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Bernoulli's equation is the cornerstone of fluid mechanics — a statement of energy conservation for a flowing fluid. It explains how aircraft wings lift, how a venturimeter measures flow, and how water pressure changes through a pipe network. This article derives it carefully and applies it to numerical examples.

The Equation

p/(ρg) + v²/(2g) + z = constant   (along a streamline)
TermNameRepresents
p/(ρg)Pressure headEnergy due to pressure
v²/(2g)Velocity (kinetic) headEnergy due to motion
zElevation (potential) headEnergy due to position

Each term has units of length (metres of fluid), and their sum — the total head — stays constant for ideal flow.

Derivation (Energy Approach)

Consider a fluid element moving along a streamline. Applying the work-energy principle (or integrating Euler's equation of motion along the streamline) between two points 1 and 2 for steady, incompressible, frictionless flow:

p₁/(ρg) + v₁²/(2g) + z₁ = p₂/(ρg) + v₂²/(2g) + z₂

The pressure work, kinetic energy change and potential energy change balance exactly because no energy is dissipated. This is Bernoulli's theorem.

Assumptions and Limitations

  • Steady flow (conditions do not change with time).
  • Incompressible fluid (constant density — valid for liquids and low-speed gas flow).
  • Non-viscous (no friction losses).
  • Flow along a single streamline; no energy added/removed by machines.

For real pipe flow we add a head-loss term hf (and pump/turbine heads) to get the engineering energy equation:

p₁/ρg + v₁²/2g + z₁ + h_pump = p₂/ρg + v₂²/2g + z₂ + h_f

The Continuity Equation (Always Used Alongside)

A₁ v₁ = A₂ v₂ = Q   (conservation of mass for incompressible flow)

Worked Example 1 — Pipe Contraction

Water flows through a horizontal pipe that contracts from 100 mm to 50 mm diameter. At the 100 mm section the velocity is 2 m/s and the pressure is 200 kPa. Find the velocity and pressure at the 50 mm section. (ρ = 1000 kg/m³)

  1. Continuity: v₂ = v₁ (A₁/A₂) = v₁ (d₁/d₂)² = 2 × (100/50)² = 2 × 4 = 8 m/s
  2. Bernoulli (z₁ = z₂): p₂ = p₁ + ½ρ(v₁² − v₂²) = 200 000 + 500 × (4 − 64) = 200 000 − 30 000 = 170 kPa

The fluid speeds up and the pressure drops — exactly as Bernoulli predicts.

Worked Example 2 — Venturimeter Discharge

A horizontal venturimeter has inlet 200 mm and throat 100 mm. The pressure-head difference between inlet and throat is 1.5 m of water. Find the theoretical discharge. (Cd = 0.98)

  1. A₁ = π/4 × 0.2² = 0.03142 m²; A₂ = π/4 × 0.1² = 0.007854 m²
  2. Q = Cd · (A₁A₂)/√(A₁² − A₂²) · √(2g·Δh)
  3. √(A₁² − A₂²) = √(9.87×10⁻⁴ − 6.17×10⁻⁵) = √(9.25×10⁻⁴) = 0.03042
  4. √(2 × 9.81 × 1.5) = √29.43 = 5.425
  5. Q = 0.98 × (0.03142 × 0.007854 / 0.03042) × 5.425 = 0.98 × 0.008113 × 5.425 = 0.0431 m³/s ≈ 43 L/s

Applications

  • Flow measurement: venturimeter, orifice meter, Pitot tube.
  • Aerofoil lift and the curveball effect.
  • Pipe network analysis (with head-loss terms).
  • Spillways, nozzles and jets.

Common Mistakes

  • Applying Bernoulli across a pump, turbine or region of strong turbulence without the extra energy/loss terms.
  • Forgetting to use continuity to relate the two velocities.
  • Mixing gauge and absolute pressures inconsistently between the two points.