Bernoulli's Equation

Equivalent forms

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

Energy conservation written in three pressure-equivalent terms.

Symbol	Name	SI unit	Dimension	Range
$P$	Pressureoutput Static pressure of the fluid	Pa	M·L⁻¹·T⁻²	0 – 200000
$ρ$	Density Mass density of the fluid	kg/m³	M·L⁻³	1 – 13600
$v$	Velocity Flow speed along streamline	m/s	L·T⁻¹	0 – 100
$g$	Gravity Gravitational acceleration	m/s²	L·T⁻²	9.8 – 9.8
$h$	Height Elevation above reference	m	L	0 – 100

Unit systems

Symbol	SI	CGS	Imperial
$P$	Pa	Ba	psi
$ρ$	kg/m³	g/cm³	lb/ft³
$v$	m/s	cm/s	ft/s
$g$	m/s²	cm/s²	ft/s²
$h$	m	cm	ft

Where it holds

Inviscid, incompressible, steady flow along a single streamline. Fails when viscosity, turbulence, or compressibility (M > 0.3) become significant.

Dimensional analysis

$[P] = [\rho v^{2}] = [\rho gh] \to M\cdot L^{-1}\cdot T^{-2} = (M\cdot L^{-3})(L^{2}\cdot T^{-2}) = (M\cdot L^{-3})(L\cdot T^{-2})(L) \checkmark (all$ energy density)

Discovery

Daniel Bernoulli · 1738

Published in Hydrodynamica, derived from energy conservation in flowing fluids — though the modern form was completed by Euler decades later.

Try this

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.

Research status: stable

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)

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

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.

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:

(P_{1} - P_{2})dV = \tfrac{1}{2}dm(v_{2}^{2} - v_{1}^{2}) + dm\cdot g(h_{2} - h_{1})

.

Dividing by dV and using

dm/dV = \rho

gives

P_{1} + \tfrac{1}{2}\rho v_{1}^{2} + \rho gh_{1} = P_{2} + \tfrac{1}{2}\rho v_{2}^{2} + \rho gh_{2}

.

Limiting cases

v = 0

⟶

P + \rho gh

= constStatic fluid — recovers hydrostatic pressure.

h =

const⟶

P + \tfrac{1}{2}\rho v^{2}

= constHorizontal flow — pure pressure-velocity tradeoff (Venturi effect).

P =

const (free surface)⟶

\tfrac{1}{2}v^{2} + gh

= constFree-surface flow — leads directly to Torricelli's law.

What if…

What if the flow is highly viscous (like honey)?

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.

What if the air speed exceeds Mach 0.3?

Compressibility matters and density changes along the streamline. Use the compressible Bernoulli (with enthalpy) or full gas dynamics equations.

What if you blow between two sheets of paper?

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.

1

Lift on an airplane wing

Given ·

ρ:: 1.2
v top:: 80
v bot:: 70

Find ·

\Delta P

Steps

Apply Bernoulli at same height: $P_bot + \tfrac{1}{2}\rho v_bot^{2} = P_top + \tfrac{1}{2}\rho v_top^{2}$
$\Delta P = P_bot - P_top = \tfrac{1}{2}\rho (v_top^{2} - v_bot^{2})$
$\Delta P = 0.5 \times 1.2 \times (6400 - 4900) = 0.5 \times 1.2 \times 1500 = 900\,\mathrm{Pa}$

Result ·

\Delta P = \tfrac{1}{2}\rho (v_top^{2} - v_bot^{2}) = \tfrac{1}{2}(1.2)(6400 - 4900) = 900\,\mathrm{Pa}

2

Venturi meter pressure drop

Given ·

ρ:: 1000
v 1:: 1
v 2:: 4

Find · P_1 - P_2

Steps

Same height: $P_{1} + \tfrac{1}{2}\rho v_{1}^{2} = P_{2} + \tfrac{1}{2}\rho v_{2}^{2}$
$P_{1} - P_{2} = \tfrac{1}{2}\rho (v_{2}^{2} - v_{1}^{2})$
$= 500 \times 15 = 7500\,\mathrm{Pa}$

Result ·

P_1 - P_2 = \tfrac{1}{2}\rho (v_2^{2} - v_1^{2}) = \tfrac{1}{2}(1000)(16 - 1) = 7500\,\mathrm{Pa}

Continuity Equation→Torricelli's Law→Hydrostatic Pressure→