Tag Archives: Example Sheet

by

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.

by

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.

by

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.

by

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.

by

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.