Flow(#2) – Plateau- Rayleigh Instability

August 9, 2019December 2, 2020 / 153armstrong / Leave a comment

When you wake up in the morning and open/close your faucet when you brush your teeth, you might have noticed that it undergoes a transition between a smooth jet to a dripping flow like so :

When the velocity of the fluid exiting the faucet is high, it appears smooth for a longer time before it breaks into droplets:

But when you make the velocity of the fluid exiting the faucet low, it seems to form droplets much earlier than before.

Here’s the water breaking into smaller droplets shot in slow motion:

Notice that just by changing the exit velocity of the water you can control when the droplets form.

You can also control the nature of the droplets that form by changing fluids. Here’s how it looks like if you use water as the fluid (left), pure glycerol(center) and a polymeric fluid(right).

What is causing a jet of fluid to form droplets?

A simple answer to this is perturbations on the surface of the fluid. What does that mean?

Initially the fluid is just falling under the influence of gravity. And velocity of any freely falling object increases as it falls:

But the surface tension of the fluid holds the molecules of the fluid together as they fall down.

Therefore depending on the initial velocity of fluid, the surface tension of the fluid and the acceleration you get a characteristic shape of the jet as it falls down:

This is what you observe as the fluid exits the faucet.

Perturbations

Just after exiting the faucet, there are tiny perturbations on the surface of this fluid as it falls down. This is apparent when you record the flow at 3000fps:

Source: engineerguy

Those tiny perturbations on the surface of the fluid grow as the fluid falls down i.e the jet becomes unstable.

And as a result the fluid jet breaks down to smaller droplets to reach a more thermodynamically favorable state. This is known as the Plateau-Rayleigh instability.

Source

It takes different fluids different time scales to reach this instability. This depends on the velocity of the fluid, the surface tension and the acceleration it experiences.

And some viscous fluids like honey are also able to dampen out these perturbations that occur on their surface enabling them to remain as fluid thread for an extended time.

A note on inkjet printers

By externally perturbing the fluid instead of making the fluid do its own thing, you can make droplets of specific sizes and shapes.

This engineerguy video explains how this is used in inkjet printers in grand detail. Do check it out.

We started today by trying to understand why water exiting a faucet behaves the way it does. Hopefully this blog post has gotten you a step closer to realizing that. Have a great day!

Sources and more:

* This is a topic that is home to a lot of research work and interesting fluid dynamics. If you like to explore more take a look into the mathematical treatment of this instability here.

Note on average density and why ships do not sink

July 28, 2019December 2, 2020 / 153armstrong / 1 Comment

Let’s ask a very generic question: I hand you an object and ask you to predict whether the object would float or sink. How would you go about doing that ? Well, you can measure the mass of the object and the volume of the object and can derive this quantity called Average Density ( $\rho_{avg}$ )

$\rho_{avg} = m_{object}/V_{object}$

It is the average density of the entire object as a whole. If this object is submerged in a fluid of density $\rho_f$ , then we can draw the following force diagram:

If $\rho_{avg} > \rho_{f}$ , we note that this generic object would sink and if $\rho_{avg} < \rho_{f}$ it would float!. Therefore in order to make any object float in water, you need to ensure its average density is less than the density of the fluid its submerged in!

Why does a ship stay afloat in sea?

A ship is full of air! Although it is made from iron which sinks in water but with all the air that it is full of, it’s average density ( $m_{ship}/V_{ship}$ ) drops down such that $\rho_{avg-ship} < \rho_{sea-water}$ .

Fun Experiment:

If you drop some raisins in soda, you will notice that they raise up and fall down like so (Try it out!):

This is because air bubbles that form on the top of the raisin decrease its average density to the point that its able to make the raisin raise all the way from the bottom to the top. BUT once it reaches the top all the air bubbles escape into the atmosphere (its average density increases) and the raisin now falls down.

Questions to ponder:

Why do people not sink in the dead sea ?
How are submarines/divers able to move up and down the ocean ? How would you extend the average density argument in this case.
Why do air bubbles in soda always want to raise up ?
If the total load that needs to on a ship is 25 tons. What should be the total volume of the ship in order to remain afloat if the density of sea water is 1029 kg/m3,

How do hot air balloons work?

June 28, 2019December 2, 2020 / 153armstrong / 1 Comment

This post follows the concepts from the previous post on Buoyancy.

One of the most surprising things about air that may not be intuitive is that it is a fluid and like any other fluids exerts a pressure on objects.

Standing on earth with layers of air above us, we are constantly being ‘weighed down’ by a pressure of ~1atm at all times.

All objects in air are also assisted by a buoyant force that is caused from the pressure difference between the top and bottom surfaces.

Let’s now consider the hot-air balloon in particular. A force diagram is probably the best way to start:

*Envelope is the the actual fabric which holds the air inside the balloon*

Therefore we see that in order for the hot air balloon to float, we need to have the buoyant force compensate for the net weight from the balloon, the load on it and the air inside it:

$F_B > w_{envelope} + w_{load} + w_{hot-air}$

Say you built a modest hot-air balloon that just barely managed to get off the ground. How would you make it go higher ?

You don’t want to play around with $w_{load}$ because you don’t want to throw out any of your passengers or your supplies. You can’t really play around with $w_{envelope}$ without re-making the balloon again.

BUT you can reduce the weight of the hot air inside the balloon ( $w_{hot-air} = m_{hot-air} g = (\rho_{hot-air} V_{hot-air}) g$ ) ! How ?

Recall the ideal gas law $P = \rho R T$ . Assuming that the pressure and the volume of the hot-air balloon does not change we note that:

$\rho_{hot-air} \propto 1/T_{hot-air}$

meaning that if we want to reduce the density of air, we just need to crank up the temperature.

And with a spectacular burner onboard our hot air balloon, we can easily increase the temperature of the air inside the balloon!

That’s pretty much how hot-air balloons work! You increase the temperature of the air inside the balloon to go up, decrease the temperature to go down and skillfully adjust the temperature to hover.

Questions to ponder:

When you increase the temperature of air, its density decreases. What do you think happens to all the molecules that were previously inside ? Do they exit the balloon ?
Different shaped hot-air balloons is a common sight. Do you think that the Buoyant force changes for each shape? (Review the formula for Buoyant force discussed in the previous post)
How do you think Helium balloons work ?

Origins of the Buoyant force

June 27, 2019December 2, 2020 / 153armstrong / 2 Comments

Check out the video description on YouTube on the details of how buoyancy is related to each clip in the video

Useful Preliminaries :

$\rho = \frac{m}{V}$

$w = mg = \rho Vg$

$p = \frac{F}{A}$

$p = p_0 + \rho g h$

Pressure depends on depth:

In a tank filled with a fluid, the bottom most part of the tank experiences a higher pressure than the top most part of the tank.

This is simply because if you are at the bottom, there are more layers of the fluid above your head “weighing you down” than compared to the top where you have fewer layers.

Fact: Marina Trench is 1000 times the pressure at the sea level.

As a result of the pressure depending on depth, any object placed in a fluid experiences a pressure difference between its top and bottom surfaces.

The top is at a lower pressure and the bottom at a higher pressure. Therefore, there is an upward force called the ‘Buoyant Force’ that acts on the object when you submerge it in a fluid.

Anything that you submerge underwater will feel lighter than it actually is because the Buoyant force acts upward on the objects an help you counteract the effect of gravity.

$F_{net} = mg - F_B$

(i) If $mg>F_B$ , object sinks

(ii) If $mg<F_B$ , object floats

(i) If $mg=F_B$ , object remains stationary. (It is neutrally buoyant)

Note on the Buoyant force:

Air is also a fluid and also offers a Buoyant force on any object.
The Buoyant force is a property of the fluid and has nothing to do with the nature of the object that you submerge. A 1 $m^3$ of Iron, Styrofoam, Lead, etc all would feel the same Buoyant force.

An equation to represent Buoyant force:

So far, all of this has been qualitative. Let’s try to obtain an expression for the Buoyant force that we can work with.

Consider an object submerged in a fluid with density $\rho_f$ as follows:

$F_B = (P_{high} - P_{low}) A$

$F_B = \left[ p_0 + \rho_f g (h+d) - (p_0 + \rho_f g h ) \right] A$

where $\rho_f$ -> Density of the fluid.

$F_B = \rho_f g (d A )$

$F_B = \rho_f g V_{object}$

Notice that the Buoyant force only depends on the volume occupied by the object and not it’s density and as a result all objects with the same volume irrespective of its density experience the same Buoyant force!

We can make this even better by realizing that $V_{object} = V_{fluid- displaced}$ because when an object submerges in water it pushes away all the fluid that was already there to occupy it for itself.

How much fluid does it need to displace ? For submerged objects it’s exactly how much volume it needs to accommodate the object in the fluid. Therefore we can rewrite the above formula like so:

$F_B = (\rho_f V_{fluid-displaced}) g$

$F_B = m_f g$ (since $\rho = m/V )$

This is the Archimedes principle. It reads that the Buoyant force experienced by any object is equal to the weight of the fluid displaced.

Eureka!

Things to explore:

Powering a 13W CFL light bulb using a 3V battery

April 14, 2019December 20, 2020 / 153armstrong / Leave a comment

Recently I stumbled upon this cheap high voltage converter on Amazon which claims a boost from 3-6V to 400kV. Although really skeptical about the 400kV claim, a lot of comments indicated that it did boost atleast to 10kV so I got one of these to test it out.

Schematic diagram for lighting up a CFL using the high voltage converter

Using a 1.5V battery to power the circuit

And boom! There we go, that’s how you light up a CFL light bulb using a 3V battery!

If you do have access to a plasma globe or a tesla coil, things become a little bit more simpler:

The way CFL light bulbs works is by exciting the electrons in the lamp and when they return to the ground state they radiate ultraviolet light. This emitted light is converted to visible light when it strikes the fluorescent coating on the glass.

So it really does not matter how you decide to excite the electrons to the higher energy state. It might be a high voltage converter, a tesla coil, a plasma globe, etc but all you need is a device that will kick those electrons inside the bulb from their ground state to the higher excite state. That’s all you need!

How to photograph shock waves ?

March 9, 2019May 1, 2022 / 153armstrong / 1 Comment

fyfd:

This week NASA released the first-ever image of shock waves interacting between two supersonic aircraft. It’s a stunning effort, requiring a cutting-edge version of a century-old photographic technique and perfect coordination between three airplanes – the two supersonic Air Force T-38s and the NASA B-200 King Air that captured the image. The T-38s are flying in formation, roughly 30 ft apart, and the interaction of their shock waves is distinctly visible. The otherwise straight lines curve sharply near their intersections.
Fully capturing this kind of behavior in ground-based tests or in computer simulation is incredibly difficult, and engineers will no doubt be studying and comparing every one of these images with those smaller-scale counterparts. NASA developed this system as part of their ongoing project for commercial supersonic technologies. (Image credit: NASA Armstrong; submitted by multiple readers)

How do these images get captured?

It may not obvious as to how this image was generated because if you have heard about Schlieren imaging what you have in your head is a setup that looks something like:

But how does Schelerin photography scale up to capturing moving objects in the sky?

Heat Haze

When viewing objects through the exhaust gases emanating from the nozzle of aircrafts, one can observe the image to be distorted.

Hot air is less dense than cold air.
And this creates a gradient in the refractive index of the air
Light gets bent/distorted

Method-01 : BOSCO ( Background-Oriented Schlieren using Celestial Objects )

You make the aircraft whose shock-wave that you would like to analyze pass across the sun in the sky.

You place a hydrogen alpha filter on your ground based telescope and observe this:

Notice the ripples that pass through the sunspots

The different air density caused by the aircraft bends the specific wavelength of light from the sun. This allows us to see the density gradient like the case of our heat wave above.

We can now calculate how far each “speckle” on the sun moved, and that gives us the following Schlieren image.

Method-02: Airborne Background Oriented Schlieren Technique

In the previous technique how far each speckle of the sun moved was used for imaging. BUT you can also use any textured background pattern in general.

An aircraft with camera flies above the flight like so:

The patterned ground now plays the role of the sun. Some versions of textures that are commonly are:

The difficulty in this method is the Image processing that follows after the images have been taken.

And one of the main reasons why the image that NASA has released is spectacular because NASA seems to have nailed the underlying processing involved.

Have a great day!

* More on Heat hazes

** More on BOSCO

*** Images from the following paper : Airborne Application of the Background Oriented Schlieren Technique to a Helicopter in Forward Flight

**** This post obviously oversimplifies the technique. A lot of research goes into the processing of these images. But the motive of the post was to give you an idea of the method used to capture the image, the underlying science goes much deeper than this post.

Jackson’s Laplacian in spherical Coordinates

February 18, 2019December 30, 2020 / 153armstrong / Leave a comment

If you took a look at one of the previous posts on how to remember the Laplacian in different forms by using a metric, you will notice that the form of the Laplacian that we get is:

$\nabla^2 \psi = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2 \sin(\theta)} \frac{\partial}{\partial \theta} \left( sin(\theta) \frac{\partial \psi}{\partial \theta} \right) + \frac{1}{r^2 \sin^2(\theta)} \frac{\partial^2 \psi}{\partial \phi^2}$

But in Jackson’s Classical Electrodynamics, III edition he notes the following:

20190218_135719

This is an interesting form of the Laplacian that perhaps not everyone has encountered. This can obtained from the known form by making the substitution $u = r \psi$ and simplifying. The steps to which have been outlined below:

20190218_135737_1 20190218_135743_1

A note on the Hydrogen spectrum

February 9, 2019December 20, 2020 / 153armstrong / Leave a comment

The emission spectrum of atomic hydrogen is given by this amazing spectral series diagram:

Let’s take a closer look at only the visible portion of the spectrum i.e the Balmer series.

If a hydrogen lamp and a diffraction grating just happen to be with you, you can take a look at the hydrogen lamp through the diffraction grating, these lines are what you would see:

Source

These are known emission lines and they occur when the hydrogen atoms in the lamp return to a state of lower energy from an excited energy state.

Representation of emission and absorption using the Bohr’s model

Here’s another scenario that could also happen:

You have a bright source of light with a continuous spectrum and in between the source and the screen, you introduce a gas (here, sodium)

Source: Harvard Natural sciences

The gas absorbs light at particular frequencies and therefore we get dark lines in the spectrum.

This is known as absorption line. The following diagram summarizes what was told thus-far in a single image:

The absorption and emission spectrum for hydrogen look like so :

Stars and Hydrogen

One of the comments from the previous post was to show raw spectrum data of what was being presented to get a better visual aid.

Therefore,the following spectrum is a spectrum of a star taken from the Sloan Digital Sky Survey (SDSS)

Plot of wavelength vs median-flux

Here’s the spectrum with all the absorption lines labelled:

Source: SDSS

You can clearly see the Balmer series of hydrogen beautifully encoded in this spectrum that was taken from a star that is light-years away.

And astronomers learn to grow and love these lines and identify them immediately in any spectrum, for they give you crucial information about the nature of the star, its age, its composition and so much more.

Source: xkcd

Have a great day!

*If you squint your eyes a bit more you can find other absorption lines from other atoms embedded in the spectrum as well.

On the strong 5577Å spectrum line

January 13, 2019December 20, 2020 / 153armstrong / 1 Comment

The above is a plot of the Wavelength(in Å) in the x-axis vs the flux of some objects from the Sloan digital survey ( consists of galaxies, young stars, Quasars, etc)

But there is one strong peak in all of those plots that seems to stand out: 5577 Å

And if you like, the color that it represents is the above (Made with Stanford’s color matcher app)

A nightmare for the astronomer

This line at 5577.338 is what astronomers refer to as a ‘skyline emission’ or a ‘mesospheric night-glow’ and arises from the recombination of atomic oxygen in the mesosphere.[2]

Source

This line is of no significance to an astronomer who is looking to find out properties about a far away astronomical object. Yet, this line pops up in every spectrum of any object that you look at in outerspace!

In addition since the line is so strong, it contaminates the nearby pixels making the nearby data unusable and also messes up the scaling of the plot.

Example of contaminated pixel columns in an image because of bright object

Wavelength Calibration

What do you do with something that is always there but has no use for you? – Re-purpose it!

Having noticed that this peak was consistent at 5577.338, Astronomers decided that they would use this peak line in the data as a reference to calibrate their actual data. (known as ‘zero-point correction’).

This ensures that all the spectrum lines in the data are aligned and any errors that might have occurred during observation are corrected for.

Other lines ?

There are other lines at 6300,6363, etc which are sometimes as bright or brighter than the 5577 line that are also used for calibration.

If you are interested in learning more, the following are three papers that this post was inspired from and they dive deeper into more technical details that underlie this fascinating topic:

[1] Night-Sky High-Resolution Spectral Atlas of OH and O2 Emission Lines for Echelle Spectrograph Wavelength Calibration
[2] Mesospheric nightglow spectral survey taken by the ISO Spectral Spatial Imager on ATLAS 1
[3] Variability of the mesospheric nightglow sodium D2/D1 ratio

Have a great day!

Feynman’s trick applied to Contour Integration

December 18, 2018December 19, 2020 / 153armstrong / Leave a comment

A friend of mine was the TA for a graduate level Math course for Physicists. And an exercise in that course was to solve integrals using Contour Integration. Just for fun, I decided to mess with him by trying to solve all the contour integral problems in the prescribed textbook for the course [Arfken and Weber’s ‘Mathematical methods for Physicists,7th edition” exercise (11.8)] using anything BUT contour integration.

You can solve a lot of them them exclusively by using Feynman’s trick. ( If you would like to know about what the trick is – here is an introductory post) The following are my solutions:

All solutions in one pdf

Arfken-11.8.4*

Arfken-11.8.6 & 7 – not applicable

Arfken-11.8.21 & Arfken-11.8.23* (Hint: Use 11.8.3)

Arfken-11.8.25*

*I forgot how to solve these 4 problems without using Contour Integration. But I will update them when I remember how to do them. If you would like, you can take these to be challenge problems and if you solve them before I do send an email to 153armstrong(at)gmail.com and I will link the solution to your page. Cheers!

Ecstasy Shots

A poetry of logical ideas

Flow(#2) – Plateau- Rayleigh Instability

What is causing a jet of fluid to form droplets?

Perturbations

A note on inkjet printers

Note on average density and why ships do not sink

How do hot air balloons work?

Origins of the Buoyant force

Powering a 13W CFL light bulb using a 3V battery

How to photograph shock waves ?

How do these images get captured?

Heat Haze

Method-01 : BOSCO ( Background-Oriented Schlieren using Celestial Objects )

Method-02: Airborne Background Oriented Schlieren Technique

Jackson’s Laplacian in spherical Coordinates

A note on the Hydrogen spectrum

Stars and Hydrogen

On the strong 5577Å spectrum line

A nightmare for the astronomer

Wavelength Calibration

Other lines ?

Feynman’s trick applied to Contour Integration