https://file.aiquickdraw.com/mathbot/file/1726786054547u3nj1ogj.jpg

To solve Question 2 in detail, let's break it down step-by-step:

### a. Principle of Operation of a Transformer (4 marks)
A transformer operates on the principle of electromagnetic induction. It consists of two coils, the primary and the secondary, wound on a common core. When an alternating current flows through the primary coil, it creates a time-varying magnetic field in the core. This magnetic field induces a voltage in the secondary coil. The voltage induced in the secondary depends on the ratio of the number of turns in the primary to the number of turns in the secondary. This allows the transformer to step up or step down the voltage as needed, while ideally maintaining the same power level (assuming no losses).

### b. Derivation of \( E_{\text{rms}} = 4.44 f \Phi_{\text{m}} N \) (6 marks)
Given that:
\[ \Phi = \Phi_{\text{m}} \sin \omega t \]
And,
\[ E = -N \frac{d\Phi}{dt} \]

1. **Find \( \frac{d\Phi}{dt} \):**
   \[ \frac{d\Phi}{dt} = \frac{d}{dt} (\Phi_{\text{m}} \sin \omega t) = \Phi_{\text{m}} \omega \cos \omega t \]

2. **Calculate \( E \):**
   \[ E = -N \cdot \Phi_{\text{m}} \omega \cos \omega t \]
   Here, \( \omega = 2 \pi f \), so:
   \[ E = -N \cdot \Phi_{\text{m}} (2 \pi f) \cos \omega t \]

3. **RMS Value:**
   The RMS value of \( E \) is given by:
   \[ E_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T E^2 \, dt} \]
   where \( T = \frac{1}{f} \).

4. **Using Standard Formula:**
   For a sinusoidal waveform, the RMS value is \( E_{\text{rms}} = \sqrt{2} \cdot E_{\text{peak}} \). Substituting \( E_{\text{peak}} = N \Phi_{\text{m}} \omega \), we get:
   \[ E_{\text{rms}} = 4.44 f \Phi_{\text{m}} N \]

### c. Calculation for the Transformer (10 marks)
Given:
- Power rating = 50 kVA
- Voltage = 11 kV/415 V
- Single phase
- Cross-sectional area = 20 cm × 20 cm = 0.04 m²
- Maximum flux density = 1.3 Tesla
- Stacking factor = 0.85

#### i. Number of Primary and Secondary Turns per Phase (6 marks)
1. **Determine the maximum flux \( \Phi_{\text{m}} \):**
   \[ \Phi_{\text{m}} = B \times \text{Area} \times \text{Stacking factor} \]
   \[ \Phi_{\text{m}} = 1.3 \times 0.04 \times 0.85 = 0.0442 \text{ Wb} \]

2. **Number of turns using \( E_{\text{rms}} = 4.44 f \Phi_{\text{m}} N \):**
   - For primary side, \( E_{\text{primary}} = 11,000 \text{ V} \).
     \[
     11,000 = 4.44 \times 50 \times 0.0442 \times N_{\text{primary}}
     \]
     \[
     N_{\text{primary}} = \frac{11,000}{4.44 \times 50 \times 0.0442} \approx 1120 \text{ turns}
     \]

   - For secondary side, \( E_{\text{secondary}} = 415 \text{ V} \).
     \[
     415 = 4.44 \times 50 \times 0.0442 \times N_{\text{secondary}}
     \]
     \[
     N_{\text{secondary}} = \frac{415}{4.44 \times 50 \times 0.0442} \approx 42 \text{ turns}
     \]

#### ii. EMF per Turn (4 marks)
1. **EMF per turn calculation:**
   \[
   \text{EMF per turn} = 4.44 \times f \times \Phi_{\text{m}}
   \]
   Substituting the values:
   \[
   \text{EMF per turn} = 4.44 \times 50 \times 0.0442 = 9.8 \text{ V}
   \]

This completes the detailed solution of Question 2. Let me know if you need further details or clarifications.

### Related Questions:
1. What are the different types of losses in a transformer, and how do they affect efficiency?
2. How can the efficiency of a transformer be calculated and improved?
3. What is the significance of the turns ratio in a transformer, and how does it affect voltage levels?
4. What role does the core material play in transformer performance?
5. How does the frequency of the AC supply affect the operation of a transformer?

**Tip:** Always ensure the magnetic core of the transformer is not saturated as it can lead to distortion in the output voltage and increased losses.

a. Briefly explain the principle of operation of a transformer. b. Given that ϕ=ϕmsinωt and E = -N(dϕ/dt), show that Erms = 4.44fΦmN. c. The core of 50 kVA, 11 kV/415 V, single phase, core type transformer has a cross-section of 20cm x 20cm. Find the: i. Number of primary and secondary turns per phase. ii. E.M.F per turn if the maximum core density is not to exceed 1.3 Tesla. Assume a stacking factor of 0.85.

This detailed solution explains the principle of operation of a transformer, derives the EMF equation, and calculates the primary and secondary turns per phase, as well as the EMF per turn for a 50 kVA, 11 kV/415 V transformer. The solution includes step-by-step calculations, suitable for undergraduate electrical engineering students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation