Unsteady Bernoulli equation

For time-dependent irrotational flow with $u=\nabla \varphi$, constant mass density, and no body force, the Euler equations for an inviscid fluid integrate toA time-dependent change of gauge in $\varphi$ may set $C(t)=0$.

This is the unsteady potential-flow version of Bernoulli's principle, with a time-dependent potential gauge.

A spherical balloon feeding an inviscid uniform tube plug, with entrance excess pressure $2\gamma /R$, satisfies $L\dot{u}=2\gamma /(\rho R)$ and $4\pi R^2\dot{R}=-\pi a^{2}u$. Eliminating $u$ yields the displayed equation. Its first integral is $R^4{\dot{R}}^2+K{R}^2=C$. From rest at radius $R_0$, $C=K{R}_0^2$; integration gives $t=R_0^2{K}^{-1/2}(\pi /4-\theta /2+\sin (2\theta )/4)$, where $\theta =\arcsin (R/R_0)$. The ideal emptying time is $\pi R_0^2/(4\sqrt{K})$. Independently prescribed initial flow changes $C$ and therefore this time.

For a slender tube of length $L$ supplied from a large reservoir with excess pressure $P$, negligible gravity and entrance inertia, the Unsteady Bernoulli equation gives $L\dot{U}+U^2/2=P/\rho$. A stationary initial liquid column has the displayed speed and volume flux $\pi a^{2}U$ for tube radius $a$. The inertial term controls the early linear acceleration; the kinetic-energy term sets the terminal inviscid speed. This is a leading slender-tube model while the reservoir remains large and the tube full.

An inviscid uniform tube plug driven by entrance excess pressure $KV$ obeys $\rho L\dot{u}=KV$ and $\dot{V}=-\pi a^{2}u$. From rest, the first zero of $V$ occurs after a quarter harmonic period, independent of its initial volume. The maximum speed is proportional to that initial volume. Entrance losses, a reservoir kinetic head or independently prescribed initial flow change this ideal model.