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While studying terminal velocity, I came across these formulas:

$$v = \sqrt{\frac{2mg}{\rho A c_d}}$$, and

$$\frac{2 r^2 g(\rho- \sigma)}{9 \eta}$$

These two formulas seem to show a different relationship with the radius. In the first, terminal velocity is proportional to $\frac{1}{A}$ (because $A$ will lead to $r^2$). In the second, the terminal velocity is proportional to $r^2$. This seems like a contradiction.

So, what is the difference in these formulas and what exactly is the relationship between terminal velocity and radius (or area in this case)?

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  • $\begingroup$ What is $c_d$ in the first expression? $\endgroup$ Feb 5 '20 at 5:27
  • $\begingroup$ Drag co-efficient $\endgroup$ Feb 5 '20 at 5:30
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The two formulas you mentioned are for different flow regimes where the ratios of inertial and viscous damping forces characterised by the Reynolds number are different. Both are "correct" if applied in the appropriate frame: The first one is a crude approximation valid for any object with a projected area $A_{proj}$ that is moving at high Reynolds numbers through a medium with significantly lower density (no buoyancy) while the second is derived rigorously for spherical particles for very low Reynolds numbers also considering buoyancy and as a consequence is much more limited but exact in the limit $Re \to 0$ where the viscosity dominates.


Flow regimes and Reynolds number

A moving fluid is like a collection of interacting particles that have a certain inertia, that tries to keep the fluid moving at its characteristic speed $U$, and a counter-acting mechanism of internal friction, characterised by the viscosity $\nu$, that tries to slow down the fluid. Depending on which mechanism dominates, the behaviour of the fluid is very different. Just think of stirring water compared to stirring a glass of oil or honey: While for the former is easily stirred, you have to put significant more effort to stir the glass of honey - the fluid sticks together: A layer of fluid that is moved has a larger tendency to drag neighbouring layers along, it has a higher viscosity, a higher degree of internal friction.

Every wall will introduce friction (dissipation) but as pointed out in this post the dimensionless Reynolds number

$$ Re := \frac{ U L }{ \nu } \propto \frac{\text{inertial forces}}{\text{viscous damping forces}}$$

can be used to characterise this behaviour in the bulk. A large Reynolds number is characterised by large velocities and/or low viscosities (e.g. stirring water) while small Reynolds numbers belong to small velocities and/or large viscosities (e.g. oil or honey).


Terminal speed

You are certainly familiar with Newton's laws. Forces that are not compensated by another counter-acting force lead to an acceleration of the object into the direction of the corresponding force.

$$\vec F_{net} = m \, \vec a = \sum_i \vec F_i$$

For most systems we can assume that these forces will balance after some time and the object won't accelerate further. The final velocity it will reach in such an equilibrium is called terminal velocity $\vec v_{term}$.

$$\vec F_{net}( \vec v_{term}) = \vec 0$$

If you consider an object falling inside a fluid you have two main forces: The acceleration due to gravity and a certain slow-down due to the fluid that depends on several mechanisms. The fluid will flow around the object and depending on the precise flow field (boundary layer, separation, vortices, compressibility, ...) will lead to a pressure around the object thus to a force. If you wanted to obtain a precise idea you would have to conduct experiments or at least numerical simulations but lumping the precise flow field into certain forces might help us estimate the terminal speed. We will use a driving gravitational force $F_g$ and a corresponding drag force $F_d$ caused by the surrounding fluid:

$$\vec F_{net} = \vec F_g + \vec F_d$$


Terminal speed for low Reynolds-number (Stokes') flow and spherical particles

For very small Reynolds numbers $Re \lesssim 1$ (creeping or Stokes' flow) the friction characterised by the viscosity dominates. Only for the simple case of a spherical particle we can derive the drag analytically by summing up the shear stresses along the surface (Stokes' law)

$$F_d = 6 \pi \mu R U$$

where $\mu = \nu \rho$ is the dynamic viscosity. For the buoyancy and weight we can find

$$F_g = \Delta m g = \Delta \rho V g = (\rho_s - \rho_f) \frac{4}{3} \pi R^3 g$$

and as a result the terminal velocity of a sphere with density $\rho_s$ and radius $R$ in another fluid with density $\rho_f$ can be found by solving

$$ 0 = F_g + F_d = (\rho_s - \rho_f) \frac{4}{3} \pi R^3 g - 6 \pi \mu R U_{term}$$

to obtain

$$U_{term} = (\rho_s - \rho_f) \frac{4}{18 \mu} R^2 g$$


Terminal speed for turbulent flow with quadratic air resistance

For turbulent flows (higher Reynolds number $Re \gtrsim 1000$) the resistance in a moving fluid is generally assumed to scale quadratically with speed, the ratio of drag force and its corresponding dynamic pressure times the projected area is assumed to be approximately constant:

$$c_d = \frac{2 \, F_d}{\rho U^2 A_{proj}}$$

This so called drag coefficient $c_d$ might still change in magnitude depending on the precise Reynolds numbers as the flow conditions might change and thus separation points might move. Additionally it changes often drastically with flow direction where the projected area and flow conditions vary greatly.

As a consequence the terminal velocity can be calculated as

$$0 = F_g + F_d = m g - \frac{c_d \rho U_{term}^2 A_{proj}}{2}$$

which results in the terminal velocity

$$U_{term} = \sqrt{\frac{2 m g}{c_d \rho A_{proj}}}.$$

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  • $\begingroup$ Thank you so much for your time and effort :) $\endgroup$ Feb 5 '20 at 13:49
  • $\begingroup$ @Curiouscase You are welcome! Stay curious! :) $\endgroup$
    – 2b-t
    Feb 5 '20 at 13:49
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The first formula applies when the fluid in which the object is falling (e.g. air) is not too viscous, and no account is taken of buoyancy.

The second formula applies when the fluid is more viscous, and buoyancy is taken into account.

The Wikipedia article on terminal velocity and this blog post on terminal velocity provide more information. I found the Wikipedia article a bit confusing because of ordering, but ultimately what they seem to do is: (1) derive your first equation, assuming no buoyancy; (2) derive an equation including buoyancy and then substitute a value for $C_d$ that reflects high viscosity ("laminar flow"), giving your second equation.

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  • $\begingroup$ Ah ok, thank you! $\endgroup$ Feb 5 '20 at 13:49

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