Viscosity: friction inside a fluid
When a liquid flows over a surface, the layer touching the surface sticks to it and stays still. The next layer slides over it a little faster, and so on. The layers rub against each other. This internal friction is viscosity.
The force needed to keep a layer of area A moving is proportional to A and to how fast the speed changes with distance (the velocity gradient dv/dx):
F = η A (dv/dx)
η (eta) is the coefficient of viscosity. SI unit: N s m⁻² = Pa s (also called poiseuille, Pl). CGS unit: poise; 1 Pl = 10 poise. Water at 20 °C: η ≈ 1 × 10⁻³ Pa s. Honey: about 10 Pa s.
Temperature: liquids get less viscous when heated (hot oil flows easily). Gases get more viscous when heated, because their molecules move faster and mix momentum more.
Stokes' law
A small sphere of radius r moving slowly at speed v through a fluid of viscosity η feels a drag force:
F = 6π η r v
Drag grows with speed. It holds for small, slow spheres in streamline flow (like dust in air, small raindrops, or a steel ball in glycerine). You can check it by dimensions: η has [ML⁻¹T⁻¹], so η r v has [MLT⁻²], a force.
Terminal velocity
A ball of density ρ falls in a fluid of density σ. Three forces act:
- Weight = (4/3)πr³ρg, down
- Buoyancy = (4/3)πr³σg, up
- Viscous drag = 6πηrv, up, growing with v
At first the ball speeds up. Drag grows until the forces balance. Then acceleration is zero and speed stays fixed at the terminal velocity:
6πηr vt = (4/3)πr³(ρ − σ)g ⇒ vt = 2r²(ρ − σ)g / (9η)
Notice vt ∝ r². Big drops fall faster than tiny ones. If σ > ρ (air bubble in water), vt is negative: the bubble rises at a steady speed.
Streamline and turbulent flow; critical velocity
In streamline (steady) flow, every particle passing a given point has the same velocity. The paths are called streamlines; they never cross. Close streamlines mean fast flow.
When the speed is too high, the flow becomes turbulent: eddies and swirls appear, and velocity at a point keeps changing. Smoke rising from an incense stick is smooth first, then curls: streamline turning turbulent.
The Reynolds number decides which: Re = ρvd/η (d = pipe diameter). It has no unit.
- Re < 1000 (about): streamline
- Re > 2000 (about): turbulent
- In between: unsteady
The speed at which flow just turns turbulent is the critical velocity: vc = Rec η / (ρd). Thick, viscous liquids in thin pipes stay streamline up to higher speeds.
Equation of continuity
For a liquid that does not compress, the volume entering a pipe each second must leave it: A₁v₁ = A₂v₂. So where the pipe is narrow, the liquid moves faster. That is why pressing a garden hose's mouth makes the water shoot far.
Bernoulli's theorem
For steady, streamline flow of an ideal (non-viscous, incompressible) fluid, along a streamline:
P + ½ρv² + ρgh = constant
Each term is energy per unit volume: pressure energy, kinetic energy and potential energy. It is simply the work–energy theorem for a moving fluid: the work done by pressure at the ends turns into changes in kinetic and potential energy.
Derivation idea: in time Δt, a volume ΔV enters at section 1 and leaves at section 2. Work by pressure = (P₁ − P₂)ΔV. Change in KE = ½ρΔV(v₂² − v₁²). Change in PE = ρgΔV(h₂ − h₁). Setting work = ΔKE + ΔPE and dividing by ΔV gives Bernoulli's equation.
At the same height: fast flow ⇒ low pressure. A venturimeter uses this to measure flow speed from the pressure drop in a narrow throat.
Torricelli's law (speed of efflux)
A tank open at the top has a small hole at depth h below the surface. Both the surface and the jet are at air pressure P₀. The surface moves very slowly (wide tank). Bernoulli gives P₀ + ρgh = P₀ + ½ρv², so:
v = √(2gh)
This is the same speed as a stone falling freely through height h. If the tank is closed with gas pressure P above the liquid: v = √(2(P − P₀)/ρ + 2gh).
Dynamic lift
Dynamic lift is the force on a body moving through a fluid because of the flow around it.
- Aeroplane wing (aerofoil): its curved shape and small tilt make air flow faster over the top. Lower pressure on top, higher below: upward lift. Lift force ≈ ΔP × wing area.
- Spinning ball (Magnus effect): a spinning ball drags air around with it. On one side the spin adds to the airflow (faster, lower pressure); on the other it opposes (slower, higher pressure). The ball swerves. This is how a cricket or football curves.
Try it: the paper strip and the two glasses
1) Hold a paper strip under your lower lip and blow across the top: it rises (fast air, low pressure). 2) Drop a marble into a glass of water and another into a glass of cooking oil. Count how many seconds each takes to reach the bottom. The oil one is slower: more viscosity, smaller terminal velocity. In the 3D, step 3, find the speed at which the flow turns turbulent (Re ≈ 2000).
Key formulas and definitions
- F = η A (dv/dx); unit of η: Pa s
- Stokes' law: F = 6π η r v
- Terminal velocity: vt = 2r²(ρ − σ)g / (9η)
- Reynolds number: Re = ρ v d / η
- Continuity: A₁v₁ = A₂v₂
- Bernoulli: P + ½ρv² + ρgh = constant
- Torricelli: v = √(2gh)
- Lift ≈ ½ρ(v₁² − v₂²) × A
Worked examples
1. A metal plate of area 0.1 m² slides at 0.1 m/s over a 1 mm layer of oil on a table. The force needed is 0.2 N. Find η.
Step 1: dv/dx = 0.1 / 10⁻³ = 100 s⁻¹. Step 2: η = F / (A dv/dx) = 0.2 / (0.1 × 100). Answer: η = 0.02 Pa s.
2. Water flows through a pipe of area 20 cm² at 2 m/s. The pipe narrows to 5 cm². Find the speed in the narrow part.
Step 1: A₁v₁ = A₂v₂. Step 2: v₂ = 20 × 2 / 5. Answer: v₂ = 8 m/s.
3. Find the speed of water from a hole 5 m below the surface of an open tank. (g = 10 m/s²)
Step 1: v = √(2gh). Step 2: = √(2 × 10 × 5) = √100. Answer: 10 m/s.
4. A raindrop of radius 0.3 mm falls in air. η(air) = 1.8 × 10⁻⁵ Pa s, ρ(water) = 1000 kg/m³, ignore air density. Find the terminal velocity. (g = 9.8 m/s²)
Step 1: vt = 2r²ρg / (9η). Step 2: r² = 9 × 10⁻⁸ m². Step 3: numerator = 2 × 9 × 10⁻⁸ × 1000 × 9.8 = 1.764 × 10⁻³. Step 4: denominator = 9 × 1.8 × 10⁻⁵ = 1.62 × 10⁻⁴. Answer: vt ≈ 10.9 m/s. (Real drops this big are slower, because Stokes' law fails at such speeds, but the method is the one to learn.)
5. Water (η = 1 × 10⁻³ Pa s) flows at 0.1 m/s in a pipe of diameter 2 cm. Find Re. Is the flow streamline?
Step 1: Re = ρvd/η. Step 2: = 1000 × 0.1 × 0.02 / 10⁻³. Answer: Re = 2000. It is on the border; any faster and the flow turns turbulent.
6. Water flows horizontally in a pipe at 2 m/s where pressure is 1.5 × 10⁵ Pa. The pipe narrows and the speed becomes 6 m/s. Find the pressure there.
Step 1: Same height, so P₁ + ½ρv₁² = P₂ + ½ρv₂². Step 2: P₂ = P₁ − ½ρ(v₂² − v₁²) = 1.5 × 10⁵ − ½ × 1000 × (36 − 4). Step 3: = 1.5 × 10⁵ − 16000. Answer: P₂ = 1.34 × 10⁵ Pa.
7. Air flows over the top of a wing at 70 m/s and under it at 60 m/s. Wing area is 20 m², air density 1.2 kg/m³. Find the lift force.
Step 1: ΔP = ½ρ(v₁² − v₂²) = 0.6 × (4900 − 3600). Step 2: ΔP = 0.6 × 1300 = 780 Pa. Step 3: Lift = ΔP × A = 780 × 20. Answer: 15600 N, enough to hold up about 1.6 tonnes.
Common mistakes
- Thinking viscosity of every fluid falls with temperature. It falls for liquids but rises for gases.
- Writing vt ∝ r instead of vt ∝ r². Doubling the radius makes the terminal velocity four times.
- Using Bernoulli's equation for turbulent or very viscous flow. It is for steady, streamline, non-viscous, incompressible flow.
- Thinking fast flow means high pressure. At the same height, faster flow means lower pressure.