16/02/2021

All day Efficiency

 

ALL DAY EFFICIENCY

  • The performance of the Distribution Transformer cannot be judged by ordinary efficiency. 
  • As the primary winding of the transformer is energized throughout the day ( 24 hours ) , iron loss occurs continuously whether the transformer is loaded or unloaded. 
  • The copper loss occurs only when the transformer is loaded therefore the distribution transformer is designed such that the core losses are very small. 
  • The performance of such transformers can accurately judge by following equation.

All day efficiency  = Output in ( kWh ) / Input in ( kWh )

  • Although the all day efficiency depends on kWh not in kW, its value depends upon load cycle     ( from hour to hour ) during the day. 
  • The all day efficiency is always is less the ordinary efficiency because the distribution transformer does not supply rated load throughout the day.

Example

Find out all-day efficiency of 600 kVA transformer having core loss and full load copper losses are 3 kW and 4 kW. The transformer is loaded as shown below during the day.

Load

Power factor

Hours during day

450 kW

0.8

5

400 kW

Unity

6

300 kW

0.5

6

No load

 

Rest of the day

 

Solution

Data given

600 kVA transformer

Wi = 3 kW

Wcu = 4 kW

To find out

All day efficiency = ?

Output Energy = kW1 h1 + kW2 h2 + kW3 h3

                        = (450 × 5) + (400 × 6) + (300 × 6)

                        = 6450 kWh

Iron losses throughout the day = 3 × 24 = 72 kWh

The copper loss varies as the square of the load kVA therefore

kVA1 = kW1 / Cos Φ1

           = 450 / 0.8

           = 562.5

kVA2 = kW2 / Cos Φ2

          = 400 / 1

          = 400

kVA3 = kW3 / Cos Φ3

          = 300 / 0.5

          = 600

Copper losses throughout the day

= {[ kVA1 / kVA ]2 × Wc × h1}

    +{ [ kVA2 / kVA ]2 × Wc × h2 }

    +{ [ kVA3 / kVA ]2 × Wc × h3 }

= { [ 562.5 / 600 ]2 × 4 × 5 } + { [ 400 / 600 ]2 × 4 × 6 } + { [ 600 / 600 ]2 × 4 × 6 }

=52.25 kWh

All day efficiency = { Output energy ( 24 hours ) / Input energy ( 24 hours ) } × 100%

= { Output energy /  [ Output energy + Iron losses + Copper losses ] }× 100%

= 6450 / [ 6450 + 52.25 + 72 }× 100%

= 98.10%

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