Bernoulli's Equation — Venturi Tube

A pipe narrows in the middle. Conservation of mass says the fluid has to speed up; Bernoulli says the pressure has to fall. Watch both effects together.

Visualization

Input

D_1

Inlet diameter

\mathrm{cm}

D_2

Throat diameter

\mathrm{cm}

Q

Volumetric flow rate

\mathrm{L/s}

p_1

Inlet gauge pressure

\mathrm{kPa}

\Delta h

Outlet elevation gain

\mathrm{m}

\rho

Fluid density

\mathrm{kg/m^{3}}

g

Gravity

\mathrm{m/s^{2}}

Show the streamline + cross-section view

Two ideas talking to each other

A venturi tube — a pipe that gets narrower in the middle — is the cleanest way to see two of fluid mechanics' biggest ideas working together.

The first is conservation of mass (continuity). Whatever flow rate $Q$ enters the pipe has to come out the other side, so the velocity has to change as the cross-section changes. With area $A_1$ at the inlet and $A_2$ at the throat:

Q = A_1 v_1 = A_2 v_2 \quad\Rightarrow\quad v_2 = \frac{A_1}{A_2}\,v_1

A throat half the diameter has a quarter the area, so the fluid moves four times as fast through it. Source: Wikipedia — Venturi effect.

The second is conservation of energy (Bernoulli's equation). Along a streamline of an inviscid, incompressible, steady flow, the sum of pressure, kinetic energy per volume, and gravitational potential energy per volume is constant:

p + \tfrac{1}{2}\rho v^{2} + \rho g h = \text{const}

Source: OpenStax University Physics §14.6; Wikipedia — Bernoulli's principle.

Velocity up, pressure down

Apply Bernoulli's equation between the inlet (subscript 1) and the throat (subscript 2):

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

Solving for the throat pressure:

p_2 = p_1 - \tfrac{1}{2}\rho\,(v_2^{2} - v_1^{2}) - \rho g\,\Delta h

The first correction term is the Venturi pressure drop: kinetic energy has to come from somewhere, and that "somewhere" is the static pressure. The second is the gravity term — when fluid climbs ( $\Delta h > 0$ ), it pays more pressure to gain potential energy. Try the elevation slider to see them trade off.

If $p_2$ goes below atmospheric, you'll get cavitation — vapour bubbles form in the low-pressure region. The diagram flags it.

The fire-hose example

OpenStax §14.6 Example 14.7 is a fire hose narrowing from 6.4 cm to a 3.0 cm nozzle, lifting water 10 m, with a 1.62 MPa source pressure pushing 40 L/s. The textbook reports $v_1 \approx 12.4$ m/s, $v_2 \approx 56.6$ m/s, and $p_2 \approx 0$ at the nozzle (atmospheric — exactly what you want at the open end). Load that preset and watch the model reproduce all three.

Show your work

Inlet cross-sectional area: $A_1 \;=\; \dfrac{\pi D_1^{2}}{4} \;=\; 0.003217\;\mathrm{m^{2}}$
Throat cross-sectional area: $A_2 \;=\; \dfrac{\pi D_2^{2}}{4} \;=\; 7.069 \times 10^{-4}\;\mathrm{m^{2}}$
Inlet velocity (continuity): $v_1 \;=\; \dfrac{Q}{A_1} \;=\; 12.43\;\mathrm{m/s}$
Throat velocity (continuity): $v_2 \;=\; \dfrac{Q}{A_2} \;=\; 56.59\;\mathrm{m/s}$
Kinetic-energy pressure drop (½ρ Δv²): $\Delta p_{\text{kin}} \;=\; \tfrac{1}{2}\,\rho\,(v_2^{2} - v_1^{2}) \;=\; 1.524 \times 10^{6}\;\mathrm{Pa}$
Gravity pressure drop (ρ g Δh): $\Delta p_{\text{grav}} \;=\; \rho\,g\,\Delta h \;=\; \rho\,\textcolor{#dc4124}{\left(9.8\right)}\,\Delta h \;=\; 98000\;\mathrm{Pa}$
Outlet pressure (Bernoulli): $p_2 \;=\; p_1 - \tfrac{1}{2}\rho(v_2^{2} - v_1^{2}) - \rho g\,\Delta h \;=\; p_1 - \tfrac{1}{2}\rho(v_2^{2} - v_1^{2}) - \rho \textcolor{#dc4124}{\left(9.8\right)}\,\Delta h \;=\; -1823\;\mathrm{Pa}$

Worked Example

OpenStax University Physics §14.6 Example 14.7 (Fire Hose Nozzle): hose diameter 6.40 cm, nozzle diameter 3.00 cm, flow 40.0 L/s. Continuity gives v₁ ≈ 12.4 m/s and v₂ ≈ 56.6 m/s (the source's reported values).

Quantity	Expected	Computed	Δ
v1	12.43	12.43	0.00398
v2	56.59	56.59	0.001576

Source: OpenStax, University Physics Vol. 1 §14.6 Example 14.7 (Fire Hose Nozzle) — velocities

Worked Example

Same OpenStax Example 14.7: with p₁ = 1.62 MPa gauge and a 10.0 m height gain at the nozzle, Bernoulli predicts the nozzle exit pressure p₂ ≈ 0 (atmospheric). With unrounded velocities the model evaluates to p₂ ≈ −1.82 kPa — still essentially atmospheric; the small residual is the rounding propagating from OpenStax's 3-significant-figure intermediate velocities.

Quantity	Expected	Computed	Δ
p2_pa	-1820	-1823	2.949

Source: OpenStax, University Physics Vol. 1 §14.6 Example 14.7 (Fire Hose Nozzle) — outlet pressure

Sources

[1] OpenStax. University Physics Volume 1 — §14.6 Bernoulli's Equation, eq. p + ½ρv² + ρgh = constant; p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂ (openstax.org)
[2] Wikipedia — Bernoulli's principle, eq. ½ρv² + ρgz + p = constant; assumptions: steady, incompressible, inviscid, along a streamline (en.wikipedia.org)
[3] Wikipedia — Venturi effect, eq. p₁ − p₂ = ρ/2·(v₂² − v₁²); Q = v₁A₁ = v₂A₂ (continuity) (en.wikipedia.org)

Assumptions

Steady flow — flow parameters at any point do not change with time (OpenStax §14.6).
Incompressible flow — fluid density ρ is constant (OpenStax §14.6).
Inviscid flow — viscous (frictional) forces are negligible (OpenStax §14.6; Wikipedia: Bernoulli's principle).
Bernoulli's equation applies along a streamline (Wikipedia: Bernoulli's principle).
The pipe contains a single fluid (no two-phase or compressible-gas effects).
Pressure is gauge pressure relative to atmospheric.