In the right side of the Bernoulli equation you should add the Energy Loss, Delta h. The equation should be read as total energy at first point is equal to the total energy at the second point plus the energy (head) loss. Therefore, the actual energy at the two sections is not the same. Energy must be lost in the direction of flow.

May be that my answer is to simple: Bernoulli's equation is a statement of energy conservation for a material fluid parcel (under the assumption that the motion is adiabatic and temporal presssure variations are negligible). Whether there flow or not is determined the momentum balance, ie if there are unbalanced pressure or external forces acting on the fluid parcel. Take for simplicity the harmonic oscillator: total energy is constant by the thing is moving.

I understand the difference between (and orthogonality between) equipotential lines and streamlines (in an isotropic medium), which is why I'm afraid I'm still confused. I am unable to find any definitions/derivations of the Bernoulli equation that mention constant fluid energy along equipotential lines... only constant fluid energy along streamlines. How can there be flow along a streamline if the fluid energy is constant along the streamline?

From Whitaker, Introduction to Fluid Mechanics:

"We will derive Bernoulli's equation along a streamline. We do so by forming the scalar product of the unit tangent vector to a streamline and the equation of motion, thus yielding the scalar component of the equations of motion in the direction of the streamline."

From Fetter, Applied Hydrogeology:

"For steady flow of a frictionless, incompressible fluid along a smooth line of flow (i.e. a streamline), the sum of the three components [of energy per unit mass] is a constant."

Constant energy along a streamline would result in no flow in that direction, would it not?

The most general derivation of the Bernoulli equation is (to my knowlwegde) in our book 'Ocean Dynamics':

http://link.springer.com/book/10.1007%2F978-3-642-23450-7#section=1063984&page=2&locus=0

The attached pages are from the proof.

Thanks Mostafa. I was unable to let go of my "groundwater" perspective, in which an energy gradient is required to sustain flow by overcoming the resistance of the porous media.

And I thank you for your answers!

The force on fluid particle do not remain constant for Bernoulli flow. For example, there is a pressure gradient in contracting and expanding flows. However the forces only create energy transfer between mechanical energies (potential and kinetic).

Any real fluid, of course, does not conserve mechanical energy, due to friction and

turbulence. Bernoulli's equation is adapted to real flows by adding energy loss terms, such as friction and other loss factors, with names such as the Darcy–Weisbach equation or Fanning equation (depending upon in which discipline one studies this).

In the right side of the Bernoulli equation you should add the Energy Loss, Delta h. The equation should be read as total energy at first point is equal to the total energy at the second point plus the energy (head) loss. Therefore, the actual energy at the two sections is not the same. Energy must be lost in the direction of flow.

Thank you Mostafa for the Clarification. However, my understanding is that there will also be energy loss along the stream line even if the flow is steady and uniform. For example, when applying Bernoulli equation for one dimensional steady uniform flow in an open channel along the stream line you will still have energy loss in the direction of the flow. In such a case, the average velocity and water depth are pressure remain the same and hence the specific energy (energy measured from the bed as a datum) will remain constant. The head loss in such a case will be equal to the difference in elevation of the bed at the two points under consideration (dz). Therefore, dh = dz. It is important to differentiate between total energy and specific energy.

For more details, please refer to the relevant chapters of co-authored textbook,

Cruise J., Sherif M.M., and Singh V.P., "Elementary Hydraulics", hard cover, Thomson Nelson, Toronto, Ontario, Canada, 2007, 560p., ISBN: 0-534-49483-8.

This appears to be a valid question. The crux is possibly in the word 'frictionless' that is used by Fetter (above). The Bernoulli-equation holds only when friction can be neglected.

With friction, energy must still be conserved, but we have to consider an increase in internal energy of the fluid as indicated by a rise of temperature. I believe it was Joule who traveled to Schaffhausen around 1845 to measure the rise in temperature over the falls: it was tiny but observable.

Dear Wes Christensen. I want to let know you that Bernoulli's equation as you write it, represents a potential energy, i.e. defines the hydraulic load total in an aquifer, each component is expressed in units of (L, in ft or m). This means raising the groundwater level under a datum, in steady state.

Dear Wes Christensen. I want to let know you that Bernoulli's equation as you write it, represents a potential energy, i.e. defines the hydraulic head total in an aquifer, each component is expressed in units of length (ft or m). This means raising the groundwater level under a datum, in steady state.

This is less surprising than you may think. Consider a mass m in the earth gravitational field. We let it drop in vacuum (hence, no friction) from a height h. The mass starts with potential energy mgh and zero kinetic energy. During its fall, it will gain Ekin and loose Epot, but the sum of these two remains constant:

1/2 mv^2 + mgz = const.

Divide by m and we get something quite similar as Bernoulli's law:

v^2/2 + gz = const

Again, no gradient of total energy along the path, but of course a gradient in the potential energy and thus a force to accelerate the mass.

Actually, Bernoulli based his ideas on the above, realising that in fluids there is also something like "pressure energy".

"something like 'pressure energy'" is Pv flow work - the work energy transferred in

a fluid flowing through a control volume (it is also the Pv part of enthalpy, which

arises in an energy balance if including thermal energy in a flow system).

The question as first posed by Wes already implies that p/rho is an energy-term, but this one is seldom mentioned in lists of forms of energy. This is addressed in my paper 'Energy-balance of the Joule-Thomson experiment: Enthalpy change at decompression', retrievable at this site.

Wes, the Bernoulli equation derives from an approximation where all sources of dissipation are neglected (viscous friction, interfacial irreversibilities, heat conduction, and so on and so forth).

Therefore, Bernoulli's equation is CONSERVATIVE in the sense of classical mechanics: The total energy of any fluid particle remains constant along its streamline. Only the distribution of that energy in its various possible form (the three terms of the equation) may change.

In a way, you can think of a Bernoulli flow as a flow whose motion is driven only by pressure forces, gravity or inertia. There is no need for energy loss with these three sources (although, once again, this is but an approximation of reality).