Continuity, Bernoulli and the Energy Equation

Flow rate and velocity, Reynolds number, laminar and turbulent flow, the Bernoulli equation, the energy equation with a pump and losses.

Flow rate and continuity

The volume flow rate Q (m³/s) is the volume that passes a cross-section each second, and the mean velocity is v = Q ÷ A. For an incompressible fluid in a pipe that has no leaks or branches, what goes in must come out, which is the continuity equation.

Q = A v    A₁ v₁ = A₂ v₂

Worked example: flow narrowing

  • 10 L/s (0.010 m³/s) flows in a DN80 pipe (inside diameter 77.92 mm, area 47.69 cm²): v = 0.010 ÷ 0.00477 = 2.10 m/s.
  • The pipe reduces to a 25 mm outlet: v = 0.010 ÷ 0.000491 = 20.4 m/s. The area ratio is (78 ÷ 25)² ≈ 9.7, so the speed rises by the same factor.
  • In practice liquids are kept to about 1 to 3 m/s in pipework (see Pipe Systems: Series, Parallel and Networks): faster flow costs pressure and risks noise and erosion.

Reynolds number: laminar or turbulent?

Flow in a pipe is either smooth and layered (laminar) or chaotic and mixing (turbulent). The Reynolds number, the ratio of inertial to viscous forces, tells which.

Re = ρ v D ÷ μ = v D ÷ ν

Laminar and turbulent velocity profilesLaminar (Re < 2 300)u_max = 2 × mean speedTurbulent (Re > 4 000)u_max ≈ 1.2 × mean speedTurbulent flow mixes the fluid, so the speed is nearly uniform across the pipe, but the shear at the wall is muchhigher, which is why friction loss grows faster than in laminar flow.
Velocity profiles in a pipe. Turbulent flow is flatter and has a steeper gradient at the wall.
CaseReRegime
Water at 20 °C, 2.10 m/s in DN80162 784turbulent
ISO VG 46 oil (46 mm²/s) at 0.5 m/s in the same DN80 pipe847laminar
Air at 5 m/s in a 50 mm duct16 630turbulent
Very slow water: 4 mm/s in a 150 mm tube598laminar

Almost all water, air and process flow in practice is turbulent. Viscous oils, small bores, very slow flows and some food products are laminar, and need a different friction law (see Pipe Friction and the Moody Diagram).

The Bernoulli equation

Along a streamline of a steady, frictionless, incompressible flow, the total of pressure, kinetic and potential energy per volume stays constant. It is conservation of energy for a fluid.

p + ½ ρ v² + ρ g z = constant    or in head form    p ÷ (ρ g) + v² ÷ (2 g) + z = H

It applies where friction is small and no energy is added: a nozzle, a short converging section, flow into a pipe entrance, an open tank draining. With friction or a pump you need the energy equation below.

Draining a tank (Torricelli)

The speed of liquid leaving a small hole at depth h below the free surface comes straight from Bernoulli (the surface speed is negligible and both are at atmospheric pressure): v = √(2 g h), the same as a stone dropped through h.

Worked example: a hole 2 m below the surface

  • v = √(2 × 9.807 × 2.0) = 6.26 m/s.
  • A real jet contracts, so the flow is Q = Cd A v with a discharge coefficient Cd ≈ 0.62 for a sharp-edged hole: through a 25 mm hole Q = 0.62 × 4.91 cm² × 6.26 m/s = 1.91 L/s.
  • A 1 m × 1 m tank holding 1.5 m of water empties in about 13.1 minutes at that initial rate; it takes longer in reality because the head falls as the tank drains.

The Venturi: speed up, pressure down

When a pipe narrows, continuity makes the flow faster, and Bernoulli says the pressure must fall to pay for the extra kinetic energy. The effect is the basis of the Venturi meter, the carburettor and the aspirator, and it is why a high-velocity jet can suck.

Continuity and Bernoulli in a Venturienergy line (no loss)Δh = 4.96 mv₁ = 2.55 m/sv₂ = 10.19 m/sQ = 20 L/s through 100 mm narrowing to 50 mm: the speed rises four times and the pressure falls by ½ρ(v₂² −v₁²) = 48.5 kPa.
A Venturi: the narrow throat is faster and has a lower static pressure, shown by the lower water column.

Worked example: a Venturi

  • 20 L/s through 100 mm narrowing to 50 mm: v₁ = 2.55 m/s and v₂ = 10.19 m/s (four times faster, because the area is four times smaller).
  • Pressure drop (Bernoulli, no friction): Δp = ½ ρ (v₂² − v₁²) = 0.5 × 998.2 × (103.75 − 6.48) = 48.5 kPa (4.96 m of water).
  • Measuring that pressure difference and reversing the calculation gives the flow rate: see Flow Measurement. Real Venturi meters apply a discharge coefficient of about 0.98.

The energy equation with a pump and losses

Real systems have friction losses, fittings and pumps. Add them to Bernoulli between two points 1 and 2:

p₁ ÷ ρg + v₁² ÷ 2g + z₁ + Hpump = p₂ ÷ ρg + v₂² ÷ 2g + z₂ + hL

Worked example: pumping between two tanks

  • Water at 20 °C is lifted 12 m from one open tank to another through 100 m of DN80 steel pipe at 10 L/s (v = 2.10 m/s, velocity head 0.224 m) with fittings totaling ΣK = 8.45.
  • Both tank surfaces are at atmospheric pressure with negligible speed, so p and v drop out: Hpump = Δz + hL.
  • Pipe friction: f = 0.0196 (Re = 162 784, see Pipe Friction and the Moody Diagram), so hf = f (L ÷ D) v² ÷ 2g = 5.63 m. Fittings: ΣK v² ÷ 2g = 1.89 m.
  • Hpump = 12 + 5.63 + 1.89 = 19.5 m. The hydraulic power is ρ g Q H = 1 911 W, and at 70 percent efficiency the shaft needs 2.73 kW. The Pump Power and Sizing Calculator does these steps.

A pitot tube applies the same Bernoulli idea to measure speed: the stagnation pressure minus the static pressure equals ½ ρ v². A 250 Pa reading in air (ρ = 1.204) means v = √(2 × 250 ÷ 1.204) = 20.4 m/s; in water, 3 m/s gives 4.49 kPa.

Related

FAQ

Why does pressure drop where the fluid goes faster?

Energy is conserved: kinetic energy per volume ½ρv² goes up, so the pressure energy p must go down. The fluid is not “sucked”; the higher pressure behind pushes it into the narrow section.

When can I not use Bernoulli?

When there is friction, a pump or a turbine, heat transfer, compressible flow at high speed (above about Mach 0.3), or unsteady flow. Use the energy equation with a loss term instead.

What is the critical Reynolds number?

About 2 300 for pipe flow. Below it the flow is laminar; above about 4 000 it is turbulent; in between it flips unpredictably. Careful experiments in very smooth pipes can keep laminar flow to much higher Re, but real piping does not.

Which diameter do I use for a non-round duct?

The hydraulic diameter Dh = 4 A ÷ P (area over wetted perimeter), for example 2 a b ÷ (a + b) for a rectangular duct. It is accurate enough for turbulent flow.

References

  • White, F. M., Fluid Mechanics, McGraw-Hill (Bernoulli, energy equation, Reynolds number).
  • Munson, B. R., et al., Fundamentals of Fluid Mechanics, Wiley.
  • Crane Co., Flow of Fluids Through Valves, Fittings and Pipe, Technical Paper No. 410.
  • Idelchik, I. E., Handbook of Hydraulic Resistance.

Last reviewed: October 2026.