EMF equation and Torque equation of a DC machine

EMF equation of a DC generator

Consider a DC generator with the following parameters,

P = number of field poles
Ø = flux produced per pole in Wb (weber)
Z = total no. of armature conductors
A = no. of parallel paths in armature
N = rotational speed of armature in revolutions per min. (rpm)

  • Average emf generated per conductor is given by dΦ/dt (Volts) ... eq. 1
  • Flux cut by one conductor in one revolution = dΦ = PΦ ….(Weber),
  • Number of revolutions per second (speed in RPS) = N/60
  • Therefore, time for one revolution = dt = 60/N (Seconds)
  • From eq. 1, emf generated per conductor = dΦ/dt = PΦN/60 (Volts) …..(eq. 2)
Above equation-2 gives the emf generated in one conductor of the generator. The conductors are connected in series per parallel path, and the emf across the generator terminals is equal to the generated emf across any parallel path.

Therefore, Eg = PΦNZ / 60A

For simplex lap winding, number of parallel paths is equal to the number of poles (i.e. A=P),
Therefore, for simplex lap wound dc generator, Eg = PΦNZ / 60P

For simplex wave winding, number of parallel paths is equal to 2 (i.e P=2),
Therefore, for simplex wave wound dc generator, Eg = PΦNZ / 120

Torque equation of a DC motor

When armature conductors of a DC motor carry current in the presence of stator field flux, a mechanical torque is developed between the armature and the stator. Torque is given by the product of the force and the radius at which this force acts.
  • Torque T = F × r (N-m) …where, F = force and r = radius of the armature
  • Work done by this force in once revolution = Force × distance = F × 2πr    (where, 2πr = circumference of the armature)
  • Net power developed in the armature = word done / time
    = (force × circumference × no. of revolutions) / time
    = (F × 2πr × N) / 60 (Joules per second) .... eq. 2.1
But, F × r = T and 2πN/60 = angular velocity ω in radians per second. Putting these in the above equation 2.1
Net power developed in the armature = P = T × ω (Joules per second)

Armature torque (Ta)

  • The power developed in the armature can be given as, Pa = Ta × ω = Ta × 2πN/60
  • The mechanical power developed in the armature is converted from the electrical power,
    Therefore, mechanical power = electrical power
    That means, Ta × 2πN/60 = Eb.Ia
  • We know, Eb = PΦNZ / 60A
  • Therefore, Ta × 2πN/60 = (PΦNZ / 60A) × Ia
  • Rearranging the above equation,
    Ta = (PZ / 2πA) × Φ.Ia (N-m)
The term (PZ / 2πA) is practically constant for a DC machine. Thus, armature torque is directly proportional to the product of the flux and the armature current i.e. Ta ∝ Φ.Ia

Shaft Torque (Tsh)

Due to iron and friction losses in a dc machine, the total developed armature torque is not available at the shaft of the machine. Some torque is lost, and therefore, shaft torque is always less than the armature torque.

Shaft torque of a DC motor is given as,
Tsh = output in watts / (2πN/60) ....(where, N is speed in RPM)