The background to Bernoulli's Theorem.

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Overview

Fig 1. Venturi Tube - Figure courtesy Wikipedia \url [http://en.wikipedia.org/]
Fig 1. Venturi Tube - Figure courtesy Wikipedia http://en.wikipedia.org/

Bernoulli's principle states that in a steady, streamlined, incompressible flow of a fluid, the pressure of the fluid is inversely proportional to its velocity. If a fluid is flowing through a horizontal pipe of varying cross-sectional area, as in Figure 1 for example, the fluid speeds up in constricted areas so that the pressure the fluid exerts is least where the cross section is smallest. This phenomenon is also called the Venturi effect. The Venturi effect has several applications including that associated with generating lift atop an aerofoil, making all forms of flight on aeroplanes, gyroplanes, and helicopters possible.

Note: The Bernoulli's theorem is also the law of conservation of energy, i.e. the sum of all energy in a steady, streamlined, incompressible flow of fluid is always a constant.

Theorem

Proof

Consider the motion of a fluid down a steam tube.

22109/img_0002.png

Consider the motion of an isolated volume a,b,c,d. After a time \delta\,t it is at position a',b',c',d',and since the volume a',b',c,d, is equal in to both, then the net change is equivalent to moving the mass of fluid from a,b,b',a', to c,d,c',d'.

The work done by the pressure on AB in time dt = P_1 a_1 v_1\,\delta\,t

All the work is expended in:

  • a) Doing work against the pressure at CD i.e. P_2 a_2 :v_2\:dt
  • b) Raising the weight of A,B,B',A', to C,D,C',D', i.e. w\,a_1\,v_1\,dt
  • c) Increasing the kinetic energy of A,B,B',A', to that of C,D.C',D',

i.e. \frac{1}{2}\left(\frac{\rho a_1 v_1}{g}} \right)dt (v_2^2 - v_1^2)

Equating (#1) to the work expended P_1=P_2+\rho(z_2-z_1)+\frac{1}{2} \frac{\rho}{g} (v_2^2 - v_1^2) \therefore\;\;\;P_1+\frac{1}{2} \frac{\rho}{g} v_1^2 + \rho z_1 = P_2 + \frac{1}{2} v_2^2 + \rho z_2 = C where C = Constant

This equation can be expressed in three ways:

  • P+\frac{1}{2} \rho v^2+\rho g z=C
  • \frac{P}{g}+\frac{1}{2} \frac{\rho v^2}{g}+\rho z=C
    (2)
  • \frac{P}{\rho g}+\frac{v^2}{2g}+Z=C

where C is constant

Notice: that the above equations make no allowance for a head lost due to Friction and it is normal to write Bernoulli'\b{s Equation} as: \frac{p_1}{\rho g}+\frac{v_1^2}{2g}+Z_1=\frac{p_2}{\rho}+\frac{v_1^2}{2g}+Z_2+h_L where h_L is the head lost due to friction.

See Also