Proof of Bernoulli's theorem

Consider a fluid of negligible viscosity moving with laminar flow, as shown in Figure 1.


Let the velocity, pressure and area of the fluid column be v₁, P₁ and A₁ at Q and v₂, P₂ and A₂ at R. Let the volume bounded by Q and R move to S and T where QS = L₁, and RT = L₂. If the fluid is incompressible:

A₁L₁ = A₂L₂

The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume. Now:

Work done = force x distance = p x volume
Net work done per unit volume = P₁ - P₂
k.e. per unit volume = ½ mv² = ½ Vρ v² = ½ρv² (V = 1 for unit volume)

Therefore:

k.e. gained per unit volume = ½ ρ(v₂² - v₁²)

p.e. gained per unit volume = ρg(h₂ – h₁)

where h₁ and h₂ are the heights of Q and R above some reference level. Therefore:

P₁ - P₂ = ½ ρ(v₁² – v₂²) + ρg(h₂ - h₁)
P₁ + ½ ρv₁² + ρgh₁ = P₂ + ½ ρv₂² + rgh₂

Therefore:

P + ½ ρv² + ρgh is a constant

For a horizontal tube h₁ = h₂ and so we have:

P + ½ ρv² = a constant

This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa.

No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.