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Bernoulli's Principle

Intuition

A fluid particle sliding along a streamline is like a bead on a frictionless wire: it trades one form of energy for another, never losing any. Speed up → pressure drops. Climb → pressure or speed must pay for the height. Bernoulli's equation is just energy book-keeping per unit volume of moving fluid — valid exactly when nothing (viscosity, shocks, heat) can steal from the account.

The equation

For steady, inviscid, incompressible flow along a streamline:

\[\boxed{\,p + \tfrac{1}{2}\rho v^2 + \rho g z = \text{constant}\,}\]

Three terms, three energies per unit volume: pressure work, kinetic, gravitational potential. See the equation page for variables, assumptions and applications in detail.

Derivation (Newton along a streamline)

Apply \(F = ma\) to a fluid element of length \(ds\) and cross-section \(dA\) along a streamline: pressure difference \(-dP\,dA\) and gravity component \(-\rho g\,dA\,ds\, \frac{dz}{ds}\) drive acceleration \(V\frac{dV}{ds}\):

\[-dP - \rho g\, dz = \rho V\, dV \;\Longrightarrow\; \frac{dP}{\rho} + \tfrac{1}{2}d(V^2) + g\,dz = 0\]

For constant \(\rho\), integrate → Bernoulli. The same result follows from integrating Euler's equation along a streamline.

Compressible extensions

If \(\rho\) varies, keep \(\int dP/\rho\) and specify thermodynamics:

  • Isothermal ideal gas (\(P = \rho R T\)): \(\;RT\ln P + \tfrac{1}{2}V^2 + gz = \text{const.}\)
  • Isentropic (\(P/\rho^\gamma\) const.): \(\;\dfrac{\gamma}{\gamma - 1}\dfrac{P}{\rho} + \tfrac{1}{2}V^2 + gz = \text{const.}\)

The isentropic form is the workhorse of compressible aerodynamics below shock strength; across a shock Bernoulli fails (entropy jumps).

Worked example: tank discharge, optimal hole

Water drains from a tank filled to height \(H\) through a small hole at height \(h\). Where should the hole be for the jet to land farthest?

Bernoulli from the surface to the hole: \(v = \sqrt{2g(H - h)}\) (Torricelli). Projectile fall time: \(t = \sqrt{2h/g}\). Range:

\[x = vt = 2\sqrt{h(H - h)}\]

Maximizing: \(\frac{dx}{dh} = 0 \Rightarrow H - 2h = 0 \Rightarrow \boxed{h = H/2}\), with maximum range \(x = H\). Elegant symmetry: holes at \(h\) and \(H-h\) hit the same spot.

More classics solved on linked pages

Common mistakes

  • Applying it across streamlines — the constant is per-streamline unless the flow is irrotational (potential flow), where it becomes global.
  • Using it in viscous regions (boundary layers, pipes with friction) or across shocks — energy is dissipated there.
  • The "equal transit time" airfoil myth. Lift does follow from pressure differences consistent with Bernoulli, but the premise that parcels split at the leading edge must reunite at the trailing edge is simply false.
  • Forgetting the elevation term in anything taller than a bench-top apparatus.

Knowledge graph position

Prerequisites: Pressure, Material derivative, Continuity. Leads to: Potential flow, Kelvin–Helmholtz analysis (perturbed Bernoulli at the interface).

Quiz

Q1 (computational). Wind of speed 30 m/s (\(\rho = 1.2\ \text{kg m}^{-3}\)) blows over a flat roof. Estimate the lift pressure (suction) relative to still air inside.

Answer

\(\Delta p = \tfrac12 \rho v^2 = 0.5\times1.2\times900 = 540\) Pa — about 54 kg-force per square metre. This is why roofs peel upward in storms.

Q2 (conceptual). Two identical tanks drain through identical holes, one at the bottom, one halfway up. Which jet travels farther on level ground at the tank base?

Answer

The halfway hole (\(h = H/2\) is optimal; the bottom hole has maximum speed but zero fall time, so range → 0).

Q3 (multiple choice). Which assumption is not required for the basic Bernoulli equation?

  • (a) steady flow (b) incompressible (c) irrotational everywhere (d) inviscid
Answer

(c). Along a single streamline, rotationality doesn't matter. Irrotationality only upgrades the constant from per-streamline to global.