Question

A 16 pole, 3 phase star connected alternator armature has 12 slots with 24 conductors per slot and flux per pole is 0.1 wb, sinusoidally distributed. Calculate line emf generated at 50 Hz.

Answer 1

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**.

Answer 2

To calculate the line emf generated by a 16-pole, 3-phase star-connected alternator with the given parameters, you need to follow these steps:

### Given Data:
- Number of poles (\(P\)) = 16
- Number of phases = 3
- Number of slots = 12
- Number of conductors per slot = 24
- Flux per pole (\(\Phi\)) = 0.1 Wb
- Frequency (\(f\)) = 50 Hz

### Step-by-Step Calculation:

1. **Determine the Number of Conductors in the Alternator:**
   - Total number of conductors = Number of slots × Number of conductors per slot
   - Total number of conductors = 12 × 24 = 288

2. **Calculate the Number of Conductors per Phase:**
   - In a star-connected alternator, the total number of conductors is equally divided among the three phases.
   - Number of conductors per phase = Total number of conductors / Number of phases
   - Number of conductors per phase = 288 / 3 = 96

3. **Determine the Number of Poles and Slots per Phase:**
   - Since the alternator has 16 poles and 12 slots, the number of slots per pole per phase can be calculated by dividing the total number of slots by the number of poles.
   - Slots per pole = Number of slots / Number of poles
   - Slots per pole = 12 / 16 = 0.75 slots per pole (not directly used but confirms the slot distribution)

4. **Calculate the Electrical Frequency of the Alternator:**
   - The electrical frequency \(f_e\) is given by the formula \( f_e = \frac{P \times f}{2} \)
   - For a 16-pole machine, \( f_e = \frac{16 \times 50}{2} = 400 \text{ Hz} \)

5. **Calculate the Emf per Phase:**
   - The emf per phase can be calculated using the formula:
     \[
     E_{\text{phase}} = \frac{4.44 \times f \times \Phi \times N}{P}
     \]
     where:
     - \( f \) = Frequency of the alternator = 50 Hz
     - \( \Phi \) = Flux per pole = 0.1 Wb
     - \( N \) = Number of conductors per phase = 96
     - \( P \) = Number of poles = 16
     
     Plugging in the values:
     \[
     E_{\text{phase}} = \frac{4.44 \times 50 \times 0.1 \times 96}{16}
     \]
     \[
     E_{\text{phase}} = \frac{4.44 \times 50 \times 0.1 \times 96}{16} = \frac{21.312}{16} = 1.332 \text{ V}
     \]

6. **Calculate the Line Emf:**
   - In a star-connected alternator, the line voltage \( E_{\text{line}} \) is related to the phase voltage by:
     \[
     E_{\text{line}} = \sqrt{3} \times E_{\text{phase}}
     \]
     
     Plugging in the value of \( E_{\text{phase}} \):
     \[
     E_{\text{line}} = \sqrt{3} \times 1.332 \approx 1.732 \times 1.332 \approx 2.3 \text{ V}
     \]

### Summary
The line emf generated by the alternator at 50 Hz is approximately **2.3 V**.