Percentage energy loss of 25 kVA transformer

Percentage energy loss = (core energy loss + copper energy loss) / energy sold × 100. Copper energy loss uses the loss of load factor.

Assumption: month = 30 days = 720 h.

Step 1: Load factor

$$LF = \frac{\text{Energy}}{P_{max} \times T} = \frac{4800}{20 \times 720} = 0.3333$$

Step 2: Loss of load factor

$$\begin{aligned} LLF &= 0.3\,LF + 0.7\,LF^2 \\ &= 0.3(0.3333) + 0.7(0.3333)^2 \\ &= 0.1000 + 0.0778 = 0.17778 \end{aligned}$$

Step 3: Copper loss at peak load

$$\begin{aligned} S_{peak} &= \frac{20}{0.85} = 23.529 \text{ kVA} \\ P_{cu,peak} &= 300 \times \left(\frac{23.529}{25}\right)^2 = 300 \times 0.8858 = 265.74 \text{ W} \end{aligned}$$

Step 4: Monthly energy losses

$$\begin{aligned} E_{cu} &= 0.26574 \times 0.17778 \times 720 = 34.02 \text{ kWh} \\ E_{core} &= 0.072 \times 720 = 51.84 \text{ kWh} \\ E_{loss} &= 34.02 + 51.84 = 85.86 \text{ kWh} \end{aligned}$$

Step 5: Percentage energy loss

$$\%E_{loss} = \frac{85.86}{4800} \times 100 = 1.79\%$$

(Based on energy input $4800 + 85.86 = 4885.86$ kWh, it is $85.86/4885.86 = 1.76\%$.)

Answer: Percentage energy loss ≈ 1.79% of energy sold (monthly loss ≈ 85.9 kWh: 51.84 kWh core loss + 34.02 kWh copper loss).

Note that core loss is the larger part, because the load factor is low (0.33): the transformer is energised all month but carries peak load only briefly.