Identity 3.5.1.
Note the following identity:
\begin{equation}
(\bu \cdot \nabla) \bu = \nabla \left(\frac{1}{2}|\bu|^2\right) + (\nabla \times \bu)
\times \bu.\tag{3.5.1}
\end{equation}
Section 3.5 Bernoulli’s equation
There is a reformulation of the momentum equations that proves to be useful. In essence, it emerges from attempting to integrate the momentum equation (3.4.5) and yields the famous Bernoulli equation(s).
We will first need a vector identity.
It will also be useful for us to introduce the notion of vorticity.
Definition 3.5.2. Vorticity of a vector field.
The vorticity, \(\omega\text{,}\) of a vector field is defined by
\begin{equation}
\omega \equiv \nabla \times \bu.\tag{3.5.2}
\end{equation}
The vorticity is a measure of the local rotation of the flow.
Let us recall the definition of a conservative force.
Definition 3.5.3. Conservative forces.
A force \(\bF\) is a conservative force if and only if there exists a potential \(\chi\text{,}\) such that
\begin{equation*}
\bF = -\nabla \chi,
\end{equation*}
in a simply connected region where the quantities are defined.
Note the distinction about a simply-connected neighbourhood. The above is not quite the definition of a conservative force (typically defined to be a force for which the work done on an object between two points is independent of path). For now, this is not an important distinction since we will focus on fluid regions that are free of holes. Until told otherwise, we will always assume that the fluid is simply connected and therefore the above serves as a definition of conservative forces.
Theorem 3.5.4. Bernoulli’s equation for steady flow.
Bernoulli’s equation (theorem) for steady flow states that
\begin{equation}
B(\bx) \equiv \frac{p}{\rho} + \frac{1}{2} |\bu|^2 + \chi = \textrm{constant along
streamlines},\tag{3.5.3}
\end{equation}
where we have assumed that the body force is conservative, i.e. it can be written as
\begin{equation}
\bg = -\nabla \chi,\tag{3.5.4}
\end{equation}
for a potential function, \(\chi\) .
Proof.
Start from the momentum equation (3.3.1),
\begin{equation*}
\pd{\bu}{t} + \bu \cdot \nabla\bu = - \frac{1}{\rho}\nabla p - \nabla \chi
\end{equation*}
and use the vector identity (3.5.1) gives
\begin{equation}
\pd{\bu}{t} + (\nabla \times \bu) \times \bu = - \nabla \left(\frac{1}{\rho}\nabla p +
\frac{1}{2} |\bu|^2 + \chi\right).\tag{3.5.5}
\end{equation}
Next, the flow is steady, and therefore we can zero the first term. This leaves
\begin{equation*}
\bomega \times \bu = -\nabla B.
\end{equation*}
We now take the dot product of both sides of the equation with \(\bu\text{.}\) We use the fact that,
\begin{equation*}
\bu \cdot (\bomega \times \bu ) = 0,
\end{equation*}
since it is a triple scalar product with two repeated entries. It therefore results in the fact that
\begin{equation*}
\bu \cdot \nabla B = 0.
\end{equation*}
Notice that this is the steady component that comes from the material derivative, \(\DD{B}{t} = 0\text{.}\) So we conclude that \(B\) is constant along streamlines of the flow.
Remark 3.5.5.
For the typical gravity force directed in the \(-z\) direction, we can write \(\bg =
-g[0, 0, 1] = -g \be_z\text{.}\) So in this case, the (gravitational) potential is \(\chi = gz\) .
The proof of Theorem 3.5.4 introduced a useful form of the momentum equation using the vorticity function, resulting in (3.5.6). This leads to the so-called vorticity equation form of the momentum equation.
Theorem 3.5.6. The vorticity equation for incompressible flow.
The momentum equation for incompressible flow, reposed in terms of the vorticity is:
\begin{equation}
\pd{\bomega}{t} + (\bu \cdot \nabla)\bomega = (\bomega \cdot \nabla)\bu,\tag{3.5.7}
\end{equation}
or, equivalently,
\begin{equation}
\DD{\bomega}{t} = (\bomega \cdot \nabla)\bu,\tag{3.5.8}
\end{equation}
and is known as the vorticity equation.
In the proof, it is useful to use the vector identity for the curl of a cross product:
\begin{equation}
\nabla \times (\bu \times \bv) = (\nabla \cdot \bv)\bu - (\nabla \cdot \bu)\bv + (\bv
\cdot \nabla)\bu - (\bu \cdot \nabla)\bv.\tag{3.5.9}
\end{equation}
Proof.
Crucially, we recall the result that "curl grad equals zero" for a vector field. So from (3.5.6), we take the curl of both sides to obtain
\begin{equation*}
\nabla \times \left(\pd{\bu}{t} + \bomega \times \bu\right) = 0.
\end{equation*}
Expanding the outer cross product, the first term can be simplified by swapping differentiation in space with differentiation in time. The second term requires the cross product identity (3.5.9). For the second term, we get
\begin{equation*}
-\nabla \times (\bu \times \bomega) = 0 - 0 + (\bomega \cdot \nabla)\bu - (\bu \cdot
\nabla)\bomega,
\end{equation*}
with the first zero resulting in the fact that "div curl equals zero", while the second results from incompressibility.
Thus we have,
\begin{equation*}
\pd{\bomega}{t} + (\bu \cdot \nabla)\bomega = (\bomega \cdot \nabla)\bu,
\end{equation*}
which is the main result.
You will prove this result from first principles in Exercise 3.6.7.
Subsection 3.5.1 Potential flow
We are interested in the simplest scenarios that will result in reducing the governing equations. Previosly, we introduced the notion of incompressible flows, which results in the Euler equations in Definition 3.4.5 and in particular, the reduction of the mass conservation equation to \(\nabla \cdot \bu = 0\text{.}\)
We attempt to reduce further by assuming that the fluid is irrotational. This lends to the following definition.
Definition 3.5.7. Irrotational flows.
A flow is said to be irrotational if the vorticity is identically zero:
\begin{equation}
\textrm{flow is irrotational} \Longleftrightarrow \bomega = \nabla \times \bu \equiv
0.\tag{3.5.10}
\end{equation}
Theorem 3.5.8. Equivalence of irrotational and potential flow.
If a flow with velocity vector \(/bu(/bx,t)\) is irrotational, there exists a function \(\phi(\bx,t)\) such that
\begin{equation*}
\bu = \nabla\phi.
\end{equation*}
The function \(\phi\) is called the potential.
Proof.
Any function \(\bf(\bx,t)\) whose curl is everywhere zero \(\nabla\times\bf=0\) can be written as a gradient of a scalar function. In the case of irrotational flow, the velocity field is such a function.
For an irrotational and incompressible flow, there is a more powerful version of Bernoulli’s equation.
Theorem 3.5.9. Bernoulli’s equation for steady irrotational flow.
For an incompressible and irrotational flow, Bernoulli’s theorem states that
\begin{equation}
B(\bx) \equiv \frac{p}{\rho} + \frac{1}{2} |\bu|^2 + \chi = \textrm{constant
everywhere},\tag{3.5.11}
\end{equation}
where \(\bg = -\nabla \chi\) is the body force.
Proof.
The proof of this relies on returning to the proof of the unsteady Bernoulli’s equation in Theorem 3.5.4. In the proof, we arrived at the result in (3.5.3) that
\begin{equation*}
\pd{\bu}{t} + \bomega \times \bu = -\nabla B.
\end{equation*}
For steady flow, the first term disappears. If we have the additional assumption that the flow is irrotational, then we have that \(\bomega = 0\text{,}\) and therefore we have
\begin{equation*}
\nabla B = 0.
\end{equation*}
But the only way that all spatial derivatives is zero is if the function is constant.
You will practice an application of Bernoulli’s equation in Exercise 3.6.8.
Example 3.5.10. Airflow into or out of a room.
During the lecture, just for fun, we will play this video by Matthias Wandel about airflow out of a room. This is somewhat related to Bernoulli’s equation (the connection between pressure and velocity), though the situation is considerably more difficult.
Example 3.5.11. Pascal’s bursting barrel.
For fun during the lecture, we will play this video demonstration done at Princeton demonstrating the power of hydrostatic pressure.
Example 3.5.12. Urination across species.
During the lecture, we will play this video describing the work by Yang et al. on "Duration of urination does not change with body size". The title is not quite accurate, and the article describe that for most mammals tested greater than about 3kg, this is true. Bernoulli’s equation is used.
