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I keep wondering about the Solar Sinter project by Markus Kayser * What if instead of manufacturing artwork, it manufactured a mechanical energy storage, large

I keep wondering about the Solar Sinter project by Markus Kayser * What if instead of manufacturing artwork, it manufactured a mechanical energy storage, large enough to store either daily variations or even yearly insolation? Perhaps glass blocks (filled with sand) as weights for storing as potential energy (how deep would a shaft have to be to store daily or yearly insolation, assuming the weight is ideally half as tall as the shaft?) Perhaps storing compressed (even liquified?) air? How airtight is the resulting sintered glass? Perhaps making 1 upper reservoir and somehow digging a narrow shaft (making the walls glass while digging/blowing up the sand through the already vitrified upper part of the shaft? Then somehow building the lower underground reservoir from the inside out, and store potential energy in the closed system containing water (condensed from air?) Solar panels, wiring, cheap plastic Fresnel lenses, control electronics would need to keep getting supplied to attach to newly built energy storages. But as much as possible, especially things that wear out should be renewably built from glass onsite (for example scoops to move and redistribute sand) In example consider an NxN square of already completed storage, sites then there is ~ 4N adjacent sites under construction so the project either could speed up by actually building ~N/4 circumference layers at a time, or start delivering ~(N-4)N of renewable on-demand energy (since the storage site height was chosen to have enough capacity to store daily, weekly or yearly insolation) * https://kayserworks.com/ https://kayserworks.com/

DoctorOetker 8y ago
ok I calculated the height/depth for a shaft housing a glass weight half the size (which is the optimal height for the weight in any shaft of fixed height, I assume you understand this is optimum, if you want a derivation I can provide it upon request) For clear skies (typical of deserts), we can use the daily insolation at sea level: 21,6 MJ/m^2 (I will ignore inefficiency of solar panel, the less efficient, the less deep or tall we need to store the energy generated, and one time construction of the storage is preferable over reconstructing the weights and shafts once more efficient panels become available) So if we want to store it as potential energy of a glass weight under the solar panel that generated the energy we have for each square meter of panel: a glass weight with cross section one scquare meter, and height l = L/2 where L is the height of the shaft. glass weights 2500 kg/m^3, so we have mass of the weight: m = l * 1 m^2 * 2500 kg/m^3 = l * 2500 kg/m = L/2 * 2500 kg/m We want the weight to be at all times contained in the shaft so it can be closed and sand can not blow under the weight or into the mechanisms. the height difference for the glass weigh at the top -but still completely inside the shaft- to the bottom -but also still completely inside the shaft- is h = L - l/2 - l/2 = L -l = L - L/2 = L/2 The potential energy difference for a weight of mass m and vertical travel distance h is E = m * g * h Plugging everything together: E = ( L/2 * 2500 kg/m * 9.81 N/kg * L/2 ) Solving for L = 2 * sqrt ( E / (2500 * 9,81 N/m)) Since the assumed insolation is 21.6 MJ/m^2 then one square meter of (ideal futuristic) solar panel will simply generate 21.6 MJ = 21,6 MNm Plugging into the equation for L we get L = 59,35 m I seeriously underestimated potential energy I guess, I thought it would have been larger. If your panels are less efficient (say 10%) or you have a large unused area you can of course make the structure less tall/deep.

DoctorOetker 8y ago

ok I calculated the height/depth for a shaft housing a glass weight half the size (which is the optimal height for the weight in any shaft of fixed height, I assume you understand this is optimum, if you want a derivation I can provide it upon request) For clear skies (typical of deserts), we can use the daily insolation at sea level: 21,6 MJ/m^2 (I will ignore inefficiency of solar panel, the less efficient, the less deep or tall we need to store the energy generated, and one time construction of the storage is preferable over reconstructing the weights and shafts once more efficient panels become available) So if we want to store it as potential energy of a glass weight under the solar panel that generated the energy we have for each square meter of panel: a glass weight with cross section one scquare meter, and height l = L/2 where L is the height of the shaft. glass weights 2500 kg/m^3, so we have mass of the weight: m = l * 1 m^2 * 2500 kg/m^3 = l * 2500 kg/m = L/2 * 2500 kg/m We want the weight to be at all times contained in the shaft so it can be closed and sand can not blow under the weight or into the mechanisms. the height difference for the glass weigh at the top -but still completely inside the shaft- to the bottom -but also still completely inside the shaft- is h = L - l/2 - l/2 = L -l = L - L/2 = L/2 The potential energy difference for a weight of mass m and vertical travel distance h is E = m * g * h Plugging everything together: E = ( L/2 * 2500 kg/m * 9.81 N/kg * L/2 ) Solving for L = 2 * sqrt ( E / (2500 * 9,81 N/m)) Since the assumed insolation is 21.6 MJ/m^2 then one square meter of (ideal futuristic) solar panel will simply generate 21.6 MJ = 21,6 MNm Plugging into the equation for L we get L = 59,35 m I seeriously underestimated potential energy I guess, I thought it would have been larger. If your panels are less efficient (say 10%) or you have a large unused area you can of course make the structure less tall/deep.