How to derive Bernoulli's theorem for an elastic fluid from Lamb's equation

Question

Here I state Lamb's version of the Euler's equation for an elastic fluid: $$ \partial_t \vec v + \vec \Omega \times \vec v + \vec \nabla \left( \frac{\|\vec v\|^2}{2} + \frac{p}{\rho} + \psi + \phi_b\right) = 0$$ where $\vec\Omega$ is the vorticity of the fluid, $p$ the pressure, $\rho$ the density, $\vec b$ is a potential body force such that $\vec b = \vec \nabla \phi_b$ and $\psi$ is Helmholt's density of energy that in our course has been defined as the fuction $\psi = \psi(\rho)$ such that $$p(\rho) = \rho^2 \frac{\partial \psi}{\partial \rho}$$

Using Lamb's equation we stated this version of Bernoulli's theorem:

Theorem: Consider an elastic eulerian fluid, then in steady flow the quantity $\frac{\|\vec v\|^2}{2} + \frac{p}{\rho} + \psi + \phi_b$ is constant along the flow lines. (Usually I've been told that the quantity $w :=\frac{p}{\rho} + \psi$ is called Gibbs' free energy)

Proof: It sufficies to show that $\vec v \cdot \vec \nabla \left( \frac{\|\vec v\|^2}{2} + w + \phi_b\right) = 0$ but this is obvious from Lamb's equation.

My questions are:

• Why It sufficies to show that $\vec v \cdot \vec \nabla \left( \frac{\|\vec v\|}{2} + w + \phi_b\right) = 0$ to prove that theorem? (The obviousness of the next passage is clear to me)

• How this theorem can be linked to the "classical" Bernoulli statement: $p = -\rho \frac{\|\vec v \|^2}{2} + k$ where $k$ is a constant?

Answer 1

A Streamline (flow line) is a curve with parametric representation $\mathbf{x}_S(s)$ at some fixed time $t$ such that

$$\mathbf{x}_S(s) \times \mathbf{v}(\mathbf{x}_S(s),t) = 0$$

The velocity field is, therefore, tangential to the streamline at any point, and

$$\frac{d \mathbf{x}_S}{ds}(s) = \alpha(s) \mathbf{v}(\mathbf{x}_S(s),t) $$

Defining the quantity $H (\mathbf{x},t) = \frac{\| \mathbf{v}\|^2}{2} + w + \phi_b$, we have

$$\mathbf{v}(\mathbf{x}_S(s),t) \cdot \nabla H(\mathbf{x}_S(s),t) = 0,$$

and, by the chain rule,

$$\frac{\partial}{\partial s} H(\mathbf{x}_S(s),t) = \nabla H(\mathbf{x}_S(s),t)\cdot \frac{d \mathbf{x}_S}{ds}(s) = \alpha(s) \mathbf{v}(\mathbf{x}_S(s),t)\cdot \nabla H(\mathbf{x}_S(s),t) = 0 $$

Therefore, $H$ is constant along a streamline, where value of this constant can be different for each streamline.

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What you refer to as the "classical" Bernoulli statement,

$$\tag{1} p + \frac{1}{2} \rho \|\mathbf{v}\|^2 = k,$$

where $k$ is constant at all points in the domain is valid for steady inviscid, incompressible and irrotational flow.

For a proof, begin with the Euler equations governing steady, inviscid flow,

$$\tag{2}\rho (\mathbf{v} \cdot \nabla) \mathbf{v} = - \nabla p,$$

along with the incompressibility condition $\nabla \cdot \mathbf{v} = 0$.

By the general vector identity

$$\nabla (\mathbf{a} \cdot \mathbf{b}) = (\mathbf{a} \cdot \nabla) \mathbf{b} + (\mathbf{b} \cdot \nabla) \mathbf{a} + \mathbf{a} \times (\nabla \times \mathbf{b}) + \mathbf{b} \times (\nabla \times \mathbf{a}),$$

we get with $\mathbf{a} = \mathbf{b} = \mathbf{v}$,

$$\nabla (\|\mathbf{v}\|^2) = \nabla (\mathbf{v} \cdot \mathbf{v}) = 2(\mathbf{v} \cdot \nabla) \mathbf{v} + 2\mathbf{v} \times (\nabla \times \mathbf{v})$$

For irrotational flow, where $\nabla \times \mathbf{v} = 0$, this reduces to

$$\tag{3}(\mathbf{v} \cdot \nabla) \mathbf{v} = \nabla (\frac{1}{2}\|\mathbf{v}\|^2)$$

Since $\rho$ is constant we obtain after substituting (3) into (2),

$$\nabla (\frac{1}{2}\rho\|\mathbf{v}\|^2) = -\nabla p, $$

whence $\nabla (p + \frac{1}{2}\rho\|\mathbf{v}\|^2) = 0$ and (1) must hold where $k$ is a global constant.

Answer 2

I show an intuitive argument. The velocity field defines some integral curves. These are referred to as flow lines and have the property of having the velocity field as tangents. Clearly any quantity F with a gradient always orthogonal to the velocity field will have the gradient orthogonal to the integral curves as well. In such a situation F must be constant along the integral curves.