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The Bernoulli equation for unsteady flow is given by: $$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle} \p{\phi}{t}+\f{1}{2}\vec u\cdot \vec u+gh+\f{p}{\rho}=0$$ In steady state the first term is zero with the equation been a statement of the conservation of energy:

  • 2nd term is kinetic energy
  • 3rd term is potential energy
  • 4th term is pressure energy (described e.g. here)

Assuming that in the unsteady case the equation still represents a statement of the conservation of energy, what type of energy/phenomenon does: $$\p{\phi}{t}$$ represent?

Qmechanic
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Quantum spaghettification
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3 Answers3

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There is a hypothesis left over here: the flow is irrotational, i.e. the so-called vorticity

$$\vec{\omega}=\nabla\times\vec{v}$$

is zero. As a result, the velocity field is a gradient,

$$\vec{v}=\nabla\phi,$$

for some scalar field $\phi$: the field appearing in the Bernoulli equation you quoted. Note that the right-hand side is not necessarily zero but it can be taken as a constant, which does not depend either on position or on time.

This is a simple consequence of (i) Euler equation for an incompressible fluid,

$$\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot\nabla) v + \nabla\left(\frac{p}{\rho}+\psi\right)=0,$$

where $\psi=gh$ but it could be the potential for any other conservative force field; and (ii) the identity

$$\vec{v}\times\vec{\omega} = \nabla\left(\frac{v^2}{2}\right) - (\vec{v}\cdot\nabla) v.$$

Then Euler equation reads

$$\nabla\left(\frac{\partial\phi}{\partial t}+\frac{v^2}{2}+\frac{p}{\rho}+\psi\right)=0.$$

Therefore the term between parentheses is function of time only but we can absorb any time dependence into $\phi$, and therefore we end up with

$$\frac{\partial\phi}{\partial t}+\frac{v^2}{2}+\frac{p}{\rho}+\psi=C$$

where $C$ is constant over space and time. This is therefore a stronger result than the traditional Bernoulli theorem where the constancy is only along each streamline, with a priori a different constant for each streamline. Here the constancy is over the entire fluid.

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The question starts with the assumption that the Bernoulli Equation is an energy equation. In general, it is NOT!! The Bernoulli Equation is the integral of fluid momentum along a streamline. The energy equation is the integral of fluid displacement in the direction of the force on it. The fluid displaces along its trajectory or "pathline". For steady flow, streamlines and pathlines are the same. So the Bernoulli Equation "looks" like an energy equation. For unsteady flow, streamlines and pathlines are generally different. So there is NO energy interpretation of the unsteady Bernoulli term $\rho \, \partial \phi/ \partial t$.

Hyperon
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James Grotberg
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If we write the Euler equation along a streamline, we obtain,

$$ \frac{\partial V_s }{\partial t} + V_s \frac{\partial V_s }{\partial s} = -\frac{1}{\rho} \frac{\partial p }{\partial s} + g_s$$

Moreover, assuming constant $\rho$ and integrating this along a streamline between points 1 and 2 we obtain,

$$ \frac{p_1}{\rho} + \frac{V_1^2}{2} + g z_1 = \frac{p_2}{\rho} + \frac{V_2^2}{2} + g z_2 + \int_1^2 \frac{\partial V_s}{\partial t} \ ds$$

In this form it is easier to see what exactly the new unsteady term physically represents. This term is obtained from integrating the momentum equation and has units of energy per unit mass. Therefore, we would say that $\int_1^2 \partial V_s / \partial t \ ds$ is the work involved in the rate of change of momentum of the fluid on the streamline overtime. Note, this has the same meaning as $\partial \phi / \partial t$ in your expression.

TRF
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