3 ms·

Some maths. 1 squared meter surface times 10 km in depth yields ... 10,000 cubic m of mass to be vaporized and then "dispersed" into the atmosphere. At a dens

Some maths. 1 squared meter surface times 10 km in depth yields ... 10,000 cubic m of mass to be vaporized and then "dispersed" into the atmosphere. At a density of 5.5 kg/dm^3 it would weight about ... 55 thousands metric tonnes of rock. To vaporize 1 kg of liquid water it takes 2250 kcal of energy. If that mass was water it would require 1.23e11 kcal of energy. I suspect that both the amount of energy needed and the amount of pollution added would be unbearable. And this is for a hole that's "just" 10 km deep.

foobarian 2y ago
1 m^2 x 10,000 m = 10,000 m^3 10 km^3 = 10 x 1,000 x 1,000 x 1,000 m^3 != 10,000 m^3
- notorandit 2y ago
  Correct! I fixed my numbers too. Still a lot of stuff.
istjohn 2y ago
Your cross section area is off by two orders of magnitude. They currently are drilling a 0.008 m² hole at one meter per hour with 1 MW of power. They hope to scale that up to a 0.032 m² hole while maintaining the same rate of depth penetration and energy use per meter [0]. 0. https://m.youtube.com/watch?v=b_EoZzE7KJ0&t=500s https://m.youtube.com/watch?v=b_EoZzE7KJ0&t=500s
itishappy 2y ago
Think you're somewhat off. Quaise says they intend to allow their tech to be "plugged into existing oil rigs," and most boreholes are 5-20 in wide. 20 in down 7.6 mi is 0.2 m^2 down 12321 m. This gives 2500 m^3 or 2.5e-6 km^3 of rock. Google says "rocks are generally between 1600 kg/m^3 (sediments) and 3500 kg/m^3 (gabbro)" so going with 2500 km/m^3 that would be 6240000 kg, 6.24 Gg, or 6.24 thousand metric tons. Heat of vaporization of water is roughly 2250 kJ/kg, or 537 kcal/kg, so if the mass was water it would need 14e9 kJ, 14 TJ, or 3.35e9 kcal to vaporize. For a 5 in borehole divide all the above by 16. This ignores the heat required to actually change the temperature of the water, but I'm betting you can get a lot of that back via condensation, and for this reason I'd also assume that much of the rock will not end up in atmosphere. These assumptions also go entirely out the window of there's an inflow of material, but unfortunately I'm betting that's exactly what will happen. Edit: Fixed post after I confused radius and diameter. Someone should probably check my math as well.

foobarian 2y ago

1 m^2 x 10,000 m = 10,000 m^3 10 km^3 = 10 x 1,000 x 1,000 x 1,000 m^3 != 10,000 m^3

notorandit 2y ago
Correct! I fixed my numbers too. Still a lot of stuff.

istjohn 2y ago

Your cross section area is off by two orders of magnitude. They currently are drilling a 0.008 m² hole at one meter per hour with 1 MW of power. They hope to scale that up to a 0.032 m² hole while maintaining the same rate of depth penetration and energy use per meter [0]. 0. https://m.youtube.com/watch?v=b_EoZzE7KJ0&t=500s https://m.youtube.com/watch?v=b_EoZzE7KJ0&t=500s

itishappy 2y ago

Think you're somewhat off. Quaise says they intend to allow their tech to be "plugged into existing oil rigs," and most boreholes are 5-20 in wide. 20 in down 7.6 mi is 0.2 m^2 down 12321 m. This gives 2500 m^3 or 2.5e-6 km^3 of rock. Google says "rocks are generally between 1600 kg/m^3 (sediments) and 3500 kg/m^3 (gabbro)" so going with 2500 km/m^3 that would be 6240000 kg, 6.24 Gg, or 6.24 thousand metric tons. Heat of vaporization of water is roughly 2250 kJ/kg, or 537 kcal/kg, so if the mass was water it would need 14e9 kJ, 14 TJ, or 3.35e9 kcal to vaporize. For a 5 in borehole divide all the above by 16. This ignores the heat required to actually change the temperature of the water, but I'm betting you can get a lot of that back via condensation, and for this reason I'd also assume that much of the rock will not end up in atmosphere. These assumptions also go entirely out the window of there's an inflow of material, but unfortunately I'm betting that's exactly what will happen. Edit: Fixed post after I confused radius and diameter. Someone should probably check my math as well.