After watching https://www.youtube.com/watch?v=mmrwdTGZxGk I wondered what kind of device I would have to build in order to store enough energy to run a house for one day.

Let's consider a 100% efficiency of the energy storage system for simplicity.

A standard 40 feet container:

  • Has a mass of 3700 kilograms
  • Measures 12.035 x 2.35 x 2.385 meters (exterior)
  • Has an internal volume of 67.5 m³

Let's say we fill this container with steel (volumetric mass density of 7800 kg / m³).

The mass of steel is: $$m = \rho * V = 7800 * 67.5 = 526500 \textrm{ kg} $$ The total mass of the filled container is 526500 + 3800 = 530200 kg (530 tons)

I need to store 10 kWh of energy using gravity. I have

$$E = m *g * h $$

  • E being the energy (in Joules)
  • m being the mass (in kilograms)
  • g being the gravitational acceleration
  • h being the height of the object in meters (or the difference of height)

10 kWh corresponds to 36 000 000 Joules.

$$h = \frac{E}{m * g} = \frac{36000000}{526500 * 9.81} = 6.92 \textrm{ m}$$

This results in a height of 6.92 meters. Is my computation correct?


closed as off-topic by Aaron Stevens, alephzero, rob Feb 7 at 15:24

This question appears to be off-topic. The users who voted to close gave these specific reasons:

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  • $\begingroup$ With the same formulas and a container filled with water: 51.7 meters, filled with wet sand: 29.3 meters, granite: 20.5 meters. I can't find a volumetric mass / price graph so it's hard to tell what's the best material for a mass / price compromise. $\endgroup$ – Victor Lamoine Feb 7 at 14:33
  • 1
    $\begingroup$ In its present form (v2), this is a check-my-work question. $\endgroup$ – rob Feb 7 at 15:25
  • $\begingroup$ Agreed, I will post again into a more appropriate place. $\endgroup$ – Victor Lamoine Feb 7 at 15:29