How to calculate the counter-electromotive force of a motor?

[cabrera]

Dear Forum,

I would like to get familiar with motors. I "think" I understand the basic principles of DC motor:

1) Fix magnetic field (stator), Bf
2) Rotating magnetic field( rotor), Br
3) Force (Torque) proportional to intensities of Bf, Br, angle
4) Commutation of loops to keep Br at constant angles with respect to Br, hence constant torque.
5) Counter electro-motive force, CEMF, generated by the speed ( I know "lenz's law) of the motor. I am trying to understand this point better
.
(Please, correct the list above if either there is anything wrong or I have omitted something)I am having a few questions:

A) In a motor, how is the CEMF calculated and, hence, how is the Kv obtained?

B) I am having problems visualising what happens to the coils of the rotor when they are switched off/on (commutation) ? I a normal circuit current is rectified (forced to follow a different path), but in a DC motor the coils with current are disconnected and a new coils is connected. My guess is that the field generated, "Br" is kept by current "jumping" to the new coil connected to the power source (external voltage source).

B.1) What would happen to the EMF if the rotors have coils with different turns...and the torque?

 

[CWatters]

Perhaps google... How to calculate voltage of DC generator.

For example...

https://circuitglobe.com/emf-equation-of-dc-generator.html

 

[Delta2]

Perhaps google... How to calculate voltage of DC generator.

For example...

https://circuitglobe.com/emf-equation-of-dc-generator.html

Does that calculation include the EMF due to the self induction of the rotor?

Anyway I would expect that the EMF due to the induction of the stator with the rotor to be $E=K\omega\sin({\omega t})$ where $\omega$ the angular velocity of the rotor and $K$ a constant that depends on how the rotor is constructed(number of coils, number of turns per coil, cross area per turn e.t.c) and on the intensity of magnetic field of the stator.

 

[essenmein]

For a brushed DC machine, the field is typically not rotating, the permanent magnets in the outer shell are fixed, so the field generated by the rotor must also be fixed in relation to those permanent magnets to allow operation. So the commutators essentially turns on the correct coils to push or pull against the permanent magnets as the rotor spins, but the fields are stationary (relative to the outer shell magnets).

Your questions
A) back emf is the result of wires exposed to a changing magnetic field (ie rotating past poles), so if you know the structure of the machine and all the characteristics of the magnets etc you can calculate the motor constants, but since this mechanical information is usually not readily available, its usually much easier to measure it, spin it at a known speed and measure the voltage.
B) Since the permanent magnets are not moving, each coil must be turned "on" when that coil is at the correct position relative to that permanent magnet. That is the only thing the commutator does, so the current has to be applied to each coil successively as it spins past its magnetic pole.
B1) A-t (amp turns) is what makes MMF (Magneto motive force) and this is what makes the forces in the machine, so if this number is the same, then the torque produced is the same, 100A x 1turn = 100A-t, 50A x 2turns also is 100A-t, both of those make the same torque. Now 1 turn in a changing field makes a voltage, double the turns in that same changing field makes twice the voltage. So doubling the turns on a machine means you need half the current but twice the voltage to reach a given operating point, which you might notice is also the same power (it should be).

 

[cabrera]

Thanks for your help