The efficiency discussed so far is the ordinary or commercial efficiency which is given by the ratio of output power to input power, that is,
Commercial efficiency,
The load on certain transformers fluctuate throughout the day. The distribution transformers are energized for 24 h, but they deliver very light loads for major portion of the day. Thus, iron losses occur for whole day, but copper losses occur only when the transformer are loaded. Hence, the performance of such transformers cannot be judged by the commercial efficiency, but it can be judged by all-day efficiency also known as operational efficiency or energy efficiency which is computed on the basis of energy consumed during a period of 24 h.
The all-day efficiency is defined as the ratio of output in kWh (or Wh) to the input in kWh (or Wh) of a transformer over 24 h.
∴
To find this all-day efficiency, we have to know the load cycle on the transformer.
Example 10.33
A 20 kVA transformer on domestic load, which can be taken as of unity power factor, has a full-load efficiency of 94.3%, the copper loss then being twice the iron loss. Calculate its all-day efficiency on the following daily cycle; no-load for 10 h, half load for 8 h, and full load for 6 h.
Solution:
Full-load output = 20 × 1 = 20 kW
Full-load input = 
Total losses, Pi + Pc = 20.986 − 20 = 0·986 kW
Now, Pc = 2Pi (given) Pi + 2Pi = 0.986 kW
Or lron losses, Pi = 0.3287 kW
Full-load copper losses = 2 × 0.3287 = 0.6574 kW
kWh output in 24 h = 
× 20 × 8 + 1 × 20 × 6 = 200 kWh
Iron losses for 24 h = 0.3287 × 24 = 7.89 kWh
Copper losses for 24 h = cu. losses for 8 h at full load + cu. losses for 6 h at full load
input in 24 h = kWh output in 24 h + iron and cu losses in kWh for 24 h
= 200 + 7.89 + 5.259 = 213.149 kWh
All day efficiency, 
Example 10.34
A 5 kVA single-phase transformer has a core loss of 50 W and full-load ohmic loss of 100 W. The daily variation of load on the transformer is as follows:
| 7 am to 1 pm | 3 kW at power factor 0.6 lagging. |
| 1 pm to 6 pm | 2 kW at power factor 0.8 lagging. |
| 6 pm to 1 am | 5 kW at power factor 0.9 lagging. |
| 1 am to 7 am | No-load |
Determine the all-day efficiency.
(P.T.U., Dec. 2006)
Solution:
Transformer rating = 5 kVA; Pi = 50 W; Pc = 100 W
Load variation in tabulated form is given below:
kWh output in 24 h = 3 × 6 + 2 × 5 + 5 × 7 + 0 × 6 = 63 kWh
Iron losses in 24 h = Pi in kW × 24 = 
Copper losses in 24 h = 
= 0.6 + 0.125 + 0.8625 + 0 = 1.5875 kWh
All-day efficiency, 
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