---
title: "Annual savings of 90% vs 85% efficient motor"
subject: "Industrial Power Distribution and Illumination"
chapter: 10
key: "semester_6/industrial_power_distribution_and_illumination:motor-efficiency-cost-savings"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/motor-efficiency-cost-savings
site: BCT Engineering
---

# Annual savings of 90% vs 85% efficient motor

Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry

- Asked in: 2072 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/motor-efficiency-cost-savings
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10

## Question

Calculate the annual cost savings of a 25 hp motor operating at 90% efficiency compared to 85% efficiency, if the operating hours is 8000 hrs/year and the energy cost is Rs 7.5 per kWh.

## Short answer

Annual cost saving ≈ Rs 73,137, found from output 18.65 kW, input powers 21.94 kW (85%) and 20.72 kW (90%), giving 1.219 kW saving and 9752 kWh/year at Rs 7.5/kWh.

## Answer

Energy saving comes from the lower input power of the more efficient motor for the same shaft output. Assume the motor runs at full rated output (25 hp) for all 8000 h.

**Output power:**
```text
Po = 25 hp × 0.746 kW/hp = 18.65 kW
```

**Input power at each efficiency (Pin = Po / η):**
```text
At 85 %: Pin1 = 18.65 / 0.85 = 21.941 kW
At 90 %: Pin2 = 18.65 / 0.90 = 20.722 kW
```

**Power saved:**
```text
ΔP = 21.941 − 20.722 = 1.219 kW
```
(Formula form: ΔP = hp × 0.746 × (1/η1 − 1/η2).)

**Annual energy saved:**
```text
ΔE = 1.219 kW × 8000 h = 9751.6 kWh/year
```

**Annual cost saving:**
```text
Saving = 9751.6 × 7.5 = Rs 73 137 per year
```

| Item | 85 % motor | 90 % motor |
|---|---|---|
| Output | 18.65 kW | 18.65 kW |
| Input | 21.94 kW | 20.72 kW |
| Energy/year | 175 529 kWh | 165 778 kWh |
| Cost/year | Rs 1 316 471 | Rs 1 243 333 |

If the high-efficiency motor costs, say, Rs 50 000 more, the simple payback is 50 000 / 73 137 ≈ 0.7 year, which shows why energy-efficient motors are recommended in energy audits.

**Answer: Annual saving ≈ 9752 kWh, i.e. about Rs 73 140 per year (at full load).**

## Check yourself

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

1. Why does the 90% efficient motor save energy compared with the 85% one for the same 25 hp output?
   A. It delivers more shaft output
   B. It draws less input power for the same output
   C. It runs fewer hours
   D. It has a lower energy tariff
   Answer: B. Output is the same; input = output / efficiency is lower for the more efficient motor, so less energy is drawn. (Key point: Source of saving)

2. Which expression gives the input power saving for the same output Po?
   A. Po x (eta2 - eta1)
   B. Po x (1/eta1 - 1/eta2)
   C. Po x (eta1 / eta2)
   D. Po / (eta1 + eta2)
   Answer: B. Pin = Po/eta, so the saving is Po/eta1 - Po/eta2 = Po(1/eta1 - 1/eta2). (Key point: Input power formula)

3. With Po = 18.65 kW, inputs of 21.94 kW and 20.72 kW, 8000 h and Rs 7.5/kWh, what is the annual saving?
   A. About Rs 73 140
   B. About Rs 9 750
   C. About Rs 1 316 000
   D. About Rs 7 314
   Answer: A. Saved power 1.219 kW x 8000 h = 9752 kWh, and 9752 x 7.5 is about Rs 73 140 per year. (Key point: Numerical result)

4. How is the 25 hp rating converted to output power in the solution?
   A. 25 x 0.746 = 18.65 kW
   B. 25 / 0.746 = 33.5 kW
   C. 25 x 0.9 = 22.5 kW
   D. 25 x 1.34 = 33.5 kW
   Answer: A. 1 hp = 0.746 kW, so 25 hp = 18.65 kW of shaft output. (Key point: Hp to kW conversion)

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