The Household Energy Audit and the Oversized Fuse
A household in Coimbatore has the following appliances on a 230 V supply: a 2 kW water heater used 45 minutes every day, a 75 W ceiling fan run for 8 hours, a 40 W fluorescent tube light run for 6 hours, a 1.5 kW air conditioner run for 4 hours, and a set-top box that draws 5 W continuously on standby. The tariff is ₹4.50 per unit plus a fixed charge of ₹60 per month; assume 30 days. The house has separate 5 A lighting and 15 A power circuits. The owner wants to cut the monthly bill by 20% and proposes only two changes: replace the 40 W tube with a 20 W LED and switch off the set-top box at the plug. He also asks whether he can run the 2 kW heater from the 5 A lighting socket by fitting a 15 A fuse.
1. Calculate the monthly energy consumption in kWh and the monthly bill. Then evaluate whether the proposed LED replacement and standby switch-off can achieve a 20% reduction.
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Convert each appliance to kWh per day, then multiply by 30:
- Water heater: 2 kW × 0.75 h = 1.5 kWh/day → 45 kWh/month.
- Fan: 0.075 kW × 8 h = 0.6 kWh/day → 18 kWh/month.
- Tube light: 0.04 kW × 6 h = 0.24 kWh/day → 7.2 kWh/month.
- Air conditioner: 1.5 kW × 4 h = 6 kWh/day → 180 kWh/month.
- Standby: 0.005 kW × 24 h = 0.12 kWh/day → 3.6 kWh/month. Total = 45 + 18 + 7.2 + 180 + 3.6 = 253.8 kWh/month. Energy charge = 253.8 × 4.50 = ₹1142.10; with the fixed charge the bill is ₹1202.10. A 20% cut means saving about 50.8 kWh, or about ₹228 in energy charge. The LED saves (40 − 20) W × 6 h × 30 = 3.6 kWh/month. Switching off the standby saves 5 W × 24 h × 30 = 3.6 kWh/month. Together they save only 7.2 kWh, about 2.8%, worth ₹32.40. The proposal is far short of the target. The air conditioner is the dominant load: reducing its daily use by 1 hour saves 1.5 × 30 = 45 kWh, which nearly meets the 20% target by itself.
2. The owner will not reduce air-conditioner use. Suggest a combination of realistic measures that would meet the 20% target, with calculations, and state any assumption you make.
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The target is about 50.8 kWh/month. The LED and standby changes give 7.2 kWh, leaving about 43.6 kWh to find. Practical options on this load pattern include:
- Reduce the water heater from 45 to 30 minutes per day: saving 0.5 kW × 0.75 h? Careful: 2 kW for 0.25 h less = 0.5 kWh/day → 15 kWh/month.
- Reduce the air conditioner by 30 minutes per day: 0.75 kWh/day → 22.5 kWh/month.
- Replace the 75 W fan with a 50 W efficient fan for 8 h: saving 0.025 × 8 × 30 = 6 kWh/month. These three together save 15 + 22.5 + 6 = 43.5 kWh, and with the 7.2 kWh from the original proposal the total is 50.7 kWh, essentially the target. This assumes the appliances are used for the stated times every day and that the replacements perform the same duty. A more robust approach is to measure actual units over a month and compare with the calculated figure, because thermostat duty cycles, voltage variation and seasonal use can change the air-conditioner consumption substantially.
3. The owner proposes to run the 2 kW heater from the 5 A lighting socket by replacing the 5 A fuse with a 15 A fuse. Analyse the currents and wiring implications, and explain why this is unsafe.
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The heater current at 230 V is I = P/V = 2000/230 = 8.70 A. That is already above the 5 A rating of the lighting circuit, so the heater does not belong on that socket even before considering starting or tolerance. Lighting circuits are wired with small conductors, commonly 1.0 or 1.5 mm², and their sockets, switches and ceiling roses are rated for 5 A. Replacing the 5 A fuse with a 15 A fuse does not upgrade the wiring; it removes the protection that limits current to the cable's safe value. If a fault or overload occurs, the 1.5 mm² cable can carry more than its rating, overheat, damage insulation and start a fire before the 15 A fuse operates. The correct action is to run the heater from the 15 A power circuit, which should have 2.5 mm² conductors and a proper 15 A or 16 A protective device, and to ensure the socket and plug are rated accordingly. A fuse must always be the weakest link, chosen to protect the cable, not merely to stop nuisance blowing.