Question

Find the all-day efficiency of a 500 KVA distribution transformer whose copper loss and iron loss at full load are 4.5 KW and 3.5 KW respectively. During a day of 24 hours, it is loaded as under: No of hours Loading in KW Power factor 6 400 0.8 10 300 0.75 4 100 0.8 4 0 -.

Answer 1

To find the all-day efficiency of a distribution transformer, we'll follow these steps:

1. **Calculate the Total Energy Input to the Transformer:**

   The energy input to the transformer is the total energy supplied to it over the day, accounting for the load and the losses.

2. **Calculate the Total Energy Output from the Transformer:**

   The energy output is the useful energy delivered to the load over the day, which is the energy minus the losses.

3. **Calculate the Efficiency:**

   Efficiency is the ratio of the useful energy output to the total energy input, expressed as a percentage.

### Step-by-Step Solution:

#### 1. Calculate Total Energy Input

**Total Losses**: The total losses in the transformer are the sum of copper losses and iron losses.

- Copper Losses (\(P_c\)) at full load: 4.5 kW
- Iron Losses (\(P_i\)) at full load: 3.5 kW

Since the iron losses are constant (they do not depend on the load), we need to consider these losses along with varying copper losses throughout the day.

**Copper Losses Variation**:

Copper losses vary with the square of the load. Therefore, we need to calculate the average copper losses over the day.

The transformer operates at different loads for different hours:

- **6 hours** at 400 kW with a power factor of 0.8
- **10 hours** at 300 kW with a power factor of 0.75
- **4 hours** at 100 kW with a power factor of 0.8
- **4 hours** at no load

**Load Calculation for Copper Losses**:

1. **For 400 kW load**:
   - Copper Loss = \(\frac{400^2}{500^2} \times 4.5 \text{ kW} = \frac{160000}{250000} \times 4.5 \text{ kW} = 2.88 \text{ kW}\)
   - Energy Loss = 2.88 kW \(\times\) 6 hours = 17.28 kWh

2. **For 300 kW load**:
   - Copper Loss = \(\frac{300^2}{500^2} \times 4.5 \text{ kW} = \frac{90000}{250000} \times 4.5 \text{ kW} = 1.62 \text{ kW}\)
   - Energy Loss = 1.62 kW \(\times\) 10 hours = 16.2 kWh

3. **For 100 kW load**:
   - Copper Loss = \(\frac{100^2}{500^2} \times 4.5 \text{ kW} = \frac{10000}{250000} \times 4.5 \text{ kW} = 0.72 \text{ kW}\)
   - Energy Loss = 0.72 kW \(\times\) 4 hours = 2.88 kWh

4. **For no load**:
   - Copper Loss = 0 kW
   - Energy Loss = 0 kWh

**Total Copper Loss** = 17.28 kWh + 16.2 kWh + 2.88 kWh = 36.36 kWh

**Total Iron Loss**:
- Iron Loss = 3.5 kW \(\times\) 24 hours = 84 kWh

**Total Losses**:
- Total Losses = Copper Losses + Iron Losses = 36.36 kWh + 84 kWh = 120.36 kWh

**Total Energy Input** = Energy Output + Total Losses

#### 2. Calculate Total Energy Output

Energy output is calculated based on the load and duration:

1. **For 400 kW load**:
   - Energy Output = 400 kW \(\times\) 6 hours = 2400 kWh

2. **For 300 kW load**:
   - Energy Output = 300 kW \(\times\) 10 hours = 3000 kWh

3. **For 100 kW load**:
   - Energy Output = 100 kW \(\times\) 4 hours = 400 kWh

4. **For no load**:
   - Energy Output = 0 kWh

**Total Energy Output** = 2400 kWh + 3000 kWh + 400 kWh = 4800 kWh

#### 3. Calculate Efficiency

The total energy input is the sum of the total energy output and the total losses.

**Total Energy Input** = 4800 kWh + 120.36 kWh = 4920.36 kWh

**Efficiency** = \(\frac{\text{Total Energy Output}}{\text{Total Energy Input}} \times 100\)

**Efficiency** = \(\frac{4800}{4920.36} \times 100 \approx 97.55\% \)

### Conclusion

The all-day efficiency of the 500 KVA distribution transformer is approximately **97.55%**.

Answer 2

To find the all-day efficiency of the distribution transformer, we need to determine both the total input power and the total output power over a 24-hour period and then use these to calculate the efficiency. Here's a step-by-step process to find the all-day efficiency:

### 1. **Determine the Total Output Power:**

Let's calculate the output power for each loading condition and then find the total output energy.

**For each loading condition:**

1. **Condition 1:**
   - Load: 400 kW
   - Duration: 6 hours
   - Power Factor: 0.8

   Output Power = Load × Power Factor
   \[
   \text{Output Power}_1 = 400 \text{ kW} \times 0.8 = 320 \text{ kW}
   \]
   Energy Output = Output Power × Duration
   \[
   \text{Energy Output}_1 = 320 \text{ kW} \times 6 \text{ hours} = 1920 \text{ kWh}
   \]

2. **Condition 2:**
   - Load: 300 kW
   - Duration: 10 hours
   - Power Factor: 0.75

   Output Power = Load × Power Factor
   \[
   \text{Output Power}_2 = 300 \text{ kW} \times 0.75 = 225 \text{ kW}
   \]
   Energy Output = Output Power × Duration
   \[
   \text{Energy Output}_2 = 225 \text{ kW} \times 10 \text{ hours} = 2250 \text{ kWh}
   \]

3. **Condition 3:**
   - Load: 100 kW
   - Duration: 4 hours
   - Power Factor: 0.8

   Output Power = Load × Power Factor
   \[
   \text{Output Power}_3 = 100 \text{ kW} \times 0.8 = 80 \text{ kW}
   \]
   Energy Output = Output Power × Duration
   \[
   \text{Energy Output}_3 = 80 \text{ kW} \times 4 \text{ hours} = 320 \text{ kWh}
   \]

4. **Condition 4:**
   - Load: 0 kW
   - Duration: 4 hours

   Output Power = 0 kW
   \[
   \text{Energy Output}_4 = 0 \text{ kWh}
   \]

**Total Energy Output:**
\[
\text{Total Energy Output} = \text{Energy Output}_1 + \text{Energy Output}_2 + \text{Energy Output}_3 + \text{Energy Output}_4
\]
\[
\text{Total Energy Output} = 1920 \text{ kWh} + 2250 \text{ kWh} + 320 \text{ kWh} + 0 \text{ kWh} = 3490 \text{ kWh}
\]

### 2. **Determine the Total Input Power:**

To find the total input power, we need to account for both copper losses and iron losses.

**Copper Losses:**
   - Full Load Copper Loss: 4.5 kW
   - Since copper losses are proportional to the square of the load, we need to calculate the losses for each load condition.

1. **Condition 1:**
   \[
   \text{Copper Loss}_1 = 4.5 \text{ kW} \times \left(\frac{400}{500}\right)^2 = 4.5 \text{ kW} \times \frac{16}{25} = 2.88 \text{ kW}
   \]
   Total Loss Energy = Copper Loss × Duration
   \[
   \text{Loss Energy}_1 = 2.88 \text{ kW} \times 6 \text{ hours} = 17.28 \text{ kWh}
   \]

2. **Condition 2:**
   \[
   \text{Copper Loss}_2 = 4.5 \text{ kW} \times \left(\frac{300}{500}\right)^2 = 4.5 \text{ kW} \times \frac{9}{25} = 1.62 \text{ kW}
   \]
   Total Loss Energy = Copper Loss × Duration
   \[
   \text{Loss Energy}_2 = 1.62 \text{ kW} \times 10 \text{ hours} = 16.2 \text{ kWh}
   \]

3. **Condition 3:**
   \[
   \text{Copper Loss}_3 = 4.5 \text{ kW} \times \left(\frac{100}{500}\right)^2 = 4.5 \text{ kW} \times \frac{1}{25} = 0.18 \text{ kW}
   \]
   Total Loss Energy = Copper Loss × Duration
   \[
   \text{Loss Energy}_3 = 0.18 \text{ kW} \times 4 \text{ hours} = 0.72 \text{ kWh}
   \]

4. **Condition 4:**
   \[
   \text{Copper Loss}_4 = 4.5 \text{ kW} \times \left(\frac{0}{500}\right)^2 = 0 \text{ kW}
   \]
   Total Loss Energy = Copper Loss × Duration
   \[
   \text{Loss Energy}_4 = 0 \text{ kWh}
   \]

**Iron Losses:**
Iron losses are constant regardless of the load. For a full day:
   \[
   \text{Iron Losses} = 3.5 \text{ kW} \times 24 \text{ hours} = 84 \text{ kWh}
   \]

**Total Loss Energy:**
\[
\text{Total Loss Energy} = \text{Loss Energy}_1 + \text{Loss Energy}_2 + \text{Loss Energy}_3 + \text{Loss Energy}_4 + \text{Iron Losses}
\]
\[
\text{Total Loss Energy} = 17.28 \text{ kWh} + 16.2 \text{ kWh} + 0.72 \text{ kWh} + 0 \text{ kWh} + 84 \text{ kWh} = 118.2 \text{ kWh}
\]

**Total Input Energy:**
\[
\text{Total Input Energy} = \text{Total Energy Output} + \text{Total Loss Energy}
\]
\[
\text{Total Input Energy} = 3490 \text{ kWh} + 118.2 \text{ kWh} = 3608.2 \text{ kWh}
\]

### 3. **Calculate the All-Day Efficiency:**

Efficiency is given by the ratio of the total output energy to the total input energy:
\[
\text{Efficiency} = \frac{\text{Total Energy Output}}{\text{Total Input Energy}} \times 100\%
\]
\[
\text{Efficiency} = \frac{3490 \text{ kWh}}{3608.2 \text{ kWh}} \times 100\% \approx 96.7\%
\]

So, the all-day efficiency of the transformer is approximately **96.7%**.