---
title: "All-day efficiency of 150 kVA transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:all-day-efficiency-150kva"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/all-day-efficiency-150kva
site: BCT Engineering
---

# All-day efficiency of 150 kVA transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2081 Chaitra (new course), 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/all-day-efficiency-150kva
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

Daily loading of a 150 kVA transformer having iron loss of 400 W and full load copper loss of 1700 W is given below.

| Time | Load (kW) | Power factor |
|---|---|---|
| 6:00 – 10:00 | 50 | 0.9 |
| 10:00 – 17:00 | 60 | 0.85 |
| 17:00 – 21:00 | 70 | 0.8 |
| 21:00 – 6:00 | 10 | 0.95 |

Determine the all-day efficiency of the transformer.

## Short answer

Total energy output is 990 kWh. Iron loss over 24 h is 9.6 kWh, total copper loss 5.957 kWh, sum 15.557 kWh. All-day efficiency = 990 / (990 + 15.557) = 98.45%.

## Answer

All-day efficiency = (energy output in 24 h) / (energy output + energy losses in 24 h). Iron loss lasts all 24 hours; copper loss varies as the square of kVA load.

**Data:** 150 kVA, $P_i = 0.4$ kW, $P_{cu,fl} = 1.7$ kW.

#### Energy table

Copper loss at each load $= 1.7\times(\text{kVA}/150)^2$.

| Period | Hours | kW | pf | kVA | Cu loss (kW) | Cu energy (kWh) | Output (kWh) |
|---|---|---|---|---|---|---|---|
| 6–10 | 4 | 50 | 0.90 | 55.56 | 0.2332 | 0.933 | 200 |
| 10–17 | 7 | 60 | 0.85 | 70.59 | 0.3765 | 2.635 | 420 |
| 17–21 | 4 | 70 | 0.80 | 87.50 | 0.5785 | 2.314 | 280 |
| 21–6 | 9 | 10 | 0.95 | 10.53 | 0.0084 | 0.075 | 90 |
| **Total** | 24 | | | | | **5.957** | **990** |

#### Losses and efficiency

$$
\begin{aligned}
\text{Iron energy} &= 0.4\times24 = 9.6\ \text{kWh} \\
\text{Total losses} &= 9.6 + 5.957 = 15.557\ \text{kWh} \\
\eta_{all\text{-}day} &= \frac{990}{990 + 15.557}\times100 = 98.45\%
\end{aligned}
$$

**Answer: all-day efficiency ≈ 98.45%.**

## Check yourself

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

1. How is all-day efficiency defined?
   A. Energy output in 24 h divided by energy output plus energy losses in 24 h
   B. Output kW over input kW at full load
   C. Average of efficiencies at each load
   D. Maximum efficiency during the day
   Answer: A. It uses energy (kWh), not power. (Key point: Definition)

2. How is iron loss energy counted?
   A. Only during loaded hours
   B. 0.4 kW for all 24 hours = 9.6 kWh, as the core is always excited
   C. 0.4 kW × 9 h
   D. Zero when load is light
   Answer: B. Iron loss is constant while the transformer is energised. (Key point: Iron loss)

3. How is copper loss at each load found?
   A. 1.7 × (kW/150) kW
   B. 1.7 kW at every load
   C. 1.7 × (kVA/150)² kW, using kVA = kW/pf
   D. 1.7 × (kVA/150) kW
   Answer: C. Copper loss varies as the square of load kVA relative to rating. (Key point: Copper loss)

4. What is the all-day efficiency?
   A. About 97.2%
   B. About 99.0%
   C. About 96.5%
   D. About 98.45%, from 990/(990 + 9.6 + 5.957)
   Answer: D. Output 990 kWh, losses 15.557 kWh. (Key point: Result)

## Related in Electrical Machines I

- [Efficiency and loading of 50 kVA transformer at 40 kW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A50kva-200-2200-40kw-overload) — Electrical Machines I, chapter 2
- [Full-load and half-load efficiency of 100 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A100kva-1200w-960w-efficiency) — Electrical Machines I, chapter 2
- [Full-load efficiency of 500 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A500kva-90pc-half-load-full-load) — Electrical Machines I, chapter 2
- [Iron loss and maximum efficiency of 20 kVA 220/2200 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A20kva-iron-loss-max-efficiency-current) — Electrical Machines I, chapter 2, asked 2×
- [Iron loss and maximum efficiency of 20 kVA 250/2500 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A20kva-250-2500-iron-loss-max) — Electrical Machines I, chapter 2, asked 2×
- [Efficiency of 5 kVA transformer at 2 kW load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A5kva-1100-110-max-efficiency-40a) — Electrical Machines I, chapter 2

## Related in other subjects

- [150 kVA transformer: max efficiency load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-11000-230-max-efficiency-kva) — Electrical Machines, chapter 2
- [500 kVA transformer efficiency full/half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-500kva-efficiency-full-half) — Electrical Machine, chapter 2
- [500 kVA transformer efficiency, full and half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-6600-400-efficiency) — Electrical Machines, chapter 2
- [100 kVA transformer efficiency at full and half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-100kva-efficiency-full-half-load) — Electrical Machines, chapter 2