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-   -   Universal motor and the second law of thermodynamics (http://www.physicsforums.com/showthread.php?t=672025)

 TopiRinkinen Feb15-13 05:06 PM

Universal motor and the second law of thermodynamics

Hi,

I noticed that universal electric motors have peculiar electrical <=> mechanical transfer function, which seems to break the second law of thermodynamics.

Can anyone show where did I do the mistake ?

**

Universal motor (series wounded) is a motor where rotor and stator currents always equal on magnitude.

The transfer function between mechanical and electrical domains is (ideally):

T = k*I^2,
which can be expressed also by:
U/I = k*w.
(T = torque, w = angular velocity, I = current, U = voltage, k = motor constant)

This says that a motor rotating clockwise (in this example, k>0) can act as a motor only (the sign of the torque equals to the sign of angular velocity), and not as a generator.
And the same motor rotating counter-clockwise can act as a generator only (the sign of torque differs from the sign of angular velocity), and not as a motor.

Now we take a very small universal motor rotating clockwise (k>0), and connect the terminals to a resistor; and we heat up the whole system to high (and uniform) temperature. Any electrical thermal noise presented in the electrical system (resistor, cables, coils) is seen as fluctuating AC-current in the circuit. And if this noise-current has any effect on rotation, it can only increase the angular velocity.
Also any mechanical noise on rotor (AC component of w) cannot change the sign of U*I ( =sign(U/I) ) as long as w>0, which is the requirement for transferring mechanical energy to electrical domain; so mechanical noise energy can not be transferred to electrical noise energy.

So for me it looks like any thermal energy in the resistor is transferred to thermal electrical noise which is then transferred to mechanical energy.
Another way to see is that the noise temperature of hot universal motor approaches zero Kelvins, thus breaking the second law of thermodynamics.

And this is not restricted to rotating machines only (requiring sparky/noisy commutators).
Another kind of universal motor is a plain wire loop. Any current, not depending on the sign of it, on the loop creates a magnetic field which tries to maximize the area of the loop. In any moment, the force on small part of wire is k*I^2 (k might vary during loop expansion), which classifies this as a universal motor.

I definitely missed something, but cannot pinpoint it.

BR, -Topi

 Drakkith Feb15-13 09:15 PM

Re: Universal motor and the second law of thermodynamics

Thermal motion of the charges in a conductor do not create a magnetic field because the motion is entirely random. Overall each charge will cancel out other charges, with the net result being no magnetic field. There is no current flow generated by the heat, as that requires that the motion not be random, but have a net flow in a certain direction.

 TopiRinkinen Feb16-13 11:50 AM

Re: Universal motor and the second law of thermodynamics

I don't quite buy that, yet.

If you connect a resistor to antenna, the antenna will radiate thermal energy originating from the resistor.
Ideally, if you point the antenna towards sky, and insulate everything, the resistor shall cool down to sky temperature. Or if the sky is hotter than the resistor, the resistor shall heat up until equilibrium is reached.
And you can also connect the resistor to antenna through a transformer and the result is still the same (assuming ideal components). And in the transformer, the noise power is transported by means of magnetic field.

So I don't see any reason given why electromagnet would repel electricity originating from thermal noise, and accept everything else.

BR, -Topi

 Drakkith Feb16-13 12:00 PM

Re: Universal motor and the second law of thermodynamics

Do you have a reference for any of that? The only way I know of for any of what you said to happen is through thermal radiation, which is not the same thing as what you are saying. However I am no expert and I could very well be mistaken, so if you know of something that supports your claims please say so.

 TopiRinkinen Feb16-13 01:33 PM

Re: Universal motor and the second law of thermodynamics

http://first-quantum.net/forStudents...e/chapter4.pdf
Page 14, last sentence:

"If the conductors R1 and R2 are at different temperatures, µ1 and µ2, the
net heat °ow should be proportional to the temperature di®erence µ1 ˇ µ2 and thus the
power spectrum Sv(!) is proportional to the temperature µ."

BR, -Topi

 DaleSpam Feb16-13 01:51 PM

Re: Universal motor and the second law of thermodynamics

Quote:
 Quote by TopiRinkinen (Post 4271515) Universal motor (series wounded) is a motor where rotor and stator currents always equal on magnitude. The transfer function between mechanical and electrical domains is (ideally):
What are the assumptions that go into this ideal transfer function? Are any of those assumptions violated in thermal noise? In particular, are thermal fluctuations in rotor and stator currents always equal in magnitude?

 TopiRinkinen Feb16-13 02:13 PM

Re: Universal motor and the second law of thermodynamics

Assumptions I can find are:

1. The bandwidth of thermal noise is upto several hundreds of GHz in room temperature. When the wavelength of the current is shorter than the total coil length, the current shall have opposite direction in rotor and stator. But this can be circumvented by band-pass limiting the system.

2. Practical universal motors have commutator (bushes) which cause excessive noise and discontinuities (in time) of currents and voltages. And this can be circumvented by using the plain-loop-setup I described in the original post.

3. The noise power is such a tiny value, that to run any motor with it one needs either extremely high temperatures, or extremely small motors (the noise-power of any resistor is constant (in constant temperature), so it does not scale with size), but the motor scales down with size.

The rotor and stator are serially connected, so any current (having frequency low enough to avoid phase shift) are equal in magnitude and direction. And it does not depend on the source of the current (whether noise or power supply).

One way to think of this is to build a device which:

a) has band-pass between resistor and motor, to filter of 50 Hz..60 Hz to mimic utility electricity.
b) heating the whole system to (very high) temperature which generates 1 Volts AC RMS when measured after the filter.
c) by replacing the resistor with 1 Volt AC 55 Hz power supply.

It should not matter whether the band-passed power comes from thermal resistor or power supply. If the motor runs with power-supply-supplied electricity, it should also run with thermal-generated electricity.

BR, -Topi

 256bits Feb16-13 03:39 PM

Re: Universal motor and the second law of thermodynamics

Nice question or observation.
Thermal noise or white noise is present in electrical components and the random agitation of electrons is temperature dependant. It also is called Johnson noise. So any electrical component will have a random voltage across it.

Would not attempting to harness this voltage as a power source, have the same objections as Maxwells Demon? A parallel would be Brownian motion, yes/no.

It would be interesting to see how someone familiar with fluctation-dissipation theory or Johnson–Nyquist noise can answer this.

 DaleSpam Feb16-13 05:33 PM

Re: Universal motor and the second law of thermodynamics

Quote:
 Quote by TopiRinkinen (Post 4272516) Assumptions I can find are: 1. The bandwidth of thermal noise is upto several hundreds of GHz in room temperature. When the wavelength of the current is shorter than the total coil length, the current shall have opposite direction in rotor and stator. But this can be circumvented by band-pass limiting the system. 2. Practical universal motors have commutator (bushes) which cause excessive noise and discontinuities (in time) of currents and voltages. And this can be circumvented by using the plain-loop-setup I described in the original post. 3. The noise power is such a tiny value, that to run any motor with it one needs either extremely high temperatures, or extremely small motors (the noise-power of any resistor is constant (in constant temperature), so it does not scale with size), but the motor scales down with size.
I don't see the connection between any of those assumptions and the equation that you gave for the transfer function. You also didn't describe how well those assumptions hold in this example.

I don't know enough about universal motors to understand the resolution to your specific problem, nor do I clearly see the link between thermal noise, the transfer function you listed, and a violation of thermodynamics, but in general you must be making some assumption which is wrong.

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