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The background to Bernoulli's Theorem

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Theorem

Consider the motion of a fluid down a steam tube

Consider the motion of an isolated volume a, b, c, d. After a time $\delta\,t$ it is position a',b',c',d',and since the volume a',b',c ,d, is to both. 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 = $p_1a_1v_1\,\delta\,t$

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.

$\frac{1}{2}\left(\frac{w\,a_1\,v_1}{g}}\, \right)dt\:(v_2^2\,-\:v_1^2)$

(1)

Equating 1 to the work expended

$p_1\:=\:p_2\:+\:w(Z_2\:-\:Z_1)\:+\:\frac{1}{2}\;\frac{w}{g}\:(V_2^2\,-\,V_1^2)$

(2)

$\therefore\;\;\;p_1\:+\:\frac{1}{2}\;\frac{w}{g}\:V_1^2\:+\:wZ_1\:=\:p_2\:+\:\frac{1}{2}\:V_2^2\:+\:wZ_2\;=\:Constant$

(3)

The equation can be expressed in three ways:

$p\:+\:\frac{1}{2}\;\frac{w\:v^2}{g}\:+\:wZ\;=\:Constant$
(4)
$p\:+\:\frac{1}{2}\;\rho\,v^2\:+\:\rho\,g\,Z\;=\:Constant$
(5)
$\frac{p}{w}\:+\:\frac{v^2}{2g}\:+\:Z\;=\:Constant$
(6)