4 ms·
Dig a 15 meter pit 5 meters across and suspend a 5 meter cube of ferro-concrete on a cable feeding onto a drum above it connected to a gearbox and motor/generat
by moocowduckquack 13y ago
Dig a 15 meter pit 5 meters across and suspend a 5 meter cube of ferro-concrete on a cable feeding onto a drum above it connected to a gearbox and motor/generator. Attach this to your PV and then relax.
- csense 13y agoGoogle says density of concrete is 2400 kg / m^3. Multiply that by 5x5x5 m^3 and you have 300,000 kg of concrete (300 metric tons). m = 300000 kg g = 9.8 m/s^2 h = 10 m The total amount of potential energy you can store is: m * g * h = 29400000 J = 29.4 MJ The more customary unit of energy in electric circles is the kW hr. Converting a kW hr to joules gives: 1 kW hr = 1 kJ hr / s = 1000 J hr / s * (60 s / min) * (60 min / hr) = 3600000 J = 3.6 MJ So the potential energy stored in the parent's system is: 29.4 MJ * (kW hr / 3.6 MJ) ~ 8.17 kW hr. Probably needs to be bigger to supply a decent sized household. Not to mention issues with inefficient conversion of potential energy to electricity, maintenance, and a backup plan for cloudy days.
- moocowduckquack 13y agoThanks for that, I thought it would give around a couple of days of backup, however I had completely fudged the numbers somewhere along the line.
- URSpider94 13y agoI don't have time to look for the cite now, but I've read a proposal to build electric traction railways up some steep slopes in SoCal, then load up cars of rocks, mine tailings, slag, etc. The trains would head up the mountain at night when electricity is in surplus, then back down in the daytime when demand is near peak. Seems outlandish, but the calculations predicted very high efficiency and comparatively low cost per MWh of storage.