Fluids

A steel ship weighing many thousands of tonnes floats, while a steel bolt sinks at once. A mechanic lifts a car with one hand on a pump handle. Both follow from one idea: a fluid pushes back with pressure, and pressure depends on depth and area in ways this unit makes exact.

The differences between solids, liquids, and gases come from how strongly their particles interact. A fluid 流体 (liquid or gas) has no fixed shape – it flows because its particles move past one another. Density 密度 is mass per unit volume:

It depends on the substance and its state. An object sinks or floats depending on how its density compares with the surrounding fluid's – less dense floats, more dense sinks. An ideal fluid 理想流体 is incompressible 不可压缩 (constant density, whatever the pressure) and has no viscosity 黏度 (internal friction) – the model AP uses throughout.

Worked example (density). A $0.27\ \text{kg}$ metal block measures $5.0\ \text{cm}\times4.0\ \text{cm}\times5.0\ \text{cm}$. Its volume is $100\ \text{cm}^3=1.0\times10^{-4}\ \text{m}^3$, so $\rho=\dfrac{0.27}{1.0\times10^{-4}}=2700\ \text{kg/m}^3$: aluminium. Cut the block in half and the density stays the same: density belongs to the material, not to the size of the piece.

Pressure 压强 is the perpendicular 垂直 force per unit area, a scalar 标量 measured in pascals 帕斯卡 (Pa):

In a fluid at rest, pressure increases with depth 深度 because of the weight of fluid above: where $P_0$ is the pressure at the surface and $h$ is the depth. Pressure acts equally in all directions at a point and pushes perpendicular to any surface.

Where pressure comes from. A fluid is made of particles moving in all directions. Each one that hits a wall bounces off and gives the wall a tiny push. Pressure is the total effect of all those collisions, spread over the area. Because the pushes come from every direction, pressure has no direction of its own: it is a scalar, and the force it makes on a surface is always perpendicular to that surface.

Squeezing a liquid hardly changes its volume. An ideal (incompressible) fluid keeps the same volume and density whatever the pressure, which is why $\rho$ in $\rho gh$ can be treated as constant all the way down.

Distinguish the two pressures AP asks about: the gauge pressure 表压 is pressure above the surrounding atmospheric pressure 大气压强. For a stationary liquid with its surface open to the atmosphere, $P_{\text{gauge}}=\rho g h$. The absolute pressure 绝对压强 is measured from vacuum, $P=P_{\text{atm}}+P_{\text{gauge}}$. A tyre gauge reading "$200\ \text{kPa}$" is gauge pressure; the air inside is really at about $300\ \text{kPa}$ absolute.

Worked example. Find the total pressure on a diver $10\ \text{m}$ below the surface of water ($\rho=1000\ \text{kg/m}^3$, surface pressure $P_0=1.0\times10^{5}\ \text{Pa}$):

Every $10\ \text{m}$ of water adds roughly one extra atmosphere of pressure. Notice the pressure does not depend on the shape or width of the container, only on the depth.

Worked example (a hydraulic lift). Pressure applied to an enclosed fluid is passed on equally throughout it (Pascal's principle 帕斯卡原理). A hydraulic lift has a small piston of area $0.0010\ \text{m}^2$ and a large piston of area $0.10\ \text{m}^2$. What force on the small piston holds up a $1500\ \text{kg}$ car?

The force is 100 times smaller, but nothing is free: to raise the car $0.10\ \text{m}$, the small piston must move $10\ \text{m}$. The work done on each piston is the same, $1470\ \text{J}$.

An object in a fluid feels an upward buoyant force 浮力 equal to the weight of the fluid it displaces – Archimedes' principle 阿基米德原理:

Combine this with Newton's laws: the object floats when buoyancy balances weight, sinks when weight wins, and rises when buoyancy wins. A floating object displaces exactly its own weight of fluid.

Worked example. A block of density $600\ \text{kg/m}^3$ floats in water ($1000\ \text{kg/m}^3$). What fraction is under the surface? For floating, the buoyant force equals the weight, so $\rho_{\text{fluid}}\,g\,V_{\text{sub}}=\rho_{\text{object}}\,g\,V$:

So $60\%$ sits below the water – the same reason most of an iceberg (density $\approx 900$) hides underwater.

Worked example (apparent weight). A $2.0\ \text{kg}$ rock with volume $8.0\times10^{-4}\ \text{m}^3$ hangs from a spring scale, fully under water. Its weight is $19.6\ \text{N}$. The buoyant force is $(1000)(9.8)(8.0\times10^{-4})=7.8\ \text{N}$. The rock is at rest, so the scale reads $19.6-7.8=11.8\ \text{N}$: this is its apparent weight 视重. The rock seems lighter, though its weight has not changed.

The buoyant force is not a new kind of force. The fluid particles push on every surface of the object, and the pushes on the bottom are larger than those on the top, because the bottom is deeper. The net result of all those pushes is an upward force: the buoyant force.

Worked example (a block released under water). A swimmer holds a block of mass $m$ and volume $V$ fully under water of density $\rho$ and lets go. Two forces act on it: the weight $mg$ down and the buoyant force $\rho Vg$ up. Newton's second law, with up positive:

In salt water, which is denser than fresh water, the buoyant force is larger but the weight is the same. So the net upward force is larger, and the block accelerates upward faster: $a_{\text{salt}}>a_{\text{fresh}}$. The formula agrees: a larger $\rho$ gives a larger $a$.

For an ideal fluid flowing steadily, two conservation ideas apply:

• Fluid flows because of a pressure difference. Fluid moves from a region of higher pressure toward lower pressure, just as an object accelerates toward the side with the smaller force on it.
• Continuity 连续性 (conservation of mass 质量守恒): for steady incompressible flow, the volume flow rate 体积流量 is constant, so $A_1 v_1 = A_2 v_2$. A narrower pipe forces faster flow.
• Bernoulli's equation 伯努利方程 (conservation of energy 能量守恒 per volume): for steady, incompressible, non-viscous flow along a streamline 流线,
  $$P+\tfrac{1}{2}\rho v^2+\rho g y = \text{constant}.$$

Together they explain why fluid speeds up and its pressure drops where a pipe narrows or where flow is fastest.

Worked example. Water flows at $2.0\ \text{m/s}$ through a pipe of cross-section $0.010\ \text{m}^2$, then enters a narrower section of $0.0040\ \text{m}^2$. By continuity the speed there is

By Bernoulli's equation this faster stream is at lower pressure – the effect that lifts an aeroplane wing and pulls two passing ships together.

Why continuity holds. In a pipe that is full of an incompressible fluid, every kilogram that enters one end must leave the other: the fluid cannot pile up or thin out inside. In time $\Delta t$, a length $v\,\Delta t$ of fluid passes any point, a volume $Av\,\Delta t$. So the volume flow rate is $Av$, and the mass flow rate $\rho Av$, and both are the same everywhere along the pipe.

Worked example (a fountain). Water leaves a nozzle of radius $r_0$ at angle $\theta_0$ above the horizontal, and a droplet rises a height $h_1$. Find the volume flow rate.

• At the top, the vertical velocity is zero, so the vertical part of the launch velocity is $v_0\sin\theta_0=\sqrt{2gh_1}$.
• Launch speed: $v_0=\dfrac{\sqrt{2gh_1}}{\sin\theta_0}$.
• Flow rate: $Q=Av_0=\dfrac{\pi r_0^2\sqrt{2gh_1}}{\sin\theta_0}$.

If the nozzle is made narrower with the same flow rate, the water must leave faster ($Av$ is fixed and $A$ is smaller), so it rises higher: $h_2>h_1$. This is why putting your thumb over the end of a garden hose makes the water spray farther.

Continue the example with Bernoulli's equation. The pipe is level, so the height terms cancel:

The pressure in the narrow section is about $10\ \text{kPa}$ lower, roughly a tenth of an atmosphere.

Worked example (a draining tank). Water leaves a small hole $1.8\ \text{m}$ below the surface of an open tank. The surface and the jet are both at atmospheric pressure, and the surface barely moves, so Bernoulli's equation reduces to $\rho g h=\tfrac12\rho v^2$:

This is Torricelli's result 托里拆利定理: the water leaves as fast as if it had fallen freely from the surface.

Exam skill. Choose the right law by what changes. If the pipe changes width, start with continuity ($A_1v_1=A_2v_2$) to get the speeds; if you then need a pressure, feed those speeds into Bernoulli. Watch the height term $\rho g y$ only when the pipe also changes level.

Testing a model with a straight-line graph. Many lab questions give two models and ask which one fits. A group measures the force $F_{\max}$ that breaks plastic rods of different radius $r$. Model A says $F_{\max}\propto r$; model B says $F_{\max}\propto r^2$ (proportional to the cross-sectional area $\pi r^2$).

• Plot $F_{\max}$ against $r$: the points curve upward, so model A does not fit.
• Plot $F_{\max}$ against $r^2$: the points lie on a straight line through the origin, which is exactly what model B predicts.
• In words: doubling the radius makes the rod about four times as strong, not twice.

To get good data, test several rods of each radius and average, measure each radius with calipers in a few places, and pull slowly with a force sensor so the reading at the moment of breaking is not missed.

Answer each before reading the answers below. Take $g=9.8\ \text{m/s}^2$ and $P_0=1.0\times10^5\ \text{Pa}$.

1. Find the gauge pressure and the absolute pressure $25\ \text{m}$ below the surface of sea water, density $1030\ \text{kg/m}^3$.
2. Ice of density $920\ \text{kg/m}^3$ floats in sea water of density $1030\ \text{kg/m}^3$. What fraction of the ice is under the surface?
3. A hydraulic press has pistons of area $2.0\ \text{cm}^2$ and $400\ \text{cm}^2$. What force can a $50\ \text{N}$ push on the small piston produce?
4. Water flows through a pipe whose radius halves at a narrowing. By what factor does the speed change?
5. A student says: "Steel is denser than water, so a steel ship should sink." Correct the student.

1. Gauge: $(1030)(9.8)(25)=2.5\times10^5\ \text{Pa}$. Absolute: $1.0\times10^5+2.5\times10^5=3.5\times10^5\ \text{Pa}$.
2. $920/1030=0.89$, so about $89\%$ is under the surface.
3. $F_2=F_1\dfrac{A_2}{A_1}=50\times\dfrac{400}{2.0}=1.0\times10^4\ \text{N}$.
4. Halving the radius divides the area by 4, so by continuity the speed is 4 times larger.
5. What matters is the average density of the whole ship, including the air inside the hull. The hull pushes aside a large volume of water; the ship floats when the weight of that displaced water equals its own weight. A solid steel bolt cannot displace enough water, so it sinks.

• Pressure with depth is $P=P_0+\rho g h$ — it depends on depth only, not the container's shape or width.
• Buoyant force = weight of fluid displaced ($\rho_{\text{fluid}}\,gV_{\text{disp}}$); a floating object displaces its own weight, so the fraction submerged is $\rho_{\text{object}}/\rho_{\text{fluid}}$.
• Compare densities to predict floating vs sinking; a floating object is in equilibrium 平衡 (buoyancy = weight), not weightless.
• Use continuity $A_1v_1=A_2v_2$: a narrower pipe means faster flow.
• Bernoulli: where a fluid flows faster its pressure is lower (wing lift, spray).

Common mistakes

• Using the volume of the whole object in $F_b = \rho g V$ for a floating object. Use the submerged volume, and the density of the fluid, not of the object.
• Mixing gauge and absolute pressure. $\rho g h$ alone is the gauge pressure; add $P_0$ for the absolute pressure.
• Making the pressure at a depth depend on the width or shape of the container. Only the depth below the surface (and the fluid's density) matter.
• Writing $A_1 v_1 = A_2 v_2$ with radii instead of areas. Area goes as $r^2$, so halving the radius makes the flow 4 times faster.
• Saying that faster-moving fluid has higher pressure. In Bernoulli's equation, where the speed rises (same height), the pressure falls.
• Saying a hydraulic lift multiplies energy. It multiplies force; the small piston moves further, so the work in equals the work out.
• Saying a denser fluid pushes a submerged block up with the same force. The buoyant force is $\rho Vg$, so it is larger in a denser fluid, while the weight stays the same.
• Thinking a narrower nozzle gives less water. With the same flow rate, a smaller area means a faster jet, not less water per second.
• Giving pressure a direction. Pressure is a scalar; the force it makes on a surface is perpendicular to that surface.