# Applying unsteady Bernoulli (to a balloon attached to a tube)

Using Bernoulli’s equation for an unsteady irrotational motion of an inviscid incompressible ﬂuid, solve the following:

Water of density $\rho$ is driven through a horizontal tube of length $L$ and internal radius $a$ from a water-ﬁlled balloon attached to one end of the tube. Assume that the pressure exerted by the balloon is proportional to its current volume (in excess of atmospheric pressure). Also assume that water exits the tube at atmospheric pressure, and that gravity may be neglected. Show that the time for the balloon to empty does not depend on its initial volume. Find the maximum speed of water exiting the pipe.

Bernoulli is: $\rho \frac{\partial\phi}{\partial t} + \frac{1}{2}\rho|\bigtriangledown\phi|^2 + p + \chi$ is constant along streamlines.

First of all, I take a streamline to be a line from one end of tube to the other and apply Bernoulli to both ends, 1 and 2.

1. But how do I express the "pressure exerted by the balloon is proportional to its current volume" condition? How is the volume changing over time?
2. How do I express the conservation of mass? Let $u_1, u_2$ be velocities at ends. $\rho u_1*$ (area at 1) =$\rho u_2*$ (area at 2). But aren't both these areas $\pi a^2/4$? So velocities equate?
3. Equating Bernoulli expressions, how do I work with the fact that I have $\phi(x=0)$ and $\phi(x=L)$?

Generally you would have $p(t) = p_0 + \beta V(t)$. Yes, the velocities at both end are equal but you should conservation of mass to get a relation between volume of balloon and the velocity. As for your last question I would say not using the potential form of unsteady Bernoulli equation is more intuitive but the velocity potential is $\phi(x,t)=u(t)x$ so $\dot\phi(0,t)=0$ and $\dot\phi(L,t)=\dot u(t)L$

But I prefer this form here: $$p_1+{1\over2}\rho u_1^2+\rho g z_1=p_2+{1\over2}\rho u_2^2+\rho g z_2+\rho\int_1^2 {\partial u \over \partial t}ds$$

Because the velocity (and therefore ${\partial u \over \partial t}$) does not depend on the position, the last term simply is $\rho \dot u L$

• So does mass conservation give $\frac{dV(t)}{dt}=\frac{\pi a^2}{4}u(t)$? If so, what is the intuition behind this? (I was just matching dimensions.) I then get exponential decay for the volume and velocity. Does this make sense physically? May 20, 2015 at 9:34
• @timni Change of volume should be equal to volume flow rate. I think it's pretty much intuitive. The assumption that pressure at the surface of balloon and mouth of the tube is equal is valid when we could neglect effects of the velocity in the balloon. Apart from that it's make sense.