# How Do You Calculate Buoyant Energy?

1. Dec 27, 2012

### genergy

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?

2. Dec 27, 2012

### Staff: Mentor

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?

3. Dec 27, 2012

### Staff: Mentor

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.

4. Dec 27, 2012

### genergy

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.

5. Dec 27, 2012

### Staff: Mentor

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.

6. Dec 27, 2012

### genergy

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?

7. Dec 27, 2012

### genergy

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

8. Dec 28, 2012

### Staff: Mentor

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.

9. Dec 28, 2012

### genergy

I used 39 seconds.

Does that seem good?

10. Dec 28, 2012

### Staff: Mentor

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.

11. Dec 28, 2012

### genergy

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.

12. Dec 28, 2012

### Staff: Mentor

I am quite sure that an object with that size and velocity in water will not follow Stoke's law.

13. Dec 28, 2012

### genergy

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?

14. Dec 29, 2012

### vin300

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 $(\rho_{object}h +\rho_{liquid}x)g$, where x is the distance from the datum, -h in your case.

Last edited: Dec 29, 2012
15. Dec 29, 2012

### Staff: Mentor

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.

16. Dec 29, 2012

### genergy

Ouch! I had no idea of Stoke's Law limitation! Can you please direct me to some specific formulas?

17. Dec 31, 2012

### genergy

Of course, it is even more disconcerting because you know the more agitated that you get the more oxygen you burn.

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