Tag Archives: pressure

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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 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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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