Mini Physics
Part of Mini EducationSwitch subject

Fluids: Bernoulli, Continuity & Scaling (IPhO Mechanics)

IPhO mechanics lesson on continuity, Bernoulli, control-volume momentum, and scaling arguments for fluid flow.

  • International Physics Olympiad preparation

Reading settings

On this page

IPhO fluids questions often reward one skill: state a clean idealisation (steady, incompressible, negligible viscosity), apply continuity and Bernoulli consistently, and use momentum flux when the question asks for forces.

Learning objectives

By the end, you should be able to:

  • IPHO-M-FLU-01: state the regime and select the conservation law that matches the requested quantity;
  • IPHO-M-FLU-02: analyse pressure forces and momentum flux across a labelled control surface; and
  • check units, pressure convention and the wide-reservoir or equal-area limit.

The official hydrodynamics syllabus includes pressure, buoyancy, continuity, Bernoulli, surface tension, surface energy and capillary pressure. This lesson covers continuity and Bernoulli deeply and adds control-volume momentum as training for force questions; it does not yet cover the full official list.

Venturi flow with a control volume and labelled state variablesA horizontal pipe narrows from area A one to A two. Flow arrows point right. The wider inlet has speed v one and pressure p one; the narrow outlet has larger speed v two and lower pressure p two for ideal steady incompressible flow. A dashed control volume encloses the contraction.control surfaceA₁, v₁, p₁A₂, v₂, p₂wide inletnarrow outletsteady + incompressiblev₂ > v₁; p₂ < p₁
Scroll diagram horizontally to read all labels.
Continuity gives A₁v₁ = A₂v₂. Bernoulli relates p and v; a force question instead requires the pressure forces and momentum flux across the dashed control surface.
Prerequisites (quick refresh)

1. Definitions

  • Incompressible flow: density ρ is approximately constant (often valid for liquids at modest speeds).
  • Steady flow: flow variables do not depend explicitly on time at a fixed location.
  • Streamline: curve tangent to the velocity field vector v.
  • Volume flow rate: Q = dV/dt. For uniform pipe flow, Q = Av.
  • Continuity (steady, incompressible): Av is constant along a pipe.
  • Bernoulli along a streamline (ideal flow):
    p + (1/2)ρ v² + ρ gh = constant.
  • Dynamic pressure: (1/2)ρ v².
  • Stagnation pressure: p₀ = p + (1/2)ρ v².
  • Reynolds number: Re = (ρ v L)/η (inertia compared to viscosity).

2. Key ideas

  • Use continuity first to relate speeds, then use Bernoulli to relate pressure and height.
  • Bernoulli requires stated assumptions (steady, incompressible, negligible viscosity) and points connected by the same streamline (or justified as nearly irrotational).
  • For forces on nozzles, bends, and plates, use a control volume and momentum flux:
    vector Fₑₓₜ = ∑ₒᵤₜ ρ Q vector v - ∑ᵢₙ ρ Q vector v
    for steady 1D flows (plus pressure forces on faces if needed).
  • Scaling decides what to ignore: viscosity, compressibility, gravity, or surface tension.
Bernoulli checklist (say this before using it)
  • Steady and incompressible?
  • Viscous losses negligible in the region you connect?
  • Same streamline (or justified)?
  • No pumps, turbines, or strong dissipation between points?

3. Detailed explanation

3.1 Continuity in one line

For steady incompressible pipe flow:

Q = Av = constant ⇒ A₁v₁ = A₂v₂.

If density changes, use ρ₁A₁v₁ = ρ₂A₂v₂.

3.2 Bernoulli as energy per unit volume

Each term has units of pressure:

  • p is pressure energy density,
  • (1/2)ρ v² is kinetic energy density,
  • ρ gh is gravitational potential energy density.

This makes it a good equation for unit checks and for quick scaling estimates.

3.3 When you should switch to momentum

Bernoulli relates scalar quantities. If the question asks for a force (for example, a jet hitting a plate or a nozzle thrust), momentum flux is the direct method.

4. Common mistakes

  • Applying Bernoulli between points not connected by a streamline without justification.
  • Forgetting that a free surface open to air has pressure pₐₜₘ.
  • Mixing gauge pressure and absolute pressure mid-solution.
  • Neglecting ρ gh even when it is the only driver.
  • Assuming the reservoir surface speed is exactly zero when the reservoir is not much larger than the outlet.

5. Problem-solving tips

  1. Label two points clearly and write (p,v,h) at each.
  2. Use continuity first; it usually removes one unknown immediately.
  3. If you use Bernoulli, state the assumptions in a sentence.
  4. For force questions, draw a control volume and list pressure forces plus momentum fluxes.
  5. Check units: p and ρ v² are both pascals.

6. Worked examples

1) Draining a tank: Torricelli speed and drain time

A large tank of cross-sectional area A drains through a small hole of area a at the bottom. Let the water height above the hole be h(t).

(1) Efflux speed (ideal flow): compare the free surface to the hole. Both are at atmospheric pressure, and the surface speed is negligible:

ρ gh = (1/2)ρ v² ⇒ v = square root of 2gh .

(2) Drain time: outflow rate is Q = av, so

-Adh/dt = a square root of 2gh .

Separate variables:

dh/(square root of h) = -(a/A) square root of 2g dt.

Integrate from h₀ to h₁:

2(square root of h₁ - square root of h₀) = -(a/A) square root of 2g t

so

t = (2A/(a square root of 2g))(square root of h₀ - square root of h₁).
2) Venturi meter: pressure drop and flow speed

A horizontal pipe narrows from area A₁ to A₂. The fluid is incompressible with density ρ. Let pressures be p₁ and p₂. Define Δ p = p₁-p₂.

Continuity:

A₁v₁ = A₂v₂ ⇒ v₂ = (A₁/A₂)v₁.

Bernoulli (same height):

p₁ + (1/2)ρ v₁² = p₂ + (1/2)ρ v₂² ⇒ Δ p = (1/2)ρ(v₂²-v₁²).

Substitute:

Δ p = (1/2)ρ v₁²[(A₁/A₂)²-1].

Therefore

v₁ = square root of ((2Δ p/ρ)/((A₁/A₂)²-1)), v₂ = (A₁/A₂)v₁.
3) Jet stopped by a plate: force from momentum flux

A horizontal water jet of area A, speed v and density ρ strikes a large stationary plate normally. After impact, the water spreads sideways, so its mean outgoing x-velocity is zero. Take + x towards the plate and use gauge pressure, so atmospheric pressure on the free jet and exposed control surface contributes zero net gauge-pressure force.

The mass flow rate is m dot = ρ Av. For the water inside a steady control volume,

∑ Fₓ = m dot (v_(x,out)-v_(x,in)) = ρ Av(0-v) = -ρ Av².

This is the force of the plate on the water. By Newton’s third law, the water pushes the plate in the + x direction with magnitude

F_(water on plate) = ρ Av².

Check: ρ Av² has units kg/m³ × m² × m²/s² = N, and the force vanishes as either A or v tends to zero.

7. Extension

1) Siphon height limit: why you cannot lift water arbitrarily high

Consider a siphon drawing water from a reservoir. Let the siphon crest be height h above the reservoir free surface.

Apply Bernoulli between the reservoir surface (pressure pₐₜₘ, speed negligible) and the crest (pressure p_c, speed v):

pₐₜₘ = p_c + (1/2)ρ v² + ρ g h.

So

p_c = pₐₜₘ - ρ g h - (1/2)ρ v².

As h increases, p_c drops. If p_c approaches the liquid’s vapour pressure pᵥ, vapour cavities form and the continuous liquid column can fail.

Requiring p_c ≥ pᵥ gives

h ≤ (pₐₜₘ-pᵥ-(1/2)ρ v²)/(ρ g).

Ignoring vapour pressure and the kinetic term gives only the rough sea-level scale hₘₐₓ∼ pₐₜₘ/(ρ g) ≈ 10 m for water. The practical limit is lower and depends on temperature, flow speed and losses.

8. Practice and evidence

Practice
  • Do one Bernoulli problem where you justify neglecting viscosity using a scaling argument.
  • Do one jet-force problem using a control volume and momentum flux.
  • Treat IPHO-M-FLU-01/02 as passed only if both solutions label the state variables, pressure convention, assumptions and a units or limit check.
  • Then continue via the IPhO Mechanics Hub.
Syllabus and review details

No official syllabus alignment is listed for this lesson.

Last reviewed: