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Suppose we have a long cylinder with diameter $D$, then we move it parallel to its axis with a velocity $V$ in a medium such that the Reynolds number of the flow is in the range $250<R_e<2\times10^5$. As a result, vortex shedding will takes place with a frequency

$f=\frac{S_tV}{D} $


Where $S_t$ is the Strouhal number

This formula looks so simple, except for the Strouhal number. So I think maybe there are some simple explanations for it.

Can anybody provide a simple model to explain at least the proportionality of the vortex shedding frequency with $V/D$? (Without using dimensional analysis of course)

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2 Answers 2

up vote 4 down vote accepted

I think @Killercam is right, I'll try to explain the same thing a little more elaborately.

Firstly. in the case considered, since the fluid and the cylinder is chosen, increase in velocity directly translates to increase in the Reynolds number as $R_e = \frac{\rho V D}{\mu}$.

Before considering flow in the range $250 < R_e < 2\times 10^5$ , lets first observe what happens in the region where the viscous force dominates over inertial forces i.e $R_e <<1$. The fluid slowly "crawls" over the surface of the cylinder. There are 2 stagnation points on the leftmost and rightmost parts of the cylinder. enter image description here

From the solution for inviscid flow over cylinder (superposition) we can note that tangential velocity is maximum at mid-section and decreases as and decreases as we proceed "downhill"

This can be extended to viscous flow and two important things are to be noted here:

  1. The shear stress is maximum at the mid-section, which is implicit as a higher velocity gradient is created because of the larger value of tangential velocity.
  2. The static pressure starts to increase after the mid-section. i.e the pressure is increasing in the direction of flow $\frac{\partial P}{\partial x}>0$ which is called as an adverse pressure gradient

As $R_e$ is increased (i.e velocity is increased), the inertial forces start to dominate over viscous forces.The flow velocity is zero at the surface and the particles very close to the boundary have a very low momentum since they experience very strong viscous forces. On the right part of the cylinder, the fluid particles close to the cylinder not only experience strong viscous force, but also adverse pressure gradient which eventually forces the fluid particles to stop/reversed, causing the neighboring particles o move away from the surface. This is called as flow separation. It results in the creation of a free shear layer which ultimately rolls up to form a vortex.

@Killercam said:

The velocity of the flow divided by the diameter of the cylinder is the typical crossing time of the fluid, hence is directly related to the frequency of the observed oscillations for a specific Reynolds number.

After a vortex is shed, the fluid particles behind have to undergo the same process i.e it takes the same distance to come to rest and then causing the neighboring particles to separate. Since this distance is a small part of the cylinder, $dist \propto D$ and hence the time interval between 2 vortices shed from the same side(top/bottom) of the cylinder $time \propto \frac{dist}{V}$ i.e $time \propto \frac{D}{V}$.

The time interval is exactly the time period of vortex shedding and hence the frequency of shedding is $f=\frac{1}{T}\propto \frac{V}{D}$

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Thank you, things are clearer to me now. But I need to think about it for a while.. –  Emitabsorb Jul 15 '12 at 6:38
@Emitabsorb, glad to help :) I'm yet to figure out the relation between $S_t$ and the shedding frequency $f$. Do post it here in case you figure it out. –  u6844 Jul 16 '12 at 9:56

The velocity of the flow divided by the diameter of the cylinder is the typical crossing time of the fluid, hence is directly related to the frequency of the observed oscillations for a specific Reynolds number. It is as simple as that. Clearly, this time scale is then correlated with observation to provide the Strouhal number for this particular phenomenon.

I hope this helps.

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Thanks for answering, but I don't think it's that simple. The frequency in the formula corresponds to how fast the wake "moves" back and forth. It doesn't directly related to how fast the fluid passes through the cylinder right? It's much more complicated. Take a look at this –  Emitabsorb Jul 13 '12 at 17:38
Of course the phenominon is more complex. You are asking how you can think about the $V/D$ componant of this simple formula - neglecting the Strouhal number. Well neglecting the Strouhal number you are also neglecting the details, as it is this dimensionless number that classifies the oscillating flow mechanism. $V/D$ here provides the flow crossing time which does have a direct bearing on the shedding frequency... I hope this helps. –  Killercam Jul 13 '12 at 18:03
I don't say that I would neglect the Strouhal number. The fact that Strouhal number is almost constant in this region is part of the question, it is one of the reason why the formula is so simple. Wait a second, what do you mean be the flow crossing time? Can you explain more clearly how does it related to how fast the wake switch from one side to the other? –  Emitabsorb Jul 13 '12 at 18:26
The flow crossing time is the time taken for fluid to cross the dyameter of the cylinder... The faster the flow the faster the higher the frequencies which the eddies are being shed (for a constant Reynolds number, hence constant $S_{t}$ number). –  Killercam Jul 13 '12 at 18:50
There is a leap in logic here "The faster the flow the faster the higher the frequencies which the eddies are being shed". Suppose at a moment, the flow is building a vortex in one direction and then after a while it starts creating another vortex in the opposite direction. You are just talking about how fast one vortex is getting bigger. What matters here is how long does it takes to change the direction of the wake, am I right? –  Emitabsorb Jul 13 '12 at 19:02

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