haya.Learning made clear
Unit III · Properties of fluids (lesson 9)
12 min readConcept sheetSection A objective (2 marks), often a short passage on a sprayer or a pipe; Section B short answer (3 marks) or long answer (5 marks) on Bernoulli's theorem and its uses; also a NEET favourite

Equation of continuity and Bernoulli's theorem

In steady flow A₁v₁ = A₂v₂: fluid speeds up where the pipe narrows. Bernoulli: where speed is high, pressure is low; P + ½ρv² + ρgh is constant.

Equation of continuityA₁v₁ = A₂v₂ for an incompressible fluid in streamline flow; Av is the volume flowing per second (m³ s⁻¹)
Bernoulli's equationP + ½ρv² + ρgh = constant along a streamline: pressure energy + kinetic energy + potential energy, each per unit volume
AssumptionsThe fluid is incompressible and non-viscous, and the flow is streamlined (steady)
Speed of effluxWater leaves a hole at depth h below the surface with speed v = √(2gh)

Derivation: equation of continuity

A fluid flows steadily through a tube. At end A the tube is wide (area A₁, speed v₁); at end B it is narrow (area A₂, speed v₂).

  1. 1
    In one second, the fluid at A moves a distance v₁. Volume entering per second = A₁v₁.
  2. 2
    Mass entering per second = ρA₁v₁, where ρ is the density.
  3. 3
    In the same way, the mass leaving at B per second = ρA₂v₂.
  4. 4
    No fluid piles up inside or leaks out, so mass in = mass out: ρA₁v₁ = ρA₂v₂.
  5. 5★
    So, A₁v₁ = A₂v₂: the speed is greater where the tube is narrower. This is the equation of continuity, a form of conservation of mass.

★ key step

Bernoulli's theorem

A flowing fluid has three kinds of energy: pressure energy, kinetic energy and potential energy. Per unit volume, they are P, ½ρv² and ρgh. Bernoulli's theorem: for an incompressible, non-viscous fluid in streamline flow, the sum P + ½ρv² + ρgh stays constant along a streamline. In simple words: where the speed is high, the pressure is low, and where the speed is low, the pressure is high. Your NIOS lesson writes density as d: P + ½dv² + hdg = constant. You need to state and use the theorem; its derivation is not set.

Uses of Bernoulli's theorem

UseWhat happensWhy (Bernoulli)
Aircraft wing (aerofoil)The plane is lifted as it speeds along the runwayThe curved top makes streamlines crowd above the wing. Air is faster above, so the pressure is lower above than below
Sprayer, atomiser, spray gunLiquid is drawn up a thin tube and sprayed outA fast jet of air across the top of the tube lowers the pressure there. The liquid is pushed up and broken into drops
Venturi meterMeasures the rate of flow of a liquid in a pipeThe liquid speeds up in the narrow throat, so the pressure drops there; the pressure difference gives the speed
Bunsen burnerAir enters through a side hole and mixes with the gasThe fast gas jet lowers the pressure near the nozzle, so outside air rushes in
Roof blown off in a stormA tin roof lifts off in a strong windFast wind above lowers the pressure there, while still air below keeps the normal pressure
Swing of a spinning ballA spinning cricket ball curves in the airSpin makes the air faster on one side and slower on the other; the pressure difference pushes the ball sideways

How to draw a Venturi meter

  1. 1
    Draw a horizontal pipe with a wide part on the left, labelled A₁, v₁, P₁. It narrows to a throat on the right, labelled A₂, v₂, P₂.
  2. 2
    Draw a vertical glass tube rising from the top of the wide part, and another from the throat.
  3. 3
    Show the liquid level higher in the tube over the wide part and lower over the throat. Mark the height difference h.
  4. 4
    Draw arrows inside the pipe for the direction of flow, closer together in the throat.
  5. 5★
    Working: P₁ − P₂ = ½ρ(v₂² − v₁²) and P₁ − P₂ = hρg. With A₁v₁ = A₂v₂, the volume flowing per second, A₁v₁, is proportional to √h.

★ key step

Worked example: water in a narrowing pipe

Problem

Water flows through a level pipe whose diameter narrows from 4.0 cm to 2.0 cm. Its speed in the wide part is 1.5 m s⁻¹. Find (a) the speed in the narrow part, (b) the pressure difference between the two parts. (ρ = 1000 kg m⁻³)

Working
  1. (a) A₁v₁ = A₂v₂, and area ∝ (diameter)², so v₂ = v₁ × (d₁/d₂)².
  2. v₂ = 1.5 × (4.0 ÷ 2.0)² = 1.5 × 4 = 6.0 m s⁻¹.
  3. (b) The pipe is level, so P₁ − P₂ = ½ρ(v₂² − v₁²).
  4. P₁ − P₂ = ½ × 1000 × (6.0² − 1.5²) = 500 × (36 − 2.25) = 500 × 33.75 Pa.
  5. P₁ − P₂ = 16875 Pa ≈ 1.7 × 10⁴ Pa.
Answer

(a) 6.0 m s⁻¹; (b) about 1.7 × 10⁴ Pa, with the lower pressure in the narrow part

Activity: fractional rate of flow from a burette (an analogy for radioactive decay)

This experiment shows that water draining from a burette behaves like a decaying radioactive sample. You need a 50 mL burette, a thistle funnel, a rubber tube, two stands and a stop clock.

  1. 1
    Clamp a 50 mL burette and a thistle funnel on two stands. Join their lower ends with a rubber tube.
  2. 2
    Set the mouth of the funnel level with the bottom mark of the burette, so only the water above that mark drives the flow.
  3. 3
    Close the stopcock and fill the burette a little above the top mark.
  4. 4
    Open the stopcock slightly, so the flow is slow and smooth, not turbulent. Start the stop clock as the level passes 50 mL.
  5. 5
    Without touching the stopcock again, record the time at every 5 mL fall, and at every 2 mL fall below 20 mL.
  6. 6
    Plot volume V (y-axis) against time t (x-axis). The curve falls fast at first and then more slowly, like a decay curve.
  7. 7
    From the graph, read the times for V to fall from 40 mL to 20, 10, 5 and 2.5 mL. Each halving takes about the same time, a 'half-life' of the flow.
  8. 8
    Draw tangents at these points to find the rate of flow, ΔV/Δt, and divide it by V. This fractional rate of flow comes out about the same at every stage.
  9. 9★
    Why: the rate of flow is proportional to the pressure head, and so to the volume left. In the same way, the decay rate of a radioactive sample is proportional to the number of nuclei left.

★ key step

In real life

When you press the mouth of a garden hose, the water shoots farther, because the narrower opening makes it flow faster. An aeroplane gets lift because air rushes faster over the curved top of its wings, lowering the pressure there.

Exam angle

3 marks: “State Bernoulli's theorem. Using it, explain how an aircraft wing gets lift.” P + ½ρv² + ρgh is constant along a streamline (incompressible, non-viscous fluid, streamline flow). The curved top of the wing makes air flow faster above it, so the pressure above is lower than below. This difference gives the upward lift.

2 marks: “State the equation of continuity. Why does the jet reach farther when you partly cover the end of a garden hose with your thumb?” A₁v₁ = A₂v₂ for the steady flow of an incompressible fluid. Covering part of the end reduces the area, so the water leaves faster and is thrown farther.

2 marks: “Water flows out of a small hole 5.0 m below the water level in a large tank. Find its speed. (g = 9.8 m s⁻²)” By Bernoulli's theorem, v = √(2gh) = √(2 × 9.8 × 5.0) = √98 ≈ 9.9 m s⁻¹.

Remember it

Narrow pipe, fast flow (Av stays constant); fast flow, low pressure (Bernoulli).