# How big would a gravity storage need to be for a city’s electric needs?

An alien planet with gravitational acceleration of 8.9m/s$$^2$$ backs up its wind power with gravity storage on a hillside. The hill grade is a steep 65% and the hill is 450m at the peak.

# Consumption

The city consumes wind energy at 25MW, they want to prepare for up to two days of low wind. (count earth days in this case, the time frame is the equivalent, or 24 hours = 1.2GWh energy)

# Storage

Granite: They have access to coarse rock granite and can forge buckets of nickel-iron to hold the loads.

Lead: They use molten lead to carry wind energy, which could be used with the difficulty of building an elevated reservoir.

How tall and wide would this device be if they wanted it to be as narrow as possible?

• Sorry to pick it to pieces, but do you mean 25MW mean consumption over 24 hours? Are you set on a cone-shape... as a narrow pipe with a cylindrical bowl on top would maintain regulation of pressure more easily? Mar 11, 2022 at 2:48
• Yes mean consumption from wind is 25MW daily, the shape is not limiting Mar 11, 2022 at 2:54
• Several glaring problems: units MW/day is J/s^2. So you want a solution that is increasing storage per day? A value of Joules or MWh would be more typical.. m/s is speed gravity is an acceleration or m/s^2. This looks like a physics question solve for mas of gravitational potential energy Energy = (mass)*(acceleration)*(height) then determine reservoir shape based upon incomplete information. Mar 11, 2022 at 19:28
• @GaultDrakkor that is the amount of their daily consumption, sorry. It is not increasing. Mar 11, 2022 at 20:48
• @VogonPoet How else can J/s^2 be interpreted? That is what MW/day is (neglecting some decimal shifting) I would have expected something like daily consumption is 25MWh. 1Joule = 1Ws or 1J/s = 1W. Watts are to liters per second as watt seconds are to liters. Mar 11, 2022 at 21:10

To store 1.2GWh (or 4.32TJ) of energy in a gravity battery that had a vertical drop of 450m in a gravitational field providing an acceleration of 8.9m/s2 (or so I assume... you've missed the power-of-two bit out of your question) you'd need a total weight of about 1.8 million tonnes.

If that mass were granite (density ~2.7g/cm3), it would have a volume of about 400000 cubic metres. If that were a cube, it would be about 73.7 metres to a side (~17-18 storeys tall). If you made the mass out of iron (density ~7.874g/cm3), it would have a volume of ~137000m3 (or a cube ~51.5m to a side).

Of course, that assumes you have the ability to perfectly convert the descent of the block to electrical power, and in practise that's going to be challenging and I'm assuming your peeps don't have ready access to room-temperature superconductors and the like.

Real-world gravity batteries have efficiencies of around 80-90%, but notably they use pumped liquids or vertical drops and so would not have the problem of friction losses over several hundred metres of track. I suspect the best thing to do would be to cut a shaft down from the top of the mountain, and the excavated material could be used as the working load of the battery.

You probably wouldn't want to drop the entire load down at once, but one (or a few) large lumps at a time to minimise the necessary strength of the infrastructure, or the destructive effects of an accident. With a descent of 1m/s, a load of about 3000 tonnes would provide you with the required 25MW power output. You'd need 600 of these blocks to provide power for the full two days. Down a 65% slope (~33 degrees) it'd be moving at about 1.8m/s to get the requisite downward velocity.

How tall and wide would this device be if they wanted it to be as narrow as possible?

Your question is under-constrained, because you haven't said what your peep's engineering capabilities are. Clearly, when dealing with thousands of tonnes of rock moving at a time, you can't really make things that small. A cube of granite that size is about 10 metres to a side. You'd want at least two of these, so as one brakes its load to a halt at the bottom the other can start going. Two 10-metres slides (or shafts) are probably wanted, with enough space at the top and bottom to move the loads out of the way ASAP. The fewer slides you have, and the narrower your system, the harder it will be to get the weights into place at the top and out of the way at the bottom, and that means more losses and more inefficiency. TANSTAAFL, etc.

• +1 for use of TANSTAAFL. Mar 21, 2022 at 21:17

Since the construction of the storage is under specified I assume pumped hydro.

Given

• Average rate of energy consumption of $$25$$MW.
• Storage of two days full load. $$E_p=25 \text{MW}*48\text{h} = 1200$$ MWh
• Upper reservoir height of $$h=450$$ m
• Local gravity $$g=8.9 \text{ms}^{-2}$$
• Water is $$\frac{1 m^3}{1 \text{Mg}}$$

Calculate the mass of reservoirs which can then be converted to volume. $$\text{E}_p= \text{m} \times \text{g} \times \text{h}$$ solve for mass $$\text{m}= \frac{\text{E}_p}{\text{g} \times \text{h}}$$ Since working mass is water the mass in tonnes is equivalent to cubic meters. A wide and narrow reservoir with no excess capacity would be something like 800m by 60m by 23m.

This is of course not considering losses. Nor considering future growth. Nor the fact that the wind doesn't blow all the time. So being able to handle when the wind is blowing strong, probably three 30MW pump/generators are needed to keep up with charge demand if not more. Since the city would likely need in excess of 100MW of nameplate wind generation. Assumption each 3MW nameplate of wind turbines will produce 1MW on average.

Nameplate:= the amount that can be produced in best conditions.

• (1) Two days is 48 hours. (2) Units of measurement are to be set upright and separated from the numbers by a space. It is 25 MW, or, if you feel a need to use LaTeX, $25\,\text{MW}$, but never $25MW$. Mar 22, 2022 at 20:27