Does conservation of energy for unsteady ideal fluid hold?


by dEdt
Tags: conservation, energy, fluid, hold, ideal, unsteady
dEdt
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#1
Mar5-12, 07:24 PM
P: 249
My text was able to show that for an ideal (incompressible and inviscid) and steady fluid in a gravitational field, the energy density [itex]E=\frac{1}{2}\rho u^2 + \rho\chi+P[/itex] is constant for any fluid element, where [itex]\chi[/itex] is the gravitational potential. That is
[tex]\frac{DE}{Dt}=\frac{\partial E}{\partial t} + \mathbf{u}\cdot\nabla E=0[/tex].Does this hold for an unsteady ideal fluid? If not, what causes the change in the mechanical energy of the fluid element?
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Jano L.
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#2
Mar5-12, 08:53 PM
P: 1,030
Hello dEdt,
beware, energy density is [itex]\epsilon = \frac{1}{2}\rho u^2 + \rho\chi[/itex]. Pressure does not contribute to energy density. For example, consider incompressible liquid; there can great pressure in it but no work is required to produce it, since it does not change its volume. The equation you wrote is supposed to mean that the right-hand side is constant along the streamline, even if the fluid is moving. This is true for incompressible liquid without friction. For compressible liquid there is enthalpy density there instead of pressure.

Jano
dEdt
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#3
Mar5-12, 10:40 PM
P: 249
Sure, but the force on a fluid element of volume [itex]\delta V[/itex] due to pressure is [itex]-\nabla P \delta V[itex]. So for that fluid element, it's possible to regard P as a potential energy (per unit volume). It might not be true energy, but I still think it's valid to regard it as a form of potential energy, at least for the purposes of calculations, no?

Jano L.
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#4
Mar5-12, 11:49 PM
P: 1,030

Does conservation of energy for unsteady ideal fluid hold?


It is not correct to regard pressure of liquid as density of energy. You would get wrong total energy of the liquid.

Consider water in a tank.

The buoyant force on a small volume of fluid is indeed [itex]\mathbf F = -\frac{\nabla p}{\rho} \Delta m[/itex], so you can view [itex] p/\rho[/itex] as a potential of the buoyant force. But you should not ascribe this potential energy to the fluid element, because then if you sum up these energies, you will get [itex]\int_V pdV = \int \rho g h dV[/itex], which however is already accounted for by the potential energy in gravitational field. So counting pressure as energy actually accounts for this energy twice, which is wrong.
dEdt
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#5
Mar6-12, 08:41 AM
P: 249
Quote Quote by Jano L. View Post
But you should not ascribe this potential energy to the fluid element, because then if you sum up these energies, you will get [itex]\int_V pdV = \int \rho g h dV[/itex], which however is already accounted for by the potential energy in gravitational field. So counting pressure as energy actually accounts for this energy twice, which is wrong.
I don't see the problem with recognizing two sources of potential energy, one from gravity and the other from pressure. In your example, [itex]K+\int{\rho ghdV} + \int{\rho ghdV}[/itex] is conserved if [itex]K+\int{\rho ghdV}[/itex] is conserved, because h doesn't change. In other situations, such as in unsteady flow, where the total gravitational potential energy changes, it is only the first expression that is conserved, because although the walls of the container (ie the source of the pressure) don't do any work, pressure does do work when the fluid expands, which is the only time gravitational potential energy will ever change.

At any rate, you can frame my original question this way: in steady, ideal flow, there's a mysterious quantity [itex]E=\frac{1}{2}\rho u^2+\rho\chi+P[/itex] which is conserved for any fluid element. Is this quantity still conserved in unsteady ideal flow?
Jano L.
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#6
Mar6-12, 11:52 AM
P: 1,030
Think of it this way. Energy is defined in terms of work. How much work do you need to fill a tank of height [itex] h[/itex] and base area [itex] S[/itex] with water (density [itex] \rho[/itex]), if the water is initially in the lake at the same level as the base of the tank?

This work defines energy of the system. You can calculate this by adding the small amount of work needed to raise the level of the water by [itex]\Delta h[/itex]. You will see the work is [itex]\frac{1}{2} S\rho gh[/itex] which defines the total energy of the system. There is no trace of pressure energy, precisely because the liquid is incompressible and it does not take any work to increase the pressure in the liquid.

To your question: in nonstationary flows the expression is not constant.

If the nonstationary flow is potential (means [itex]\mathbf v = \nabla\phi[/itex]), the equation reads

[tex]
\frac{\partial \phi}{\partial t} + \frac{1}{2}v^2 +\chi + p/\rho = f(t)
[/tex]

for some function [itex]f(t)[/itex].
dEdt
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#7
Mar6-12, 11:24 PM
P: 249
Okay, I've understood. Thanks for taking the time to explain.
Jano L.
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#8
Mar7-12, 04:16 AM
P: 1,030
Glad to be of help,
Jano


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