Bearnoulli's equation

1. Jan 24, 2005

vinter

Consider a U- tube filled with water. The water in one arm of it is pushed down and left, result- the full water column will start oscillating.

Now, consider an instant when the water levels in the two arms are not same and water in one arm is going up with a velocity v, water in the other arm is going down with the same velocity. Apply Bernoulli's equation to the points on the two water surfaces open to air, one on each, i.e, one point on the surface of water column in the left arm and one point on the surface of water column in the right arm.
You will have
pressure + half * (rho) * (v^2) + (rho ) *g * h = a similar expression for the second point.

This creates all the problem. The pressures at the two points are same, so that term will cancel out. the velocities are same, so the half rho v squared term will go. the remaining term is the rho*g*h term which cannot go since the heights are different in the two columns, that means the above inequality cannot hold in such a situation. But that is the Bernoulli's thrm!

What's wrong here?

Is Bernoulli's equation wrong, or we have not yet learnt to apply it properly?

2. Jan 24, 2005

vincentchan

you should not apply your whatever equation in this problem... check your text book in what situation your equation hold and tell me why can't you use it in this problem

3. Jan 24, 2005

arildno

You can certainly use the NON-STATIONARY version of Bernoulli's equation here.
But it is rather tricky..

4. Jan 24, 2005

vinter

Why is the stationary, non-viscous, non-rotational version not applicable here?

(I hope you all understood that equation I have written. Or do I need to use latex? I have not learnt it's tags and stuff yet, so for now, I will just tell you that rho was the famous greek letter that represents density of water here.)

-vinter

5. Jan 24, 2005

Q_Goest

Interesting question... Bernoulli's equation is for steady state, whereas the example you gave has an accelerating fluid.

6. Jan 24, 2005

arildno

Hi, vinter:
Let us write the Euler equations as:
$$\frac{\partial\vec{v}}{\partial{t}}+\nabla(\frac{1}{2}\vec{v}^{2})+\vec{c}\times\vec{v}=-\frac{1}{\rho}\nabla{p}-g\vec{k}$$
I've used the vector identities:
$$(\vec{v}\cdot\nabla)\vec{v}=\nabla(\frac{1}{2}\vec{v}^{2})+\vec{c}\times\vec{v},\vec{c}\equiv\nabla\times\vec{v}$$

Now, let us multiply our equation with the streamline tangent $$d\vec{s}$$.
Since the streamline tangent is parallell to the velocity field, we gain, by integrating between two points:
$$\oint_{s_{0}}^{s_{1}}\frac{\partial\vec{v}}{\partial{t}}\cdot{d}\vec{s}+\oint_{s_{0}}^{s_{1}}\nabla(\frac{1}{2}\vec{v}^{2}+\frac{p}{\rho}+gz)\cdot{d}\vec{s}=0$$
(s is a scalar variable running along the streamline, (x,y,z) are therefore functions of s)

The second integral is easy to evaluate, since we have a gradient field:
$$(\frac{1}{2}\vec{v}^{2}+\frac{p}{\rho}+gz)\mid_{s=s_{1}}-(\frac{1}{2}\vec{v}^{2}+\frac{p}{\rho}+gz)\mid_{s=s_{0}}=-\oint_{s_{0}}^{s_{1}}\frac{\partial\vec{v}}{\partial{t}}\cdot{d}\vec{s}$$
Note that in the stationary case, the right-hand side is zero; i.e, you get the usual Bernoulli equation.
However, in your case the right-hand-side is NOT zero.

The non-stationary version resolves your dilemma:
The height difference equals the curve integral on your right-hand side.

Last edited: Jan 24, 2005
7. Jan 25, 2005

vinter

Thanks Arildno.
I am not very much familiar with the del operators and the advanced fluid dynamics. But considering what you just said and what I know, I still have a doubt..

Let's consider a simple Bernoulli equation application example - water coming out of a container through a hole in it. This example is there in almost all the books on elementary fluid dynamics(eg = Sears and Zemansky "University Physics" by Addison-Wesly) They readily use the stationary version of Bernoulli's equation. And they use the continuity equation (amount of fluid crossing through any cross section is the same) to get different velocities at different points depending on the area of cross section. But since the velocities ARE different at different points, the motion has to be accelerated. And if it is, how can the integral on the right hand side of your equation be zero? In that case, are we allowed to use the stationary Bernoulli's equation? If yes and if it doesn't give any absurd results there, it shouldn't do that here too.

8. Jan 25, 2005

arildno

An emptying tank should, strictly speaking, always be regarded as a case of non-stationary flow.
However, if we think of a container in which the top area of the container (i.e, where the fluid surface is) is a lot bigger than the drain hole (a rather typical situation), two features follows:
1) The fluid layer by the top area is practically "at rest" (at least over quite some time)
2) The height difference between the surface area and the drain hole remains practically constant.
This justifies the stationary treatment of the situation, in which we say that at a fixed spatial position, the velocity profile does not change with time.

I don't know which specific examples Spears&Zemansky uses; so I can't say whether they have told which approximations they've used.

Note:
So, it is the area ratio which makes the problems different:
In the emptying tank case, the surface area is a lot bigger than the exit area;
in the U-tube case, the "inlet"/"outlet" areas are equal.

Last edited: Jan 25, 2005
9. Jan 26, 2005

vinter

OK thanks for the help