Bernoulli and continuity

Why it matters

Water mains, fuel lines, ventilation ducts, carburettors, flow meters and even blood vessels share two simple rules: mass is conserved (the continuity equation), and along a streamline energy is traded between pressure, speed and height (Bernoulli's equation). With just these two, you can find the velocity in a pipe, the pressure drop through a nozzle or the reading of a Venturi meter.

Continuity: what goes in must come out

For steady flow of an incompressible fluid (liquids, and gases at low speed), the volume flow rate is the same at every section of a pipe:

Q = A1 · V1 = A2 · V2

• Q is volume flow rate in m³/s.
• A is the cross-section area: for a round pipe, A = π D² / 4.
• V is the average velocity in m/s.

Halve the diameter and the area drops to a quarter, so the velocity is four times larger. For compressible flow, use mass flow rate instead: ṁ = ρ1 A1 V1 = ρ2 A2 V2.

Bernoulli's equation

Along a streamline in steady, incompressible, frictionless flow, the total mechanical energy per unit volume stays constant:

P1 + ½ ρ V1² + ρ g z1 = P2 + ½ ρ V2² + ρ g z2

The three terms are static pressure, dynamic pressure and elevation (hydrostatic) pressure. Engineers often divide by ρg and write it in head form, in metres of fluid:

P/(ρg) + V²/(2g) + z = constant

The assumptions matter. Bernoulli does not include friction (next lesson), pumps or turbines (which add or remove energy), or heat transfer. Use it between two points on the same streamline in a short, smooth section.

Worked example: a pipe reducer

Water (ρ = 1,000 kg/m³) flows at Q = 0.02 m³/s through a horizontal pipe that narrows from D1 = 100 mm to D2 = 50 mm. The pressure upstream is P1 = 150 kPa (gauge). What is the pressure in the narrow section?

Step 1, continuity:

A1 = π (0.10)² / 4 = 0.007854 m²     V1 = 0.02 / 0.007854 = 2.546 m/s
A2 = π (0.05)² / 4 = 0.001963 m²     V2 = 0.02 / 0.001963 = 10.186 m/s

Step 2, Bernoulli (horizontal, so z1 = z2):

P2 = P1 + ½ ρ (V1² − V2²)
   = 150,000 + 500 × (6.484 − 103.75)
   = 150,000 − 48,634 ≈ 101.4 kPa (gauge)

The fluid speeds up, so its static pressure falls by about 48.6 kPa. That pressure drop is exactly how a Venturi meter measures flow: measure P1 − P2, and you can solve backwards for Q.

Worked example: draining a tank (Torricelli)

Water drains from a small hole 2 m below the surface of a large open tank. Take point 1 at the surface (P = atmospheric, V ≈ 0, z = 2 m) and point 2 in the jet (P = atmospheric, z = 0):

V2 = √(2 g h) = √(2 × 9.81 × 2) ≈ 6.26 m/s

The jet leaves at the same speed as a stone dropped 2 m. Real jets are a bit slower, because of friction and the jet contracting just outside the hole (engineers apply a discharge coefficient of about 0.6 for a sharp-edged hole).

Common mistakes

• Using diameter instead of area in continuity. Area scales with D².
• Mixing gauge and absolute pressures between the two points. Either works, but use the same one on both sides.
• Applying Bernoulli across a pump, turbine or long rough pipe without adding the energy terms.
• Forgetting units. ½ρV² with ρ in kg/m³ and V in m/s gives Pa, not kPa.
• Ignoring cavitation. If the absolute pressure at P2 falls below the fluid's vapour pressure, the liquid can flash into vapour (cavitation), which damages pumps and valves.

Recap

• Continuity: A1 V1 = A2 V2; a smaller pipe means faster flow.
• Bernoulli: P + ½ρV² + ρgz is constant along a streamline, without friction or machines.
• Faster flow means lower static pressure, which is the principle behind Venturi meters.
• These are idealised; real piping systems are sized to codes and checked by licensed engineers.

Try it

On the Do tab, use continuity and Bernoulli in Python to find the pressure in a pipe reducer.