I am a student revising for an exam and are stumped by a question:

Using dimensional analysis, show the the relationship between the frictional drag co-efficient, Cf, and the Reynolds’ Number Re = 2rVρ/ μ, Between a sphere or radius r, moving in a fluid dynamic viscosity μ and density ρ with relative velocity v between the fluid and the sphere is:

Cf = 24/Re

The Stokes’ constant of proportionality is 6π and Cf is the frictional force divided by face area and velocity pressure

I dont understadn even how to start it?? any ideas??

You presumably know the units of the drag coefficient, and the units of the Reynolds number is given. Compare the 2 and multiply by appropriate constants (in high energy physics we'd be talking about c and h bar, I don't know what you'd use for fluid mechanics) so they have the same units and then you can form an equation. Hopefully, you'll get the answer thats given.

In fact, having just read the question properly, you are given the dimensions of both the coefficient and reynolds number, so all you need to know is what constants to use. Do you know the dimensions of stokes' proportionality constant?

Maybe I'm vastly underestimating the problem, but both the Drag Coefficient and Reynold's Number are dimensionless (hence why they're so useful), as is the product 24/Re.

I'm not sure if that's the answer you were looking for, but dividing a constant by a dimensionless number is still a dimensionless number.