04 Lesson 3: Flow Rate and the Continuity Equation
3.1 Volume flow rate and mass flow rate
Consider steady flow of liquid of density ρ in a tube, and take a cross-section of area A at which the speed is v.
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In a time Δt the liquid at that section advances a distance Δx = vΔt, so the volume crossing the section is ΔVol = AΔx = A v Δt.
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Volume flow rate: the volume of liquid flowing through the section per unit time:
Qv = A v (unit m³/s)
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Mass flow rate: the mass of liquid passing per unit time:
Qm = ρ Qv = ρ A v (unit kg/s)
Useful consequences:
- Volume flowing in time t: Vol = Qv t = A v t = π r² v t (for a circular pipe)
- Mass flowing in time t: M = Qm t = ρ A v t = π r² v ρ t
3.2 The continuity equation
- Take two cross-sections of the tube, area A₁ with speed v₁ and area A₂ with speed v₂.
- Because the liquid is incompressible and cannot accumulate or disappear in the tube, the mass passing each section per second is the same (conservation of mass):
ρ A₁ v₁ = ρ A₂ v₂ → A₁ v₁ = A₂ v₂ (the continuity equation)
or v₁/v₂ = A₂/A₁.
Derivation (by volumes). In a time Δt the liquid at section 1 moves Δx₁ = v₁Δt, so the volume entering is A₁v₁Δt. The same volume must leave through section 2 (incompressible liquid): A₂v₂Δt. Therefore A₁v₁Δt = A₂v₂Δt, i.e. A₁v₁ = A₂v₂.
Meaning. The speed of the liquid at any point is inversely proportional to the cross-sectional area of the tube at that point:
- Wide section (large A) → slow flow, streamlines far apart.
- Narrow section (small A) → fast flow, streamlines close together.
For circular pipes A = πr² = πd²/4, so v₁/v₂ = (r₂/r₁)² = (d₂/d₁)². Halving the diameter makes the speed four times larger.
Both Qv (m³/s) and Qm (kg/s) are constant along the tube.
3.3 Everyday examples
- Putting a finger over the end of a garden hose makes the water spray faster and farther: the exit area is smaller, so the speed is greater.
- A river flows fast where it is narrow and slowly where it is wide and deep.
- Blood flows fast in the aorta and slowly in the capillaries, because the total cross-section of all the capillaries is far larger than that of the aorta. The slow flow allows time for exchange of oxygen, carbon dioxide and nutrients with the tissues.
3.4 Worked examples
Example 1 (pipe). Water flows at 1.5 m/s in a pipe of diameter 4 cm that narrows to a diameter of 2 cm. (a) Find the speed in the narrow part. (b) Find the volume and mass of water passing per minute. (a) A₁v₁ = A₂v₂: (4)² × 1.5 = (2)² × v₂ → v₂ = 16 × 1.5 / 4 = 6 m/s. (b) A₁ = π(0.02)² = 1.257 × 10⁻³ m²; Qv = A₁v₁ = 1.257 × 10⁻³ × 1.5 = 1.885 × 10⁻³ m³/s. Volume per minute = 1.885 × 10⁻³ × 60 = 0.113 m³. Mass per minute = 1000 × 0.113 = 113 kg. (Qm = 1.885 kg/s.)
Example 2 (hose and nozzle). Water leaves a hose of inner diameter 1.8 cm at 2 m/s through a nozzle of diameter 0.6 cm. Speed at the nozzle: (1.8/0.6)² × 2 = 9 × 2 = 18 m/s.
Example 3 (blood). The aorta has radius 0.5 cm and blood flows in it at 0.20 m/s. It branches finally into 5000 capillaries (ignoring intermediate vessels) of radius 0.05 cm each. A₁ = π(0.5)² = 0.25π cm²; A₂ = 5000 × π(0.05)² = 12.5π cm². v₂ = A₁v₁/A₂ = 0.25 × 0.20/12.5 = 0.004 m/s. The blood slows down 50 times.
Example 4 (time to fill). A tap delivers water through a pipe of cross-section 1.5 cm² at a speed of 4 m/s. Qv = 1.5 × 10⁻⁴ × 4 = 6 × 10⁻⁴ m³/s = 0.6 L/s. A 60 L tank fills in 60/0.6 = 100 s.
Example 5 (area from flow rate). Water flows at a steady 0.003 m³/s in a pipe in which its speed is 6 m/s. A = Qv/v = 0.003/6 = 5 × 10⁻⁴ m² = 5 cm².
3.5 Enrichment: Bernoulli's principle
Bernoulli's principle
- relates the pressure of a fluid to its speed: where the speed of a fluid increases, its pressure decreases. Application: the wing of an aeroplane has a curved upper surface and a flat lower surface, so air flows faster over the top than below it.
- The pressure above the wing (P₁) is therefore lower than below (P₂); the pressure difference gives an upward lift force that supports the aircraft.