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

W_i = 3 kW

W_cu = 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 × W_c × h₁}

    +{ [ kVA₂ / kVA ]² × W_c × h₂ }

    +{ [ kVA₃ / kVA ]² × W_c × 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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