To calculate the line emf generated by a 16-pole, 3-phase star-connected alternator, we'll use the following data:
- Number of poles (\(P\)): 16
- Number of phases: 3
- Number of slots: 12
- Conductors per slot: 24
- Flux per pole (\(\Phi\)): 0.1 Wb
- Frequency (\(f\)): 50 Hz
### Step-by-Step Solution
1. **Determine the Number of Conductors per Phase:**
In a star-connected alternator, the total number of conductors in the armature winding can be calculated as:
\[
\text{Total Conductors} = \text{Number of Slots} \times \text{Conductors per Slot}
\]
\[
\text{Total Conductors} = 12 \times 24 = 288
\]
Since the alternator is star-connected and each phase has an equal number of conductors, the conductors per phase are:
\[
\text{Conductors per Phase} = \frac{\text{Total Conductors}}{\text{Number of Phases}}
\]
\[
\text{Conductors per Phase} = \frac{288}{3} = 96
\]
2. **Calculate the Number of Phases in a Single Slot:**
Each slot contains conductors for all three phases. Therefore, the number of conductors per slot per phase is:
\[
\text{Conductors per Slot per Phase} = \frac{\text{Conductors per Slot}}{\text{Number of Phases}}
\]
\[
\text{Conductors per Slot per Phase} = \frac{24}{3} = 8
\]
3. **Determine the Winding Factor:**
The winding factor \( k_w \) accounts for the distribution of the windings and is calculated using the following formula:
\[
k_w = \text{Distribution Factor} \times \text{Pitch Factor}
\]
- **Distribution Factor** (\( k_d \)): This factor considers the spreading of the winding over several slots. For a 16-pole machine with 12 slots, the distribution factor can be approximated as:
\[
k_d = \frac{\sin\left(\frac{p \cdot \alpha}{2}\right)}{p \cdot \sin\left(\frac{\alpha}{2}\right)}
\]
where \( p \) is the number of poles and \( \alpha \) is the slot angle in radians:
\[
\alpha = \frac{2 \pi}{\text{Number of Slots}} = \frac{2 \pi}{12} = \frac{\pi}{6}
\]
Hence,
\[
k_d = \frac{\sin\left(\frac{16 \cdot \frac{\pi}{6}}{2}\right)}{16 \cdot \sin\left(\frac{\frac{\pi}{6}}{2}\right)} = \frac{\sin\left(\frac{8\pi}{6}\right)}{16 \cdot \sin\left(\frac{\pi}{12}\right)}
\]
Simplify to get \( k_d \approx 0.98 \) (approximately).
- **Pitch Factor** (\( k_p \)): The pitch factor depends on the coil span. For full-pitched windings (coils span the full pole pitch), \( k_p = 1 \).
Combining these:
\[
k_w \approx 0.98 \times 1 = 0.98
\]
4. **Calculate the EMF Per Phase:**
The emf per phase (\( E_{\text{phase}} \)) can be calculated using:
\[
E_{\text{phase}} = \frac{4.44 \cdot f \cdot N \cdot \Phi \cdot k_w}{P}
\]
where \( N \) is the number of conductors per phase, \( \Phi \) is the flux per pole, and \( P \) is the number of poles:
\[
E_{\text{phase}} = \frac{4.44 \cdot 50 \cdot 96 \cdot 0.1 \cdot 0.98}{16}
\]
\[
E_{\text{phase}} \approx \frac{4.44 \cdot 50 \cdot 9.6 \cdot 0.1 \cdot 0.98}{16}
\]
\[
E_{\text{phase}} \approx \frac{4.44 \cdot 50 \cdot 0.96}{16}
\]
\[
E_{\text{phase}} \approx \frac{213.12}{16} = 13.32 \text{ V}
\]
5. **Calculate the Line EMF:**
In a star-connected system, the line emf (\( E_{\text{line}} \)) is related to the phase emf by:
\[
E_{\text{line}} = \sqrt{3} \cdot E_{\text{phase}}
\]
\[
E_{\text{line}} = \sqrt{3} \cdot 13.32 \approx 23.1 \text{ V}
\]
### Summary
The line emf generated by the alternator at 50 Hz is approximately **23.1 V**.