Proof of common sense when maths is lacking concept

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,

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.

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

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

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

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.

Are you sure about that?

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

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.

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,

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. Best, I think, to just do it systematically. (Ugh)

Ratch

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toneboy1

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,

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.

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.

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

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 dont 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,

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

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

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,

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

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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,

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.