Venturi Effect — Physica

Equivalent forms

v_2 = v_1\frac{A_1}{A_2}

Q = A_1 v_1 = A_2 v_2

p_2 = p_1 - \frac{1}{2}\rho v_1^2\left(\frac{A_1^2}{A_2^2} - 1\right)

Squeeze the flow, drop the pressure — the engine of carburetors, atomizers, and aspirators.

Symbol	Name	SI unit	Dimension	Range
$p_1$	Inlet Pressure Static pressure at the wide section	Pa	M·L⁻¹·T⁻²	0 – 500000
$\rho$	Density Fluid mass density	kg/m³	M·L⁻³	1 – 2000
$v_1$	Inlet Velocity Flow speed at the wide section	m/s	L·T⁻¹	0 – 20
$v_2$	Throat Velocity Flow speed at the constriction	m/s	L·T⁻¹	0 – 80
$p_2$	Throat Pressureoutput Static pressure at the constriction	Pa	M·L⁻¹·T⁻²	-100000 – 500000

Unit systems

Symbol	SI	CGS	Imperial
$p_1$	Pa	dyn/cm²	psi
$\rho$	kg/m³	g/cm³	slug/ft³
$v_1$	m/s	cm/s	ft/s
$v_2$	m/s	cm/s	ft/s
$p_2$	Pa	dyn/cm²	psi

Where it holds

Valid for steady, incompressible, low-loss flow along a streamline (negligible friction). Real meters apply a discharge coefficient

(\sim 0.95-0.98)

for viscous losses; compressibility corrections are needed above Mach 0.3.

Dimensional analysis

$[\tfrac{1}{2}\rho v^{2}] = (M\cdot L^{-3})(L^{2}\cdot T^{-2}) = M\cdot L^{-1}\cdot T^{-2} \checkmark$ (pressure)

Discovery

Giovanni Battista Venturi · 1797

Venturi, an Italian physicist and protégé of Lagrange, experimentally documented how constricting a flow lowers its pressure. Clemens Herschel later used the principle to build the first practical flow meter in 1887.

Try this

Why does a fast jet of air between two ships pull them together — and how does a carburetor sip fuel?

Water flows through a pipe that narrows from area 0.01 m² to 0.0025 m². If the inlet velocity is 1 m/s and inlet pressure is 200 kPa, what is the pressure at the throat? (ρ = 1000 kg/m³)

Research status: stable

Real-world applications

Venturi flow meters in pipelines
Carburetors and aspirators (drawing fuel or liquid into a fast airstream)
Atomizers and spray nozzles
Bunsen burner and jet-pump (eductor) entrainment

Common misconceptions

The fluid does not 'suck' — lower pressure at the throat is a consequence of higher kinetic energy, not a pulling force
Total mechanical energy is conserved in the ideal case; the pressure drop is recovered in the diffuser if losses are small
It is the same physics as an airplane wing only in part — circulation and the Kutta condition also matter for lift

Experimental verification

Venturi meters are standard industrial flow instruments calibrated against gravimetric standards to within 1%. Wind-tunnel and water-channel pressure taps confirm the throat pressure minimum predicted by Bernoulli.

Derivation

Combine continuity

(A_{1}v_{1} = A_{2}v_{2})

with Bernoulli

(p_{1} + \tfrac{1}{2}\rho v_{1}^{2} = p_{2} + \tfrac{1}{2}\rho v_{2}^{2})

for a horizontal tube.

Substituting

v_{2} = v_{1}

(A_{1}

/

A_{2})

gives the pressure drop

p_{1} - p_{2} = \tfrac{1}{2}\rho v_{1}^{2}[

(A_{1}

/

A_{2})^{2} - 1]

.

Measuring that drop yields the flow rate, which is the basis of the Venturi meter.

Limiting cases

A_

2 \to A_1

(no constriction)⟶

v_2 \to v_1

,

p_2 \to p_1

With no narrowing there is no speed-up and no pressure drop.

A_2 ≪ A_1 (severe constriction)⟶ p_2 can fall below vapor pressureExtreme throat speeds drop pressure enough to boil the liquid — cavitation.

What if…

What if the throat is made even narrower?

Throat velocity rises and pressure keeps falling; below the liquid's vapor pressure it cavitates, forming bubbles that collapse violently and erode the pipe.

What if the fluid is a gas at high speed?

Above Mach 0.3 you must include compressibility; near Mach 1 the throat can choke and the simple Bernoulli relation fails.

1

Throat pressure in a 4:1 constriction

Given ·

p 1:: 200000
\rho:: 1000
v 1:: 1

Find · p_2

Steps

Continuity: $v_{2} = v_{1} \times A_{1}$ / $A_{2} = 1 \times (0.01/0.0025) = 4\,\mathrm{m/s}$
Bernoulli: $p_{2} = p_{1} - \tfrac{1}{2}\rho (v_{2}^{2} - v_{1}^{2})$
$p_{2} = 200000 - \tfrac{1}{2}(1000)(16 - 1) = 192$ ,500 Pa

Result ·

v_{2} = v_{1}

(A_{1}

/

A_{2}) = 1 \times 4 = 4\,\mathrm{m/s}

;

p_{2} = p_{1} - \tfrac{1}{2}\rho (v_{2}^{2} - v_{1}^{2}) = 200000 - \tfrac{1}{2}\times 1000\times (16 - 1) = 200000 - 7500 = 192

,500 Pa

Bernoulli's Equation→Continuity Equation→Torricelli's Law→