Question

A 500kVA distribution transformer having copper and iron losses of 5kW and 3kW respectively on full load. The transformer is loaded as shown below: Loading (kW) Power Factor (Lag) No. of hours 400 0.8 06 300 0.75 12 100 0.8 03 No load - 03 Calculate the all-day efficiency.

Answer 1

To calculate the all-day efficiency of the transformer, you need to determine the total energy losses and the total energy output over a 24-hour period. Here's a step-by-step process to do this:

### 1. Calculate the Energy Output

**Energy Output for Each Loading Condition:**

1. **400 kW Load:**
   - Power Factor (PF) = 0.8
   - Load = 400 kW
   - Number of hours = 6 hours

   \[ \text{Output Energy} = \text{Load} \times \text{Hours} = 400 \text{ kW} \times 6 \text{ hours} = 2400 \text{ kWh} \]

2. **300 kW Load:**
   - Power Factor (PF) = 0.75
   - Load = 300 kW
   - Number of hours = 12 hours

   \[ \text{Output Energy} = \text{Load} \times \text{Hours} = 300 \text{ kW} \times 12 \text{ hours} = 3600 \text{ kWh} \]

3. **100 kW Load:**
   - Power Factor (PF) = 0.8
   - Load = 100 kW
   - Number of hours = 3 hours

   \[ \text{Output Energy} = \text{Load} \times \text{Hours} = 100 \text{ kW} \times 3 \text{ hours} = 300 \text{ kWh} \]

4. **No Load:**
   - No Load = 0 kW
   - Number of hours = 3 hours

   \[ \text{Output Energy} = 0 \text{ kW} \times 3 \text{ hours} = 0 \text{ kWh} \]

**Total Energy Output:**

\[ \text{Total Output Energy} = 2400 \text{ kWh} + 3600 \text{ kWh} + 300 \text{ kWh} = 6300 \text{ kWh} \]

### 2. Calculate the Energy Losses

**Losses Calculation:**

1. **Copper Losses (Variable Losses):**
   - Copper losses are proportional to the square of the load.
   - For each loading condition, calculate the energy loss as follows:

     \[ \text{Copper Losses} = \text{Load}^2 \times \text{Loss per kW} \times \text{Hours} \]

     Given copper losses on full load (500 kW) = 5 kW, loss per kW = \(\frac{5 \text{ kW}}{500 \text{ kW}} = 0.01 \text{ kW/kW}\).

     - **400 kW Load:**

       \[ \text{Energy Loss} = (400^2 \times 0.01 \text{ kW}) \times 6 \text{ hours} = (1600 \text{ kW}) \times 6 = 9600 \text{ kWh} \]

     - **300 kW Load:**

       \[ \text{Energy Loss} = (300^2 \times 0.01 \text{ kW}) \times 12 \text{ hours} = (900 \text{ kW}) \times 12 = 10800 \text{ kWh} \]

     - **100 kW Load:**

       \[ \text{Energy Loss} = (100^2 \times 0.01 \text{ kW}) \times 3 \text{ hours} = (10 \text{ kW}) \times 3 = 30 \text{ kWh} \]

     - **No Load:**

       \[ \text{Energy Loss} = (0 \text{ kW}) \times 3 \text{ hours} = 0 \text{ kWh} \]

   - **Total Copper Losses:**

     \[ \text{Total Copper Losses} = 9600 \text{ kWh} + 10800 \text{ kWh} + 30 \text{ kWh} = 20730 \text{ kWh} \]

2. **Iron Losses (Constant Losses):**
   - Iron losses are constant regardless of load.

   - For each condition:

     \[ \text{Iron Losses} = 3 \text{ kW} \times \text{Hours} \]

     - **400 kW Load:**

       \[ \text{Energy Loss} = 3 \text{ kW} \times 6 \text{ hours} = 18 \text{ kWh} \]

     - **300 kW Load:**

       \[ \text{Energy Loss} = 3 \text{ kW} \times 12 \text{ hours} = 36 \text{ kWh} \]

     - **100 kW Load:**

       \[ \text{Energy Loss} = 3 \text{ kW} \times 3 \text{ hours} = 9 \text{ kWh} \]

     - **No Load:**

       \[ \text{Energy Loss} = 3 \text{ kW} \times 3 \text{ hours} = 9 \text{ kWh} \]

   - **Total Iron Losses:**

     \[ \text{Total Iron Losses} = 18 \text{ kWh} + 36 \text{ kWh} + 9 \text{ kWh} + 9 \text{ kWh} = 72 \text{ kWh} \]

**Total Energy Losses:**

\[ \text{Total Energy Losses} = 20730 \text{ kWh} + 72 \text{ kWh} = 20772 \text{ kWh} \]

### 3. Calculate the All-Day Efficiency

The all-day efficiency is given by:

\[ \text{All-Day Efficiency} = \frac{\text{Total Output Energy}}{\text{Total Output Energy} + \text{Total Losses}} \times 100 \]

Substitute the values:

\[ \text{All-Day Efficiency} = \frac{6300 \text{ kWh}}{6300 \text{ kWh} + 20772 \text{ kWh}} \times 100 \]

\[ \text{All-Day Efficiency} = \frac{6300}{27072} \times 100 \approx 23.3\% \]

So, the all-day efficiency of the transformer is approximately **23.3%**.

Answer 2

To calculate the all-day efficiency of a transformer, you need to determine both the energy delivered and the energy lost over a 24-hour period. Here’s a step-by-step guide on how to perform the calculation.

### Given Data:

- **Transformer Rating**: 500 kVA
- **Copper Losses (Full Load)**: 5 kW
- **Iron Losses (No Load)**: 3 kW

### Load Profile:

1. **400 kW, Power Factor 0.8, Duration 6 hours**
2. **300 kW, Power Factor 0.75, Duration 12 hours**
3. **100 kW, Power Factor 0.8, Duration 3 hours**
4. **No Load, Duration 3 hours**

### Calculation Steps:

#### 1. **Calculate Total Energy Delivered**

Energy delivered is calculated as:

\[ \text{Energy Delivered (kWh)} = \text{Load (kW)} \times \text{Duration (hours)} \]

- For 400 kW load for 6 hours:
  \[
  \text{Energy}_{400} = 400 \text{ kW} \times 6 \text{ hours} = 2400 \text{ kWh}
  \]

- For 300 kW load for 12 hours:
  \[
  \text{Energy}_{300} = 300 \text{ kW} \times 12 \text{ hours} = 3600 \text{ kWh}
  \]

- For 100 kW load for 3 hours:
  \[
  \text{Energy}_{100} = 100 \text{ kW} \times 3 \text{ hours} = 300 \text{ kWh}
  \]

- For No Load for 3 hours (No load means no power delivered, but losses still occur):
  \[
  \text{Energy}_{\text{No Load}} = 0 \text{ kWh}
  \]

Total energy delivered:
\[
\text{Total Energy Delivered} = 2400 + 3600 + 300 = 6300 \text{ kWh}
\]

#### 2. **Calculate Total Losses**

Losses include both copper losses and iron losses. Copper losses vary with the load, while iron losses are constant.

- **Copper Losses**: These are proportional to the square of the load. To calculate average copper losses, we need to calculate them for each load condition.

**Load Conditions**:

- For 400 kW load:
  \[
  \text{Copper Loss}_{400} = \left(\frac{400 \text{ kW}}{500 \text{ kVA}}\right)^2 \times 5 \text{ kW} = (0.8)^2 \times 5 \text{ kW} = 0.64 \times 5 \text{ kW} = 3.2 \text{ kW}
  \]

- For 300 kW load:
  \[
  \text{Copper Loss}_{300} = \left(\frac{300 \text{ kW}}{500 \text{ kVA}}\right)^2 \times 5 \text{ kW} = (0.6)^2 \times 5 \text{ kW} = 0.36 \times 5 \text{ kW} = 1.8 \text{ kW}
  \]

- For 100 kW load:
  \[
  \text{Copper Loss}_{100} = \left(\frac{100 \text{ kW}}{500 \text{ kVA}}\right)^2 \times 5 \text{ kW} = (0.2)^2 \times 5 \text{ kW} = 0.04 \times 5 \text{ kW} = 0.2 \text{ kW}
  \]

**Average Copper Losses** over 24 hours:
\[
\text{Average Copper Loss} = \frac{(3.2 \text{ kW} \times 6 \text{ hours}) + (1.8 \text{ kW} \times 12 \text{ hours}) + (0.2 \text{ kW} \times 3 \text{ hours})}{24 \text{ hours}}
\]
\[
= \frac{19.2 + 21.6 + 0.6}{24} = \frac{41.4}{24} \approx 1.725 \text{ kW}
\]

- **Iron Losses**: Constant at 3 kW for the entire 24 hours.

Total losses:
\[
\text{Total Losses} = (\text{Average Copper Loss} + \text{Iron Losses}) \times \text{Time}
\]
\[
= (1.725 \text{ kW} + 3 \text{ kW}) \times 24 \text{ hours}
\]
\[
= 4.725 \text{ kW} \times 24 = 113.4 \text{ kWh}
\]

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

The all-day efficiency is the ratio of the energy delivered to the total energy input. The total energy input is the sum of the energy delivered and the losses.

\[
\text{All-Day Efficiency} (\%) = \left(\frac{\text{Energy Delivered}}{\text{Energy Delivered} + \text{Total Losses}}\right) \times 100
\]

\[
\text{All-Day Efficiency} = \left(\frac{6300}{6300 + 113.4}\right) \times 100
\]

\[
= \left(\frac{6300}{6413.4}\right) \times 100 \approx 98.2\%
\]

### Summary

The all-day efficiency of the transformer is approximately **98.2%**.