The Lodge's 'Hot Water by Three O'Clock' Claim
A lodge outside Maun advertises 'hot water from the sun by three o'clock'. The system is a flat, unglazed black 200 L tank mounted on the roof, uninsulated, feeding one tap. On a clear day the water is at 18 degrees Celsius at 08:00, and the owner claims it reaches 55 degrees Celsius by 15:00. The maintenance worker proposes fitting a 3 kW electric element in the same tank and running it for three hours as a backup, plugged into the lodge's only 240 V, 15 A circuit, which already supplies the kitchen fridge and the lights. Local average solar intensity is about 800 W/m2, and the tank's exposed collecting area is about 2.0 m2. Specific heat capacity of water = 4200 J/(kg degrees Celsius); density of water 1.0 kg/L. The owner wants to know, with reasons, whether the solar claim is believable and whether the electric backup is a sensible plan.
1. Calculate the energy needed to raise 200 L of water from 18 to 55 degrees Celsius, and the time a 3 kW element would take.
Read the worked answer
Mass of water: 200 L x 1.0 kg/L = 200 kg.
Temperature rise: 55 - 18 = 37 degrees Celsius.
Energy: Q = mc(delta T) = 200 x 4200 x 37
- 200 x 4200 = 840 000 J per degree
- 840 000 x 37 = 31 080 000 J, about 31 MJ
Time for a 3 kW element: P = 3000 J/s, so t = Q/P = 31 080 000/3000 = 10 360 s, about 2.9 hours.
Judgement: three hours is only just enough, and that assumes no losses at all - no heat escaping through the tank walls, no water drawn off, no cold water entering. A real uninsulated tank would lose heat continuously, so a 3 kW element in three hours would fall short of 55 degrees Celsius. The element is not a comfortable backup; it is borderline even on paper.
Cost note: 3 kW for 3 hours is 9 kWh of electricity - a significant daily cost for a lodge, and it would have to run every day.
2. Test the solar claim by comparing the energy needed with the energy arriving, and state whether 55 degrees Celsius is believable.
Read the worked answer
Energy arriving: power = intensity x area = 800 x 2.0 = 1600 W.
Over 08:00 to 15:00 is 7 hours = 25 200 s, so energy = 1600 x 25 200 = about 40 MJ.
Required efficiency: 31/40 = about 77%.
Is 77% believable? No. A flat black tank is a poor collector:
- The black surface absorbs well, but an unglazed tank loses heavily by convection to the air and by radiation, especially once it is hot.
- There is no insulation underneath or around the sides, so heat leaks into the roof and the night air.
- Water evaporating from an open or vented tank carries away latent heat.
- Not all 7 hours delivers 800 W/m2; early morning and late afternoon are much weaker, and the sun is not perpendicular to the surface.
A realistic yield for such a tank is perhaps 40-50%, i.e. 16-20 MJ, which raises the water by 20 x 10^6/(200 x 4200) = about 19-24 degrees Celsius, giving roughly 37-42 degrees Celsius - warm, but not 55.
Judgement: the claim is not credible with this equipment. A glazed, insulated collector of larger area could reach 55 degrees Celsius easily; the fault is the tank, not the sunshine. The lodge should either insulate and glaze, or advertise honestly.
3. Assess the proposed electric backup against the lodge's electrical supply, and recommend what to do instead.
Read the worked answer
Current drawn by the element: I = P/V = 3000/240 = 12.5 A.
The circuit: a 15 A circuit already supplying a fridge (about 0.6 A) and lights (perhaps 1-2 A). Adding 12.5 A gives roughly 14-15 A, right at the breaker's rating.
Why this fails:
- A 15 A breaker trips somewhere near its rating, and running continuously at the limit makes it trip on a warm day. The kitchen would lose power along with the water heating.
- A 12.5 A continuous load should not be placed on a circuit already in use; fridges also draw a surge when the compressor starts, which can push the total over 15 A.
- The element is a fixed, high-power appliance and should be on its own dedicated circuit with correctly sized cable, an isolator switch and an earth connection - not plugged into an existing general circuit.
Recommendation:
- Keep the solar tank but insulate it and add glazing, and if possible increase the collector area - this is far cheaper than 9 kWh a day of electricity.
- If an electric backup is essential, have a dedicated 15/16 A circuit installed by a qualified electrician for the element only, or fit a smaller element (e.g. 1.5 kW) that can run overnight on off-peak power.
Judgement: the solar claim is optimistic and the electric plan is electrically unsafe as described. The cheapest genuine improvement is insulation, not a bigger heater.