Let me calculate: Australia consumes 5882 PJ of electricity per year [1] = 1.63 PWh/y = 4.46 TWh/day. A Nickel-Iron battery stores 30 Wh/L [2]. They would nee

Let me calculate: Australia consumes 5882 PJ of electricity per year [1] = 1.63 PWh/y = 4.46 TWh/day. A Nickel-Iron battery stores 30 Wh/L [2]. They would need ~150 GL (giga liters) of NiFe batteries to store all the power consumed in a day (they would need much less just for the night) = 150 million m3 (cubic meters). A stack of cells 3m high (+ frames) would occupy 50 million m2 (sq. meters) or a square with ~7 km long sides (without service roads), so let's say battery storage industrial zone 10 km by 20 km in total, split around major cities. Cost: ~ 1T$ ( tera-dollars :) without the land and the roof ) I hope I didn't miss any 0s. [1] https://www.energy.gov.au/energy-data/australian-energy-statistics/energy-consumption https://www.energy.gov.au/energy-data/australian-energy-stat... [2] https://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery https://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery