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)
| Term | Name | Represents |
|---|---|---|
| p/(ρg) | Pressure head | Energy due to pressure |
| v²/(2g) | Velocity (kinetic) head | Energy due to motion |
| z | Elevation (potential) head | Energy 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)
- Continuity: v₂ = v₁ (A₁/A₂) = v₁ (d₁/d₂)² = 2 × (100/50)² = 2 × 4 = 8 m/s
- 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)
- A₁ = π/4 × 0.2² = 0.03142 m²; A₂ = π/4 × 0.1² = 0.007854 m²
- Q = Cd · (A₁A₂)/√(A₁² − A₂²) · √(2g·Δh)
- √(A₁² − A₂²) = √(9.87×10⁻⁴ − 6.17×10⁻⁵) = √(9.25×10⁻⁴) = 0.03042
- √(2 × 9.81 × 1.5) = √29.43 = 5.425
- 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.