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Speed Regulation =
                     nNl - nFL   x 100   %
                        nFL
Where
nNL = speed at no load
nFL = speed at full load
Regulation values of 5.6 per cent
indicate good speed regulation

 100 per cent regulation is very poor
and means the motor stops altogether
when placed on load.
Example:
A DC shunt-connected motor runs at 1050
rpm (rev/per/min) on no load and slows to
990 rpm on full load. Find the speed
regulation.
nNl - nFL x 100   %
   nFL
1050-990
Speed regulation =        x 100 =6%
                    990
pΦIZ
T=
   2πa
Where
T = Torque in Newton-metres (Nm)
p = Number of poles
Φ = Flux per pole in weber
I = Total armature current
Z = Number of armature conductors
a = Number of parallel paths in the armature
  (two parallel paths for wave and equal to the
    number of poles for a lap winding).
Example:
A four pole dc motor has a lap-wound
armature of 30 coils, each with 20
conductors. If the flux per pole is 0.02 Wb
and the armature current is 19 A calculate
the torque produced.
pΦIZ
T=
   2πa
4 x 0.02 x 19 x 20 x 30
T=                         = 36.3 Nm
       2 x 3.146 x 4

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11.3.4 Speed Regulation

  • 1.
  • 2. Speed Regulation = nNl - nFL x 100 % nFL
  • 3. Where nNL = speed at no load nFL = speed at full load
  • 4. Regulation values of 5.6 per cent indicate good speed regulation 100 per cent regulation is very poor and means the motor stops altogether when placed on load.
  • 5. Example: A DC shunt-connected motor runs at 1050 rpm (rev/per/min) on no load and slows to 990 rpm on full load. Find the speed regulation.
  • 6. nNl - nFL x 100 % nFL
  • 8.
  • 9.
  • 10.
  • 11. pΦIZ T= 2πa
  • 12. Where T = Torque in Newton-metres (Nm) p = Number of poles Φ = Flux per pole in weber I = Total armature current Z = Number of armature conductors a = Number of parallel paths in the armature (two parallel paths for wave and equal to the number of poles for a lap winding).
  • 13. Example: A four pole dc motor has a lap-wound armature of 30 coils, each with 20 conductors. If the flux per pole is 0.02 Wb and the armature current is 19 A calculate the torque produced.
  • 14. pΦIZ T= 2πa
  • 15. 4 x 0.02 x 19 x 20 x 30 T= = 36.3 Nm 2 x 3.146 x 4