Question

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.

Answer 1

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;

$E = \frac{\theta ZN}{60}$ × $\frac{P}{A}$

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:

$Z = 144$ × $2$ × $3$

Z = 864

Substituting the given parameters into the formula, we have;

$520 = \frac{\theta (864)(660)}{60}$ × $\frac{4}{2}$

$520 = \theta (864)(11)$ × $2$

$520 = 19008 \theta \\\\\Theta = \frac{520}{19008}$

Magnetic flux = 0.0274 Weber.

Therefore, the magnetic flux per pole is 0.0274 Weber.

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