# Bernoulli equation

1. Mar 20, 2008

### optrix

P + (1/2)mv^2 + mgh = constant.

So obviously (1/2)mv^2 is the kinetic energy, and pgh is the potential energy.

1.) Is this equation basically a statement of conservation of energy?

2.) If yes, then how exactly is pressure, 'energy'? I understand that a force acting over a distance is energy (work), and that pressure is force acting over an area, but in the case of a fluid in a pipe, where exactly is the distance that the force acts over?

3.) Also why does the equation only work when the flow is incrompressible (constant density), steady, and the thing thats got me most is non-viscous? I understand that from boundary layer effects, you can get a velocity gradient across the pipe (laminar flow), but why should this effect Bernoulli's equation?

I would really appreciate any help you guys can give. Thanks alot

2. Mar 20, 2008

### Snazzy

Your equation is wrong, substituate m for rho (density).

3. Mar 20, 2008

### TanGeng

I think you're missing a term in the equation.

PV + (1/2)mv^2 + mgh = constant : where V is the volume of the fluid and P, density of the mass constant over the entire volume

PV is an energy term

but it's more generally

P + (1/2)pv^2 + pgh = constant : where p is the local density of the fluid

this equation is true at every single point. Integrate the Bernoulli equation over a volume V and you get the first equation.

4. Mar 20, 2008

### optrix

yeh sorry I mean to put rho, and not m, for mass. When you put rho, in you do not have worry about having PV, only P. I think this is because you then have the pressure/kinetic/potential energy PER VOLUME?

Can you explain/link me to a proper derivation of this equation? I know a little multivariable calc, but I'm no expert. Do you consider infinitesimally small volumes, for which the equation hold, and then integrate that?

Also what about the requirement for a NON-VISCOUS fluid? Why should fluid viscosity make a difference? You will still have a steady flow, in that the streamlines/flow pattern stays constant, and that the velocity is constant at any point, but the laminae are simply travelling at different constant speeds with respect to one another (which is in fact a form of steady flow), so there is no acceleration.

5. Mar 20, 2008

### Staff: Mentor

Since the units are of pressure, obviously it isn't an energy equation, but it is a similar (the same, really) concept.

The reason for the requirement of inviscid flow is the loss of energy due to viscous friction. In a real situation (such as an air conditioning duct), the total pressure decreases.

Last edited: Mar 20, 2008
6. Mar 21, 2008

You can develop bernoulli's equation from the first law of thermodynamics and newtons second law. (maybe there are more)

If you develop this equation by yourself you will notice that to get to Bernoulli's Equation you need to make to many assumptions, like no shear, incompressible and steady state.
This means that Bernoulli's Equation have to many restrictions and is a simplistic equation.

7. Mar 21, 2008

### optrix

Russ, Thanks I read late last night that the fluid would have to provide energy to overcome the friction in a viscous flow, so that explains it. I suppose for a viscous flow we would have to accomodate for this too as a term in the equation.

I believe the equation IS actually a statement of the conservation of energy. It contains the changes in the kinetic and potential energy wehn the fluid has some external work done on it. This work is given by $$F(1)\Delta$$x - $$F(2)\Delta$$y, where each of the forces can be given as the respective pressure at those points, times the area. So actually it is really an energy term in the equation.

Link, thanks I've managed to follow a proper derivation of the equation. Since this is only applicable to steady, inviscid, incompressible flows, then what is the next step in expanding this formula to more complicated flows? I'm assuming this extension requires some calculus? (I hear that modified VERSIONS of bernoulli principle can be applied to more complicated flows)

Cheers!

8. Mar 21, 2008

Yes there is a modified Bernoulli Equation.

But you can't expand the "simple" Bernoulli's Equation into the modified version. The way to get into the modified bernoullis eq is to drop a couple of assumptions, like invicid and uniform flow.
This modified equation includes losses from fiction and work (turbine and pumps), also for non-uniform flows the velocity term have an coefficient.

9. Mar 21, 2008

### optrix

Thanks, Can you exsplain to me, or link me to an application of Bernoulli's equation to an incompressible, viscous, steady flow? I think i've really got hang of the equation now, and i want to move to the next stage and look at viscous flows. I have looked at Poissuielles law and see that its the rate of flow of volume in a viscous flow, but I don't think this is actually the same thing as what bernoulli's equation is saying.

10. Mar 21, 2008

### TanGeng

Poissuilles Law is applies in a very specific situation. It gives the flux of an incompressible, viscous liquid through a circular tube. The conduit for the liquid has to be perfectly circular. It's derived from equations of drag forces of laminar flow when applied to a circular tube.

Bernoulli's equation is about energy density at steady state. At steady state, energy density doesn't change and thus the constant on the right hand side of the equation. Bernoulli assumes a perfect liquid. You cannot apply it to a viscous liquid. You might think of it as the trivial case of a much more general equation that takes into account compression, viscosity, and other factors of fluid dynamics.

Last edited: Mar 21, 2008
11. Mar 22, 2008

12. Mar 22, 2008

### optrix

Thanks link, I appreciate the link, but I can't read the equations. I found a version of the bernoullis equation for compressible flows, and this is the kind of extension I was talking about. I'm not sure if the equation you gave takes into account compressibility and/or viscous forces, cause I can't read it (From the text it looks like it takes into account viscosity though). :)

I found similar pages anyway. They look like they are getting into divergence theorem, and more advanced calculus, like the Navier stokes equations do. This is good, cause eventually I want to understand them too, but i'm gonna work my way through this stuff first.

Thanks for the help

13. Mar 24, 2008

### wongdaisiu

Bernoulli's equation assumes, inviscid (no friction between wall and fluid, and between fluid and fluid), incompressible flow (constant density). The link that Link referenced was a modified form of Bernoulli's equation. However, the added terms are just losses based upon empirical data. I digress.

I will try to explain it physically to you. You have a liquid (since those substances are basically incompressible) coming in with certain inlet conditions (p, v, A). p = pressure, v = velocity, A = cross sectional area. We will assume there is no change in height to make it simple. Now if the outlet is smaller, the velocity would have to increase (this is because of conservation of mass, no accumulation of mass within the control volume). So to compensate, the fluid pressure would have to decrease. The opposite is true. If the outlet area is bigger, then the velocity would decrease due to conservation of mass (can't have a vacuum), and the pressure would increase. This also applies to gases at low velocities.

Don't think too much into compressiblity of fluid flow for now.

Hope this helps.

14. Mar 25, 2008

### optrix

Thanks for the explanation, wong I think I already understand that stuff now. i'm making good progress with the fluid dynamics, and have looked at some other theorems for idealised, inviscid, incompressible, streamlined flows. it's all making perfect sense to me now. I'm looking at some dimensional analysis too.