Fluid Mechanics

Big Picture

Fluids at rest obey simple pressure-depth rules and buoyancy principles. Fluids in motion obey conservation of mass (continuity) and conservation of energy (Bernoulli's equation). The key insight: pressure in a fluid acts in all directions, and a fluid cannot support a shear stress at rest — it deforms continuously, which is why dams leak if cracked and why a small force can lift a car with a hydraulic jack.

ELI-10: Explain It Like I'm 10

Water is a fluid. Air is a fluid. Fluids flow and take the shape of their container. When you dive into a pool, your ears hurt because water pressure increases with depth — all that water above you is pushing down. A boat floats because it pushes water out of the way, and the water pushes back up. Fast-moving air has lower pressure, which is how airplane wings generate lift and how a perfume sprayer works.

7.1 Fluid Statics — Pressure and Depth

Core Idea

Pressure in a static fluid increases linearly with depth because the fluid above exerts a weight.

Physics and Mathematics

• Density: ρ= m/V. SI unit: kg/m³. Water: ρ ≈ 1000 kg/m³.
• Pressure: P = F/A. SI unit: pascal (Pa). 1 Pa = 1 N/m².
• Pressure with depth: P = P₀ + ρgh, where P₀ is the pressure at the surface and h is depth (positive downward).

Atmospheric pressure at sea level: P_atm ≈ 1.01 × 10⁵ Pa = 101 kPa.

ELI-10: Explain It Like I'm 10

Swim to the bottom of a deep pool and your ears feel the pressure. Every 10 meters of water depth adds about one atmosphere of pressure on top of the air pressure already pushing down. The deeper you go, the more water is stacked above you, and the higher the pressure.

Key Takeaway

Pressure increases with depth: P = P₀ + ρgh. Pressure at a given depth is the same in all directions.

Worked Example: Pressure at the Bottom of a Lake

A freshwater lake is 25 m deep. What is the absolute pressure (total pressure) and the gauge pressure (pressure above atmospheric) at the bottom?

Given: h = 25 m, ρ_water = 1000 kg/m³, P_atm = 1.01 × 10⁵ Pa, g = 9.8 m/s².

Gauge pressure (pressure from the water alone):

P_gauge = ρg h = (1000)(9.8)(25) = 2.45 × 10⁵ Pa ≈ 2.4 atm

Absolute pressure (water + atmosphere):

P = P₀ + ρgh = 1.01 × 10⁵ + 2.45 × 10⁵ = 3.46 × 10⁵ Pa ≈ 3.4 atm

Why this matters: A dam wall at 25 m depth must withstand roughly 3.4 times the force per unit area that a wall at the surface experiences. This is why dam walls are thicker at the bottom — the pressure increases linearly with depth, so the structural demands grow the deeper you go.

7.2 Pascal's Principle

Core Idea

Pressure applied to an enclosed fluid is transmitted undiminished to all parts of the fluid.

P₁ = P₂   ⇒  F₁A₁ = F₂A₂

Hydraulic lift: A small force on a small piston creates a large force on a large piston because pressure is constant but area differs. The trade-off: the small piston must move a larger distance (A₁ d₁ = A₂ d₂, since fluid volume is conserved).

ELI-10: Explain It Like I'm 10

A hydraulic car lift uses Pascal's principle. You push down on a narrow cylinder with a small force. The pressure travels through the fluid to a wide cylinder under the car. Because the wide cylinder has a much bigger area, the upward force on the car is multiplied. You push down 50 cm; the car rises 5 cm. Same fluid volume moved, but a much bigger force where it counts.

Worked Example: Hydraulic Lift

A hydraulic lift has a small piston of radius r₁ = 2.0 cm and a large piston of radius r₂ = 15 cm. What force must be applied to the small piston to lift a 1200 kg car?

Step 1 — Compute piston areas:

A₁ = πr₁² = π(0.020)² = 1.26 × 10^-3 m² A₂ = πr₂² = π(0.15)² = 7.07 × 10^-2 m²

Step 2 — Force needed on the large piston:

F₂ = mg = (1200)(9.8) = 11 760 N

Step 3 — Apply Pascal's principle F₁/A₁ = F₂/A₂:

F₁ = F₂ A₁A₂ = 11 760 × 1.26 × 10^-37.07 × 10^-2 = 11 760 × 0.0178 = 209 N

That is about the weight of a 21 kg object — a force a person can easily apply.

Step 4 — Distance trade-off. To lift the car by d₂ = 5.0 cm, how far must the small piston move?

A₁ d₁ = A₂ d₂   ⇒  d₁ = d₂ A₂A₁ = 5.0 × 7.07 × 10^-21.26 × 10^-3 = 5.0 × 56.1 ≈ 280 cm = 2.8 m

Key insight: The hydraulic lift multiplies force by the area ratio (56× here) but divides displacement by the same factor. Work in = work out (ignoring friction): F₁ d₁ ≈ F₂ d₂. You trade distance for force — the same principle behind levers, just using fluid pressure instead of a rigid bar.

7.3 Archimedes' Principle — Buoyancy

Core Idea

A fluid exerts an upward buoyant force on any object immersed in it. The buoyant force equals the weight of the fluid displaced by the object.

F_B = ρ_fluid  V_displaced  g

• Object floats: F_B = mg_object.
• Object sinks: ρ_object > ρ_fluid.
• Object floats: ρ_object < ρ_fluid.

Apparent weight of a submerged object = true weight − buoyant force.

ELI-10: Explain It Like I'm 10

Lower a rock into water and it feels lighter. The water is pushing up on it. The upward push equals the weight of the water the rock pushed out of the way. If the object is less dense than water (like wood), the upward push is stronger than its weight, and it floats. If it is denser (like a rock), the upward push is weaker and it sinks. A steel ship floats because its hollow shape pushes aside a huge volume of water — its average density (steel + air) is less than water.

Worked Example: Fraction Submerged

A block of wood has density ρ_wood = 650 kg/m³ and floats in water (ρ_water = 1000 kg/m³). What fraction of its volume is submerged?

At equilibrium, the buoyant force balances the weight:

F_B = mg ρ_water  V_submerged  g = ρ_wood  V_total  g

Cancel g on both sides and solve for the fraction:

V_submergedV_total = ρ_woodρ_water = 6501000 = 0.65

Answer: 65% of the block is underwater. This is a general result — for any floating object at equilibrium, the fraction submerged equals the ratio of the object's density to the fluid's density.

Why "tip of the iceberg" is literal: For an iceberg (ρ_ice ≈ 917 kg/m³) floating in seawater (ρ_seawater ≈ 1025 kg/m³):

V_submergedV_total = 9171025 ≈ 0.895

About 90% of an iceberg is underwater. Only ~10% is visible above the surface — hence "the tip of the iceberg."

Apparent weight of a submerged object: A 5.0 kg rock (ρ_rock = 2700 kg/m³) is submerged in water. What does it appear to weigh?

First, find the rock's volume: V = m/ρ= 5.0/2700 = 1.85 × 10^-3 m³.

Buoyant force: F_B = ρ_water V g = (1000)(1.85 × 10^-3)(9.8) = 18.1 N.

True weight: mg = (5.0)(9.8) = 49.0 N.

Apparent weight: 49.0 - 18.1 = 30.9 N — the rock feels about 37% lighter underwater.

7.4 Fluid Dynamics — Continuity and Flow Rate

Ideal Fluid Assumptions

The equations in this chapter model an ideal fluid — a simplification that makes the math tractable while still capturing the essential physics for many real-world flows. An ideal fluid has four properties:

1. Incompressible — Density ρ is constant and does not change with pressure. This is an excellent approximation for liquids (water barely compresses) and holds for gases at low speeds (well below the speed of sound). At high speeds or with large pressure changes, gases compress and this assumption breaks down.

1. Nonviscous — The fluid has zero internal friction. Viscosity is a fluid's resistance to flow: honey is highly viscous, water has low viscosity, air has even less. Viscosity dissipates mechanical energy into heat. When viscosity matters (oil pipelines, blood flow in capillaries), Bernoulli's equation must be modified with a dissipation term.

1. Laminar flow — The fluid moves in smooth, orderly layers (streamlines) with no mixing between adjacent layers. Each particle follows a well-defined path. The opposite is turbulent flow, characterized by chaotic eddies, swirls, and unpredictable mixing. Turbulence dramatically increases flow resistance and invalidates Bernoulli's equation. The transition from laminar to turbulent flow is predicted by the Reynolds number — a dimensionless parameter that depends on fluid speed, density, viscosity, and the characteristic length scale of the flow.

1. Steady flow — At any given point in space, the fluid velocity does not change with time (∂v/∂t = 0). The flow pattern may vary from point to point, but at each point it stays constant. This means streamlines don't shift or wiggle over time.

When these assumptions hold (e.g., water through a smooth pipe at moderate speed, air flowing gently over a wing), the simple equations work remarkably well. When they break down (turbulent airflow behind a truck, viscous crude oil in a pipeline, compressible flow at supersonic speeds), more advanced fluid dynamics — the Navier-Stokes equations — is required.

Core Idea

For an incompressible fluid flowing through a pipe, the volume flow rate is constant.

Flow rate: Q = Av, where A is cross-sectional area and v is fluid speed. SI unit: m³/s.

Continuity equation: A₁ v₁ = A₂ v₂

Where the pipe narrows, the fluid speeds up. This is conservation of mass — the same amount of fluid must pass each point per unit time.

ELI-10: Explain It Like I'm 10

Put your thumb partly over a garden hose. The water speeds up and shoots farther. The same amount of water has to get through a smaller opening each second, so it must move faster. This is the continuity equation: narrow = fast, wide = slow.

Worked Example: Continuity in a Garden Hose

Water flows through a garden hose of diameter d₁ = 2.0 cm at a speed of v₁ = 1.5 m/s. The nozzle narrows to diameter d₂ = 0.80 cm. What is the exit speed?

Step 1 — Compute cross-sectional areas (or skip straight to the ratio):

A₁ = πr₁² = π(0.010)² = 3.14 × 10^-4 m² A₂ = πr₂² = π(0.0040)² = 5.03 × 10^-5 m²

Step 2 — Apply the continuity equation:

A₁ v₁ = A₂ v₂   ⇒  v₂ = v₁ A₁A₂ = 1.5 × 3.14 × 10^-45.03 × 10^-5 = 1.5 × 6.25 = 9.4 m/s

Step 3 — Check with diameter-ratio shortcut. Since A ∝ r² ∝ d²:

A₁A₂ = (d₁d₂)² = (2.00.80)² = (2.5)² = 6.25 v₂ = 6.25 × 1.5 = 9.4 m/s

Same answer without computing either area individually. The water exits the nozzle over six times faster than it entered — which is why a nozzle turns a gentle stream into a powerful jet.

7.5 Bernoulli's Equation

Core Idea

Bernoulli's equation expresses conservation of energy for an ideal fluid (incompressible, non-viscous, laminar flow) along a streamline.

P₁ + 12ρv₁² + ρg h₁ = P₂ + 12ρv₂² + ρg h₂

Interpretation:

• P = pressure energy per unit volume
• 12ρv² = kinetic energy per unit volume
• ρg h = gravitational potential energy per unit volume

Key consequences:

• Where speed is high, pressure is low (and vice versa) — the Venturi effect.
• Airplane wings: faster air over the curved top → lower pressure above → lift.
• The equation assumes no viscosity and incompressible flow — it does not apply in turbulent or viscous situations.

ELI-10: Explain It Like I'm 10

Blow across the top of a piece of paper. The paper rises. Fast-moving air has lower pressure than still air, so the higher pressure underneath pushes the paper up. This is Bernoulli's principle in action: faster fluid → lower pressure. Airplane wings, atomizers, and curveballs in baseball all use this idea.

Worked Example: Torricelli's Law (Tank with a Hole)

A large open tank of water has a small hole in its side a distance h = 3.0 m below the water surface. The tank is open to the atmosphere at the top. Assuming ideal fluid behavior, with what speed does water exit the hole?

Apply Bernoulli between point 1 (water surface) and point 2 (hole):

P₁ + 12ρv₁² + ρg h₁ = P₂ + 12ρv₂² + ρg h₂

Simplify using the problem conditions:

• Both the surface and the hole are open to the atmosphere: P₁ = P₂ = P_atm. These cancel.
• The tank is "large," meaning the water surface area is much greater than the hole area. By continuity, the surface drops very slowly: v₁ ≈ 0.
• Define h = h₁ - h₂ = 3.0 m (the vertical distance from the surface to the hole).

Substituting and canceling:

ρg h = 12 ρv₂²

Cancel ρ and solve for v₂:

v₂ = √2gh = √2(9.8)(3.0) = √58.8 ≈ 7.7 m/s

This is Torricelli's law: v = √2gh. Remarkably, the water exits at the same speed as an object that fell freely from height h. The density canceled — water, mercury, or any ideal fluid would exit at the same speed from the same depth.

Real-world caveat: In practice, the exit speed is slightly lower due to viscosity and the vena contracta effect — the fluid jet narrows just after exiting the hole, reducing the effective area. The ideal Torricelli result is an upper bound; actual speeds are typically 5–15% lower.

Proportional Reasoning with Torricelli's Law

Since v ∝ √h in Torricelli's law:

• Doubling the depth multiplies exit speed by √2 ≈ 1.41 (a 41% increase).
• Quadrupling the depth doubles the exit speed (√4 = 2).
• Halving the depth reduces speed to about 71% of the original (√0.5 ≈ 0.707).

This square-root scaling is the same pattern seen in free fall: v = √2gh for an object dropped from rest. Torricelli's law is a direct consequence of energy conservation — gravitational potential energy at the surface converts to kinetic energy at the exit.

Common Misconception: The "Longer Path" Theory of Lift

A widely repeated but physically incorrect explanation of airplane lift goes like this: air molecules split at the leading edge of a wing, the ones going over the curved top must travel a longer path than those going under the flat bottom, so to "meet up" at the trailing edge they must go faster, and by Bernoulli faster air means lower pressure — producing lift. This is the "equal transit time" or "longer path" fallacy.

Why this is wrong:

• There is no physical law requiring air molecules that split at the leading edge to reunite at the trailing edge. In fact, wind tunnel measurements and computational simulations show that air traveling over the top of a wing typically reaches the trailing edge before the air traveling underneath — they do not "meet up."
• Symmetric airfoils (identical top and bottom curvature) generate lift when angled upward (positive angle of attack). The equal-transit-time theory cannot explain this, because both sides have the same path length.
• Inverted flight is possible — a plane can fly upside down. If lift depended solely on the upper surface being more curved, inverted flight would push the plane downward, not upward.
• The real mechanism of lift is more nuanced: the wing deflects air downward (Newton's third law — the wing pushes air down, air pushes the wing up), and the pressure distribution around the wing is shaped by both its curvature and its angle of attack. Bernoulli's principle does contribute — faster airflow over the top does lower pressure — but the reason the air goes faster is not because it has a longer path to cover. The wing's shape and angle of attack alter the entire flow field, creating a circulation pattern that accelerates air over the top and decelerates it underneath.

What to remember for physics class: Bernoulli's principle says faster fluid → lower pressure, and that is correct and testable. The flawed part is only the equal-transit-time explanation for why the air goes faster. For most introductory physics problems, it is enough to know that airspeed differences create pressure differences, and those pressure differences generate a net upward force on the wing.

Important Limitation

Bernoulli's equation is an idealization. Real fluids have viscosity (internal friction), which dissipates energy and makes Bernoulli inaccurate over long distances. For introductory physics, treat Bernoulli as a model that works well for smooth, short flows.

7.6 Proportional Reasoning in Fluid Mechanics

Proportional reasoning — analyzing how a quantity changes when its inputs scale — is one of the most powerful shortcuts in fluid mechanics. Instead of recomputing from scratch every time, you can often extract the answer from how variables relate.

Pressure with Depth: P = P₀ + ρg h

• If depth doubles, gauge pressure (ρg h) doubles — it scales linearly with h.
• If the fluid is replaced with one of twice the density (e.g., saltwater ρ ≈ 1030 kg/m³ vs. freshwater), gauge pressure at the same depth increases by the same factor ( × 2).
• Absolute pressure does not double when depth doubles because the constant atmospheric term P₀ is still there. Gauge pressure doubles; absolute pressure less than doubles.

Pascal's Principle: F₁/A₁ = F₂/A₂

• Force multiplication equals the area ratio A₂/A₁. If the large piston has 10× the radius, it has 10² = 100 ×  the area, so F₂ = 100 F₁.
• The distance trade-off is the inverse: d₁ = (A₂/A₁) d₂. The small piston moves 100 ×  farther than the large one.

Buoyancy: F_B = ρ_f V_disp g

• The fraction submerged for a floating object depends only on the density ratio: V_sub/V_total = ρ_obj/ρ_fluid. Neither the object's mass nor volume matters individually — only the density ratio.
• If the fluid density doubles (replace water with a much denser fluid), the fraction submerged halves for the same floating object.

Continuity: A₁ v₁ = A₂ v₂

• Speed is inversely proportional to area: v ∝ 1/A. If area halves, speed doubles.
• Since area scales as radius squared (A ∝ r²), halving the radius quarters the area ( × 1/4) and quadruples the speed ( × 4).
• Diameter-ratio shortcut: v₂/v₁ = (d₁/d₂)². No need to compute areas.

Bernoulli / Torricelli: v = √2 g h

• Exit speed scales as the square root of depth: v ∝ √h. This is a sub-linear relationship — doubling depth increases speed by only ~41%.
• Fluid density cancels out: mercury and water exit at the same speed from the same depth.

The general pattern: identify whether the relationship is linear (y ∝ x), inverse (y ∝ 1/x), quadratic (y ∝ x²), or square-root (y ∝ √x), and scale accordingly. This skill is transferable far beyond fluid mechanics — it is essential across all of physics.

Topic Summary

• Pressure = force/area. Increases with depth: P = P₀ + ρgh.
• Pascal's principle = pressure applied to an enclosed fluid is transmitted equally everywhere. Force multiplication: F₂/F₁ = A₂/A₁.
• Archimedes' principle = buoyant force equals the weight of displaced fluid. Floating objects: V_sub/V_total = ρ_obj/ρ_fluid.
• Continuity equation A₁ v₁ = A₂ v₂: narrower → faster (inversely proportional to area).
• Bernoulli's equation = energy conservation along a streamline. Higher speed → lower pressure.
• Torricelli's law: v = √2gh — fluid exits a hole at the same speed as an object in free fall from the same height.
• Ideal fluid assumptions (incompressible, nonviscous, laminar, steady) are required for these equations to hold exactly.
• Real fluids have viscosity, which dissipates energy and limits Bernoulli's applicability. The "equal transit time" explanation of airplane lift is a common misconception — the physics of lift is more about flow deflection and pressure distribution than path-length differences.

Essential Equations

Concept Check

1. Why does a steel ship float when a solid steel bar sinks?
2. If a pipe narrows to half its diameter, by what factor does the fluid speed change?
3. Two identical cups are filled — one with water, one with saltwater. In which liquid is the buoyant force on an identical submerged object greater? Why?
4. Explain why a roof can lift off a house in a hurricane using Bernoulli's principle.
5. Why does Bernoulli's equation not accurately describe the flow of honey through a narrow tube?
6. A dam is 40 m tall and holds back a reservoir. Why does the dam wall need to be thicker at the bottom than at the top? Use the pressure-depth equation in your answer.
7. A hydraulic lift has pistons with radii of 3 cm and 18 cm. By what factor is the force multiplied? How far must the small piston move to lift the large one by 1 cm?
8. A block of density 800 kg/m³ floats in water. What fraction of it is submerged? If it were placed in oil (ρ= 900 kg/m³), would more or less be submerged?
9. A water tank has a hole punched at a depth of 1.0 m. If the hole is moved to 4.0 m depth, by what factor does the exit speed change? (Use proportional reasoning.)
10. Briefly explain why the "equal transit time" explanation of airplane lift is physically incorrect.

Open Educational References

• OpenStax, College Physics, Chapters 11–12: Fluid Statics, Fluid Dynamics
• OpenStax, University Physics, Volume 1, Chapter 14: Fluid Mechanics