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Bernoulli Equation Calculator

Solve pressure, velocity and height in a flowing fluid

Updated · Free, no signup

kPa

Absolute or gauge — use the same reference for P₂.

m/s
m
m
cm
cm

Answer

P₂ = 121.12 kPa

Pressure at point 2

121.125 kPa

Velocity at point 2

8 m/s

Pressure change (P₂ − P₁)

-78.875 kPa

Volumetric flow rate

15.708 L/s

Shown when pipe diameters are used.

Fluid density

998 kg/m³

Total head (energy per weight)

20.639 m

  • The fluid speeds up from 2 to 8 m/s, so 29.94 kPa of pressure is converted into kinetic energy.

Energy per unit volume at each point (kPa)

About the Bernoulli Equation Calculator

This Bernoulli equation calculator applies energy conservation to a fluid flowing along a streamline. Enter the pressure, speed and height at point 1, the height at point 2, and either the speed at point 2 or the two pipe diameters, and it returns the pressure at point 2. Switch modes to solve for the outlet velocity instead — useful for problems like water draining from a tank or flow through a nozzle.

It is aimed at physics and engineering students, and at anyone sizing a simple pipe run, siphon or venturi. When you give the two pipe diameters, the calculator uses the continuity equation (A₁v₁ = A₂v₂) to find the outlet velocity automatically and also reports the volumetric flow rate. A stacked chart compares the pressure, velocity and elevation terms at each point so you can see how energy shifts from one form to another.

Bernoulli’s equation assumes steady, incompressible, frictionless (inviscid) flow along a single streamline with no pumps or turbines in between. Real pipes lose some pressure to friction, so treat results as the ideal case; for air the incompressible assumption holds well below about Mach 0.3 (roughly 100 m/s). Pressures can be entered as absolute or gauge, as long as both points use the same reference.

How to use the bernoulli equation calculator

  1. 1Choose whether to solve for the pressure or the velocity at point 2.
  2. 2Pick the fluid, or enter a custom density.
  3. 3Enter pressure, velocity and height at point 1 and the height at point 2.
  4. 4Give the speed at point 2 directly, or the two pipe diameters to use continuity.
  5. 5Read the result and use the chart to see how energy moves between terms.

Formula and method

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂ A₁v₁ = A₂v₂

Bernoulli's equation says that for steady, incompressible, frictionless flow along a streamline, the sum of static pressure P, dynamic pressure ½ρv² and hydrostatic term ρgh stays constant. Each term has units of pressure (energy per unit volume). Solving for the downstream pressure gives P₂ = P₁ + ½ρ(v₁² − v₂²) + ρg(h₁ − h₂); solving for the downstream speed gives v₂ = √(v₁² + 2(P₁ − P₂)/ρ + 2g(h₁ − h₂)).

The continuity equation expresses conservation of mass: the same volume flows through every cross-section each second, so v₂ = v₁ × (d₁/d₂)² for circular pipes. Dividing every term by ρg gives the total head in metres, a convenient way to compare pressure, velocity and elevation. g is standard gravity, 9.80665 m/s².

P
Static pressure (Pa)
ρ
Fluid density (kg/m³)
v
Flow speed (m/s)
h
Height above a reference level (m)
g
Gravitational acceleration, 9.80665 m/s²
A, d
Pipe cross-sectional area and diameter

Worked examples

Water pipe narrowing and rising 5 m

Halving the diameter from 10 cm to 5 cm quadruples the speed from 2 to 8 m/s. P₂ = 200,000 + ½ × 998 × (4 − 64) + 998 × 9.80665 × (0 − 5) = 121,125 Pa, so the pressure falls by about 78.9 kPa. The flow rate is π/4 × 0.1² × 2 = 15.7 L/s.

Water draining from a tank 10 m deep (Torricelli)

Both the surface and the outlet are open to the atmosphere, so the pressures cancel. v₂ = √(2 × 9.80665 × 10) = 14.0 m/s — the same speed as an object falling 10 m, which is Torricelli’s law.

Air speeding up over a surface

Air at 1.204 kg/m³ accelerating from 10 to 30 m/s loses ½ × 1.204 × (900 − 100) = 481.6 Pa of static pressure. This pressure drop over a faster-moving surface is part of how wings and spoilers generate force.

Frequently asked questions

What does Bernoulli’s equation state?+

It states that along a streamline in steady, incompressible, frictionless flow, P + ½ρv² + ρgh is constant. When a fluid speeds up its pressure drops, and when it rises its pressure or speed must fall.

Why does pressure drop when a pipe narrows?+

Continuity forces the fluid to move faster through the smaller area. The extra kinetic energy has to come from somewhere, so static pressure falls — the venturi effect used in carburettors, flow meters and aspirators.

Should I use absolute or gauge pressure?+

Either works because the equation only involves differences, but both points must use the same reference. If you get a negative absolute pressure, the real fluid would cavitate or the flow could not exist.

When is Bernoulli’s equation not accurate?+

It ignores friction, turbulence losses, pumps, turbines and compressibility. Long pipes, valves and fittings lose head to friction (use Darcy–Weisbach), and gases above about Mach 0.3 need compressible-flow equations.

What is the continuity equation?+

For an incompressible fluid the volume flow rate Q = A × v is the same at every cross-section. For round pipes this means v₂ = v₁ × (d₁/d₂)², so halving the diameter makes the fluid four times faster.

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