Question

In a 20kVA, 1000/400V, 1f 50Hz transformer, iron and full-load copper losses are 300 W and 500 W respectively. Calculate its efficiency at ¾ full load at unity power factor.

Answer 1

To calculate the efficiency of the transformer at \(\frac{3}{4}\) full load, we'll follow these steps:

1. **Determine the full-load output power:**

   The transformer's rated power is 20 kVA. At full load, the output power is:
   \[
   P_{\text{out, full-load}} = 20 \text{ kVA} \times 1 \text{ (unity power factor)} = 20 \text{ kW}
   \]

2. **Calculate the output power at \(\frac{3}{4}\) full load:**

   \[
   P_{\text{out, } \frac{3}{4}\text{ full-load}} = 20 \text{ kW} \times \frac{3}{4} = 15 \text{ kW}
   \]

3. **Determine the copper losses at \(\frac{3}{4}\) full load:**

   Full-load copper losses are given as 500 W. Copper losses vary with the square of the load. Therefore, at \(\frac{3}{4}\) full load, the copper losses are:
   \[
   \text{Copper losses}_{\frac{3}{4}\text{ full-load}} = 500 \text{ W} \times \left(\frac{3}{4}\right)^2
   \]
   \[
   \text{Copper losses}_{\frac{3}{4}\text{ full-load}} = 500 \text{ W} \times \frac{9}{16} = 281.25 \text{ W}
   \]

4. **Calculate the total losses at \(\frac{3}{4}\) full load:**

   The total losses are the sum of iron losses and the adjusted copper losses:
   \[
   \text{Total losses} = \text{Iron losses} + \text{Copper losses}_{\frac{3}{4}\text{ full-load}}
   \]
   \[
   \text{Total losses} = 300 \text{ W} + 281.25 \text{ W} = 581.25 \text{ W}
   \]

5. **Calculate the input power at \(\frac{3}{4}\) full load:**

   The input power is the output power plus the total losses:
   \[
   P_{\text{in}} = P_{\text{out, } \frac{3}{4}\text{ full-load}} + \text{Total losses}
   \]
   \[
   P_{\text{in}} = 15,000 \text{ W} + 581.25 \text{ W} = 15,581.25 \text{ W}
   \]

6. **Calculate the efficiency of the transformer:**

   Efficiency is given by:
   \[
   \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%
   \]
   \[
   \text{Efficiency} = \frac{15,000 \text{ W}}{15,581.25 \text{ W}} \times 100\%
   \]
   \[
   \text{Efficiency} \approx 96.3\%
   \]

So, the efficiency of the transformer at \(\frac{3}{4}\) full load and unity power factor is approximately **96.3%**.

Answer 2

To calculate the efficiency of a transformer at \( \frac{3}{4} \) full load with unity power factor, we need to understand how to handle the iron losses, copper losses, and how they relate to the transformer's operation at different load levels. Here's a step-by-step explanation:

### 1. **Identify the Given Data:**
- **Rated power of the transformer (S\(_{rated}\))**: 20 kVA
- **Primary voltage (V\(_{primary}\))**: 1000 V
- **Secondary voltage (V\(_{secondary}\))**: 400 V
- **Frequency**: 50 Hz
- **Iron (core) losses**: 300 W
- **Full-load copper losses**: 500 W
- **Load**: \( \frac{3}{4} \) of full load
- **Power factor**: Unity (1.0)

### 2. **Calculate the Load Power at \( \frac{3}{4} \) Full Load:**

The full-load power of the transformer is given as 20 kVA. At \( \frac{3}{4} \) full load, the actual load power is:

\[
S_{load} = \frac{3}{4} \times 20\ \text{kVA} = 15\ \text{kVA}
\]

At unity power factor, the real power (P\(_{load}\)) is equal to the apparent power (S\(_{load}\)):

\[
P_{load} = 15\ \text{kW}
\]

### 3. **Calculate Copper Losses at \( \frac{3}{4} \) Full Load:**

Full-load copper losses are given as 500 W. Copper losses are proportional to the square of the load. At \( \frac{3}{4} \) full load, the copper losses are:

\[
P_{cu\ (load)} = 500\ \text{W} \times \left(\frac{3}{4}\right)^2
\]

\[
P_{cu\ (load)} = 500\ \text{W} \times \frac{9}{16}
\]

\[
P_{cu\ (load)} = 281.25\ \text{W}
\]

### 4. **Calculate Total Losses:**

Total losses in the transformer include iron losses and copper losses. Since iron losses are constant regardless of the load, total losses at \( \frac{3}{4} \) full load are:

\[
\text{Total Losses} = \text{Iron Losses} + \text{Copper Losses}
\]

\[
\text{Total Losses} = 300\ \text{W} + 281.25\ \text{W}
\]

\[
\text{Total Losses} = 581.25\ \text{W}
\]

### 5. **Calculate Efficiency:**

Efficiency (\(\eta\)) of the transformer is given by:

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

The output power is the load power, which is 15 kW or 15,000 W. Therefore:

\[
\eta = \frac{15,000\ \text{W}}{15,000\ \text{W} + 581.25\ \text{W}}
\]

\[
\eta = \frac{15,000}{15,581.25}
\]

\[
\eta \approx 0.9626
\]

\[
\eta \approx 96.26\%
\]

### Conclusion:

The efficiency of the transformer at \( \frac{3}{4} \) full load and unity power factor is approximately **96.26%**.