What power is needed to get to mach 3 constant velocity for a supercativating submarine ? A Supercativating submarine could in theory achieve about 3600 miles per hour but powering the propulsion is a technical challenge.
Goatguy provides the energy for water displaced, times its density, times ½, times its outward radial velocity squared would be the amount of energy invested every second in the slipstream bubble's frontal profile. Maybe more, but this is kind of a minimum. The drag for supercativiation is 200,000 times less so the water displacement is an approximation.
The 1,000 meter per second bullet with a 20° cone uses 1,500,000 J/s (watts) of energy to perpetually keep its 1000 m/s velocity. That is due entirely to the displaced water, and its outward radial velocity dependent on the angle-of-attack of the frontal cone. There's no provision for invested-energy recycling (allowing the collapse of the bubble to propel the tail, to whatever degree water dynamics allows such action), but I'm betting the situation isn't much better at that end, either. Maybe what, get back 70% of the energy? That'd be nice. So, maybe 500,000 W to keep the bullet flying.
And that's a bullet with a cross section of π × 0.0045² m². Now wait a moment: here's the bad news … at the cross-section of a useful human scale sub (5 meters across, which is pretty cramped, considering all that nuclear reactor equipment and such needs also to be on board), the ratio is (D/d)² or (5 ÷ 0.0045)² = 300,000× larger. 300,000 × 500,000 W = 150 billion watts. Assuming that you get back 70% of the invested displacement energy.
German Type VII U-boat submarines were 4.7 meter across at the beam for their pressurized hull
The longest submarine was the USS Triton which were 136 meters long.
A MUCH pointier cone (4°, it only requires 4 gigawatts of motive energy. Well, at 1°, where the ship is 100× longer than it is wide (500 meters long, for a 5 meter wide ship.), you're still looking at 500 megawatts of energy to keep the thing chugging along at Mach3.
I do not think the 1° submarine is unreasonable.
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Goatguy provides the energy for water displaced, times its density, times ½, times its outward radial velocity squared would be the amount of energy invested every second in the slipstream bubble's frontal profile. Maybe more, but this is kind of a minimum. The drag for supercativiation is 200,000 times less so the water displacement is an approximation.
The 1,000 meter per second bullet with a 20° cone uses 1,500,000 J/s (watts) of energy to perpetually keep its 1000 m/s velocity. That is due entirely to the displaced water, and its outward radial velocity dependent on the angle-of-attack of the frontal cone. There's no provision for invested-energy recycling (allowing the collapse of the bubble to propel the tail, to whatever degree water dynamics allows such action), but I'm betting the situation isn't much better at that end, either. Maybe what, get back 70% of the energy? That'd be nice. So, maybe 500,000 W to keep the bullet flying.
And that's a bullet with a cross section of π × 0.0045² m². Now wait a moment: here's the bad news … at the cross-section of a useful human scale sub (5 meters across, which is pretty cramped, considering all that nuclear reactor equipment and such needs also to be on board), the ratio is (D/d)² or (5 ÷ 0.0045)² = 300,000× larger. 300,000 × 500,000 W = 150 billion watts. Assuming that you get back 70% of the invested displacement energy.
German Type VII U-boat submarines were 4.7 meter across at the beam for their pressurized hull
The longest submarine was the USS Triton which were 136 meters long.
A MUCH pointier cone (4°, it only requires 4 gigawatts of motive energy. Well, at 1°, where the ship is 100× longer than it is wide (500 meters long, for a 5 meter wide ship.), you're still looking at 500 megawatts of energy to keep the thing chugging along at Mach3.
I do not think the 1° submarine is unreasonable.
Read more »
