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?
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1The pump is irrelevant. Just calculate the potential energy stored in that amount lifted that high (m * g * h) (I'm not doing that in an answer because the non-metric quantities are too complicated for me). Note that you can never fully reclaim that energy, because of conversion losses.– user2451Aug 30, 2017 at 7:59
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Thanks. I finally worked through it. The theoretical energy value (68.9 w) is quite a bit more than it takes to power the pump (24 w). That leaves 44.9 w– Anthony ScottiAug 31, 2017 at 17:15
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Anthony, I'm not sure where/how you got 68.9, but w (I assume W) is Watts and is a rate of energy transfer (Joules/second) not a measure of energy itself (which Joules is). As per LShaver's answer, the total stored energy is ~18.3kJ. A 1W pump would draw (1*60*60*24=) 86.4kJ over 24h. You have thus used 86.4kJ of electrical energy to store 18.3kJ of potential energy — making your pump (18.3/86.4=) 21% efficient. As Jan says, you can never, ever, get more energy out of a system than you put into a system due to losses and the Law of Conservation of Energy.– TimOct 11, 2017 at 10:06
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1 Answer
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
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54gal == 204.4L; 30' == 9.14m; Ep = 204.4*9.81*9.14 = 18,327J == 5.1Wh; 5.1/24 = 0.21W– TimOct 11, 2017 at 9:43