Fluid Mechanics

Dive to the bottom of a pool and your ears hurt. Nothing is touching them except water, yet the water presses harder the deeper you go. Why, and by how much?

Fluids#

In solids, strong forces between molecules keep both shape and size fixed. In liquids those forces are weaker: the shape changes easily, though the volume (and density) of a given mass is hard to change. In gases the forces are tiny, and both shape and density change easily. Liquids and gases together are called fluids: things that flow.

We will mostly treat liquids, and model them as

• incompressible: the density stays the same whatever the pressure;
• nonviscous: neighbouring layers exert no sideways (tangential) forces on each other, so one part of the liquid pushes on another only perpendicular to the surface between them.

Pressure at a point#

In a fluid at rest this doesn’t depend on how the little area is tilted, so pressure has a value at each point but no direction: it is a scalar. Its SI unit is the newton per square metre, called the pascal (1 Pa = 1 N/m²).

How pressure changes with height#

Take a thin horizontal slab of fluid of area ΔS and thickness dz, with pressure P on its lower face and P + dP on its upper face. Three vertical forces act on it: P ΔS upward from below, (P + dP) ΔS downward from above, and its weight ρ(ΔS dz)g. In equilibrium

Going up by dz lowers the pressure by ρg dz. If the density is the same everywhere, integrating over a depth h gives

Now take two points at the same level. A thin horizontal cylinder of fluid joining them feels P₁ΔS one way and P₂ΔS the other, and nothing else horizontally. In equilibrium $P_1 = P_2$:

A mechanic lifts a car with one hand on a lever. The trick is a liquid that passes on every bit of extra pressure, undiminished, to every corner of the machine.

Pascal’s law#

We just showed that the pressure difference between two points in a liquid at rest is ρgz, depending only on their height difference z. Suppose we somehow raise the pressure at one point. Since the liquid is incompressible, z doesn’t change, so the pressure difference between that point and any other must stay ρgz. The pressure everywhere else must therefore rise by the same amount.

Fit a piston of area A to a flask full of liquid and push it with force F. The pressure just under the piston rises by F/A, and so does the pressure at every other point in the flask.

The hydraulic lift#

Two cylinders of different cross-sections, A₁ (small) and A₂ (large), are joined by a pipe and filled with oil; each has a piston. A valve lets oil pass from the small cylinder into the large one. Push the small piston with F₁: the pressure everywhere rises by F₁/A₁, so the large piston feels an extra upward force

With A₂ much larger than A₁, a small push supports a heavy car. Hydraulic presses, car brakes and excavator arms all work this way.

Above you stands a column of air hundreds of kilometres tall. Its weight presses on everything: on this page, on your skin, on the surface of every lake. We just don’t notice, because the push is the same from all sides.

The weight of the air#

The atmosphere extends to roughly 200 km above the Earth. The force per unit area it exerts on any surface (always perpendicular to that surface) is the atmospheric pressure. At the top of the atmosphere it is zero; going down, it builds up as $\int\rho g\,dz$. Air is so light that over everyday height differences the pressure hardly changes: we can speak of “the” atmospheric pressure of a town without worrying whether we measure it at street level or on a rooftop.

Torricelli’s barometer#

Fill a glass tube about a metre long with mercury, close the open end with a thumb, turn it upside down into a dish of mercury and remove the thumb. The mercury falls a little and stops, leaving a vacuum at the top of the tube.

At the top of the column (A) the pressure is zero (vacuum). A point B inside the tube at the level of the dish’s surface is at the same height as the open surface C, where the pressure is atmospheric, so $P_B = P_0$. If B is a height H below A,

Atmospheric pressure is often quoted as the height of this mercury column. At sea level it is about 76 cm:

If the tube were shorter than 76 cm, the mercury would not fall at all; the glass would press down on it to make up the difference.

The manometer#

To measure the pressure of a gas in a closed vessel, connect it to one arm of a U-tube containing liquid, leaving the other arm open to the air. Points at the same level in the liquid have the same pressure, so

A steel ship weighing thousands of tonnes floats, while a steel nail sinks. Lift someone in a swimming pool and they feel lighter. Both come down to one principle, which follows directly from the pressure forces we already understand.

The buoyant force#

When a body is dipped partly or wholly into a fluid, the fluid pushes on every bit of its surface, perpendicular to the surface, with a force equal to the local pressure times the area. Deeper parts are pushed harder. The resultant of all these contact forces is the buoyant force, B.

Archimedes’ principle#

This isn’t a separate law of nature; it follows from Newton’s laws. Imagine removing the body and filling the space it occupied with the same fluid. The fluid is now uniform and at rest, so this “substitute” fluid is in equilibrium. Only two forces act on it: its weight mg and the push B of the surrounding fluid. Hence

The surrounding fluid neither knows nor cares what fills the space: the substitute fluid and the original body have exactly the same boundary, so the pressure forces on them are identical. So the body feels the same B.

Floating#

If the buoyant force equals the body’s weight, the body stays in equilibrium: it floats. A body whose average density is less than the fluid’s floats with only part of it submerged, just enough to displace its own weight of fluid:

A ship floats because its hull encloses lots of air, making its average density much less than water’s.

Carry a full cup of tea while walking quickly and it slops over the back edge. Our hydrostatic results assumed the liquid was at rest in an inertial frame. What changes when the container accelerates?

A liquid in a lift#

Pressure difference

A beaker of liquid rides in a lift accelerating upward at a₀. Take a vertical cylinder of liquid of cross-section ΔS between a point A and a point B a height z above it. The forces on it are $P_A\Delta S$ up, $P_B\Delta S$ down and its weight $(\Delta S)z\rho g$ down, and it accelerates upward with the lift:

Buoyant force

Replace an immersed body by liquid again. The substitute liquid accelerates up with everything else, so $B - mg = ma_0$:

A liquid accelerating sideways#

Now let the container accelerate horizontally at a₀. Take two points A and B on the same horizontal line, a distance l apart along the acceleration (A behind, B in front). The horizontal cylinder of liquid between them must be pushed forward:

So points at the same level no longer have equal pressure. There is no vertical acceleration, so vertically $P = P_0 + \rho gh$ still holds: the liquid must be deeper above A than above B. Writing the depths h₁ and h₂,

A waterfall crashing onto rocks, smoke curling from a wood fire, clouds drifting: every droplet obeys Newton’s laws, yet predicting each one is hopeless. To make progress we focus on the simplest kind of flow.

Steady and turbulent flow#

Watch the liquid flowing slowly through a glass tube and concentrate on one point A. If every particle arriving at A has the same speed and direction, and every particle that passes A goes on to the same point B, the flow is steady (or streamline). The velocity at each point stays the same over time, though it may differ from point to point.

Push the liquid through faster and the flow can become turbulent: particles passing the same point have different, erratically changing velocities. A fast river or a high waterfall is turbulent.

Streamlines and tubes of flow#

Two streamlines can never cross: a particle at the crossing would have to move in two directions at once. Draw streamlines through every point on the edge of a small area S, and they enclose a tube of flow. Fluid never crosses its walls, even though no physical wall exists. A pipe carrying a slow, steady flow is itself one tube of flow.

The ideal fluid#

To keep the mathematics simple we assume the fluid is

• incompressible: constant, uniform density (very good for liquids);
• nonviscous: no internal friction between layers, so no mechanical energy is lost;
• in irrotational flow: the fluid particles have no net spin (unlike water stirred round in a bucket, or the small whirlpools in a river).

The equation of continuity#

Fluid entering a tube of flow at one cross-section must leave through any other, since none crosses the walls. At section 1 (area A₁, speed v₁), in a time Δt the fluid in a cylinder of length v₁Δt passes through: a volume A₁v₁Δt. Similarly A₂v₂Δt leaves at section 2. For an incompressible fluid these are equal:

Blow between two hanging sheets of paper and they swing together, not apart. Fast-moving fluid is at lower pressure. Bernoulli’s equation makes this precise, and it is nothing more than the work–energy theorem applied to a flowing liquid.

Setting up#

Take steady, irrotational flow of an ideal (incompressible, nonviscous) liquid of density ρ through a tube whose cross-section and height vary. Follow the liquid between sections A and B. At A: area A₁, speed v₁, pressure P₁, height h₁. At B: A₂, v₂, P₂, h₂. In a short time Δt this liquid moves forward so that it lies between A′ and B′, where $AA' = v_1\Delta t$ and $BB' = v_2\Delta t$. By continuity the slices AA′ and BB′ have the same volume and the same mass:

Work done on the liquid#

• The liquid behind pushes forward with P₁A₁ through a distance v₁Δt: $W_1 = P_1A_1v_1\Delta t = P_1\Delta m/\rho$.
• The liquid ahead pushes back with P₂A₂: $W_2 = -P_2\Delta m/\rho$.
• Gravity: the net effect is as if the slice Δm were lifted from h₁ to h₂, so $W_3 = \Delta m\,g h_1 - \Delta m\,g h_2$.
• The walls push perpendicular to the flow and do no work.

Change in kinetic energy#

The flow is steady, so the liquid between A′ and B (common to both moments) has the same kinetic energy before and after. The change is just that of the slice moving from AA′ to BB′: $\tfrac12\Delta m\,v_2^2 - \tfrac12\Delta m\,v_1^2$.

Setting total work equal to the change in kinetic energy and dividing by Δm/ρ:

One equation, many uses: it recovers hydrostatics, predicts how fast a tank drains, measures flow in a pipe, powers a perfume spray, and explains why a cricket ball swings.

Hydrostatics as a special case#

Put v₁ = v₂ = 0 in Bernoulli’s equation:

exactly the pressure–depth law of a fluid at rest.

Speed of efflux: Torricelli’s theorem#

A wide tank of cross-section A₁, open at the top, has a small hole of area A₂ a depth h below the free surface. Both the surface and the jet are at atmospheric pressure P₀. Bernoulli between them:

With continuity $A_1v_1 = A_2v_2$ and $A_2 \ll A_1$, the surface barely moves (v₁ ≈ 0), giving

The venturi meter#

A venturi tube has a smooth constriction (the throat). Liquid speeds up in the throat and its pressure drops. Two vertical side tubes, one at the wide part (area A₁) and one at the throat (A₂), show a level difference h, with $P_1 - P_2 = \rho gh$. Bernoulli (same height) gives $P_1 - P_2 = \tfrac12\rho(v_2^2 - v_1^2)$, so

The aspirator (sprayer)#

In a sprayer, air pushed by a piston along a barrel rushes through a narrow nozzle. Its speed there is high, so its pressure is low. A thin tube dips from the nozzle into the liquid below; atmospheric pressure on the liquid’s surface pushes liquid up the tube into the fast airstream, which breaks it into a fine spray. Perfume atomisers, paint sprayers and old car carburettors work this way.

Why a spinning ball swings#

Swing bowlers make a cricket ball drift sideways out of its original plane of motion. Work in the frame of the ball: air streams past it. If the ball spins, it drags the air near it round. On one side the drag adds to the airflow, so the air there moves faster and its pressure drops; on the other side it opposes the airflow, so the air is slower and the pressure higher. The pressure difference pushes the ball sideways, bending its path.