120V from both legs

Or pretends he has knowledge

where

------------- NO. IE =E^2/Z in an impedance load. Ohm's law is alive and well. It applies to E'I where E' =(E-Eg) but not to EgI except in special cases. In a motor, any kind, there is a back emf dependent on speed. There is a torque developed which depends on current. The relationship between Eg and I cannot, except in special cases, be practically represented by a Z (as the Z would have to be derived from Eg and I and would vary widely with load making it impractical to use and also not giving any useful information that is not already known from Eg and I.-so why bother) .

Power transfer from electrical to mechanical side depends on Torque times speed =Back emf times current (ignoring phase relationships) Generally the relationship between Eg and I is non-ohmic and cannot be practically modelled by an impedance.

I don't think that I have changed my tune on this - when You asked about EI=E^2/Z I answered in terms of an impedance model-where Z is useful, not a motor model where an overall Z generally doesn't mean anything. If I misled you, I am sorry. I didn't intend the EI=E^2/Z discussion to apply to motors.

There are cases where it is useful - one is that of a speaker where, if the mechanical parameters (mass, spring, damping) are independent of velocity and force, it is possible to use a "transformer" type of model to link electrical and mechanical sides. as in this case Eg/I =Velocity/force =1/mechanical impedance. I have said this before. Speakers are generally modelled in this manner rather than as a "motor". The basic equations are the same but sometimes ezpressing them differently gives more convenient models for analysis.

I noted the smiley and, frankly, I get the feeling that you are bloody sharp-I will strive for more clarity. Goodonyou,

Snipping

Likewise.

No problem.

So then, staying with the speaker motor, do you have a term other than E^2/Z for electrical power in eq.A above? TIA

Northstar

Could it be that in our age of more severe cyclic on-off switching in power systems, could it be that the term "back emf" takes on different meanings? Don't flame me, power is not my specialty, but power electronics and power systems are not what they used to be 20 or 30 years ago, so I'm just asking. I, too, think of "back emf" in the older motor sense of it being a component of the total applied voltage, and not a transmission line phenomenon which is "sent" back to the generator. However, the word "older" is in there, and systems are not the same as in the "old" days.

When in doubt, IEEE STD 100. All of your questions regarding the definitions of electrical engineering terms will be answered. And to answer your question, no, emf still means what it always has.

Charles Perry P.E.

Maybe you are right Phil, However, From my experience, the changes from "recent" innovations in Electronics are mainly applied in the "Control" side of things, not just motion control but also in power monitoring and analysis.

One of the points I emphasise when teaching electricians and engineers about the more advanced control systems we now supply to customers, is that OHMS LAW is still valid and that equations for AC and DC motors are still 100% Correct.

I Do and say this because, these days, the young as well as the older (or should I say "Experienced") electricians seem to want to Blame the recent technology thingo's first and then go looking at the traditional things secondly when trying to find something that has gone wrong. From Experience it is the same old things that go wrong now that did 25 years ago. The art of "Card Swapping" seems to have evolved in place of good solid analytical Deduction.

I am not saying that AC and DC Drives, or PLC's do not fail, They Do, But mostly the problems in the industrial applications that I am involved with are outside of the control system.

I know I am geting "off Subject" a little here, but I firmly stick to the traditional theories of Counter EMF, Harmonic Current content, Power factor and Motor Theory and they are yet to be proven wrong.

If I may ask, how do you and/or other posters here define counter emf? TIA

Nothstar

----------- For the speaker under the assumption that the mechanical load can be expressed in terms of constant mass, spring and damping terms as is assumed in conventional small signal speaker models. . -----Then-------------, at resonance, Z=((K*phi)^2)/D and is resistive so E^2/Z is a measure of power input However, at any other frequency, it is not- you need to account for phase. The total power input is R(I^2) +Eg*I* cos (phase angle between Eg and I)= R(I^2) + Eg^2/"Z"cos (previous angle) where "Z" is the effect of the mechanical impedance as seen from the electrical side. This is R(I^2) +[((K*phi)^2)/D}(I^2) -- the second term being the real power transferred to the mechanical side- of which part is useful.

I have the feeling of going around in circles. I also have the feeling that you could take a book such as Kinsler's "Acoustics" and follow it quite well. :)

------------ My two bits worth- Faraday and Lenz's Laws give e=-d(N*phi)/dt in an N turn coil with flux phi. This can be broken into two terms: a) a transformer voltage b) a speed voltage In a motor or generator (b) is the dominant mechanism and the voltage so produced is the generated voltage in the generator and, in the motor, this is what is called the "counter" or "back" emf. It is still simply a speed "generated" voltage. It is "counter" or "back" in that it is of the same polarity as the applied voltage and opposes it.

The only essential difference between a motor and a generator is in the direction of power flow. Yes, Tom's old fashioned motor concepts still apply

At resonance (Bl)^2/D is the motional impedance as seen by the electrical source. It is only part of the total electrical impedance.

The total electrical power input cannot be E^2/motional impedance.

R(I^2) +Eg*I* cos (phase angle between Eg and I)

This is the *output*, not the input.

R(I^2) +[((K*phi)^2)/D}(I^2)

Here you would be correct by using your questionable eq.B above as mechanical impedance as seen by the source. The correct form is

Mechanical power = I^2 [(Bl)^2/Zmech cos angle] which would be valid for any frequency.

Blessed are those who go around in circles, for they shall be called big wheels. :)

Leaving the question "what is the apparent electrical power input IE?" still unanswered.

Northstar

Sorry, misplaced ]

Mechanical power = I^2 [(Bl)^2/Zmech] cos angle which would be valid for any frequency.

Northstar

Quite right. We used a style of motor-generator consisting of an AC synchronous motor driving a DC generator to charge ship's battery. Whenever we needed to reduce the load on the ship's normal AC generators, we just decreased the DC generator field currently slightly. The DC generator power flow would drop to zero and reverse. The AC synchronous motor's power angle would smoothly drop to zero and rise on the other side of the bus voltage and power would begin flowing from the ship's battery to the AC busses. No relay contacts, switching, or anything. Just a slight reduction in the field current on the DC machine and it smoothly transitions power flow from one direction to the other.

And the problems today of harmonics are not because of some mystical 'sending back' to the generating station the harmonics of non-linear loads. They are quite easily explained and analyzed by studying the non-linear current flows and the impedance in the supply. The lower the supply impedance, the less a distorted voltage drop is created by a non-linear load's non-sinusoidal current.

daestrom

Sorry - You are right- that last was wrong. I also jumped from K(phi) to Bl - needed sleep.

-----------

No. This is the electrical input power. RI^2 is the coil copper loss.

snipping...............

Your eq.C relates to output, since Eg is the back emf voltage generated by and coming from the motor itself. The *input* is voltage generated by and coming from the power station. The input expression uses applied voltage E as in IE, not Eg. This can be confusing since power output must equal real power input.

Thank you for the motor equation analysis. Looks good.

Northstar

-------------- I disagree:

E=ZI +Eg where Z is the electrical impedance of the voice coil itself as measured with the speaker held at standstill. (not the total equivalent input impedance)

|EI | =|(ZI^2) +EgI| is the "apparent power input" ( |x| is magnitude of term x) taking the real part of both sides EI cos (angle between E and I) =total real power input = real electrical power lost in Z + real power transferred to the mechanical/acoustic side.

RI^2 is the real electrical loss EgIcos (angle between Eg and I ) is the real power transferred

Together they are equal to the electrical input power.

So: EIcos (angle between E and I) =Pinput =RI^2 + [EgI cos( angle between Eg and I)] which is what I said.

Sorry, I cannot accept that. The generated back voltage Eg of the motor cannot help supply energy to drive its own self (the motor). The driving voltage is generated entirely by the power source (amplifier). The numbers work out (for a reason), but the equation makes no sense.

Northstar

---------- Nowhere in the equation is there any implication that the motor is supplying energy to drive itself. The equation simply says: Power from source =power loss in electrical side of the speaker + power transferred TO the mechanical side. or, in terms of energy

I am sorry, but I don't see where you get the idea that Eg is helping to supply energy to drive the motor. It actually looks like a sink of energy from the electrical side which is all that the equation is dealing with. (Eg the driving voltage and damn well better do so or conservation of energy is all shot to rat shit). I see nothing wrong with the equation.

Could you give me your reasoning?

Another example: Take a battery supplying a lightbulb through a wire of resistance R. The source voltage is E and the voltage across the bulb is Eb E=RI +Eb applies EI =RI^2 +EbI also applies. The wiring loss is RI and the power delivered to the bulb is EbI

This doesn't imply that the bulb is helping to supply power to itself. EbI simply indicates how much power is delivered TO the bulb. Electrically the bulb is a sink for energy.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

The original question was regarding the applied power I*E*cosine angle between voltage and current. I ask if I=E/Z or I=E/R. In other words what the input *consists* of, not what it was equal to, i.e. what other than I*E*cosine angle between voltage and current can we put on the left-hand side of the equation. Therefore back emf cannot go on the left-hand side. Certainly your equation is correct in that the input is equal to the output.

Could you give me your reasoning? TIA

Northstar

----------- The question of I=E/Z vs I=E/R was answered before. It is I=E/Z. This is NOT Ohm's Law. - However, the question of what is Z then arises. This has also been covered. Do we represent the motor load as an equivalent Z =Eg/I ? This is possible but unless the mechanical parameters are independent of force or velocity, this is rather useless - why bother? In small signal analysis of a speaker, this can be done so, in that case, it is convenient. Ohm's law is fine for (linear) impedance loads. The fundamental relationships are Kirchoff's Laws- not Ohm's Law.

As to the left side of the equation The basic equation is: (KVL) sum of voltage drops around a closed loop =0 In otherwords, electrically, if you go around the block Taking voltage drops in the direction of the assumed current flow

0= (-E) +Vcoil +Eg

-E is a voltage rise in the direction of the current Add E to both sides of the equation

0+E =E-E +Vcoil +Eg or E=Vcoil +Eg The coil impedance Zc is R+jX = root(R^2 +X^2) with an associated angle tan^-1 (X/R) This impedance is constant. We can use Ohm's Law to say that Vcoil =ZcI The phase angle of I with respect to Vcoil is - tan^-1 (X/R) Eg is a voltage dependent on the speed (U) of the mechanical side Eg =BlU. Nothing in this equation gives a relationship between Eg and I. The equation is now

E =ZcI +Eg

Now multiply both sides by I EI =ZcI +EgI We want the real part of this (some knowledge of phasors would be useful) Real part of EI = real part of ZcI +real part of EgI This becomes EI cos (angle between E and I) =I^2R + EgIcos(angle between Eg and I) (Eq. Crepeated)

Now it would be nice to represent the second term as some (I^2)Requivalent and in the small signal speaker model, this can be done and is useful.

Note that I did NOT put Eg on the left side. Where did you get this idea?

Also note that I could do so if I follow the rules of arithmetic and say E-Eg =ZcI This is perfectly valid.

As to what the input consists of : That is apparent. It is EI (volt amps or "apparent" power) and the real part of it is what I have given on the left side of Eq. Crepeated. Input power is EI cos(angle between E and I) The equation Crepeated simply gives the equivalence in other terms - that's what equations do - the key is the = sign. All that I have put on the left hand side is IE cos (angle between E and I) (Pinput) One side of an equation without the other is meaningless. (saying power input =EI cos... is an equation). Saying EIcos... leaves one hanging - what is it?

--------------- If we take a positive charge from one potential to a higher potential, we are doing work on the charge- in fact potential difference or voltage is defined on this basis. All our polarity conventions are also based on this. Now consider a circuit consisting of an ideal battery (E=10V) and a resistor of 1000 ohms. We can write Kirchoff's voltage law around the closed circuit and note that the sum of voltage drops is 0. Defining a voltage drop in the direction of current and using conventional current (in direction of positive charge flow).

0=-10 +(1000)I (sum of voltage drops) Representing the battery as a voltage rise: 10 =1000I We are raising each unit of charge to a higher potential in the battery. This means that there is an energy input to the electrical system. On the right side (1000I) each unit of charge is dropping to a lower potential so that energy is being delivered elsewhere- heat in this case. It is leaving the electrical system- a zero sum game as far as energy is concerned. Instead of using units of charge, we use amperes which is the time rate of change of charge. A power balance is: 0 =-10I + 1000I^2 or 10I =1000I^2 In the battery a positive charge is raised in potential - work to do this is coming in from somewhere- where? It is a result of conversion of chemical energy to electrical energy and in this case the power input is 10I =10*0.1 =1 watt. (1 joule/second) In 1 second the enery converted from chemical to electrical is 1 Joule. Where does this energy go?

The equation 0= -10I + 1000I^2 implies that the total power into the system is 0. The power input from the battery is 1 Joule/second (1 watt which is energy /second) Then there must be an power output from the system of 1 watt. In 1 second the energy out is 1 Joule. Where does this go? It is converted to some other form of energy - eventually all goes to heat - each watt-second or Joule of energy becomes heat or light or mechanical work - or whatever. Note that on this basis a drop of potential exists in the 1000 ohm load - each unit of charge loses energy - which ends up as heat. etc. If one want's to use "electron current" then one can use potential rises in the direction of the electron current and come up with the same result. The problem is that the definition of voltage is based on work to move a + charge so that using electron movement means careful application of sign changes to correct. This is a pain in the butt- stick to the conventional current and this is not necessary.

An analogy to the electric circuit model is a hydraulic one: A pump pumps water up a hill to a reservoir, increasing its gravitic potential energy. The water is released through a penstock and a turbine. There will be some pressure drop in the penstock and energy is lost due to this- ending up as heat. There will be (hopefully) most of the pressure drop across the turbine and the flow rate of water times the pressure drop is the power transferred to the turbine and converted to mechanical power. Tack a generator on the end and there is a further conversion to electrical energy.>