---
title: "Energy loss of 100 kVA transformer in third year"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:transformer-energy-loss-third-year"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/transformer-energy-loss-third-year
site: BCT Engineering
---

# Energy loss of 100 kVA transformer in third year

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

- Asked in: 2079 Chaitra, 8 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-energy-loss-third-year
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

A 3-phase transformer of 100 kVA supplies a load of 75 kVA in its first year of installation and addition 10 kVA for subsequent two years. Calculate the energy loss for third year with following parameters given: Voltage level = 11/0.4 kV, Frequency = 50 Hz, Load Factor = 0.8, No load losses = 200 W, Copper loss = 1100 W, Loss of load factor = 0.3×LF + 0.7×(LF)².

## Short answer

Third-year load 95 kVA, LF=0.8 gives LLF=0.688. Peak copper loss = 1100×(95/100)² = 992.75 W. Core loss = 200×8760 = 1752 kWh; copper loss = 992.75×0.688×8760 ≈ 5983 kWh. Total annual energy loss ≈ 7735 kWh.

## Answer

Energy loss = core loss for all hours + copper loss at peak × loss of load factor × hours.

**Assumptions:** "additional 10 kVA for subsequent two years" means the load is 85 kVA in year 2 and 95 kVA in year 3. The given copper loss (1100 W) is at rated load (100 kVA). The year has 8760 h.

#### Step 1: Third-year peak load

$$
S_3 = 75 + 10 + 10 = 95 \text{ kVA}
$$

#### Step 2: Loss of load factor

$$
\begin{aligned}
LLF &= 0.3\,LF + 0.7\,LF^2 = 0.3(0.8) + 0.7(0.8)^2 \\
&= 0.24 + 0.448 = 0.688
\end{aligned}
$$

#### Step 3: Copper loss at 95 kVA

$$
P_{cu} = 1100 \times \left(\frac{95}{100}\right)^2 = 1100 \times 0.9025 = 992.75 \text{ W}
$$

#### Step 4: Annual energy losses

$$
\begin{aligned}
E_{cu} &= 0.99275 \times 0.688 \times 8760 = 5983.19 \text{ kWh} \\
E_{core} &= 0.200 \times 8760 = 1752.00 \text{ kWh} \\
E_{total} &= 5983.19 + 1752.00 = 7735.19 \text{ kWh}
\end{aligned}
$$

**Answer: Energy loss in the third year ≈ 7735 kWh** (copper 5983 kWh + core 1752 kWh).

Because the load factor is high (0.8) and the transformer is loaded at 95%, copper loss is the dominant part of the energy loss.

## Check yourself

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

1. Third-year peak load is:
   A. 75 kVA
   B. 85 kVA
   C. 95 kVA
   D. 105 kVA
   Answer: C. 75 + 10 + 10 = 95 kVA. (Key point: Load year 3)

2. For LF = 0.8, LLF is:
   A. 0.688
   B. 0.64
   C. 0.8
   D. 0.24
   Answer: A. 0.3(0.8) + 0.7(0.64) = 0.24 + 0.448 = 0.688. (Key point: LLF)

3. Copper loss at 95 kVA is:
   A. 1100 W
   B. 992.75 W
   C. 1045 W
   D. 200 W
   Answer: B. 1100 x 0.95^2 = 992.75 W. (Key point: Copper loss)

4. Annual energy loss is:
   A. 1752 kWh
   B. 5983 kWh
   C. 7735 kWh
   D. 9500 kWh
   Answer: C. Copper 5983.19 plus core 1752 = 7735 kWh. (Key point: Total)

## Related in Transmission and Distribution System Design

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