To solve this, we can proceed with each part step by step. Here's the approach:

Given Data:

• Number of poles, $P = 12$
• Speed, $N = 450 \, \text{rpm}$
• Flux per pole, $\phi = 25 \, \text{mWb} = 25 \times 10^{-3} \, \text{Wb}$
• Star-connected winding
• Number of slots, $S = 120$
• Conductors per slot, $Z = 8$
• Turns per phase $T = \frac{S \times Z}{6} = \frac{120 \times 8}{6} = 160$ (since it’s a 3-phase machine)

i. Frequency $f$

The frequency of a 3-phase alternator is given by: $f = \frac{P \times N}{120}$ Substituting the values: $f = \frac{12 \times 450}{120} = 45 \, \text{Hz}$

ii. EMF per phase $E_{\text{phase}}$

The EMF per phase for a star-connected alternator is calculated using the formula: $E_{\text{phase}} = 4.44 \times f \times T \times \phi$ Substituting the values: $E_{\text{phase}} = 4.44 \times 45 \times 160 \times 25 \times 10^{-3}$ $E_{\text{phase}} = 799.2 \, \text{V}$

iii. Line Voltage $E_{\text{line}}$

For a star-connected system, the line voltage $E_{\text{line}}$ is given by: $E_{\text{line}} = \sqrt{3} \times E_{\text{phase}}$ Substituting $E_{\text{phase}} = 799.2 \, \text{V}$: $E_{\text{line}} = \sqrt{3} \times 799.2$ $E_{\text{line}} \approx 1384.76 \, \text{V}$

Summary of Answers:

1. Frequency (f): $45 \, \text{Hz}$
2. EMF per phase (E_phase): $799.2 \, \text{V}$
3. Line Voltage (E_line): $1384.76 \, \text{V}$

Would you like further details or have any questions?

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Related Questions:

1. How would the calculations change if the alternator was delta-connected instead?
2. What happens to the frequency if the number of poles increases?
3. How does changing the speed affect the EMF generated in the alternator?
4. What would be the impact on EMF if the flux per pole is doubled?
5. How can we reduce the EMF per phase without changing the speed or frequency?

Tip:

Remember, for star-connected alternators, the line voltage is $\sqrt{3}$ times the phase voltage.