PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2020-12-04, 05:32:35
News: Check out the Benches; a place for people to moderate their own thread and document their builds and data.
If you would like your own Bench, please PM an Admin.
Most Benches are visible only to members.

Pages: [1] 2
Author Topic: Average Power Computation - MEAN or RMS? - Participant Posts Only  (Read 20322 times)
Group: Guest
"What method of computation should be applied to a Power Waveform Trace to obtain the Average Power of the sample set under consideration?"



Lawrence Tseung stands on the side of RMS while Poynt99 stands on the side MEAN

Who are the participants?

Lawrence C.N. Tseung:
Born in Hong Kong and recieved secondary schooling via the now famous Tiger Mom method where every task had to be done to the very best and every mathematics problem in the textbook was completed. He holds a B.Sc in Physics and a M.Sc in Aeronautics from Universities in England. He authored "Kinetic Theory of Gases in Motion".  His work history includes working for Digital Equipment Corporation in both England and USA and his own company which he reports failed. He holds the distinction of titles such as Instructor, Consultant and International Manager. He was also awarded a patent for Many-to-Many data communications, one of three which were instrumental in the development of the Internet. The other two being awarded to AT&T and IBM. He is the unfortunate recipient of two strokes and an early retirement but credits his recovery to blessings from the Lord and an experimental drug. From 2002 onward he has focused attention to Alternative Energy Research. He has worked on several projects that have been presented or demonstrated at such places as Beijing Aeronatutical University and Inno Tech Expo 2009 - Hong Kong. His projects are centered on his concept of Lead-Out/Bring-In Energy relationships.



These are the only participants in this debate and all others should post their comments and observations in the Audience thread.

« Last Edit: 2011-02-04, 02:53:10 by Harvey »
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
My opening post in this debate (assuming that Lawrence has accepted it) will refer to two posts previously made by me.

This issue of debate makes no reference to any particular wave form, therefore the sinusoidals used, and my examples given, are offered as supporting evidence for my stance that a MEAN computation is the only type that can produce the correct result for the average power value of an instantaneous power trace.

The first post illustrates the average power in a very simple zero-phase 60Hz AC circuit as measured with digital meters. The resulting average power is 1W, and is used as the reference power for all other resulting computations performed by an oscilloscope.

It is clearly shown that taking the MEAN of the power trace produces the correct result in every case, while taking the RMS of the power trace produces the incorrect result in every case. The degree of error resulting from a RMS computation of the power trace is magnified immensely when the power trace exhibits excursions both above and below the 0-Watt reference line.

A power trace which takes a sinusoidal wave form that deviates equally above and below the 0-Watt line for example, exhibits or contains no net power whatsoever. In this case, there is an equal amount of power going back in to the source (battery for eg.) as there is coming out. It is possible to demonstrate such phenomenon in real life, although there will be slight losses in the process.

Taking the RMS of the above example produces a result being; 0.707 * Pp (peak power), in this case resulting in a value of 0.707W. Clearly a computation that produces a net result of 0.707W (or any wattage other than 0) can not be correct, when there is no net average power contained in the power trace. See this screen capture of the results.

The second post illustrates what happens when there is a DC offset in the voltage and current traces used to obtain the power trace. The discrepancy between the correct MEAN computation, and incorrect RMS computation is widened.

.99
   
Hero Member
*****

Posts: 931
Let me start my side in congratulating that Poynt99 is absolutely correct in the special case when the Instantaneous Power Curve is all positive.

The exact area integration method - integrating the Instantaneous Power Values against time will yield the CORRECT energy over that period.

The following diagrams show the concept clearly.

But that only applies to the special case!

I shall discuss the other cases later.  It is the Chinese New Year.  My friends and I have celebrations.
 :)
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
A friendly reminder to Lawrence and Harvey:

So far, and assuming what is to follow will be along the same train of thought, it does not appear that you are addressing the topic of the debate. You are speaking of integration (a different topic of discussion), and no mention at all is made of average power.

.99
   
Group: Guest
As a reminder to both participants, the topic of this debate is centered on "Average Power Computation"

It may be a good idea for one or both to post an industry accepted definition for 'Average Power' as a basis for further discussion

Moderator
   
Hero Member
*****

Posts: 931
Let us examine another Instantaneous Power Curve 2

In this Instantaneous Power Curve 2, we can see that the curve cuts across the Zero axis.  There will be a positive area and a negative area.  Such a curve can easily be produced with any LCR circuit when the Instantaneous Voltage and the Instantaneous Current are out of Phase.

The FLEET devices under verification show such characteristics.  FLEET devices definitely have significant inductance (L) in the torord and some capacitance (C). 

The correct area integration should yield the SUM of A1 plus A2.  Thus, for accurate results in this case, the area integration must be applied.  The use of the Mean Power value will give the DIFFERENCE of A1 and A2.  That is totally wrong!

If we want a quick comparison of different FLEET prototypes, we can use the RMS Power Value derived from the Instantaneous Power Curve.  This is closer to the true value than that provided by the Mean Power Value.  The RMS value is not correct but closer to the correct value. 

Thus the graph shown by PhysicsProf on reply 350 showing the RMS value is more accurate and meaningful than a Mean value.

*** My understanding in this case is – average is the same as mean.  The sign of the numbers will be taken into account.  For example, the average of 5 and -2 is 1.5 = (5-2)/2


More to come later.
   
Group: Guest
Lawrence,

Could you please substantiate your statement "The correct area integration should yield the SUM of A1 plus A2" and clearly show the relationship between this and "Average Power".


Note to both participants:
References to specific devices should be avoided except where they are used as accepted examples in support of a given stance. It is hoped that the knowledge shared in this debate can be applied to any device under consideration and not be restricted to any particular device.

   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
As a reminder to both participants, the topic of this debate is centered on "Average Power Computation"

It may be a good idea for one or both to post an industry accepted definition for 'Average Power' as a basis for further discussion

Moderator

Assuming periodic wave forms, the average power is the effective or net heating power in one complete cycle.

Ham Radio operator Roy Lewallen said it best when he wrote the following (I've posted this document a number of times, and it supports my position in this debate):


RMS Power

The RMS value of a power waveform can be calculated just like the RMS value
of any other waveform, although it doesn’t represent heating power or any other
useful quantity. I’ll do it just to show how it would be done.

Recall that the power was:



Squaring it, we get:


The average of the square of the power is clearly 50 watts, and the square root
of that is 7.0711. . . watts. We found earlier that the equivalent heating power of
our circuit – the average power -- was 5 watts, not 7. The RMS value of power
is not the equivalent heating power and, in fact, it doesn’t represent any
useful physical quantity.
The RMS and average values of nearly all waveforms
are different. A notable exception is a steady DC waveform (of constant value),
for which the average, RMS, and peak values are all the same.


It should be noted that the term “RMS power” is (mis)used in the consumer audio
industry. In that context, it means the average power when reproducing a single
tone, but it’s not actually the RMS value of the power.


Summary

I've shown that:
* The equivalent heating power of a waveform is the average power.
* The RMS power is different than the average power, and therefore isn’t the
equivalent heating power. In fact, the RMS value of the power doesn’t
represent anything useful.
* The RMS values of voltage and current are useful because they can be used
to calculate the average power.

For those who are interested, mathematical derivations are readily available and
can be found on various web sites and in textbooks. What I’ve tried to do here is
to present the basic concepts with as little mathematics as possible. Questions
are welcome – send them to w7el@eznec.com. And I’d appreciate very much if
anyone finding an error would notify me so it can be corrected.

Roy Lewallen, W7EL

Modified 2004-11-18.
Thanks to Steve Nosko, K9DCI, for suggestions.
Modified 2009-09-11 (typo correction): Thanks to Blaine Newman.


For the full document, go here.

.99
« Last Edit: 2011-02-05, 03:59:39 by poynt99 »
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
Let us examine another Instantaneous Power Curve 2

The correct area integration should yield the SUM of A1 plus A2.  Thus, for accurate results in this case, the area integration must be applied.  The use of the Mean Power value will give the DIFFERENCE of A1 and A2.  That is totally wrong!
"The integration should yield the simple SUM of A1 and A2." Agreed 100%, because that represents the average power in one complete cycle. Note that the SUM of A1 and A2 is equivalent to this: A1 + A2, and since A2 is actually -A2, the SUM becomes A1 - A2. Therefore in cases where the power waveform crosses the x-axis in the negative direction, we know that some of the power is returning back to the power supply. The true average power then is the net difference between the power being sourced by the supply and the power returning to the supply. A MEAN or AVG computation on the wave form incorporates this SUM function.

.99
   
Hero Member
*****

Posts: 931
Lawrence,

Could you please substantiate your statement "The correct area integration should yield the SUM of A1 plus A2" and clearly show the relationship between this and "Average Power".


Note to both participants:
References to specific devices should be avoided except where they are used as accepted examples in support of a given stance. It is hoped that the knowledge shared in this debate can be applied to any device under consideration and not be restricted to any particular device.



The correct area integration:

1.   We have the Instantaneous Power Curve derived the Instantaneous Voltage and the Instantaneous Current.  With most of the Oscilloscopes, Channel 1 and Channel 2 have internal common ground.  This means that the trigger points for both voltage and current will be started at the same time.
2.   Integration is equivalent to taking a sampled or calculated value and multiplies it by a tiny internal.  In this case, it is the Instantaneous Power Value x Time.  That is the Energy detected at the interval.
3.   When the Power Waveform Curve cuts across the zero reference axis, there will be positive and negative areas.  Positive area is easy to understand as it represents the energy supplied or consumed during that period.
4.   Negative area is the concept requiring more explanation.  It does NOT represent negative energy.  We have to look back at the initial voltage and current waveforms.  We would find that either the Voltage or Current (or both) cut their zero reference axis. 
5.   Thus the negative area simply means the voltage or current is in the opposite direction (as common or a well-known feature in AC).  The negative area still represents energy supplied or consumed.
6.   Thus these two areas must be added to get the actual energy.

The exact Average Power can be derived by Area (A1 + A2)/ total time interval considered.
It cannot be derived by Area (A1-A2)/ total time interval considered.

The concept of any device under consideration:

1.   We should look at the Instantaneous Power Curve carefully.  If it is all positive, we can apply the Poynt99 method of Mean power.  That result will be CORRECT and correspond to the area integration method.
2.   However, if the waveform cuts the zero reference axis (have both positive and negative values), we must NOT apply the Poynt99 method of Mean power.
3.   The actual device is immaterial.  The key is the shape of the Instantaneous Power Curve.  Once we detect negative area; the Poynt99 Mean Power method fails.

More on RMS later.
   
Hero Member
*****

Posts: 931
"The integration should yield the simple SUM of A1 and A2." Agreed 100%, because that represents the average power in one complete cycle. Note that the SUM of A1 and A2 is equivalent to this: A1 + A2, and since A2 is actually -A2, the SUM becomes A1 - A2. Therefore in cases where the power waveform crosses the x-axis in the negative direction, we know that some of the power is returning back to the power supply. The true average power then is the net difference between the power being sourced by the supply and the power returning to the supply. A MEAN or AVG computation on the wave form incorporates this SUM function.

.99

Poynt99,

One interpretation is that Power is returning back to the power supply.

However, the cutting of voltage and current across the zero reference line when they are out of phase occurs on all AC waveforms!

Thus another interpretation is that the Power is the standard AC Power that provides us with the needed energy!

I do not know the various oscilloscopes to know how their integral function works.  My ATTEN from China does not have that function.  However, the waveforms are supposed to be exportable so that an external maths function can be applied to do the integration.

From the Professors in Electrical Engineering in Hong Kong, they told me to use the sum of A1 and A2 to get the correct Energy consumption.  I used that technique manually.  The eyeball technique showing the difference in area of the following is obvious and convincing – even to the eyes of an old man.

Lawrence
   
Group: Guest
Participants,

Each of you have well established your stance on the issue.

However, as of yet your statements can be considered as opinion rather than fact.

What is missing is substantiation in support of your respective stances.

Poynt99 has provided an internet quote of Roy Lewallen as a form of substantiation. While this quote tells us what Average Power is not, it fails to provide a definition of what Average Power is nor does it discuss the use of MEAN in a calculation.

Roy does offer a clue when he states in his summary:
"* The RMS values of voltage and current are useful because they can be used to calculate the average power."

The debate still needs a verifiable definition for "Average Power"

Moderator
   
Hero Member
*****

Posts: 931
Understanding RMS values

Quote

From Roy Lewallen

The RMS value of power is not the equivalent heating power and, in fact, it doesn’t represent any useful physical quantity.


I believe Mr. Roy Lewallen took an extreme position.  In order to address the issue properly, we need to re-examine the whole concept of rms value carefully.

Let me start with the top slide.  In this slide, I show the sinusoidal Voltage curve on top.  I show the sinusoidal Current curve in the middle.  I separated this middle curve into section X where the Voltage and Current are in Phase.  There is a section Y where they are out of phase.

The last curve is the product of the instantaneous voltage values multiplied by the instantaneous current values.  Note that in section X, the product (area) will always be positive.  (Negative voltage time negative current will become positive number).

In section Y, the product will sometime produce positive value and sometimes negative value.  Please examine the P curve in section Y carefully to confirm it.

The question is – if the correct area can be obtained by flipping the negative area, why should the Industry introduce an RMS value?

The answer lies in slide 2 and slide 3 from http://en.wikipedia.org/wiki/Root_mean_square
In power measurements, we can still use the concept of Instantaneous P = Instantaneous V x Instantaneous I.  Instantaneous V can still be represented by Instantaneous I x R where R is a fixed resistance.  Thus the Instantaneous P = R * I * I.

The mathematics shown on slide 3 shows the usefulness of the RMS value in calculating Power.  The negative sign is automatically guaranteed to be positive with the square function.  Any integration will involve positive numbers only.

The rms Power by itself is not useful in providing an accurate power value.  However, if there are a number of prototypes to be ranked, this rms Power value will provide a good ranking order.  This can be confirmed with a number of prototypes (say 5) with different true COP > 1 values (obtained by the exact area integration method).  These prototypes are again ranked with the (Output rms Power value divided by Input rms Power value).  The ranking order will be the same.

Thus the Tseung FLEET Comparison Index has very solid scientific backing and must not be dismissed because someone does not understand or care to understand it.

To summarize,
1.   If the Instantaneous Power Curve is all positive, the Poynt99 method of using a Mean Power Value for calculation is 100% correct.  The result will be the same as with the exact area integration method.
2.   If the Instantaneous Power Curve has positive and negative areas, the Poynt99 method MUST not be used.
3.   The RMS method is useful because it automatically changes all numbers into positive for integration.  And the value is directly relevant for Power Calculation because the square of I or V is involved.
4.   The rms Power value is useful in providing a ranking order for different prototype.  However, if the more exact area integration results are available, such results should be used.

More to come on Pulsing waveforms, importance of scope sampling rates and difference between DC and AC power concepts.
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
Inherent in any power wave form, there is a "DC Component".

It is this DC component that represents the average power (effective heating power) of the wave form. In cases where there is equal area above and below the x-axis in one cycle, the DC component is zero.



In order to obtain the correct value when computing the DC component (i.e. average value of the power), an even number of complete cycles must be used, but one cycle is sufficient. However, using an oscilloscope set either to provide the MEAN of the power trace, or if the wave form itself is averaged, the scope will provide a running average of the wave form shown on the screen, and generally, at least 10 cycles should be displayed in order for the scope to give an accurate reading of this running average.

.99
« Last Edit: 2011-02-05, 04:02:12 by poynt99 »
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
A definition of "average power" can be found in this document by the high-tech company "Newport".

Quote
Definition of average power:
Rate of energy flow in every pulse. Rate of energy flow averaged over one full
period (recall that f = 1/T ).

.99
   
Hero Member
*****

Posts: 931
Inherent in any power wave form, there is a "DC Component".

It is this DC component that represents the average power (effective heating power) of the wave form. In cases where there is equal area above and below the x-axis in one cycle, the DC component is zero.

.99

Dear Poynt99,

In cases where there is equal area above and below the x-axis in one cycle, the DC component is zero.

If we use the above definition you proposed or supported, does it mean all in phase AC has zero DC Components and cannot heat anything – even pure resistors?  In phase AC means the AC Voltage and AC Current have no phase lag - +ve peak V coincides with +ve peak I etc.

Please clarify. 
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
Dear Poynt99,

In cases where there is equal area above and below the x-axis in one cycle, the DC component is zero.

If we use the above definition you proposed or supported, does it mean all in phase AC has zero DC Components and cannot heat anything – even pure resistors?  In phase AC means the AC Voltage and AC Current have no phase lag - +ve peak V coincides with +ve peak I etc.

Please clarify.  


Lawrence,

Please refer to the above diagram which illustrates an instantaneous power trace (P) derived from the product of the shown instantaneous voltage (e) and the instantaneous current (i). The "DC Component" indicated on the diagram pertains only to the power trace (P), and in this case is illustrating a net positive average power.

The diagram shows an example of a pure resistive scenario, hence the presence of zero phase angle between the v and i traces. May I remind all, that the issue of this debate is in regards to any given power trace. As such, my point made above and in your quote, pertains to a power trace. Although not pertinent to this debate, it should also be understood that the said statement applies to all wave forms; voltage, current, and power. However, for this particular argument, allow me to repeat and clarify my statement:

In cases where there is equal area above and below the x-axis in any given power trace, and in one cycle, the DC component, and hence, "average power", is zero.

.99
« Last Edit: 2011-02-04, 20:48:31 by poynt99 »
   
Hero Member
*****

Posts: 931

In cases where there is equal area above and below the x-axis in any given power trace, and in one cycle, the DC component, and hence, "average power", is zero.

.99

Dear Poynt99,

I think we finally narrowed our debate to the key disagreement issue.  You believe that the “average power” for a sinusoidal power trace is zero. The negative area under the axis should cancel out the positive area above the axis.  Am I correct that this is your assumption?

My viewpoint is different.  The negative area under the axis still represents energy (Power x time).  Energy cannot be negative.  The reason that the area is negative is because one of the voltage or current value happens to be negative while the other remains positive. A negative number multiplied by a positive number will always be negative.

Thus my interpretation of the negative area is that either the voltage or current turned negative.  This occurs with all AC waveforms that are out-or-phase.  We cannot interpret this as energy flowing back to recharge the battery or the power source.  We must still treat that as energy provided or spent.

The analogy is a man running in the +X direction.  He then turns back and runs in the –X direction.  Energy is spent in both cases.  We must add the energy spent. We must NOT subtract one from the other because of the sign.

Thus I strongly disagree that: In cases where there is equal area above and below the x-axis in any given power trace, and in one cycle, the DC component, and hence, "average power", is zero.

The average power should be the sum of the positive areas PLUS the sum of the negative areas divided by the time interval being considered.  The negative sign must be dropped in the addition.  The correct average power is given by the total energy provided or spent divided by the time.  Or the sum of Positive and Negative areas divided by the time interval under consideration.  The correct area integration method should yield such a value.

Lawrence
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
I think we finally narrowed our debate to the key disagreement issue.  You believe that the “average power” for a sinusoidal power trace is zero. The negative area under the axis should cancel out the positive area above the axis.  Am I correct that this is your assumption?
Yes, I hope that we indeed now have a mutual understanding of the key issue.

I agree with your statement "You believe that the “average power” for a sinusoidal power trace is zero." if and only if the sinusoid is symmetrical about the x-axis. You saw in the example diagram I posted that the power trace was sinusoidal, but the average power was not zero.

Quote
My viewpoint is different.  The negative area under the axis still represents energy (Power x time).  Energy cannot be negative.

Thus my interpretation of the negative area is that either the voltage or current turned negative.  This occurs with all AC waveforms that are out-or-phase.  We cannot interpret this as energy flowing back to recharge the battery or the power source.  We must still treat that as energy provided or spent.

The analogy is a man running in the +X direction.  He then turns back and runs in the –X direction.  Energy is spent in both cases.  We must add the energy spent. We must NOT subtract one from the other because of the sign.

Thus I strongly disagree that: In cases where there is equal area above and below the x-axis in any given power trace, and in one cycle, the DC component, and hence, "average power", is zero.

The average power should be the sum of the positive areas PLUS the sum of the negative areas divided by the time interval being considered.  The negative sign must be dropped in the addition.  The correct average power is given by the total energy provided or spent divided by the time.  Or the sum of Positive and Negative areas divided by the time interval under consideration.  The correct area integration method should yield such a value.

Lawrence

I strongly urge Lawrence to carefully read and understand this Wikipedia page on Power Factor, and in particular the 3 pictures at the right panel. Power can and does flow back to the source. Pay particular attention to the Average Power trace in each case, and how it relates to where the power trace is positioned.

.99



   
Group: Guest
There have been some strong assertions from both sides of the debate that still lack substantiation.

Lawrence claims that a negative power is impossible but has not substantiated this claim.

Poynt99 claims that the application of a MEAN function to an instantaneous power trace will give the "Average Power"

I find the definition for average power supplied by Newport with reference to Pulsed Lasers wanting with respects to this debate.

What we really need is a definition related to LRC circuits which comprise the fundamental components of typical arbitrary loads where "Average Power" is analyzed.

To help in this regard I offer the following quotes:

=======================
Fundamentals of Electrical Engineering Second Edition - Leonard S. Bobrow

Page 213, Subheading "Average Power"

. . . for the arbitrary element ( or interconnection of elements) . . . For the case that the voltage is a sinusoid, say v(t) = V cos(ωt + Φ1), the current will also be sinusoidal, say i(t) = I cos(ωt + Φ2). Then the instantaneous power absorbed by the element is p(t) = [V cos(ωt + Φ1][I cos(ωt + Φ2] . . . If this is the instantaneous power absorbed by the the element, what is the average power P absorbed by the element? . . . Thus the average power absorbed by the element is P = ½VI cos (Φ1 - Φ2) = ½VI cos (θ) where θ = Φ1 - Φ2 is the difference in phase angles between voltage and current.

Page 219, Subheading "Effective Value of a Sinusoid"  (with regards to a pure resistance)
We have seen that for a resistor R with a sinusoidal current through it, . . . the average power absorbed by the resistor is ½RI2. What dc current Ie would yield the same power absorption by R? . . . we could say that I/ √2 is the effective value of the sinusoid whose amplitude is i . . . then the effective value of v(t) is Ve = V/ √2. Furthermore, the average power absorbed by R is PR = RI2/2 =(V// √2) (I/ √2) = VeIe . . . For an arbitrary element P = ½VI cos (θ) = VeIe cos (θ)

Page 220 subheading "RMS Values"
Consider for a resistor R, the case that the current (or voltage) is nonsinusoidal. . . . Summing the instantaneous power from time t1 to t2 and dividing by the width of the interval t2 - t1 yields the average power absorbed by the resistor in that interval . . . the expression for Ie, the effective value of i(t), is Ie = √(1/Tt0t0+T i2(t) dt [ Moderators note: this is the equation for RMS]

Page 223, 4.5 Important Power Concepts
. . . if the load is not purely resistive, even though it might "appear" at first glance that the average power absorbed is VeIe, the average power absorbed is [ P = ½VI cos (θ) = VeIe cos (θ)
] We call VeIe apparent power and use the unit volt-ampere (VA) to distinguish this quantity from the actual power (the average power) that is absorbed by the load. . . . We see that the quantity cos (θ), called the power factor (pf), is the ratio of average power to apparent power, that is,  pf = cos (θ) =  P / VeIe = Average Power / Apparent Power
The angle θ is referred to as the power-factor angle

Pages 227 - 228 subheading Complex Power
. . . we see that the average power absorbed by the load is equal to the real part of some complex quantity. Defining Vrms and Irms, respectively, by Vrms = Vee1 and Irms = Iee2, then P = Re[ VrmsI*rms ] In other words, the average power absorbed by the load is the real part . . .
. . . where P = VeIe cos (θ) is the average power - also called the real power - absorbed by the load
==============

It should be readily recognized from the foregoing quotes that all of the RMS treatment associated with computing the Average Power must be applied to the voltage and current prior to taking the product.

Are the participants now able to ascertain the correct method of using the oscilloscope to obtain the Average Power?
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
There have been some strong assertions from both sides of the debate that still lack substantiation.

Lawrence claims that a negative power is impossible but has not substantiated this claim.
Agreed. When power factor considerations are understood, it becomes obvious that it can not be substantiated. The Wikipedia page I referenced above was the last nail in the coffin so to speak.

Quote
Poynt99 claims that the application of a MEAN function to an instantaneous power trace will give the "Average Power".
Not only is the claim true and correct, but it has been shown to be absolutely irrefutable based on the simulations referenced early on by me in this debate. Has the moderator not carefully reviewed those posts and simulations?

Quote
It should be readily recognized from the foregoing quotes that all of the RMS treatment associated with computing the Average Power must be applied to the voltage and current prior to taking the product.

Are the participants now able to ascertain the correct method of using the oscilloscope to obtain the Average Power?
Using the product of the RMS voltage and current is one way to compute the average power. There is another method. However, the moderator seems to have forgotten what the debate is about; and indeed it is regarding the application of one type of computation to an already-existing power wave form trace.

In order to refresh everyone's memory, let's again examine the introductory question which is the basis for this debate:

"What method of computation should be applied to a Power Waveform Trace to obtain the Average Power of the sample set under consideration?"


.99
   
Hero Member
*****

Posts: 931
Harvey showed an interesting XLS on:

http://www.overunityresearch.com/index.php?topic=711.msg10396#msg10396

If that Spreadsheet is correct and applicable to any waveform, does that show the running mean and running rms will yield the same value?

If that were so, there is no more point to continuing the debate.  Let us confirm that carefully.

Poynt99 and PhysicsProf, both of you have FLEET prototypes and oscilloscopes, could you check that?

I have to admit that I learned much from the debate.  Poynt99 is great and the debate has been kept on scientific grounds.

   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
Harvey showed an interesting XLS on:

http://www.overunityresearch.com/index.php?topic=711.msg10396#msg10396

If that Spreadsheet is correct and applicable to any waveform, does that show the running mean and running rms will yield the same value?

If that were so, there is no more point to continuing the debate.  Let us confirm that carefully.

Poynt99 and PhysicsProf, both of you have FLEET prototypes and oscilloscopes, could you check that?

I have to admit that I learned much from the debate.  Poynt99 is great and the debate has been kept on scientific grounds.



Actually, Harvey's spreadsheet can be somewhat deceiving in the way the graphs are labeled, if you are not careful about your assumptions.

What Harvey is trying to convey, is that VRMS * IRMS = PAVE = MEAN[v(t) * i(t)], and he is correct. However, as I pointed out above, Harvey has apparently lost sight of the debate topic, and this is not a good thing.

Let's stay on topic guys, as this is becoming rather tedious.  :-\

.99
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
"What method of computation should be applied to a Power Waveform Trace to obtain the Average Power of the sample set under consideration?"
   

Group: Administrator
Hero Member
*****

Posts: 2916
It's not as complicated as it may seem...
A bit more information for you Lawrence, on reactive power and how power can be both sourced and returned to a supply (from Circuit Analysis - Theory and Practice, 3ed.)




.99
« Last Edit: 2011-02-05, 04:04:41 by poynt99 »
   
Pages: [1] 2
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2020-12-04, 05:32:35