russ_watters said:
Also, no need to multiply and then divide by 12. Just use one hour. Anyway, not a huge difference, so...
That was because, at night it is usually dark for 12 hours, and an LED light needs about 10 watt.
The hill behind my house is 30 m. I need 300 watt for 12 hours = 43,200 sec, to run my base load through the night. I have solar to re-energise the system during the day.
How much mass do I need? 300 watt * 43200 sec = 12.96 MJ.
Mass = 12.96e6 / ( 9.8 * 30 ) = 44.1 tonne.
If I use a massive 50 tonne trolley on rails, the rope drum and gearbox will be inefficient, and it will be dangerous.
But it could be done with 45 m³ of water. I need at least another 5 m³ to cover the inefficiency. If I use water, then pumps and turbines have poor efficiency, but I can select the flow per revolution that drives the alternator.
Is it worth the effort? 300 W for 12 hours is 3.6 kW⋅hr. My buy price per night is about AU$1.20 or AU$440 PA. It would take 10 years to just pay off the tanks. The pumped storage system is probably "not worth the candle".
The real pity is that, over the last 40 years I have paid the State to build a hydro-electric system. I could store my solar energy during the day by delaying the fall of water in the State's system, but the sale price of my energy during the day is 25% of my buy price which makes that storage system less than 25% efficient. For each dollar I save, I must give them three.