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Re armature of a 4 pole DC generator is required to generate an emf of 520v on open circuit when revolving at a speed of 660rpm. Calculate the magnetic flux per pole required if the armature has 144 slots with 2 coil sides per slot each coil consisting of 3 turns. The armature is wave wound.

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Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

Given the following data:

  • Emf = 520 Volts
  • Speed = 660 r.p.m
  • Number of armature conductors = 144 slots
  • Number of poles = 4 poles
  • Number of parallel paths = 2

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

[tex]E = \frac{\theta ZN}{60}[/tex] × [tex]\frac{P}{A}[/tex]

Where:

  • E is the electromotive force in the DC generator.
  • Z is the total number of armature conductors.
  • N is the speed or armature rotation in r.p.m.
  • P is the number of poles.
  • A is the number of parallel paths in armature.
  • Ф is the magnetic flux.

First of all, we would determine the total number of armature conductors:

[tex]Z = 144[/tex] × [tex]2[/tex] × [tex]3[/tex]

Z = 864

Substituting the given parameters into the formula, we have;

[tex]520 = \frac{\theta (864)(660)}{60}[/tex] × [tex]\frac{4}{2}[/tex]

[tex]520 = \theta (864)(11)[/tex] × [tex]2[/tex]

[tex]520 = 19008 \theta \\\\\Theta = \frac{520}{19008}[/tex]

Magnetic flux = 0.0274 Weber.

Therefore, the magnetic flux per pole is 0.0274 Weber.

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