#### Sohan

When we evaluate the bernoulli's equation-p/(density*gravity)+v^2/(2*gravity)+z=constant.
we get a value whose unit is metre.how am i supposed to actually interpret this result???

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#### entorm

Wouldn't that depend on how you interpret the rest of the quantities you put in?
I mean, you must have had a reason to use this equation.

#### minger

The essence of the equation is that the quantities will always be equal to a constant. That means that "energy" can be transferred from pressure to velocity, or potential to velocity, etc, but the summation will always be the same (in absense of frictional losses of course).

However, in some applications, that number can be referred to as 'head', particularly when relating to pumps. A pump might be rated at 30 feet of head. Since each term in the equation is in units of feet, you can get appropriate quantities. For example, if you want to know how high you can pump a fluid, then you have a situation of lets say a vertical tube. At the top, the only term present will be potential, and you find that you can pump it 30 feet high.

If you want to know maximum pressure obtained, then you get a closed flow solution where velocity and potential are zero, then you can get pressure from density of the fluid; likewise you can get maximum velocity.

#### stewartcs

When we evaluate the bernoulli's equation-p/(density*gravity)+v^2/(2*gravity)+z=constant.
we get a value whose unit is metre.how am i supposed to actually interpret this result???
Sounds like you've found the elevation change.

CS

#### minger

Sounds like you've found the elevation change
It's not necessarily an elevation change. The equation is written so that the terms are in units of length. That constant on the right if anything could be multiplied by specific weight to get the total pressure (the summation of static and dynamic pressure). In order to get a change in elevation, one would need two separate conditions. If you have velocity and pressure at two instances, then you can solve for $$z_2 - z_1$$ or $$\Delta z$$ which is the change in elevation.

#### stewartcs

It's not necessarily an elevation change. The equation is written so that the terms are in units of length. That constant on the right if anything could be multiplied by specific weight to get the total pressure (the summation of static and dynamic pressure). In order to get a change in elevation, one would need two separate conditions. If you have velocity and pressure at two instances, then you can solve for $$z_2 - z_1$$ or $$\Delta z$$ which is the change in elevation.
Sorry I misread the equation the OP wrote, I thought the OP had point 2 on the RHS (i.e. an energy balance), which is how the equation is normally used in practice.

The version of the Bernoulli equation the OP listed expresses the total energy at any particular point above some arbitrary horizontal datum plane given as total head in meters of fluid.

The fact the RHS is set equal to a constant is to emphasis that this is an ideal situation in which there is no energy loss (or addition) such as friction (or pumps). If it is not ideal, which in real life it isn’t, these losses must be included and the RHS is not constant. Hence, the Bernoulli equation is normally written as an energy balance to include point 2.

CS

#### ank_gl

Ideally it is the height to which the fluid ll be raised from the reference in a vertical tube, if bought to rest.

#### Sohan

Ok.thanks a lot everybody,i think i have got it....... By the way i have a little confusion over how the euler's equation for fluids was derived..(from which bernoulli's equation was obtained by integration considering the fluid is incompressible) Can anyone please post a link where the derivations is properly explained??

#### minger

The derivation of the Navier-Stokes and subsequent Euler equations (if those are the Euler equations that you're thinking of) are quite complex and lengthy. When I learned them, the full derivation took a few weeks of class and like 12 pages of notes.

http://www.ae.gatech.edu/labs/windtunl/classes/hispd/hispd06/ns_eqns.html [Broken]
as what looks like a decent reference. After the N-S are derived, one obtains the Euler equations by removing viscous effects. This has the effect of removing the troublesome right-hand side of the equations as well as IIRC some of the left hand side terms.

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#### ank_gl

Ok.thanks a lot everybody,i think i have got it....... By the way i have a little confusion over how the euler's equation for fluids was derived..(from which bernoulli's equation was obtained by integration considering the fluid is incompressible) Can anyone please post a link where the derivations is properly explained??
euler equation can also be derived simply by a mechanics analysis on a fluid element neglecting viscous forces.

#### Sohan

The link posted was too complex for me to understand.. i have recently taken up this course..
I wanted a derivation where in only pressure and gravity effect is taken into consideration.(streamline flow )
moreover I don't know how exactly to apply partial differentiation in problems related to practical physical situations like the ones encountered throughout in fluid mechanics.. are there any books which would help me in this aspect?

#### minger

I realize that the link would probably be too complex for you to grasp at this time which I why I mentioned that the derivation is quite complex and lengthy. Bernoulli's Equation is then derived from the full equations by eliminating terms which we choose to assume are negligible (read: engineers way of saying ignore).

Even if you don't wish to understand the derivation of the full equations or want to sit down and do it yourself, it's still important to understand them so you can see where the derived forms come from.

They look hard, but essentially they are the fundamental laws of physics, put into fluid perspective: Conservation of Mass, Conservation of Momentum, and Conservation of Energy. Assuming no viscous effects lets you ignore LOTS of terms, which gives you the Euler Eqeuations.

One can then further reduce down terms, maybe the problem is 1D, letting you ignore any $$\frac{\partial}{\partial y}$$ and $$\frac{\partial}{\partial z}$$ terms, etc etc.

As far as partial differentiation, typically it's used as a starting point for a numerical analysis (at least for me). When numerically solving fluid flows, we look at the equations and see them in the form of:

$$\frac{\partial \vec{Q}}{\partial t} + \frac{\partial \vec{E}}{\partial x} + \frac{\partial \vec{F}}{\partial y} + \frac{\partial \vec{G}}{\partial z} = S$$

We then say that if we can just find spatial derivatives of the flux vectors E,F,G, then we can "march" in time by computing the time derivative to obtain the new "Vector of Conserved Variables", the Q vector which is mass,momentum and energy. Starting with some initial condition and proceeding in time until the solution converges gives us our answer.

As far as analytic solutions, I learned them in a class called Conduction. It seems like it would be Heat Transfer II, but rather it's a class about solving PDE's, due to the fact that they appear so much in heat transfer. Solving these equations analytically are quite difficult and require a significant amount of prerequisite math before attempting IMHO.

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