Fluid flow and Bernoulli's theorem

Hold two sheets of paper a few centimetres apart and blow between them — instead of flying apart, they snap together. A perfume atomiser sprays liquid upward using nothing but a puff of air across a tube. A curved aircraft wing, moving through still air, generates enough upward force to lift a jumbo jet. All three tricks come from the same idea: a moving fluid trades pressure for speed. This article builds that idea from scratch — starting with the simple fact that fluid squeezed through a narrower pipe must speed up (the equation of continuity), and then applying energy conservation along a streamline to get Bernoulli's theorem, $P + \frac12\rho v^2 + \rho gh = \text{constant}$ . You will see the theorem derived from the work-energy theorem (a derivation your board can ask for directly), applied to a real narrowing pipe and to lift on a wing, and used to explain the Venturimeter, the atomiser, and the Magnus effect that makes a cricket ball swing.

The core insight

For an ideal fluid in steady streamline flow, mass conservation forces speed up where a pipe narrows (continuity, $A_1v_1=A_2v_2$ ), and energy conservation then forces pressure down wherever speed is up (Bernoulli, $P+\frac12\rho v^2+\rho gh=\text{constant}$ ).

The core idea

Ideal fluid, streamline flow. Both results below assume the fluid is incompressible (constant $\rho$ ), non-viscous (no internal friction), and flowing in a steady, irrotational way — every particle passing a given point follows the same path, called a streamline. Real liquids like water are close enough to this ideal for these results to work well in practice.

Equation of continuity. Picture a pipe of varying cross-section with an incompressible fluid flowing through it. No fluid is created or destroyed inside the pipe, so the volume flowing per second through any cross-section must be the same everywhere along it. If a section of area $A_1$ carries fluid at speed $v_1$ , and a narrower section of area $A_2$ carries it at speed $v_2$ :

A_1 v_1 = A_2 v_2

where: $A$ is cross-sectional area (m²) and $v$ is flow speed (m/s) at that section; the product $Av$ , the volume flow rate (m³/s), stays constant along the tube. The direct consequence: fluid speeds up wherever the pipe narrows, and slows down wherever it widens.

Bernoulli's theorem. Now bring in energy. As a fluid element travels along a streamline from a point with pressure $P_1$ , speed $v_1$ , height $h_1$ to a point $P_2, v_2, h_2$ , the work-energy theorem applied to that element (worked out in full below) gives:

P_1 + \frac12\rho v_1^2 + \rho g h_1 = P_2 + \frac12\rho v_2^2 + \rho g h_2

Since points $1$ and $2$ were any two points on the same streamline, this holds everywhere along it:

P + \frac12\rho v^2 + \rho g h = \text{constant}

where: $P$ is pressure (Pa), $\rho$ is fluid density (kg/m³), $v$ is flow speed (m/s), $g = 9.8\ \text{m/s}^2$ , and $h$ is height above a reference level (m). Each term has the dimensions of energy per unit volume — pressure energy, kinetic energy, and gravitational potential energy — so the theorem is simply energy conservation for a flowing fluid. The headline consequence, used in every application below, is that at the same height, higher speed goes with lower pressure.

Figure

A horizontal pipe narrowing from a wide section to a narrow section, with flow lines bunching together, speed and pressure labels showing low speed and high pressure in the wide part and high speed and low pressure in the narrow part

Applications, all from the same rule. A Venturimeter constricts a horizontal pipe so the fluid speeds up there; the resulting pressure drop, measured directly, gives the flow rate. An atomiser blows air fast across the top of a tube dipped in liquid, lowering the pressure there so atmospheric pressure pushes the liquid up to be sprayed. Dynamic uplift on an aerofoil comes from air moving faster over the curved top than the flatter bottom, so pressure below exceeds pressure above and the wing is pushed up. The Magnus effect — the physics behind a swinging cricket ball or a curving football — comes from a spinning ball dragging a layer of air with it, speeding air (and lowering pressure) on one side and slowing it (raising pressure) on the other, producing a sideways force.

Derivation ABernoulli's theorem from the work-energy theorem

The goal is to derive $P+\frac12\rho v^2+\rho gh=\text{constant}$ for an ideal fluid element moving along a streamline, by applying the work-energy theorem (work done by all forces on the element equals its change in kinetic energy) to a small plug of fluid pushed from a wide, low section of a tube to a narrow, high section, exactly as the syllabus requires you to reproduce.

Step 1. Set up the fluid element. Consider a tube of flow of varying cross-section, and follow a small element of fluid of volume $V$ and mass $m = \rho V$ as it moves from section 1 (area $A_1$ , pressure $P_1$ , speed $v_1$ , height $h_1$ ) to section 2 (area $A_2$ , pressure $P_2$ , speed $v_2$ , height $h_2$ ) in a short time $\Delta t$ . By continuity, the volume $V = A_1(v_1\Delta t) = A_2(v_2\Delta t)$ entering at 1 equals the volume leaving at 2.

Step 2. Compute the work done by pressure forces. At section 1, the fluid behind pushes the element forward with force $P_1A_1$ , doing positive work $W_1 = P_1A_1(v_1\Delta t) = P_1V$ as the element advances by $v_1\Delta t$ . At section 2, the fluid ahead pushes back on the element with force $P_2A_2$ , doing negative work $W_2 = -P_2A_2(v_2\Delta t) = -P_2V$ . The net work done by pressure is $W_{\text{pressure}} = P_1V - P_2V$ .

Step 3. Compute the work done against gravity. Lifting the mass $m = \rho V$ from height $h_1$ to height $h_2$ requires work against gravity of $W_{\text{gravity}} = -mg(h_2-h_1) = -\rho V g(h_2-h_1)$ , since this work is stored as gravitational potential energy rather than appearing as kinetic energy.

Step 4. Apply the work-energy theorem. The total work done on the element equals its gain in kinetic energy: $W_{\text{pressure}} + W_{\text{gravity}} = \Delta KE$ , where $\Delta KE = \frac12 m v_2^2 - \frac12 m v_1^2 = \frac12\rho V(v_2^2-v_1^2)$ . Substituting: $P_1V - P_2V - \rho Vg(h_2-h_1) = \frac12\rho V(v_2^2-v_1^2)$ .

Step 5. Divide through by $V$ and rearrange. Every term now carries units of energy per unit volume (pressure). Collecting the section-1 quantities on one side and section-2 quantities on the other gives $P_1 + \frac12\rho v_1^2 + \rho gh_1 = P_2 + \frac12\rho v_2^2 + \rho gh_2$ , which is Bernoulli's theorem, since sections 1 and 2 were any two points on the streamline: $P + \frac12\rho v^2 + \rho gh = \text{constant}$ along that streamline.

Worked example

A horizontal pipe carries water and narrows from a wide section of area $A_1 = 4.0\times10^{-3}\ \text{m}^2$ to a Venturi-style narrow throat of area $A_2 = 1.0\times10^{-3}\ \text{m}^2$ . The speed in the wide section is $v_1 = 1.5\ \text{m/s}$ , and $\rho_{\text{water}} = 1000\ \text{kg/m}^3$ . Find (a) the speed in the throat, (b) the pressure drop between the two sections, and (c) the volume of water delivered by the pipe in one minute.

Step 1. Use continuity to find the throat speed. $A_1v_1 = A_2v_2 \Rightarrow v_2 = \dfrac{A_1v_1}{A_2} = \dfrac{(4.0\times10^{-3})(1.5)}{1.0\times10^{-3}} = 6.0\ \text{m/s}$ . The narrow throat carries the same volume per second, so it must move four times faster, matching the four-fold drop in area.

Step 2. Apply Bernoulli's theorem for horizontal flow ( $h_1 = h_2$ , so the height terms cancel). $P_1 + \frac12\rho v_1^2 = P_2 + \frac12\rho v_2^2 \Rightarrow P_1 - P_2 = \frac12\rho(v_2^2-v_1^2) = \frac12(1000)(6.0^2-1.5^2) = 500(36-2.25) = 16875\ \text{Pa} \approx 1.69\times10^4\ \text{Pa}$ . The faster, narrower section is at the lower pressure, as the theorem predicts.

Step 3. Find the volume delivered in one minute using the (constant) volume flow rate. $Av = A_1v_1 = (4.0\times10^{-3})(1.5) = 6.0\times10^{-3}\ \text{m}^3/\text{s}$ . In $60\ \text{s}$ : $V = (6.0\times10^{-3})(60) = 0.36\ \text{m}^3 = 360\ \text{litres}$ .

Answer: throat speed $= 6.0\ \text{m/s}$ , pressure drop $\approx 1.69\times10^4\ \text{Pa}$ , and $360$ litres pass through the pipe every minute.

Common traps

Applying $A_1v_1 = A_2v_2$ or Bernoulli's equation to a fluid that is compressible, viscous, or in turbulent flow — both results strictly assume an ideal, incompressible, streamline-flowing fluid.
Forgetting the height term $\rho gh$ when the two points compared are at different elevations — it can be dropped only for horizontal flow.
Thinking higher speed always means higher pressure — Bernoulli's theorem says the opposite: at the same height, higher speed goes with lower pressure.
Mixing up which section of a pipe has the higher speed — by continuity, speed is higher where the cross-sectional area is smaller, not larger.

Ask the tutor: Want a walkthrough of the aircraft-wing lift calculation, or more practice turning Venturimeter pressure readings into flow rates?

Why does blowing between two sheets of paper make them come together instead of apart?

Blowing air between the sheets speeds it up in that gap. By Bernoulli's theorem, at the same height, higher speed means lower pressure — so the air pressure between the sheets drops below the ordinary atmospheric pressure pushing on their outer faces. That imbalance pushes the sheets inward, together. It is the same principle as an atomiser or a Venturimeter, just seen at very small scale.

Does Bernoulli's theorem mean pressure and speed always trade off, everywhere in any fluid?

Only under the theorem's assumptions: an ideal fluid (incompressible, non-viscous) in steady, streamline flow, and points compared on the same streamline. Real fluids have viscosity, which dissipates energy as heat and slightly breaks the strict conservation, and turbulent flow does not follow smooth streamlines at all. For gentle flows of low-viscosity fluids like air and water, the ideal-fluid approximation works well enough to explain atomisers, Venturimeters, and wings accurately.

Is the equation of continuity a separate law, or part of Bernoulli's theorem?

They are separate, though related. Continuity, $A_1v_1=A_2v_2$ , comes purely from conservation of mass (volume, for an incompressible fluid) — it says nothing about pressure. Bernoulli's theorem comes from conservation of energy along a streamline and does involve pressure. In practice you almost always use them together: continuity tells you how speed changes when the pipe's area changes, and Bernoulli's theorem then tells you how pressure changes because the speed changed.

Fluid flow and Bernoulli's theorem

The core idea

Worked example

Common traps

Before this

Where this leads