✈️ Bernoulli's Equation

Part 1 of 7 — Energy Conservation for Fluids

Bernoulli's equation is one of the most powerful and beautiful results in fluid dynamics. It connects pressure, speed, and height in a single elegant equation — and explains everything from airplane lift to why shower curtains billow inward.

Deriving Bernoulli's Equation

Bernoulli's equation is really just conservation of energy applied to a flowing fluid.

Consider a small parcel of fluid moving through a pipe:

• Kinetic energy: $\frac{1}{2}\rho v^2$ per unit volume
• Gravitational PE: $\rho g h$ per unit volume
• Pressure energy: $P$ (pressure acts like energy per unit volume — it does work on the fluid)

Energy conservation says the total energy per unit volume is constant along a streamline:

$\boxed{P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}}$

Or between two points on the same streamline:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$

Each Term's Role

All three terms have units of pressure (Pa = N/m² = J/m³).

When Does Bernoulli Apply?

Bernoulli's equation requires the same ideal fluid conditions as continuity:

• ✅ Incompressible fluid
• ✅ Non-viscous (no friction losses)
• ✅ Steady flow
• ✅ Along a streamline (you can't compare random points!)

Common Special Cases

Case 1: Horizontal flow ($h_1 = h_2$):

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

Faster flow → lower pressure (the Bernoulli effect!)

Case 2: Static fluid ($v = 0$):

$P_1 + \rho g h_1 = P_2 + \rho g h_2$

This reduces to our hydrostatic equation $\Delta P = \rho g \Delta h$!

Case 3: Open to atmosphere:

At any point open to air, $P = P_{\text{atm}}$ (atmospheric pressure).

Bernoulli Concept Check

Bernoulli Basics ($\rho_w = 1000$ kg/m³, $g = 10$ m/s²)

Water flows horizontally through a pipe. At point 1: $P_1 = 200{,}000$ Pa, $v_1 = 3.0$ m/s. At point 2 (same height): $v_2 = 5.0$ m/s.

1. Dynamic pressure at point 1: $\frac{1}{2}\rho v_1^2$ (in Pa)
2. Dynamic pressure at point 2 (in Pa)
3. Pressure at point 2 (in Pa)

Exit Quiz

🌀 The Venturi Effect

Part 2 of 7 — Fast Flow, Low Pressure

The most important consequence of Bernoulli's equation: where fluid speeds up, pressure drops. This counterintuitive result explains everything from atomizers to carburetors.

The Venturi Tube

A Venturi tube is a pipe with a constriction (narrow section). Let's analyze it:

Setup

• Wide section: area $A_1$, speed $v_1$, pressure $P_1$
• Narrow section: area $A_2 < A_1$, speed $v_2$, pressure $P_2$
• Horizontal pipe: same height ($h_1 = h_2$)

Continuity gives:

$v_2 = \frac{A_1}{A_2} v_1 > v_1$

Bernoulli gives:

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

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

Since $v_2 > v_1$: $P_2 < P_1$

The Pressure Drop

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

This pressure difference can be measured with a U-tube manometer connected between the wide and narrow sections — this is a Venturi meter, used to measure flow speed!

Venturi Effect Quiz

Applications of the Venturi Effect

Perfume Atomizer / Spray Bottle

Blowing air across the top of a tube creates low pressure above the tube. Higher atmospheric pressure at the liquid surface pushes fluid up the tube, where it gets caught in the airstream and atomized into a fine mist.

Carburetor (Older Cars)

Air flows through a Venturi tube in the carburetor. The low-pressure region draws gasoline from a reservoir and mixes it with the air — no pump needed!

Prairie Dog Burrows

Prairie dogs build mounds at different heights around their burrow entrances. Wind over the tall mound creates lower pressure (Venturi effect), while the flat entrance has higher pressure → natural ventilation!

Shower Curtain Mystery

Hot shower water heats the air → air rises → creates faster airflow inside the shower. By Bernoulli, faster flow → lower pressure inside → the curtain gets pushed inward by higher-pressure air outside.

Venturi Meter

By measuring the pressure drop between wide and narrow sections, you can calculate the flow speed — this is how many industrial flow meters work!

Venturi Tube Drill ($\rho_w = 1000$ kg/m³)

A horizontal Venturi tube has wide diameter 8.0 cm and narrow diameter 4.0 cm. The pressure difference between the two sections is 9000 Pa.

1. Area ratio $A_1/A_2$
2. Express $v_1$ in terms of the pressure difference (solve for $v_1$, in m/s)
3. Flow rate through the tube (in L/s)

Round all answers to 3 significant figures.

Exit Quiz

🏺 Torricelli's Theorem

Part 3 of 7 — The Speed of Draining Fluid

What happens when you poke a hole in a tank? How fast does the water come out? Torricelli answered this in 1643 — and his answer is beautifully simple.

Deriving Torricelli's Theorem

Setup

A large tank is filled with water to height $h$. A small hole is opened at the bottom.

Point 1: The surface of the water (top of tank) Point 2: The hole (bottom of tank)

Assumptions

• Both the surface and the hole are open to atmosphere: $P_1 = P_2 = P_{\text{atm}}$
• The tank is large compared to the hole: $v_1 \approx 0$ (surface barely moves)
• Take the hole as the height reference: $h_1 = h$, $h_2 = 0$

Applying Bernoulli

$P_{\text{atm}} + \frac{1}{2}\rho(0)^2 + \rho g h = P_{\text{atm}} + \frac{1}{2}\rho v_2^2 + \rho g(0)$

$\rho g h = \frac{1}{2}\rho v_2^2$

$\boxed{v = \sqrt{2gh}}$

The Beautiful Result

This is exactly the speed an object would have if it fell freely from height $h$! The water exits at the same speed as if it had simply fallen from the surface level. This is Torricelli's Theorem.

Torricelli Concept Check

Torricelli Drill (use $g = 10$ m/s²)

A large water tank has a small hole 5.0 m below the water surface. The hole has area $2.0 \times 10^{-4}$ m².

1. Exit speed of the water (in m/s)
2. Volume flow rate from the hole (in L/s)
3. How far horizontally the water lands if the hole is 1.2 m above the ground (in m)

Round all answers to 3 significant figures.

Variations on Torricelli

Hole Not at the Bottom

If the hole is at height $h_{\text{hole}}$ above the bottom, and the surface is at height $H$:

$v = \sqrt{2g(H - h_{\text{hole}})}$

Only the height above the hole matters.

Pressurized Tank

If the tank is sealed and pressurized to gauge pressure $P_g$ above atmospheric:

$P_{\text{atm}} + P_g + \rho g h = P_{\text{atm}} + \frac{1}{2}\rho v^2$

$v = \sqrt{\frac{2(P_g + \rho g h)}{\rho}}$

The extra pressure makes the water exit faster — like a pressurized water gun!

Maximum Range Problem

Classic AP problem: At what height should you make a hole to maximize the horizontal range of the water jet?

The exit speed increases with depth ($v = \sqrt{2gh}$), but a deeper hole means less fall height. The optimal hole position is at half the tank height — it maximizes the product of exit speed and fall time.

Exit Quiz

🛫 Lift and Aerodynamics

Part 4 of 7 — Why Airplanes Fly

Bernoulli's equation helps explain how wings generate lift, though the full story is more nuanced than textbooks sometimes suggest. Let's get it right.

Lift on an Airfoil

The Bernoulli Explanation

An airplane wing (airfoil) is shaped so that air flows faster over the top surface and slower under the bottom.

By Bernoulli: faster flow → lower pressure

$P_{\text{top}} < P_{\text{bottom}}$

This pressure difference creates a net upward force — lift!

$F_{\text{lift}} = (P_{\text{bottom}} - P_{\text{top}}) \times A_{\text{wing}}$

The Full Picture

The Bernoulli explanation is partially correct but incomplete. Lift also involves:

• Angle of attack: The wing is tilted, deflecting air downward. By Newton's 3rd law, the air pushes the wing up.
• Circulation: The wing creates a circulation pattern that increases speed above and decreases it below.
• Coandă effect: Air "sticks" to the curved upper surface.

The combination of pressure differences (Bernoulli) and momentum change (Newton) gives the complete picture. Both are important!

Key Point for AP

AP Physics 2 focuses on the Bernoulli explanation: faster air on top → lower pressure → net upward force.

Lift Quiz

More Bernoulli Applications

⚾ Curve Balls

A spinning baseball creates faster air on one side (spin adds to airflow) and slower air on the other (spin opposes airflow). The pressure difference curves the ball's path — this is the Magnus effect.

• Topspin: Ball curves downward (faster air below → low pressure below → net force downward)
• Backspin: Ball curves upward (faster air above → low pressure above → net force upward)
• Sidespin: Ball curves left or right

🏠 Wind and Roofs

Hurricane-force winds blow over a roof. By Bernoulli, the fast-moving air creates low pressure above the roof. The still air inside the house has atmospheric pressure (higher). The roof gets pushed upward and can be ripped off!

$F = (P_{\text{inside}} - P_{\text{outside}}) \times A_{\text{roof}}$

🚗 Race Car Spoilers

Race cars use inverted airfoils (spoilers) — faster air goes underneath, creating low pressure below. This pushes the car down onto the road (downforce), improving traction.

At high speeds, a Formula 1 car generates enough downforce to drive upside down on a ceiling!

Hurricane Roof Problem ($\rho_{\text{air}} = 1.2$ kg/m³)

A hurricane with wind speed 50 m/s blows over a flat roof. The air inside the house is still.

1. Dynamic pressure of the wind: $\frac{1}{2}\rho v^2$ (in Pa)
2. Net upward pressure on the roof (in Pa)
3. If the roof area is 200 m², the upward force (in kN)

Exit Quiz

🧮 Bernoulli Problem-Solving Workshop

Part 5 of 7 — Multi-Step AP Problems

Let's tackle the full-blown Bernoulli problems that combine continuity + Bernoulli + height changes — the kind AP loves.

Problem-Solving Strategy

Step-by-Step Approach

1. Draw the system — identify points 1 and 2 (on the same streamline)
2. List knowns: $P$, $v$, $h$ at each point
3. Apply continuity if areas are given: $A_1 v_1 = A_2 v_2$
4. Apply Bernoulli: $P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$
5. Solve for the unknown

Common Boundary Conditions

Problem 1: Pipe with Height Change ($\rho = 1000$ kg/m³, $g = 10$ m/s²)

Water flows through a pipe that rises 8.0 m. At the bottom: area $A_1 = 40$ cm², speed $v_1 = 5.0$ m/s, pressure $P_1 = 300{,}000$ Pa. At the top: area $A_2 = 20$ cm².

1. Speed at the top, $v_2$ (in m/s)
2. Pressure at the top, $P_2$ (in Pa)

Problem 2: Pitot Tube ($\rho_{\text{air}} = 1.2$ kg/m³)

A Pitot tube on an airplane measures the stagnation pressure (air brought to rest) and the static pressure. Stagnation pressure: 102,000 Pa. Static pressure: 100,000 Pa.

1. Pressure difference (in Pa)
2. Air speed of the airplane (in m/s)
3. This speed in km/h (multiply m/s by 3.6)

Round all answers to 3 significant figures.

Problem 3: Conceptual Challenge

Challenge Problem ($\rho = 1000$ kg/m³, $g = 10$ m/s², $P_{\text{atm}} = 100{,}000$ Pa)

A large open tank has water 10 m deep. A pipe at the bottom carries water horizontally to a nozzle that is 3.0 m above the bottom of the tank. The nozzle exit area is very small (open to atmosphere).

1. Speed of water exiting the nozzle (in m/s)
2. If the nozzle area is $1.0 \times 10^{-4}$ m², the flow rate (in L/s)

Exit Quiz

🔧 Measurement Devices & Real-World Systems

Part 6 of 7 — Bernoulli in Practice

Bernoulli's equation powers many practical devices for measuring speed, flow, and pressure. Let's see how engineers apply the physics.

Pitot Tube (Airspeed Indicator)

Every airplane has a Pitot tube — a forward-facing tube that measures airspeed.

How It Works

• Forward opening: Air is brought to rest (stagnation) → measures total pressure $P_0 = P + \frac{1}{2}\rho v^2$
• Side ports: Measure static pressure $P$ (unaffected by airflow)
• Difference: $P_0 - P = \frac{1}{2}\rho v^2$

$v = \sqrt{\frac{2(P_0 - P)}{\rho}}$

Why Pitot Tubes Are Critical

Pitot tubes are essential for flight safety. If they become blocked (ice, insects), the airspeed reading fails — this has caused aviation accidents. Modern planes have heated Pitot tubes and redundant systems.

Venturi Meter (Flow Measurement)

A Venturi meter measures flow rate using the pressure drop in a constriction.

Combining Continuity + Bernoulli

At the wide section (1) and narrow section (2):

Continuity: $v_2 = (A_1/A_2)v_1$

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

Substituting:

$P_1 - P_2 = \frac{1}{2}\rho v_1^2 \left[\left(\frac{A_1}{A_2}\right)^2 - 1\right]$

$v_1 = \sqrt{\frac{2(P_1 - P_2)}{\rho\left[(A_1/A_2)^2 - 1\right]}}$

Then: $Q = A_1 v_1$

Advantages

• No moving parts
• Very reliable
• Works for any incompressible fluid

Measurement Devices Quiz

Real-World Bernoulli Systems

Water Supply in Buildings

Water pressure at the street main is typically $\sim 400$ kPa. As water rises through the building:

• Pressure decreases by $\rho g h$ per meter of height
• Flow through narrow pipes increases speed → further pressure drop

At floor $n$ (height $h$): $P_n = P_{\text{main}} - \rho g h - \frac{1}{2}\rho(v_n^2 - v_{\text{main}}^2)$

Tall buildings need pumps to maintain adequate pressure on upper floors!

Medical Applications

IV Drip: The IV bag is raised above the patient. Bernoulli explains the flow:

• Point 1: Fluid surface in bag (height $h$, $v \approx 0$, $P = P_{\text{atm}}$)
• Point 2: Needle tip in vein ($h = 0$, pressure = venous pressure)

$P_{\text{atm}} + \rho g h = P_{\text{vein}} + \frac{1}{2}\rho v^2$

The bag must be high enough that $\rho g h$ overcomes the venous pressure.

Fire Hydrants

Fire trucks connect to hydrants ($P \sim 300$ kPa). The nozzle converts pressure energy to kinetic energy:

$v = \sqrt{\frac{2P}{\rho}} = \sqrt{\frac{2(300{,}000)}{1000}} = 24.5 \text{ m/s}$

Building Water Pressure ($\rho = 1000$ kg/m³, $g = 10$ m/s²)

Street main pressure: 400,000 Pa. The pipe (constant area) goes up to the 10th floor, 30 m above street level. (Same pipe area, so $v$ stays constant.)

1. Pressure loss due to height (in Pa)
2. Pressure at the 10th floor (in Pa)
3. Maximum building height where water pressure reaches 0 gauge (in m)

Exit Quiz

🎯 Bernoulli Synthesis & AP Review

Part 7 of 7 — Complete Fluids Review

This final part ties together everything from all four fluid mechanics topics: density & pressure, buoyancy, continuity, and Bernoulli's equation.

Complete Fluids Concept Map

The Four Pillars of Fluid Mechanics

Decision Tree: Which Equation(s) to Use

1. Fluid at rest? → Hydrostatic pressure ($P = P_0 + ho g h$), buoyancy ($F_B = ho_f V_d g$)
2. Fluid moving, area changes? → Continuity ($A_1 v_1 = A_2 v_2$) first, then Bernoulli
3. Fluid moving, same area? → Bernoulli only (speed is the same by continuity)
4. Open surface or hole in tank? → Torricelli ($v = \sqrt{2gh}$)
5. Horizontal flow? → Simplified Bernoulli ($P + \frac{1}{2}\rho v^2 =$ const)

Top 5 AP Mistakes

Comprehensive Quiz

AP Comprehensive Problem ($\rho = 1000$ kg/m³, $g = 10$ m/s²)

A fire hose (diameter 6.0 cm) is connected to a hydrant at ground level with pressure 400,000 Pa. The hose goes up 10 m to a nozzle (diameter 2.0 cm) that is open to the atmosphere ($P_{\text{atm}} = 100{,}000$ Pa).

1. If the speed in the hose at ground level is $v_1$, find $v_2$ in terms of $v_1$ (ratio $v_2/v_1$)
2. Using Bernoulli (with continuity), find $v_1$ (in m/s) [Hint: $v_2 = 9v_1$]
3. Speed of water exiting the nozzle (in m/s)

Round all answers to 3 significant figures.

AP FRQ Practice — Full Problem

Setup

A large open tank is filled to height $H = 5.0$ m with water. A small circular hole of radius $r = 0.50$ cm is opened at the bottom.

(a) Find the speed of water exiting the hole. $v = \sqrt{2gH} = \sqrt{2(10)(5)} = 10 \text{ m/s}$

(b) Find the volume flow rate. $Q = \pi r^2 v = \pi(0.005)^2(10) = 7.85 \times 10^{-4} \text{ m}^3\text{/s} \approx 0.79 \text{ L/s}$

(c) The tank sits on the ground. A hole is made at height $h = 1.0$ m on the side (not at the very bottom). The stream exits horizontally. How far from the tank does the water land?

Water depth above hole: $H - h = 5.0 - 1.0 = 4.0$ m. $v = \sqrt{2g(H-h)} = \sqrt{2(10)(4)} = 8.94 \text{ m/s}$ $t = \sqrt{2h/g} = \sqrt{2(1)/10} = 0.447 \text{ s}$ $x = v \cdot t = 8.94(0.447) = 4.0 \text{ m}$

(d) If the hole is moved to height $h = 2.5$ m (middle of tank), will the range be greater or less?

The range is $x = v \times t = \sqrt{2g(H-h)} \times \sqrt{2h/g} = 2\sqrt{h(H-h)}$.

At the middle ($h = H/2 = 2.5$ m), this product is maximized! Range = $2\sqrt{2.5 \times 2.5} = 5.0$ m — greater than from the 1.0 m hole (4.0 m).

Final Exit Quiz