# Bernoulli's equation and fluid movement

1. Apr 19, 2015

### sciman

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hello!

I think Bernoulli's equation it's just an equation of conservation of energy. Am I right?

but I don't understand how to apply the conservation of energy principle exactly.

I get it that if the fluid moves, it has kinetic energy which is 1/2mV^2.

I also get that potential energy of a fluid due to its weight, is mgh.

I don't get what is potential energy due to the hydrostatic pressure, how do I calculate that?

and last, I just sum all these three at the begining of the experiment and at the end of the experiment and I equate them?

thanks!

2. Apr 19, 2015

### Simon Bridge

3. Apr 19, 2015

### sciman

yes, I saw that pressure is energy per unit volume

so to find the pressure energy and apply energy conservation, I just multiply pressure with volume? but which volume?
or I divide all the rest energies (potential due to weight and kinetic) with volume, so I end up with ρ instead of volume? but is all the volume the same?

4. Apr 19, 2015

### Simon Bridge

Hint: look at the derivation ... what is A, what is d?

Note: The energy (stored as hydrostatic pressure) is W=PV, so the energy per unit volume is ... ?
For Bernoullis equation: energy per unit volume before = energy per unit volume after (because fluid is incompressible).

5. Apr 19, 2015

### BvU

Hello SciMan,

The link Simon gave you says it all. However, because I remember it took me an awfully long time to get my head around this concept, maybe you want to look at it from these points of view:

p times V represents an energy. If nothing else happens, p times V is a constant (Boyle's law) and a (small) Δp is accompanied by a (small) ΔV such that VΔp + pΔV = 0 .
If one of the two changes and the other remains constant, something has to change: work is done, either on or by the system under consideration.

p times V represents an energy. Imagine an empty, fixed volume V in deep space and start filling it with gas molecules. A pressure inside builds up and you have to do work to get more molecules in, to the tune of VΔp to get from some p to some p+Δp . Total amount of work to achieve a final pressure pf is the sum of all these VΔp. Fancy writing: $$W = \Sigma_0^{p_f} V \Delta p = V \Sigma_0^{p_f} \Delta p = Vp_f$$ Even fancier: $W = \int_0^{p_f} Vdp= V \int_0^{p_f} dp= Vp_f$

p times V represents an energy. combustion engines, pumps, compressors, turbines: all of them have to do with this.

(the repetition isn't meant as condescending, I just like it )

 slow typist -- this was in addition to post #2

6. Apr 19, 2015

### sciman

thanks but I am talking about bernoulli, so not gases but liquids

you also introduced maths that I don't comprehend

I am 100% that you can explain without fancy maths and this is what I am after

my question is simple:

Instead of memorizing the bernouli equation, I want to apply energy conservation instead
can I do that? in other words, is bernouli basically energy conservation equation?
if yes, how do I apply energy conservation equation exactly? I know that the sum of potential energy and kinetic energy is stable
I also know that kinetic energy is 1/2mV^2
but which procedure I follow to apply energy conservation to come up with bernouli?

thanks!

PS: keep please the maths simple

7. Apr 19, 2015

### Simon Bridge

Yes.

Yes... along with the assumption that the fluid is incompressible.

The same way you always apply energy conservation. Energy in = energy out.
Since the fluid is incompressible, the energy density out = energy density in.

See ink post#2 for details - it provides what you are after: yes - for incompressible fluids (note: gasses are also fluids).
Since you don't want to understand the maths you need, you may need a link from that page.

8. Apr 20, 2015

### sciman

I want to understand the maths, but not advanced maths, it's an overkill to use calculus if it can be avoided

but let's go again to the topic

1) in that website, it divides kinetic energy and potential energy by V
for this to be correct, we must refer to the very same V! but is this the case? how can the V be the same, when part of the liquid moves out of a orifice? does the kinetic energy apply for ALL the liquid in that container? because part of the fluid has much velocity (it runs out of the orifice, and the rest of it inside the container very slow). And we calculate as V the velocity of the liquid going out from the orifice! can you elaborate on that please?

2) in that website, it says about fluid potential energy, is this the potential energy due to weight or the potential energy due to pressure?
because in bernoulli, I think we have two different potential energies, the one due to the gravitational field and the other due to the pressure of the liquid
so can you write me please the energy conservation equation? and then, I will divide it by V to find bernoulli's equation

9. Apr 20, 2015

### Simon Bridge

The V is the same volume if you say that it is.
Just say: for a volume V the energy density is... it can help if that is a small volume compared with the overall setup ... say, a volume over which things like the crossectional area and the speed of the fluid are reasonably the same. If you used calculus, the volume can be infinitesimal.

While part of the liquid moves out of the orifice, another part moves in at the other end.

Of course not - that's why you also track the other energy densities: only the total energy is conserved. That's how you normally do conservation of energy problems. Just pick a small volume and track it as it moves through the system. It will be fast in some parts and slow in other parts, it will be higher in some places than in others, and it's pressure will vary. Pick a volume that fills the area of the pipe but varies in thickness... like the diagrams you see in text books considering the derivation.

"V" would be the volume, not the velocity.
The fluid will have different speeds in different parts of the system ... you can relate the speed, gravitational potential energy, and pressure in each part of the system to each other.

It tells you - sometimes it is talking about gravitational PE and sometimes about pressure and it says which it means. Sometimes you have to use context.

OK - I'll set you up: for a given volume $V$ of liquid the energy stored as pressure is $H=PV$, the energy stored as bulk motion in the fluid is $K=\frac{1}{2}mv^2$ and the energy stored lifting it a height h above some reference height is $U=mgh$ where $m= \rho V$ ... The quantity E=H+K+U is conserved.

Now how do you normally do conservation of energy problems?

10. Apr 20, 2015

### sciman

ok, so let's take a container with water, of height h, mass m, volume V, and we open a hole at its bottom so that water comes out
we take as reference of zero potential energy the botton of the container, is that okay?
as for the reference of pressure energy, we take the top of the water level, is that okay?

can you help me apply energy conservation before we open the hole and after we open the hole?

what is the potential and pressure energy of the water before the hole opens? kinetic energy is obviously 0
what is the after opening the hole sum?

11. Apr 20, 2015

### Simon Bridge

If you do not try some of the work yourself, I cannot help you.

12. Apr 20, 2015

### sciman

I have difficulty in finding what these energies equal to, that's what I am asking

Potential energy of what? the molecule at the top of the liquid or at the botton at the liquid?
Kinetic energy: what mass and what velocity?
Pressure energy: where? at the top or at the bottom of the liquid?

and if you write the conservation of energy equation as a per Volume, you need to be sure that the Volume you will divide all the parts of the equation with, is the SAME Volume!

aren't my questions valid?

13. Apr 20, 2015

### BvU

Check out this one

Re your post #6: let me try:

U shaped tube with liquid. Equilibrium: both sides same level, call it h.
Massless plunger ('cork' in the pic) at left is pushed down from h to 0. Costs work.
How much work ? Complicated math again (sorry). Bernoulli to the rescue !

Potential energy: gain (A to A') = loss (B to B')

Pressure energy wrt zero line:
In the left picture: average pressure $\rho g {h\over 2}$, volume $h\times{\rm Area}$ in each leg $\Rightarrow$ pressure energy in both legs combined $E_p = \rho g h^2 \times {\rm Area}$

Right picture: average pressure $\rho g {2h\over 2}$, volume $2h\times{\rm Area} \ \ \Rightarrow \ \ E_p = 2\rho g h^2 \times {\rm Area}$.

Difference $\Delta E_p = \rho g h^2 \times {\rm Area} = p V$, exactly the outcome of the complicated math for Force times displacement.

(Not for you but for others who may correct me if I am wrong:
$$F = \rho g h_r \times {\rm Area} \ \ \ \ \ h_r = 2 h_l \quad \Rightarrow$$ $$\int_0^h F\; dh_l = 2 \rho g \;{1\over 2} h^2 \times {\rm Area} = pV$$ )

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14. Apr 20, 2015

### sciman

thanks for this

can you answer my #10 and #12 questions?

15. Apr 20, 2015

### BvU

Your questions are valid, don't worry. For us it's difficult to explain in such a way that you can pick it up, that's all.

And the math IS pretty complicated.

Bernoulli follows from the equations of motion (basically a momentum balance), the equation of continuity (while 'going with the flow') and the equation of state (for compressible fluida). Far too much to worry about, nobody does. I quote from a famous book: "These equations in their complete form are seldom used to set up flow problems. Usually restricted forms of the equations of motion are used for convenience".

First restrictions: $\rho$ constant, viscosity constant. Gives the Navier-Stokes equation. Widely and intensely used !
Next steps: going over to energies, viscosity zero, no rotation, external force comes from (gravitational) potential
to get (via Euler equation) to Bernoulli.

 I typed up the above -- slowly -- but decided not to post it. Torn between agreeing with Simon (wom I highly respect -- you should do something about post #11 !) and wanting to help you -- assuming you are doing the best you can just as well as Simon.

You're welcome. I don't know if it helped much. By the way: the Bernoulli equation is not only for liquid flows, but also for gas flows at low Mach number (see hyperphysics: baseball curve, airfoil, )

Re #10: I thought I did already: " Check out this one "

Re #12: You want to consider the energy density along a flow line . The entire deviation is set up "going with the flow". That's why this energy density is so much more useful than the energy itself.

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