Work that can be done with a buoy, not the work it does the one I can do with it

[rnegrete]

Hi.

I need to find out how much work I can do with a buoyant object. By intuition, if I give it the largest possible load, it will not move, and if it dont put any load on it, it will move but it will move nothing, so no work is performed in either case. I want to understand the loads, the movement and the inertial forces involved. My first approach was to simulate a wave, then go to cfd-post and get the wave height on a transient simulation at the point where the buoy would stand. I then calculated the mass proportional to the buoyancy force generated if the buoy sank half of the traveled distance of the wave surface during each timestep. My expectation was that it would sink half the distance and travel half the distance in every timestep plus/minus the inertial forces but results didnt make much sense as the inertial forces needed would be excessive to get those results. My next approach is to make a UDF so that it helps me put the load necessary on the buoy so that it travels half the distance change from the wave surface. My intuition is that this will give me the work I can perform, regardless if it comes from buoyancy force or inertia. I´m I on the right path? is there something I´m overlooking? Is there a simpler way to do this? I have no experience with UDFs, can I make a function to accomplish this? what would it look like?

Many thanks! 

[Mark O]

The maximum work you can do with it is the net work associated with the motion. The power is, by definition, the vector dot product of the net force and the buoy velocity. On the up cycle buoyancy dominates and on the down cycle gravity dominates, but the power calculation is the same.

 

For a free floating buoy the fluid force and gravity are in balance and so there is no power. You have to restrict the motion in some way to obtain some power and ideally you should restrict it in the way the real device operates. You restrict the motion by applying an additional load to the rigid body. You can do this with DEFINE_SDOF_PROPERTIES.  The net force is then the buoyancy force plus gravity plus the load plus the fluid drag force on the walls. As a crude example, for a simple mechanical linkage imagine the bouy has a saw tooth extension linked to a cog which is linked to a dynamo. The dynamo generates a resistance torque on the cog with results in a resistance force on the buoy. The resistance force always acts oppositely to the bouy's motion. On the up cycle it is pulling down. On the down cycle it is pushing up. Though it is also possible but less efficient, depending on the design, for the device to only extract energy on half the cycle with no load on the other half.

 

The calculated power will be the maximum. Some will be lost due to frictional resistance forces from any kind of mechanical linkage which will be lost as heat.

 

For a crude hand calc, you could imagine the bouy is moving freely on the up cycle with no load, then locked for 1/4 cycle for the wave to pass below, then released for 1/2 cycle. That way you can ignore buoyancy and the fluid drag forces by assuming it is fully above the water assuming a large wave height. Then you can calculate what load must be applied so that it falls by some distance (s) still above the wave in 1/2 cycle, with F=ma and s=1/2*a*t^2. The load would have to evaluate to opposing the motion or else the whole approach would not be practical.

 

Regards

 

Mark

 

[Mark O]

PS. Actually, I think the maximum exctractable power is the power associated with the load. The load is the connection to whatever divice is being used to generate electricity.