Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The Law:

  • The size of a solid material that can be carried by a (water) stream is proportional to the sixth power of its velocity.

I have just discovered that I read this law long time ago (it has a name starting with N or P or something), and I was required to construct a convincing model that naturally lead to the sixth power. Can anyone give me an interesting or simple model?

share|improve this question

1 Answer 1

up vote 14 down vote accepted

Suppose you have a spherical particle being pushed up a slope of angle $\theta$ by the current:

Sixth power law

Assume that the system is dominated by inertial forces not viscous forces, in which case the force on the particle is equal to the momentum change per second of the fluid striking it. If the flow velocity is $v$ then the amount of water hitting the particle per second is a cylinder of area $\pi r^2$ and length $v$, where $r$ is the particle radius. If the liquid density is $\rho_f$ the mass of water per second is $\rho_f\pi r^2 v$ and the momentum change is:

$$ mv = A\rho\pi r^2 v^2 $$

where $A$ is a fudge factor that gives the percentage of the liquid's momentum that gets transferred to the particle. So the force F pushing the particle up the slope is:

$$ F_{up} = A\rho_f\pi r^2 v^2 \cos\theta $$

The force down the slope is $mg\sin\theta$, and if $\rho_p$ is the particle density the force in terms of the radius is:

$$ F_{down} = \tfrac{4}{3}\pi r^3 (\rho_p - \rho_f) g \sin\theta $$

Remember to subtract off the liquid density to allow for the upthrust due to the fluid displaced. Now just equate the two forces to get the velocity at which the particle is in balance and we get:

$$ \tfrac{4}{3}\pi r^3 (\rho_p - \rho_f) g \sin\theta = A\rho_f\pi r^2 v^2 \cos\theta $$

or:

$$ r = \tfrac{3}{4}A\frac{\rho_f}{\rho_p - \rho_f} \frac{1}{g\tan\theta} v^2 $$

So for the particle that can just be moved by the flow we find:

$$ r \propto v^2 $$

and therefore:

$$ m \propto r^3 \propto v^6 $$

Your question is actually slightly wrong as the size is proportional to the velocity squared. It's the mass of the particle that is proportional to the sixth power of the velocity.

share|improve this answer
1  
Congratz for having reached the 100k reputation, very well deserved! Always look forward to your new posts. Cheers! –  Phonon Aug 26 at 19:31
    
Uhh noob here, what is the fish symbol between $m$ and $r^3$? –  corsiKa Aug 26 at 20:52
1  
@corsiKa It means "proportional to", its a way of supressing the irrelevant constants (here $A, \rho_f, \rho_p, \theta$), to allow the reader to focus on the main relationships between $m, r$ and $v$. –  Dave Aug 26 at 21:23
    
When I say size I mean volume :) I thought it would be related to vicosity. Great answer! –  Mr.T Aug 27 at 7:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.