From Water Jet to Little Droplets

A: Hey B, you accidentally left the water faucet not fully closed!

B: Really??? Cause I don’t hear anything even though the faucet is just behind this door.

A: Well I didn’t hear it, I saw it!

B can’t say a word after that..

So is there anything wrong with B’s hearing? Perhaps no, we can’t blame her/him; sometimes a small water stream can make a lot of noise, while some other times it is really quiet. The quiet stream is identical with a cylindrical column of water giving a continuous disturbance to the sink. However it may happen that the column of water reaches a critical length and it breaks up into tiny droplets and falls in disjoined cluster. If the water jet breaks up into droplets before hitting the sink, the disturbance will be more concentrated, hence it will give rise to louder and more annoying sounds. Let’s call the point where the water jet breaks up into droplets point x. I observed this point x is also moving up and down in somewhat jerky manner or probably the motion is actually periodic but it is too fast that our eyes won’t be able to catch up.

Okay let’s play with it experimentally! Take a plastic cup and make a small hole at its bottom, I used a hot nail to melt a hole on it. Then put some water on it, and observe the breaking point. It is hard to observe where exactly the point x is, so I illuminated the water with a flashlight. Since the water column won’t scatter many photons, this way we can clearly distinguish it with water droplets which can scatter many photons. This is how the set up looks like

But just a moment, there is an unwanted effect, which is the water vortex. If the water is spinning before exiting the hole, the resulting jet won’t be cylindrical and it can breaks into droplets much easily and it will be crazy and complicated. Here is one simple way to minimize this effect:

Yupee it works! Ok good, now the experiment can be more peaceful. The length of the water jet depends of the height of water in the container at that instance. Thus I measured the length of water jet as function of the height of water in the container and I found that they are proportional. The higher the height of water column, the longer the jet length will be. And further observation shows that the length of the jet is longer for a larger orifice diameter. Then I think for a while, go to pee, realized that the breaking up occurs independent on the direction of gravity, and decided to stop the measurement. Later I confirmed it using holed water bottle to make a parabolic water jet, and it really turned out to be independent of the direction of gravity.

The first plausible explanation that came to my mind is:

Suppose we drop one ball per second from the roof of a building, initially the distance between two consecutive balls is . We know that at any time, a ball is always faster than the ball above it, and is always slower than the other ball below it, it means that the distance between two consecutive balls will increase as the balls go down. Similarly this effect may also happens in water stream, as it goes down, it tends to move apart.
The independent of gravity direction argument obviously crushes this explanation.. At least this effect might not be dominant.

Let’s start over and contemplate at this problem more carefully. First it would be awesome to get rid of viscosity, the problem will be much simpler without it. Let’s compare it with surface tension.

Where is the viscosity of water, is the surface tension of water, and is the characteristic velocity between adjacent layers of water, I assumed it to be equals to the jet velocity which is about (I am a bit too generous here). At , , and .

So the viscosity is no match for surface tension. The surface tension has no alibi! Let’s get him!

A liquid desires to be in a minimal energy state, any movement that will lead to reduction of surface area is favorable by surface tension.  On a level of quasi-static motion it would thus be desirable to collect all fluid into one sphere, corresponding to the smallest surface area. Evidently it does not happen. The surface tension has to work against inertia, which opposes fluid motion over long distances.

The cross section of water jet with circular shape has the lowest energy, in other words the cross section of the jet tends to be circular. In general, the faucet’s hole is not circular, thus it will induce a cross sectional oscillation. Initially the cross section of the water jet is not circular and it tends to be circular. When it reach the circular form, the water still has some kinetic energy (inertia), therefore it overshoots and the cross section continues to evolve further until all the kinetic energy goes into the surface tension potential energy after that it tends to be circular again, and so on. I suspect that this disturbance is amplified as it is transmitted down the jet until it breaks up into drops at a point. Then I take a fork and using its tip I give a small perturbation on the jet, I could see the perturbation propagates upwards and downwards like a wave.  The perturbation that goes upward decays and eventually vanished, but the perturbation that goes downward grows in size, and it moves the point x higher or the water jet shorter. Then I did another experiment, If I put the disturbance higher the point x moves down, if I put the disturbance lower the point x moves up. Which proves my hypothesis that the disturbance amplifies as it goes down. The longer the disturbance travels down the jet the stronger it will be. Thus the measurement of the length of water jet as function of the height of water in the container is pointless because it depends on how uncircular the orifice is. It just proves that for higher water height in the container, the velocity of jet is higher thus it allows the jet travels more distance before the disturbance goes sufficiently large to breaks into drops. And it also proves that for smaller orifices radius, the uncircularness becomes more effective.

The last thing that I tried: Dimensional analysis. I wonder that the skill of using dimensional analysis might be useful for me in the future. The argument that sounds like “the only <insert a dimension here> that can be formed by combining the relevant alone is” might be useful, it may give us some physical insights, the meaning of that unique dimensional combinations. Some physics problems can be solved easily using this method, e.g. a hoop of rope problem; to find the acceleration of the end point of the rope if the rope is allowed to slide down through a hole. Now back to the case, The only time scale that can be formed from fluid parameters alone is

But what does it means? I don’t know, it looks like some inertial terms divided by the surface tension term. The get some idea, let’s try to plug in the numerical values

We get

This is sooo small, it is not possible that this is the time it takes from the orifice to breaking point. I think it might be the time scale of the point x’ periodic motion, probably it is the time it takes to form a single droplet.

Look at this, it is really so fast

Physics of Walking While Holding a Plastic Bag

When we are walking in a straight line, while holding a plastic bag on our hand, with an acceleration much smaller than . There are two possible cases that may happen….

Sometimes the plastic bag would swing so peacefully that it makes almost no difference whether it is there or not….

But sometimes it may swing with a large amplitude, annoying enough to force us to make a conscious attempt to stop it.

Then how should we walk in order to avoid the latter case?

Let’s take a look at the plastic bag:

Yes, that is how it looks like in the physics world!

We can neglect the mass of the plastic bag relative to the total mass of the stuffs inside the bag. Due to centrifugal force, the stuffs inside will distribute themselves as far as possible from the pivot(finger holding the bag), so it is quite likely that they will end up having moment of inertia close to relative to the pivot. And since the plastic bag has a large cross sectional area, we cannot underestimate the power of air drag. Let’s say it is proportional to the velocity of the bag, so we can write the torque due to the drag force as . Further I will assume that the mass of the plastic and the stuffs inside is small enough that we can neglect the transient effect. Yet another assumption, we may say that while walking human’s center of mass translates in approximately sinusoidal motion so that we can write the acceleration of center of mass as . Since we are working in the hand’s frame, actually the drag force will change the equilibrium position so that it is not vertical anymore, but we will assume that this effect is small.

Okay now let’s write down the equation of motion, here is the torque about pivot equation:

Forget about the homogeneous solution, the steady state solution for this differential equation is:

The amplitude of oscillation

where

So the high amplitude motion will occurs at resonance, when the of walking pace close to

But it is not that simple.. I measured my own standard walking pace using my ipod, the average time it takes for me to do ten steps is , hence the average time between steps is or the frequency of pace is . And the natural frequency of oscillation of a random plastic bag filled with snacks and drinks can be calculated using , and the result is for . Thus resonance won’t occur at around the standard walking pace.

But we may pass through the resonance frequency during the process of building up speed. During the process, the frequency pace is increased from zero to , at some point it will hit . Now the thing is during the process how long will it stay inside the “near resonance region” which is roughly ? Where the amplitude of the oscillation is at least half the maximum amplitude. The longer it stay in that region the more effective the resonance will be. We know that the characteristic time required for the oscillations to die out is . Therefore our walking pace must pass through this “near resonance region” in time more than to get significant excitation of the resonance. The value of gamma can be calculated by comparing the natural frequency of oscillation with the experimental frequency of oscillation which turns out to be . We get

The average walking velocity is related to the pace frequency as follows

Where is the pace length. It can be measured by wetting my sandals before walking on the dry road, and then I came back and measure the distance between two  footsteps using a ruler.

The average distance turns out to be . Assuming constant pace length, the change in velocity corresponds to the change in as follows:

Thus the time required to pass through the resonance width is

Which must be larger than the damping time in order to excite an effective resonance. At the end we get

Therefore to avoid an annoying high amplitude swinging plastic bag, we must accelerate faster than , at least for this plastic bag

Childhood Question Finally Solved?

When you look at a light source (of course a glowing one!) you will notice something awesome, don’t you?

The bright point of the light will appear to be surrounded by a spray of radial light rays!

This thing is so cool that one may not be able to sleep because he or she can’t stop playing with it.

But what are those rays???

It is one of the biggest questions that I have since I was a child…

You may say, “Maybe it is due to scattering”

OK now look at the light source again, and rotate you head, what do the rays do?

They rotate with you!

Then rotate the light source if you can.

Notice that the rays do not rotate!

So it is something to do with our eyes? Perhaps something related to our eyelids, eye liquid, brain perception or such things?

Not sure yet? Ok now choose an opaque object around you, it could be your hand, your Gundam toy, a book or etc..

Hold that thing between your eyes and the light source.

Try to block the rays without blocking the light source.

Notice that even if you hold the thing very close to your eyes the rays are always magically appears between the

the thing and your eyes. And also the rays will vanish if you cannot see the light bulb.

So it is a “biological” effect? But my camera does not agree, she says that she can capture that effect too!

What is Going On?

Wait a second… Look at that innocent little bottle of chili sauce over there!

Did you see the “white line” on the bottle?

Now why? Each “blob” on the bottle will produce one copy of mini light bulb, as a result we got a line of mini light bulbs. If the blobs are much smaller, our eyes won’t be able to distinguish the gaps between the mini light bulbs! I think water ripples also do the same thing:

Now look at this:

Even more epic one:

Thanks to the chili sauce bottle.. Now I have a special power to control light!! Maybe I am the chosen one!

I must protect the world from the dark forces!

Okay okay enough bullshitting.. What did I do??

When I rub the camera using my finger horizontally the light line becomes vertical and when I rub it vertically the line becomes horizontal. The light line is always perpendicular to the direction in which I rub the camera. And if I twist my finger while touching the camera, I got lines in every direction or you can say there is no line at all.

I think when I rub the camera in a certain directions, I will leave trace amounts of oil that I get on my fingers maybe from rubbing my forehead or nose. The trace is in a form of series of parallel lines, as in the chili sauce bottle case and water ripples case it will reflect light perpendicular to itself .

But most photos of such light rays that I saw have very neat six dominant lines. like this one

I think it is a completely different phenomena. Maybe it is because the camera’s light receptor is not isotropic, usually it has hexagonal symmetry.

Let’s get back to our eyes. I would assume that there are scratches on our eye lens and they are more or less randomly oriented. This scratches will act as reflectors as in above cases. The problem is if the scratches are randomly placed why are the rays radial? Only a few selected scratches glint light towards our retina.  The basic principle is a scratch will reflect light perpendicular to itself, thus only scratches perpendicular to our line of sight or “tangential” to the light source will reflect light to our eyes. Therefore it would be very unlikely to have a non radial line, the scratches must be very specially ordered. So the scratches produce lines of light radiating in all directions around the point source. We can also play with the lines by goggling and narrowing our eyes. That way we can change the shape of our eye lens, it’s like stretching the scratches vertically and thus the lines. When we goggle the lines are more “vertical”, when we narrow our eyes the lines are more “horizontal”.

There is another optical effect that we can observe. At first you can look at a light bulb without squinting. Then by squinting the eyes you can see the rays being extended. And when we look above or below the light bulb the the rays are extended further upward or downward. I think it is because when we move our eyelids some eye liquid is moved and covered the cornea , it will change the liquid’s thickness profile at some part. To prove this hypothesis, I tried to yawn several times to produce some tear before looking at the light source. And what did I get? I can see similar long line even without squinting!

Yet another interesting stuff that I discovered while playing with light:

If we look at a bright light source for a long enough  time and then we look the other way, the part where the bright light was will become dark. If we put the source at the center of our vision, it would be awesome. After the light source is removed, the dark spot will show us where we are looking at. For example I tried to read something while the dark spot still there, I found that upon reading my eyes are focused at left edge, middle, and right edge of the text repeatedly. Moreover you can also know where your eyes are focusing while staring at optical illusion picture!

How Slow is “sufficiently slow” ?

A  quasi-static process that is reversible would be one which takes an infinitely long time to complete and during which the system and the surroundings are infinitessimally close to equilibrium. Sadly both are ideal situations and are not really achievable in practice

However, if I can execute the process sufficiently slow, I should be able to get close to reversibility. So the question is:

How slow do we have to go to get the reversible quasi-static assumption to be valid?

It depends…… Depends on what???

Let’s see…

Consider a system(piston) that is to be led along the quasi-static adiabatic process.

An ideal gas with density of molecules , molar mass and temperature inside an insulating cylindrical piston with cross-sectional area and length .

The constraint are to be moved slowly step by step, the system being permitted at each step to come to a new equilibrium state while fulfilling the condition . After a slight outward motion of the piston we must wait until the system fully achieves equilibrium, then we proceed with the next movement and we wait again, and so forth. Although this is the theoretical prescribed procedure, the practical realization of the process seldom follows this prescription. In practice the constraints usually are, relaxed continuously, at some “sufficiently slow” rate.

The characteristic time tau it takes for a system to reach equilibrium called the relaxation time. Although thermodynamics is a powerful theory that has been successfully predicts the fate of so many physical systems with some environment, strictly speaking thermodynamics does not deal with what is happening during relaxation. It describes the initial and final states only.

So, if the adiabatic expansion of the gas in the cylinder is performed in times much longer than this relaxation time the expansion occurs reversibly and isentropically. If the expansion is performed in times comparable to or shorter than the relaxation time there is an irreversible increase in entropy within the system and the expansion. The external pressure is decreased too rapidly that the resulting rapid motion of the piston is accompanied by turbulence and inhomogeneous flow within the cylinder. The process is then neither quasi-static nor reversible.

To estimate the relaxation time we first recognize that a slight outward motion of the piston reduces the density of the gas immediately adjacent to the piston. If the expansion is to be reversible this local “rarefaction” in the gas must be homogenized before the piston again moves appreciably. The rarefaction itself propagates through the gas with the velocity of sound, reflects from the walls of the cylinder and gradually dissipates. The mechanism of dissipation involves both diffusive reflection from the walls and viscous damping within the gas which is the damping due to imaginary part of the refractive index of sound wave in air.

Let’s assume that the second one is more dominant

Divide the air inside the piston into thin slices of thickness that are perpendicular to the direction of sound propagation.The intensity of sound wave emerges from the layer immediately adjacent to the piston will decay as it travel through distance because some of the them have run into particles and did not make it to the next layer or end up having lower average velocity. The change in intensity due to a slice of air is

To determine , we need to know about the cross section of the particles . We assume that  is sufficiently small that one particle in the slab cannot obscure another particle in the slab when viewed along the  direction. It follows that the fraction of particles collided when passing through this slab is equal to the total opaque area of the particles in the slab, , divided by the area of the slab , which yields on . Thus the fraction of particles collided when passing through the slab is given by:

Integrating both sides we get

Where $v_{s}$  is the velocity of sound. Which is easily derived and given by

Thus the characteristic relaxation time or the time it takes for the the rarefaction pulse to be effectively dissipated is given by

So, if the adiabatic expansion of the gas in the cylinder is performed in times much longer than this relaxation time, the  quasi-static assumption is a good one.

Assume the following numerical values:

(hydrogen atom)

* the numerical value of the cross section of hydrogen is “stolen” from this paper http://arxiv.org/pdf/plasm-ph/9506003.pdf

Then we get

Which is surprisingly  small… Therefore we are safe to assume the adiabatic condition in real life

Bernoulli Part I (gas)

Without doing any initial bullshitting, let’s go directly to the problem…

One of the most ubiquitously found explanation for the lift force on an airplane is quick, simple, and gives the correct answer. But yet, I think it introduces misconception and misleadingly invokes Bernoulli’s equation. Let’s take a look at the argument…

Using the argument, one should also expect a lift for a symmetric wing profile as shown above. The argument claims that the distance from this point to the trailing edge is greater along the upper surface than along the lower surface. Using the ‘equal time’ argument it is assumed that two neighbouring fluid particles which ‘split’ at stagnation point should somehow meet again at the end point then this requires that the average flow velocity on the upper surface is greater than that on the lower surface.  Now Bernoulli’s equation is quoted, which states that larger velocities imply lower pressures and thus a net upwards pressure force is generated. The above explanation is extremely widespread, it can be found in many textbooks. The problem is why should two particles on either side of the wing take the same time to travel from A to B? I never saw any argument for that.

However, if we considers the problem from a microscopic point of view, we comes to a different conclusion: a symmetric wing profile won’t produce net lift.

Now let’s consider the case using ideal gas point of view. If the wing is stationary, the pressure on all parts of the wing is identical, i.e. there is no lift. If the wing is moving in the indicated direction, the front half of the upper wing surface experiences an increased pressure because of the increased speed  air molecules hitting it (this creates a downward force). On the other hand, the rear half experiences a reduced pressure because the of the reduced speed(creating a lift). Overall, there is consequently no lift. It is obvious that an overall lift is only achieved if the rear section of the wing has a larger area than the front section.

Bernoulli’s principle seemed to had successfully predicted or explained many strange behavior. But indeed in some of those cases Bernoulli’s principle does not give a correct explanation (e.g. blowing air between two pieces of paper, the paper appears to get closer to each other). it’s just misleadingly plausible, in many cases the “fast moving air area” has lower pressure than the “slow moving air area” has although the causes is not necessary due to the Bernoulli’s principle. In fact, for most cases the main effect of pressure decrease is due the to Coanda effect, which mainly reduce the air density in some area. The effect governed by the Bernoulli’s principle is often too small to be significant, It is beautifully shown in this demonstration:

http://www.physics.umd.edu/lecdem/services/demos/demosf5/f5-12.htm

Small Bubble, Big Trouble

Simple yet interesting:

Consider a perfectly rigid box in a gravitational field. The box is completely filled with an incompressible ideal fluid(zero viscosity) except there is a small air bubble on the bottom. Assuming that the bubble is always spherical, what will happen next??

We know that the gravitational field will produce hydrostatic pressure gradient on the water so that the pressure decreases as the altitude increase. Because of this, the bubble will feel non-uniform pressure all over the surface, and as a result of it the bubble will be pushed upwards. As the bubble goes up, it will encounter a lower pressure region, thus it will expand….

Wait!! Expand??? Didn’t I just mentioned that the box is perfectly rigid and also the fluid is incompressible?? What’s possibly wrong here?

So… the bubble’s volume is conserved.. what should we do to account for this? hhmm….

(1). While maintaining the pressure gradient, the pressure of the fluid and the pressure on the walls increase as a whole so that the decrease in pressure with height is balanced with the increase of pressure overtime. Thus the bubble will always feels the same pressure.

(2). As the bubble rises, the temperature decreases precisely as to balance with the local pressure of the fluid.

Wait wait wait… the second one is suspicious! What if initially the bubble is colder than the fluid, the colder body will cools down while the hotter will gets even hotter. Thus it will violates the second law thermodynamics right?? NO! its not a problem it doesn’t violates the 2nd thermodynamics, since there is an external work done on the system. The graviational potential energy is decreased as the bubbles rises, (some fluid goes down).

And the first one… But how does the pressure of the fluid changes so easily? Does pressure suppose to be due to thermal collisions? Let’s imagine matter as consisting of jiggling molecules, each neighboring molecules are connected with a spring. There are two types of elastic energy stored in the material:

i. The one that is related to the average distance between neighbouring molecules, this is determined by the local density.
ii. Related with the “standard deviation” of the molecules movement, this one is determined by the local temperature.

If the the interaction between molecules is super strong, as in our case, the “standard deviation” of the molecules movement is unimportant. The “spring” only needs to change it’s length a little bit to account for the kinetic energy. Thus the temperature is unimportant, and we also know that if we change the temperature of water, it’s pressure won’t change significantly. In our case the pressure of fluid will acts like a normal force, it will respond to any pressure so that the volume of the fluid doesn’t change.

Eventually I think both (1) and (2) will happen at once, and the rate of heat transfer(compared to how fast the bubble moves) and the heat capacity of the bubble(number of air molecules) will determine which one is more significant.

Why did he spin the egg before throwing it?

I’m bored cooking and eating an ordinary omelette so I try to search some videos about how to cook an omelette in youtube. Somehow I stumble on this video

Watch it! The rest of this post is pointless if you don’t watch it

I assume you already watch the video, so now you understand the title of this post. Okay let’s discuss about it..

First, we know that his goal was to cut the egg. In more detail, he wants the egg to be in horizontal orientation before hitting the rectangular knife(or rectangular what?). The orientation of the egg is described by the vector pointing from the “blunter” end to the “sharper” end. And let’s call everything inside the egg with the word “fluid”.

When the egg is spinning the centrifugal force pushes the fluid to the ends of the egg, blunt and sharp.  So Perhaps he spin the egg because he didn’t want to cut the yolk at the center of the egg? NO! I hope the purpose must be something much cooler than that, so let’s assume that this is not his goal.

Another possibility, he made the egg spinning about the vertical axis so that it becomes more stable against perturbation and more likely to remains in horizontal orientation. It’s much easier to throw and catch a stable egg. And he also needs to spin the egg for a long time because the fluid inside and the shell are not quite coupled, the friction between them is very small.

Yet another possible advantage… if we spin the egg around the vertical axis, it makes the egg harder to move in a straight line. It’s common sense! you can imagine it! Because when the egg is moving translationally it is also rolling, when the orientation change due to the rotation the egg still preserve it’s rolling angular velocity. Consequently the egg tends to move in a circle, the radius of the circle depends on the velocity of egg’s center of mass, in other words it makes the egg harder to move in a straight line. So what if the egg is harder to move in a straight line? Of course it’s very helpful! It’s very hard to throw an egg properly if it can rolls easily from the rectangular tool(or what?), it’s much easier to control when the egg is spinning.

But the pan and the rectangular tool is very oily, it hardly rolls at all, it might simply slide without rolling.. You can see that when the awesome guy was spinning the first egg, it’s clearly shown that the egg moved through a curve path. That means that the friction is significant! I also tried putting an egg on an oiled frying pan and then I tilted the pan. It turns out that although the pan is very slippery the egg was still rolling through it’s way down. But actually my egg is pretty rough, maybe his egg is not as rough as mine…

Maybe the last theory is right, but the guy’s hands are also very talented, I think he can balance the egg just by tilting the rectangular tool rapidly. I hope one day I can meet an omurice cook and ask him about the reason…