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 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.
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 = A₂v₂.
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
No comments:
Post a Comment