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.

(ii) The flow is steady.

(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 Δl_{1} and area A_{1} is Δm_{1} = ρ_{1}A_{1 }Δl_{1}Since fluid does not leave the tube of flow, this mass will later pass through a cylinder of length Δl_{2} and A_{2}.

The mass in this cylinder is Δm_{2} = ρ_{2}A_{2 }Δl_{2.}

The lengths Δl_{1} and Δl_{2} are related to the speeds at the respective locations Δl_{1} = v_{1 }Δt and Δl_{2} = v_{2 }Δt.

Since no mass is lost or gained.

Δm_{1 }= Δm_{2 }and ρ_{1}A_{1}v_{1} = ρ_{2}A_{2}v_{2}

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 ρ_{1} = ρ_{2} then A_{1}v_{1} = A_{1}v_{2}.

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

A_{1}n_{1} = A_{2}n_{2} or, (1.0cm^{2}) (2cm/s) = (20mm^{2}) v_{2}

v_{2} = (1.0cm^{2}) (2cm/s)/ (20mm^{2})

= (100 mm^{2}) (2cm/s)/ (20mm^{2}) = 10 cm/s