Struggling to understand how Back EMF works in AC Motors 2

[bizkitgto]

I am having a hard time understanding how Back EMF works in induction (where the slip is very low, say 1%) and synchronous motors. My understanding of Back EMF is that back EMF is directly proportional to the motor's speed. As the motor spins faster, the back EMF increases, opposing the applied voltage and limiting the current flow. Once the back EMF equals the applied voltage, the motor reaches its maximum speed.

In a motor where the slip is very low, let's say full load is around 1%, how can Back EMF be minimum if the speed has only dropped by 1%?
In a synchronous motor, where there is no slip, how can Back EMF vary when the motor is fully loaded at synchronous speed?
If motor speed in AC motors doesn't change much from full load to no load, how can Back EMF vary so much?

 

[waross]

Typically, full load = 1760/1800 = 40 RPM slip = 40RPM/1800RPM x 100 = 2.2%
You are on the wrong track considering voltage.
Voltage differences drive reactive current.
Normally, the phase angle of the back EMF decreases as the motor speed increases.
Phase angle difference is what drives the transfer of real power.
An induction motor will not, by itself, reach synchronous sped due to parasitic losses such as windage and bearing friction.
If an induction motor is overdriven the phase angle between applied voltage and back EMF will decrease until at synchronous speed, the back EMF phase angle with the applied voltage is zero, and there is no transfer of power. If the motor is overdriven at more than synchronous speed then power will be transfered back into the supply system.
The motor is now an induction generator.
Back EMF voltage determines reactive current and the power factor.
Back EMF phase angle determines the real current. This may be power into the motor or power out of a motor driven faster than synchronous speed.

 

[waross]

In a synchronous motor the speed is fixed.
The back EMF voltage is controlled by the excitation.
At lower levels of excitation and back EMF voltage, the motor draws reactive power.
As the excitation is raised, the back EMF rises until it equals the supply voltage.
At this level there is no reactive current and the power factor is unity or 100%.
If the excitation is raised further, reactive power is produced and exported from the motor.
This exported power may supply the reactive power drawn by equipment that draws reactive power.
This is typically induction motors and transformers.
When a synchronous motor is operated at a leading power factor to improve the system power factor, the capacity to deliver mechanical power or shaft power is reduced.
Often it is economically feasible to oversize synchronous motors so that they may drive a load as well as produce reactive power to improve the system PF.
PF penalties are expensive.