---
title: "Monthly energy loss of 50 kVA transformer"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:transformer-monthly-loss-50kva"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/transformer-monthly-loss-50kva
site: BCT Engineering
---

# Monthly energy loss of 50 kVA transformer

Transmission and Distribution System Design · Chapter 7: Distribution System Design

- Asked in: 2073 Magh, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/transformer-monthly-loss-50kva
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

A 11 kV/0.4 kV, 50 kVA distribution transformer has no-load loss of 120 W and rated copper loss of 200 W. Determine the percentage monthly energy loss of the transformer if the peak demand is 30 kVA at a power factor of 0.9 and load factor of 0.3. Given LLF = 0.3LF + 0.7LF².

## Short answer

Core loss energy = 86.4 kWh, copper loss energy = 7.93 kWh using LLF = 0.153. Total monthly loss = 94.33 kWh, which is 1.62% of the energy delivered (5832 kWh).

## Answer

A transformer has two losses: the **core (no-load) loss**, present all 720 h of the month, and the **copper loss**, which varies with the square of the load and is averaged using the loss load factor (LLF). Month taken as 30 days = 720 h.

#### Step 1: Core loss energy
$$
E_{core} = 120\times720 = 86\,400\ \text{Wh} = 86.4\ \text{kWh}
$$

#### Step 2: Copper loss at peak
Copper loss varies as $(S/S_{rated})^2$:
$$
P_{cu,peak} = 200\times\left(\frac{30}{50}\right)^2 = 200\times0.36 = 72\ \text{W}
$$

#### Step 3: Loss load factor
$$
LLF = 0.3(0.3) + 0.7(0.3)^2 = 0.09 + 0.063 = 0.153
$$

#### Step 4: Copper loss energy
$$
E_{cu} = 72\times0.153\times720 = 7931.5\ \text{Wh} = 7.93\ \text{kWh}
$$

#### Step 5: Energy delivered in the month
Peak demand in kW $= 30\times0.9 = 27$ kW.
$$
E_{out} = 27\times0.3\times720 = 5832\ \text{kWh}
$$

#### Step 6: Percentage loss
$$
\begin{aligned}
E_{loss} &= 86.4 + 7.93 = 94.33\ \text{kWh}\\
\%\,\text{loss} &= \frac{94.33}{5832}\times100 = 1.62\ \%
\end{aligned}
$$
(If expressed on input energy, $94.33/(5832+94.33) = 1.59\ \%$.)

| Item | Energy (kWh/month) |
|---|---|
| Core loss | 86.40 |
| Copper loss | 7.93 |
| Total loss | 94.33 |
| Energy supplied to load | 5832 |

Observation: at low load factor the core loss dominates (over 90 % of the loss), so for rural transformers with low LF, low no-load-loss (e.g. amorphous core) designs are preferred.

**Answer: Monthly loss ≈ 94.3 kWh, i.e. about 1.62 % of the energy delivered.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Core loss energy for 720 h at 120 W:
   A. 86.4 kWh
   B. 120 kWh
   C. 8.64 kWh
   D. 864 kWh
   Answer: A. 0.12 x 720 = 86.4 kWh. (Key point: Core energy)

2. Copper loss at 30 kVA:
   A. 200 W
   B. 120 W
   C. 72 W
   D. 36 W
   Answer: C. 200 x 0.36 = 72 W. (Key point: Cu at peak)

3. Energy delivered:
   A. 5832 kWh
   B. 6480 kWh
   C. 21600 kWh
   D. 8100 kWh
   Answer: A. 27 kW x 0.3 x 720 = 5832. (Key point: Energy out)

4. Percentage loss:
   A. 1.62%
   B. 16.2%
   C. 0.16%
   D. 3.2%
   Answer: A. 94.33/5832 = 1.62%. (Key point: Percent)

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