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Proof of common sense when maths is lacking concept

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milesyoung
#19
Nov18-12, 10:09 AM
P: 535
Hehe, yes - less power but also less current, which means less field strength and less torque. It's really not that simple.
toneboy1
#20
Nov18-12, 10:13 AM
P: 174
Quote Quote by milesyoung View Post
Hehe, yes - less power but also less current, which means less field strength and less torque. It's really not that simple.
M'mm, I'll accept that. I'm going to get my multimeter tomorrow and check what my fan and 500W distiller are, I really can't believe a fan would be so resistive.

Thanks for your help.
Ratch
#21
Nov18-12, 10:13 AM
P: 315
toneboy1,

I see what your saying,
I am not so sure about that.

...if you endulge me, so if I had two motors, both with the same guage wire, one with more windings and larger, one with less and smaller, the smaller one would have more current and thus use more power?
Are the current and voltage in phase? If you applied AC line voltage to a single coil of high inductance, you could have a tremendous amount of current, and yet dissipate a very small amount of power. That is because the voltage and current are out of phase with each other. It will take a large amount of current to build up energy in the magnetic field of the coil during a quarter cycle, but that energy will be given back to the circuit in the next quarter cycle, and repeat the process in the reverse direction. Therefore, it will be large current and small power dissipation.

You are trying to confuse AC power distribution with DC power. It doesn't work that way. You need to know the phase ramifications of AC before you can calculate AC power.

Only the resistance part of the impedance dissipates power. The more the voltage and current are in phase, the more the resistance dominates the impedance.

Ratch
toneboy1
#22
Nov18-12, 10:16 AM
P: 174
Quote Quote by Ratch View Post
toneboy1,

Are the current and voltage in phase? If you applied AC line voltage to a single coil of high inductance, you could have a tremendous amount of current, and yet dissipate a very small amount of power. That is because the voltage and current are out of phase with each other. It will take a large amount of current to build up energy in the magnetic field of the coil during a quarter cycle, but that energy will be given back to the circuit in the next quarter cycle, and repeat the process in the reverse direction. Therefore, it will be large current and small power dissipation.

You are trying to confuse AC power distribution with DC power. It doesn't work that way. You need to know the phase ramifications of AC before you can calculate AC power.

Only the resistance part of the impedance dissipates power. The more the voltage and current are in phase, the more the resistance dominates the impedance.

Ratch
YES that's more the answer I've been looking for! THANK YOU, I knew a bloody fan wasn't that resistive.
Out of curiosity, what about a DC motor, given that there is no phase change (or voltage lag) would one use more power than its ac counterpart?
toneboy1
#23
Nov18-12, 10:20 AM
P: 174
Quote Quote by milesyoung View Post
Edit: And of course as phinds pointed out, I assume we're strictly talking DC here.

Edit: You're not actually going to use this argument for anything important, right? There's a lot more to motors..
I wasn't just talking DC, yeah I know there is a lot more to motors but every analogy I use seems to make the question harder to explain, I knew there was just a simple answer, I just ended up using motors to make the point.

EDIT: DC probably would have simplified things though :P
milesyoung
#24
Nov18-12, 10:27 AM
P: 535
I was just about to write that you were probably going to come back tomorrow angry that you've measured the resistance using an ohmmeter and you were right about the resistance etc :)

Just to make my point clear - I've only been talking about DC current and ohmic resistance. If you measure the average current into your circuit for some voltage, that relationship will give you its effective DC resistance - it won't have much to do with the winding resistance you'll measure (which will be very low).
Ratch
#25
Nov18-12, 10:31 AM
P: 315
TB1,

Out of curiosity, what about a DC motor, given that there is no phase change (or voltage lag) would one use more power than its ac counterpart?
A DC motor has a commutator that switches the current direction in the rotor. That causes magnetic fields to form and collapse, thereby storing and releasing energy. This in turn causes a phase change in the voltage and current. Its behavior is complicated. See the neat animation at this link http://en.wikipedia.org/wiki/DC_motor .

Ratch
AlephZero
#26
Nov18-12, 10:42 AM
Engineering
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P: 7,293
When a DC motor is rotating it acts like a generator, and creates a "back EMF", in other words a DC voltage that is opposite to the applied voltage. If you prevent the motor from turnng, the current through it is be given the Ohm's law. That is called the "stall current" for the motor and it can be much higher than the normal working current.

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 than doing the math, and from your original question, I'm guessing you are not familiar with how to analyse AC electrical circuits yet.
Runei
#27
Nov18-12, 10:47 AM
P: 140
The fan and the heater are both connected to an AC outlet. So we have to consider the AC system.

The heater would be made up of some sort of simple ohmic resistance that can handle a lot of power.

The power dissipated by the heater would then be (V_rms^2)/R.

The fan is a coil/resistance circuit. Possibly the resistances are not very big. But the coils may have a large inductance. Because of the large inductance, the circuit will not draw very much current because it is running in AC mode. Thus, the resistances are not dissipating very much power.

Heater - Small resistance - High power.
Fan - Large indutance leads to less AC current - Low power.
Ratch
#28
Nov18-12, 11:11 AM
P: 315
milesyoung,

If you measure the average current into your circuit for some voltage, that relationship will give you its effective DC resistance - it won't have much to do with the winding resistance you'll measure (which will be very low).
Perhaps you know more about this than I do, but let me give you my take on what happens.

The average current is zero, just like it is when you apply an AC voltage across a plain resistor. DC resistance is the same as AC resistance, specifically the resistance of the motor coils and brush resistance. I don't know what effective DC resistance is. When a motor is running with no load, its voltage and current are way out of phase, and it is taking relatively little power from its voltage source. When a motor has a load, its impedance lessens because its reactance lessens, and its impedance becomes more resistive. This causes the motor to draw more power from the voltage source due to the phase change. During this time the resistance has not changed. It is only the reactance that has changed, which in turn changes the phase and increases power consumption.

Ratch
milesyoung
#29
Nov18-12, 12:38 PM
P: 535
Ratch,

I can't argue with any of that. It seemed like at the start that there were some trouble with the basics, so I was trying to stick to the definition of resistance and avoid the details of AC/motors: A DC motor would appear as having a higher resistance at its terminals when running due to back-emf.
yungman
#30
Nov18-12, 12:56 PM
P: 3,904
How can you say who use more power if there is no specification of the heater and fan? You can have a really big fan and a tiny heater!!!
milesyoung
#31
Nov18-12, 01:00 PM
P: 535
... 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
#32
Nov18-12, 01:02 PM
P: 315
yungman,


See post #9 of this thread.

Ratch
OmCheeto
#33
Nov18-12, 01:40 PM
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P: 1,435
Quote Quote by toneboy1 View Post
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
#34
Nov19-12, 02:30 AM
P: 174
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.

Quote Quote by AlephZero View Post
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 learnt 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 dont 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: http://www.physicsforums.com/showthr...=1#post4165324

I'd REALLY appreciate some help!

EDIT: What the hell, I have a Fourier question too for what it's worth:
http://www.physicsforums.com/showthr...36#post4165336
Windadct
#35
Nov19-12, 02:21 PM
P: 575
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
#36
Nov19-12, 03:32 PM
P: 315
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: http://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


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