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Bernoulli's Equation

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Bernoulli's Equation - Complete Interactive Lesson

Part 1: Bernoulli's Equation Setup

📜 Bernoulli's Equation — Setup

Part 1 of 7 — Fluids: Bernoulli's Equation

Bernoulli's Equation captures conservation of energy for a flowing fluid. It's the second pillar of AP fluids (alongside continuity), and it explains why airplanes fly, why your shower curtain pulls inward, and how a Venturi meter works.

In this lesson you will learn:

The full Bernoulli equation
The three energy-density terms (pressure, kinetic, gravitational)
The assumptions (incompressible, non-viscous, steady, along a streamline)
How to identify "two points" on a streamline for problem solving

Bernoulli's Equation

For an ideal fluid (incompressible, non-viscous, steady, irrotational) along a streamline:

$\boxed{P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g y_2}$

Equivalently: $P + \tfrac{1}{2}\rho v^2 + \rho g y = \text{constant}$ along a streamline.

Each Term Has Units of Pressure (Pa)

Term	Meaning
$P$	Static pressure (Pa)
$\tfrac{1}{2}\rho v^2$

It's just energy conservation per unit volume of fluid.

Required Assumptions

Incompressible — $\rho$ constant
Non-viscous — no internal friction
Steady — flow speed at each point doesn't change in time
Along a streamline — same fluid parcel from point 1 to point 2

Choosing Points 1 and 2 (the AP technique)

Pick where you know the most ( $P$ , $v$ , $y$ all knowable).
Reservoir surface, pipe outlets, openings to atmosphere → known $P$ ( $P_{atm}$ ) and often .

Bernoulli Concepts 🎯

Bernoulli Setup Calculations 🧮 (g = 10, $\rho_w = 1000$ , $P_{atm} = 1.0\times10^{5}$ Pa)

Bernoulli Setup Reasoning 🔍

Exit Quiz — Bernoulli Setup ✅

Part 2: Conservation of Energy in Fluids

⚡ Conservation of Energy in Fluids

Part 2 of 7 — Fluids: Bernoulli's Equation

Bernoulli's equation is just energy conservation, repackaged. Each term is an energy density. Understanding the link to mechanical energy makes Bernoulli intuitive — it's $\tfrac{1}{2}mv^2 + mgy$ from kinematics, divided by volume, plus a pressure-work term.

Part 3: Pressure-Speed Trade-off

🔄 The Pressure–Speed Trade-off

Part 3 of 7 — Fluids: Bernoulli's Equation

The most famous Bernoulli result: fast-moving fluid has lower pressure. This is what makes airfoils lift, atomizers spray, and shower curtains pull inward. AP loves to test this counter-intuitive idea.

In this lesson you will learn:

The Bernoulli effect at constant elevation
Why "lift" and "suction" happen
Real-world examples: airplane wings, chimneys, curveballs
How to predict pressure direction from speed change

Horizontal Bernoulli

For a horizontal streamline ( $y_1 = y_2$ ):

Part 4: Torricelli's Theorem

🚿 Torricelli's Theorem

Part 4 of 7 — Fluids: Bernoulli's Equation

Torricelli's Theorem is the classic Bernoulli application: water draining out of a hole in a tank flows like a free-falling object. It's a guaranteed AP Physics 1 favorite — fast and elegant.

In this lesson you will learn:

How to derive $v = \sqrt{2gh}$ from Bernoulli
The "large reservoir" approximation

Part 5: Real-World Applications

🌍 Real-World Applications

Part 5 of 7 — Fluids: Bernoulli's Equation

Bernoulli isn't just an exam formula — it explains lift, plumbing, weather, sports, and your shower curtain. AP often presents qualitative scenarios where you must identify the Bernoulli effect at work.

In this lesson you will explore:

Airfoil lift (qualitative + estimate)
Venturi meters and flow measurement
Pitot tubes (airplane airspeed)
Atomizers, chimneys, and Magnus effect

Application Catalog

1. Airfoil / Lift

Curved upper surface forces air to travel a longer path → faster speed (in the simple model).
Faster speed on top ⇒ lower $P$ on top by Bernoulli.
Pressure difference × wing area = lift force.

$F_{lift} \approx \Delta P \cdot A = \tfrac{1}{2}\rho_{air}(v_{top}^2 - v_{bot}^2)\, A$

Part 6: Combined Continuity + Bernoulli

🛠️ Combined Continuity + Bernoulli

Part 6 of 7 — Fluids: Bernoulli's Equation

The hardest AP fluids problems combine BOTH governing equations: continuity to find unknown speeds from geometry, then Bernoulli to find unknown pressures (or vice versa). This part is the workshop where you put them together.

Key combination workflow:

Identify two points on the same streamline.
Continuity: $A_1 v_1 = A_2 v_2$ → solve for the unknown speed.

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review — Bernoulli

Part 7 of 7 — Fluids: Bernoulli's Equation

You've completed the full AP Physics 1 fluids progression: density, pressure, buoyancy, continuity, and Bernoulli. This synthesis ties Bernoulli together with continuity into the integrated mental model AP expects — fast questions, big-picture reasoning, and confident calculations.

Big Ideas Recap:

$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g y_2$

Use one point where you have many unknowns? → No. Pick known points!

Calculate each TERM of Bernoulli at the given point — answer in pascals.

Surface of an open tank ( $P = P_{atm}$ , $v = 0$ , $y = 5.0$ m). The $\rho g y$ term (Pa)?
Same surface — the $\tfrac{1}{2}\rho v^2$ term (Pa)?
A point inside a pipe with $v = 4.0$ m/s. The dynamic pressure term $\tfrac{1}{2}\rho v^2$ (Pa)?

In this lesson you will learn:

The derivation: work-energy theorem applied to a fluid parcel
Why pressure × volume is "flow work"
How each term maps to a familiar mechanics concept
A units check

Derivation Sketch

Consider a fluid parcel of volume $V$ moving from point 1 to point 2 along a streamline.

The work-energy theorem says:

$W_{net} = \Delta KE + \Delta PE \quad \text{(for each unit volume)}$

Forces doing work on the parcel:

Pressure forces from neighboring fluid: $W_{pressure} = (P_1 - P_2) V$

KE change per unit volume: $\tfrac{1}{2}\rho v_2^2 - \tfrac{1}{2}\rho v_1^2$ . PE change per unit volume: .

Combining:

$(P_1 - P_2) = (\tfrac{1}{2}\rho v_2^2 - \tfrac{1}{2}\rho v_1^2) + (\rho g y_2 - \rho g y_1)$

Rearranging:

$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g y_2$

Mapping to Mechanics

Mechanics	Bernoulli (per unit volume)
KE = $\tfrac{1}{2}mv^2$	$\tfrac{1}{2}\rho v^2$

Quick Sanity Check (Units)

$[P] = \text{Pa} = \text{N/m}^2 = \text{N}\cdot\text{m/m}^3 = \text{J/m}^3$

Energy Conservation Concepts 🎯

Energy-Density Calculations 🧮 (g = 10, $\rho_w = 1000$ )

Compute kinetic energy per unit volume of water moving at 5 m/s (Pa).
Compute gravitational PE per unit volume of water at $y = 8$ m (Pa).
The total $P + \tfrac{1}{2}\rho v^2 + \rho g y$ at a point with $P = 90$ kPa, $v = 2$ m/s, $y = 4$ m (Pa)?

Energy in Bernoulli Reasoning 🔍

Exit Quiz — Energy in Fluids ✅

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

So $\Delta P = \tfrac{1}{2}\rho (v_1^2 - v_2^2)$ .

Where the fluid is faster, the pressure is lower.

Phenomenon	Why
Airplane wing (lift)	Faster air over curved top → lower pressure on top → net upward force
Atomizer/perfume spray	Fast air over a tube tip → low $P$ at top → liquid pushed up by atmospheric pressure
Shower curtain pulled in	Falling water drags air faster on inside → lower $P$ → curtain pushed in
Wind blowing roof off	Fast wind over roof → low $P$ above → high $P$ below pushes roof up
Curveball / Magnus effect	Spinning ball makes air speed asymmetric → pressure difference deflects ball

A Common Misconception

❌ "Lower pressure pulls things into the fast stream." ✅ Higher pressure on the OTHER side PUSHES things toward the fast stream. Pressure only pushes.

Linking to Continuity

$A_1 v_1 = A_2 v_2$ tells you HOW the speed changes (geometry). Bernoulli tells you HOW the pressure responds (energy). Both are needed for venturi-style problems. (Coming in Part 6.)

Pressure-Speed Trade-off 🎯

Pressure-Speed Calculations 🧮 ( $\rho_w = 1000$ , $\rho_{air} = 1.2$ )

A horizontal pipe carries water at 2 m/s where $P = 150$ kPa. At a constriction it speeds up to 8 m/s. Pressure at the constriction (kPa)?
Air flows over a wing at 50 m/s on top and 40 m/s on bottom. Pressure DIFFERENCE (Pa, top vs bottom: $\Delta P = P_{bottom} - P_{top}$ )?

Pressure-Speed Reasoning 🔍

Exit Quiz — Pressure-Speed Trade-off ✅

Time-of-flight for a horizontal jet exiting a tank

Range of the projected stream

Torricelli's Theorem

Setup: An open tank with a small hole of negligible cross-section, depth $h$ below the free surface. Both the surface (point 1) and the hole (point 2) are open to atmosphere ⇒ $P_1 = P_2 = P_{atm}$ .

Bernoulli:

$P_{atm} + \tfrac{1}{2}\rho v_1^2 + \rho g h = P_{atm} + \tfrac{1}{2}\rho v_2^2 + 0$

Atmospheric terms cancel. Since the tank is large, $v_1 \approx 0$ . So:

$\rho g h = \tfrac{1}{2}\rho v_2^2 \;\Rightarrow\; \boxed{v_2 = \sqrt{2 g h}}$

This is the same as a freely-falling object dropped from height $h$ ! That's the "Torricelli" insight.

Conditions for Validity

Hole area $\ll$ tank area (so $v_1 \approx 0$ ).
Both surfaces exposed to same external pressure.
Ideal fluid (no viscous losses).

Projectile Range of the Jet

If the hole is at height $H$ above the ground, the jet exits horizontally at speed $v = \sqrt{2gh}$ and falls under gravity:

Time to ground: $t = \sqrt{2H/g}$
Horizontal range: $R = v t = \sqrt{2 g h} \cdot \sqrt{2 H / g} = 2 \sqrt{}$

Memorable result: $R = 2\sqrt{hH}$ . Maximum range when $h = H$ (hole at half-height).

Torricelli Concepts 🎯

Torricelli Calculations 🧮 (g = 10)

Water in a large tank has surface 5.0 m above a small hole. Exit speed (m/s)?
Same tank — hole is 1.25 m above the ground. Time for the jet to hit the ground (s, 2 sig fig)?
Same tank — horizontal range of the jet (m)?

Torricelli Reasoning 🔍

Exit Quiz — Torricelli's Theorem ✅

F_{l i f t} \approx Δ P \cdot A = \frac{1}{2} ρ_{ai r} (v_{t o p}^{2} - v_{b o t}^{2}) A

2. Venturi Meter (flow measurement)

A pipe with a narrow throat and a U-tube manometer between wide and narrow.
Difference in heights of the manometer fluid → $\Delta P$ → using continuity + Bernoulli, solve for $v$ and hence $Q$ .

3. Pitot Tube (airspeed)

Has a forward-facing port (stagnation point, $v = 0$ , $P = P_{stag}$ ) and a side port (flow speed $v$ , $P = P_{static}$ ).
Bernoulli at constant height: $P_{stag} = P_{static} + \tfrac{1}{2}\rho v^2$ .
$v = \sqrt{2(P_{stag} - P_{static})/\rho}$ .

4. Atomizer / Spray Bottle

Squeezing air across a vertical tube creates fast horizontal flow → low $P$ at top.
Atmospheric pressure pushes liquid up the tube and into the air stream.

Wind across a chimney top → low $P$ at top → draws smoke up.

6. Magnus Force (curveballs)

Spinning ball drags air around it; one side speeds up, the other slows down → pressure differential → side force.

Application Concepts 🎯

Application Calculations 🧮 ( $\rho_{air} = 1.2$ , $\rho_w = 1000$ , g = 10)

A Pitot tube on an airplane reads $P_{stag} - P_{static} = 600$ Pa. Airspeed (m/s)?

Application Reasoning 🔍

Exit Quiz — Real-World Applications ✅

Bernoulli:

P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g y_2

→ solve for the unknown pressure.

Strategy Cheat Sheet

Given	Use
Areas + one $v$	Continuity for unknown $v$
Speeds + one $P$	Bernoulli for unknown $P$
Manometer reading	$\Delta P = \rho_{man} g h$ then Bernoulli
Tank with hole	Torricelli (special-case Bernoulli)
Pitot tube	$P_{stag} = P_{static} + \tfrac{1}{2}\rho v^2$

Worked Venturi

A horizontal Venturi: wide section $A_1 = 0.10$ m² has $v_1 = 2$ m/s, $_{}$ kPa; narrow throat m². Find .

Step 1 — Continuity: $v_2 = v_1 (A_1/A_2) = 2 \times 4 = 8$ m/s.

Step 2 — Bernoulli (horizontal): $P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 - v_2^2) = 200{,}000 + \tfrac{1}{2}(1000)(4 - 64) = 170{,}000 \text{ Pa}$

So $P_2 = 170$ kPa. The throat pressure dropped by 30 kPa.

Combined Pipe with Elevation Change

A pipe carries water from $A_1 = 0.04$ m², $v_1 = 1$ m/s, $y_{}$ , kPa upward to m², m. Find .

$v_2 = 4$ m/s by continuity.
Bernoulli: $P_2 = 300{,}000 + \tfrac{1}{2}(1000)(1 - 16) + 1000(10)(0 - 5) = 300{,}000 - 7500 - 50{,}000 = 242{,}500$ Pa.

Combined Application Quick MC 🎯

Combined Workshop 🧮 ( $\rho_w = 1000$ , g = 10)

Horizontal Venturi: $A_1 = 0.020$ m², $v_1 = 3$ m/s, $P_1 = 250$ kPa; $A_2 = 0.010$ m². Find $P_2$ (kPa).
A vertical pipe: $A_1 = 0.005$ m², $v_1 = 4$ m/s, $_{}$ , kPa. Upper end: m² (same area), m. Find (kPa).
Same pipe, but upper end narrows to $A_2 = 0.0025$ m². Find $P_2$ (kPa).

Combined-Reasoning Drill 🔍

Exit Quiz — Combined Workshop ✅

Faster fluid = lower pressure (at same height)

Torricelli:

v = \sqrt{2gh}

Pitot:

P_{stag} - P_{static} = \tfrac{1}{2}\rho v^2

Combined Continuity + Bernoulli unlocks Venturi-type problems

Bernoulli AP Cheat Sheet

Concept	Equation
Bernoulli (general)	$P + \tfrac{1}{2}\rho v^2 + \rho g y =$ const
Horizontal pipe	$P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$
Constant area pipe	$P_1 + \rho g y_1 = P_2 + \rho g y_2$ (just hydrostatic)
Torricelli	$v = \sqrt{2gh}$
Pitot tube	$v = \sqrt{2(P_{stag} - P_{static})/\rho}$
Lift (rough)	$F_{lift} \approx \tfrac{1}{2}\rho_{air}(v_t^2 - v_b^2) A$
Continuity (incompressible)	$A_1 v_1 = A_2 v_2$

AP Reasoning Patterns

"Pressure drops where fluid speeds up" — Bernoulli at same height.
"Tank drains faster when..." — Torricelli, larger $h$ ⇒ larger $v$ .
"Wing lift comes from..." — pressure difference from speed difference.
"Two parallel boats drift together" — Bernoulli ⇒ low $P$ between them.
"Combined speed + pressure problem" — apply continuity, then Bernoulli.

Common Pitfalls

❌ Forgetting that ALL Bernoulli terms are per unit volume (Pa).
❌ Confusing $\rho_{fluid}$ with $\rho_{air}$ in lift problems.
❌ Applying Bernoulli to viscous or turbulent flow (e.g., long thin pipes with friction).
❌ Forgetting to use continuity FIRST when speeds aren't given directly.

Final Mental Model

Bernoulli = "energy conservation per unit volume along a streamline." Find your two points, identify $P$ , $v$ , $y$ at each, and let energy do the rest.

AP Synthesis MC 🎯

AP Synthesis Calculations 🧮 ( $\rho_w = 1000$ , $\rho_{air} = 1.2$ , g = 10)

A large open tank's water surface is 10 m above a small hole. Exit speed (m/s)?
Horizontal pipe: $A_1 = 0.04$ m², $v_1 = 1$ m/s, kPa; m². Find (kPa).

AP Concept Synthesis 🔍

Exit Quiz — Bernoulli Synthesis ✅

Gravity: appears in

\Delta PE

\rho g y_2 - \rho g y_1

[\tfrac{1}{2}\rho v^2] = (\text{kg/m}^3)(\text{m/s})^2 = \text{kg}\cdot\text{m}^{-1}\text{s}^{-2} = \text{J/m}^3 ✓

[\rho g y] = (\text{kg/m}^3)(\text{m/s}^2)(\text{m}) = \text{J/m}^3 ✓

A horizontal water pipe widens. At point 1: $v_1 = 6$ m/s, $P_1 = 100$ kPa. At point 2: $v_2 = 2$ m/s. $P_2$ in kPa?

R = v t = 2 g h \cdot 2 H / g = 2 h H

An airfoil has $v_{top} = 80$ m/s, $v_{bot} = 60$ m/s. Pressure difference $\Delta P = P_{bot} - P_{top}$ (Pa)?

Wing area = 12 m². Lift force using the $\Delta P$ from Q2 (N)?

A Pitot tube on a small drone reads $\Delta P = 24$ Pa ( $\rho_{air} = 1.2$ ). Airspeed (m/s)?

Bernoulli's Equation

Derivation Sketch

Mapping to Mechanics

Quick Sanity Check (Units)

Real-World Examples

A Common Misconception

Linking to Continuity

Torricelli's Theorem

Conditions for Validity

Projectile Range of the Jet

2. Venturi Meter (flow measurement)

3. Pitot Tube (airspeed)

4. Atomizer / Spray Bottle

5. Chimney Effect

6. Magnus Force (curveballs)

Strategy Cheat Sheet

Worked Venturi

Combined Pipe with Elevation Change

Bernoulli AP Cheat Sheet

AP Reasoning Patterns

Common Pitfalls

Final Mental Model