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I am trying to figure out how to calculate the mechanical energy or power required to move a plate in water, vertically, at a certain speed and on a certain distance.

What if I want to rotate a disc in water? I am interested in the energy consumed, as a function of speed and torque (and perhaps there are other dependencies as well).

Any ideas where I could go and get read about this?

Thanks!

Pompilia

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  • $\begingroup$ What part of the calculation do you understand? What similar calculations can you do? $\endgroup$ Nov 17 '16 at 17:01
  • $\begingroup$ Well, I still have a vague idea about potential energy, kinetic energy, friction, and so on. :D $\endgroup$
    – Pompilia
    Nov 20 '16 at 5:08
  • $\begingroup$ In case I push and pull a plate in water... are there any other forces acting on the plate apart from gravitational, Archimede's, the drag force and the force I am applying to push or pull the plate? $\endgroup$
    – Pompilia
    Nov 21 '16 at 15:37
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The forces you mention in your comment are the only forces you need to consider, but calculating the 'drag force' is difficult because it has 2 causes.

Turbulent drag (aka hydrodynamic or pressure drag) dominates at relatively high speed, and depends on the cross-sectional area, which in this case is small. It is caused by having to push fluid out of the way. Viscous drag (aka skin friction) dominates at relatively low speed, and depends on the area parallel to the motion. It is caused by having to drag fluid along with the object.

In the case of a thin plate the cross-sectional area is small. So viscous drag will dominate in your case. If your plate is being dragged between or close to container walls, the viscous drag force can be calculated from the relative speed and distance and viscosity using Newton's formula. However, when the plate is far from any walls, what formula do you use? In the case of a sphere the viscous drag can be calculated from Stokes' Law. I don't know of any equivalent for a flat plate.

In Engineering the total drag from both sources is modelled by a single equation :
$$F=\frac12C_d \rho A v^2$$
where $\rho$ is fluid density, $v$ is speed through the fluid, $A$ is an appropriate area and $C_d$ is the relevant drag coefficient, which depends on the shape of the object relative to the direction of flow. As you can see, viscosity does not enter this calculation. The only difficulty is deciding what values to use for $A$ and $C_d$. (This formula is not entirely satisfactory from a physics point of view, because $C_d$ varies with Reynolds' Number (Re) which itself depends on speed $v$.)

The RoyMech website states :

... For thin flat plates and similar shaped items subject to primarily skin-friction, area A = the total surface area swept by the fluid flow (both sides).

Engineering Toolbox tabulates values for $C_d$. For a long plate it suggests $0.001$ for laminar flow and $0.005$ for turbulent flow. You should calculate Reynolds' Number ($Re$) to decide whether the flow is laminar or turbulent. (This is where viscosity comes into the calculation.) Then you can use the appropriate graph on the RoyMech page (which I think is "Drag for long plate parallel to flow and streamlined strut") to read off the value of $C_d$ for the calculated value of $Re$.

This Fluid Mechanics Tutorial contains a worked example #1 on page 3.


This is perhaps more of an Engineering question than Physics. You might get a more helpful answer on Engineering SE.

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    $\begingroup$ In your equation for the drag force it should be velocity squared. $\endgroup$
    – nrabbit
    Nov 21 '16 at 19:59
  • $\begingroup$ Hi, @sammy gerbil, That is how I had calculated the drag force, with the exception that the drag coefficient was equal to 1.46/sqrt(Re). Somehow the work/power required to pull the plate upwards resulted negative because of a high Archimedes' force. I guess that for the speed I had proposed and with such a high Archimedes' force, my motor would need to slow down the plate in order to reach the desired speed. So it is actually braking, instead of pulling - does that make any sense? For a higher speed both energies (pushing and pulling) are positive. Cheers, Pompilia $\endgroup$
    – Pompilia
    Nov 21 '16 at 21:38

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