I've developed a water pump that is powered by a 1 watt solar air pump. In 24 hours it pumps 54 gallons to a height of 30 feet. How much energy have I harvested?
Energy in the pumped water
- 54 gallons is about 200 L, equivalent to 200 kg
- 30 feet is about 9 m
- Using the formula @JanDoggen mentioned:
m ⨉ g ⨉ h = (200kg)(9.81m/s²)(9m) ≅ 18,000 kg*m²/s²
- This is the same as 18,000 joules (J) or 18 kJ
- We can convert this to watt-hours (Wh) easily:
18,000 J / (60 s ⨉ 60 m) = 5 Wh
Energy produced by the pump
Assuming a rough efficiency of 80% for the pump, this means the solar panel would need to produce about 6 Wh of energy.
Net energy output
Energy harvested: 6 Wh Energy lost: 1 Wh Useful stored energy: 5 Wh