Consider a horizontal pipe of area A1 which is suddenly enlarged to the are A₂.
Consider two section 1-1 and 2-2 before and after the expansion.

Let  P₁, V₁, A₁ is the pressure intensity, velocity and area at 1-1

P₂, V₂, A₂ is the pressure intensity, velocity and area at 2-2

P = intensity of pressure of the liquid eddies on the area (A₂ - A₁)

The resultant force between section 1-1 and section 2-2.

= P₁ A₁ – P₂ A₂ + P’ (A₂-A₁)

But it has bee found by experiment that P’ = P1

Force = P₁ A₁ – P₂ A₂ + P₁ A₂ – P₁ A₁

= (P₁ – P₂) A₂

By Newton’s second law of motion,

Force = rate of change of momentum of the mass

(P₁ - P₂)A₂  =  ρ Q (V₂ – V₁)

= γ. g Q ( V₂ - V₁)

P₁ - P₂ / γ = Q/ A₂. g (V₂ – V₁)

P₁ – P₂ / γ = V₂ (V₂ - V₁) / g

Applying modified Bernoulli’s equation between section 1 and 2.

P₁ / γ + V²₁/ 2g + Z₁ = P₂ / γ + V²₂ / 2g + Z₂ + loss due to expansion

h_c = (p₁/ γ - P₂/ γ ) + ( V¹₂ - V²₂ / 2g

h_c = V₂(V₂-V₁) / g + V¹₂ – V²₂ / 2g

h_c = 2V₂² – 2V₁ V₂ + V¹₂ + V₁² – V₂² / 2g  = V₁² – 2V₁V₂ + V₂² / 2g

h_c = (V₁ – V₂)² / 2g

The Equation gives head loss due to sudden expansion.

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