Proving Bernoulli's equation from the work–energy theorem

Step 1 of 4: Work–energy theorem for a fluid element

Intuition

The liquid pressure 'pushes' the element like hands at both ends of the pipe: the hand behind pushes harder than the hand ahead resists, and the difference is the net energy the fluid receives.

p_1 A_1\, ds_1 - p_2 A_2\, ds_2 = \Delta K + \Delta U_g

Analysis

Consider the fluid between two cross-sections $A_1$ , $A_2$ of a stream tube. Pressure at end 1 pushes the slug forward by $ds_1$ (positive work $p_1A_1ds_1$ ); pressure at end 2 opposes the motion (negative work $p_2A_2ds_2$ ). With no viscosity these are all the external works besides gravity; by the work–energy theorem, their difference equals the gain in kinetic plus gravitational potential energy of the slug.