The **all day efficiency of transformer** is related to the distribution transformer. It is used to find the efficiency of the distribution transformer.

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As we know that efficiency of a transformer is given by the ratio of output power to the input power. This is true when a transformer is subjected to the same load for all time duration. But this is not true with a distribution transformer.

The distribution transformer is kept in operation 24 hours a day with varying loads. The load varies throughout the day.

Distribution Transformers are those transformers that change the voltage level to another voltage level suitable for utilization purposes at the consumer’s premises. A distribution transformer must have its secondary directly connected to the consumer’s terminals.

For example, a distribution transformer may have practically no load or little load during a considerable period of the day but in the evening time, it may go up to the rated capacity. Hence the load on a distribution transformer varies a wide range of 24 hours a day.

As we know that losses occurring in a transformer are core losses and copper losses. As the primary of a distribution transformer is always energized, core loss will take place continuously i.e. whole day or 24 hours. This is the reason a distribution transformer is designed to have a very low value of core loss.

The copper losses are variable losses. It depends upon the load. Since load varies throughout the day. Hence copper losses also vary throughout the day.

The efficiency of the distribution transformer is calculated for a day or 24 hours. For this, all day efficiency comes in light.

## All day efficiency definition

The **all day efficiency of transformer** is defined as the ratio of energy output to the energy input taken over a 24 hour period. Mathematically, all day efficiency formula is given as

It can also be written as

It is noted that all day efficiency of transformer depends primarily on amount of load and duration of load. The all day efficiency is calculated for distribution transformer.

## All day efficiency of transformer Examples

Consider the following example.

A 100 kVA, 2000/200V, 50 Hz distribution transformer has core loss of 500 W at rated voltage and copper losses of 1200 W at full load. The distribution transformer is loaded as follows during the day.

% rated load | 0 | 20 | 50 | 80 | 100 | 120 |

Power factor | — | 0.7 lag | 0.8 lag | 0.9 lag | 1 | 0.85 lag |

Hours | 2 | 4 | 4 | 5 | 7 | 2 |

Determine the all day efficiency of this distribution transformer.

The energy output can be calculated as

E_{o} = kVA × p.f. × hours (in KWh)

OR

E_{o} = kVA × cosφ × hours (in KWh)

% rated load |
p.f. |
hours |
kVA |
E_{o} = kVA × p.f. × hours (in kWh) |

20 | 0.7 | 4 | 100 × 0.2 = 20 | 20 × 0.7 × 4 = 56 |

50 | 0.8 | 4 | 100 × 0.5 = 50 | 50 × 0.8 × 4 = 160 |

80 | 0.9 | 5 | 100 × 0.8 = 80 | 80 × 0.9 × 5 = 360 |

100 | 1 | 7 | 100 × 1 = 100 | 100 × 1 × 7 = 700 |

120 | 0.85 | 2 | 100 × 1.20 = 120 | 120 × 0.85 × 2 = 204 |

1480 |

Total energy output over 24 hours (excluding 2 hours at no load), E_{ot} = 1480 kWh

Total energy loss in the core over 24 hours (including 2 hours at no load), E_{i} = 500 W × 24 h = 12000 Wh or 12 kWh

The copper loss at any load can be calculated as

P_{cu} = x^{2} × copper loss at rated load

where x = given load/rated load

% rated load |
x |
hours |
Copper losses (kW) |
Energy loss in winding, E_{c} = kW × hours (in kWh) |

20 | 0.2 | 4 | (0.2)^{2} × 1.2 = 0.048 |
0.048 × 4 = 0.192 |

50 | 0.5 | 4 | (0.5)^{2} × 1.2 = 0.3 |
0.3 × 4 = 1.2 |

80 | 0.8 | 5 | (0.8)^{2} × 1.2 = 0.768 |
0.768 × 5 = 3.84 |

100 | 1 | 7 | (1)^{2} × 1.2 = 1.2 |
1.2 × 7 = 8.4 |

120 | 1.2 | 2 | (1.2)^{2} × 1.2 = 1.728 |
1.728 × 2 = 3.456 |

17.088 |

Total energy loss in winding for 24 hours (excluding 2 hours at no load), E_{cu }= 17.088 kWh

Total energy loss in 24 hours = E_{i} + E_{cu }= 12 + 17.088 = 29.088 kWh

The all day efficiency is given by

∴ The all day efficiency of distribution transformer is 98.07%.