Motor theory question 2

[jhyoung]

I just came across this forum and was wondering if anybody can get me straightened out.

When a mechanical load attached to an AC motor increases, the electrical load must increase in order to keep the motor running at a constant speed. Obviously the current through the motor must increase for this to happen, but where does this extra current come from? For the motor current to increase the CEMF must decrease. Since CEMF inhibits current flow, it is essentially inductive reactance. Inductive reactance is found by 2(pi) x frequency x inductance.
According to this formula, the only two things that can change the value of inductive reactance is frequency and inductance. Both of which do not change (as far as I know) when the mechanical load across a motor changes.

 

[itsmoked]

As the load increases the slip increases. More slip more current.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[davidbeach]

Look at the motor equivalent circuit and notice that rotor resistance is a function of slip.

 

[waross]

When you apply your formula to the motor you must be aware that the induction of the rotor is subject to and determined by the slip frequency which changes as the rotor slows slightly under load. This change in characteristics is reflected to the main windings and in the current drawn.
You may feel intuitively that an increase in slip frequency should result in an increase in inductive reactance and a lower current. This is an indication that there are a lot of complex interactions in an electric motor as the load is varied. A hint, the lower armature current may be providing less amp turns, resulting in a lower Counter Electromotive Force, and greater line current.
Do a little research on motor theory. Your interest is a good step in the direction of aquiring new knowledge.
respectfully

 

[electricpete]

"For the motor current to increase the CEMF must decrease"

That statement would have a straightforward interpretation for a dc motor but I'm not sure it is as applicable to induction motors.

As load increases, slip increases. The rotor draws more current. The rotor is like the secondary of a transformer... if you draw more current on the secondary (rotor), you will draw more current on the primary (stator). The rotor current does create an emf in the stator winding, and the associated stator current (load component) creates a roughly equal/opposite emf... so stator load current will match rotor current on an amp-turn basis.

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[cswilson]

"Since CEMF inhibits current flow, it is essentially inductive reactance."

This is your key mistake. CEMF is in phase with the line voltage, so it does not behave at all like an inductive reactance. At the synchronous (no-load) speed, the CEMF is equal in magnitude to the line voltage, completely inhibiting any current that is in phase with the line voltage.

In this condition, there is reactive current flow that is magnetizing the rotor, but since it is out of phase with the voltage, it cannot do any work. The motor is essentially an unloaded transformer in this case.

Now, if you add some mechanical load to the rotor shaft, you will decelerate the rotor, creating a slip frequency. The rotor is a transformer secondary that "sees" the slip frequency. The voltage induced in the rotor is proportional to slip, and for low values of slip, the current is also proportional to the rotor.

On the stator side, the reduced speed of the magnetized rotor reduces the CEMF. The difference between the line voltage and the CEMF permits current (in phase with the voltage) to flow. This current can do real work (power transfer), and creates torque at speed (mechanical power). The reactive current that does not do work is essentially constant through this process.

Curt Wilson
Delta Tau Data Systems

 

[waross]

I like your explanation cswilson.
lps
respectfully

 

[powerjunx]

good points, cswilson.









"Whenever you are asked if you can do a job, tell them, certainly I can! Then get busy and find out how to do it." Theodore Roosevelt.

 

[jhyoung]

Thank you everyone for your reply. I think I understand this concept. I was really putting too much thought into it.

One question for cswilson:

"CEMF is in phase with the line voltage, so it does not behave at all like an inductive reactance. At the synchronous (no-load) speed, the CEMF is equal in magnitude to the line voltage, completely inhibiting any current that is in phase with the line voltage."

If the CEMF is in phase with the line voltage how is the current being inhibited? Wouldn't the CEMF be opposing the line voltage, or 180 degrees out of phase with it?

 

[cswilson]

The CEMF is definitely in phase with the line voltage, not 180 degrees out of phase. It "opposes" the line voltage in the sense of "using it up", colloquially speaking.

Let's say you had a 230VAC line voltage, and your motor was spinning at almost synchronous speed, producing a 225VAC CEMF (in phase with the line). This means you have only 5VAC of "headroom" to provide torque-producing current. The closer you get to synchronous speed, the less in-phase current can be created, and the less torque can be produced. At the synchronous speed, the CEMF is equivalent to the line voltage, so there is no headroom left to provide this current -- this is what I mean by the current being inhibited.

Curt Wilson
Delta Tau Data Systems