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Drag (physics) derivations - Wikipedia, the free encyclopedia

Drag (physics) derivations

From Wikipedia, the free encyclopedia

(See Huntley 1967)The drag equation may be derived to within a multiplicative constant by the method of dimensional analysis. If a moving fluid meets an object, it exerts a force on the object, according to a complicated (and not completely understood) law. We might suppose that the variables involved under some conditions to be the speed, density and viscosity of the fluid, the size of the body (expressed in terms of its frontal area A), and the drag force. Using the algorithm of the π-theorem, one can reduce these five variables to two dimensionless parameters: the drag coefficient and the Reynolds number.

Alternatively, one can derive the dimensionless parameters via direct manipulation of the underlying differential equations.

That this is so becomes obvious when the drag force F is expressed as part of a function of the other variables in the problem:

f(F,u,A,\rho,\nu)=0. \!

This rather odd form of expression is used because it does not assume a one-to-one relationship. Here, f is some (as-yet-unknown) function that takes five arguments. We note that the right-hand side is zero in any system of units; so it should be possible to express the relationship described by f in terms of only dimensionless groups.

There are many ways of combining the five arguments of f to form dimensionless groups, but the Buckingham Pi theorem states that there will be two such groups. The most appropriate are the Reynolds number, given by

Re=\frac{u\sqrt{A}}{\nu}

and the drag coefficient, given by

C_D=\frac{F}{\rho Au^2}.

Thus the function of five variables may be replaced by another function of only two variables:

f\left(\frac{F}{\rho Au^2},\frac{u\sqrt{A}}{\nu}\right)=0.

where f is some function of two arguments. The original law is then reduced to a law involving only these two numbers.

Because the only unknown in the above equation is F, it is possible to express it as

\frac{F}{\rho Au^2}=f\left(\frac{u\sqrt{A}}{\nu}\right)

or

F=\rho Au^2f(Re). \!

Thus the force is simply ρAu2 times some (as-yet-unknown) function of the Reynolds number—a considerably simpler system than the original five-argument function given above.

Dimensional analysis thus makes a very complex problem (trying to determine the behavior of a function of five variables) a much simpler one: the determination of the drag as a function of only one variable, the Reynolds number.

The analysis also gives other information for free, so to speak. We know that, other things being equal, the drag force will be proportional to the density of the fluid. This kind of information often proves to be extremely valuable, especially in the early stages of a research project.

To empirically determine the Reynolds number dependence, instead of experimenting on huge bodies with fast-flowing fluids (such as real-size airplanes in wind-tunnels), one may just as well experiment on small models with slow-flowing, more viscous fluids, because these two systems are similar.

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