AP Physics 1 · Unit 8: Fluids ·  Lesson 8.4

Deep Dive: Fluids and Conservation Laws

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
8.4.A.1Concept

A Pressure Difference Drives Flow

Fluids don't move on their own — something has to push them. That something is a difference in pressure between two locations. Fluid flows from high pressure toward low pressure, the same way any other object accelerates in the direction of a net force.

💡This is a direct extension of Lesson 8.2 and 8.3: pressure differences are exactly what created the buoyant force. Here, the same idea explains why fluid moves through a pipe, out of a syringe, or across a pressure gradient in the atmosphere.
8.4.A.1.i–ii8.4.A.2Math

The Continuity Equation

For a tube that's open at both ends and completely filled with an incompressible fluid, matter can't pile up or disappear anywhere inside it. That means the rate at which matter enters must exactly equal the rate at which it exits — this is just conservation of mass.

The rate at which matter flows past any cross-section is proportional to that section's area and the fluid's speed there. Setting the rate in equal to the rate out gives the continuity equation:

A₁v₁ = A₂v₂

A narrower cross-section must have a faster flow speed to keep the same amount of matter moving through per second. This is exactly why water speeds up when a hose nozzle narrows.

ExampleWorked Example — Applying the Continuity Equation

Water flows through a pipe that narrows from a cross-sectional area of 0.020 m² to 0.005 m². If the fluid's speed in the wider section is 1.5 m/s, find its speed in the narrower section.

8.4.B.18.4.B.2Math

Bernoulli's Equation

A difference in a fluid's height between two locations produces a difference in gravitational potential energy — and conservation of energy says that difference has to show up somewhere else, as a difference in kinetic energy and pressure. Bernoulli's equation captures exactly this trade-off along a streamline:

P + ½ρv² + ρgh = constant

Pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume — the three terms trade off against each other, but their sum along any streamline never changes.

Shrink the narrow section and watch particles speed up exactly as much as continuity demands — and watch pressure drop right along with it.

A₁, v₁, P₁A₂, v₂, P₂
Wide area A₁ (m²)0.02
Narrow area A₂ (m²)0.01
Speed at A₁, v₁ (m/s)2.00
Pressure at A₁, P₁ (Pa)150000
v₂ = A₁v₁/A₂
5.00 m/s
P₂ (Bernoulli, level pipe)
139.5 kPa

Notice: shrinking A₂ increases v₂ (continuity) and that increase in speed drops P₂ (Bernoulli) — the two laws are working on the same pipe at once.

ExampleGuided Example — Applying Bernoulli's Equation

Water flows through a horizontal pipe (constant height) that narrows partway through. At the wide section, pressure is 1.50×10⁵ Pa and speed is 2.0 m/s. At the narrow section, the speed is 8.0 m/s. Find the pressure at the narrow section. (ρ_water = 1000 kg/m³)

Step 1Simplify Bernoulli's equation for a horizontal pipe
Since h₁ = h₂ (the pipe is level), the ρgh terms are equal on both sides and cancel: P₁ + ½ρv₁² = P₂ + ½ρv₂²
8.4.B.3Math

Torricelli's Theorem

Torricelli's theorem isn't a new law — it's Bernoulli's equation applied to one very specific, very common situation: fluid draining out of a small opening in a tank.

v = √(2gh)
ExampleWorked Example — Deriving and Applying Torricelli's Theorem

A large open tank has a small hole in its side, 2.0 m below the water's surface. Both the surface and the exiting stream are open to the atmosphere. Find the speed of the water as it exits the hole. (g = 9.8 m/s²)

⚠️Every equation in this lesson — continuity, Bernoulli, Torricelli — assumes an ideal fluid (non-viscous, incompressible) and pipes that are completely filled by the fluid, unless a problem explicitly says otherwise. That's the boundary statement for this entire topic.
← Back to Lesson 8.4That's Unit 8 — and the whole course. Nice work getting here.