Kinematic Viscosity Formula

The viscosity of a liquid is defined as a frictional force in its flow. In other words, viscosity occurs when there is relative motion between layers of fluid. It can be interpreted as the measure of a fluid’s thickness or barrier to passing items through it. There are two types of viscosity: Absolute viscosity (or dynamic viscosity) and kinematic viscosity. The internal resistance to the flow of a fluid is measured by dynamic viscosity, and then it is used to calculate the kinematic viscosity. The latter is also known as momentum diffusivity and is calculated as a ratio.

Kinematic Viscosity Formula 

The formula for kinematic viscosity is given by the ratio of absolute viscosity to the density of the fluid. It is denoted by the symbol v. Its unit of measurement is newton seconds per meter square (Ns/m²), and the dimensional formula is given by [M¹L^-1T^-1]. It is directly proportional to the absolute viscosity but inversely proportional to the density of the fluid.

v = μ/ρ

Where,

• v is the kinematic viscosity,
• μ is the absolute viscosity,
• ρ is the density of the fluid.

Sample problems

Problem 1: Calculate the kinematic viscosity of a fluid with absolute viscosity as 1 Ns/m² and density as 2.5 kg/m³.

Solution:

We have,

μ = 1

ρ = 2.5

Using the formula we have,

v = μ/ρ

= 1/2.5

= 0.4 Ns/m²

Problem 2: Calculate the kinematic viscosity of a fluid with absolute viscosity as 0.3 Ns/m² and density as 2 kg/m³.

Solution:

We have,

μ = 0.3

ρ = 2

Using the formula we have,

v = μ/ρ

= 0.3/2

= 0.15 Ns/m²

Problem 3: Calculate the kinematic viscosity of a fluid with absolute viscosity as 0.78 Ns/m² and density as 4 kg/m³.

Solution:

We have,

μ = 0.78

ρ = 4

Using the formula we have,

v = μ/ρ

= 0.78/4

= 0.195 Ns/m²

Problem 4: Calculate the absolute viscosity of a fluid with kinematic viscosity as 0.67 Ns/m² and density as 1.5 kg/m³.

Solution:

We have,

v = 0.67

ρ = 1.5

Using the formula we have,

v = μ/ρ

=> μ = vρ

= 0.67 (1.5)

= 1.005 Ns/m²

Problem 5: Calculate the absolute viscosity of a fluid with kinematic viscosity as 0.48 Ns/m² and density as 1.3 kg/m³.

Solution:

We have,

v = 0.48

ρ = 1.3

Using the formula we have,

v = μ/ρ

=> μ = vρ

= 0.48 (1.3)

= 0.624 Ns/m²

Problem 6: Calculate the density of a fluid with kinematic viscosity as 0.129 Ns/m² and absolute viscosity as 0.520 Ns/m².

Solution:

We have,

v = 0.129

μ = 0.520

Using the formula we have,

v = μ/ρ

=> ρ = μ/v 

= 0.520/0.129

= 4.1 kg/m³

Problem 7: Calculate the density of a fluid with kinematic viscosity as 0.234 Ns/m² and absolute viscosity as 0.612 Ns/m².

Solution:

We have,

v = 0.234

μ = 0.612

Using the formula we have,

v = μ/ρ

=> ρ = μ/v

= 0.612/0.234

= 2.61 kg/m³