How to work out the motor speed of a DC motor?

you can calculate it by formula

U-IR=C*(phi)*(omiga)

but the best way is measure with a meter.

to get more precision rotation speed

torque equal to

M=C*(phi)*I

c= constant depend on poles etc.

The DC motor doesn't have a speed that can be calculated like an AC motor, it's going to be dependant on the inductance/resistance of the coils, the physical loads and the commutation. Imagine a perfect motor, what would stop it spinning faster and faster? You'll need to measure it. Jeff

voltage equation for dc motor

E=(fluxperpole*no.of conductor*poles*speed*no.of parallel path)/120

Try calculating using these formulae. Note notations eq.1a...eq.2a are used here to marke equations as stated in numbering order.

1. Torque (T).

Assuming constant flux/pole of weber and an amarture current of divided over a parallel paths, the torque developed is given by

T = (Φ x Z x P x I_a) / 2 x (phi) x a Nm eq.1a

Where: Φ is flux per pole in webers.

Z is number of armature conductors.

P is number of poles.

I_a is armature current.

a is number of parallel paths on the armature.

Practically in dc motors, the Φ and I_a can be varied, hence the torque developed by the motor can be given by

T = k x Φ x I_a Nm eq.1b

Where: k is a constant depending on the physical properties of the motor.

2. Counter emf (E_g).

By generator action of a motor the counter emf is given by

E_g= kΦn eq.2a

Where: n is the speed of the dc motor.

The effective voltage at the armature is the applied voltage (terminal voltage, V_t) minus the counter emf (E_g). Then by Ohm's law the armature current (I_a) is

I_a = (V_t - E_g) / R_a eq.2b

Where: R_a is armature circuit resistance.

Multiplying eq.2b by R_a and making V_t the subject gives

V_t = E_g + I_a R_a is the fundamental dc motor equation. eq.2c

Note that

V_t = E_g + I_a R_a = kΦn + I_a R_a eq.2d

I hope this will help.