Continuity Equation

Equivalent forms

\rho_1 A_1 v_1 = \rho_2 A_2 v_2

\nabla \cdot (\rho \vec{v}) + \frac{\partial \rho}{\partial t} = 0

Mass conservation reduced to a single product rule.

Symbol	Name	SI unit	Dimension	Range
$A_1$	Inlet Area Cross-sectional area at point 1	m²	L²	0.001 – 1
$v_1$	Inlet Velocity Fluid velocity at point 1	m/s	L·T⁻¹	0 – 20
$A_2$	Outlet Area Cross-sectional area at point 2	m²	L²	0.0001 – 1
$v_2$	Outlet Velocityoutput Fluid velocity at point 2	m/s	L·T⁻¹	0 – 200

Unit systems

Symbol	SI	CGS	Imperial
$A_1$	m²	cm²	ft²
$v_1$	m/s	cm/s	ft/s
$A_2$	m²	cm²	ft²
$v_2$	m/s	cm/s	ft/s

Where it holds

Steady, incompressible flow. For compressible flow

use \rho Av

form; for unsteady flow use the differential form

\partial \rho /\partial t + \nabla \cdot (\rho v) = 0

.

Dimensional analysis

$[A_{1}][v_{1}] = [$ $A_{2}][v_{2}] \to (L^{2})(L\cdot T^{-1}) = (L^{2})(L\cdot T^{-1}) \to L^{3}\cdot T^{-1} = L^{3}\cdot T^{-1} \checkmark$ (volume flow rate)

Discovery

Leonhard Euler · 1757

Formulated as part of Euler's foundational equations of fluid dynamics, expressing mass conservation in differential form.

Try this

Why does water shoot out faster when you pinch the end of a hose?

A garden hose with cross-sectional area 4 cm² carries water at 1 m/s. You pinch the end to 1 cm². How fast does the water shoot out?

Research status: stable

Real-world applications

Garden hose nozzles and fire hoses
Carburetor and venturi meter design
River flow narrowing at canyons
Blood flow through narrowed arteries (stenosis)

Common misconceptions

Continuity does NOT say mass flow rate equals volume flow rate — only true for incompressible fluids
It applies along a streamtube, not necessarily across an entire flow with branching
Velocity here is the average over the cross-section, not the local maximum (which is higher in viscous flow)

Experimental verification

Verified by countless flow experiments with venturi meters, pitot tubes, and pipe networks. Modern PIV (particle image velocimetry) confirms continuity to high precision in liquids and low-Mach gases.

Derivation

Consider a steady flow through a tube.

In time

\Delta t

, mass entering at section

1\,\mathrm{is} \rho \cdot A_{1}\cdot v_{1}\cdot \Delta t

, and mass exiting at section

2\,\mathrm{is} \rho \cdot A_{2}\cdot v_{2}\cdot \Delta t

.

Conservation of mass requires these be equal:

\rho A_{1}v_{1} = \rho A_{2}v_{2}

.

For an incompressible fluid (constant

\rho )

, this reduces to

A_{1}v_{1} = A_{2}v_{2}

.

Limiting cases

A_

2 \to A_1

⟶

v_2 \to v_1

Same pipe diameter throughout means uniform speed.

A_

2 \to 0

⟶

v_2 \to \infty

Infinitely narrow opening would require infinite speed (unphysical — viscosity and compressibility kick in).

A_

2 \to \infty

⟶

v_2 \to 0

Wide reservoir slows the flow to nearly zero.

What if…

What if you pinch the hose to 1/10 the area?

Speed multiplies by 10. The water exits $10\times$ faster, which is why pinching makes it spray much further.

What if the fluid is compressible (like air at high speed)?

You must $use \rho _{1}A_{1}v_{1} = \rho _{2}A_{2}v_{2}$ . Density changes (e.g., across shock waves in a rocket nozzle) become essential.

What if the pipe branches into two?

The total mass flow splits: $A_{1}v_{1} = A_{2}v_{2}$ + $A_{3}v_{3}$ . Each branch gets a share proportional to the pressure-driven flow.

1

Pinched garden hose

Given ·

A 1:: 0.0004
v 1:: 1
A 2:: 0.0001

Find · v_2

Steps

Apply $A_{1}v_{1} = A_{2}v_{2}$
Solve for $v_{2} = A_{1}v_{1}$ / $A_{2}$
$v_{2} = (4\,\mathrm{cm}^{2} \times 1\,\mathrm{m/s}) / (1\,\mathrm{cm}^{2}) = 4\,\mathrm{m/s}$

Result ·

v_2 = A_1\cdot v_1

/ A_

2 = (0.0004 \times 1) / 0.0001 = 4\,\mathrm{m/s}

2

River widening into a lake

Given ·

A 1:: 50
v 1:: 2
A 2:: 500

Find · v_2

Steps

$v_{2} = A_{1}v_{1}$ / $A_{2}$
$v_{2} = 100 / 500 = 0.2\,\mathrm{m/s}$
Slowing by $10\times as$ the channel widens $10\times$

Result ·

v_2 = (50 \times 2) / 500 = 0.2\,\mathrm{m/s}

Bernoulli's Equation→Torricelli's Law→