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Old 02-15-2013, 02:51 PM   #1
TopiRinkinen
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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 = w*k.
(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 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
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Old 02-16-2013, 07:02 AM   #2
ChipB
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A few things that strike me about your system:

1. I question this: U/I = w*k, which seems to say that if voltage is positive and current is positive you get a certain positive angular velocity of the motor, and if both are negative you get the same positive angular velocity, i.e. in the same direction. But I would think that running current backwards through this motor would cause it to spin backwards - how does this motor avoid that?

2. If current is running through the resistor it creates i^2R heat, regardless of which direction the current is flowing, and the system has no way to capture that heat, so it's lost.

3. I don't understand this statement: "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." Why must noise current cause increase in angular rotation? This is tied into #1 above - doesn't it depend on which direction the noise current is flowing? And again - that noise current will result in heating and losses in the resistor.

4. Your system assumes zero losses to mechanical friction. That assumption is the basis of most perpetual motion machine proposals, but of course is not real world.
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Old 02-16-2013, 10:26 AM   #3
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Universal motor has a property that it runs in same direction with AC, DC or DC inverted polarity.
http://en.wikipedia.org/wiki/Electri...niversal_motor

If the direction of current is changed, it is changed in both rotor and stator. And any term which has form I1 x I2, keeps it's sign (direction).

Can we dismiss your point #1 ?

#2:
IČR heat is not relevant when studying thermal equilibrium problems.
If you have two resistors connected together, other being colder than the other, electrical energy shall flow from hotter to colder. Always.
IČR heat is the part of thermal energy which is not "exported" from the resistor, but it is newer 100% (unless the other resistor's resistance is 0 ohms, or infinity). If the both resistors have same resistance, then 50% of the thermal energy transferred to the other resistor (best transfer ratio available).

#3:
If we accept the AC-motor function of universal motor, this should be quite clear. Is it not?
Link to #2 is that part of thermal electric energy is heating the resistor (and cables, if they are resistive), but for me it seems that the apparent dynamic resistance of the motor is U/I (which is w*k), and the apparent temperature of the apparent resistance is zero Kelvins.
Which is to say that part of electrical thermal energy from the resistor must flow towards motor.

#4: Yes, frictionless is key to thinking perpetual machines. This however is not about perpetual machines, I guess.
Frictionless system is no problem when studying theoretical problems.

BR, -Topi
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Old 09-25-2013, 12:15 AM   #4
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thanks, and yeah im not understanding the connection either, but im being forced to work on the question........
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Old 09-25-2013, 09:38 AM   #5
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Quote:
Originally Posted by TopiRinkinen View Post
#2:
IČR heat is not relevant when studying thermal equilibrium problems.
If you have two resistors connected together, other being colder than the other, electrical energy shall flow from hotter to colder. Always.
IČR heat is the part of thermal energy which is not "exported" from the resistor, but it is newer 100% (unless the other resistor's resistance is 0 ohms, or infinity). If the both resistors have same resistance, then 50% of the thermal energy transferred to the other resistor (best transfer ratio available).
You are probably violating a number of principles here, and definitely the Second Law of Thermodynamics. (Which is usually how perpetual motion machines operate.) The Second Law is just as sound as any other Law of Thermodynamics. Heat loss is heat loss and cannot be completely put back into the system to do mechanical work ...there will always be some heat that radiates out of the system.

For future work on this motor: Look up how the entropy will change when running your motor.

Oh. I was just reading this over and I have another comment to make. You say
Quote:
If you have two resistors connected together, other being colder than the other, electrical energy shall flow from hotter to colder.
This isn't true. It is heat that flows from hot to cold, not electrical energy. Yes, electricity can be made to flow from a hot object to cold one, but it can easily be sent in the other direction. That has nothing at all to do with heat flow.

-Dan
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Last edited by topsquark; 09-25-2013 at 10:25 AM.
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Old 09-25-2013, 01:25 PM   #6
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Quote:
This isn't true. It is heat that flows from hot to cold, not electrical energy. Yes, electricity can be made to flow from a hot object to cold one, but it can easily be sent in the other direction. That has nothing at all to do with heat flow.
I disagree:
Thermal Noise
Every resistor emits very wide bandwidth white thermal noise with constant power spectral density (PSD [W/Hz]).
If two resistors are connected (electrically) together, the hotter emits higher PSD level than the colder one, which makes energy transfer (in an electrical form) from hot resistor to cold one. period.

If there is no other energy source, apart from resistor (and wires) temperature, this will always be the case.

BR, -Topi
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Old 09-25-2013, 03:14 PM   #7
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Quote:
Originally Posted by TopiRinkinen View Post
I disagree:
Thermal Noise
Every resistor emits very wide bandwidth white thermal noise with constant power spectral density (PSD [W/Hz]).
If two resistors are connected (electrically) together, the hotter emits higher PSD level than the colder one, which makes energy transfer (in an electrical form) from hot resistor to cold one. period.

If there is no other energy source, apart from resistor (and wires) temperature, this will always be the case.

BR, -Topi
I see. Thanks for the link; I've never heard of that.

Regardless we still have the issue of the 2nd Law...Specifically the Kelvin-Planck statement of the Second Law. I'll see if I can get a better link later.

-Dan
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Last edited by topsquark; 09-25-2013 at 03:21 PM.
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