Question

Problem 1

Turbulent kinetic energy per kg of fluid has unit m²/s²

If we define turbulent length scale L, time scale T, and velocity scale U, based on the dimensional argument derive a relationship between turbulent scales and :

1. production of turbulent kinetic energy Pk

2. dissipation of turbulent kinetic energy

3. turbulent kinematic viscosity

Note that turbulent energy is U2, production is the amount of energy that is created every second, dissipation is the amount of energy that is dissipated into heat every second.

Answer 1

Solution 1:

Given:

Unit of Turbulent kinetic energy / mass = m2 / s2

Or, Unit of Turbulent Kinetic Energy = m2 / s2 * m = m3 / s2

Also,

Unit of Turbulent length, L = m [As unit of length is m]

Unit of turbulent time, T = s

Unit of Turbulent velocity, U = m/s

1. Now, Production of Turbulent KE (Pk) = Energy created per second (Given)

Thus, unit of Production of Turbulent KE (Pk) =   Unit of Turbulent Kinetic Energy / second

or, unit of Production of Turbulent KE = m3 / s2 * 1/s = m3 / s3

We need to find the expression of production of turbulent KE. On the basis of dimensional analysis, both the sides of the expression should have same units.

Thus, Pk = L3 / T3 = (L/T)3 = U3

2. Now, dissipation of turbulent KE (Dk) = Energy dissipated per second

Thus, the unit of dissipation of turbulent KE = m3 / s3

Thus, the expression of dissipation of turbulent KE should have the same units but with a negative sign showing the dissipation of energy. Thus, the required expression shall be

Dk = - (L/T)3 = - U3

3. We know that the unit of turbulent kinematic viscosity (Vk) is m2 / s

Thus the expression of Vk should have the same units on both the sides

Thus, Vk = L2 / T = L/T * L = U * L