Question

A 500 KVA, 3-phase, 50 Hz transformer has a voltage ratio (line voltages) of 33/11KV and is delta/star connected. The resistance per phase is: high voltage 35 Ω low voltage 0.876 Ω and iron loss is 3050W. Calculate the value of efficiency at full load.

Answer 1

To calculate the efficiency of the transformer at full load, we will follow these steps:

1. **Determine the full load current on both sides of the transformer.**
2. **Calculate the copper losses for both high and low voltage sides.**
3. **Identify the iron losses.**
4. **Calculate the output power at full load.**
5. **Use the efficiency formula to find the efficiency at full load.**

### Step 1: Determine the Full Load Current

The transformer has a rating of 500 kVA and operates at a line voltage of 11 kV (low voltage side). The full load current (\(I_{FL}\)) can be calculated using the formula:

\[
I_{FL} = \frac{S}{\sqrt{3} \times V}
\]

Where:
- \(S = 500 \, \text{kVA} = 500,000 \, \text{VA}\)
- \(V = 11,000 \, \text{V}\)

Calculating the current:

\[
I_{FL} = \frac{500,000}{\sqrt{3} \times 11,000} \approx \frac{500,000}{19,052} \approx 26.25 \, \text{A}
\]

Now calculating the full load current on the high voltage side (33 kV):

\[
I_{FL (HV)} = \frac{S}{\sqrt{3} \times V_{HV}} = \frac{500,000}{\sqrt{3} \times 33,000} \approx \frac{500,000}{57,108} \approx 8.75 \, \text{A}
\]

### Step 2: Calculate the Copper Losses

**Copper losses** occur in the transformer due to the resistance of the windings and are calculated using the formula:

\[
P_{Cu} = I^2 \times R
\]

**Low Voltage Side:**
\[
P_{Cu (LV)} = I_{FL (LV)}^2 \times R_{LV} = (26.25)^2 \times 0.876 \approx 689.06 \, \text{W}
\]

**High Voltage Side:**
\[
P_{Cu (HV)} = I_{FL (HV)}^2 \times R_{HV} = (8.75)^2 \times 35 \approx 2,687.5 \, \text{W}
\]

### Step 3: Identify the Iron Losses

The **iron loss** (core loss) is given as 3050 W. This loss occurs irrespective of the load on the transformer and includes hysteresis and eddy current losses in the core.

### Step 4: Calculate the Output Power

The total losses in the transformer at full load consist of copper losses and iron losses:

\[
P_{loss} = P_{Cu (LV)} + P_{Cu (HV)} + P_{iron} = 689.06 + 2,687.5 + 3050 = 5,426.56 \, \text{W}
\]

The output power at full load (\(P_{out}\)) can be calculated by subtracting the total losses from the input power:

\[
P_{input} = 500,000 \, \text{W} = 500 \, \text{kW}
\]

Now, calculating the output power:

\[
P_{out} = P_{input} - P_{loss} = 500,000 - 5,426.56 \approx 494,573.44 \, \text{W}
\]

### Step 5: Calculate Efficiency

The efficiency (\(η\)) of the transformer can be calculated using the formula:

\[
η = \frac{P_{out}}{P_{input}} \times 100\%
\]

Substituting the values:

\[
η = \frac{494,573.44}{500,000} \times 100\% \approx 98.91\%
\]

### Conclusion

The efficiency of the 500 kVA, 3-phase, 50 Hz transformer at full load is approximately **98.91%**. This indicates that the transformer is quite efficient in transferring power with minimal losses.

Answer 2

To calculate the efficiency of the transformer at full load, we need to determine both the output power and the losses. The efficiency \(\eta\) of a transformer is given by:

\[
\eta = \frac{\text{Output Power}}{\text{Output Power} + \text{Total Losses}}
\]

### Step-by-Step Solution

#### 1. **Determine the Full-Load Output Power**

Given:
- Transformer rating: 500 kVA (kilo-volt-amperes)
- High Voltage (HV) side: 33 kV (line-to-line)
- Low Voltage (LV) side: 11 kV (line-to-line)

At full load, the output power in kW is equal to the transformer’s rated power (500 kVA) assuming a power factor of 1. Hence:

\[
\text{Output Power} = 500 \text{ kVA} \times \text{Power Factor} = 500 \text{ kW}
\]

#### 2. **Calculate the Copper Losses**

Copper losses occur due to the resistance of the windings. The copper losses per phase for both HV and LV windings can be calculated using:

\[
\text{Copper Loss} = I^2 R
\]

To find the current for each side, use:

\[
I = \frac{\text{Rated Power}}{\sqrt{3} \times \text{Rated Voltage}}
\]

- **HV Side Current:**

  \[
  I_{\text{HV}} = \frac{500 \text{ kVA}}{\sqrt{3} \times 33 \text{ kV}} = \frac{500 \times 10^3}{\sqrt{3} \times 33 \times 10^3} \approx 8.75 \text{ A}
  \]

- **LV Side Current:**

  \[
  I_{\text{LV}} = \frac{500 \text{ kVA}}{\sqrt{3} \times 11 \text{ kV}} = \frac{500 \times 10^3}{\sqrt{3} \times 11 \times 10^3} \approx 26.24 \text{ A}
  \]

**Copper Losses Calculation:**

- **High Voltage Side Copper Losses:**

  \[
  \text{Copper Loss}_{\text{HV}} = I_{\text{HV}}^2 \times R_{\text{HV}} = (8.75)^2 \times 35 = 3031.25 \text{ W}
  \]

- **Low Voltage Side Copper Losses:**

  \[
  \text{Copper Loss}_{\text{LV}} = I_{\text{LV}}^2 \times R_{\text{LV}} = (26.24)^2 \times 0.876 = 604.8 \text{ W}
  \]

**Total Copper Losses:**

\[
\text{Total Copper Losses} = \text{Copper Loss}_{\text{HV}} + \text{Copper Loss}_{\text{LV}} = 3031.25 \text{ W} + 604.8 \text{ W} = 3636.05 \text{ W}
\]

#### 3. **Calculate Total Losses**

Total losses include both copper losses and iron losses:

\[
\text{Total Losses} = \text{Copper Losses} + \text{Iron Losses} = 3636.05 \text{ W} + 3050 \text{ W} = 6686.05 \text{ W}
\]

#### 4. **Calculate Efficiency**

\[
\eta = \frac{\text{Output Power}}{\text{Output Power} + \text{Total Losses}} = \frac{500 \times 10^3}{500 \times 10^3 + 6686.05} \approx \frac{500000}{506686.05} \approx 0.986 \text{ or } 98.6\%
\]

### Conclusion

The efficiency of the transformer at full load is approximately **98.6%**.