Power of viscous friction on a falling sphere

Question

I have derived a simple model of a rotameter using an homogeneous solid ball in a rigid cone where a fluid flows. I consider 4 forces: Weight, Buyancy, Viscous Friction and Drag. I have written my forces balance, and I made no assumption about momentum balance:

$\sum\limits_i\vec{F}_i = \vec{0},\quad \sum\limits_i\vec{M}_i\neq\vec{0}$

I also found useful dimensionless relations in Perry's Chemical Engineering Handbook. So I am able to estimate terminal velocity of the sphere in a fluid for a given flow domain (Stokes, Allen, Newton).

My questions are the following:

1) I would like to estimate the power required to maintain the plunger at its position. Is it right to estimate it that way?

$P_\mathrm{trans}=\vec{F}_W \bullet \vec{v}_s$

Where $\vec{F}_W$ is the only driving force (Weight or may I withdraw the Buyancy) and $\vec{v}_s$ the terminal velocity of the sphere.

2) I would like to estimate the power absorbed by the plunger rotation. Can I do it the same way?

$P_\mathrm{rot} = \vec{\tau}\bullet\vec{\omega}$

If so, how can I estimate $\vec{\tau}$ if I have a good estimation of $\omega$ but I have no idea how $\vec{\omega}$ vary? It should help to underline if rotation is negligible in front of suspension.

3) Knowing those two powers, can I estimate the pressure drop $\Delta \xi$ for a given volumetric flow $\dot{V}$ with the following formula?

$P_\mathrm{tot} = \dot{V}\cdot \Delta \xi$

Answer 1

Thinking deeper and better (properly write down your question is half of the problem solved), I found solutions to almost all my questions.

1) Because Weight ($\vec{F}_\mathrm{W}$) and Buyancy ($\vec{F}_\mathrm{B}$) are central forces, they are conservative. Because Friction ($\vec{F}_\mathrm{F}$) and Drag ($\vec{F}_\mathrm{D}$) forces depend on the velocity, they are dissipative forces. Therefore to estimate power needed to sustain and rotate the sphere in the fluid, you need to evaluate the power of dissipation. Recalling the force balance, it comes:

$P_\mathrm{dis} = (\vec{F}_\mathrm{F}+\vec{F}_\mathrm{D})\bullet\vec{v}_s = -(\vec{F}_\mathrm{W}+\vec{F}_\mathrm{B})\bullet\vec{v}_s = - g V_s (\rho_s - \rho_f)v_s$

This is a convenient way to assess dissipation without bothering with dissipative terms. That is, when equilibrium is established power of conservative force exactly balances dissipations. Minus signs comes from the inner product and does mean (assuming $\rho_s > \rho_f$, we are using a classical rotameter) that power is withdrawn from system by dissipations.

2) I haven't found yet an easy way to assess the rotation only. It probably requires a more descriptive model. Question remains open.

3) That power relation stands for isochoric process (and arises when you are dealing with pump). Thus it works well for incompressible fluids. But if you are going to use gas, then it is a little bit tricky. Reminding rotameter is a measuring device, it has to withdraw a very-very-little amount of energy from the system it is measuring. So, transformation should be close to no transformation at all and can be seen as isochoric process with few error (or adiabatic, or isothermal, or -why not- isobaric). My opinion: it is a blind-way to estimate the pressure drop and result might not be accurate or even consistent.

I am open to any constructive feedback to this answer.