Super conductor motor?

Discussion in 'Homework Help' started by Dr.EMST, May 31, 2011.

1. Dr.EMST Thread Starter New Member

May 31, 2011
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Hello, I am currently apprenticing as an electric motor systems tech and I am learning about the power characteristics of three phase induction motors. I have learned that one Horse Power is equal to 746 Watts of true power and from that I can determine the Wattage consumed by a motor. Given the formula P=E(squared)/R I can then find the resistance of the windings. This suggests that HP is derived directly from resistance of a motor.
So if a motor was wound with super conductors, therefore having zero resistance and consuming all quadrature power and no true power, would the motor run and if so how would you calculate the HP output?

Thanks for any help,
Kevin.

2. beenthere Retired Moderator

Apr 20, 2004
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3. Dr.EMST Thread Starter New Member

May 31, 2011
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Thanks for the response!
So now I know super conductor motors do exist, how do you calculate the power output though?

Dec 26, 2010
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You have some basic misunderstandings here. A motor with superconducting windings might have no resistance (or "copper") loss, but other sources of loss such as hysteresis and eddy currents in the magnetic circuit material, bearing friction, and windage would still exist. These losses, and more importantly the mechanical output obtained fron the motor would all have to be supplied by in-phase power.

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5. Hagen Active Member

May 8, 2010
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With no dc resistance, does that mean the only impedance is a result of inductive reactance? If that is true, the motor is basically a pure inductive circuit. As such, would the current not lag the voltage by 90 electrical degrees, giving a power factor of 0?

6. Dr.EMST Thread Starter New Member

May 31, 2011
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Alright, I think that my mistake was that I was calculating a motor like it was a coil. So how I understand it is that when a motor is loaded more, it will start drawing more in phase current, increasing the power factor?

Last edited: May 31, 2011
7. amilton542 Active Member

Nov 13, 2010
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Calculating a motor like it was a coil would be treating the motor as if it were a pure parameter, the real transformer on load and no load gives a general understanding.

The speed of an unloaded motor (DC) will rise until the back emf is equal and opposite to the terminal voltage, the only current being drawn from the mains will be to overcome friction. If you apply an opposing load torque on the shaft the back emf will be less than the terminal voltage and more current will be drawn from the mains to balance the load which can be referred to as equilibrium. Voltage determines speed, current determines torque. If you were to weaken the field flux and increase the voltage the motor could operate above base speed, but in effect you would sacrifice torque for speed.

If the motor were a seperately excited "machine" and you increase the MMF excitation (current), you would increase the power factor.

Are you familiar with the four quadrant operation and regenerative braking?

8. Dr.EMST Thread Starter New Member

May 31, 2011
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Thank you for the information. When you are talking about increasing the MMF excitation on a dc machine, is that similar to the effects of a synchronous condenser?
Yes I am familiar with regenerative braking, like when torque is provided to that shaft on most dc machines they will produce power.

9. amilton542 Active Member

Nov 13, 2010
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Yes that is correct.

If the rotating reference frame (armature) or stationary reference frame (yoke) are being driven with excitation and you apply a driving torque to the shaft you will generate power, this is not regenerative braking.

When the motor is in equilibrium and you decrease the voltage to a negligible value the armature will resist the change in speed for a brief period because anything in motion has a tendency to stay in motion. The resultant will cause the back emf to be significantly bigger than the terminal voltage, the back emf will feed the system until the speed begins to diminish and will cease when the back emf is equal to the reduced value of terminal voltage.

10. Dr.EMST Thread Starter New Member

May 31, 2011
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I understand that when an armature is spinning, voltage is being induced into the windings in the armature and in the case of a motor is the result of the back EMF produced. When there is not supply voltage though, is this EMF induced not the same as the principal of a generator? Whether there is a prime mover on the shaft of the motor or the motor is just coasting, the energy contained in the spinning of the armature can be used for regenerative braking, dynamic braking or other uses. So is regenerative braking not just another form of generation?