# Concept of Continuum

In fact all fluids consist of discrete molecules but we consider a continuous and homogeneous fluid medium where the distribution of mass within the matter or system without empty space, is continuous. A continuum is a continuous and homogeneous medium where fluid properties such as density, viscosity, thermal conductivity, temperature, etc., can be expressed as continuous function of space and time. molecular size, mean free path are generally insignificant in order to analyse the fluid motion because the strong intermolecular cohesive forces make the entire liquid mass to behave as a continuous mass of substance. So in the study of liquid motion the concept of continuum is of great importance.

SOLVED EXAMPLES

Example 1.1. One litre of a liquid weighs 8 N. Calculate its (i) specific weight (ii) density and (iii) specific gravity.

(ii) p = w / g = 8 x 103 / 9.81 = 8.155 x 107 kg/m3

(iii) s = p / 100 = 8.155 x 102 / 103 = 0.8155

Example 1.2. AB and CD are horizontal plates. AB is fixed CD moves with a velocity of 0.6 m/s and requires a force of 4 N per square meter. The distance between the plates is 0.03 mm.

Find out the dynamic viscosity of the liquid between the plates.

Solution.                d y = 0.03 mm = 0.03 x 10-3m

du = 0.6 m/s- 0 m/s = 0.6 ml s.

t = F /A = 4N / 1 m2 = 4 N/m2

t = µ.d u / d y , or 4 = µ x 0.6 / 0.03 x 10-3

µ = 4 x 0.03 x 10 -3 / 0.6 = 0.2 x 10 -3 Ns /m2

Example 1.3. The upper plate is placed 0.2 mm apart from the lower plate. The space between

them is filled with a viscous fluid of dynamic viscosity 2 poise. The upper plate of area 2 m2 is

required to move with a speed of 0.65 m/s relative to the lower plate. Determine the necessary (i)

shear resistance or force and (ii) power required to maintain this speed.

Solution.                dy = 0.2 mm

= 0.2x 10-3m

A = 2m2

µ = 2 poise= 0.2 Nslm2

du = u2 – u1 = 0.65 m/s

t = µ.d u / d y = 0.2 x 0.65 / 0.2 x 10 -3

(i) Shear resistance or force, F = t A = 0.2 x 0.65 x 2 / 0.2 x 10 -3

= 1.3 x 103 N