The torque produced by a six-pole, lap-wound DC motor is 500Nm. If the machine has 15 conductors per slot, calculate the number of armature slots for an armature current of 55 amperes and a useful flux per pole of 0,03 webers.

Answer 1

First, we need to calculate the torque constant, Kt, of the motor using the following formula:

Kt = (60 * T) / (2 * π * Ia)

Where:
T = 500Nm (torque)
Ia = 55A (armature current)

Kt = (60 * 500) / (2 * π * 55)
Kt = 54545.45 Nm/A

Next, we can calculate the back EMF constant, Ke, using the following formula:

Ke = Kt / φ

Where:
φ = 0.03 Wb (useful flux per pole)

Ke = 54545.45 / 0.03
Ke = 1818181.82 V/rad/s

Now we can calculate the number of armature slots, S, using the formula:

S = (Ke * Zp) / (2 * p * A * B)

Where:
Zp = 15 (conductors per slot)
p = 6 (poles)
A = 2 (number of parallel paths in lap winding)
B = 1 (number of coil-sides per slot)

S = (1818181.82 * 15) / (2 * 6 * 2 * 1)
S = 6818181.82 / 24
S = 283113.41

Therefore, the number of armature slots required for an armature current of 55 amperes and a useful flux per pole of 0.03 webers is approximately 283113 slots.