Euler’s equation | math3402liamsnow

Hello again everybody, and welcome back to my blog! This week’s blog post is revisiting a topic that we covered way back at the start of the term, and that is Bernoulli’s Principle. The principle itself is defined from Bernoulli’s equation, which I will derive here as well. With that out of the way, let’s get started! 🙂

Bernoulli’s Equation, and its derivation

We start our derivation by re-introducing Euler’s Equation, a topic that I covered in my last blog post:

$\dfrac{D\vec{u}}{Dt} = \dfrac{-1}{\rho}\nabla p + \vec{F}$

If we assume that the body force is conservative (i.e. $\vec{F} = -\nabla\phi$ ) then we can rewrite Euler’s equation as follows:

$\dfrac{D\vec{u}}{Dt} = \dfrac{-1}{\rho}\nabla p -\nabla\phi$

With our knowledge of the convected acceleration (as mentioned in my blog post on flow visualisation), we can then expand the L.H.S. as follows:

$\dfrac{\partial\vec{u}}{\partial t} + \nabla\left(\dfrac{1}{2}|\vec{u}|^2\right) - \vec{u} \times \text{curl}(\vec{u}) = -\dfrac{1}{\rho}\nabla p - \nabla\phi$

If we now make another assumption in that we assume the flow to be steady (i.e. the velocity of the flow does not depend on time), then our first term vanishes, leaving:

$\nabla\left(\dfrac{1}{2}|\vec{u}|^2\right) - \vec{u} \times \text{curl}(\vec{u}) = -\dfrac{1}{\rho}\nabla p - \nabla\phi$

Just for convenience, I’ll combine all the terms that are affected by a $\nabla$ , so that we get:

$\nabla\left(\dfrac{1}{2}|\vec{u}|^2 + \dfrac{p}{\rho} + \phi\right) = \vec{u} \times \text{curl}(\vec{u})$

Now this derivation involves us finding Bernoulli’s equation along a streamline, a fact that I have not brought up until now. So now, we need to find the component of our previous equation in the direction of said streamline.

For those of you who did Vector Calculus last year, we know that we can do this by simply defining a tangential vector $\vec{s}$ (which is also defined as a unit vector) to the streamline, and then take the dot product of this with the equation above. This will then give the following:

$\vec{s}.\nabla\left(\dfrac{1}{2}|\vec{u}|^2 + \dfrac{p}{\rho} + \phi\right) = \vec{s}.\vec{u} \times \text{curl}(\vec{u})$

The R.H.S. cancels to zero, due to the fact (from Vector Calculus) that the resultant dot product of two perpendicular vectors $\left(\vec{u} \times \text{curl}(\vec{u})\right)$ must always be equal to zero. Therefore we can now simplify our above equation to:

$\dfrac{d}{ds}\left(\dfrac{1}{2}|\vec{u}|^2 + \dfrac{p}{\rho} + \phi\right) = 0$

Since the derivative equals zero, then we know that the contents in the brackets must be equal to a constant along our streamline. Thus we get streamline version of Bernoulli’s equation:

$\dfrac{1}{2}|\vec{u}|^2 + \dfrac{p}{\rho} + \phi = \text{const}$

Bernoulli’s Principle

If we now consider the case where our body forces are negligible $(\phi = 0)$ , then we write Bernoulli’s equation in the following way:

$\dfrac{1}{2}|\vec{u}|^2 = \text{const} - \dfrac{p}{\rho}$

Thus we have produced a relationship between the speed of the flow, and its pressure, which if you recall, is the basis of the observations that I spoke about in one of my first blog posts.

Therefore, by examining the relationship, we can see that:

High speed $\implies$ low pressure

Low speed $\implies$ high pressure

The various applications of Bernoulli’s Equation

During the lectures, we were shown various applications for Bernoulli’s equation. These included measuring the flow from a reservoir, a Pitot tube, and a Venturi tube, as well as more complex examples, such as forces exerted on a wall due to a vortex, and the flow around a cylinder.

In the future, I would like to revisit these applications, as a means of further research, and to help me solidify my understanding of Bernoulli’s equation for revision purposes.

Example Sheet Questions

Below you will find my scanned answers to (most of) question 2 of Example Sheet 3.

Example Sheet 3 – Question 2

Reflection

Much like the derivation of Euler’s equations, I believe that the derivation of Bernoulli’s equation is relatively straight forward, and the conclusion from the Bernoulli’s Principle is quite a simple, but yet powerful, observation.

However, I did have a few difficulties with understanding it in regards to the second question on Example Sheet 3. The change in sign within $\psi$ (once I had introduced the arctangent), its subsequent derivation, and upon checking the solutions, my inclusion of $\rho$ (that is seemingly missing in the solutions), have all caused me problems with this question. When it comes to revision, its clear that I need to spend some time swatting up on common derivatives of trigonometric functions, and perhaps getting some further clarification about the assumptions made during the calculation.

If I had more time in which to complete these blog posts, I would have done significantly more background research and revision. For this post alone, I should really do more reading into the variety of applications that Bernoulli’s equation has in practice. This is something that I’ll keep in mind and carry forward into my second term, and begin doing as soon as I have the opportunity to do so. I am pleased though with what I have done with them, even though most of my more recent posts are inconsistent in style to my older ones.

Thanks for a great first term, and I’ll see you in the New Year! 🙂

References and Sources:

1) All of the concepts talked about in this blog post are derived from notes taken from Chapter 2 of the Inviscid Flow PowerPoints on Bernoulli’s equation, and my own notes derived from the lectures.

Welcome everybody, to yet another blog post on fluid dynamics! 😀 This blog post will try and follow a slightly different format to my previous posts. Hopefully this one will be more succinct than past attempts, which should mean less sore eyes for you! 😛 Without further ado, let us begin.

Euler’s Equations, and their derivation

As you will see soon, Euler’s equations are very closely linked to the Navier-Stokes equations, which I mentioned in a previous blog post. However, we must still be aware that these equations are related to inviscid flow (as has most of the previous content I’ve covered), instead of the viscous flow that the Navier-Stokes equations cover.

Let me now describe the scenario of the derivation (but don’t worry, I’ll include a picture below from my lecture notes for clarity).

Visual overview of the derivation of Euler’s Equations, from the lecture notes.

Once again, we are examining but a small fraction of a moving fluid, which is composed of: a small volume $V$ , that has a small surface $\delta S$ (with corresponding normal vector $\vec{n})$ , all the while moving with a velocity $\vec{u}$ .

From this, we can deduce the other following properties:

(Convected) Acceleration = $\dfrac{D\vec{u}}{Dt}$
Mass = $\rho V$
Body force = $\vec{F}$
Pressure (due to force of moving fluid) = $-p\vec{n} \delta S$

By substituting these appropriately into an integral, we get:

$\dfrac{D\vec{u}}{Dt}\int_V \rho dV = \int_V \rho\vec{F} dV - \int_S p\vec{n} dS$

where the first integral equates to a total mass multiplied by the acceleration (a force), which is equivalent to the total body force minus the total surface force.

For convenience, we can now apply Gauss’ theorem to the integral over the surface, to convert it into an integral over the volume (like the other integrals), and thus produce:

$\dfrac{D\vec{u}}{Dt}\int_V \rho dV = \int_V (\rho\vec{F} - \nabla p) dV$

As stated above, we have already assumed that the volume is small, so if we now take $V \to 0$ , we can conclude that:

$\dfrac{D\vec{u}}{Dt} = \dfrac{-1}{\rho}\nabla p + \vec{F}$

The above is known as Euler’s equation (in its whole form), and is also obtained when we allow the viscosity to equal zero in the Navier-Stokes equations.

To be more explicit, here is the Cartesian version of Euler’s equation in summation form:

$\rho\left(\dfrac{\partial u_i}{\partial t} + u_j \dfrac{\partial u_i}{\partial x_j}\right) = \dfrac{\partial\rho}{\partial x_i} + \rho(F_B)_i$

A polar form is also shown similarly, but I will not state it here.

Example Sheet Questions

Below you will find my scanned answers to question 1 of Example Sheet 3.

Example Sheet 3 – Part 1

Reflection

In my honest opinion, I think that the derivation of Euler’s equations was not too bad (as I had mentioned, the process to deduce them is essentially a simplified Navier-Stokes method), although I still feel a bit apprehensive about certain details of it. In particular, I don’t think I quite understand the logic in deducing the force due to the pressure. Perhaps this is some anxiety I still hold from my experiences with similar properties in Vector Calculus last year? Either way, when it comes to revision, I will make sure that I fully understand the derivation.

Even though I’ve only attempted the first question of Example Sheet 3 here, I feel a bit more confident about using Euler’s equation now. The only problem I have with my uploaded solution is that, upon checking, it seems I have a factor of $\rho$ in front of the equation, whereas the uploaded solutions do not. If I have made a mistake, then perhaps I will clarify this with one of my peers, or through David directly.

On the whole, I am especially pleased with how this new ‘format’ went, as it seems to be much more succinct, and cuts to the chase. Of course, I still felt inclined to show a derivation, but that felt necessary in order to explain where the equations came from.

Thanks to anyone who has taken the time to read this blog post! See you soon! 🙂

References and Sources:

1) All of the concepts talked about in this blog post are derived from notes taken from Chapter 2 of the Inviscid Flow PowerPoints on Euler’s equation, and my own notes derived from the lectures.

math3402liamsnow

Nothing but a torrent of fluid dynamics.

Tag Archives: Euler’s equation

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