13.02.2013 Views

Mechanics of Fluids

Mechanics of Fluids

Mechanics of Fluids

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

this book, it is important to give attention to the fundamental assumptions<br />

on which the solution is based.<br />

First, it is assumed that the motion is such that the inertia forces on the<br />

particles <strong>of</strong> fluid (i.e. the forces required to accelerate or decelerate them)<br />

may be neglected in comparison with the forces due to viscosity. As we have<br />

seen, the ratio <strong>of</strong> inertia forces to viscous forces is represented by Reynolds<br />

number, and so this condition is met if the Reynolds number <strong>of</strong> the flow is<br />

very small.<br />

The measurement <strong>of</strong> viscosity 213<br />

Other assumptions are that no other boundary surface is sufficiently near Stokes’s law<br />

to affect the flow round the sphere, that the sphere is rigid, that the motion<br />

is steady and that there is no slip between the fluid and the sphere. Using<br />

these hypotheses, Stokes found that the drag force opposing the motion<br />

equals 3πµud, where µ represents the dynamic viscosity, u the velocity <strong>of</strong><br />

the sphere relative to the undisturbed fluid and d the diameter <strong>of</strong> the sphere.<br />

This result is now known as Stokes’s Law. It has been found that, to obtain<br />

good agreement with experimental results, the Reynolds number (expressed<br />

as ρud/µ) must be less than about 0.1. The result <strong>of</strong> this restriction is that, for<br />

ordinary fluids such as water or air, the sphere must be almost microscopic in<br />

size. If Stokes’s Law is to be valid for larger spheres, then either the viscosity<br />

must be very large or the velocity exceedingly small.<br />

If a small solid particle is falling through a fluid under its own weight, the<br />

particle accelerates until the net downward force on it is zero. No further<br />

acceleration is then possible and the particle is said to have reached its terminal<br />

velocity. This may be calculated for a small sphere on the assumption<br />

that the Reynolds number is small enough for Stokes’s Law to be valid. If the<br />

density <strong>of</strong> the fluid is ρ and the mean density <strong>of</strong> the sphere is ρs, then, when<br />

the terminal velocity has been reached, the force balance on the sphere may<br />

be stated as<br />

Downward force = Weight <strong>of</strong> sphere − Buoyancy − Drag force on sphere<br />

= π<br />

6 d3ρsg − π<br />

6 d3ρg − 3πµud = 0<br />

whence<br />

u = d2 (ρs − ρ)g<br />

18µ<br />

(6.28)<br />

For the determination <strong>of</strong> viscosity, a small solid sphere <strong>of</strong> known weight is<br />

allowed to fall vertically down the centre <strong>of</strong> a cylinder containing the liquid<br />

under test. The velocity with which the sphere falls is measured; it does not,<br />

however, quite coincide with the terminal velocity as given by eqn 6.28. In<br />

practice the liquid cannot be <strong>of</strong> infinite extent as assumed in the derivation<br />

<strong>of</strong> Stokes’s Law, so corrections are necessary to allow for the effect <strong>of</strong> the<br />

walls <strong>of</strong> the cylinder. (For this effect to be negligible, the diameter <strong>of</strong> the<br />

cylinder must be more than about 100 times the diameter <strong>of</strong> the sphere.)<br />

Moreover, the measurement <strong>of</strong> velocity must not be begun until the sphere<br />

has reached its terminal velocity, and should not be continued when the<br />

sphere nears the base <strong>of</strong> the cylinder since this influences the rate <strong>of</strong> fall <strong>of</strong><br />

objects in its vicinity.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!