 Constant-tension balloon discharge

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.