💧 Bernoulli Equation Calculator

Calculate fluid velocity or pressure at a second point using Bernoulli's equation for incompressible flow.

Inputs

Fluid Density (ρ)

Point 1

Pressure at Point 1 (P₁)

Velocity at Point 1 (v₁)

Elevation at Point 1 (h₁)

Point 2

Pressure at Point 2 (P₂)

Elevation at Point 2 (h₂)

Results

Velocity at Point 2 (v₂)0.00 m/s

Fluid Flow Process

Calculator Description

Bernoulli's equation is a foundational principle of fluid mechanics expressing conservation of energy along a flow: the trade-off between pressure, velocity, and elevation. For an ideal (incompressible, inviscid, steady) flow, it states that the total mechanical energy per unit volume of fluid stays constant along a streamline.

What this calculator finds

Given the pressure, velocity, and elevation at one point (1), it solves for an unknown quantity (such as velocity or pressure) at a second point (2) on the same streamline. It quantitatively predicts how a fluid accelerates and how its pressure changes as the pipe narrows or the elevation changes.

Why it matters

Predicting velocity and pressure changes in pipe contractions and nozzles
Understanding the operating principle of Venturi meters and Pitot tubes
Analyzing airfoil lift, atomizers, and efflux velocity (Torricelli) applications

Formula

Bernoulli's Equation

The sum of three terms (static + dynamic + hydrostatic pressure) is constant along the streamline. Each term has units of energy per unit volume, i.e. pressure.

P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2

P — Static pressure [Pa (psf)]
ρ — Fluid density [kg/m³ (lb/ft³)]
v — Fluid velocity [m/s (ft/s)]
g — Acceleration of gravity [9.81 m/s² (32.17 ft/s²)]
h — Elevation height [m (ft)]

How the formula works

Because P + ½ρv² + ρgh is constant, if one term rises another must fall.
Where the pipe narrows and velocity v rises, dynamic pressure increases and static pressure P drops — the Venturi effect.
Raising the elevation h increases the hydrostatic term, reducing pressure or velocity accordingly.

Worked example

Water (ρ = 1000 kg/m³) flows in a horizontal pipe (h₁ = h₂). At point 1, P₁ = 200,000 Pa and v₁ = 2 m/s. Where the pipe narrows so v₂ = 6 m/s: P₂ = 200000 + ½·1000·(2² − 6²) = 200000 − 16000 = 184,000 Pa — the pressure drops.

Useful Tips

Practical tips

The two points must lie on the same streamline (or flow path). Pick a convenient datum (h=0) but apply it consistently to both points.
Combine with continuity A₁v₁ = A₂v₂ to first find velocity from the change in cross-sectional area.
For low-density fluids like air, the hydrostatic term (ρgh) can often be neglected.

Limitations & cautions

It ignores viscous friction losses. With long pipes, elbows or valves, real pressure is lower than predicted — a head-loss term is needed.
The incompressible assumption breaks down for high-speed compressible flow above Mach 0.3.
Where a pump or fan adds energy, that work (head) must be added separately.