Proof of common sense when maths is lacking concept

[milesyoung]

... To reitterate the electric fan uses less power than the kettle (or water heater, air con etc.) ...

He/she did mention it in the OP..

 

[Ratch]

yungman,


See post #9 of this thread.

Ratch

 

[OmCheeto]

that seems like it's violating concervation of energy (I know it mustn't be) but you know what I mean, the more you get (like more torque or whatever) the less power, for that matter it seems like I could just put a resistor on the end of the terminal and make it use even less power (though that would probably slow the fan down). Sorry this question must be so irritating but I hope you can see why I'm confused

I cannot understand why you are confused. But I have extensive electrical training/education/background.

Electric Kettle:
rated power: 1500 watts
line voltage: 120
derived current: p/v=i --> 1500/120=12.5 amps
derived resistance: v/i=r --> 120/12.5=9.6 ohms

Electric Fan:
rated power: 100 watts
line voltage: 120
derived current: p/v=i --> 100/120=.833 amps
derived resistance: v/i=r --> 120/.833=144 ohms

I guess I don't understand where common sense comes into play with this problem. There are only equations.

hmmm... (scratches head, runs aerodynamic analogy through skull...)

Ah ha! I think I understand what you are asking. Interesting.

If mythical car A has an aerodynamic drag resistance of 144 units, it would make sense that it would consume more power than mythical car B with an aerodynamic drag resistance of 9.6 units.

(I'm ignoring back EMF and inductive resistance here, as I'm obviously interpreting your question differently than others.)

hmmm.. Where do we go from here? The mythical cars will both have to have engines. I'll give them both the same engine, rated at 120 mythical units, which I will call volts.

And what does this tell us so far? It tells us how fast the cars will go! Volts/Ohms = Amps.

Amps are the mythical unit for velocity that the cars will travel at.
(This is somewhat of a bad analogy, as real Amps are a measure of a certain number of electrons passing a certain point per second. Kind of like the number of cars driving through an intersection per second. But it kind of implies motion, and I'll stick with it. Let's just pretend that each car is negatively charged with an excess of 6.241 × 10¹⁸ electrons(or 1 Coulomb), and each cars length is equal to 1 coulomb per 0.003 miles, and see what happens)

Car A: (aka, the fan)
volts: 120
ohms: 144
amps: .833
(e-gads. now I understand the PF-EE heavyweights aversion to analogy)
length: 13 feet
velocity: 9 mph
power: (Force, aka volts)*Velocity = 1076 oomphs(new mythical name of electro-aerodynamic power unit)

Car B: (aka, the kettle)
volts: 120
ohms: 9.6
amps: 12.5
(why is length based on charge?)
length: 198 feet
velocity: 135 mph
power: 16,200 oomphs

So what does this all mean? They've both got the same motive force. But the car with the least resistance is using the most power.

scratches head again. hmmm... It still doesn't make sense, but at least the maths worked out:

16,200 oomphs / 1076 oomphs = 15.06
1500 watts / 100 watts = 15

-----------------------------
ok to delete. I enjoy nothing more than Sunday morning mental aerobics.

 

[toneboy1]

Some good points were made and I did understand everything said and why, except for maybe you cheeto, but it was a hilarious departure nevertheless.

When the motor is turning, the back EMF (which is proportional to the RPM) reduces the current. So as the motor speed increases, the current decreases and the electrical power decreases.

The amount of mechanical work being done by the motor usually increases with the RPM, and the result is that the motor runs at the speed where the electrical power equals the mechanical power. Below that speed, the electrical power "wins" and speeds the motor up. Above that speed, the mechanical work "wins" and slows it down.

The same idea applies to AC motors, but it gets complicated trying to explain it in words rather then doing the math, and from your original question, I'm guessing you are not familiar with how to analyse AC electrical circuits yet.

Actually I have some experience with AC circuits a few years ago but have learned a lot since than and you know what they say, learn something new pushes something old out of your brain. I need a refresher.
So to summerise, the more resistance of a load on a motor, the more it increases the current, which makes it use more power. So if it's an ac motor it will change the phase of the voltage or current, (which I don't know) so the peaks will be closer and I*V will be bigger, or if its DC then the back emf will decrease and it will use more power. COOL. Exactly the sort of explanation I was looking for.

thanks

EDIT: If any of you (possibly clever engineers) could, I have an algebraic nightmare (a few of them) in this thread: https://www.physicsforums.com/showthread.php?p=4165324&posted=1#post4165324

I'd REALLY appreciate some help!

EDIT: What the hell, I have a Fourier question too for what it's worth:
https://www.physicsforums.com/showthread.php?p=4165336#post4165336

 

[Windadct]

Hello Toneboy - it may be possible that if you just measure the Resistance of the motor it has less resistance then the heater - but STILL uses less power - why? Because a motor when running - is not a resistor. The structure of the motor generates a magnetic field, and as the motor rotates, it generates "back EMF" - and still a little more dynamic than just impedance - for example a DC motor generated Back EMF - but steady state DC in an inductor does not have back EMF.
Since now the motor is generating EMF ( simply put Voltage) that opposes the voltage being applied - this reduces the current.

 

[Ratch]

toneboy1,

So to summerise, the more resistance of a load on a motor, the more it increases the current, which makes it use more power.

You still don't have it quite right. Yes, you can stall the motor and draw a fuse popping, wire burning, switch melting amount of current, but if the current is not in phase with the voltage across the motor, the power will not appear. I tried to explain that with my inductor example previously.

So if it's an ac motor it will change the phase of the voltage or current,

All motors will change the phase of their voltage/current if they have inductance.

so the peaks will be closer

The peaks are determined by the line frequency. That usually does not change.

and I*V will be bigger,

The power will be greater because the voltage and current are in phase longer.

or if its DC then the back emf will decrease and it will use more power.

The "back EMF" is due to the collapsing magnetic field. Any motor type that uses inductance will have a back voltage. A motor will use more power if its voltage and current are in phase during more of the rotation cycle.

COOL. Exactly the sort of explanation I was looking for.

Are you sure about that?

If any of you (possibly clever engineers) could, I have an algebraic nightmare (a few of them) in this thread: https://www.physicsforums.com/showthr...=1#post4165324

I'd REALLY appreciate some help!

You should start a new thread in the math section for that problem.

Ratch

 

[sophiecentaur]

It would be a great idea if this problem could be sorted out for Resistive Loads first. When we've got that all straightened out then we could move on to inductive and rotating loads. These are much more complicated and certainly not to be leaped into without getting the basics first.
For resistive loads, all you need to thing of is
1) P = IV : the Power relationship
and 2) R = V/I : the way we define Resistance

Combine these two together and you get two more useful formulae:
3) P = V²/R
and
4) P = I²R

Those four formulae will give you the answer to all possible questions about resistance, power and the way it can be supplied.
3) shows you that the power delivered (and used) goes down as the resistance is increased.
4) shows how the power lost in a supply wire, for instance, will increase as the resistance of the wire increases (that's if you keep the current the same)

 

[NascentOxygen]

On the contrary, it means more ohmic resistance, but what I'm saying is that say a fan would have LESS resistance than a heater but still use LESS power, or am I wrong, is a fan actually manifold more resistive?

Correct, your fan's motor could have less resistance and yet still use less power. Indeed, as an experiment you could make your motor windings have almost no resistance by using pure gold wire, an almost perfect conductor, and the fan would still draw less current* and use less power than the heater.

This is because the current in a fast spinning fan is largely not determined by the windings' resistance. As others have pointed out, the electrical model for a fan is not simply a fixed resistance and nothing else. It's that "something else" that is the key to explaining why it is that motors use more power the harder they work. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

 

[toneboy1]

Thanks Windadct. Rach: "The power will be greater because the voltage and current are in phase longer." good way to put it, but I didn't necessarily mean stall the motor, just slow it down.

"The "back EMF" is due to the collapsing magnetic field" Ah, right, ofcourse, thanks.

As for the thread in the math section, I did, but I from what I've seen the people here stand a better chance of allowing me to figure it out.Thanks sophiecentaur, I think I've got the purely resistive relationships down and as far as I'm aware the phase relationship of an ac motor too.
The only possible area of lacking I feel might be possible is the way the EMF reduces the current in a DC motor. That said I'm still satisfied.

 

[Ratch]

TB1,

If any of you (possibly clever engineers) could, I have an algebraic nightmare (a few of them) in this thread: https://www.physicsforums.com/showthr...=1#post4165324

I'd REALLY appreciate some help!


As for the thread in the math section, I did, but I from what I've seen the people here stand a better chance of allowing me to figure it out.

You really need to get a computer program that can do equations like that. Otherwise you can spend hours and hours spinning your wheels and probably making lots of mistakes. I think there are some trial programs and freebies on the web from outfits like Wolfram, but I am not sure. Anyway, the first thing to do is rationalize both sides of the equation and see if they match. Rationalize means no complex terms in the denominator. That is a lot of work, but if they do match, then you proved they are equal and are finished. You rationalize the left side by first calculating a common denominator, and then multiply the numerator and denominator by the conjugate of each complex term in the denominator. Collect similar terms whenever you can. That is a lot of scut work, but it shows if the equations match. As you can see from the attachment, the right side should be twice what the rationalization shows to equal what the left side is, so the problem equation is wrong. You might be tempted to look for shortcuts by finding equal terms in the numerator and denominator, but for an equation that complex, you will probably just get tangled up trying to take advantage that situation. I think, to just do it systematically. (Ugh)

Ratch

 

[toneboy1]

TB1,



You really need to get a computer program that can do equations like that. Otherwise you can spend hours and hours spinning your wheels and probably making lots of mistakes. I think there are some trial programs and freebies on the web from outfits like Wolfram, but I am not sure. Anyway, the first thing to do is rationalize both sides of the equation and see if they match. That is a lot of work, but if they do match, then you proved they are equal and are finished. You rationalize the left side by first calculating a common denominator, and then multiply the numerator and denominator by the conjugate of each complex term in the denominator. Collect similar terms whenever you can. That is a lot of scut work, but it shows if the equations match. As you can see from the attachment, the right side should be twice what the rationalization shows to equal what the left side is, so the problem equation is wrong. You might be tempted to look for shortcuts by finding equal terms in the numerator and denominator, but for an equation that complex, you will probably just get tangled up trying to take advantage that situation. I think, to just do it systematically. (Ugh)

Ratch

Thanks, well I had no reason to think they wouldn't be the same, but you're saying they're not equal? That shouldn't be, are you sure?

For argument's sake if I just had the LHS and wanted to reduce it to the form of the RHS would I still have to do the conjugate times numerator and denominator? (surely not)

I was just looking for the simplest way to solve the LHS without expanding into 25 term polynomials and having to factorise cubics.

I've used wolfram on the net and I have MATLAB and I think maple (which I've never used) but I don't know what I'd use to show me how to solve anything like that.

 

[Ratch]

TB1,

Thanks, well I had no reason to think they wouldn't be the same, but you're saying they're not equal? That shouldn't be, are you sure?

Yes, I am. I used Maple, and it has never given me a wrong answer. I rationalized both the RHS and the LHS, and they don't come out equal. You can spend a lot of time trying to wrestle one expression into another, especially if they are not equal. There are a zillion alternative ways to represent an expression, but if you know the end result, it is best just to determine if they are equal and walk away.

For argument's sake if I just had the LHS and wanted to reduce it to the form of the RHS would I still have to do the conjugate times numerator and denominator? (surely not)

To do it the hard way, first rationalize it. Then multiply the numerator by a hairy factor to get what you want in the numerator. Multiply the denominator by the same factor to keep everything equal. Simplify the denominator. Then use the terms in the denominator, and factor if necessary to create your own expression. You have to use all the denominator terms, and cannot use extra terms. Of course, you can use the computer to subtract the terms you use, and factor if necessary. The choices are many.

I was just looking for the simplest way to solve the LHS without expanding into 25 term polynomials and having to factorise cubics.

You are not solving an equation. You are proving an equality.

I've used wolfram on the net and I have MATLAB and I think maple (which I've never used) but I don't know what I'd use to show me how to solve anything like that.

Sounds like you are well stocked with software.

Ratch

 

[toneboy1]

Thanks for pointing out that they're not equal, I could have wasted even more time.
I'll have to give maple a go when I get a chance. To varify.
To clarify, these two steps were in someone's working out, and I didn't know how they got from LHS to RHS, it wasn't a question of proove they're equal or not (if there's been a misunderstanding). I still don't see how I'd factorise the denominator to get it like ()^2 + ()^2 from the denominator of your (1). To do this it doesn't matter if there are j's in the denominator.

 

[Ratch]

TB1,

Thanks for pointing out that they're not equal, I could have wasted even more time.
I'll have to give maple a go when I get a chance. To varify.
To clarify, these two steps were in someone's working out, and I didn't know how they got from LHS to RHS, it wasn't a question of proove they're equal or not (if there's been a misunderstanding). I still don't see how I'd factorise the denominator to get it like ()^2 + ()^2 from the denominator of your (1). To do this it doesn't matter if there are j's in the denominator.

You can see from the attachment that I used the rationalize function of Maple. I would have thought that Matlab and the Wolf could do what Maple did. Maple can factor expressions, but that is not necessary to rationalize an expression, even if you do it manually. You need to multiply by conjugates to get rid of complex terms in the denominator. Then use the procedure I outlined previously to make any expression that is possible from the terms in the denominator.

Ratch

 

[toneboy1]

TB1,
You can see from the attachment that I used the rationalize function of Maple. I would have thought that Matlab and the Wolf could do what Maple did. Maple can factor expressions, but that is not necessary to rationalize an expression, even if you do it manually. You need to multiply by conjugates to get rid of complex terms in the denominator. Then use the procedure I outlined previously to make any expression that is possible from the terms in the denominator.

Ratch

I'm sure there is a way with them but I'm not that proficient yet (matlab wolf).
I don't see why it's necessary to eliminate all the complex numbers on the denominator or how it could be factorised to that, given the highest exponent of the solution saught will be 4, not 8. I could factorise the two term part by completing the square, but as for the 3 term polynomial squared, I wouldn't know where to start...Thanks for your help anyway.

 

[sophiecentaur]

Thanks sophiecentaur, I think I've got the purely resistive relationships down and as far as I'm aware the phase relationship of an ac motor too.
The only possible area of lacking I feel might be possible is the way the EMF reduces the current in a DC motor. That said I'm still satisfied.

Sorry for trying to teach my Grandmother to suck eggs!

Reading this thread reminds me how similar electric motors and radio antennae behave (and even loudspeakers). The up-front resistance is usually just not relevant to their performance - a dipole is an 'open circuit' but it can look like a useful 70Ω at the right frequency. Electrical resistance turns up where you'd least expect it but Energy Conservation requires it.
The Back EMF is only a 'way of looking at it', I think. There must be an alternative way of looking at it involving self inductance of a rotating machine.

 

[Ratch]

TB1,

I think I've got the purely resistive relationships down and as far as I'm aware the phase relationship of an ac motor too.
The only possible area of lacking I feel might be possible is the way the EMF reduces the current in a DC motor.

Like I said before, the back voltage is caused by the collapsing magnetic field when the current changes direction. It shouldn't be hard to imagine a "bucking" voltage going against the source voltage to reduce its value and lowering the current. It is a physical happening, and needs no alternative way at looking at it. The constant shift of electrical energy from the circuit to the inductor and back again also causes a current/voltage phase shift associated with inductive reactance.

Ratch

 

[Ratch]

TB1,

Just to follow up on your previous question about that math expression. If you look at equation (6) of the attachment, you can find an alternative form of that expression. If fact, there are many different forms. Once you have the denominator rationalized, you can use partial fraction expansions. Then the number of different expressions is only limited by your imagination.

Ratch

 

[milesyoung]

I'm going to throw in my 2 cents here because I think there's a small misconception with regards to self-inductance and its significance.

Let's consider a single winding of an AC motor which we'll model as a resistor in series with an inductor. It's true that a changing current through the winding will produce an opposing voltage in proportion to the self-inductance of the winding, but this isn't what limits the current in the circuit as the angular velocity of the rotor increases.

Be it a DC or an AC motor, as the motor speeds up, the current is limited by the voltage produced by the increasing rate of change of magnetic flux through the winding, due to the increasing angular velocity of the rotor field. It's this component of induced voltage that's usually termed 'back-EMF' in motors.

 

[Ratch]

milesyoung,

It's true that a changing current through the winding will produce an opposing voltage in proportion to the self-inductance of the winding, but this isn't what limits the current in the circuit as the angular velocity of the rotor increases.

Be it a DC or an AC motor, as the motor speeds up, the current is limited by the voltage produced by the increasing rate of change of magnetic flux through the winding, ...

I don't understand. First you say that the opposing voltage does not limit the current. Then you say that the voltage does limit the current. What's an inquiring mind to think?

Ratch

 

[milesyoung]

Give it a second read-through - in the simplest case there are two causes of changing magnetic flux in the winding. One is a change in current in the circuit - the other is a change in the angular position of the rotor. The latter is what limits the current.

Edit: And just clarify further, I'm talking about two different components of opposing voltage - that's my whole point.

 

[Ratch]

milesyoung,

Give it a second read-through

I did, several times.

in the simplest case there are two causes of changing magnetic flux in the winding. One is a change in current in the circuit

OK, I believe you are referring to the change caused by the alternating line voltage.

the other is a change in the angular position of the rotor. The latter is what limits the current.

Are you sure you don't mean the angular velocity of the rotor?

The latter is what limits the current

A big inductance coil does not have a rotor, and yet it limits current. In any case, anything that causes a conductor to cross a magnetic flux in a perpendicular manner will generate a voltage in the conductor.

Ratch

 

[milesyoung]

OK, I believe you are referring to the change caused by the alternating line voltage.

I'm talking about a change in instantaneous current.

Are you sure you don't mean the angular velocity of the rotor?

No (edit: As in, yes I'm sure :). A change in rotor position will change the magnetic flux through the winding and induce a voltage. A change in velocity will chance the rate of change of magnetic flux through the winding.

A big inductance coil does not have a rotor, and yet it limits current. ...

Yes but this isn't what limits the current as the angular velocity of the rotor increases. How could it? This voltage isn't a function of anything that has to do with the rotor. It's only a function of the rate of change of current in the winding, considering the self-inductance of the winding to be constant.

How else would you explain an increasing opposing voltage in a DC motor? For a constant current the inductive voltage drop associated with the self-inductance is clearly not present.

 

[Ratch]

milesyoung,

I'm talking about a change in instantaneous current.

Caused by?

change in rotor position will change the magnetic flux through the winding and induce a voltage. A change in velocity will chance the rate of change of magnetic flux through the winding.

A change of position is velocity, a change of velocity is acceleration.

Yes but this isn't what limits the current as the angular velocity of the rotor increases. How could it?

Partly by inductive reactance, whether the rotor is turning fast or slow.

This voltage isn't a function of anything that has to do with the rotor.

Are we still talking about the back-voltage?

It's only a function of the rate of change of current in the winding, considering the self-inductance of the winding to be constant.

Stator winding or rotor winding?

How else would you explain an increasing opposing voltage in a DC motor?

Permanent magnet DC motor? OK, then rotor winding. So both the rotation of the rotor and the switching of the commutator can change the current.

How else would you explain an increasing opposing voltage in a DC motor?

Yes, changing current with constant inductance produces voltage proportional to current change. Now, what all can change the current? Rotation and the commutator?

For a constant current the inductive voltage drop associated with the self-inductance is clearly not present

The current may be constant from the source, but the commutator changes its direction in the rotor.

Ratch

 

[milesyoung]

Caused by?

Why does it matter?

A change of position is velocity, a change of velocity is acceleration.

I wrote "change in the angular position of the rotor". If I replace "position" with "velocity", which I would argue your response would ask me to consider, the statement is false.

Partly by inductive reactance, whether the rotor is turning fast or slow.

Yes _partly_ if the instantaneous current is changing - which it isn't necessarily in a DC motor. The effect that's limiting current in your "big inductance coil" might also be present, but this is not what is the limiting factor that increases with the angular velocity of the rotor and it's not usually what would be called "back-EMF" in motors.

Are we still talking about the back-voltage?

Back-EMF is a confusing term because its used by different people to describe different things. Usually it means the part of the "back-voltage" that's dependent on the angular velocity of the rotor but some will lump the voltage produced by a change in the instantaneous current in with it.

Stator winding or rotor winding?

Stator winding. Why would it have something to do with the rotor winding?

Permanent magnet DC motor? OK, then rotor winding. So both the rotation of the rotor and the switching of the commutator can change the current.

It can certainly not change the current in the stator winding (that would make it AC), which is where the back-EMF appears.

Yes, changing current with constant inductance produces voltage proportional to current change. Now, what all can change the current? Rotation and the commutator?

Again, I have no idea why you would bring the commutator into this - we're naturally talking the stator winding.

Edit:
Ah sorry I completely missed the "Permanent magnet DC motor" part. In that case I'm referring to the part of the rotor windings on the stator side of the brushes (which sees no alternating current).

 

[sophiecentaur]

Again, I have no idea why you would bring the commutator into this - we're naturally talking the stator winding.

It would be better to consider a permanent magnet motor first, surely - there is no stator winding. A brush motor with a permanent field magnet will behave just like a dynamo. By Lenz's law, the EMF generated by such a dynamo will be 'such as to oppose its cause' i.e. it will be against the polarity of the supply volts. The commutator is necessary for this to happen and it is, of course, necessary for the motor to be turning in the first place.
If you want a stator winding then, as this passes DC (series or shunt connection) will that affect or produce any 'back EMF'? I don't think so.
The load on the motor is what limits the speed of the motor and that governs the difference between supply voltage and back EMF, which determines the current that can flow through the resistance of the rotor. The resistance that the supply 'sees' would be supply volts / rotor current.

For a series wound motor, the field current will drop as the speed builds so the speed regulation is not so good.

 

[milesyoung]

You're absolutely right, consider the simple DC-motor model which probably needs no introduction:

http://www.library.cmu.edu/ctms/ctms/simulink/examples/motor/motorsim.htm

Let the motor be loaded so the rotor has a constant angular velocity and let the current in the circuit be constant in magnitude. Would you say the back-EMF in this case has anything to do with the self-inductance L?

 

[sophiecentaur]

If you were to increase the supply volts, the current would increase significantly until the revs built up, a reducing the current to slightly more than before. That looks, to me, mighty like a big parallel capacitor, relating to the MI of rotor and load.

 

[milesyoung]

I'm not quite sure what you mean. I can't see how you could argue that the induced voltage from flux cutting the rotor windings could be attributed to self-inductance.

 

[toneboy1]

Sorry for trying to teach my Grandmother to suck eggs!

Reading this thread reminds me how similar electric motors and radio antennae behave (and even loudspeakers). The up-front resistance is usually just not relevant to their performance - a dipole is an 'open circuit' but it can look like a useful 70Ω at the right frequency. Electrical resistance turns up where you'd least expect it but Energy Conservation requires it.
The Back EMF is only a 'way of looking at it', I think. There must be an alternative way of looking at it involving self inductance of a rotating machine.

heh, it wasn't quite like that. Thanks for the interesting input.

TB1,

Just to follow up on your previous question about that math expression. If you look at equation (6) of the attachment, you can find an alternative form of that expression. If fact, there are many different forms. Once you have the denominator rationalized, you can use partial fraction expansions. Then the number of different expressions is only limited by your imagination.

Ratch

So you're saying I could force it into that form after partial fractions, interesting.
THOUGH it would take ages to even get to (1) if you didn't have a computer to aid in expanding and simplifying the fraction, wouldn't it?
Also, I don't quite see where H came from when you introduced it.

Thanks