Engineer's Notebook §1.1

Continuity

In steady flow of a liquid, whatever goes in one end of a pipe must come out of the other. Where the pipe narrows, the flow speeds up:

A1v1 = A2v2
A
cross-sectional area
m²
v
mean flow speed
m/s
D
pipe diameter
m
Q
volume flow rate
m³/s
ṁ
mass flow rate
kg/s
ρ
density
kg/m³
A pipe that changes diameter, with an arrow in each section whose length shows the flow speed. v₁ = 2 m/s v₂ = 8 m/s
Fig. 1.1-1Pipe diameters to scale; arrow lengths show speed.
A1 = πD1²/47,850 mm²
A2 = πD2²/41,960 mm²
Q = A1v115.7 L/s
v2 = v1A1/A28 m/s

Worked example

Uses the values in Fig. 1.1-1. Change them and the working updates.

Given
  • D1inlet diameter100 mm
  • D2outlet diameter50 mm
  • v1inlet speed2 m/s
Find

The speed v2 in the outlet, and the flow rate Q.

Solution
A1 = πD1²/4 = π × (0.1 m)² / 4 = 7.85 × 10⁻³ m² A2 = πD2²/4 = π × (0.05 m)² / 4 = 1.96 × 10⁻³ m² v2 = v1A1/A2 = 2 m/s × 7.85 × 10⁻³ / 1.96 × 10⁻³ = 8 m/s Q = A1v1 = 7.85 × 10⁻³ m² × 2 m/s = 0.0157 m³/s
Answer
v2 = 8 m/s

That is 15.7 L/s, the same through both sections.

Related cases

  • Volume flow rate

    Q = Av

  • Gas or other compressible flow

    ρ1A1v1 = ρ2A2v2

  • Branching pipe

    Q1 = Q2 + Q3

Assumptions
  • Flow is steady: nothing changes with time and no fluid builds up between the sections.
  • The fluid is incompressible. That suits liquids, and gases well below about a third of the speed of sound.
  • v is the mean speed over the section. The real speed is highest at the centre and zero at the wall.
  • No leaks and no branches between the sections.

Derivation
  1. In a short time Δt, the fluid entering section 1 fills a volume A1v1Δt.
  2. In the same time the fluid leaving section 2 fills A2v2Δt.
  3. Mass is conserved and none is stored between them: ρA1v1Δt = ρA2v2Δt.
  4. With ρ constant, A1v1 = A2v2.

For round pipes v2 = v1(D1/D2)²: halving the diameter makes the flow four times faster.