Bernoulli's Equation

📜 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)

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

Required Assumptions

1. Incompressible — $\rho$ constant
2. Non-viscous — no internal friction
3. Steady — flow speed at each point doesn't change in time
4. 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 $v \approx 0$.
• Use one point where you have many unknowns? → No. Pick known points!

Bernoulli Concepts 🎯

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

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

1. Surface of an open tank ($P = P_{atm}$, $v = 0$, $y = 5.0$ m). The $\rho g y$ term (Pa)?

2. Same surface — the $\tfrac{1}{2}\rho v^2$ term (Pa)?

3. A point inside a pipe with $v = 4.0$ m/s. The dynamic pressure term $\tfrac{1}{2}\rho v^2$ (Pa)?

Bernoulli Setup Reasoning 🔍

Exit Quiz — Bernoulli Setup ✅

⚡ 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.

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: $W_{pressure} = (P_1 - P_2) V$
• Gravity: appears in $\Delta PE$

KE change per unit volume: $\tfrac{1}{2}\rho v_2^2 - \tfrac{1}{2}\rho v_1^2$. PE change per unit volume: $\rho g y_2 - \rho g y_1$.

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

Quick Sanity Check (Units)

• $[P] = \text{Pa} = \text{N/m}^2 = \text{N}\cdot\text{m/m}^3 = \text{J/m}^3$
• $[\tfrac{1}{2}\rho v^2] = (\text{kg/m}^3)(\text{m/s})^2 = \text{kg}\cdot\text{m}^{-1}\text{s}^{-2} = \text{J/m}^3 ✓$
• $[\rho g y] = (\text{kg/m}^3)(\text{m/s}^2)(\text{m}) = \text{J/m}^3 ✓$

Energy Conservation Concepts 🎯

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

1. Compute kinetic energy per unit volume of water moving at 5 m/s (Pa).

2. Compute gravitational PE per unit volume of water at $y = 8$ m (Pa).

3. 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 ✅

🔄 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, shower curtains
• How to predict pressure direction from speed change

Horizontal Bernoulli

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

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

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

Where the fluid is faster, the pressure is lower.

Real-World Examples

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 🧮 ($\rho_w = 1000$, $\rho_{air} = 1.2$)

1. 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)?

2. Air flows over a wing at 50 m/s on top and 40 m/s on bottom. Pressure difference (Pa, top vs bottom: $\Delta P = P_{bottom} - P_{top}$)?

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?

Pressure-Speed Reasoning 🔍

Exit Quiz — Pressure-Speed Trade-off ✅

🚿 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
• 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: $R = v \, t = \sqrt{2gh} \cdot \sqrt{2H/g} = 2\sqrt{hH}$

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

Torricelli Concepts 🎯

Torricelli Calculations 🧮 (g = 10)

1. Water in a large tank has surface 5.0 m above a small hole. Exit speed (m/s)?

2. Same tank — hole is 1.25 m above the ground. Time for the jet to hit the ground (s, 2 sig fig)?

3. Same tank — horizontal range of the jet (m)?

Torricelli Reasoning 🔍

Exit Quiz — Torricelli's Theorem ✅

🌍 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 and chimneys

Application Catalog

1. Airfoil / Lift

• A wing's shape and its tilt (angle of attack) turn the oncoming air downward. In doing so, the air flowing over the top speeds up and the air underneath slows slightly.
• The speeds are measured or given; Bernoulli then says faster air on top ⇒ lower $P$ on top.
• Pressure difference × wing area = lift force.
• A common textbook story says air over the top must travel a longer path and "meet up" with the air underneath at the trailing edge. That is false: nothing forces the two streams to rejoin, and air over the top actually reaches the trailing edge first. Bernoulli explains lift from the speed difference, not from equal travel times.

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

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.

5. Chimney Effect

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

Application Concepts 🎯

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

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

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

3. Wing area = 12 $m^{2}$. Lift force using the $\Delta P$ from Q2 (N)?

Application Reasoning 🔍

Exit Quiz — Real-World Applications ✅

🛠️ 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:

1. Identify two points on the same streamline.
2. Continuity: $A_1 v_1 = A_2 v_2$ → solve for the unknown speed.
3. 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

Worked Venturi

A horizontal Venturi: wide section $A_1 = 0.10$ $m^{2}$ has $v_1 = 2$ m/s, $P_1 = 200$ kPa; narrow throat $A_2 = 0.025$ $m^{2}$. Find $P_2$.

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 $A_1 = 0.04$ $m^{2}$, $v_1 = 1$ m/s, $y_1 = 0$, $P_1 = 300$ kPa upward to $A_2 = 0.01$ $m^{2}$, $y_2 = 5$ m. Find $P_2$.

• $v_2 = 4$ 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)

1. Horizontal Venturi: $A_1 = 0.020$ $m^{2}$, $v_1 = 3$ m/s, $P_1 = 250$ kPa; $A_2 = 0.010$ $m^{2}$. Find $P_2$ (kPa).

2. A vertical pipe: $A_1 = 0.005$ $m^{2}$, $v_1 = 4$ m/s, $y_1 = 0$, $P_1 = 400$ kPa. Upper end: $A_2 = 0.005$ $m^{2}$ (same area), $y_2 = 10$ m. Find $P_2$ (kPa).

3. Same pipe, but upper end narrows to $A_2 = 0.0025$ $m^{2}$. Find $P_2$ (kPa).

Combined-Reasoning Drill 🔍

Exit Quiz — Combined Workshop ✅

🎯 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:

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

AP Reasoning Patterns

1. "Pressure drops where fluid speeds up" — Bernoulli at same height.
2. "Tank drains faster when..." — Torricelli, larger $h$ ⇒ larger $v$.
3. "Wing lift comes from..." — pressure difference from speed difference.
4. "Two parallel boats drift together" — Bernoulli ⇒ low $P$ between them.
5. "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)

1. A large open tank's water surface is 10 m above a small hole. Exit speed (m/s)?

2. Horizontal pipe: $A_1 = 0.04$ $m^{2}$, $v_1 = 1$ m/s, $P_1 = 200$ kPa; $A_2 = 0.01$ $m^{2}$. Find $P_2$ (kPa).

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

AP Concept Synthesis 🔍

Exit Quiz — Bernoulli Synthesis ✅