Bernoulli's Principle Simulator — Fluid Speed & Pressure

What Is Bernoulli's Principle?

Published by Daniel Bernoulli in 1738, the principle is a statement of energy conservation for flowing fluids. For an ideal fluid — incompressible, inviscid, and in steady flow — the total mechanical energy per unit volume is constant along any streamline. That total energy has three parts:

• Static pressure P — the "pushing" pressure acting on the fluid.
• Dynamic pressure ½ρv² — kinetic energy per unit volume.
• Hydrostatic pressure ρgh — potential energy per unit volume.

When any one of these increases, the others must decrease to keep the sum constant. The most dramatic consequence: faster flow means lower static pressure.

The Bernoulli Equation

The full Bernoulli equation, valid along a streamline between any two points (1 and 2) in an ideal fluid:

At the same elevation (h₁ = h₂), this simplifies to:

Rearranging for the pressure at point 2:

The assumptions — ideal, steady, incompressible — hold well for water in short pipework and for air at low Mach number (<0.3). Real fluids introduce viscous losses that are handled by the Darcy–Weisbach equation, but Bernoulli remains an excellent first approximation.

The Continuity Equation

Before applying Bernoulli you need to know the velocity at each point. The continuity equation (conservation of mass for an incompressible fluid) states:

where Q is the volumetric flow rate (m³/s) and A is the cross-sectional area. For a circular pipe of diameter d, A = πd²/4. The velocity ratio equals the inverse of the area ratio:

This is why a partially blocked garden hose nozzle produces a faster jet — the same flow rate must pass through a smaller area.

Worked Example: Venturi Pressure Drop

The simulator's default state uses water (ρ = 1000 kg/m³) at v₁ = 3 m/s through a 100 mm diameter pipe narrowing to 50 mm at the throat (Δh = 0):

1. Continuity — throat velocity:
   v₂ = v₁ × (d₁/d₂)² = 3 × (100/50)² = 3 × 4 = 12 m/s
2. Dynamic pressures:
   q₁ = ½ × 1000 × 3² = 4 500 Pa
   q₂ = ½ × 1000 × 12² = 72 000 Pa
3. Throat static pressure (P₁ = 200 kPa baseline):
   P₂ = 200 000 + 4 500 − 72 000 = 132 500 Pa = 132.5 kPa
4. Pressure drop: ΔP = 200 − 132.5 = 67.5 kPa
5. Flow rate: Q = A₁ × v₁ = π(0.05)² × 3 = 0.02356 m³/s = 23.6 L/s

All five results are visible simultaneously in the simulator's readout panel — velocity, pressures, flow rate, and Reynolds number.

Dynamic Pressure and the Pitot Tube

Dynamic pressure is the kinetic energy per unit volume of the flowing fluid:

A Pitot tube works by bringing the flow to rest at a stagnation point. All kinetic energy converts to pressure, giving the total (stagnation) pressure P_t = P_s + q. Measuring the difference between total and static pressure yields the flow velocity:

Aircraft airspeed indicators and industrial flow probes both use this principle. In the simulator, try the Pitot tube concept card to see a numerical example.

Real-World Applications of Bernoulli's Principle

Bernoulli's equation appears in a surprising range of engineering systems:

• Airplane lift — The lift force on a wing is L = ½ρv²·S·C_L. At cruise speed (v = 250 m/s), even a modest C_L on a large wing area generates hundreds of kilonewtons of upward force.
• Venturi meters — Flow rate in industrial pipework is measured by deliberately narrowing the pipe and reading the pressure drop with a manometer.
• Carburettors and atomisers — Fast air over a liquid tube creates low pressure that draws liquid up and breaks it into droplets — the basis of perfume sprayers and old-fashioned petrol carburettors.
• Siphons — A continuous column of liquid can flow upward over a barrier if the downstream pressure is low enough to "pull" the liquid.

Reynolds Number and Flow Regime

Before applying Bernoulli it is worth checking whether the flow is laminar or turbulent, since viscous losses matter more in laminar flow. Reynolds number is:

For our worked example: Re₁ = 1000 × 3 × 0.1 / 0.001 = 300 000 (strongly turbulent). At the throat: Re₂ = 1000 × 12 × 0.05 / 0.001 = 600 000. Both are well into the turbulent regime (Re > 4000), so Bernoulli applies with good accuracy. The simulator displays Re for both sections and labels the regime automatically.

Using the Simulator

Open the Bernoulli's Principle Simulator and try these experiments:

1. Fluid type: Switch from Water to Oil (ρ = 850) or Air (ρ = 1.225). Notice how lower density fluids produce lower dynamic and static pressure differences even at the same velocity.
2. Diameter ratio: Reduce d₂ progressively from 50 mm down. Watch P₂ fall rapidly — for very narrow throats it can approach zero (cavitation threshold).
3. Height offset Δh: Raise the throat above the inlet. The hydrostatic term ρgΔh adds to the pressure drop. For water rising 1 m: extra ΔP = 1000 × 9.81 × 1 = 9810 Pa ≈ 9.8 kPa.
4. Step-by-step calculator: Click the "Show Calculation" button on the canvas to see a detailed derivation for your current state.
5. Explore and Practice modes: Work through the concept cards and 12 randomised problems — Venturi, Pitot tube, flow rate, Reynolds number.

Key Takeaways

• Bernoulli's equation \(P + \tfrac{1}{2}\rho v^2 + \rho gh = \text{const}\) conserves energy along a streamline.
• The continuity equation \(A_1v_1 = A_2v_2\) links velocity to pipe area — halving diameter quadruples velocity.
• Dynamic pressure \(q = \tfrac{1}{2}\rho v^2\) scales with v². Doubling speed quadruples q.
• The Venturi pressure drop \(\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2)\) is how flow meters work.
• Reynolds number Re = ρvD/μ determines whether Bernoulli's equation applies accurately.
• Lift, Pitot tubes, carburettors, and atomisers all exploit the same speed–pressure trade-off.