Buoyancy is an artifact of gravity that moves mass up away from the center of the earth in a fluid whenever the specific gravity of the object is less than the surrounding fluid. Buoyant Force is calculable. Gravitational Potential Energy is calculated by the formula: mgh Is Buoyant Potential Energy calculated mg'h? Here is how I have calculated it so far; mass = 100,000 kg height or depth = 100 meters g' = 1 - (1000 / 1028) = 1 - 0.97276265 = 0.02723735 (The density of the solid object is 1,000 kg/m. The density of the fluid is 1,028 kg/m) By this calculation the Buoyant Potential Energy would be 272,373.54 Joules Is this correct? Could it be improved on?
The units of density would be kg/m^3, not kg/m as you show. BTW, would there be more "bouyant energy" stored if you pushed a solid object down to a depth, or pushed an air-filled object down, where the object shrinks in size as it gets to greater pressures at lower depths?
You can define g' such that the potential energy is mg'h, but this would be an odd definition as the density of the object would contribute to g'. I would change m to m' which is the difference of the mass of the object and the corresponding mass of water. If you just know the density, but not the total mass or volume of the object, all you can calculate is an energy density: ##\frac{E}{V}=\rho'gh = (\rho_{object}-\rho_{water})gh## where ρ are density values.
I did forget to put cubic meters. I don't think that there would be more energy but there might be less area to cause viscous drag as the object accelerated upwards.
One thing I've observed (as a scuba diver) is that if you inflate a baloon at depth and release it, it accelerates very slowly at first, and as it grows in size it accelerates faster. You are right that there is more drag as it gets bigger (and goes faster), but the added bouyant lift from the larger displaced water volume seems to overpower that. Aside -- it is *very* disconcerting when you are in trouble and have to pull your Bouyancy Compensator (BC) vest's ripcord, to inflate it to take you to the surface -- if you are at any depth, you barely start to accelerate to the surface at all. As you rise, you go faster and faster, which is finally reassuring.
Here is another way that I tried where I copied the formula from the web; Net Gravitational Acceleration g' -0.315072129 m/s2 1-p'/p -0.032128514 m/s2 Object Density p 996 kg/m3 Fluid Density p' 1,028 kg/m3 Buoyant Force Fb 1,013,473 N The idea was that I could back in F=MA and multiply the Buoyant Force * g' * the mass and come up with the energy potential. Should I put in m' - the differential mass - instead of the mass?
This was the last attempt that I made before posting the thread. I think that I am applying what you said but I am not quite sure. Does this method accomplish the Energy Density method of calculating Buoyant energy? Buoyant Force Calculator Gravity ("g") Force 9.80662 m/s2 Mass 100,129 kg Force 981,925 N Density 996 kg/m3 Submerged Volume 101 m3 Volume 101 m3 Fluid (Seawater) Density 1,028 kg/m3 Buoyant Force 1,013,473 N Net Buoyant Force Upwards 31,548 N Buoyant Path Distance 244.8 m Buoyant Joules (N m) 7,722,903 J Buoyant Potential Energy 2.15 kWh Buoyant Power 198 kW
Looks right, but I did not check the numbers in a calculator. You need some time (for those 244.8m) or velocity to calculate a power.
Depends on the setup. And if you let that thing float upwards, it depends on the water resistance, which depends on the shape of the object.
I used the calculation off of this page at Hyper Physics; http://hyperphysics.phy-astr.gsu.edu/hbase/lindrg.html I plugged the variables into Excel and ran the formula. Now that it is in Excel, (theoretically if I did not make any mistakes copying) I can adjust the size of the radius and the density. Is this page a standard or are there different variations of how to estimate the drag on a sphere ascending by buoyancy? I have heard of Reynolds Number, Laminar Flow, Cavitation, and a lot more variables: very complex and over my head calculations. But, I am hoping that this gives me a good ballpark estimate.
I have been doing some other work on the more advanced equations. I was hoping to get a "Ballpark" estimate of the acceleration. Do you think that the Reynolds Number will be out of line or is there some other reason that I cannot use this as a First Estimate?
That is, if you use h=0 as datum for calculating potential energy. In general, for a sufficiently small mass (call it a point mass) immersed in a liquid, the energy density at the point is [itex](\rho_{object}h +\rho_{liquid}x)g [/itex], where x is the distance from the datum, -h in your case.
Stoke's law in water is useful for falling/rising objects in the mm-range. There are formulas for turbulent flow, with force proportional to the velocity squared. You can solve this for the equilibrium velocity.
Of course, it is even more disconcerting because you know the more agitated that you get the more oxygen you burn.