Fluid velocity in free jets doubt

1. Jun 3, 2010

R Power

Hi friends
This isn't a homework poblem. However,I got this confusion when i was solving some problems.
Consider fluid flowing out of a tank as shown in figure attached.
Fluid moves out of pipe through a nozzle (generally used to accelerate the fluid). Now if we apply bernoulli's equation at point 1 and 2(shown in fig.) we find that fluid velocity comes out to be: V= (2gh)^1/2 (where h is height of tank above nozzle)
which is independent of nozzle diametre. That means if i remove the nozzle and fluid flows through a pipe of constant cross section, then also velocity will be same. This can be checked by applying bernoulli's equation.
But in general nozzle accelerates the fluid for same mass flow rate.
How will you explain this?

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2. Jun 3, 2010

Andy Resnick

You lost me here. How does a nozzle accelerate the fluid? Where is the motive force?

3. Jun 3, 2010

zeebek

This is just an estimate for the velocity. Strickly speaking it is applied only before the nozzle. Because inside the nozzle the flow has mean shear (vorticity is not zero), thus the flow is no longer potential and the bernoulli's equation does not apply.

P.S. do not read crappy wiki articles, read proper books.

4. Jun 3, 2010

Cyrus

Its accelerating the flow via the continuity equation. The force is coming in the form of reduction in static pressure.

5. Jun 3, 2010

Cyrus

Assume it's isentropic! :tongue2:

6. Jun 3, 2010

Andy Resnick

Not so fast... the force is bouyancy, acting on the open container. If the container were sealed, there won't be any flow.

The nozzle does not *accelerate* the flow... re-direct, at best.

7. Jun 3, 2010

Cyrus

If you have a nozzle, the flow velocity coming in is less than the flow going out (again, just look at the contiunity equation). So the flow had to accelerate. Also, the force acting on the open container is not bouyancy, it is hydrostatic.

To be more precise, let's just look at what the math says:

$$A_i V_i = A_e V_e$$

where e - exit, and i - inlet. Then clearly:

$$V_{e} = \frac{A_i}{A_e}V_i$$

But by definition the ratio of the areas is > 1 for a nozzle. Hence the flow speeds up. By Bernoulli, the static pressure goes down. This causes a static pressure gradient across the nozzle (a bouyancy force), which pushes the nozzle in the axial direction inline to the flow contraction. Take a garden hose and put your thumb over it, you will feel this force as it pushes your hand forward.

We correct for this type of problem in the wind tunnel for large models.

Last edited: Jun 3, 2010
8. Jun 4, 2010

R Power

I am reading FM by munson young and okishi which is one of the very good books on the subject as far as i know. And author applies bernoulli's equation as i did.

9. Jun 4, 2010

R Power

so what's the solution to my problem?

10. Jun 4, 2010

Cyrus

You applied your analysis incorrectly. Go back and do it over again, and have point two just before the nozzle, and a new point three just after the nozzle. See what you get.

11. Jun 5, 2010

R Power

if i take point 2 just before the nozzle and then i apply bernoulli, i can't find velocity at 2 because i don't know pressure at 2.
Can u help

12. Jun 5, 2010

Cyrus

I tried working this problem out, here is what I would do. Make three points, 1 (top of the tank), 2 (bottom of the tank), 3 (nozzle entrance), 4 nozzle exit.

I would solve this by finding the pressure at point 2, which is just hydrostatic. That will be the total pressure (with zero velocity). Use that as your inlet condition to the nozzle (3), and find the flow velocity at (4).

Side note: I is capitalized, and u is spelled "you". The first letter of a sentence is capitalized, and the sentence ends in a period.

13. Jun 5, 2010

R Power

Are you sure static pressure at points 2 and 3 would be same?
Since fluid moves from 2 to 3 and since elevation remains constant, so all the pressure energy in converted into kinetic energy and that's why velocity increases in pipe and so static pressure decreases there, so pressure at 2 won't be equal to that at 3.

14. Jun 5, 2010

Cyrus

Strictly speaking, no the pressure wont be the same (your observation is correct). I neglected, and thinking twice about it you probably shouldn't. Try writing out the equations, (with the pressure as an unknown). And then see if you can eliminate it later on. I.e, write the equations from 1-2, 2-3, 3-4. Then see if things will drop out. If you write said equations here, we can try and look at them together.

15. Jun 5, 2010

R Power

1-2 pgh = P2 (p = density of water) (P2 = pressure at 2)

2-3 P2 = P3 + [0.5 * p * (V3)^2] (V3 = velocity at 3)

3-4 P3 + [0.5 * p * (V3)^2] = 0.5 p (V4)^2

16. Jun 5, 2010

R Power

come on cyrus i have wrote the equations , now help me !

17. Jun 5, 2010

zeebek

I am gonna give you one more reply

if we assume the liquid has no viscosity (ideal fluid) than bernulli is applied before and after the nozzle. In this case there is veolicty slip at the wall, it does not go to zero like in reality. Then the velocity before the nozzle and after the nozzle not gonna change. By nozzle here I mean a tube of constant cross section like in your figure. If the cross section does change, then the velocity will increase in case the cross section is shrinking and vice versa like Cyrus wrote.

However, in reality there is no slip at the wall, slowly boundary layer will be develloping, the flow in the middle of the tube will accelerate. Also, since there will be definitely vorticity in the tube, bernulli is not applied. I am not gonna go into details here, but there are many reasons why velocity in the outlet not gonna be the same as near the inlet besides just varying cross section. Meanwhile, provided the nozzle is of constant diameter and not very long, this estimate from bernullli equation is reasonably correct.

P.S. Never heard about the book you mention. Proper well known books are: Batchelor - Introduction to fluid mechanics, Kundu - Fluid Mechanics, etc.

Last edited: Jun 5, 2010