# Height of a fluid in a small tube

1. Apr 10, 2013

### Ampere

1. The problem statement, all variables and given/known data

An upper tube of cross sectional area A, atmospheric pressure, and height h1 is connected to a lower tube with cross sectional area A/2 (and height 0). A fluid of uniform density rho flows through the system, with velocity v measured in the upper tube. A small open tube branches up off the lower one, with a column of fluid of height h2 inside it. Find h2.

(See attached diagram.)

2. Relevant equations

Bernoulli's equation,

P1 + pgh1 +1/2*pv12 = constant

Continuity equation,

A1v1 = A2v2

3. The attempt at a solution

Applying Bernoulli's equation between the top of fluid of height h2 and the left endpoint of the system, I got:

(point 1 = left endpoint, point 2 = top of the column with height h2)

P1 = atmospheric
P2 = atmospheric
KE1 = 1/2*rho*v2
KE2 = 0
PE1 = rho*g*h1
PE2 = rho*g*h2

Therefore h2 = h1 + v2/(2g). Is this right? I would think that a faster moving fluid pushes h2 up even higher (with h2 = h1 when v=0).

The continuity equation tells us that the speed of the fluid inside the smaller tube is 2v. But does this even matter?

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Last edited: Apr 10, 2013
2. Apr 10, 2013

### tiny-tim

Welcome to PF!

Hi Ampere! Welcome to PF!
No, you can't do that … Bernoulli's equation only applies along a streamline

and the streamline doesn't go up that little tube, does it?
Yes, because that is in the streamline!

3. Apr 10, 2013

### Ampere

Thanks. So in that case you can find the pressure right under the small tube (with height h2 - did this below). But what equation do I use to travel outside the streamline?

P1 = atmospheric
P2 = unknown
KE1 = 1/2*rho*v2
KE2 = 1/2*rho*(2v)2
PE1 = rho*g*h1
PE2 = 0

So P2 = Patm + rho*g*h1 - 3/2*rho*v2

4. Apr 10, 2013

### tiny-tim

Hi Ampere!
The fluid in the thin vertical tube is stationary, so just use pressure difference = ρgh

(technically, you can use any line in a stationary fluid as a streamline … but that doesn't mean you can then tag it onto a "real" streamline in a moving part of the fluid )

5. Apr 10, 2013

### Ampere

So I get

dP = rho*g*h2
P2 - Patm = rho*g*h2

and

h2 = h1 - 3v2/(2g)

So the faster the fluid flows, the lower the height h2 is? Is there actually a flow rate which would make h2 = 0, or even less than 0?

6. Apr 10, 2013

### tiny-tim

Bernoulli's equation does mean that when the speed is faster the pressure is lower.
work it out!

7. Apr 10, 2013

### Ampere

I found that the speed at which h2=0 is √(2gh1/3).

So if h1 = 1m, then v = 2.556 m/s. What if I had v = 3 m/s, or even 6 m/s? What would be the physical meaning of a negative height?

For example, if we go with h1 = 1m and v = 6 m/s, we get h2 = -4.51m. Huh?

8. Apr 10, 2013

### tiny-tim

It would mean you have a vacuum pump!

9. Apr 10, 2013

### Ampere

Ah so air would be sucked into the tube along with the fluid? Interesting!

Thanks, this makes sense now.