120V from both legs

Power transferred to the mechanical side will be Eg*I* cos( angle between I and Eg) That old power factor thing again. This gross mechanical power is equal to Dv^2 where v is the velocity D is the combination of mechanical and acoustic resistance of the speaker system. To separate the acoustic output from the mechanical loss requires a more detialed model. Again Beranek shows this but Kinsler has a clearer description..

Thank you. On the mechanical side then, is this the case? Mechanical power = force * velocity * cos angle between applied force and velocity i.e. direction of movement.

Northstar

Basically. Power = torque times angular velocity.

(mechanical

beforehand.)

----------- Sort of: The angle referred to is not the difference in direction as you have implied but the difference in time phase- that is a power factor angle. If Eg and I are treated as phasors, then force and velocity can also be treated as phasors.

Force and velocity in a speaker are in the same direction but need not be in time phase. If the force and velocity were not in the same direction, then the physical difference in direction would have to be taken into account as well as the power factor angle.

Note that the frequency response curves are generally done on a rms basis with a single sinusoidal input whose frequency is varied over a range.

Three things:

1.Then it is the phase difference between peak velocity and peak force? 2.If so, this would be a mechanical power factor, and not electrical. 3.Also it would be zero at mechanical resonance and otherwise away from resonance.

Northstar

-----------------

1) Yes (also phase difference between rms velocity and force) 2)Yes. Note that it is the same magnitude as the angle between Eg and I on the electrical side 3) Yes -analogous to the electrical case.

Note that mechanical "impedance" F/V appears as an admittance I/Eg as seen from the electrical side Note also that Eg*I*cos (angle) =V*F*cos (angle) gives the total power transferred - not the output power.

PS I called K spring compliance this should be stiffness (1/compliance).

OK, thank you. If real power then is apparent power times power factor IE PF, is apparent power IE = E/R*E = E^2*R or is it IE = E/Z*E = E^2/Z ? TIA Northstar

Should have been: is apparent power IE = E/R*E = E^2/R or is it IE = E/Z*E = E^2/Z ? Sorry. Northstar

----- Original Message ----- From: "Northstar" Newsgroups: alt.engineering.electrical Sent: Wednesday, October 27, 2004 8:38 PM Subject: Re: Motor models

Sorry for the delay- I inadvertently sent this to you ficticious address rather than to the group.

" Apparent power" is the magnitude of IE. and is expressed in volt amps. This leads to IE =E^2/Z using the magnitudes.

Think of a right angle triangle. Z is the hypotenuse, R is the base and X is the opposite side such that Z=root(R^2+X^2) and the angle between the base and the hypotenuse is alpha=inverse tangent of X/R

Example R=4 ohms, X=3ohms so Z=5 ohms and the angle is 36.87 degrees. Note that R/Z =0.8 in this case. Now apply 50V to this impedance. The current will be 50/5=10A and will lag the voltage by 36.87 degrees. This is the power factor angle. The component of current in phase with the voltage will be 8A and the component at right angles will be 6A. The VA will be 50*10 =500 volt amps =(I^2)*Z =100*5 =500 va The pf will be cos^-1(36.87) =0.8 lag The power will be 500*0.8 =400 watts (I^2)*R= 100*4=400 watts The reactive VA will be 500*sin(36.87)=300 watts =(I^2)X volt-amps reactive (vars) Note that VA =root(watts^2 +vars^2) again a right triangle relationship. The purpose of using complex numbers for AC steady state analysis is that both the phase and magnitudes are represented and life is made easier (particulalry if you have a calculator that handles complex numbers.

Thank you for the example. I looked it over briefly tonight. Looks good. I'll be away a day or two, but will get back to you then. Northstar

Thank you. An excellent analysis and I agree regarding heating power, but there appears a contradiction if applied to the loudspeaker motor. In a motor the I^2 R power is called the copper loss Pcu, i.e. the power lost as heat in the motor winding, in this case the voice coil. Obviously this is real power creating heat, but it provides no force or power to move the load (mass and mechanical resistance). So where does the power to move the load come from?

Northstar

---------- Your comment is good but there is no contradiction: For an ordinary motor : Vin =Z*I +Eg*I and Pmech =velocity * component of torque that is in phase with velocity= Eg*I* cos(angle between Eg and I) The power input is (I^2R) +Eg*I *(angle between Eg and I) =electrical loss + power transferred to the mechanical side. For a speaker, the same relationships hold. A speaker is a motor. It has a back emf and electrical energy is converted to mechanical energy as in other motors. The difference is the nature of the load.

Suppose an AC motor (synchronous) has an resistance of 0.1 ohms and reactance of 1 ohm. The applied voltage is 100V and the current is 5A at 1.0 pf. The total power input is 500 watts and the I^2R loss is 2.5 watts. The remaining 597.5 watts is transfered to the mechanical side - appearing as gross mechanical power. If the mechanical loss is 7.5 watts, the net useful mechanical power is 490 watts . Pelectrical =P( electrical loss)+P(mechanical loss) + P useful mechanical output) At this particular load the resistance seen by the source "appears" to be 20 ohms. Resistance in this case is rather useless as it can vary from 0.1 ohms to near infinity depending on the load.

In the case of a speaker, the power balance is : Electrical power = (I^2)R +Eg*I*(angle between Eg and I) =(I^2) + P(transfered to the mechanical side) =(I^2)R + P(mechanical loss in suspension and cabinet) +P(transferred to the acoustic side) =electrical loss + mechanical loss + acoustic loss + useful acoustic power output.

If there is no distortion or non-linearity (HAH!) the acoustic load may be represented by a resistance so we have a model with fixed resistances, inductances, masses, springs, and capacitances throughout (or at worst, a frequency dependent resistance) so that a circuit model works quite well. In this case, the total mechanical power may be represented by "loss" in an equivalent resistance {(Eg/I)cos (phase angle)} and the "loss" in this resistance includes mechanical and acoustic losses as well as useful acoustic output.

Note that acceleration of the mass is not involved in the power balance. A mass is like a capacitance so that instantaneously there is work to accelerate the mass but this work is recovered so that over a cycle the power to accelerate the mass is 0.

Circuits texts cover this aspect in electrical systems and mechanical systems exhibit the same sort of behaviour. Berenak, or, better yet, Kinsler (Acoustics) handle the speaker "models" and theory in detail..

Out of curiosity, how much electrical or energy conversion theory have you had? It is easier to answer if one knows doesn't over or under-assume a background. It is also be easier to answer directly with the use of diagrams, etc where needed as Ascii diagrams and ordinary text don't always do the job.

Thank you for the moet excellent analysis. However my question remains "So where does the power to move the load come from?" I shall be more specific: You state above: " " Apparent power" is the magnitude of IE. and is expressed in volt amps. This leads to IE =E^2/Z using the magnitudes." You then state (after giving magnitude examples): "The VA will be 50*10 =500 volt amps=(I^2)*Z =100*5 =500 va "The pf will be cos^-1(36.87) =0.8 lag" "The power will be 500*0.8 =400 watts (I^2)*R= 100*4=400 watts"

Putting this in In equation form, you say; Power real=IE pf=(I^2)*R=400 watts. where you state IE is "apparent power", and of sourse I=current and R=resistance.

Now note this(I^2)*R power is the copper loss, i.e. power generating heat in the motor winding or voice coil in this case. The law of conservation of energy says that when power is supplied for anything other than heat, there must be a second term on the right hand side of the equation other than the (I^2)*R heat term. This second term would in this case define the power driving the mechanical resistances or equivilant engendered by motion. Which is why I ask: "So where does the power to move the load come from?" To be more speficic here also, Your equation provides no power to overcome the mechanical resistances of the load, IOW it is incorrect regarding the motor you chose, i.e. a loudspeaker motor.

As to my background, I majored in ohms law :).

Northstar

----------- The statement "above" that I made applied to a impedance load where R=4 ohms and X is 3 ohms. This is NOT a motor or speaker load but was an example to show the relationship for an impedance load (R and X constant at any given frequency) Compare this to the synchronous motor example where Ohm's Law doesn't apply even though one can calculate an equivalent resistance. The actual resistance drop =R*I =0.5 Volts follows Ohm's Law. The remaining 99.5 V is not governed by Ohm's Law. At this particular operating point it appears that Rload =99.5/5 =19.9 ohms but if the voltage changes to 110V and the load and Eg stay the same (speed not changing), Ohm's Law will give nonsense values. If the voltage applied changes by x%, an equivalent resistance based on Ohm's Law will not give you the correct current and output power. An Ohm's Law model, as you have correctly surmised below, doesn't have the second term needed.

Possibly the problem is that I used "appears" in referring to this resistance and "apparent" referring to the magnitude V*I -------------

I repeat:

Note that Eg*I* cos(angle between Eg and I ) is the second term that you are looking for.

Ohm's Law applies to the ZI term but DOES NOT apply to the Eg term Ohm's law is inadequate except under limited conditions which may be approximately so in a speaker.

In terms of power the input is V*I *cos (angle between V and I) =R*I^2

+Eg*I* cos (angle between Eg and I) The coil loss is R*I^2

The power delivered to the mechanical side is Eg*I* cos (angle between Eg and I)

For a voice coil Eg= K*phi*U where U is the mechanical velocity , phi is the flux linking the coil and K is a factor of the coil geometry and number of turns. Also the force produced is F=K*phi*I Then Eg*I* cos (angle between Eg and I)=F*U*cos (angle between U andF) =mechanical power developed = mechanical power lost plus mech power converted to acoustic power = mech power lost + acoustic power lost + useful acoustic power.

If one can represent the mechanical elements and the acoustic elements as constants then one can reduce the mechanical and acoustic side to an equivalent impedance. In this case the acoustic output can be represented as an equivalent power loss in a resistance which is not the voice coil resistance. This is limited to a smaller range of acoustic power levels than the speaker/ cabinet makers may admit to. Ohm's law does not apply for an active device such as a speaker, motor or battery. It applies to a fixed impedance (originally to a fixed resistance ---incremental change of current/incremental change of voltage = constant) Active or non-linear elements blow this all to ratshit. A battery, motor or speaker load cannot be represented by Ohm's Law except as indicated in the previous paragraph.

ENOUGH ENOUGH ENOUGH Phil, The professional Engineers you so quickly denigrate here, have been posting here for a long time and have contributed selflessly to many an honest request for help. As a professional engineer myself, I go over what they write and suggest, Partly to see if there is something more for me to learn but also to see if they are correct and not bullshititng. ( I have yet to find an instance where they have seriously mislead or missinformed an honest question) Needless to say The arguments and advice they put forward are almost always based on Solid Electrical Engineering Theory and practice.

Remember that the profession of Electrical Engineering is based on the professional work and publications recognized and recomended by the various institutes of Electrical Engineers around the world. I think very little of it is based on what you can find with a "google Search" of the internet or the marketing material published in the Various electrical trades magazines.

The fact that most of us choose to ignore these long and tedious threads you create does not, in anyway, endorse what you say.

Without having the full benefit of the deleted history of these threads, I can only go on what appears today, and looking through the threads I see very true things written by Don, Charles, and Daestrom and a confusing mixture of engineering terms, Miss-quoted physics terms, gut feelings and "Buzz words" from yourself.

My suggestion to you is that you stay away from Electrical Engineering as your knowledge is embarrassingly deficient in this field. Maybe you should stick to quoting Electrical codes and leave the real engineering to those who are clearly Qualaified to do so.

Tom

But we were discussing motors. Regardless... You focus on ohms law as not applicable throughout in your reply here, as though I based my statement on ohms law. Not so. I said "As to my background, I majored in ohms law :)" in humor (note the smiley face), and it was clear my statement was in response to you inquiring about my background. I said nothing to indicate ohms law invalidated your analysis.

Regardless, you now state (which I shall label equation A):

Electrical power = (I^2)R + Eg*I*(angle between Eg and I) eq.A =(I^2) + P(transfered to the mechanical side)

I agree, but electrical power was noted as IE, and I ask if I = E^2/R or E^2/Z. You replied

" "Applied power" is the magnitude of IE, and is expressed in volt amps. This leads to IE =E^2/Z using the magnitudes"

So does IE = E^2/Z in equation A? TIA

Northstar

Well Tom... those guys were utterly full of shit on the back EMF issues fed into the power grid from users...as was laborously demonstrated by following posters...with the math.

Sorry. There is no shortage of idiot PE's. or doctors or plumbers or car salesmen...but yes on most of the basic every day issues these guys posting here have been correct.

But not many were correct on the back emf issues... and that has been amply and effectively discussed now by others as I left the conversation... sorry thats just how it is.

Now this ends my involvement on this issue. Denials beyond the point of proof are not a sign of intelligence or integrity...regardless the generally 98% accuracy of the other remarks.

Phil Scott

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WRONG again! Dang you should learn to read. You really should look up back EMF and find out what that term really means.

Reactive current is not back EMF, harmonics from adjustable speed drives is not back EMF. All of the math that was shown was done so to prove you wrong. Get over it.

Charles Perry P.E

I'm with you charles, It is just this guy seems to have his own beliefs regardless of what the majoritory of main stream Engineering has to say

Tom Grayson