Concept of Continuum deals with the continuous and homogenous fluid medium that expresses the fluid properties as continuous function. Whether it is density, viscosity, temperature or thermal conductivity, all its properties follow significant path having strong cohesive forces to behave as continuous substance.

In other words, the elements of the matter should be continuously distributed to fill the entire region it occupies. It is of great importance in finding different relationships of fluids under different conditions.

**Example 1.1:**

Liquid of one litre weighs 8 N. Determine its density, specific weight and specific gravity.

**Solution:**

**Given Data:**

V = 1 litre = 10-3 m3

W = 8N

** Density:**

We know that,

ρ = dm/ dV

or,

ρ = W/g . V (W = m . g)

ρ = 8/ 9.81 x 10-3

ρ= 8.155 x 102 kg/ m3

** Specific Weight**

We know that,

ω = ρ. g

Or,

ω = W/V

= 8/ 10-3

= 8 x 103N/m3

** Specific Gravity:**

We know that,

ωliquid = S x 9.81 x 103

Or,

S = ρ/ 100 (ρ = ω/ g)

= 8.155 x 102/ 103

= 0.8155

**Example 1.2:**

Two plates AB and CD are placed horizontally on a particular surface. When a force of 4 N/m^{2} is applied on the plates, CD moves with the velocity of 0.6 m/s. If AB is fixed and distance between these two plates is measured to be 0.03 mm. Determine the dynamic viscosity of fluid between these two plates.

**Solution:**

**Given Data:**

Since initial velocity is 0 and final velocity is 0.6 m/s, change in velocity can be given by,

du = 0.6 – 0

= 0.6 m/s

dy be the distance between two plates = 0.03 mm = 0.03 x 10^{-3} m

Shear stress = 4 N/m^{2}

We know that, shear stress is given by-

τ = μ du/dy

F/A = μ du/dy

By substituting all values from the given data, we get-

4 = μ 0.6/(0.03 x 10-3)

Hence,

**μ = 0.2 x 10-3 Ns/m2**

**Example 1.3:**

Two plates are separated from each other with a viscos fluid. The upper plate has an area of 2 m^{2} can move with a relative speed of 0.65 m/s as compare to lower plate which is kept at a distance of 0.2 mm. Calculate the force required to maintain this speed and the shear resistance.(Take dynamic viscosity of fluid as 2 poise)

**Solution:**

**Given Data:**

Area, A = 2 m^{2}

Relative speed, du = 0.65 m/s

Distance between the plates, dy = 0.2 mm = 0.2 x 10^{-3} m

Co-efficient of viscosity, = 2 poise = 0.2 Ns/m^{2}

**Shear Stress:**

We know that,

τ = μ du/dy

Substituting all values from the given data, we get-

τ = 0.2 x 0.65/ 0.2 x 10-3

= 0.62 x 103N/ m2

** Force:**

We know that,

τ = F/A

Or,

F = τ x A

Substituting the above values, we get-

F = 0.62 x 103 x 2

= 1.3 x 103N

**Links of Previous Main Topic:-**

- Ideal gas or perfect gas
- Introduction about air standard cycles
- Properties of pure substances introduction
- Vapour compression refrigeration cycle introduction
- Basic fluid mechanics and properties of fluids introduction
- Definition of fluid
- Properties of fluids
- Classification of fluids
- Kinematic viscosity
- Variation of dynamic viscosity with temperature
- Compressible fluids

**Links of Next Mechanical Engineering Topics:-**