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, we need to consider the total energy output, the total energy input, and the losses over a 24-hour period. The transformer experiences two types of losses: copper losses (variable) and iron losses (constant).

Step 1: List out the given data

1. Rated transformer capacity: 500 kVA
   

1. Copper losses (at full load): 5 kW
   

1. Iron losses (constant): 3 kW
   

1. Load profile:
   

  - 400 kW for 6 hours with a power factor of 0.8
  - 300 kW for 12 hours with a power factor of 0.75
  - 100 kW for 3 hours with a power factor of 0.8
  - No load for 3 hours

Step 2: Calculate the losses for each loading condition

Copper Losses:

Copper losses are proportional to the square of the load. Therefore, at any given load \( P_L \), the copper losses can be calculated as:

\[
\text{Copper Losses} = \left(\frac{P_L}{500}\right)^2 \times 5 \, \text{kW}
\]

Where 500 kVA is the full-load rating and 5 kW is the copper loss at full load.

Iron Losses:

Iron losses are constant at 3 kW, regardless of the load, so they will be the same for all conditions.

Step 3: Calculate the total losses and output for each period

Now, we'll break it down for each loading condition.

For 400 kW load (6 hours):

1. Copper losses at 400 kW:
   

  \[
  \text{Copper Losses} = \left(\frac{400}{500}\right)^2 \times 5 = \left(0.8\right)^2 \times 5 = 3.2 \, \text{kW}
  \]

1. Total losses = Copper losses + Iron losses = 3.2 + 3 = 6.2 kW
   

1. Output power = 400 kW
   

For 300 kW load (12 hours):

1. Copper losses at 300 kW:
   

  \[
  \text{Copper Losses} = \left(\frac{300}{500}\right)^2 \times 5 = \left(0.6\right)^2 \times 5 = 1.8 \, \text{kW}
  \]

1. Total losses = Copper losses + Iron losses = 1.8 + 3 = 4.8 kW
   

1. Output power = 300 kW
   

For 100 kW load (3 hours):

1. Copper losses at 100 kW:
   

  \[
  \text{Copper Losses} = \left(\frac{100}{500}\right)^2 \times 5 = \left(0.2\right)^2 \times 5 = 0.2 \, \text{kW}
  \]

1. Total losses = Copper losses + Iron losses = 0.2 + 3 = 3.2 kW
   

1. Output power = 100 kW
   

For no load (3 hours):

1. Copper losses at no load = 0 (since copper losses are proportional to the load)
   

1. Iron losses = 3 kW (constant)
   

1. Total losses = 3 kW
   

1. Output power = 0 kW
   

Step 4: Calculate the total energy input and output over 24 hours

1. Total energy output:
   

  - For 400 kW load (6 hours):
    \[
    \text{Energy output} = 400 \, \text{kW} \times 6 \, \text{hours} = 2400 \, \text{kWh}
    \]
  - For 300 kW load (12 hours):
    \[
    \text{Energy output} = 300 \, \text{kW} \times 12 \, \text{hours} = 3600 \, \text{kWh}
    \]
  - For 100 kW load (3 hours):
    \[
    \text{Energy output} = 100 \, \text{kW} \times 3 \, \text{hours} = 300 \, \text{kWh}
    \]
  - For no load (3 hours):
    \[
    \text{Energy output} = 0 \, \text{kW} \times 3 \, \text{hours} = 0 \, \text{kWh}
    \]

  Total energy output = 2400 + 3600 + 300 + 0 = 6300 kWh

1. Total energy input:
   

  The energy input includes the output energy plus the losses.

  - For 400 kW load (6 hours):
    \[
    \text{Energy input} = \text{Energy output} + \text{Losses} = 2400 + (6.2 \times 6) = 2400 + 37.2 = 2437.2 \, \text{kWh}
    \]
  - For 300 kW load (12 hours):
    \[
    \text{Energy input} = 3600 + (4.8 \times 12) = 3600 + 57.6 = 3657.6 \, \text{kWh}
    \]
  - For 100 kW load (3 hours):
    \[
    \text{Energy input} = 300 + (3.2 \times 3) = 300 + 9.6 = 309.6 \, \text{kWh}
    \]
  - For no load (3 hours):
    \[
    \text{Energy input} = 0 + (3 \times 3) = 9 \, \text{kWh}
    \]

  Total energy input = 2437.2 + 3657.6 + 309.6 + 9 = 6403.4 kWh

Step 5: Calculate the all-day efficiency

The all-day efficiency \( \eta_{\text{all-day}} \) is given by:

\[
\eta_{\text{all-day}} = \frac{\text{Total Energy Output}}{\text{Total Energy Input}} \times 100
\]

\[
\eta_{\text{all-day}} = \frac{6300}{6403.4} \times 100 = 98.39\%
\]

Final Answer:

The all-day efficiency of the transformer is approximately 98.39%.