Playground
Venturi tube showing pressure drop where velocity increases. Manometer tubes visualize pressure at each section.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Pressureoutput Static pressure of the fluid | Pa | M·L⁻¹·T⁻² | 0 – 200000 | |
| Density Mass density of the fluid | kg/m³ | M·L⁻³ | 1 – 13600 | |
| Velocity Flow speed along streamline | m/s | L·T⁻¹ | 0 – 100 | |
| Gravity Gravitational acceleration | m/s² | L·T⁻² | 9.8 – 9.8 | |
| Height Elevation above reference | m | L | 0 – 100 |
Deep dive
Derivation
Apply the work-energy theorem to a fluid element along a streamline. Net work done by pressure forces equals the change in kinetic plus potential energy: (P₁ - P₂)dV = ½dm(v₂² - v₁²) + dm·g(h₂ - h₁). Dividing by dV and using dm/dV = ρ gives P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂.
Experimental verification
Verified daily in venturi meters, pitot tubes (used to measure aircraft airspeed), and atomizers. Modern wind-tunnel pressure mapping on airfoils confirms Bernoulli's predictions to high precision in incompressible regimes.
Common misconceptions
- Bernoulli alone does NOT fully explain airplane lift — circulation theory and Newton's third law also matter
- It applies along a streamline, not between arbitrary points in a flow
- It requires inviscid flow — fails inside boundary layers or in turbulent wakes
Real-world applications
- Pitot tubes for aircraft airspeed measurement
- Venturi meters for flow measurement in pipes
- Carburetors and spray atomizers
- Curveball aerodynamics in baseball and soccer
- Roof damage from hurricanes (low pressure above, high pressure inside)
Worked examples
Lift on an airplane wing
Given:
- ρ:
- 1.2
- v_top:
- 80
- v_bot:
- 70
Find: ΔP
Solution
ΔP = ½ρ(v_top² - v_bot²) = ½(1.2)(6400 - 4900) = 900 Pa
Venturi meter pressure drop
Given:
- ρ:
- 1000
- v_1:
- 1
- v_2:
- 4
Find: P_1 - P_2
Solution
P_1 - P_2 = ½ρ(v_2² - v_1²) = ½(1000)(16 - 1) = 7500 Pa
Scenarios
What if…
- scenario:
- What if the flow is highly viscous (like honey)?
- answer:
- Bernoulli fails — viscous dissipation removes energy from the fluid. Use the extended Bernoulli equation with a head-loss term, or solve the Navier-Stokes equations directly.
- scenario:
- What if the air speed exceeds Mach 0.3?
- answer:
- Compressibility matters and density changes along the streamline. Use the compressible Bernoulli (with enthalpy) or full gas dynamics equations.
- scenario:
- What if you blow between two sheets of paper?
- answer:
- They get pulled together, not blown apart. Faster air between them has lower pressure, while still air outside pushes them inward — a classic Bernoulli demo.
Limiting cases
- condition:
- v = 0
- result:
- P + ρgh = const
- explanation:
- Static fluid — recovers hydrostatic pressure.
- condition:
- h = const
- result:
- P + ½ρv² = const
- explanation:
- Horizontal flow — pure pressure-velocity tradeoff (Venturi effect).
- condition:
- P = const (free surface)
- result:
- ½v² + gh = const
- explanation:
- Free-surface flow — leads directly to Torricelli's law.
Context
Daniel Bernoulli · 1738
Published in Hydrodynamica, derived from energy conservation in flowing fluids — though the modern form was completed by Euler decades later.
Hook
How does an airplane wing generate lift from moving air?
Air flows over a wing at 80 m/s on top and 70 m/s underneath. Find the pressure difference (with ρ ≈ 1.2 kg/m³) that creates lift.
Dimensions: [P] = [ρv²] = [ρgh] → M·L⁻¹·T⁻² = (M·L⁻³)(L²·T⁻²) = (M·L⁻³)(L·T⁻²)(L) ✓ (all energy density)
Validity: Inviscid, incompressible, steady flow along a single streamline. Fails when viscosity, turbulence, or compressibility (M > 0.3) become significant.