---
title: "Energy saving calculation: lighting and motor"
subject: "Energy Resources"
chapter: 9
key: "semester_6/energy_resources:lighting-motor-energy-saving-case"
source: https://www.bctengineer.com/q/semester_6/energy_resources/lighting-motor-energy-saving-case
site: BCT Engineering
---

# Energy saving calculation: lighting and motor

Energy Resources · Chapter 9: Energy audit

- Asked in: year not given, 8 marks
- Video: [2.10.3 Motor Efficiency Calculation | EX703(B) |](https://www.youtube.com/watch?v=ipwmLGBdqqk) — RGPV Official
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/lighting-motor-energy-saving-case
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-9

## Question

An audit of a small factory finds (i) 100 fluorescent tube fittings of 36 W tubes with 8 W ballast loss each, operating 10 h/day for 300 days a year; (ii) a 15 kW, 88 % efficient motor loaded at 80 % and running 6000 h/year. The tariff is Rs 12 per kWh. Replacing each fitting with an 18 W LED lamp (cost Rs 650 each, no ballast) and the motor with a 93 % efficient energy-efficient motor (cost Rs 160 000) are proposed. Calculate the annual energy and money savings and the simple payback period for each measure and the combined measure.

## Short answer

Lighting: replace 100 × (44 W – 18 W) × 3000 h = 7800 kWh/yr saved, Rs 93,600/yr, payback 0.69 years (8.3 months) for Rs 65,000 investment. Motor: old input 12 kW/0.88 =13.64 kW, new 12/0.93=12.90 kW, saving 0.733 kW × 6000 h = 4399 kWh/yr, Rs 52,786/yr, payback 3.03 years. Combined saving: 12,199 kWh/yr, Rs 146,386/yr, total investment Rs 225,000, payback 1.54 years.

## Answer

#### (i) Lighting retrofit
Existing load per fitting $= 36 + 8 = 44$ W; proposed $= 18$ W.

$$
\begin{aligned}
\Delta P &= 100 \times (44 - 18) = 2600\ \text{W} = 2.6\ \text{kW} \\
\text{Operating hours} &= 10 \times 300 = 3000\ \text{h/yr} \\
\Delta E &= 2.6 \times 3000 = 7800\ \text{kWh/yr} \\
\text{Saving} &= 7800 \times 12 = \text{Rs } 93\,600/\text{yr}
\end{aligned}
$$

Investment $= 100 \times 650 = \text{Rs } 65\,000$.

$$
\text{Payback} = \frac{65\,000}{93\,600} = 0.69\ \text{yr} = 8.3\ \text{months}
$$

#### (ii) Motor replacement
Shaft output $= 15 \times 0.80 = 12$ kW.

$$
\begin{aligned}
P_{in,old} &= \frac{12}{0.88} = 13.64\ \text{kW} \\
P_{in,new} &= \frac{12}{0.93} = 12.90\ \text{kW} \\
\Delta P &= 0.733\ \text{kW} \\
\Delta E &= 0.733 \times 6000 = 4399\ \text{kWh/yr} \\
\text{Saving} &= 4399 \times 12 = \text{Rs } 52\,786/\text{yr}
\end{aligned}
$$

$$
\text{Payback} = \frac{160\,000}{52\,786} = 3.03\ \text{yr} = 36.4\ \text{months}
$$

#### Combined
$$
\Delta E = 7800 + 4399 = 12\,199\ \text{kWh/yr}, \quad \text{Saving} = 93\,600 + 52\,786 = \text{Rs } 146\,386/\text{yr}
$$

$$
\text{Investment} = 65\,000 + 160\,000 = \text{Rs } 225\,000, \quad \text{Payback} = \frac{225\,000}{146\,386} = 1.54\ \text{yr}
$$

**Answer:** lighting saves 7800 kWh (Rs 93 600), payback 8.3 months; motor saves 4399 kWh (Rs 52 786), payback 3.0 years; combined 12 199 kWh (Rs 146 386) per year, payback 1.54 years (about 18 months). The lighting measure should be done first.

## Check yourself

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

1. What is the power saving per fitting in the lighting retrofit?
   A. 36 - 18 = 18 W
   B. (36 + 8) - 18 = 26 W
   C. 8 W
   D. 44 W
   Answer: B. The existing fitting draws tube plus ballast (44 W); the LED draws 18 W. (Key point: Lighting load difference)

2. What are the annual energy saving and payback for lighting?
   A. 7800 kWh, Rs 93,600, payback about 8.3 months
   B. 2600 kWh, Rs 31,200, payback 2 years
   C. 7800 kWh, Rs 65,000, payback 3 years
   D. 15,600 kWh, Rs 187,200
   Answer: A. 2.6 kW x 3000 h = 7800 kWh; at Rs 12 gives Rs 93,600; Rs 65,000 / 93,600 = 0.69 yr. (Key point: Lighting savings)

3. How is the motor saving calculated?
   A. Same shaft output (15 x 0.80 = 12 kW) divided by each efficiency, difference times hours
   B. 15 kW x (0.93 - 0.88)
   C. Rated power minus 12 kW
   D. Using 0.93/0.88 as ratio of cost
   Answer: A. Input power = output / efficiency: 13.64 kW vs 12.90 kW, saving 0.733 kW x 6000 h = 4399 kWh. (Key point: Motor saving method)

4. What is the combined payback?
   A. Rs 225,000 / Rs 146,386, about 1.54 years
   B. Average of 0.69 and 3.03 years
   C. Sum of 0.69 and 3.03 years
   D. Rs 160,000 / Rs 93,600
   Answer: A. Combine total investment and total saving, not the average of paybacks. (Key point: Combined payback)

## Related in Energy Resources

- [Energy audit types and audit procedure](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Aenergy-audit-types-procedure) — Energy Resources, chapter 9
- [Energy audit instruments and energy balance](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Aenergy-audit-instruments-measurements) — Energy Resources, chapter 9
- [Preparing and presenting an energy audit report](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Aenergy-audit-report-preparation-presentation) — Energy Resources, chapter 9
- [Wind turbine power, tip speed ratio and energy](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Awind-turbine-power-energy-numerical) — Energy Resources, chapter 4

## Related in other subjects

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- [Electronic ballast replacement savings with ToD tariff](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aballast-replacement-tariff) — Industrial Power Distribution and Illumination, chapter 10
- [EVA; 10 days, 100,000, day 6, spi and cpi](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeva-numerical-day6-spi-cpi-b80) — Project Engineering, chapter 4
- [Energy loss of 100 kVA transformer in third year](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atransformer-energy-loss-third-year) — Transmission and Distribution System Design, chapter 7