List four methods used to reduce leakage flux. 2. Calculate the transformation ratio for a 30 kVA, 2200 V / 250 V transformer. 3. Calculate the full-load secondary current.

Step 1: List FOUR methods used to reduce leakage flux (3.1)
Four methods used to reduce leakage flux in a transformer are:

• Interleaving windings: Winding primary and secondary coils in sections and placing them alternately to ensure closer magnetic coupling.
• Concentric windings: Placing one winding (e.g., the low voltage winding) inside the other (e.g., the high voltage winding) to reduce the distance between them and improve coupling.
• Shell-type core construction: This construction provides a double magnetic path, which inherently reduces leakage flux compared to core-type transformers.
• Reducing the length of the mean turn: By designing the core window to be smaller, the average length of the magnetic path for the windings is reduced, which helps in reducing leakage flux.

Step 2: Calculate the transformation ratio (3.2)
Given:
Transformer rating $S = 30 \text{ kVA}$
Primary voltage $V_1 = 2200 \text{ V}$
Secondary voltage $V_2 = 250 \text{ V}$ (rated no-load)

The transformation ratio $a$ is:

$$a = \frac{V_1}{V_2} = \frac{2200 \text{ V}}{250 \text{ V}} = 8.8$$

Step 3: Calculate the full-load secondary current
The full-load secondary current $I_{2, \text{FL}}$ is calculated from the transformer rating:

$$S = V_2 I_{2, \text{FL}}$$ $$30 \times 10^3 \text{ VA} = 250 \text{ V} \times I_{2, \text{FL}}$$ $$I_{2, \text{FL}} = \frac{30000 \text{ VA}}{250 \text{ V}} = 120 \text{ A}$$

Step 4: Calculate the equivalent resistance and reactance referred to the secondary side
Primary winding: $R_1 = 1.3 \Omega$, $X_1 = 4.6 \Omega$
Secondary winding: $R_2 = 0.2 \Omega$, $X_2 = 0.4 \Omega$

Equivalent resistance referred to the secondary side ($R_{eq2}$):

$$R_{eq2} = R_2 + \frac{R_1}{a^2}$$ $$R_{eq2} = 0.2 \Omega + \frac{1.3 \Omega}{(8.8)^2}$$ $$R_{eq2} = 0.2 \Omega + \frac{1.3 \Omega}{77.44}$$ $$R_{eq2} \approx 0.2 \Omega + 0.016787 \Omega \approx 0.2168 \Omega$$

Equivalent reactance referred to the secondary side ($X_{eq2}$):

$$X_{eq2} = X_2 + \frac{X_1}{a^2}$$ $$X_{eq2} = 0.4 \Omega + \frac{4.6 \Omega}{(8.8)^2}$$ $$X_{eq2} = 0.4 \Omega + \frac{4.6 \Omega}{77.44}$$ $$X_{eq2} \approx 0.4 \Omega + 0.05939 \Omega \approx 0.4594 \Omega$$

Step 5: Calculate the voltage drop at full load
The power factor is 0.8 lagging.

$$\cos \phi = 0.8$$ $$\sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - (0.8)^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6$$

The approximate voltage drop ($\Delta V_2$) for a lagging power factor is:

$$\Delta V_2 = I_{2, \text{FL}} (R_{eq2} \cos \phi + X_{eq2} \sin \phi)$$ $$\Delta V_2 = 120 \text{ A} \times (0.2168 \Omega \times 0.8 + 0.4594 \Omega \times 0.6)$$ $$\Delta V_2 = 120 \text{ A} \times (0.17344 + 0.27564)$$ $$\Delta V_2 = 120 \text{ A} \times (0.44908 \Omega)$$ $$\Delta V_2 \approx 53.89 \text{ V}$$

Step 6: Calculate the secondary voltage at full load
The no-load secondary voltage ($V_{2, \text{NL}}$) when the primary voltage is $2200 \text{ V}$ is $250 \text{ V}$.
The secondary voltage at full load ($V_{2, \text{FL}}$) is:

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