# About how fast can a small fish swim before experiencing turbulent flow around its body? [closed]

About how fast can a small fish swim before experiencing turbulent flow around its body? An Engineering Problem! Please go through this question step by step. :D

• First tell us what have you tried!(you should do your homework first.) – Mhmd Jan 14 '14 at 19:27
• I've got no idea to be honest! This came up in an Engineering interview. I don't even know where to start! – Vaishnavi Jan 14 '14 at 19:37
• Think of it the other way: how fast can a flow be on a rigid surface before the flow is turbulent? – Kyle Kanos Jan 14 '14 at 19:38
• Doesn't this problem say anything about the diameter of the pipe the fish is going through? – Mhmd Jan 14 '14 at 19:41
• @user689 Why must the fish be in a pipe for turbulence to occur? – tpg2114 Jan 14 '14 at 20:19

Let's make some assumptions. First, assume the fish is rigid. Second, let's assume he's not flapping. Third, I guess let's assume it's a male fish since I said "he."

We'll also assume this is 2D because we're looking for an approximation. I would approximate the fish as an airfoil. NACA airfoils are a pretty good choice because they are analytically defined and very well studied. We'll take the NACA 0015 airfoil shown below (from Wikipedia, and only because this was the airfoil they had a picture of... really any NACA 00XX airfoil is fine so long as XX isn't too big).

The Reynolds number is what we're after here. Transition to turbulence over an airfoil occurs at roughly $R \approx 3 \times 10^6$ (any intro to viscous flow book will show you this) and the Reynolds number is defined as:

$$R = \frac{v L}{\nu}$$

where $\rho$ is the fluid density, $v$ is the object velocity, $L$ is the length scale describing the body and $\nu$ is the kinematic viscosity of the fluid. So you take $R$ to be $3 \times 10^6$ and you need to know the density and kinematic viscosity of water. Then you need to estimate the length of the fish, or you can report your answer in "velocity-fish length" units, i.e. $vL$. It's really just a simple manipulation to compute that.

• I sense gender bias here ! :P atleast leave poor fishes alone, maybe call it, it ! Btw, very well formed answer and more accurate too ! – Rijul Gupta Jan 14 '14 at 20:40

Maybe this can work,

Reynold's number is defined for a pipe as ${(\rho v d)}/{\eta}$, so lets assume your fish is spherical with a circular front of diameter $d$ ( ugly "ugly" fish).
Now instead of seeing a fish flowing in a straight line with velocity $v$ lets see from the fish's perspective, it seems that water is flowing towards us with velocity $v$. This seems to be like a normal flow of water through a pipe of cross sectional diameter $d$, Assuming that the fish is not disturbing the water on its side but only that on its front (too idealistic !)

Then reynolds no can be calculate as

$R_e = {(\rho v d)}/{\eta}$
For turbulence, $R_e > 4000$; Calculate for v.
$\rho$ is density of liquid, $\eta$ is viscosity of liquid

• No for turbulence flow Re > 4000. – Mhmd Jan 14 '14 at 20:08
• Yes, it is thanks for pointing out, I wrote the transitional one from memory which also happens to be 2300 :P – Rijul Gupta Jan 14 '14 at 20:11
• A fish is probably more like a NACA airfoil, let's just say NACA 0012 for the sake of argument, in which case the turbulent transition is really $R_e \approx 3e6$. None of your analysis is wrong of course, but I would choose a much better shape than a spherical fish. – tpg2114 Jan 14 '14 at 20:21
• @tpg2114 what shape would you suggest? – Mhmd Jan 14 '14 at 20:24
• @rijulgupta Sure, but why make a bad simplification when an equally simple one is a better approximation and introduces zero additional complexity? – tpg2114 Jan 14 '14 at 20:39