3 years ago

The simply supported beam in the Figure has a rectangular cross-section 150 mm wide and 240 mm high.

Engineering

1 answer:

8_murik_8 [283]3 years ago

4 0

Same question idea but different values... I hope I helped you... Don't forget to put a heart mark

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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.

bija089 [108]

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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