120V from both legs

--------- Why? I have no problem with this. (Bl^2)/Re is the electrical resistance as seen by the mechanical system.. Similarly, Rmr is an acoustic resistance as seen by the mechanical system. The model simply expresses everything as seen from the mechanical side. This is no different that representing the rotor quantities as seen from the stator side of an induction motor, or that of representing the secondary load of a transformer as seen from the primary side. No big deal. Please note that nowhere does Beranek call (Bl^2)/Re an "actual" mechanical resistance. We have been through this before- repeatedly - and nothing that I have said contradicts Beranek, Kinsler, or what you have quoted from Small. I am not in conflict with them-either the model, circuit analysis or the physics. I am in conflict with your interpretation.

-----

---------- You are accusing Beranek of being stupid. He knows better. Of course he is using "complex" notation. In fact, throughout Ch.3 he uses e, i, z as phasors (vectors in his language) When he wants to represent a scalar magnitude he uses |e|, |i|, |u| etc. In Equation 3.53 he specifies e* as the conjugate of the source voltage. This alone indicates that he is treating e as a phasor. Similarly he discusses, on the same page, Z* as the conjugate of the impedance (and impedance is a vector or phasor concept) and the use of |e| to indicate the rms magnitude of the voltage. Kinsler also uses phasor notation but is clearer.

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

They ARE, in fact, using complex notation. I agree with them. I disagree with your interpretation. Beranek uses Ze as the blocked rotor "electrical impedance" which is, neglecting inductance, the same as Re +j0 He uses Zet (3.62) as what we have been calling Ze =Re +Zmot. Again, as throughout this chapter, he is using complex notation. What I am saying is that I and Eb are not in phase so IEb is the apparent power, not the real power. IEb=0.1087 in magnitude, not 0.0328 or 0.0079. The relative phase angle is not 72.4 degrees or 59.6 degrees but 85.8 degrees. Then EbIcos (85+) =0.0079 This agrees with (Rms +2Rmr)|v|^2 as it should.

----------

--------- What I have said, FU =EbI is, in fact, true. A reactive balance as well as a real power balance must exist. The "transformer" in the model is ideal so that the power and reactive on each side must be the same. See transformer models. FU =EbI includes both real and reactive parts- for both of which, the relationship holds. The real part reduces to Fv*cos angle = BlI*v*PF as you have said. This is 0.0079 as shown above. This is not 0.0328 for the reason that I gave - the angle of 72.44 that you have used is based on a false assumption. Note that the reactive part EbI sin(85.8 ) =0.108 (Xmd-1/wCms)|U|^2 =32.78|U|*2 =0.0108 and EI sin(26.56)=0.108 Both of which agree.

-------- Probably true- but not for the reasons that you have assumed.

My mathematics has been checked and counterchecked by approaching the same thing from independent directions. It is consistent. The results using your data |Ze| =8.198, Re =7.09 (+j0), and |Zmot| =(11.01^2)/32.99 agree (within expected numerical accuracy) with the results from a direct solution of the simultaneous complex equations.

1.41 =7.09I +11.01U 0=-11.01I +(2.4+j32.87)U where I and U are complex quantities and 2.4+j32.78 =(Rms +2Rmr) +j(Xmd-1/wCms) =Zmec Beranek's 7.1 gives the same U and I is found from (2.4+j32.78)U/11.01

You have not found any flaw in my analysis, except that it disagrees with your analysis which is very definitely flawed because of an implicit assumption which is not warranted. This basic flaw has been pointed out to you. Think about it. Don't try to pretend that Beranek and Kinsler are not using a phasor approach - because it it blatently obvious that they do use such an approach. They have done their work correctly.

Yes we have been through this, but you have yet to explain satisfactorily why in eq.7.1 and 7.2, if (Bl)^2/Re is not considered a mechanical resistance by Beranek, then why does he add (Bl)^2/Re to the other two mechanical resistances Rms and 2Rmr to obtain the total mechanical resistance Rm. That he does this, giving Rm = (Bl)^2/Re + Rms + 2Rmr can only mean he considers (Bl)^2/Re as a mechanical resistance. Also as I noted in my last post, Small includes (Bl)^2/Re in total resistance.

As I recall, you noted earlier that (Bl)^2/Re as used in the velocity equation 7.1 did not apply to power, or wording thereabouts. However this is my reasonong: Since Beranek includes (Bl)^2/Re in his velocity equation as a resistance, that means (Bl)^2/Re is impeding velocity, such that the source is having to supply power to overcome (Bl)^2/Re. Therefore your exclusion of (Bl)^2/Re in your total mechanical resistance when calculating mechanical power needed is in error according to both Beranek and Small. Sorry, but until you include (Bl)^2/Re in your mechanical resistance, or give sound reasoning not to, I cannot accept your analysis as entirely correct.

Northstar

It is an equivalent mechanical resistance. Physically Re is on the electrical side. However, the two basic equations E=ReI +BlU and BlI=ZmecU , are coupled- one affects the other. In solving these equations for U, the term (Bl^2)/Re appears as the force BlI =Bl(E-Eb)/Re =BlE/Re -[(Bl^2)/Re]U so that all that is happening is that the (Bl^2)/Re can be treated as a mechanical resistance for purposes of calculating the velocity as that is what it looks like. Beranek uses a circuit diagram approach to get to the same place. Note that nowhere does he state that Rm is "the" mechanical resistance. You have also used Zmec (32.99) which doesn't include this term.

----------

---------- Beranek uses Pin =R|I^2| +(Rms +2Rmr)|U|^2 which does not include the (Bl^2)/Re term for two reasons. He does not use (Bl^2)/Re for power calculations for two good reasons. I would suggest that Small also does not use it for power calculations.

a) Re is already accounted for.

b)The electrical equation is E-ReI =Eb We have an actual voltage source with an actual series resistance feeding a transformer of ratio Bl/1 as seen in Beranek's Fig 7.2. b. Fig. 7.2d eliminates the "transformer" by referring all quantities to the mechanical side. Now we have a velocity source E/Bl in series with Re/(Bl^2) feeding a mechanical impedance Zmec which is shown by a parallel "admittance" circuit. The equation could be written as E/Bl =Re/(Bl^2)*F +Zmech U. treating it as if it were an electrical circuit. where Zmech =(Rms+2Rmr)+j(Xmd-1/wCms) The force current in Zmec is the force current through Re so that the equation can be changed to E/Bl =[Re/(Bl^2) +1/Zmec]F and solved for F. Once F is found, U can be found.

1.41/11.01 =(7.09/11.01^2 +1/(2.4+j32.78) F giving F =1.887 @-26.56 degrees Then U =F/Zm =(1.887 @-26.56)/32.87 @-85.81 =0.0574 @ 59.26 degrees The power in is (1.41/11.01)*1.887*cos 26.56 =0.216 ((7.09)/11.01^2)|1.887|^2 =0.208 which is the same as |I|^2Re U^2(Rmec)=0.0574^2(2.4) =0.008 Sum =0.216 as it should be. This is one way to solve the circuit model. It does represent the equivalent of Re correctly and powers do work out correctly. Beranek shows this circuit model in Fig. 7.2d.

However, it is somewhat easier to use a Norton Equivalent and this is what Beranek decided to use. He did this in Fig. 7.3 and replaced the velocity source (E/Bl) in series with Re/(Bl^2) with a current source of BlE/Re in parallel with an "resistance " of (Bl^2)/Re This appears in parallel with Zmec and can easily be added to get a force source (BlE/Re) driving what looks like from a circuit perspective, Zmec and (Bl^2)/Re in parallel with a common voltage. Hence U=(BlE/Re)/{(Bl^2)/Re +Zmec} which is the same as Beranek's Eq.7.1. The model gives the correct results for U and the actual mechanical power is correct. It doesn't give the correct internal values and is not intended to do so. In 7.3, he treats Re as being internal to the source as seen from the mechanical system.

He modifies, unnecessarily, the model by using the concept of a dual, to get a series circuit model as in 7.4. At the time that he wrote his book, there was more comfort, paticularly for a non-EE person in using such a circuit instead of the circuit of Fig 7.2d modified as shown in fig. 7.3b (nodal model)

There is no problem the conversion to a force source or in using the dual EXCEPT that the conversion from a non-ideal velocity/voltage source to an equivalent Norton force/current source will give correct results ONLY beyond the terminals of the source. The Re term is treated as "internal" to the source in this case and (|U|^2)(Bl^2)/Re will not give the correct power loss corresponding to I^2Re. Zmec is external and the correct velocity and losses in Zmec will be found. That is the nature of the model- The source of fig 7.3b is equivalent to that of 7.3a ONLY TO THE RIGHT of the terminals. It is convenient to use this Norton equivalent but it is incorrect to give any real significance to losses in this element.

Here is an example of what I mean. A battery has a voltage of 6V and an internal resistance of 1 ohm. It is connected to a load of 5 ohms

6V =1*I +5*I =6*I so I =1A and the load voltage is 5V Source internal power =6*1 =6watts I^2(1)=1 watt and load power =5 watts

Now replace the non ideal voltage source by a non-ideal current source of (E/Rinternal) =6A shunted by a resistance of 1 ohm. (as done by Beranek in Fig 7.3 The model is then a 6A current feeding the 1 ohm resistor in parallel with the 5 ohm resistor

-------------------o----- ^ | | |5A 1 ohm 5 ohms

------------|-------o-----|

Equivalent resistance =1*5/6 =5/6 ohm voltage across this is 6A(5/6) =5V (and is the voltage across the current source) Load current =5/5 =1A OK power in load =5W OK These are correctly represented. This model cannot be distinguished from the original battery by any measurements on the load side- at any load. That is why it or a Thevenin equivalent is used.

Current in the 1 ohm resistor =5V*1 =5A which is as expected. Psource =6A*5V =30watts losses in the 1 ohm resistor =5V*5A =25 watts.

Although the difference does give the power in the load (5 watts), neither of the above two powers is true.

The current source model gives the correct conditions but the internal conditions are generally not correct. This is what is happening in the circuit model that Beranek uses. He recognises this so he does not use (Bl^2)/Re in power calculations as doing so is meaningless. Note that (1.41*11.01/7.09)(0.0574)cos 59.26 =0.0642 (11.09^2)*0.0574^2/7.09 =0.0563 Neither of these are meaningful but the difference = 0.0079 does have meaning. This is the same situation as in the example.

Note that in moving the mechanical elements to the electrical side, this source change is not used so this problem doesn't occur.

I hope this helps - Siskind should cover Norton/Thevenin equivalents and their limitations.

I just won't say anything about this post, not that i could contribute much, all i've done with motors is,,, not of impact.

I am clearing up some Uncharged Mental Space from what I've Sampled, here ~|)

Have a nice week gents ~>I'll be installing

sorry I can't complete my homework either,

they did not include the pages you obviously are reffering from into my study module, I'm just cockling from what I know already......

if they send me the missing text I'll go do it and leave you all alone to your gadgetry :)

The motors spin back and forth and are reciprocated, note. The armatures are pushed by the electrons, which are confused and jump back and forth, like AC. The armatures push this and that, but mostly some form of Mr. Newtons mass, The mass = F/a such that it gets heavy if Force F becomes big. The acceleration a don't have much say, being told what to do mostly, note. Much like a henpecked husband, as specific analogy.

Hope this helps.

Northwest

We are down to the wire here Mr. Kelly (I have snipped below but am saving your analysis for reference).

You still have not addressed my point and have made a mis-statement. First your mis-statement. You say above, referring to Beranek:

" Note that nowhere does he state that Rm is "the" mechanical resistance. " Not true. He defines terms up front in his book, including Rm: Page 55 " Rm is the mechanical resistance in mechanical ohms " Page 55, " Mechanical Resistance Rm obeys the following physical law " Page 54, paragraph headed as: " Mechanical Resistance Rm. "

Then Beranek gives total mechanical resistance Rm in eq.7.2 as

Rm = [(Bl)^2/Rg+Re] + Rms + 2Rmr

with Rg = 0 as we have assumed

Rm = [(Bl)^2/Re] + Rms + 2Rmr

Now to Small:

Small gives his equation 8, defining Rat as, and I quote " total acoustic resistance "

Rat = Ras + [(Bl)^2 / (Rg + Re) Sd^2]

Transferred to mechanical by multiplying by Sd^2

Rm = Rms + [(Bl)^2 / (Rg + Re)]

Then with Rg = 0 as we have assumed

Rm = Rms + [(Bl)^2/Re]

Same as Beranek by noting that Small says he combined other resistance with Rms, which of course means Rms includes air load resistance 2Rmr.

*** Without question *** Beranek and Small include (Bl)^2/Re as part of total mechanical resistance, therefore it *IS* a resistance that dissipates power from the supply source, and you are wrong to omit (Bl)^2/Re from your power equation Pmec + Pa = v^2 (Rms + 2Rmr) = 0.0079 watt.

I have explained the derivation and why (Bl)^2/Re is a resistance that dissipates power before, and shall repost the info if you wish.

Let me explain in an intuitive way regarding Beraneks eq. 7.1 and 7.2:

v = (Bl E / Re) / [(Bl)^2/Re + Rms + 2Rmr] + jXm = 0.0574

Bl)^2/Re is in the denominator impeding velocity v, just as Rms and 2Rmr impede velocity, therefore Bl)^2/Re requires power from the source to overcome, meaning power is dissipated into Bl)^2/Re, just as in Rms and 2Rmr.

Sorry, but your analysis and phase angles contain errors due to your use of the wrong value of mechanical resistance.

Hope the above is of help.

Northstar

meaning The Electrons cause the Push Off for the Armature within the Coils or Stator., the Flux Proportional in any which direction of said Motor you mentioned,

but quit testing me, Someone already went and Nearly Burnt Down an Office Building... on the Neutral Shunt (or wire) to Ground Buss caper over there in Compliance.

you got that down to the mechanical transfer, daaayum!., is that a necessity for a Masters or the PHD ?

well: if Fa is too big to handle and gets heavy, you can boost the frequency and with a little lubrication in the right places they'll continue to spin merrily.... not not to increase A when doing this it should decrease exponentially.

The llittle Electronitoes whether fed (-) or (+) or both charges can handle it as long as the armature meets Harmonic Balance with the Signal on your Electric Wave and don't ask how i would obtain this :-)

It's there in your Sol for Z equation somewhere

Like the Output Stage of your Audio Equipment and anything Power Related that you want to Match Up or Connect in Series or Parallel into It's Configuration.

your audio system is the Specific Juice or Supply and your amp is the Specific Motor Design. ~ pretty much like that but smaller or bigger Q sometimes really really small:) nanon't know anymore:)

your rubbrband theory works fine too It Popped and Their Fingers }:-o

Sophomoric.

If you would have looked, I used a Johnson Equivilant curcuit, having disposed of it in Norton's theory. It gives the angles you wouldn't beleive !!! These are then superimposed and the friction is sifoned off with the phasors (lubrication not needed, note). The angles are then indisposed and can be adjusted to suit. BTW, I have a surplus of these, let me know which size you need.. Otherwise for best motor power, you can calculate the cosines and signs, adjust phase accordingly and apply the cosmorotic theory on the incoming electrons. This gives maximum power of P=urne.

Again, hope this helps.

kindest regards Northwest

satisfactorily

=BlE/Re -[(Bl^2)/Re]U

excessive detail. Do you disagree with the reasons- if so-why? and on what basis other than your interpretation of Beranek and Small?

---------------------- I should have noted that he does not call it that in Chapter 7. Thanks for the correction. I noticed that in Ch.3. it is not the Rm used in Chapter 7. Note that where he is referring to this in ch3. , he is dealing with an individual "mechanical Resistance" element and its physical behaviour. That is, as in Eq.3.7 he is dealing with the behaviour of an actual mechanical element which he calls Rm, He also deals with an Mm and a Cm in the same manner. However, in Ch. 7. he uses Rm, Xm and Zm as local definitions. This is obvious from the context and wording. This is a typical convention in use where Rm and Xm are used, out of the blue (not in his list of parameters) in 7.1 and immediately defined by "Where: Rm=... (7.2) Xm=..." (7.3) This is done so that Eq. 7.1 doesn't run off the page or take several lines. No other reason for this.

I HAVE NEVER DENIED that (Bl^2) is part of Rm in Eq.7.2 This I have said before. It is due to the coupling of the electrical and mechanical parts of the device. Physically it is due to the presence of the electrical parameter Re - not due to any actual mechanical resistance. That is all that I have tried to say. That is a fact.

However, I DID ADDRESS YOUR POINT in the snipped out part and have not responded to yet. . .

--------- Does this imply that (Bl^2)?(Rg+Re)Sd^2 is an actual acoustic resistance?

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

--------- Now it changes from an acoustic resistance to a mechanical resistance

----------

------- Lets see, you have a mechanical force source of magnitude BlE/Re =2.186 The power in is then 2.186(0.0574) cos (angle) and i will assume an angle of 59.26 from Beranek's Zm so the real power in =2.186*0.0574)*0.511=0.0642 The loss in (Bl^2)/Re is 17.1(0.0574^2) =0.0563 What is left over is 0.0079 which is the actual power into Rms +2Rmr This is EXACTLY the situation that I mentioned in the example I sent you. You are using the equivalent resistance but not the equivalent source. Note that the values of power in and the loss are nonsense but the difference is correct. This is the nature of the model used.

Certainly (Bl^2)/Re is there- it is the electrical resistance transferred to the mechanical side where it appears as an equivalent mechanical resistance. Note also that the source is transferred. Similarly the acoustic resistance Rat can be transferred to the mechanical side but you seem to have no problem recognising that it too can be treated as an equivalent mechanical resistance - for which only part of the total is actual mechanical resistance. Similarly again, transferring to the electrical side we get Re+Zmot =Re

+(Bl^2)/Zmece where Zmec =32.879 @ 85.8 (or if you prefer , 32.99 @72.44) Are you then saying that Zmot is an actual electrical impedance rather than the mechanical impedance reflected to the electrical side as an equivalent electrical impedance? That is what you seem to be saying about (Bl^2)/Re.

I gave you an analysis based on Beranek's Fig.7.2. Note the terminals indicated by o with the mechanical -acoustic elements to the right and the electrical quantities, referred to the mechanical side to the left. Note also that in this model, the power lost is (Re/Bl^2)|F|^2 +Rms|U|^2

+(zmr/2)|U|^2 =Re|I|^2 +(Rms +2Rmr)|U|^2 as it should be. The use of (Bl^2)/Re as an effective mechanical resistance and including it in Zm in Chapter 7 is quite OK. There is no need to harp back on this as I agree. I don't agree with it for calculation of power unless you also use the equivalent force source BlE/Re in the power calculation as above. They go together.

Eq. 7.1 is simply a rewriting of BlE/Re -[(Bl^2)/Re]U =F =[(Rms+2Rmr)+j (Xmd-1/wCms)] U

where all the terms are gathered together as Rm+jXm and U is found. No big deal. The use of it to calculate power is not. Did you actually read and try to analyse what I said in the snipped parts? It is a valid approach and it did give the correct values of magnitude, phase and power.

Another way to look at it is that the actual force acting is F=BlI (I and F in phase) F=BlI =ZmecU and you have used this to get Zmec =BlI/U =32.99 (magnitude) which is fine- this Zmec doesn't include (Bl^2)/Re Power into the mechanical system is |U||F| cos (angle between U and F) Now I =(E-Eb)/Re (phasor not scalar arithmetic.) F =Bl(E-Eb)/Re =(BlE/Re -[(Bl^2)/Re]U)=(11.01*1.41)/7.09 -(17.1)(0.0574 @-59.26) =1.887 @ 26.56 degrees ACTUAL FORCE

|F||U| cos (-59.26-26.56) =1.887*0.0574*0.0729 =0.0079

Note that the U*(Bl^2)/Re term IS accounted for in the actual force that is produced. It exists because the motion of the coil causes a counter emf. The actual force is that applied to the actual mechanical elements and power into the mechanical elements depends on this actual force and the actual velocity. and the relative phase between them The power into the mechanical system becomes |U||{BlE/Re-(Bl^2)U/Re}| cos (same angle)=0.0079

THE TERM (Bl^2)U/Re IS ALREADY ACCOUNTED FOR IN THE ACTUAL FORCE F. There is no need to make the error of accounting for it twice.

Now, as to your angles etc. You have made what appears to be the assumption that Ze is real as you calculate Rmot =8.198-7.09 =1.108 This is inherent in the calculation which is an arithmetic difference. However Zmot =(11.01^2)/32.99 =3.674 so it seems that you obtained an angle from arccos 1.108/3.764 =72.44 degrees

Error 1: Assumption, whether explicit or implicit is that Ze is real doesn't have any mathematical or physical basis

Of course,IF the angle is 72.44 degrees, then Xmot =3.50 and this leads to:

Error 2: You have ignored Xmot completely = treating it as 0 -no rational mathematical or physical reason for doing so exists

- this agrees with assumption 1 for the magnitude of Ze but does run into a problem with your phase angle. If you had taken this into account, you would have a Ze which would be root(8.198^2 +3.5^2) =8.91 and this would have a phase angle of 23 degrees. However that would mean that Ze is not real (or the right magnitude) Inconsistencies which imply that your assumption is incorrect. I gave you a process which avoids such assumptions - that is use Zmot =Rmot +jXmot , and this process does not use complex math. Did you actually go over it? I also sent a vector diagram as well.

On the basis that your angle of 72.44 degrees is correct, you have found the mechanical resistance =9.95 which you have tried to explain away as not being measurable or some such unfounded reason. You have also ignored the fact that for this to be true, the mechanical reactance must be less than the actual mechanical reactance (Xmd-1/wCms). More inconsistencies. Including the problem that you have effectively taken Beranek's Rm and Xm and thrown them out the window as they don't match. Could it be that the assumed angle of 72.44 degrees is incorrect? Consider that possibility. Another is that the reactive input is 0 but there is a "reactive power" term on the mechanical side which is non-zero.

Error 3: You have said that Re/Ze as a power factor. However, Ze is resistive and Re is also resistive so there is no phase angle other than 0. I have pointed this out before (for this to be a true measure of pf then it would require Rmot =0 and Xmot =4.11 which doesn't fit facts).

There are other errors but they all stem from your evaluation of Zmot =REmot and to hell with Xmot, and the angle that you have found. . They are generally correct on that basis but the original premise is wrong. IF you had properly taken into account the existence of Xmot and its effect on the phase of Zmot and the current, you would, provided that you properly account for the phase in your calculations, (eg. E-ReI is a vector equation), you would have the same results that I have obtained. So far, you obviously have simply ignored these points when presented to you in detail. Why? They are valid comments. I have said nothing that is in conflict with Beranek or other sources.

In other words: Please deal with the following.

Why do you treat Ze as real? (and I as real follows from that) Why do you ignore Xmot? The two go together.

With curiosity

Please take your dried frog pills. You are babbling.

you know, I think somebody put something in my brownies:)

Okay., take the N from NAND and you get an AND gate operating an LED when it should'nt be working at all.

I'll suggest nothing that way again ! Frogs? for christsakes };-)

where'd you get that from.....?

You said " Please note that nowhere does Beranek call (Bl^2)/Re an "actual" mechanical resistance. "

I replied " you have yet to explain satisfactorily why in eq.7.1 and 7.2, if (Bl)^2/Re is not considered a mechanical resistance by Beranek, then why does he add (Bl)^2/Re to the other two mechanical resistances Rms and 2Rmr to obtain the total mechanical resistance Rm. "

You replied " Note that nowhere does he state that Rm is "the" mechanical resistance. "

I replied: " Not true. He defines terms up front in his book, including Rm: Page 55 " Rm is the mechanical resistance in mechanical ohms " Page 55, " Mechanical Resistance Rm obeys the following physical law " Page 54, paragraph headed as: " Mechanical Resistance Rm. "

You replied: " I should have noted that he does not call it that in Chapter 7. Thanks for the correction. I noticed that in Ch.3. it is not the Rm used in Chapter 7. "

Sorry, a typo.

Clearly then, mechanical resistance (Bl)^2/Re will dissipate electrical power

In the third line (Bl)^2/ should be (Bl)^2/Re

Northstar

Thank you - I missed that. It is the total effective mechanical resistance, which includes the effect of the electrical resistance as seen from the mechanical side. The actual mechanical resistance, as opposed to this "total" resistance is due to the mechanical and acoustic parameters. Rms

+2Rmr.

----------- Yes, I read this before. What he says is quite correct - including the "virtually equivalent" Note that he breaks things down to two forces BlE/Re and Bl(BlU)/Re which are in opposition. He is simply using superposition (Find F1 due to BlE/Re and F2 =Eb/Re in opposition so F=F1-F2 =BlE/Re -(Bl^2)/Re is the actual net force---which is what I have said all along. ) I realise that he is working in the time domain but for steady state sinusoids, phasors may be used. >

------- Certainly -that is quite right- it is a damping term.

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

Thank you for the correction. Since I have taken into account I^2Re , why should I try to take it into account a second time in (Bl^2)/Re (U^2) , particularly where this gives strange results.? The expression Power input is I^2Re +(Rms+2Rmr)U^2 " is essentially the same as that given by Berenak in the Eq.3.49 where the rm is clearly not including any Re related term. I believe that Kinsler also uses basically the same statement. Again, using the model of Fig 7.2d it is easy to solve for U, I, force, losses etc and in this model we have a source E/Re supplying the mechanical elements (treated as if an admittance) through a resistance of Re/(Bl^2). Solution of this model gives the same answer as solution of Fig.7.4 but doesn't use the source change (and source of the problem) of Fig 7.3. Note that in Fig 7.2d the loss in the Re element is {Re/(Bl^2)] |F|^2 =|I|^2Re -- that is it is correctly represented.

D> >

First, summarizing:

Your first assertion: " Please note that nowhere does Beranek call (Bl^2)/Re an "actual" mechanical resistance. "

Your second (after my rebuttal): " Note that nowhere does he state that Rm is "the" mechanical resistance. "

Your third (after my rebuttal): " I should have noted that he does not call it that in Chapter 7. Thanks for the correction. I noticed that in Ch.3. it is not the Rm used in Chapter 7. "

Your fourth (after my rebuttal): " Thank you - I missed that. It is the total effective mechanical resistance, which includes the effect of the electrical resistance as seen from the mechanical side. "

Now... you might want to read all my comments following before responding to the individual ones.

Beranek defines his Rm as the "total mechanical resistance" and you say his Rm is "total effective mechanical resistance". Then noting Beraneks total mechanical resistance Rm = 19.497 and your total mechanical resistance Rm = 2.4, this leaves you saying he defines total mechanical resistance Rm wrong. Sorry, but I think Beranek is correct, and you are in error.

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

In any event, the validity of your or my analysis hinges on the correct AC power input, the difference is that you see no power dissipated in (Bl)^2/Re in Beraneks Rm where Rm = [(Bl)^2/Re] + Rms + 2Rmr = 19.497. Note (Bl)^2/Re is what I define as the retarding motor impedance Zmtr due to electromagnetic damping from back emf Eb

Zmtr = (Bl)^2/Re = (Bl Eb/Re) / v = 17.097

Therefore, please consider the following regarding power from the source being required to overcome the back emf.

Columbia Electronic Encyclopedia, 6th ed.

Lenz's law,discovered by the German scientist H. F. E. Lenz in 1834, states that the electromotive force (emf) induced in a conductor moving perpendicular to a magnetic field tends to oppose that motion. When an electric motor is in operation, the armature is turning in a magnetic field, and an emf is thus induced in it. Lenz's law requires that this emf, called back emf or counter emf, oppose the motion of the armature and also the original emf, causing the motor to operate. As a result, the speed of the motor changes in such a way that the energy supplied by the original voltage source less the energy required to overcome the back emf is always exactly equal to the sum of the energy used to drive the mechanism to which the motor is attached and the energy lost as heat within the motor. Lenz's law may thus be seen as a consequence of the law of conservation of energy (see conservation laws, in physics).

---------

Clearly, in order for an ac motor to do useful work the external emf must be able to overcome the back-emf induced in the motor.

---------

Note Edgar Villchur (referenced in Beranek) and many others state in effect that the energy source must overcome back emf to do useful work. Please consider this carefully: This process does not take place mechanically

*per se*, but in the magnetic field (leaving your analysis mostly intact overall considering *mechanical* only). The process is between the drive voltage and back emf, or drive current and back current, if you will. In other words, the **electromagnetic damping** from back emf Eb or back current requires energy and power from the source to overcome it.

Northstar

----------- With regard to Beranek's "total" mechanical resistance, I have no problem in using this as the damping term or in determining the velocity as per q.7.1 -WHICH I DEVELOPED SEPARATELY BEFORE SEEING BERANEK. You are hung up on the fact that I have used (Bl^2)/Re +Zmec as the total or equivalent mechanical resistance. where Zmec =(Rms +2Rmr) +j(Xmd -1/wCms ) Why , I don't know, as it is exactly the the same.

In the same way, Ze =Re +Zmot includes an "equivalent electrical impedance" due to the mechanical resistance and reactance. So?

As for power dissipation- note that both I and Beranek DO account for all the power but NOT through (Bl^2)/Re but by determining the coil loss plus the loss in the actual mechanical elements.

Please READ what I said and deal with what I have said. You have given no indication that you have done that.

I repeat- If you treat U^2(Bl^2)/Re as a loss- the result is 0.0573 watts. Note that this alone is much larger than the value of 0.0328 that you call Pa +Pmec. Your mechanical power corresponds to a total mechanical resistance of 9.95 ohms which has nothing to do with (Bl^2)/Re (or with reality). It is BLATENTLY EVIDENT that YOU haven't accounted for power in the (Bl^2)/Re "resistance" You have correctly accounted for the loss in Re by using (I^2)Re

You can use a loss U^2(Bl^2)/Re as long as you also use the equivalent source power (BlEU/Re)* cos (angle of U with respect to E). I did this - again, you obviously haven't read it.

In addition, I have given you the results of calculations using the circuit of Fig. 7.2 where we have a resulting phasor, not scalar equation E/Bl=[Re/(Bl^2)]F +U where U=F/Zmec or E/Bl =[Re/(Bl^2)]F +F/Zm

This is equivalent to E=I [Re +(Bl^2)/Zm] =I[Re +Zmot] the rest follows.

Whatever way you cut it - the results are the same as what I get.

As for Lenz law- I don't need your references- This is something that I have known for over 50 years and have dealt with almost daily for most of that time. As for Villcher, this is true but somewhat simplified. Of course the back emf acts in a direction to oppose motion. (actually -tries to decrease the current and force causing motion. In fact- if this weren't true, all DC motors would overspeed to destruction, or, giving a shaft a little spin would cause a voltage aiding motion and perpetual motion would be a reality) This is simply in line with conservation of energy. Of course the magnetic field is the coupling means. In this device the magnetic field is constant so, in any time interval, the change in magnetic field energy would be 0 leading to the change in electrical input = change in mechanical output +change in losses. Thus FU =EbI The coupling relationships Eb= BlU and F=BlI implies the magnetic coupling in this simple structure. This coupling is inherent in the equations 3.59 and 3.60. Most devices are much more complex and E=BlU and F=BlI is a completely inadequate approach. Again, 50 years dealing successfully with these concepts.

I reallly don't know why you keep harping on this. I agree with Beranek et al. I may use a different notation and emphasise the electrical origin of (Bl^2)/Re . You have no problem with others using a different notation or names.such as "virtual equivalent mechanical resistance".

However, this is not the problem, however much you want it to be.

The difference between your results and mine have nothing to do with the naming of (Bl^2)/Re or with the power dissipated in this resistance (you don't account for it either! ) but with the assumption, on your part that Ze=Re +Rmot and Xmot =0 (yet Zmot has a phase angle?). If there is a rationale for this , please say so.

You say that Ze-Re =8.198-7.09 =1.11=Rmot as a SCALAR arithmetic operation, implying that all quantities are real in the mathematical sense. WHY? Then since Zmot =(11.01^2)/32.99 =3.67 in magnitude, you say that Rmot/Zmot =0.302 so that Zmot has a phase angle of 72.4 degrees. You then throw away the reactive part Xmot =3.67 sin72.4 =3.50 (round off is deliberate). WHY? It exists and does affect Ze.

When I did the work (as I showed you) correctly, in terms of vectors, without assuming any more than the following

Ze=Re +Zmot (vector equation) or in terms of magnitudes Ze^2 = (Re+Rmot)^2

+Xmot^2 and Zmot=Rmot -jXmot -in terms of magnitudes Zmot^2 =Rmot^2 +Xmot^2

the results were in agreement with the results that I got from 3 different approaches. You have all the results. Did you actually try to go through step by step.

Please deal with this. You are evading the issue. If that's what you want to do- so be it.