# Equation of Continuity

In order to describe the motion of a fluid, in principle one might apply Newton’s laws to a particle (a small volume element of fluid) and follow its progress in time. This is difficult approach. Instead, we consider the properties of the fluid, such as velocity pressure, at fixed points in space.

In order to simplify the discussion we make several assumptions:

(i) The fluid is non-viscous. There is no dissipation of energy due to internal friction between adjacent layers in the fluid.

(iii) The flow is irrotational.

A tiny paddle wheel placed in the liquid will not rotate. In rotational flow, for example, in eddies, the fluid has angular momentum about a given point. In general the velocity of a particle will not be constant along a streamline. The density and the cross-sectional area of a tube of flow will also change. Consider two sections of a tube of flow, as shown in the figure.

The mass of fluid contained in a small cylinder of length Δl1 and area A1 is Δm1 = ρ1A1 Δl1 Since fluid does not leave the tube of flow, this mass will later pass through a cylinder of length Δl2 and A2.

The mass in this cylinder is Δm2 = ρ2A2 Δl2.

The lengths Δl1 and Δl2 are related to the speeds at the respective locations Δl1 = v1 Δt and Δl2 = v2 Δt.

Since no mass is lost or gained.

Δm1 = Δm2 and ρ1A1v1 = ρ2A2v2

This is called the equation of continuity. It is statement of the conservation of mass.  If the fluid is incompressible, its density remains unchanged.  If ρ1 = ρ2 then A1v1 = A1v2.

The product Av is the volume rate of flow (m³/s) and is also called velocity flux. Figure shows a pipe whose cross section narrows. From equation of continuity we conclude that the speed of a fluid is greatest where the cross-sectional area is the least. Notice that the streamlines are close together where the speed is higher. Example 1: Figure shows a liquid being pushed out of a tube by pressing a piston. The area of cross-section of the piston is 1.0 cm² and that of the tube at the outlet is 20 mm2. If the piston is pushed at a speed of 2 cm/s, what is the speed of the outgoing liquid? Solution: From the equation of continuity

A1n1 = A2n2 or, (1.0cm2) (2cm/s) = (20mm2) v2

v2 = (1.0cm2) (2cm/s)/ (20mm2)

= (100 mm2) (2cm/s)/ (20mm2) = 10 cm/s