Bernoulli Equation Calculator
Apply Bernoulli's principle: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂ to find unknown pressure or velocity.
Enter three known values at point 1 and two at point 2 — the calculator finds the unknown.
What Is Bernoulli's Equation?
Bernoulli's principle is a statement of energy conservation for ideal fluid flow: P + ½ρv² + ρgh = constant along a streamline. Here P is pressure (Pa), ρ is fluid density (kg/m³), v is fluid velocity (m/s), g = 9.8 m/s², and h is height (m). When velocity increases, pressure decreases — the fundamental insight behind airplane lift, Venturi meters, and carburetors.
Bernoulli's equation applies to inviscid (frictionless), incompressible, steady, irrotational flow along a streamline. Real fluids deviate from these idealizations due to viscosity and turbulence, but the equation provides excellent approximations for many engineering applications. The Venturi effect is Bernoulli applied to a constricting pipe: narrower cross-section → higher v → lower P.
The three terms represent energy per unit volume: P is flow work (pressure energy), ½ρv² is kinetic energy density, and ρgh is gravitational potential energy density. Their sum is constant. Squeezing fluid through a narrow pipe converts pressure energy to kinetic energy (lower P, higher v). Releasing through a wide section restores pressure at the expense of velocity.
Applications span all scales: pressure differences contribute to lift on wings and sails, carburetors use a Venturi to draw fuel into an airstream, pitot tubes compare stagnation and static pressure, atomizers use a pressure difference to draw liquid upward, and river water can speed up as a channel narrows. For aerodynamic lift, Bernoulli describes the pressure field but does not by itself determine the circulation or flow pattern around the wing.
Formula Reference Table
| Solve For | Formula | Notes |
|---|---|---|
| Bernoulli's equation | P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂ | Energy conservation for fluids |
| Find P₂ | P₂ = P₁ + ½ρ(v₁²−v₂²) + ρg(h₁−h₂) | Pressure decreases with velocity |
| Find v₂ | v₂ = √(v₁² + 2(P₁−P₂)/ρ + 2g(h₁−h₂)) | From Bernoulli rearranged |
| Venturi meter | ΔP = ½ρ(v₂²−v₁²) | Measures flow via pressure drop |
| Stagnation pressure | P_total = P_static + ½ρv² | Pitot tube: P_total = P₀ + ½ρv² |
| Torricelli's theorem | v = √(2gh) | Efflux speed from hole at depth h |
3 Worked Examples
Water (ρ=1000) flows at v₁=2 m/s in 10 cm pipe, narrows to v₂=8 m/s. P₁=200,000 Pa.
- P₂ = P₁ + ½ρ(v₁²−v₂²)
- P₂ = 200,000 + ½×1000×(4−64)
- P₂ = 200,000 + 500×(−60) = 200,000 − 30,000 = 170,000 Pa
Pitot tube: stagnation P = 115,000 Pa, static P = 101,325 Pa. Air density ρ = 1.225 kg/m³.
- ΔP = P_stagnation − P_static = 13,675 Pa = ½ρv²
- v² = 2×13,675/1.225 = 22,327 m²/s²
- v = 149.4 m/s = 538 km/h (aircraft airspeed)
Tank with water 2 m deep. Find drain velocity from hole at bottom.
- Torricelli: v = √(2gh) = √(2 × 9.8 × 2)
- v = √39.2 = 6.26 m/s
- For 2 cm diameter hole: Q = Av = π×(0.01)²×6.26 = 1.97×10⁻³ m³/s ≈ 1.97 L/s
Real-World Applications
Common Mistakes to Avoid
Bernoulli assumes inviscid (frictionless) fluid. In viscous flow (blood, oil, water in pipes), friction losses must be included (extended Bernoulli with head loss: h_L = f·L·v²/(2gD)).
Bernoulli is valid along a single streamline, not across streamlines. Pressure distribution across a streamline requires radial momentum equations.
Use consistent pressure type (both gauge or both absolute). Gauge pressure = absolute − atmospheric. Mixing them gives wrong pressure differences.
For air speeds above Mach 0.3, compressibility effects become significant. Standard Bernoulli assumes incompressible flow. Compressible Bernoulli uses different forms involving enthalpy.
Bernoulli: energy conservation (pressure, velocity, height). Continuity: mass conservation (A₁v₁ = A₂v₂). They work together — continuity gives velocity ratio; Bernoulli gives pressure change.