What are the annual energy saving and payback for lighting?

7800 kWh, Rs 93,600, payback about 8.3 months

2600 kWh, Rs 31,200, payback 2 years

7800 kWh, Rs 65,000, payback 3 years

How is the motor saving calculated?

Same shaft output (15 x 0.80 = 12 kW) divided by each efficiency, difference times hours

Using 0.93/0.88 as ratio of cost

What is the combined payback?

Rs 225,000 / Rs 146,386, about 1.54 years

Practice · 8 marks

Energy saving calculation: lighting and motor

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.

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.

(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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Energy saving calculation: lighting and motor

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