Question

The velocity ( ) of a stream –line flow of a liquid in a small uniform tube depends on

the radius, (r) of the tube, the density of the tube (ρ) and coefficient of viscosity (η) of the fluid. Use dimensional analysis to obtain an expression for the velocity (Vo) of the
stream-line and use your results to obtain a constant K if = 2ρ/4η.

Answer 1

To obtain the expression for the velocity of the stream-line flow, we can use dimensional analysis. Let's consider the following variables:

Velocity (v)

Radius (r)

Density (ρ)

Coefficient of viscosity (η)

The dimensions of these variables are:

[v] = L/T

[r] = L

[ρ] = M/L^3

[η] = M/LT

Using the Buckingham Pi theorem, we can form the following dimensionless groups:

π1 = v/(r^a ρ^b η^c)

where a, b, and c are unknown exponents that we need to determine. To do this, we can equate the dimensions of both sides of the equation:

L/T = L^1-a M^b L^-3b T^-c

Equating the dimensions of length, mass, and time, we get:

1 - a = 0 -> a = 1

b = 0

-c = 1 -> c = -1

Therefore, the dimensionless group becomes:

π1 = v/(r ρ η^-1)

We can rewrite this as:

v = K r ρ η^-1

where K is a constant that we need to determine. To find K, we can use the given equation:

K = 2ρ/4η = 1/2η

Substituting this value of K in the expression for v, we get:

v = (r/2) (η/ρ)

Therefore, the expression for the velocity of the stream-line flow is:

v = (r/2) (η/ρ)

where v is the velocity, r is the radius of the tube, η is the coefficient of viscosity of the fluid, and ρ is the density of the fluid.