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Physics · Fluids

Bernoulli Equation Calculator

Solve for the pressure at one point on a streamline from the speed and height at another.

Pa
Pressure at point 1.
m/s
Flow speed at point 1.
m
Elevation at point 1.
m/s
Flow speed at point 2.
m
Elevation at point 2.
kg/m³
Fluid density (water ≈ 1000).
Example flows — tap to load
Pressure at point 2 (P₂)
184,000Pa

Pressure change (P₂ − P₁): -16,000 Pa — the faster/higher point has lower pressure.

Pressure at point 2 falls as its speed rises (Bernoulli)
Downward parabola showing pressure at point 2 falling as its flow speed rises202 kPa130 kPav₂ = 0v₂ = 12 m/s

Bernoulli’s equation says pressure, speed, and height trade off along a streamline: where a fluid speeds up, its pressure drops. For water (ρ = 1000 kg/m³) speeding from 2 m/s to 6 m/s at the same height, starting at 200,000 Pa, the pressure falls by 16,000 Pa to 184,000 Pa.

What Bernoulli’s equation describes

For a steady, incompressible, non-viscous flow, the sum of pressure energy, kinetic energy, and potential energy stays constant along a streamline. So if one term grows, another must shrink. This calculator holds that total fixed between two points and solves for the pressure P₂ at the second point given the pressure, speed, and height at the first. To go the other way — force per unit area on a wall — use the pressure calculator.

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

P = pressure (Pa), ρ = density (kg/m³), v = speed (m/s), g = 9.81 m/s², h = height (m)

Worked example

Water (ρ = 1000 kg/m³) flows through a horizontal pipe that narrows, speeding up from v₁ = 2 m/s to v₂ = 6 m/s. The upstream pressure is P₁ = 200,000 Pa and the height is unchanged (h₁ = h₂):

  1. 1
    Rearrange for the unknown pressure. P₂ = P₁ + ½ρ(v₁² − v₂²) + ρg(h₁ − h₂).
  2. 2
    Compute the kinetic term. ½ × 1000 × (2² − 6²) = 500 × (4 − 36) = 500 × (−32) = −16,000 Pa.
  3. 3
    Compute the potential term. Height is unchanged, so ρg(h₁ − h₂) = 1000 × 9.81 × 0 = 0 Pa.
  4. 4
    Add the terms to get P₂. P₂ = 200,000 + (−16,000) + 0 = 184,000 Pa — 16,000 Pa lower because the water sped up.

The three energy terms per unit volume

Each term has units of pressure (Pa = J/m³); Bernoulli keeps their sum constant along a streamline.

TermExpressionRepresents
PressurePStatic pressure energy of the fluid
Kinetic½ρv²Energy of motion (dynamic pressure)
PotentialρghGravitational energy from elevation

Assumptions and intuition

Bernoulli’s equation only holds when the flow is incompressible (constant density), non-viscous (no friction losses), and steady (not changing in time), and only along a single streamline. Real pipes have friction, so downstream pressure is usually a little lower than the ideal result here.

The speed–pressure trade-off is the whole story behind a Venturi meter, where a constriction speeds the fluid and the pressure drop reveals the flow rate, and behind aerodynamic lift, where faster flow over a curved surface leaves lower pressure above it. Squeeze a flow faster and its pressure must fall to keep the total energy fixed.

Why does faster flow mean lower pressure?
Along a streamline the total energy — pressure plus ½ρv² plus ρgh — is fixed. When speed rises, the kinetic term ½ρv² grows, so the pressure term must fall to keep the sum constant. The fluid trades pressure energy for speed.
What are the assumptions behind Bernoulli’s equation?
It assumes steady flow, an incompressible fluid (constant density), no viscosity (no friction losses), and that the two points lie on the same streamline. Break any of these — turbulence, long rough pipes, compressible gas at high speed — and the result drifts from reality.
What units should I use?
SI throughout: pressure in pascals (Pa), speed in m/s, height in metres, and density in kg/m³, with g = 9.81 m/s². Every energy term then comes out in pascals, so P₂ is in pascals too.
What density should I enter?
The density of the flowing fluid: about 1000 kg/m³ for water, 1025 for seawater, 1.225 for air at sea level, and roughly 800–900 for many oils. The calculator requires ρ > 0.
Does it work for a pipe that changes height?
Yes. Set h₁ and h₂ to the two elevations and the ρg(h₁ − h₂) term accounts for the height change. Raising a fluid or slowing it both cost pressure; only their combined effect sets P₂.
Can I use it for a gas like air?
Only at low speeds — well below about a third of the speed of sound — where the gas behaves as effectively incompressible. Enter the gas density (≈ 1.225 kg/m³ for air). At high Mach numbers compressibility breaks the equation.