# Hourglass fluid mechanics issue with Bernoulli's equation

I know that I'm thinking about this problem wrong, but I've been pondering it for a while and can't understand something.

On my fluid mech. exam I had a problem that asked to determine the radius of the top half of an hourglass as a function with respect to vertical coordinate (z), such that the level of the hourglass (z) lowered constantly at some specific rate.

So based on continuity, A1V1=A2V2 where A1V1 is the area at the current height of fluid in the hourglass * its downward velocity, and A2V2 is the area* downward velocity at the small hole where it would exit to the bottom half of the hourglass. Both sides of the fluid are exposed to atmospheric pressure.

So obviously the velocity must be faster at the smaller cross section, and relations can be set up to solve for the radius at height z. My answer included finding an increase in fluid velocity due to elevation difference (z) through Bernoulli's equation, then relating that with continuity. This got me thinking though, what would happen if the cross section does not change, and you have vertical flow of fluid in some pipe with an obvious change in elevation, but then had both the entrance and exit at atmospheric pressure. The energy equation, which is supposed to remain constant for a continuous fluid without losses, would have a differing height term without differing pressure or velocity terms. Is it because it's not a continuous fluid, or because friction from the wall eventually accounts for head loss of the elevation differential after the fluid speeds up enough?

Sorry if this is poorly worded, it just isn't making sense in my head. Thanks a lot.

## 2 Answers

This is a really good question. What you are wondering about is why, when the cross section is constant, the Bernoulli equation doesn't predict free fall.

It doesn't have anything to do with viscosity if the fluid is considered inviscid. The reason that the Bernoulli equation cannot be extended to the case of constant cross section is that the usual form of the Bernoulli equation is applicable only to steady flow, and this flow is not a steady state flow. Because the level of fluid is changing, the fluid velocity at any constant elevation within the tank is varying with time. But the usual Bernoulli equation does not take this part of the fluid acceleration into account. It only includes the advective part of the acceleration. If the exit hole is small, the error in using the usual steady state form of the Bernoulli equation is not too large. But, obviously for the case of constant diameter, it isn't. To get a better prediction, a time dependent modification to the Bernoulli equation must be used which properly includes the missing part of the acceleration.

This question is very timely, because, on another forum site (not stackexchange), I have currently been collaborating with another member in an analysis that quantifies this effect. We will be entering the details of our analysis, together with our results and conclusions shortly.

• I'm giving this +10 because it correctly works in the acceleration effect. As it happens I was working on a model of a syphon emptying a tank of constant cross-section (the syphon's constant pipe diameter is considered small to the tank's cross-section). As a first approximation Torricelli can be invoked to estimate the syphon's flow speed. But if the fluid is not inviscid then head loss in the syphon can not be avoided and has a definitive effect on the total emptying time of the tank. – Gert Apr 26 '16 at 13:47
• @Gert Check out the latest post at the following site: physicsforums.com/threads/velocity-of-efflux.868030 – Chet Miller Apr 26 '16 at 23:22
• It's true: the books always neglect the accelerations, assuming they're small and thus Bernoulli applies. My syphon model also assumes that. It would be interesting to see how your model works out if you add a head loss term for flow through a pipe rather than through a sharp-edged opening. Will follow that thread. :-) – Gert Apr 26 '16 at 23:50

This got me thinking though, what would happen if the cross section does not change, and you have vertical flow of fluid in some pipe with an obvious change in elevation, but then had both the entrance and exit at atmospheric pressure. The energy equation, which is supposed to remain constant for a continuous fluid without losses, would have a differing height term without differing pressure or velocity terms. Is it because it's not a continuous fluid, or because friction from the wall eventually accounts for head loss of the elevation differential after the fluid speeds up enough?

It's the latter. For a liquid falling through a pipe of constant cross-section, Bernoulli doesn't make sense anymore and leads to a mathematical impossibility in the form:

$$g\Delta z=0$$

That's because both speeds at both ends are the same (due to continuity) and both pressures are the same (both sides are at atmospheric pressure).

In reality the flowing fluid experiences drag (given basically by the Darcy-Weisbach equation). This causes a head loss, so that:

$$\frac{\Delta p_f}{\rho}=g\Delta z$$

Where $\Delta p_f$ is the pressure loss due to viscous losses in the pipe.