »

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.
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Langston Holland

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Re: electricity, energy, work, etc.
« Reply #38 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?
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Ryan Lantzy

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Re: electricity, energy, work, etc.
« Reply #39 on: October 18, 2005, 01:00:03 pm »

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?


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.
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Re: electricity, energy, work, etc.
« Reply #39 on: October 18, 2005, 01:00:03 pm »


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