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Fluids and Conservation Laws

Lesson ~11 min read 8 MCQs

In simple terms: In simple terms, this topic is about how a fluid's speed and pressure change when it flows through pipes of different sizes and heights, all while conserving mass and energy.

Why this matters

Have you ever used a garden hose and put your thumb over the end to make the water spray farther? You've already got an intuitive feel for today's topic. When you block part of the opening, the water shoots out much faster. But why? It seems like you're making it harder for the water to get out, yet it speeds up.

This isn't just a backyard curiosity. It's the principle behind everything from how airplanes fly to how a cardiologist in Dallas understands blood flow through an artery. We're going to explore the two big conservation laws that govern moving fluids. By the end of this lesson, you'll be able to predict exactly how a fluid's speed and pressure will change as it moves.

Diagram

Fluid Flow: Continuity and Bernoulli's Principle A diagram showing a fluid-carrying pipe that is wide and low on the left (Point 1) and narrow and high on the right (Point 2). The diagram illustrates how fluid speed increases and pressure changes as it moves through the pipe, with labels for pressure (P), velocity (v), area (A), and height (y) at both points. The continuity and Bernoulli equations are displayed. y=0 Point 1 P₁, v₁, A₁, y₁ v₁ y₁ A₁ Point 2 P₂, v₂, A₂, y₂ v₂ y₂ A₂ A₁v₁ = A₂v₂ P₁ + ρgy₁ + ½ρv₁² = P₂ + ρgy₂ + ½ρv₂²
This diagram shows a pipe carrying a fluid. The pipe is wider and lower on the left (labeled Point 1) and becomes narrower and higher on the right (labeled Point 2). Key equations for fluid dynamics, the continuity equation and Bernoulli's equation, are displayed at the top.

Concept map

flowchart TD
    A[Start: Fluid Flow Problem] --> B{Is pipe area changing?};
    B -->|Yes| C[Use Continuity Eq:<br>A₁v₁ = A₂v₂<br>to find new speed];
    B -->|No| D{Is height or speed changing?};
    C --> D;
    D -->|Yes| E[Use Bernoulli's Eq:<br>P + ρgy + ½ρv² = const<br>to find new P or y];
    D -->|No| F[Static Fluid<br>(Previous Topic)];
    E --> G[End: Solved];
    F --> G;

Core explanation

Alright, let's dive into the physics of moving fluids. We'll start with a simple idea and build up to one of the most important equations in this unit.

Conservation of Mass: The Continuity Equation

Imagine a river. Where the river is wide and deep, the water flows slowly and calmly. Where it narrows into a gorge, the water rushes through at a high speed. This is the core idea of the continuity equation.

For an incompressible fluid—one whose density (ρ) doesn't change, like water—the amount of mass flowing past any point in a pipe per second must be constant. If it weren't, fluid would either be magically appearing or disappearing inside the pipe!

Let's think about the volume flow rate, which is the volume of fluid (V) that passes a point in a certain amount of time (t). We can write this as V/t.

Now, picture a small "cylinder" of water moving through a pipe.

  • The volume of this cylinder is its cross-sectional area (A) times its length (L). So, V = A * L.
  • The speed of the fluid (v) is the length it travels divided by time, v = L/t.

If we substitute these into our flow rate, we get: Flow Rate = V/t = (A * L) / t = A * (L/t) = A * v

So, the volume flow rate is simply Area × speed.

Now, consider a pipe that narrows, like the one in our diagram. Let's call the wide part "point 1" and the narrow part "point 2". Since mass is conserved, the volume flow rate must be the same at both points.

Flow Rate at point 1 = Flow Rate at point 2

This gives us the Equation of Continuity:

A₁v₁ = A₂v₂

Where:

  • A₁ and v₁ are the area and speed at point 1.
  • A₂ and v₂ are the area and speed at point 2.

This simple equation is powerful. It mathematically confirms our garden hose intuition: if the area A₂ gets smaller, the speed v₂ must get larger to keep the product constant.

Conservation of Energy: Bernoulli's Equation

Now for the main event. We know from mechanics that in the absence of non-conservative forces like friction, total mechanical energy (KE + PE) is conserved. A similar principle applies to fluids, but with an extra term: pressure.

Bernoulli's equation is essentially a statement of conservation of energy for a moving fluid. It relates pressure, speed, and height between two points in a fluid's path.

The full equation looks intimidating at first, but we'll break it down.

P₁ + ρgy₁ + ½ρv₁² = P₂ + ρgy₂ + ½ρv₂²

This means the sum of these three terms is constant everywhere along a streamline in the fluid. Let's look at each piece. They represent energy per unit volume (Joules/m³).

  • P: This is the static pressure. It's the pressure you'd measure if you were moving along with the fluid. Think of it as the energy related to the work done by the fluid.
  • ρgy: This is the gravitational potential energy density. It's just like the mgh you know and love, but divided by volume (since ρ = m/V). It accounts for the energy a fluid has due to its height (y).
  • ½ρv²: This is the kinetic energy density. It's like ½mv², but again, divided by volume. It represents the energy the fluid has due to its motion (v).

Faster fluid flow means lower static pressure.

Think of it like an energy budget. If the fluid "spends" more energy on moving faster (kinetic energy), it has less energy left to exert as pressure. In our garden hose example, the fast-moving water at the nozzle is actually at a lower pressure than the slow-moving water back in the hose.

A Special Case: Torricelli's Theorem

Let's use Bernoulli's equation to solve a classic problem. Imagine a large, open water tank with a small spigot near the bottom. How fast does the water shoot out of the spigot?

Let's apply Bernoulli's equation between the top surface of the water (point 1) and the spigot's opening (point 2).

  • Point 1 (top surface)
    • The tank is open to the air, so P₁ = P_atm (atmospheric pressure).
    • The tank is large, so the water level drops very slowly. We can approximate its speed as v₁ ≈ 0.
    • Let's set the height of the spigot as y₂ = 0, so the height of the top surface is y₁ = h.
  • Point 2 (the spigot)
    • The water is exiting into the air, so P₂ = P_atm.
    • Its height is y₂ = 0.
    • Its speed v₂ is what we want to find.

Now, plug these into Bernoulli's equation: P₁ + ρgy₁ + ½ρv₁² = P₂ + ρgy₂ + ½ρv₂² P_atm + ρgh + ½ρ(0)² = P_atm + ρg(0) + ½ρv₂²

The P_atm terms on both sides cancel out. The zero terms drop away. We're left with: ρgh = ½ρv₂²

The density ρ also cancels! Solving for v₂, we get:

v₂ = √2gh

This is Torricelli's Theorem. Notice something amazing? This is the exact same speed an object would have if you dropped it from rest from a height h. It's a beautiful example of energy conservation connecting two different areas of physics.

Worked examples

Let's put these principles into practice with a few examples.

Example 1

The Continuity Equation

Problem: Water flows through a horizontal pipe at a speed of 2.0 m/s. The pipe has a diameter of 10 cm. The pipe then narrows to a diameter of 5.0 cm. What is the speed of the water in the narrow section?

Solution:

  1. 1
    Identify the principle
    The pipe's area is changing, so the speed must change. This is a job for the continuity equation: A₁v₁ = A₂v₂.
  2. 2
    List your knowns
    • v₁ = 2.0 m/s
    • d₁ = 10 cm = 0.10 m
    • d₂ = 5.0 cm = 0.05 m
  3. 3
    Calculate the areas
    The pipe is circular, so A = πr². Be careful! The problem gives diameters, not radii.
    • r₁ = d₁ / 2 = 0.05 m
    • r₂ = d₂ / 2 = 0.025 m
    • A₁ = π(0.05 m)² = 0.0025π m²
    • A₂ = π(0.025 m)² = 0.000625π m²
  4. 4
    Solve for the unknown (v₂):
    • Rearrange the continuity equation: v₂ = v₁ * (A₁ / A₂)
    • v₂ = (2.0 m/s) * (0.0025π m² / 0.000625π m²)
    • The π terms cancel. v₂ = (2.0 m/s) * 4
    • v₂ = 8.0 m/s

Why this makes sense: The diameter was halved, which means the radius was halved. Since area depends on the radius squared (A = πr²), the area decreased by a factor of four. To keep the flow rate constant, the speed must increase by a factor of four.

Example 2

Bernoulli's Equation (Horizontal Pipe)

Problem: In the same pipe from Example 1, if the pressure in the wider section (P₁) is 180,000 Pa, what is the pressure in the narrower section (P₂)? (Assume the density of water, ρ, is 1000 kg/m³).

Solution:

  1. 1
    Identify the principle
    We have changes in speed and pressure, but not height. This is a classic Bernoulli's equation problem. P₁ + ρgy₁ + ½ρv₁² = P₂ + ρgy₂ + ½ρv₂².
  2. 2
    Simplify for a horizontal pipe
    Since the pipe is horizontal, y₁ = y₂. This means the ρgy terms are equal on both sides and can be canceled out.
    • P₁ + ½ρv₁² = P₂ + ½ρv₂²
  3. 3
    List your knowns (from both examples)
    • P₁ = 180,000 Pa
    • v₁ = 2.0 m/s
    • v₂ = 8.0 m/s (we found this in Example 1)
    • ρ = 1000 kg/m³
  4. 4
    Solve for the unknown (P₂):
    • Rearrange the simplified equation: P₂ = P₁ + ½ρv₁² - ½ρv₂²
    • P₂ = P₁ + ½ρ(v₁² - v₂²)
    • P₂ = 180,000 Pa + ½(1000 kg/m³)((2.0 m/s)² - (8.0 m/s)²)
    • P₂ = 180,000 Pa + 500(4 - 64)
    • P₂ = 180,000 Pa + 500(-60)
    • P₂ = 180,000 Pa - 30,000 Pa
    • P₂ = 150,000 Pa

Why this is important: Notice that the pressure decreased from 180 kPa to 150 kPa. This confirms the counterintuitive rule: where the fluid is moving faster, the pressure is lower.

Try it yourself

Ready to try on your own? Don't just jump to the answer; think through the steps and the principles.

  1. A firefighter in Chicago is using a hose with a 9.0 cm diameter. The water flows at 1.5 m/s. The nozzle at the end of the hose has an opening with a 3.0 cm diameter. a. What is the speed of the water as it exits the nozzle? b. If the pressure in the hose is 400,000 Pa, what is the pressure at the nozzle? (Assume the hose is level).

    Hint: This is a two-step problem. First, find the exit speed using continuity. Then, use that speed in Bernoulli's equation to find the new pressure.

  2. A large water barrel at a farm in rural Georgia has a small leak 1.8 meters below the water's surface. How fast is the water leaking out of the hole?

    Hint: What can you assume about the pressure at the water's surface and at the leak? This is a perfect chance to use a special case we discussed.

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