Continuity and Bernoulli's equation

By the end of this topic you will be able to

Use continuity (A1v1 = A2v2) to relate velocity to pipe area.
Apply Bernoulli's equation between two points in a steady flow.
Explain how a Venturi meter measures flow rate.

Continuity: what goes in must come out

For a steady flow of an incompressible liquid in a pipe, the volume that passes each cross-section every second is the same. This is the continuity equation:

A_1 v_1 = A_2 v_2 = Q

where $Q$ is the volume flow rate (m³/s). A narrower pipe means a faster flow, in the same way that your thumb over a hose makes the water speed up.

Bernoulli: energy along a streamline

For steady, frictionless flow of an incompressible fluid, the sum of the pressure head, velocity head and elevation head is constant along a streamline:

\frac{p}{\rho g} + \frac{v^2}{2g} + z = \text{constant}

Each term has units of metres ("head"). It is a statement of conservation of energy: pressure energy, kinetic energy and potential energy can be exchanged, but their sum per unit weight stays the same.

The Venturi meter

A Venturi meter is a pipe with a smooth narrowing (the throat). Applying continuity and Bernoulli between the inlet ① and throat ② (same height):

p_1 - p_2 = \tfrac12\rho\left(v_2^2 - v_1^2\right)

Measuring the pressure drop, from the height difference of two columns, gives $v_1$ , and hence the flow rate $Q = A_1 v_1$ . In the model, the two vertical tubes show the pressure heads. The dashed line is the total head, which stays constant in this ideal flow.

Interactive modelSqueeze the throat or change the flow rate and watch the water columns.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example

Water flows at 10 L/s through a pipe that narrows from 100 mm to 50 mm diameter. Find the velocities, the pressure drop and the difference in water level between the two tubes.

Areas. $A_1 = \pi\times0.1^2/4 = 7.854\times10^{-3}$ m²; $A_2 = 1.963\times10^{-3}$ m². $Q = 0.010$ m³/s.

Velocities. $v_1 = Q/A_1 = 1.273$ m/s and $v_2 = Q/A_2 = 5.093$ m/s (four times faster, since the area is four times smaller).

Pressure drop. $p_1 - p_2 = \tfrac12\times1000\times(5.093^2 - 1.273^2) = 12.16$ kPa.

Head difference. $\Delta h = \dfrac{12\,160}{1000\times9.81} = 1.24$ m.

Practice problems

Water flows through a 150 mm pipe at 2 m/s. Find the flow rate in L/s.

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$Q = \pi\times0.15^2/4\times2 = 0.0353$ m³/s $= 35.3$ L/s.
The pipe narrows to 75 mm. What is the velocity in the narrow section?

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The area ratio is 4, so $v = 8$ m/s.
What is the pressure difference between those two sections (ideal flow)?

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$\Delta p = \tfrac12\times1000\times(8^2 - 2^2) = 30$ kPa.
Why can the throat pressure drop below atmospheric pressure, and what problem can this cause?

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At high flow the throat pressure can fall below atmospheric and approach the vapour pressure, so bubbles form and collapse (cavitation), which damages the pipe.

Key formulae

Continuity (incompressible): $A_1 v_1 = A_2 v_2 = Q$
Bernoulli's equation: $\frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2$
Venturi pressure drop: $p_1 - p_2 = \frac{1}{2}\rho\left(v_2^2 - v_1^2\right)$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (45 min). 5 min: hose-and-thumb demo; 10 min: continuity; 15 min: Bernoulli as energy per unit weight; 15 min: Venturi worked example and model.

Activity. Ask students to find the narrowest throat that keeps the throat pressure head above zero for the default inlet head.

Discussion. Why does a car pass a lorry and feel a pull toward it?

Assessment idea. A tank draining through a nozzle (Torricelli).