Bernoulli equation for an ideal fluid

An ideal fluid is incompressible and has no internal friction, so along a streamline the energy of each volume element is conserved.

Pressure $p = F/A$ (from the prerequisite topic) is the static pressure, while the term $\rho g h$ describes the pressure created by the weight of the fluid column above a given point.

For steady flow along a streamline, Bernoulli's equation holds:

$$p + \frac{1}{2}\rho v^{2} + \rho g h = \text{const}.$$

Every term has the units of energy per unit volume: $\mathrm{J/m^{3}} = \mathrm{Pa}$.

The term $\frac{1}{2}\rho v^{2}$ is called the dynamic pressure: it grows as the speed $v$ increases.

Because the sum is constant, wherever the flow speeds up the static pressure drops, and an increase in height $h$ also lowers the pressure.

Water ($\rho = 1000\ \mathrm{kg/m^{3}}$) flows through a horizontal pipe. At cross-section 1 the speed is $v_1 = 1.5\ \mathrm{m/s}$ and the pressure is $p_1 = 2.0\cdot10^{5}\ \mathrm{Pa}$. At cross-section 2 the speed is $v_2 = 3.0\ \mathrm{m/s}$. Find the pressure $p_2$.

Water flows out of an open tank through a small hole located $h = 2.0\ \mathrm{m}$ below the water surface. Find the speed of the outflowing jet. Atmospheric pressure acts both on the water surface and on the outflowing jet.

Water ($\rho = 1000\ \mathrm{kg/m^{3}}$) flows through a pipe that rises upward. At cross-section 1 the area is $A_1 = 4.0\cdot10^{-4}\ \mathrm{m^{2}}$, the speed is $v_1 = 2.0\ \mathrm{m/s}$, and the pressure is $p_1 = 1.5\cdot10^{5}\ \mathrm{Pa}$. At cross-section 2 the area is $A_2 = 2.0\cdot10^{-4}\ \mathrm{m^{2}}$, and the height is greater by $1.0\ \mathrm{m}$: $h_2 = h_1 + 1.0\ \mathrm{m}$. Find the pressure $p_2$.

Self-check in a flow simulation. Open an interactive fluid pressure and flow simulation (for example, PhET "Fluid Pressure and Flow") and work only inside it — do not build homemade pressurized vessels and do not use open flames or glass tubes.

1. Choose a horizontal pipe with a varying cross-sectional area. Keep the height fixed and record the speed $v$ and the static pressure $p$ for 5–6 cross-sections.
2. Plot $p$ against $v^{2}$. The points should lie on a straight line; estimate its slope and compare it with the theoretical value $-\rho/2$.
3. For each pair of cross-sections, check that the sum $p + \frac{1}{2}\rho v^{2}$ is constant within the simulation's accuracy.
4. Tilt the pipe upward and verify that now the full sum $p + \frac{1}{2}\rho v^{2} + \rho g h$ stays constant.

Show that the sum of the static, dynamic, and hydrostatic terms stays constant along a streamline while the terms are redistributed among themselves.

A diagram: a horizontal pipe with wide and narrow sections, drawn streamlines, velocity arrows (longer in the narrow section), and pressure gauges at two cross-sections. Below the diagram, a bar chart of the three terms $p$, $\frac{1}{2}\rho v^{2}$, and $\rho g h$ for both cross-sections shows that their sum is the same.

A garden hose with inner cross-sectional area $A_1 = 3.1\times10^{-4}\ \text{m}^2$ delivers water at $v_1 = 1.2\ \text{m/s}$. A nozzle with area $A_2 = 8.0\times10^{-5}\ \text{m}^2$ is fitted on the end. From continuity, $v_2 = v_1 A_1/A_2 \approx 4.7\ \text{m/s}$. The height is unchanged, water density is $\rho = 1000\ \text{kg/m}^3$, so the pressure drop inside the nozzle is $\Delta p = \tfrac12\rho\left(v_2^2 - v_1^2\right) \approx 0.5\cdot1000\cdot(22 - 1.4)\,\text{Pa} \approx 1.0\times10^4\ \text{Pa}$, about $0.1\ \text{bar}$. The jet leaves faster, but the pressure inside the nozzle falls.

This exercises the kinetic term $\tfrac12\rho v^2$ at constant height: along a streamline the sum $p + \tfrac12\rho v^2 + \rho g h$ is conserved, so the increase of $\tfrac12\rho v^2$ — the speed grows by a factor $4.7/1.2$, so $v^2$ by about $15$ — must be paid for by a drop in $p$.

A ball flies at $25\ \text{m/s}$, and because of its spin the air moves past one side at $22\ \text{m/s}$ and past the other at $20\ \text{m/s}$. Air density is $\rho = 1.2\ \text{kg/m}^3$, so the pressure difference between the sides is $\Delta p = \tfrac12\rho\left(22^2 - 20^2\right) = 0.5\cdot1.2\cdot84\,\text{Pa} \approx 50\ \text{Pa}$. The ball's cross-sectional area is $A \approx 0.038\ \text{m}^2$, so the sideways force is $F = \Delta p\,A \approx 1.9\ \text{N}$. With the ball's mass $0.43\ \text{kg}$ the acceleration is $a \approx 4.4\ \text{m/s}^2$; over a $0.8\ \text{s}$ flight the ball drifts sideways by about $\tfrac12 a t^2 \approx 1.4\ \text{m}$ — enough to curl past a goalkeeper.

The pressure–speed link in the $\tfrac12\rho v^2$ term is used: where the air moves faster the pressure is lower, so the ball is pushed towards the low-$p$ side. Height change along the trajectory is neglected, and the ideal-fluid model itself ignores viscosity, so the real curvature is smaller than this estimate.

A wing of area $A = 16\ \text{m}^2$ meets air of density $\rho = 1.2\ \text{kg/m}^3$. Over the wing the speed is $70\ \text{m/s}$, under it $60\ \text{m/s}$. The pressure difference is $\Delta p = \tfrac12\rho\left(70^2 - 60^2\right) = 0.5\cdot1.2\cdot1300\,\text{Pa} = 780\ \text{Pa}$, and the lift is $F = \Delta p\,A \approx 1.2\times10^4\ \text{N}$. That equals the weight of an aircraft of about $1.2\ \text{t}$ — the scale of a light trainer such as a Cessna 172 (take-off mass $\approx 1.1\ \text{t}$, weight $\approx 1.1\times10^4\ \text{N}$).

Bernoulli's equation is applied along a streamline over the wing and one under it: the wing is thin, so the $\rho g h$ difference between the upper and lower streamlines is neglected and the whole pressure difference comes from the $\tfrac12\rho v^2$ difference. The estimate is right in order of magnitude; exact lift requires accounting for the downward deflection of the flow.