8.4 Fluids and Conservation Laws

Fluids and conservation laws connect two familiar ideas to moving fluids: conservation of mass through the continuity equation ($A_1v_1 = A_2v_2$) and conservation of mechanical energy through Bernoulli's equation. When a pipe narrows, the fluid speeds up, and faster fluid means lower pressure.

Why This Matters for the AP Physics 1 Exam

This is the final topic in the fluids unit, and it ties together force and energy reasoning from earlier in the course. Fluids count for about 10 to 15 percent of the exam, so these tools show up often.

On the exam you may need to:

• Predict how speed and pressure change as a fluid moves through a pipe with different cross-sections.
• Use energy bar charts and conservation reasoning to explain why a fluid speeds up or slows down.
• Justify claims about pressure differences using correct vocabulary, which matters most on the free-response section where written explanation earns many points.
• Connect fluid behavior back to familiar ideas like conservation of mass and conservation of mechanical energy.

Because this unit reuses representations and models from across the course, getting comfortable here also reinforces work, energy, and momentum reasoning.

Key Takeaways

• A pressure difference between two locations is what drives a fluid to flow.
• The continuity equation, $A_1v_1 = A_2v_2$, expresses conservation of mass: narrower areas force faster speeds in an incompressible fluid.
• Bernoulli's equation, $P_1 + \rho g y_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g y_2 + \frac{1}{2}\rho v_2^2$, expresses conservation of mechanical energy along a streamline.
• Faster fluid has lower pressure; slower fluid has higher pressure (at the same height).
• Torricelli's theorem, $v = \sqrt{2g\Delta y}$, comes straight from energy conservation and gives the exit speed of draining fluid.
• Assume fluids are ideal (incompressible, no viscosity) and pipes are completely filled unless a problem says otherwise.

Flow of Incompressible Fluids

Pressure Differences Drive Flow

A fluid flows when there is a pressure difference between two locations. The fluid moves from the higher-pressure region toward the lower-pressure region, much like an object moving from high to low potential energy.

• In a tube open at both ends, the rate at which matter enters one end must equal the rate at which matter exits the other end.
• The flow rate depends on both the cross-sectional area and the fluid speed. Larger area and faster speed mean more volume passing each second.
• Volume flow rate is given by $\frac{V}{t} = Av$, where $V/t$ is the volume crossing a section each second.
• Pressure differences can come from elevation changes, pumps, or atmospheric conditions.

Continuity Equation

The continuity equation is conservation of mass written for fluid flow. For an incompressible fluid, density stays constant, so the same volume must pass every cross-section each second.

$A_{1} v_{1}=A_{2} v_{2}$

• Mass flow rate is $\dot{m} = \rho A v$. With constant density, only area and speed matter.
• This applies to steady flow, where conditions at any point stay the same over time.
• When the pipe narrows, the fluid speeds up. When it widens, the fluid slows down.
• This explains why water shoots out faster from a garden hose nozzle and helps you analyze pipes, blood vessels, and similar systems.

Energy Differences in Fluid Flow

Gravitational Potential Energy in Fluids

A difference in gravitational potential energy between two points in a fluid produces differences in kinetic energy and pressure, following conservation of energy.

• Fluid higher up has more gravitational potential energy than fluid lower down.
• As fluid flows downward, gravitational potential energy converts to kinetic energy, so its speed increases.
• Pressure also changes with depth because of the weight of the fluid column above a point.
• Along a streamline in ideal, steady flow, the total of pressure, kinetic, and gravitational energy stays constant.

Bernoulli's Equation

Bernoulli's equation is conservation of mechanical energy applied to fluid flow. It links pressure, speed, and height at two points along a streamline.

$P_{1}+\rho g y_{1}+\frac{1}{2} \rho v_{1}^{2}=P_{2}+\rho g y_{2}+\frac{1}{2} \rho v_{2}^{2}$

• Each term is an energy per unit volume: pressure ($P$), gravitational ($\rho g y$), and kinetic ($\frac{1}{2}\rho v^2$).
• It assumes steady, incompressible, inviscid flow along a streamline.
• At the same height, faster fluid has lower pressure and slower fluid has higher pressure.
• These pressure differences show up in applications like airflow over wings, venturi meters, and pitot tubes.

Torricelli's Theorem

Torricelli's theorem describes how fast a fluid exits an opening, based on the height of fluid above that opening. It comes directly from conservation of energy.

$v=\sqrt{2 g \Delta y}$

• It assumes no viscous losses and that the speed at the fluid's top surface is negligible.
• Deeper holes give higher exit speeds because more gravitational potential energy converts to kinetic energy.
• The exit speed equals the speed an object would reach falling from the same height, which shows the energy connection.

How to Use This on the AP Physics 1 Exam

Problem Solving

When a problem involves a pipe with changing area, start with the continuity equation to find the unknown speed, then use Bernoulli's equation if pressure or height also changes.

• Identify which principle the question needs: conservation of mass (continuity) or conservation of energy (Bernoulli).
• For horizontal pipes, the $\rho g y$ terms cancel because $y_1 = y_2$, which simplifies Bernoulli's equation.
• Watch your units. Convert centimeters to meters and kilopascals to pascals before plugging in.
• Remember that $1\ \text{Pa} = 1\ \text{kg/(m·s}^2)$, which helps you check that pressure terms work out.

Free Response

• State the principle you are using before the math, such as conservation of mass or conservation of mechanical energy.
• Use precise vocabulary. Keep "mass," "volume," "weight," "size," and "density" distinct, since mixing them can cost points.
• When explaining why pressure drops in a narrow section, connect speed and pressure through energy, not just memorized phrasing.

Common Trap

A narrowing pipe raises the speed but lowers the pressure. Students often expect higher speed to mean higher pressure, but Bernoulli's equation shows the opposite at the same height.

Practice Problem 1: Continuity Equation

Apply the continuity equation, where the product of area and velocity stays constant for an incompressible fluid.

Given:

• Diameter at point 1: $D_1 = 8.0$ cm = $0.080$ m
• Diameter at point 2: $D_2 = 4.0$ cm = $0.040$ m
• Velocity at point 1: $v_1 = 3.0$ m/s
• Velocity at point 2: $v_2 = ?$

Step 1: Find the areas at each point. $A_1 = \pi (D_1/2)^2 = \pi (0.080 \text{ m}/2)^2 = \pi(0.040 \text{ m})^2 = 0.00503$ m² $A_2 = \pi (D_2/2)^2 = \pi (0.040 \text{ m}/2)^2 = \pi(0.020 \text{ m})^2 = 0.00126$ m²

Step 2: Apply the continuity equation. $A_1 v_1 = A_2 v_2$

Step 3: Solve for $v_2$. $v_2 = \frac{A_1 v_1}{A_2} = \frac{0.00503 \text{ m}^2 \times 3.0 \text{ m/s}}{0.00126 \text{ m}^2} = 12.0$ m/s

The water speed at the narrower end is 12.0 m/s, which is 4 times faster than at the wider end. This matches the expectation that when the diameter is halved, the cross-sectional area drops by a factor of 4, so the velocity must increase by a factor of 4 to keep the same flow rate.

Practice Problem 2: Bernoulli's Equation

Use Bernoulli's equation together with the continuity equation.

Given:

• Diameter at point 1: $D_1 = 10.0$ cm = $0.100$ m
• Diameter at point 2: $D_2 = 5.0$ cm = $0.050$ m
• Pressure at point 1: $P_1 = 150$ kPa = $150,000$ Pa
• Velocity at point 1: $v_1 = 2.0$ m/s
• Pressure at point 2: $P_2 = ?$
• The pipe is horizontal, so $y_1 = y_2$ (no change in height)

Step 1: Find the velocity at point 2 using the continuity equation. $A_1 v_1 = A_2 v_2$ $\pi (D_1/2)^2 v_1 = \pi (D_2/2)^2 v_2$ $v_2 = v_1 \times (D_1/D_2)^2 = 2.0 \text{ m/s} \times (10.0/5.0)^2 = 2.0 \text{ m/s} \times 4 = 8.0$ m/s

Step 2: Apply Bernoulli's equation, noting that $y_1 = y_2$ for a horizontal pipe. $P_1 + \rho g y_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g y_2 + \frac{1}{2}\rho v_2^2$ $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

Step 3: Solve for $P_2$. $P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2$

$P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2)$

Taking the density of water as $\rho = 1000$ kg/m³: $P_2 = 150,000 \text{ Pa} + \frac{1}{2} \times 1000 \text{ kg/m}^3 \times ((2.0 \text{ m/s})^2 - (8.0 \text{ m/s})^2)$

$P_2 = 150,000 \text{ Pa} + 500 \text{ kg/m}^3 \times (4 \text{ m}^2/\text{s}^2 - 64 \text{ m}^2/\text{s}^2)$

$P_2 = 150,000 \text{ Pa} + 500 \text{ kg/m}^3 \times (-60 \text{ m}^2/\text{s}^2)$ $P_2 = 150,000 \text{ Pa} - 30,000 \text{ kg/m}\cdot\text{s}^2$

Since $1 \text{ kg/m·s}^2 = 1 \text{ Pa}$: $P_2 = 150,000 \text{ Pa} - 30,000 \text{ Pa} = 120,000 \text{ Pa} = 120 \text{ kPa}$

The pressure in the narrower section is 120 kPa. This demonstrates Bernoulli's principle: as fluid velocity increases, pressure decreases.

Practice Problem 3: Torricelli's Theorem

Use Torricelli's theorem, which relates exit velocity to the height of fluid above the opening.

Given:

• Height of water above the hole: $h = 5.0$ m
• Acceleration due to gravity: $g = 9.8$ m/s²

Step 1: Apply Torricelli's theorem to find the exit velocity. $v = \sqrt{2gh}$ $v = \sqrt{2 \times 9.8 \text{ m/s}^2 \times 5.0 \text{ m}}$ $v = \sqrt{98 \text{ m}^2/\text{s}^2}$ $v = 9.9$ m/s

The water exits the hole at about 9.9 m/s. This is the same speed an object would reach if dropped from 5.0 m, showing how Torricelli's theorem connects to conservation of energy. The gravitational potential energy of the water at the surface converts to kinetic energy as it exits the hole.

Common Misconceptions

• Faster fluid means higher pressure. It is the reverse. At the same height, faster fluid has lower pressure, which is exactly what Bernoulli's equation predicts.
• Continuity says the same speed everywhere. It says the same volume flow rate everywhere. Speed actually changes when the area changes.
• Bernoulli's equation works for any flow. It assumes steady, incompressible, inviscid flow along a streamline. Outside those conditions it does not apply cleanly.
• Heavier objects sink and lighter objects float. Floating and sinking depend on density and buoyancy, not just on which object feels heavier.
• A wider pipe always has more pressure or more flow. Volume flow rate is set by the system. A wider section has slower fluid, and pressure depends on speed and height together.
• Torricelli's theorem needs the hole size. The exit speed depends on the height of fluid above the opening, not on how big the hole is.

Related AP Physics 1 Guides

• 8.1 Internal Structure and Density
• 8.2 Pressure
• 8.3 Fluids and Newton's Laws