Dimensional Analysis

Question

Consider Stoke´s problem: a small sphere falling under gravity in a viscous fluid. The sphere´s velocity v is assumed to depend on the sphere's radius r, the gravity acceleration g, the dynamic viscosity  of the fliud, and the sphere's density .

a. Use dimensional analysis to identify two nondimensional numbers: the Reynolds number Re that is proportional to , and the Froude number Fr that is proportional to .

b. Stoke's analysis showed that this problem can be described by only one nondimensional number: the ratio Fr / Re. Find the relation for v that is implied by this conclusion.

Answer 1

Dimensional analysis is a very powerful tool. Remember to use when you can.

They ask to find a dimensionless quantity that involves the viscosity µ

So the first question you should have is what are the dimensions of µ

[µ] = [mass]/( [length] [time])

So we need to find a combination of variables {g, ρ, r, v} that cancel all this units. So first note.

[ρ] = [mass]/[length]³

[g] = [mass] [length]/[time]²

[r] = [length]

[v] = [length]/[time]

I would first start with the "untangled" units. Meaning is you had a variable that no other variable has. In this case we don't have that scenario. Length is everywhere, so I would start with time or mass. Let's cancel the mass

(1) [µ]/[g] = [time]/[length]² or (2) [µ]/[ρ] = [length]²/[time]

Let's continue with (1), the only other source of time is [v] so we are going to use it

( [µ] [v] /[g] ) = 1/[length], and finally to get rid of the length

( [µ] [v] [r] ) / [g]

Now for (2), we have two sources of time, but on this parameter one introduces [mass] again so I am not going to use it

[µ]/( [ρ] [v] ) = [length], and finally to get rid of the length

[µ]/( [ρ] [v] [r] ) = [length]

From the what the exercise told us, we might identify (1) as Froude number, while (2) is the inverse of Reynolds. (2) = Reynolds, but the first is not true sadly. If we use [ρ] on (1) to eliminate the mass dependence, we do get Froude number

[ρ]/[g] = [time]²/[length]⁴

[ρ] [v]²/[g] = 1/[length]²

(3) [ρ] [v]²/ ([g] [r]²)

(3)/(2)

Fr/Re = { [ρ] [v]²/ ([g] [r]²) } / { ( [ρ] [v] [r] )/[µ]} = µ v /(g r³ ρ)