Tag Archives: fluid

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Fluid Dynamics – Week 8: Bernoulli’s Equation and Principle

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

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.

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Fluid Dynamics – Week 7: Euler’s Equations

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.

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

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.

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Fluid Dynamics – Week 6: Dimensional Analysis

Hello everybody and welcome back, once again, to another instalment in my fluid dynamics blog! 🙂 Without further ado, let’s get started!

Dimensional Analysis

Dimensional analysis is a means by which we can understand the properties of physical quantities, but instead of looking at their exact units, we can measure them with regards to their actual dimensions.

Physical quantities are typically composed of mass M, length L, and time T in fluid dynamics, but to varying powers. Other dimensions do exist, but we won’t consider them here.

This leads on nicely to…

Dimensional Homogeneity

This is where we examine an equation composed of physical quantities, and determine whether it is valid or not. The validity of the equation itself is based on whether both sides of the equation equate to having the same dimensions.

For example, from the lecture notes, the following equation was given:

L = C\rho Uw

where L = lift force, C = circulation, \rho = density, U = speed, and w = wing length.

We don’t know the dimensions of circulation off the top of our heads, but we can calculate it based on the various dimensions of the physical quantities that we do know.

Therefore, by rearranging the equation above to make C the subject, we get:

C = \dfrac{L}{\rho Uw}

which we can then write in terms of dimensions as:

C = \dfrac{(MLT^{-2})}{(ML^{-3})(LT^{-1})(L)}

Upon inspection of the dimensions, we can see that the mass factor cancels. After simplification of the powers, we can see that circulation must have dimensions of:

C = L^2 T^{-1}

Based on the units of the other physical quantities in the equation,  it would then be possible to deduce the units of the circulation.

Buckingham’s Pi Groups

In the previous example, it was relatively easy to calculate the dimensions, since we already knew the equation, and thus the proportions of components involved, based on their powers. But what if we knew the physical properties of the components, but not their proportions?

This is where the idea of dimensionless groups and Buckingham’s \pi theorems come into play.

Let us first consider the force F on a propeller, as a function of the following:

F = \phi(d, u, \rho, N, \mu)

where d = diameter, u = forward velocity of the propeller, \rho = density, N = rotation rate (in revolutions per second) and finally \mu = viscosity.

This can subsequently be written as:

\phi_1(F, d, u, \rho, N, \mu) = 0

where \phi, \phi_1 are currently unknown functions. Since we can now combine the components of \phi_1 in such a way that they equal zero, we have created our first dimensionless group.

These dimensionless groups are commonly found using Buckingham’s \pi theorems. The two theorems are stated below, as taken directly from the lecture materials:

Buckingham’s first \pi theorem:

‘A relationship between m physical variables can be expressed as a relationship between m-n non-dimensional groups of variables (called \pi groups), where n is the number of fundamental dimensions (such as mass, length and time) required to express the variables.’

Buckingham’s second \pi theorem:

‘Each dimensionless group is a function of n governing or repeating variables plus one of the remaining variables.’

Oh, and one last thing before we start with the example. As the theorems mentioned, we have n repeating variables. For most fluid dynamics cases, we consider \rho, u \text{ and } d as the repeating variables, although we may sometimes also consider the viscosity over the density in certain situations.

From our equation for \phi_1, we can deduce that n (the number of repeating variables) = 3, and that m = 6 (the total number of physical variables). Therefore, from the first theorem, we have 3\pi groups, such that:

\phi_1(\pi_1, \pi_2, \pi_3) = 0

From this, we can write our 3\pi groups in terms of the repeating variables, and one of the non-repeating variables, like so:

\pi_1 = \rho^A u^B d^C F

\pi_2 = \rho^A u^B d^C N

\pi_3 = \rho^A u^B d^C \mu

where the A’s, B’s and C’s are all different in each case.

As these groups are dimensionless, then they must have a dimension of M^0 L^0 T^0, and thus, using dimensional homogeneity, we can write the groups in terms of the powers of their dimensions:

\text{For } \pi_1 : M^0 L^0 T^0 = (ML^{-3})^A (LT^{-1})^B (L)^C MLT^{-2}

Due to the power equality that is evident on both sides, we get:

A = -1, B = -2, \text{ and } C = -2

and thus we get:

\pi_1 = \dfrac{F}{\rho u^2 d^2}

And for completeness, here are the results of the other \pi groups:

\pi_2 = \dfrac{Nd}{u}

\pi_3 = \dfrac{\mu}{\rho ud}

Example Sheet Questions

Below you will find my scanned answers to question 1 of the Dimensional Analysis Example Sheet.

Dimensional Analysis Example Sheet - Part 1

Dimensional Analysis Example Sheet – Part 1

Dimensional Analysis Example Sheet - Part 2

Dimensional Analysis Example Sheet – Part 2

Reflection

I feel that I understand the initial concepts of dimensional analysis, particularly dimensional homogeneity, that are raised here, mostly because I covered this last year briefly as part of the Vector Calculus module.

Although I must say that I have struggled with the concepts of Buckingham’s \pi groups, due to not thoroughly covering the subject material over the course of the consolidation week. This was mostly due to imminent deadlines that I decided to focus on, rather than the self-taught material. However, this did cause problems during the lectures (particularly the quizzes) when I really had no idea what was going on.

Upon going back over this material though, I feel a lot more confident that I understand the material, and I will continue to work on this topic during the revision that lies ahead. I’ve only shown the first question of the Dimensional Analysis here, but it is evidence that I have shown progress in my understanding of the topic. I hope I will be able to further this understanding when I cover the ideas discussed in the lecture notes on similarity during revision.

I’m fairly pleased overall with this blog post, but with the deadline looming, I’m going to carry on working on drastically simplifying future blog posts, as well as improving my older ones marginally.

Thanks once again everybody! And before I forget, have a fantastic Christmas break! 🙂

References and Sources:

1) The vast majority of the concepts talked about in this blog post are derived from notes taken from Chapter 1 of the Inviscid Flow PowerPoints on dimensional analysis, the supplementary material, and my own notes derived from the lectures.

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Fluid Dynamics – Week 5: The Stream Function and the Principle of Superposition

Hello again everybody, and welcome back for another exciting blog post on your favourite subject; fluid dynamics! 🙂 For once, I don’t really have much to say right here, so let’s just get stuck in, shall we?

Stream functions

In my last post, I introduced the topics of flow visualisation, featuring important aspects such as pathlines and streamlines. In this post, we extend this discussion, by firstly looking at the topic of stream functions.

Stream functions, as you might be able to infer from their name, are related closely to streamlines. In fact, by utilising the fact that streamlines are parallel to the velocity, we can deduce a new streamline by defining a stream function during its derivation. Subsequently, this leads to another feature of the stream function, which is that they are constant along a streamline.

Stream functions also tell us about the flow speed. The volume flow rate between two points joined in a curve can be calculated by simply working out the difference between the values of their respective stream functions, at those points.

So how do we go about defining these stream functions? For a 2-dimensional flow, that is incompressible and has a constant density, we can define the stream function simply as follows:

\dfrac{\partial\psi}{\partial x} = -v, \dfrac{\partial\psi}{\partial y} = u

Similarly, for polar coordinates:

\dfrac{\partial\psi}{\partial r} = -v_{\theta}, \dfrac{1}{r}\dfrac{\partial\psi}{\partial\theta} = v_r

where (u, v) are the components of a Cartesian velocity, and (v_r,v_{\theta}) is the same, but for a polar system. It should be noted that you should also check these derived results, to make sure that they are consistent with the definition of incompressibility.

The Principle of Superposition

The Principle of Superposition, as awesome as it sounds, is really just an expressive way of saying that we can add stream functions together. By combining two velocity fields, which each have respective stream functions, then the combined velocity field has a stream function that is defined by just adding up the respective stream functions together. In layman’s terms, we’ve produced an entirely new flow by just adding the stream functions! Fancy, ain’t it? 😉

During the lectures, this effect was explained to us through a series of diagrams that looked at two sources. I’ll re-interpret this below, as although the concept is quite logical, it can be a little tricky to follow with just words alone, so a visual aid is always beneficial. Not to mention the fact that a question based on drawing a graph for this could appear in an exam, so it’s always good for practice!

The example considered was that of two sources, with the following stream function properties:

\text{(A)} \qquad \psi_1 = m\theta_1, \quad -\pi < \theta_1 \le \pi, \quad \text{with centre at } (a,0)

\text{(B)} \qquad \psi_2 = m\theta_2, \quad -\pi < \theta_2 \le \pi, \quad \text{with centre at } (-a,0)

By using our newly acquired knowledge of superposition, we can now combine these separate stream functions to create a new one, as follows:

\psi = \psi_1 + \psi_2 = m(\theta_1 + \theta_2)

Since we know that stream functions are constant along the streamlines, then we know that:

\theta_1 + \theta_2 = \text{constant}

We can now draw a diagram that represents how the combination of the sources’ stream functions (and their streamlines) turn into a new stream function (with new streamlines). In the diagram, the source A is shown by the blue streamlines, whereas the source B is shown by black streamlines.

Streamlines of two sources based symmetrically around x-axis.

Streamlines of two sources based symmetrically around x-axis.

As you can see, the streamlines drawn are representations of increments of m\alpha, where \alpha = \dfrac{\pi}{12}. This is due to our initial definitions of the sources’ stream functions.

Since we are interested in generating a new stream function and its streamlines, we can now observe from the graph that we can join the components of the sources together. This is done by identifying points on the graph where the sum of the components of the stream functions are the same.

New streamlines have been produced due to superposition.

New streamlines have been produced due to superposition.

For example, on my graph, we can see that at the points P_1, P_2, P_3, P_4, the sum is equal to 13m\alpha, whereas the sum of the points at Q_1, Q_2, Q_3, Q_4 is equal to 11m\alpha.

By extending this idea with the other components, we can gradually draw up a picture of what the new streamlines will look like for our combined stream function.

By extension, we can begin to observe how the streamlines of the new stream function will look.

By extension, we can begin to observe how the streamlines of the new stream function will look.

Example Sheet Questions

Below you will find my scanned answers to question 1 of Example Sheet 2.

Example Sheet 2 - Part 1

Example Sheet 2 – Part 1

Example Sheet 2 - Part 2

Example Sheet 2 – Part 2

Reflection

I feel that I understood the concept of the stream functions quite well, as they seem to be fairly similar concepts to previously looked at material, such as pathlines and streamlines. The overall objective of the principle of superposition makes sense to me, although I must say that it took me awhile to get around the idea of the example that I have re-specified above. I feel that this was mostly due to the number of streamlines coming out of the defined sources, and so it made it a bit difficult to keep track of what streamlines I should have been adding up.

Although I only have the time right now to cover the first question, I feel that I understand the calculations just fine. Like I specified last week, I feel that I will make best use of the questions that I did not cover when it comes to revision for the exam.

I am still learning from the advice that David has given to the class, and I’m trying to be more succinct in my blog posts, while still covering the fundamentals of the lessons. Compared to some of my earlier attempts however, I think the result is much better. Also, the time it has taken me to get around to uploading this has been a little better than the time between my previous two blog posts. I can only hope that any posts that follow will also have more brevity and be more punctual as well!

Thank you to everybody who took the time to read this! Expect more posts soon as the first term begins to wrap up! 🙂

References and Sources:

1) The vast majority of the concepts talked about in this blog post are derived from notes taken from Chapter 1 of the Inviscid Flow PowerPoints on continuity and stream functions, and my own notes derived from the lectures.

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Fluid Dynamics – Week 4: Flow Visualisation, Streamlines and Pathlines, and the Convective Derivative

Hello again everybody, and welcome back to another addition in my blog! 😀 Before we start, I just need to clarify a few things:

1) No, I wasn’t on vacation, as delightful as that would have been for me! I’m afraid that the workload of uni caught up with me, and as such, I didn’t put as much time into my blogs as I should’ve done. Don’t worry though, you’ll hopefully be seeing a lot more of my blog posts in the next few days, so stay tuned!

2) As I mentioned in my last post, the rest of the content that is to be featured in my blog will be all about inviscid flow, as opposed to the viscous flow that we have been looking at previously.

With that out of the way, let us begin!

Flow Visualisation

It makes sense for us to be able to describe the properties of fluids as they flow, since we can see this natural phenomena around us all the time. The wind swaying the branches in the trees, and the raging torrents of a river, are all naturally occurring examples of fluid flows. The ability for us to mathematically deduce the properties of these types of flow, and how we can visualise them, can have a significant impact on our understanding of complex problems, such as modelling the weather.

Two such ways that we can represent the flows of fluids is through pathlines and streamlines (technically, there is another way through streaklines, but we won’t go into more depth with them here.)

Pathlines

We can think of pathlines as representing the trajectory of particles through the flow. For a 3-dimensional fluid flow, we can define the pathlines in the form of the ODE:

\dfrac{d\vec{x}}{dt} = \vec{u} = (u,v,w)

In this equation, the position vector \vec{x} is defined as (x(t),y(t),z(t)), while the time derivative of the position vector is simply the velocity vector \vec{u}.

Pathlines are similarly defined in polar coordinates, but with the position vector defined as (r(t),\theta(t),z(t)) and the velocity vector as (u_{r},u_{\theta},u_{z}).

Streamlines

Streamlines can be thought of as integral curves of the velocity field, which crucially means that a streamline is simply everywhere that is parallel to the velocity. From a physical perspective, a streamline is a snapshot of the flow of particles within the fluid at a fixed time, t.

From some basic manipulation of the definition of the pathlines above, we can define the streamlines as the following:

\dfrac{dx}{u} = \dfrac{dy}{v} = \dfrac{dz}{w}

which can be rewritten as

\dfrac{dy}{dx} = \dfrac{v}{u}, \dfrac{dz}{dx} = \dfrac{w}{u}, \dfrac{dz}{dy} = \dfrac{w}{v}

Like the pathlines, streamlines can also be written in the form of polar coordinates. Once again, the time aspects of the equation are eliminated, to reveal the definitions that we want, i.e. (for 2-dimensions):

\dfrac{1}{r}\dfrac{dr}{d\theta} = \dfrac{u_{r}}{u_{\theta}} \text{ or } r\dfrac{d\theta}{dr} = \dfrac{u_{\theta}}{u_{r}}

Convected Derivative

In my introduction to viscous flow, I defined what it meant for a fluid to be incompressible, which relied upon the definition of a new form of derivative; the convected derivative. Using the definition of the convected derivative, we can now consider the acceleration of the fluid flow, since we can translate it as a convected time derivative of the velocity vector, like so:

\dfrac{D\vec{u}}{Dt} = \dfrac{\partial\vec{u}}{\partial t} + \vec{u}.\nabla\vec{u}

Alternatively, the convected acceleration can also be written as follows:

\dfrac{D\vec{u}}{Dt} = \dfrac{\partial\vec{u}}{\partial t} + \dfrac{1}{2} \nabla (|\vec{u}|^2) - \vec{u} \text{ x curl(}\vec{u}\text{)}

As another extension from viscous flow, it can be quite easily observed that the left hand side of the Navier-Stokes equations is simply the density of the fluid (\rho), multiplied by our time derivative and convected acceleration. The same can also be said in polar coordinate form.

Example Sheet Questions

Below you will find my scanned answers to questions 1 and 2 of Example Sheet 0 and question 1 of Example Sheet 1.

Example Sheet 0 - Part 1

Example Sheet 0 – Part 1

Example Sheet 0 - Part 2

Example Sheet 0 – Part 2

Example Sheet 0 - Part 3

Example Sheet 0 – Part 3

Example Sheet 1 - Part 1

Example Sheet 1 – Part 1

Example Sheet 1 - Part 2

Example Sheet 1 – Part 2

Reflection

The general principles highlighted during this particular week’s worth of lectures were fairly intuitive to understand, although I do need to make sure that I understand the physical interpretations of the flow visualisations, such as the pathlines and streamlines. I feel that the equations for these visualisations are straight-forward, so it should be a case of memorisation when it comes to revision.

In terms of the exercises, all of the processes were fairly repetitive, and as such, there weren’t many hidden surprises. In fact, the only aspect of the questions that I struggled with, was the interpretation of the first part of question 1, of Example Sheet 0. I was not confident with the graph, and I only convinced myself to put down what I did for it after experimentation with the equation in Maple.

Ideally,  I would have liked to have covered more of the questions to further my understanding of the topic. However, with the current levels of workload, I will instead use these additional questions as part of my revision for the exam at the end of the year. I am glad though that I was able to significantly shorten my blog post (particularly after my last post) without leaving out the crucial details, which will be extremely useful when it comes to revision.

As I also mentioned at the start of this post, I do apologise sincerely for not uploading this sooner, but the workload has been getting to me recently. If I have to take away anything from this term so far, it is that I have to make significant improvements to my overall time management, and not get too focused on a few activities! Oh, and that I really need to start on work as soon as I receive it, if it can be helped! 😛

Thank you to anybody who took the time to read this post! I hope you’ll be seeing more of this sooner rather than later! 🙂

References and Sources:

1) The vast majority of the concepts talked about in this blog post are derived from notes taken from Chapter 1 of the Inviscid Flow PowerPoints on flow visualisation, continuity and stream functions, and my own notes derived from the lectures.

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Fluid Dynamics – Week 3: Stress, Pressure and the Navier-Stokes equations

Hello everybody, and welcome back for the fourth instalment on my blog about fluid dynamics! 🙂 Following on from my previous blog post introducing viscous flow, we will continue our look at the properties of viscous fluids (yes, including that momentum conservation equation!), and we will finally derive the famous Navier-Stokes equations, which we will use extensively in the second term of the module for solving a variety of viscous flow systems.

Oh, and please don’t be misled by the title of this particular blog post. It most certainly isn’t going to be a gushing recount of my hardships at university, since I try to avoid drama whenever possible! 😉 With that out of the way, let the “fun” begin!

Definition of the Momentum Conservation Equation

As I mentioned previously, the deduced momentum conservation equation is vital in the derivation of the Navier-Stokes equations. As such, I feel that it is important that I introduce it now, especially since that I did not produce the result last time. Without further ado, here it is:

\dfrac{d}{dt}\int_V \rho\vec{u} dV = -\int_S \rho\vec{u}(\vec{u}.\vec{n}) dS +\int_V \rho\vec{F}_B dV +\int_S \vec{\sigma} dS

I did warn you last time that it wasn’t pretty, so I apologise now if you have nightmarish visions of this engraved in your conscience. To make it somewhat less painful to read, let me try and explain what some of the individual terms mean, especially in regards to our expression of the principle in layman’s terms.

The first term (the component of the equation before the equals sign) relates to ‘the rate of change of momentum inside the volume’. Since momentum is defined as being the mass times the velocity, then thus the rate of change of momentum will have units matching those of mass times by acceleration (the same units as a force). From checking this first term, we can verify the units are correct.

As for the second term, the integrand represents ‘the net rate of inflow of momentum into the volume’. The term is also negative because we have to consider that the normal is acting outwards of the surface of the volume, when we are considering what is occurring inside the volume.

The final two components of the equation represent ‘the total force acting on the fluid inside the volume’, the first of which refers to \vec{F}_B, which is the body force (per unit mass) acting on the fluid. For most applications we will look at, the body force is normally gravity. The latter term of the pair refers to stress (the force per unit area) acting upon the surface of the volume (represented by \vec{\sigma}), and it arises due to the motion of the surrounding fluid.

Thus that summarises the equation for momentum conservation. See, perhaps it’s not so bad after all? Well OK, maybe it is a bit. Regardless, the final term in this equation neatly leads on to the next aspect of the derivation of the Navier-Stokes equations, and that is by taking a closer look at what we mean by the stress.

Specifying stress and the stress tensor

As specified above, the stress given off by the fluid is the force that the fluid is exerting upon a particular surface. We can define the stress as a continous function with components of x, the position in space, t, the time, and n, which is the normal vector to the surface element in question, i.e.

\vec{\sigma} = \vec{\sigma}(\vec{n},\vec{x},t)

The stress is similarly defined for an inviscid fluid, where \vec{\sigma} has a non-zero component of n only (in other words, \vec{\sigma} and n are parallel), and therefore

\vec{\sigma}(\vec{n},\vec{x},t) = -p(\vec{x},t)\vec{n}

where the scalar quantity p(\vec{x},t) is the mechanical pressure of the fluid. This pressure is considered to be a positive quantity when acting inwards on the surface of the fluid, hence why there is a minus sign present in the equation.

An important observation to take away from the definition of the stress of an inviscid fluid above is that, for a viscous fluid, \vec{\sigma} will not in general, be defined in the same direction as the normal vector to the surface, n. This is simply because the force acting upon on the surface is unlikely to be perpendicular to that particular element.

Due to this observation that \vec{\sigma} and n will generally be in different directions, we have to consider a concept known as the stress tensor to further clarify this point.

The stress tensor is initially defined as

\sigma_{ij}(\vec{x},t) = \sigma_i(\vec{N}_j,\vec{x},t)

where \vec{N}_j is defined as being the unit vectors in the directions of the coordinate axes. An explanation of the above is that \sigma_{ij} is the i-th component of the stress vector \vec{\sigma} at x and t, on an element whose normal is in the j direction. As there are three axes, both i and j are summed from one to three respectively.

This initial derivation is found by considering properties of a tetrahedron (a triangular-based pyramid, if you will), but I will not cover it here. Instead, I will consider the main points of this derivation only, and how important factors that are deduced from it, will further our understanding of what we mean by stress.

From defining the stress tensor as we have, we now need to consider two directions to help us define the overall stress imposed by the fluid:

1) The direction in which the stress acts, and

2) The direction of the normal of the surface element on which the stress is acting.

From here, when taking geometrical considerations into account, we were able to determine that having knowledge of the stress tensor was sufficient enough to write down the stress vector on a surface of arbitrary orientation (the full derivation is in the notes for those interested.)

Also of note is that the stress tensor is often displayed as

\sigma_{ij} = \begin{pmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33}\end{pmatrix}

although it should be made clear that the stress tensor is not a matrix!

Using the same derivation, we can equally determine the stress tensor for an inviscid fluid to be

\sigma_{ij} = -p\delta_{ij}

where \delta_{ij}, as you all know by now, is the Kronecker delta, where

\delta_{ij} = \begin{cases} 0 & \text{if i} \neq \text{j} \\ 1 & \text{if i = j}\end{cases}

For the sake of completeness, we can then express the stress tensor of the inviscid fluid like so

\sigma_{ij} = \begin{pmatrix} -p & 0 & 0 \\ 0 & -p & 0 \\ 0 & 0 & -p\end{pmatrix}

Just quickly, it should also be noted that the conservation of angular momentum does come into play, as it shows that the stress tensor is symmetric, i.e.

\sigma_{ij} = \sigma_{ji}

however, we are mostly just interested in this as a result. As such, no rigorous derivation was given for this result.

The Constitutive Equation for an Incompressible Newtonian Fluid

The individual components of the stress tensor depend on what is known as the rate of deformation of the fluid. In fact, what distinguishes one fluid from another is the precise form of said dependence. As this particular subtitle states, we are looking at Newtonian fluids, which satisfy what is known as Newton’s Law of Viscosity. For an incompressible fluid, this law is represented mathematically as

\sigma_{ij} = -p\delta_{ij} + \mu\left(\dfrac{\partial u_i}{\partial x_j} + \dfrac{\partial u_j}{\partial x_i}\right)

where \mu is known as the dynamic viscosity and is constant, provided that the temperature and pressure are fixed, for any given Newtonian fluid. You should also clearly see that the inviscid form for the stress tensor is found if you allow for \mu = 0.

No attempt was given within the notes to justify the above form for the stress tensor, but a simple analysis was used to hint towards the argument from which the above equation is derived.

To round off this particular subsection of the blog, I will include details of two important properties of any fluid, that is the aforementioned dynamic viscosity, and the kinematic viscosity.

A change in the pressure will do little to vary dynamic viscosity, but changing the temperature can have quite a dramatic effect on its value. As I mentioned way back in my first post, viscosity (I should have specified dynamic viscosity at the time) has S.I. units of kilogram per metre-seconds (\text{kg m}^{-1}\text{s}^{-1}), which is equivalent to Pascal-seconds (Pa.s). This translates, as you can see, to dimensions of \dfrac{\text{mass}}{\text{length} \times \text{time}}.

The kinematic viscosity is related to the dynamic viscosity by this equation

\nu = \dfrac{\mu}{\rho}

Since the density, \rho, is defined as \dfrac{\text{mass}}{\text{volume}}, then it follows simply that it must have dimensions of \dfrac{\text{mass}}{\text{length}^{3}}. From this, it is pretty easy to deduce that the kinematic viscosity, \nu, has dimensions of \dfrac{\text{length}^{2}}{\text{time}}. In S.I. units, this is equivalent to metres-squared per second (\text{m}^{2}s^{-1}).

The Navier-Stokes Equations

At last, we’ve finally got here! It took us long enough, didn’t it? We’ve now defined just about everything we need in order to deduce the equations we need to finally solve those pesky viscous flow systems! 😀

Annoyingly though, the derivation is still quite long, so I’ll try and summarise the main points below:

1) Do you remember that momentum conservation equation? Yep, the one at the start of this blog. Sorry to bring that up again, but we need it! Thankfully we’ve already defined that, so I won’t post it again here, but just keep it in mind.

2) Since the volume of our fluid is fixed (i.e. it does not vary with time), and only the density and velocity vector do change with respect to time, then we can switch the order of the first term of the momentum conservation equation, so that the derivative with respect to time is brought inside the integrand. We also know that the stress vector can be represented as a product of the stress tensor and the normal vector to the surface element, so the final term is replaced with this product.

3) Now using Gauss’ Divergence Theorem, and some clever manipulation, we can change all of our surface integrals into integrals over the volume. Now every single integral is being integrated over the volume, we can now combine all these integrands into one large integral, and move them all onto the left hand side of the equation. Thus the right hand side of the equation now equals zero. Since we are considering our volume to be arbitrary, then it must mean that our integrand must be identically equal to zero.

4) Now by re-arranging our equation so that the stress tensor and the body force terms are on the right hand side, we have now successfully expressed the momentum conservation principle for a general fluid! Congratulations! Well OK, we aren’t quite done yet.

5) By inserting the definition of the stress tensor in terms of Newton’s Law of Viscosity, we will eventually arrive at the Navier-Stokes equations for an incompressible Newtonian fluid. However, there is some algebraic manipulation involved for this. Just for convenience though, I’ll state the results below.

Here are the Navier-Stokes equations for an incompressible, constant viscosity fluid using the summation notation (this form of the equation is particularly useful for re-defining it in terms of Cartesian co-ordinates.)

\rho\left(\dfrac{\partial u_i}{\partial t} + u_j\dfrac{\partial u_i}{\partial x_j}\right) = -\dfrac{\partial\rho}{\partial x_i} + \mu\dfrac{\partial^2 u_i}{\partial (x_j)^2} + \rho(F_B)_i

The above can be expressed in a more condensed form in terms of vectors

\rho\dfrac{D\vec{u}}{Dt} = -\nabla p + \mu\nabla^2\vec{u} + \rho\vec{F}_B

That’s it! Hallelujah! 🙂 With the above defined equations, and the continuity equation that I defined in my last post, we can now solve systems of viscous fluids (subject to appropriate boundary and initial conditions of course).

Just a final note though on boundary and initial conditions. It is strongly advised to consider applying the no-slip boundary condition, which requires that at a solid surface, the fluid velocity should be equal to the surface velocity. If the flow is unsteady (i.e. the flow is time-dependent), then initial conditions are also required.

Reflection

Although it took a long time to get there, it was fairly clear why we were doing what we were doing; to get to the Navier-Stokes equations, which we will use extensively throughout the second term of the module. However, there are a lot of components to consider for its derivation, with lots of subtleties, especially in relation to the stress vector and tensor, and the momentum conservation equation. I feel more comfortable with the aspects of the derivation that were covered in last week’s lectures.

Everyone knows that the best way to practice mathematics is by doing it, so I’ll have to wait until next term when we cover this material again, before I can say with any degree of certainty as to whether I understand it or not.  Although it should be stated that, while we need to have a grasp of the derivation, the vast majority of the second term will be to do with applying the appropriate Navier-Stokes equations and continuity equation to multiple scenarios.

While from a revision point of view, I’m glad that I’ve covered a significant amount of material in this post, I feel that I’ve ultimately sacrificed brevity for additional detail and clarification. Just for future reference to any readers, I won’t be writing to this level of depth in my future blog posts. The time investment was far more significant that was expected of me, and simply put, a blog post of such length can get a little tiring for most people. At least I’m uploading on a more regular schedule now, so progress has been made there! 🙂

Now that we are moving back onto the inviscid flow part of the module, maybe, just maybe, my blog posts might be slightly shorter? Well, we can always hope! 😉 Thanks for reading all of this! If you didn’t…well…I hope it was good enough to send you into a peaceful night’s sleep. See you next time!

References and Sources:

1) The vast majority of the concepts talked about in this blog post are derived from notes taken from Chapter 1 and 2 of the MATH3402 Fluid Dynamics – Viscous Flow lecture notes.

2) I have found and used this particularly useful book on \LaTeX, in order to help me write appropriate matrices and piecewise functions as part of this blog post. Perhaps you’ll find something useful too? Here’s the link: http://en.wikibooks.org/wiki/LaTeX/Mathematics

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Fluid Dynamics – Week 2: Introduction to Viscous Flow

Hello everybody, and welcome back to another instalment on my blog about fluid dynamics! 🙂

Clarifications (about change to planned upload, and differences in flow)

To those of you who are familiar with the plan of action for this module, you may note that there is an inconsistency with the subject of the title. Following on from my previous blog post, I should now be talking to you about the foundations of inviscid flow, but now I am writing to you about viscous flow.

The reason for this change is that, unfortunately, our first term lecturer David (who I also mentioned in my last blog post) has fallen ill, so our second term lecturer Jason has decided to take over for the time being. This means we will be looking at the foundations for the properties of viscous flow and why this is different to inviscid flow.

In fact, before I continue, I feel I should clarify perhaps the most obvious difference between a viscous and an inviscid flow. Also, in my first blog post, I mentioned viscosity, but I did not really give a definition of it. Thankfully, defining it now makes sense when comparing the two main types of flow.

The viscosity of a fluid is simply a means of measuring a fluid’s resistance to gradual deformation, which in the lecture notes given, we use the stress caused by shearing as a primary example (colloquially and in everyday life, we often refer to the viscosity of a fluid as its ‘thickness’).

Therefore, a viscous fluid is a fluid where viscosity has to be taken into consideration in order to understand it. An inviscid fluid, on the other hand, is often referred to as an ‘idealised fluid’, simply because we are making the assumption that the fluid has no viscosity. In reality however, there are no truly inviscid fluids since every fluid has some sort of viscosity, but there are still applications for when this assumption is useful.

Now that we are clear on our definitions for both viscous and inviscid flows, let us move onto some of the founding principles that make up our theory on viscous flow.

The Continuum Hypothesis

The Continuum Hypothesis is an assumption that allows us to regard the fluid as if it were continuous, no matter how small a portion of the fluid we are looking at. In layman’s terms, this means we can completely ignore the molecular, atomic (and sub-atomic) structure of the fluid in question.

In reality, we know that there are technically spaces (such as the spaces between individual fluid molecules, or on a more extreme level, the spaces between sub-atomic particles, such as electrons), and there could well be nothing there at any given point we are defining. This is problematic if we want to define physical properties, such as density, pressure, velocity and temperature. Therefore, by taking this assumption into account, we can now define the physical properties in terms of space, x, and time, t, since we are making sure that the fluid is continuously defined, regardless of where we are in the fluid.

Of course, by taking this assumption (and the others that follow) into question, we are only able to define an approximation, as opposed to an exact result, of a viscous flow system. This is a repeating occurrence within modern mathematics, and particularly in areas where we have to solve a system of equations.

This is because very few real world applications can be solved in a ‘simple’ manner (i.e. analytically) and therefore we need to use a combination of numerical methods and assumptions in order to produce reasonable solutions. However, these calculations should be fairly accurate to the exact solution, or what is observed naturally, since the assumptions and numerical methods are defined rigorously from appropriate theory.

Incompressibility

By assuming that our fluids are incompressible, we can ignore any and all velocity induced changes in the density, which we will write as \rho(\vec{x},t). In other words, the density of the fluid remains unchanged when any given volume in the fluid moves under any velocity.

In a similar manner to how, for the Continuum Hypothesis, our assumption does not obviously match up with that is observed in reality, the same occurs for incompressibility. This is because all fluids are somewhat compressible, however without our assumption, it would just make our solution unneccessarily more complicated without providing a significant benefit to our solution.

The incompressibility assumption is written in this form mathematically

\dfrac{D\rho}{Dt} = 0

where \frac{D}{Dt} denotes the convected derivative, which represents the rate of change following the motion of a fluid particle.

This can be defined as

\dfrac{D\rho}{Dt}\equiv\dfrac{\partial\rho}{\partial t} + \vec{u}.\nabla\rho = 0

which can then be expanded in Cartesian coordinates as follows

\dfrac{D\rho}{Dt}\equiv\dfrac{\partial\rho}{\partial t} + u\dfrac{\partial\rho}{\partial x} + v\dfrac{\partial\rho}{\partial y} + w\dfrac{\partial\rho}{\partial z} = 0

where \vec{x} = (x,y,z) is the position vector in space, \vec{u} = (u,v,w) is the velocity vector and \nabla or ‘nabla’ is the gradient operator derived from vector calculus.

Mass Conservation and the Continuity Equation

Mass conservation is a simple, and yet powerful, assumption: mass can neither be destroyed nor created. In the context of fluid dynamics, it means that our fluid can not appear out of nowhere, and it can not simply vanish.

In order to obtain what is known as the continuity equation for an incompressible fluid, we must first state that ‘the rate of change of mass inside an arbitrary, but fixed in space, volume = the net rate of inflow of mass into the arbitrary volume’.

If we were to state the above mathematically, and then apply Gauss’ Divergence Theorem to it (which is also derived from vector calculus), we can deduce the following continuity equation for a compressible fluid

\dfrac{\partial\rho}{\partial t} + \text{div}(\rho\vec{u}) = 0

where the divergence operator \text{div} upon \rho\vec{u} is defined as

\text{div}(\rho\vec{u}) = \nabla . \rho\vec{u}

By then applying the product rule for the divergence operator, this continuity equation can then be written in terms of the convected derivative

\dfrac{D\rho}{Dt} + \rho\text{div}(\vec{u}) = 0

Since we know for an incompressible fluid that the convected derivative must equal zero, and that the density of the fluid can never be equal to zero (since if a fluid had no density, then it would have no mass or volume), then we can further simplify the above to deduce the continuity equation for an incompressible fluid

\text{div}(\vec{u}) = 0

Just for completeness, the continuity equation can then be simply written in Cartesian coordinates as

\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y} + \dfrac{\partial w}{\partial z} = 0

or in suffix notation, where the double suffix notation convention applies, as

\dfrac{\partial u_i}{\partial x_i} = 0

Momentum conservation

The concept of momentum conservation is fundamentally the same as mass conservation, but with momentum being the deciding factor to consider. In fact, the equation for momentum conservation is just an expression of Newton’s Second Law defined for a continuous medium (which is what we have assumed our fluids to be, due to the Continuum Hypothesis).

Therefore, we know that momentum, like mass, can not be created nor destroyed. In a more mathematical sense, we can state that ‘the rate of change of momentum inside an arbitrary volume = the net rate of inflow of momentum into the volume + the total force on the fluid inside the volume’. As before, an equation can be deduced from this statement, but I do warn you though, the equation is most certainly not a pretty one.

In order for us to make full use of the momentum conservation equation, we will first need to do some more physical reasoning in regards to the specification of stress. Secondly, we will then need to manipulate said equation, in order to get our ultimate goal of being able to derive what are called the Navier-Stokes equations, which fundamentally enables us to solve systems where a viscous flow is being considered.

Since this particular blog post is already quite long, I will deduce these elements when I start work on next week’s material, from the second chapter of the lecture notes.

Reflection

I feel that although I’ve grasped these founding principles well, and that they are intuitive in nature, I still feel that I struggle with the initialisation of the conservation assumptions in a mathematical sense, as outlined in the lecture notes. One assumption’s derivation that is confusing at first glance, is the derivation of the equation for momentum conservation, simply due to the number of terms and what it all means for the system as a whole. I will go over these notes again at a later point, to make sure that I fully understand the arguments being put forward.

Annoyingly, I can’t really comment on whether or not I understand the inviscid and viscous flows right now. I feel that this is a question that I may only be able to answer once I have gained enough experience in both particular fields, and that may not be until I am sufficiently into my second term of this module, or perhaps until near its completion.

In terms of my upload schedule, I’ve made significant progress in being able to upload this just the day after my previous post! 🙂 However, I’ve also taken a step backwards in that this blog post is significantly longer than my previous posts. As such, I am aware that I need to be more careful about the depth that I go into, for future reference. Maybe one day I’ll get the combination of brevity and punctuality right!

Hopefully this *not quite so* brief introduction to the concepts surrounding viscous flow was both useful and informative for you. Don’t worry, we’ve barely reached the tip of the iceberg here when it comes to the mathematics front! Just wait until you see that momentum conservation equation!

If you got this far, well, you’re a trooper. Thanks for reading, and I’ll see you in my next blog post! 😀

References and Sources:

1) Most of the concepts talked about in this blog post are derived from notes taken from Chapter 1 of the MATH3402 Fluid Dynamics – Viscous Flow lecture notes.

2) Another thank you goes to Darren Mooney again, this time for his helpful guide on writing \LaTeX into WordPress, and for providing me with the necessary inspiration needed to write blog posts! You might find \LaTeX useful too for your mathematical needs, so why not give it a try? Here’s the link: http://math3402darrenmooney.wordpress.com/2013/10/25/latex-in-wordpress-a-simple-guide/

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Introduction to Fluid Dynamics: Bernoulli’s Principle

Hello everybody, and welcome back to my blog on fluid dynamics! 🙂 In this particular blog post, I will be going over the content covered in our first lecture on the subject, which took place on the 23rd of September.

As this was our first lecture on the subject, David was kind enough to not overload us with anything too technical, so please don’t expect anything substantial in the way of equations, derivations and definitions (in this blog post as least! ;)). With that out of the way, let the fun begin!

Experiments and their results

In this particular lecture, David showed us a variety of experiments which demonstrated an important aspect of fluid dynamics, although for our amusement, this was initially veiled through the use of straws, old juice bottles and ping pong balls.

The ball and the straw observation

The experiments were inherently simple in nature, and yet when conducted, seemed to produce results that were naturally counter-intuitive to what we would expect. The first example of this was David blowing through an angled straw, with some force I might add, on to a ping pong ball, that had a green line painted across the centre.

What we would expect, naturally, is for the ball to go flying away if it comes into contact with the stream of air coming out of the straw. The ball would just fall to the ground under gravity otherwise. However, we observed something totally different! Given that if David could provide enough puff, the ball actually stayed suspended in the air while rotating in the direction of the air flow. The actual rotation itself would have been hard to spot, if it were not for the green line. This then lead on to the next ‘experiment’…

The ping pong ball is suspended in the air due to Bernoulli's Principle.

The ping pong ball is suspended in the air due to Bernoulli’s Principle. [1]

The bottle observation

For our next observation, we were asked in what direction the bottles would move if David were to blow through the middle of them. Naturally, most people assumed that the bottles would move away from each other. However, once again, this proved not to be the case! As you can see from the diagram below, the bottles actually ended up being moved towards each other!

By this point, most of us were mildly amused by David running out of breath from attempting to blow on various objects, but a lot of us were also thinking “Well, what’s going on then? Why are these things not acting in the way we expected them to?”. As this was near the end of the lecture, David then mentioned in a few words that this effect was known as Bernoulli’s Principle, and what it actually meant in layman’s terms.

Counter-intuitively, the bottles move towards each other, rather than away.

Counter-intuitively, the bottles move towards each other, rather than away. [2]

The Bernoulli Principle

The Bernoulli Principle, as put forward by Daniel Bernoulli in his book Hydrodynamica in 1738 [3], states that if the pressure of a fluid decreases, then inversely, the velocity must increase. Of course, by extension and observation, the opposite is also true. Indeed, it is this simple statement that defines the seemingly counter-intuitive behaviour that we observed.

By looking back at the experiments, the reasons are clear. In our first experiment, we can now deduce that since the air flow is moving with greater velocity, then we must have an area of low pressure to compensate. In line with this, we must know that at the base of the ball, there must be an area of higher pressure, since the air flow is not as strong here. It is the difference in the pressure on the ping pong ball that causes the phenomenon we observe: the ball is forced to move from the area of high pressure to the area of low pressure, and thus with the constant stream of air, the ball is suspended in the air while spinning rapidly.

In the second experiment, a similar observation is now made. The stream of air flows quickly between the two bottles, and thus causes low pressure. On the outer sides of the bottles, there is little air flow and thus there is an increase in the pressure. Therefore, the areas of high pressure on the outsides of the bottles push the bottles inward, as we know from our observation.

Reflection

The concepts detailed here, although now seemingly quite simple in nature, were initially a bit confusing to observe. However, David’s explanation of Bernoulli’s Principle makes sense upon reflection, and his rather amusing displays will certainly stick in my mind for awhile. Hopefully I’ll be able to understand the more complicated topics, that will be covered, in a similar manner.

I took my own advice from last time, and I’m glad that I managed to get this uploaded faster than I had done my first post. However, I’m still not uploading new blog posts as fast as I should be, so this will have to improve over the coming weeks.

Hopefully this [somewhat] brief but informative post about Bernoulli’s Principle was useful to you. Many more interesting ideas and observations will be discussed in later blog posts, as we delve further into the mathematics of fluid dynamics. Who knows, maybe we will encounter a more strict definition of this observation at a later point in the module? Many thanks to those of you who got this far!

References and Sources:

[1] [2] Many thanks to my friend Darren Mooney who gave me permission to re-use his images. He’s also got a blog on fluid dynamics! Why don’t you check him out? Here’s the address: http://math3402darrenmooney.wordpress.com/

[3] http://www.britannica.com/EBchecked/topic/62591/Daniel-Bernoulli#ref200813