Physics of daily life

Transcription

1 Physics of daily life prof.dr. Robert F. Mudde Kramers Laboratorium voor Fysische Technologie Delft University of Technology Pr. Bernhardlaan BW Delft, The Netherlands November 30, 2005

2 Contents 1 Introduction Examples Sky light and alike Thickness of the atmosphere Atmospheric absorption Blue sky Distant mountain Color of smoke, fog or clouds Color of the sun Flattening of the low sun and moon Water and light The color of water Glitter Caustic network The rainbow Bubbles The sound of mountain streams The sound of boiling water Surface tension The sound of a fresh, hot cup of instant coffee Speed of sound sound in air A bubble bath Effect of bubble expansion Effect of evaporation Bernoulli Bernoulli Shower curtain Boat trip through the channels of Delft Water flowing over an elevation in the river bottom Why can air planes fly? Top and back spin Emptying container Tsunami i

3 CONTENTS 1 6 Heat Chinese egg roll Frozen slice of bread Black or white radiator? Sipping Bird Milk in the coffee and leaves in the tea Freezing lakes and channels Mechanics at work Catastrophe Toy Bungee jumping What is the shape of a suspended cable? Problem formulation Solution Lowest point of the rope Intermezzo: the suspended rope as Variational Principle High tide, low tide Fire hose Falling Pencil Capillary pump

4 2 CONTENTS

5 Chapter 1 Introduction Who loves nature observes her the same way he/she breathes and lives: from a native inner drive. This is the openings sentence of Prof. Dr. M. Minneart s series of three books about the physics of our environment. Minneart (a Belgium Physicist who died in 1970) published his books in 1937 for the first time. His books have been translated in many languages and are still in press, showing that a basic understanding of the physics of the world around us is still popular. This can also be seen in the weekly science sections of the better newspapers: many of them have a column that deals with questions from its readers asking for an answer to some remarkable observation. In many cases the answer is found in physics of every day life. Some of the questions are for a physicist rather obvious, other are difficult to answer. In many cases the observation seems trivial, but the answer can push one deeper and deeper into the world of physics. And as always, with an increased level of understanding the phenomena in question gains in beauty. This course is set up with the idea that the world around us is full of beauty and that an appreciation of it is deepend as the understanding in terms of physical reasoning is brought to a higher level. Furthermore, it is just fun to think about what we see, hear, feel and try to find an explanation. Since Prof. Minneart s book saw the light many others about the physics of daily life have appeared. Some relatively superficial, others just fun reading but also quite a few that try to give answers and insight based on sound physical reasoning. To name a few De Natuurkunde van t Vrije Veld, 1, 2 & 3 Dr. M. Minnaert B.V. W.J. Thieme & Cie -Zutphen Clouds in a Glass of Beer - Simple experiments in atmospheric physics C.F. Bohren John Wiley, 1987 Color and Light in Nature D.K. Lynch and W. Livingston Cambridge University Press,

6 4 CHAPTER 1. INTRODUCTION De Wetten van de Vliegkunst - over stijgen, dalen, vliegen en zweven H. Tennekes Aramith, 1992 The flying Circus of Physics - with answers J. Walker John Wiley, 1977 Alledaagse Wetenschap K. Knip Uitgeverij Contact Amsterdam, 2000 Finally, I would like to mention the Lectures on Physics by Richard Feynman. Although this is not written as a series about physics in daily life, it has some nice side steps to it but foremost it is excellent in explaining and providing inside of the basics of physics. 1.1 Examples It is easy to come up with a set of examples of questions concerning daily life observations that involve some degree of physics in answering. Some may at first sight be only remote connected to physics, like the question: Why do (African) elephants have such enormous ears? Part of the answer can only be given when considering the balance between production of heat and the desired temperature of the animal. All healthy mammals have to maintain a body temperature of about 37 C. However, even in rest the animal will use energy which is partly converted into heat within its body. Obviously, to arrive at a steady state as far as its temperature is concerned, the animal will have to transport this heat to the environment as can be seen from a very elementary (steady state) heat balance: 0 = heat flow from environment - heat flow to environment + + internal heat production (1.1) The heat production is proportional to the volume of the elephant. The heat flows are more complicated as there are various ways to transport heat from an object to its surroundings. Generally, we split these in three groups: heat transport by conduction, by convection or by radiation. The relative importance of these three depends on the particular circumstances. For instance, if the elephant is standing in the bright sunshine, it is obvious that he will receive radiation from the sun that will count as an inflow of heat for its heat balance. If there is wind, the elephant might be cooled by the wind if the air is colder than the outside of the elephant or heated if the air temperature is higher than its skin temperature.

7 1.1. EXAMPLES 5 Furthermore, for animals there is the very important possibility to loose heat due to evaporation of water, i.e. by sweating. For all these flows of heat it holds that they are proportional to the surface area of the elephant. So, if we look at the heat balance of the elephant, we see that the internal production of heat is proportional to the volume of the elephant and that the net balancing heat loss is to a good extend proportional to its surface. This gives us a good part of the explanation for the big ears: increase of (cooling) surface without increase of (heat producing) volume. Figure 1.1: Heat production and flows from and to an elephant. How does a dolphin in cold sea water prevent itself from too much heat losses via its flippers? This question can be investigated along the same lines as in the above example. Now we focus on the flippers. It will be clear that the blood that flows through them will be cooled substantially, being surrounded by cold sea water. Nature has found a clever way of dealing with this, that humans (especially engineers) since the industrial revolution started to utilize frequently: the heat exchanger. In the flippers of the dolphin the veins carrying blood that flows from the body into the flippers are kind of twisted around the return veins that carry the blood back into the body (see fig.(1.2)). What happens is that the warm blood coming from the dolphin body exchanges heat with the blood that return to the body. The latter is obviously cooled by the cold surrounding sea. The driving force for the heat flow from the hot to the cold flow is the temperature difference between the two blood streams itself. Hence, this is a passive system that does not cost the dolphin any extra energy and furthermore is sensitive to the heat loss of the blood by its nature: the colder

8 6 CHAPTER 1. INTRODUCTION Figure 1.2: Schematic representation of the veins in the dolphin flippers. the returning blood the more heat is transferred from the hot blood to the cold. Consequently, the colder the hot blood gets before being exposed to the cold sea. In this way the system is self regulating and the temperature of the blood entering the dolphin s body is as high as possible without the input of extra energy by the animal. In many technical applications heat exchangers (or mass exchangers for that matter) can be found that use a similar arrangement. What is the speed of returning bullet when shooting in the air This question was raised in the eight-o clock news at the time freedom fighters in Albania were celebrating their victory. The estimates that followed on television and in the newspapers were all dangerously high: several hundreds of meters per second up to a kilometer per second. This would certainly be deadly when being hit by such a returning bullet. The velocity in reality is much lower as we shall see. Before doing the calculation, let s first discuss what the velocity determines. Let s assume that the bullets are small steel spheres with a diameter of 5mm. Then gravity will act on them for sure. If this was the only force, things would be easy. Assuming the bullet to be fired perfectly vertically upward the only parameter determining the return velocity is the velocity with which the bullet leaves the rifle. Just for the sake of the argument let s assume that this is 600m/s. As only gravity acts, the bullets total energy (i.e. kinetic and potential energy) is conserved. Hence, if the bullet returns, its potential energy is the same as when it left and consequently its kinetic energy is at its starting value. Thus the velocity when returning would indeed be 600m/s. But of course gravity is not the only force. On earth an object always travels through air (or water). Thus we should include the frictional force exerted by the air on the bullet. In a steady state the gravitational force is balanced by the drag (see fig.(1.3)). This friction is a function of the shape of the object and its velocity and other parameters. For practical calculations this force, called drag force (F D ), is written in

9 1.1. EXAMPLES 7 Figure 1.3: Gravity and drag acting on a falling sphere the form: F D = C D A 1 2 ρ airv 2 (1.2) with v the (stationary) velocity of the object, ρ air the density of the air, A the area of the projection of the object on a plane perpendicular to its velocity, and C D the so-called drag coefficient. The latter is a function of (amongst other parameters) the velocity of the object. Actually, the velocity in eq.(1.2) is the relative velocity between the object and the surrounding air. Obviously, the direction of the drag force is always the opposite of the relative velocity. In the case of the returning bullet this means that the drag force tries to slow down the bullet. If we assume that the gravitational force is balanced by the drag we have that the net force on the bullet is zero and thus its velocity constant. This follows from 1 0 = F D F g = C D A 2 ρ airv 2 mg (1.3) If we now insert for A = π 4 d2 b and m = πρ 6 bd 3 b we find for the terminal velocity of the bullet: 4 ρ b gd b v b = (1.4) 3 ρ air C D where d b denotes the bullet diameter and ρ b the density of the bullet. Putting in numbers: d b = 5mm, ρ b = kg/m 3, ρ air = 1.2kg/m 3 and taking for the drag coefficient a value of order 0.5, we find for the velocity: 31m/s. Still a hit would hurt, but it is probably no longer fatal! Notice that the initial velocity at which the bullet was fired is irrelevant here (provided of course it was well above the velocity we just calculated).

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11 Chapter 2 Sky light and alike 2.1 Thickness of the atmosphere The earth is a special planet. It has the right conditions to make life possible. It has water in enormous quantities and is blessed with an atmosphere containing oxygen and other gases. It has the right distance from its star, our sun, to allow for a temperature that leaves water liquid. The earth radius is about 6378km. The atmosphere forms only a thin layer around the earth. How thick is this layer? Let s make a guess. We do know that the pressure at the earth surface is 1 atm. Just like in water, this pressure can be explained as hydrostatic pressure: the total weight of the atmosphere is lifted by the pressure. So, a first rough guess could be: assume the density of the air is constant (1.2kg/m 3 ). Then, a simple force balance over a vertical column with cross-sectional area A of the atmosphere gives: p A F grav = 0 = p A ρahg = 0 = H = p ρg = 8.6km (2.1) Obviously, this answer must be wrong as air planes routinely cruise at a height of 10km. Thus, we need to improve our model. And it goes without saying that the constant density is a very dubious assumption. Of course, going up in the atmosphere, the density decreases. We can use the ideal gas law to relate the density to the pressure. The latter is also decreasing with height as the pressure at height z does not have to carry what is below z. But the density is also a function of the temperature. Let s try isothermal conditions, hence we set T = 273K. Now we need to set up a force balance that takes into account the variation of the density with height. Therefore, consider a small slice out of a vertical column of the atmosphere, between z and z + z (see fig.(2.1)). The weight of this slice is: ρa zg. Three forces act on this slice: at the bottom the pressure at that position pushes upwards: p(z)a; at the top the pressure pushed downwards: p(z + z)a and, of course, gravity. Thus for a steady state we have according to Newton that the sum of the forces is zero: If we use a Taylor expansion on p(z + z) p(z)a p(z + z)a ρa zg = 0 (2.2) 9

12 10 CHAPTER 2. SKY LIGHT AND ALIKE Figure 2.1: Forces on a small slice of the atmosphere. p(z + z) = p(z) + dp z + h.o.t. (2.3) dz we can simplify eq.(2.2) to dp = ρg (2.4) dz For isothermal conditions the density and pressure are linked according to: pv = nrt p = ρ RT (2.5) M with M the molar mass of air (= kg/m 3 ). Combination of the last two equations gives: 1 dρ ρ dz = gm RT (2.6) Thus, we have for the density profile in the atmosphere (taking z = 0 ρ = ρ 0 ): ( ρ(z) = exp gm ) ρ 0 RT z (2.7) The characteristic length scale is thus: Λ = RT = 8.0km. Now if we define rather gm arbitrarily the end of the atmosphere as the position where the density has dropped to ρ 0 /1000, we find that the atmosphere has a thickness of 55km. Of course, we know that the atmosphere is not isothermal. Instead, we could try the assumption that the atmosphere behaves adiabatic. This means that if a portion of air moves up or down, it can adjust itself to the new surrounding without exchanging any heat. Of course, the ideal gas law is still valid, but the temperature is now no longer a constant with respect to height. In stead, it changes such that pv γ is constant (with γ = cp c V = 1.4). If we use this relation in eq.(2.4), we find: ρ γ 2dρ dz = ργ 0g γp 0 (2.8) with ρ 0 and p 0 the atmospheric density and pressure at ground level. The solution of the above equation reads as:

13 2.2. ATMOSPHERIC ABSORPTION 11 ρ(z) = ( ρ γ 1 0 γ 1 γ ρ γ ) 1 γ 1 0g z p 0 (2.9) If we use the last equation to find z = H at which ρ = 0, we find H = 30km. A good value for the actual thickness is some 50km. There, the pressure has dropped to about 1Pa. 2.2 Atmospheric absorption Light from the sun (and stars) will have to travel through the atmosphere before reaching the ground level. On its way it will be subject to absorption and scattering. How much of the sun light is absorbed? Obviously, this depends on the time of the day, as at noon the sun is overhead and travels a much smaller distance through the air to us. The absorption is due to the air molecules and to dust particles floating in the atmosphere. Figure (2.2) shows the difference between sun light outside the atmosphere and at ground level as observed at sea level with the sun at 20 altitude. Figure 2.2: Intensity of sun light outside the atmosphere ( ) and at sea level ( +, from [1]). As can be seen, the absorption is a function of the wave length of the light. Not surprising as atoms and molecules absorb light in specific bands. Of course, macroscopically this can not be seen. 2.3 Blue sky Absorption is not the entire story. Sun light (or light from the moon) also gets scattered by the air and the dust. This effect is so common, that hardly anybody

14 12 CHAPTER 2. SKY LIGHT AND ALIKE pays attention to it. Nevertheless, it is an important feature of the atmosphere. When you look on a clear day into the sky its color is blue, everybody knows that. But few people know why. The reason is found in the scattering properties of the molecules: the probability of light being scattered by an air molecule is proportional to the wave length of the light to the power -4, or rephrased: proportional to f 4 (f the frequency of the light, the theory of molecular scattering was given first given by Lord Rayleigh). Thus, blue light of a wavelength of 450nm is compared to red light (λ = 650nm) (650/450) 4 = 4.4 times more likely to be scattered. Consequently, the blue end from the (white) sun light has a reduced probability to reach our eye directly in comparison with the red end. And thus most of the scattered light that reaches us is blue: the sky is blue. This is illustrated in fig.(2.3), that compares the spectrum of light coming directly from the sun to light coming from the sky (called sky light). Figure 2.3: Comparison of spectrum of sun light ( ) and of sky light. The sun is at an altitude of 24. The spectra of the sky light are taken at various angular angles from the sun. The intensity of all spectra have been rescaled to the same value at λ = 500nm. Figure taken from [1]. The sky does not have a uniform brightness. This is easily verified when looking on a bright day into the sky: the sky above us is deep blue. Towards the horizon the sky gets lighter. This is seen in fig.(2.4). The effect is caused by the difference in the length of the line of sight when looking at different angles into the sky. Directly overhead (this is called the Zenith) the distance through the atmosphere is a minimum, see fig.(2.5). When looking at an angle θ with the vertical we see through more air. Thus, more light will be scattered into our direction and the sky looks brighter. As a first approximation the ratio of the amount of air viewed in direction θ compared to θ = 0

15 2.3. BLUE SKY 13 Figure 2.4: Photo of the entire sky, showing that the sky is light close to the horizon. The hand in the photo is from the photographer, who is blocking the sun. Taken from [1]. Figure 2.5: Geometry of the line of sight through the atmosphere when looking towards space. increases as 1/cos(θ). This rule brakes down at large angles, when the curvature of the earth becomes important. The ratio is plotted in fig.(2.6). Notice in fig.(2.4) that close to the horizon the sky is almost white rather then blue. The reason is that we wrongly assumed that the more air is on the line of sight the more (blue) light gets scattered to us. We have ignored multiple scattering. And this causes a very thick atmosphere to be opaque. A simple model for the brightness of the sky itself can be constructed as follows. The line of sight is in the direction x, making an angle θ with the vertical. For simplicity, assume that the sun is directly above us (at θ = 0), then what we see in the direction x is scattered sun light, i.e. no direct sun light. In fig.(2.7) the situation we want to analyse is shown. Consider a part of the line of sight between {x,x+dx}. The intensity of the light is called I, the initial intensity outside the atmosphere is I. The light that is coming along the x-axis into the observer s direction has at position x an intensity I x, when it is at position x+dx its intensity is changed to I x+dx. This change is due to scattering.

16 14 CHAPTER 2. SKY LIGHT AND ALIKE Figure 2.6: Length through the atmosphere as a function of the zenith angle θ, normalized by the length at θ = 0 Figure 2.7: Schematic representation of sky light received by the observer. In a good approximation the loss is proportional to the intensity of the incoming light, so µi x is scattered in different directions and will no longer reach the observer. But, we shouldn t forget the light that is added to our line of sight from direct sun light that is scattered into our direction. Again this is proportional to the intensity of the sun light at position x. We can safely ignore that this is, due to scattering, no longer the original intensity I and we formally write for this gain-term: µi 0. Note, that we did not use in this last expression I. This is because we now deal with scattering of light by the molecules on the line segment dx precisely into our direction, whereas for the losses all we need is that the light originally moving in our direction gets scattered away from the line of sight, without having to specify into which direction. So, I 0, is taking all this into account and is a lumped (unspecified, but constant) parameter. Now obviously, we should also take into account that the loss and gain term are also proportional to the length of the line segment we are considering, i.e. proportional to dx since the longer dx, the more air molecules will participate in the scattering. If we add everything up, we find: If we use a Taylor expansion on I x+dx : 0 = I x I x+dx µidx + µi 0 dx (2.10) I x+dx = I x + di dx (2.11) dx

17 2.3. BLUE SKY 15 we can reduce eq.(2.10) to The general solution is: di dx = µ (I 0 I) (2.12) ln (I 0 I) = µx + C (2.13) The integration constant C is found from the boundary condition: at position x = 0, which is the outer boundary of the atmosphere, no sun light is yet reflected into our line of sight as no molecules are present there to do so. Thus, we find C = lni 0 and the solution for the brightness of the atmosphere in direction x is: I(x) I 0 = 1 e µx (2.14) Notice that µ has a dimension of m 1, i.e. it is the reciprocal of a characteristic distance Λ. The observer will see, of course, the intensity at position x max which denotes ground level. As we have seen before, the length through the atmosphere changes on the angle, θ with the vertical as H/cos(θ) (H is the thickness of the atmosphere above us). This ignores the curvature of the earth for which we can correct. So, the sky light intensity that we see when looking into different directions is given by (the earth-atmosphere is approximated as a flat, infinitely wide system) I(θ) I 0 = 1 e H Λ cosθ (2.15) For θ = 0 we have I(0)/I 0 = 0.1 (see [1]). Thus H = ln(0.9). The brightness of Λ the sky is plotted in fig.(2.8): the dashed line is the flat atmosphere approximation, the full curve the one corrected for the earth curvature. Figure 2.8: Brightness of the sky for an observer at sea level, viewing at angle θ with respect to the vertical. Now we understand the change in brightness of the sky, but we haven t answered the question why the sky close to the horizon looses its blue color. The answer is now simple. The sky light intensity saturates at large angles; it will do so for all wave length, thus no longer is blue the dominating color but all other colors mix in and they do so at comparable intensities. Hence, the sky turns white.

18 16 CHAPTER 2. SKY LIGHT AND ALIKE Distant mountain Another consequence of the scattered sun light that is related to the whiter color of the horizon, is seen when looking at a distant mountain ridge. The mountains furthest away seem lighter, those more in the foreground look darker, see fig.(2.9). Figure 2.9: Distant mountains appear lighter than those close by (from [1]). The explanation is that the further away, the more scattered sun light mixes in with the light that is directly coming from the mountains. Since a mountain close by hides some of the scenery behind it, the distance to that mountain top is significantly less than the scenery seen just above the ridge. In other words there is a discontinuity in the distance when looking from the mountain to the scenery behind it. This is illustrated in fig.(2.10). Consequently, the intensity from the scattered light changes rather abruptly. Figure 2.10: Discontinuity in brightness caused by discontinuity in distance seen. The sky light can be so bright that mountains at great distances can not be seen during the day. But at sunrise or sunset they become visible as dark areas against a red sky, see fig.(2.11).

19 2.3. BLUE SKY 17 Figure 2.11: Distant mountains (190km away) not visible during the day, but present at sunset (from [1]) Color of smoke, fog or clouds The color of cigarette smoke is different when being inhaled first compared to coming directly from the cigarette. Why? Again this can be explained by the scattering properties of the smoke. When rising up directly from the cigarette, the smoke particles are very small. Many of them are in fact smaller than the wave length of visible light. This means that they have scattering properties quite similar to the air molecules discussed before. Therefore, blue light is preferentially scattered and the smoke looks bluish to us. If on the other hand, the smoke is first inhaled it contains relatively large water droplets. These are larger than the wave length of visible light and will therefore have different scattering properties (that can be described by the Mie theory, by Gustav Mie ( ), explaining the scattering of light by spherical particles of any size). It means that the preference for scattering of higher frequencies is lost. The scattering is for the large droplets much closer to ordinary reflection and refraction. This makes the smoke look white. Of course, this argument will also hold for fog or clouds. One might wonder what makes it different: small or big water droplets. After all, all the scattering is done by the water molecules which are the same in any type of droplet. Or perhaps, equally puzzling now that we start to think about it: why are clouds seen anyhow? Before the cloud is formed the water vapor is present already at the location of the cloud-to-form. All that is needed is a sufficient decrease in temperature (and some nuclei to start the condensation process). The total number of water molecules in the line of sight hasn t changed at all! The difference is in the scattering of single isolated molecules (the vapor) and an agglomerate of atoms (a droplet). In the droplets the atoms are packed together at inter-molecule distances much smaller than the wave length of the light. Now remember that light is nothing but electro-magnetic waves. The molecules respond to the passing electro-magnetic field. In case of the isolated molecules they do so individually and the scattering is proportional to N, the number of molecules. But when two molecules are very close together, they start to respond in phase to the electro-magnetic field. This gives a summation of the contributions of each of the two molecules to the amplitude of the scattered electro-magnetic field, which means that the energy of the scattered light is four times higher, as the energy is proportional to the square of the wave amplitude.

20 18 CHAPTER 2. SKY LIGHT AND ALIKE Thus, the linearity in N is broken. The conclusion is that lumps of water molecules are able to scatter more efficiently than based on the number of individual molecules. Thus, water vapor is invisible to the naked eye, but droplets are easily seen. For a more complete description of the scattering take a look at The Feynman Lectures on Physics, part I Ch.32 ( [2]). Clouds: floating droplets As everybody knows, clouds are floating water droplets. What is suspending them in the air? Obviously, gravity is pulling them down. And the upwards directed buoyancy force of the air (Archimedes law) is insufficient to keep the droplets up. So, apparently there is some force acting on the droplets in the upward vertical direction. Actually, we came across this force when dealing with the returning bullets example. There we found that the force on the bullets is proportional to the velocity of the bullet squared, the density of the surrounding air and the area of the projection of the bullet perpendicular to its velocity. We need to make this a little bit more general: the force is proportional to the square of the magnitude of the relative velocity, v r. This is the difference between the velocity of the droplet, v d, and that of the surrounding air, v air, i.e.: v r = v d v air (2.16) Furthermore, this drag force is in the opposite direction of the relative velocity. So we have: F D = C D A 1 2 ρ air v r 2 v r v r (2.17) As the vertical component of the water droplets in a fog or cloud is zero (on average), we have that the vertical component of the relative velocity is equal to v air,z. If we assume further that the horizontal velocity of the air and the droplets is the same, we have a non-zero vertical component of the drag force: F D,z = C D A 1 2 ρ air v air 2 v air,z v air,z (2.18) Furthermore, gravity is pulling the droplets downward, so the drag force must have a vertical component that is pointing upwards. This can only be so if the vertical component of the air velocity is pointing upwards! Apparently, the motion of air in a cloud has an upward component that is sufficient to keep the water droplets floating : the drag force exerted by the air flowing around the droplet balances the pull of gravity. In fig.(2.12) the air velocity needed is given as a function of the size of the water droplets. The atmosphere is always in motion. It is by and large transparent for sun light. So, if there are no clouds most of the sun light reaches the ground and warms it up. The ground in turn heats up the air. This process obviously induced temperature and density differences difference, causing e.g. upward motion of hot air. In a cloud the droplets may grow until finally their size is such, that the force of gravity, which is proportional to the diameter to the third power is larger than the drag force that goes as the square of the diameter. Then the droplets will fall and we have rain, or hail etc. Even during their fall to the earth the droplets might grow. Furthermore,

21 2.4. COLOR OF THE SUN 19 Figure 2.12: Required air velocity for suspending a water droplet as a function of the droplet diameter. the upward velocity of the air close to ground level could be small or zero. In the latter case we find that the velocity with which a droplet with a size of 3mm is on the order of 8m/s. 2.4 Color of the sun When, on a clear day, we would look into the sun at the middle of the day, we would find that the sun is white. It is almost as white as it is seen in space. Even though some of the visible sun light is scattered by the atmosphere and some is absorbed, most is reaching us. The spectrum is not affected to much and thus the sun s color is white. However, when the sun is low, the sun light has to pass through a much thicker layer of the atmosphere. This means scattering, which is especially effective on the high frequency part of the spectrum (remember the frequency to the fourth power in the scattering). Thus, the sun light we receive from a low sun has lost quite some energy over all wave length and the sun is less bright. But in addition the blue end of the visible spectrum is weakened more: the sun starts to look yellow. On top of the scattering is absorption by water vapor and ozone, that absorbs extra in the blue and green. The lower the sun, the longer the path through the atmosphere. So, finally the scattering of the yellow is also so effective, that the sun turns red. Here also small floating particles, dust, smoke or small droplets (all called aerosols) help in scattering. If their size is on the order of 100nm, they are especially efficient. As their concentration (i.e. number per unit volume) differs from day to day, so is every setting sun different in color. Twilight arch Just after sun set, the horizon colors yellow over a wide angle (see fig.(2.13)). This is called the twilight arch. As the sun is just below the horizon, no direct sun light can reach the observer. The sky above the horizon, however, is still illuminated by the sun. The observer will see this light indirectly, i.e. via the scattering. The light path that the sun light has to travel through the atmosphere before illuminating the sky above the observers horizon is quite long. So, most of the blue light will be scattered

22 20 CHAPTER 2. SKY LIGHT AND ALIKE before it reaches that part of the atmosphere (see fig.(2.14)). Figure 2.13: Twilight arch, just before sun rise (from [1]). This means that the scattered light is mostly yellow and the twilight arch is yellow. As the sun sets more, the path through the upper sky above the horizon gets longer and thus the twilight arch turns redder and redder. At the same time the portion of the visible sky that is illuminated by the sun gets smaller and smaller: it gets dark. Notice in fig.(2.13) that the twilight arch is brightest just above the horizon and gets redder when moving upward. This also is a consequence of the length of the light path to the illuminate sky as well as the density of air molecules that scatter the light. Figure 2.14: Light path length to the illuminated sky above the horizon increases as the sun sets more below the horizon. Furthermore, a smaller portion of the upper atmosphere can be seen by the observer. Due to symmetry, the twilight arch can of course also be seen just before sun rise. The part of the horizon opposite to the sun will be dark just before sun rise or just after sun set. This is caused by the shadow of the earth itself, that blocks the sun from illuminating the sky just above that part of the horizon. So, we will see a dark band just above the horizon. This is known as the antitwilight arch. Purple Light When the sun is about 5 below the horizon, the sky some 45 above the sun s position may show a purplish glow, see fig.(2.15). This is called the purple light. It is most likely caused by the scattering of light in the stratosphere (about 20-25km high) by dust particles. Most of this light that reaches the observer is blue. However, it gets mixed by light that is coming from the lower atmosphere. This light is red, as discussed above. So, the observer sees a mixture of red and blue, i.e. purple. The

23 2.5. FLATTENING OF THE LOW SUN AND MOON 21 different light paths are shown in fig.(2.16). The particle number density of dust (i.e. the number of particles per unit volume) in the stratosphere varies from time to time. Volcanic eruptions add dust to the stratosphere, and so do meteorites from space. This causes the occurrence and strength of the purple light to change. Figure 2.15: Purple light above the twilight arch (from [1]). Figure 2.16: Different paths of sun light to the observer lead to a mixture of red light that follows a path through the lower atmosphere and blue light scattered from the stratosphere. 2.5 Flattening of the low sun and moon If the sun gets very close to the horizon, its shape seems to change from a sphere to an ellipse which has a height to width ratio of less than one. This is caused by refraction of sun light in the atmosphere. The refraction index, n, of the atmosphere is not a constant, but depends on the density and on the type of molecules. In vacuum (e.g. outer space) its value is 1. In the atmosphere it gradually increases from a value of one at the edge of the atmosphere towards a value of about at ground level (it depends on the wavelength and is slightly higher for the blue end of the visible

24 22 CHAPTER 2. SKY LIGHT AND ALIKE spectrum than for the red end). The change might look small but the effects this has are not negligible. For instance, when the sun is seen to touch the horizon it actually is below the horizon! This can be easily understood if we think about the path of light beams through a medium that is increasing its refraction index along the beam. When a light beam passes an interface separating two layers with different refraction indices, we have according to Snellius law sin α i sin α o = n o n i (2.19) with α i and α o the angle of the incoming and outgoing light with the normal to the interface at the point of intersection and n i and n o the refraction indices at the incoming and outgoing side of the interface, respectively. This is illustrated in fig.(2.17). Figure 2.17: Refraction of a light beam at an interface separating two layers of different index of refraction. As the refraction index of the atmosphere will change continuously, we will have to set up a differential equation that describes the changes of the angles in Snellius law from place to place. This is complicated by the fact that the atmosphere is folded around the earth, hence its geometry is curved as well. Let s for simplicity assume, that the atmosphere is build of concentric layers, each with a thickness r and a index of refraction n(r). A light beam will hit the outer surface at position r at an angle α i and move on with an angle α o. These angles are defined with respect to the normal at the interface at position r. As the beam follows its path, it will hit the inner surface of the same layer, which is located at position r r. Even though the path of the beam in this layer is a straight line, it will not hit the interface at r r with the same angle α o as it has left the interface at r. This is caused by the curvature of the layer. In a flat geometry this would be the case and the analysis would be relatively easy. The two can be seen in fig.(2.18). For the curved situation we still have Snellius law at each interface, but it is not true that α o = α i. The light beam intersects the inner surface at a position where the normal is rotated over a small angle compared to the normal at the intersection point with the outer surface. The relation between the two angles is found from fig.(2.19): α i = α o + φ (2.20) We can use that r is small, thus φ is small and thus the curvature of the line segment l can be neglected. Therefore, it holds that:

25 2.5. FLATTENING OF THE LOW SUN AND MOON 23 Figure 2.18: Path of light through a medium with a changing index of refraction: (a) flat geometry, (b) curved geometry. tan φ = tanα o l r r = l r (2.21) Figure 2.19: The two normals make a small angle due to the curved geometry. Thus, eliminating l we have a relation between α o and φ: tan φ tanα o = r r r = r r + h.o.t. φ = tanα o r r (2.22) So, in principle, we have now connected α i to α o (via Snellius law) and α o to α i and r. At this stage it is important to realise that α i = α i (r r). Thus, we can write eq.(2.20), using the above equation, as: α i (r r) = α o + tanα o r r (2.23)

26 24 CHAPTER 2. SKY LIGHT AND ALIKE We can expand: α i (r r) = α i (r) dα i r and take the sinus of both sides of dr the above equation: ( sin α i dα ) ( ) i dr r r = sin α o + tanα o r (2.24) Remember that this equation gives us the change of the direction of the light beam. If we use the summation rule for the sinus: sin(α +β) = sin(α) cos(β)+cos(α) sin(β) we can write: ( sin α i dα ) i dr r ( ) ( ) dαi = sin(α i ) cos dr r dαi cos(α i ) sin dr r = ( ) r = sin(α o ) cos tanα o + r ( ) r cos(α o ) sin tanα o r (2.25) Next, we will simplify this equation by using that r is small. For small x we have sin(x) = x and cos(x) = 1. Thus, the above equation can be written as: ( ) dαi sin(α i ) cos(α i ) dr r ( ) r = sin(α o ) + cos(α o ) tanα o r = sin(α o ) + sin(α o ) r r Now we are ready to eliminate α o by using Snellius law: (2.26) sin α o = n i n o sin α i = = = n(r) n(r r) sin α i n(r) n(r) dn r sin α i dr ( ) dn n dr r sin α i (2.27) Finally, if we now combine eq.(2.26) and eq.(2.27) we have the differential equation that describes the evaluation of the angle α i of the light beam we were looking for: cos α i dα ( i 1 dr = dn n dr + 1 ) sin α i (2.28) r The above equation is easily solved. We use as boundary condition that at the edge of the atmosphere the refraction index equals 1 and the incoming angle depends on what light beam we are trying to trace. It may be from the sun or any star. So, we have at r = R + H (R denotes the earth radius and H the thickness of the atmosphere): n = 1 and we will call the incoming angle α in. With this the solution is:

27 2.5. FLATTENING OF THE LOW SUN AND MOON 25 sin α i = 1 R + H sin α in n(r) r (2.29) Although we now know the angle a light beam makes with each normal on its way from space to the observer (at e.g. ground level), we still haven t solved everything. We don t know the direction of all those normals, as they depend on where the light beam is in the atmosphere. What needs to be done is reconstruct the entire beam path and look for that particular beam from the sun, whose path will be such that it reaches the observer. As the sun is much bigger than the earth, half of the atmosphere is always hit by the sun s light beams. These are, due to the large distance from sun to earth, to a good approximation parallel to one another. However, their incoming angle changes with the position at which they enter the atmosphere, as the normal at the point of incidence changes from point to point. We can describe the light path better in cylinder coordinates {r,φ}, where we define φ = 0 as the direction of the vertical (= zenith). Further, our observer is at ground level and thus has coordinates {R,0}. The relation between φ and r is already given in eq.(2.22): dφ dr = tanα o r Furthermore, α o is coupled to α i via Snellius law: (2.30) sin α i sin α o = n(r r) n(r) = 1 1 n dn r (2.31) dr Thus we have: α o = α i + O( r). Putting this into eq.(2.30) gives us the relation between φ, r and α i (r). The latter is a known function of r and n(r), i.e. eq.(2.29). Putting it all together gives: dφ dr = tanα i r = 1 sin α i r 1 sin 2 α i = R + H nr 2 sin α in ( 1 sin 2 α R+H ) (2.32) 2 in nr The last equation does look complicated, but it can be simplified by introducing the coordinate s r : sin α in (R+H) dφ ds = 1 s s 2 n 2 1 (2.33) This equation describes the trajectory of a light beam through the atmosphere. It is now a good point to test whether or not our description makes any sense. We will analyze the situation in which the refraction index of the atmosphere is 1, i.e. what an observer would experience on the moon. Furthermore, we will assume that the light beams enters the atmosphere parallel to the horizon. Obviously, we expect

28 26 CHAPTER 2. SKY LIGHT AND ALIKE Figure 2.20: Straight path of light through an atmosphere with a refraction index of 1. that the light beam will continue on a straight line, as there is nothing to deflect it from its path. This situation is drawn in fig.(2.20). From simple geometrical considerations we see that the straight line of the beam is described by: r = (R + H) sin α in cosφ = (R + H) cos φ 0 cos φ (2.34) Let s see whether eq.(2.33) gives the same result, when taking n(r) = 1. Thus, we have to solve dφ ds = 1 s s 2 1 The general solution of this equation is: (2.35) φ = arccos 1 s + Const 1 s = cos (φ C) (2.36) For the boundary condition, we take again a look at fig.(2.20). The beam enters at {r in = R+H,φ 0 }, making an angle α in with the normal at that point. From fig.(2.20) r it is clear that sinα in = cos φ 0. Thus we have that s 0 in = 1 sin α in (R+H) sin α in = 1 cos φ 0. If we put this boundary condition into eq.(2.36) it is found that the integration constant C equals 2kπ. Thus the solution for s(φ) is: s = 1 cosφ r = (R + H) sin α in cos φ (2.37) which is exactly the same result as we have found above from the geometrical consideration! Feeling confident with our solution, we now can turn to solving the path of a light beam in the atmosphere of the earth. For simplicity we assume that the refractive index of the atmosphere changes linearly from 1 at the outer edge to at ground level: n(r) = R+H r. It is easy to see, that light rays coming directly H out of the zenith are not bend but continue on a straight line. We expect the biggest changes when light enters close to the horizon. Thus, we will analyse what the apparent position of the sun is when the upper edge of the sun just touches the horizon;

29 2.5. FLATTENING OF THE LOW SUN AND MOON 27 the sun has just set. That is to say, it would do so if there was no atmosphere. We have now ask the question: which of all the sun beams that come in parallel to the horizon and enter the atmosphere over its entire height is seen by the observer and under what angle with the horizon? From a numerical solution of eq.(2.33) we find that this angle is Hence, the observer still sees a large part of the sun above the horizon! In reality, the apparent lift of the sun s position close to the horizon is about 39 arc minutes. As the diameter of the sun is only 30 arc minutes, the sun is actually completely below the horizon when we see it touching it with its lower edge. This also means that the apparent motion of the sun slows down, when the sun approaches the horizon. This is a simple way of showing that the earth is a curved object rather than a flat one, as for flat geometries the apparent motion is different than for curved ones. Furthermore, the bending of the light paths is stronger for beams that enter the atmosphere closer to the horizon than those higher up. Consequently, the shape of the sun or moon gets distorted when they approach the horizon. The lower edge gets lifted up more than the upper edge: the sun (or moon) start to look ellipsoidal. as mentioned, normally the sun s diameter is 30, but close to the horizon it looks like its vertical dimension has shrunken to 24. The sun s portrait is no longer a circle but it has become ellipsoidal.

30 28 CHAPTER 2. SKY LIGHT AND ALIKE

31 Chapter 3 Water and light 3.1 The color of water In the previous chapter we focussed on atmospheric phenomena, especially those that had to do with light. Now we will discuss all kind of phenomena associated with water and light. Water is one of the most common substances to us. Life as we know it is unthinkable without water. We use water in enormous quantities: for drinking, cooking, washing and cleaning, for cooling or heating, for transportation, even for pleasure. Water can be beautiful, peaceful but also devastating. When we look at water, we actually look at various light sources. The light we see can come from reflections at the surface: we look at an image of what is behind the water. It may also come from the bottom of the water pool. Then the light has travelled from outside to the bottom and back. This light presents a filtered and distorted image from what is behind the water. But light can of course also be scattered from within the water: it may by the water molecules themselves or by particles floating in the water. The latter gives muddy waters their color. Figure 3.1: Transmission (T) of water for the visible light spectrum for various thicknesses of water (from [1]). Like the atmosphere, water has its own transmission characteristics for visible 29

32 30 CHAPTER 3. WATER AND LIGHT light. A thin layer of water looks perfectly transparent, but if we increase the thickness we will see that water absorbs more of the red part of the visible spectrum than of the blue one. The transmission of water is shown in fig.(3.1). It is clear from this figure that the color of water is blue. This knowledge is used in swimming pools, where the tiles are white or blue, to enhance the water color. The color of water can be seen in an ice cave. An example is given in fig.(3.2). Figure 3.2: Blue color of an ice cave. The light seen is light making its way through the ice. This causes the yellow and red to be absorbed (from [1]). 3.2 Glitter In most cases when one looks at water, the intrinsic blue color is hidden. One of the most important reasons for that is reflection of light at the water surface. Especially at low sun on clear, calm days, the ocean (or any large lake) is sparkling. Into the direction of the sun, a very bright strip of water is seen, reflecting sun light directly to us. The illumination of this strip seems to be fluctuating as we see a glittering path rather than one with a uniform brightness. This is called the glitter. Fig.(3.3) illustrates this. Obviously, the illumination of the water by the sun is uniform. The glitter is caused by waves on the water surface, in many cases induced by a small breeze. Let s assume for simplicity that the water surface is sinusoidal. The waves are moving into our direction, with wave fronts perpendicular to their direction of motion. Furthermore, we assume that the sun is seen in the direction opposite to the wave motion. So, sun light is approaching and hitting the waves as shown in fig.(3.4).

33 3.2. GLITTER 31 Figure 3.3: Glitter: bright strip on a water surface directed towards the sun (from [1]). Figure 3.4: Reflection of parallel sun beams on a wavy water surface. (a) high sun, (b) low sun). An observer at the shore will see reflected sun light only if it is send precisely into its direction. Thus is obvious, but a consequence is that of the wavy surface only a fraction can contribute: of each wave only those parts with the right orientation of the surface will reflect light into the direction of the observers eye. Furthermore, the observer must be able to see this part: some parts of the wave are hidden by the previous crest. Both these are schematically shown in fig.(3.5).

34 32 CHAPTER 3. WATER AND LIGHT Figure 3.5: Direction of reflected light must be towards the observer and the point of reflection should not be in the shadow of the previous crest. It is easy to see that only those parts of the wave facing the observer or those just beyond a crest contribute to the light visible (except for a vary high view point of the observer). As the shadowing will further create gaps in the reflecting image, we see that the glitter should be organized in dark and light bands perpendicular to the line of sight. Furthermore, a low sun with not too high waves is preferred. It is relatively simple to make a model of the glitter, as all that needs to be done is tracing the portions of the wavy surface that can contribute. 3.3 Caustic network When we look into an outdoor pool at a sunny day we see a network of constantly moving twisting lines of light. These are called the caustic network. They have the same origin as the glitter. But here, light is refracted into the water by the wavy water surface. The waves now focus the light into thin lines. And, as the waves move irregularly through one another, so do the lines. The light paths are shown schematically in fig.(3.6). Figure 3.6: Creation of the caustic network by focussing of light by the surface waves Notice that in this figure also reflections are shown. This causes the light to form a similar caustic network on e.g. sailing ships in the harbor. 3.4 The rainbow Rainbows are another example of the interplay between light and water. Here, water droplets present in the atmosphere are refracting light. As the refractive index is a

35 3.4. THE RAINBOW 33 (weak) function of the wave length of light, different colors get refracted over different angles. The rainbow is obviously only seen with the sun in the back. This is so, because the rain drops in the atmosphere internally refract the light back and sent it back. The paths of light are shown in fig.(3.7). Figure 3.7: Light rays (of a single wave length) refracted through a water droplet. It is easy to show that for a spherical droplet the maximum angle between the incoming and outgoing light is 42. Consider fig.(3.8). Light, coming in, is making an angle α in with the normal on the droplet surface. According to Snellius law we have that the light beam is refracted towards the droplets normal, making an angle α out. It hits the droplet surface from the inside at an angle φ with the normal. Part of the light is reflected back into the droplet, obviously also at an angle φ with the normal. Then, it reaches the surface again, now at an angle α in with the normal and is refracted out of the droplet with an angle α out. The angle describing the total deviation is called α. Figure 3.8: Geometry of a refracted light beam through a water droplet. From fig.(3.8) it is clear, that the two triangles, IMP and OMP, within the droplet are equivalent, hence α out = α in. Consequently, α in = α out. Furthermore, φ = α out as the sides IM and MP in the triangle IMP are equal. So, for the total deviation of the light beam path we find: α = 4α out 2α ( in ) sin αin = 4 arcsin 2α in (3.1) n

36 34 CHAPTER 3. WATER AND LIGHT The incoming angle, α in, varies from 0 to π. The deviation, α, shows a maximum 2 at for n= This is the value for the refraction index for (red) light with a wave length of 600nm. In fig.(3.9) the deviation angle is shown. Figure 3.9: Deviation of light beam that is refracted and reflected by a spherical water droplet: (a) as a function of the incoming angle α in, (b) as a function of x/r. As the figure shows, there is region for x/r between say 0.8 and 0.9 where all the incoming light is focussed in a small region of the deviation angle. This obviously means that this portion of light is seen with a much greater intensity by the observer. This light forms the rainbow. The maximum deviation angle is depending on the refraction index, which in turn is a (weak) function of the wave length. This dependence is plotted in fig.(3.10). Figure 3.10: Maximum deviation against wave length. Thus, we see that the colors of the rainbow run from red at the outside to blue at the inside. Notice further, that the separation between the colors blue and red is 1.5. Considering that the effective diameter of the sun is less than 0.5 we find that this separation is sufficient to have a rainbow with separate colors, a fact we obviously new.

37 3.4. THE RAINBOW 35 Secondary rainbow The rainbow discussed above is caused by light that is reflected once inside the droplet. There is, however, the possibility of multiple internal reflections. These generate different deviation angles and thus more rainbows can be formed simultaneously. Under the right circumstances, a secondary rainbow can be seen. The light path through a droplet for a secondary rainbow is shown in fig.(3.11). Figure 3.11: Geometry of a refracted light beam that forms the secondary rainbow. Now there are two internal reflections and of course the beam is refracted twice. As shown in the figure, all internal angles are equal. The deviation angle, α s, is connected to the incoming angle, α in : ( ) sin αin α s = π 6 arcsin + 2α in (3.2) n This is plotted in fig.(3.12), together with the result we had for the primary rainbow both for a wave length of 600nm. Figure 3.12: Deviation of light beam that is refracted and reflected by a spherical water droplet. There are a few points to be noticed in this figure.

38 36 CHAPTER 3. WATER AND LIGHT The deviation shows a minimum, rather than a maximum, at α = 51. There is a forbidden band in the deviation angles from about 42 to 51 The second point means that there is a dark band between the primary and secondary rainbow, where no light refracted by droplets is sent into the direction of the observer. This is known as Alexander s band, in honor of the Greek Alexander of Aphrodisias, who first pointed out that the area between the two red bands of the primary and secondary bows is darkened rather than brightened. In fig.(3.13) the dependence of the deviation angle with the wave length of the light is plotted for both the primary and the secondary rainbow. Figure 3.13: Maximum deviation against wave length. We see, that the secondary rainbow is reverted with respect to the primary one: now the colors run from red at the center towards blue at the outside. Of course, the intensity of the primary rainbow is greater than that of the secondary. This has two main reasons. On the first place, the light intensity decreases after each (internal) reflection. Secondly, the secondary rainbow is broader than the primary: the focussing of light along the minimum angle is less. Of course, tertiary and higher rainbows are present each with one more internal reflection than the previous one. Obviously, each higher order rainbow gets fainter and broader and therefore becomes less visible. Very few reliable reports exist about the observations of the tertiary and quaternary rainbows. Higher orders are not observed. In table(3.1) for the first 5 orders the mean angle, width and intensity relative to the primary are listed.

39 3.4. THE RAINBOW 37 order angle width rel. intensity ( ) ( ) Table 3.1: Primary and higher order rainbows

40 38 CHAPTER 3. WATER AND LIGHT

41 Chapter 4 Bubbles 4.1 The sound of mountain streams Every small stream that is flowing not too smooth but also not too wild seems to make the same sound, a characteristic kind of babbling, whispering. Where does this come from? Why do large rivers in low lands not make the same noise? The sound is caused by small bubbles that are captured in the streams for a while before they escape again. These bubbles are air trapped by fluctuations of the water surface or tiny water falls. In order to understand this we have to think about bubbles as a source of sound. Why or how do bubbles in water generate sound? If the pressure inside the bubble is in equilibrium with its surrounding and no perturbation is present, no sound will be generated. However, the bubble-water system can be seen as a mass-spring system. If the spring is not at its equilibrium length, it will exert a force on the mass and cause the mass to move. Obviously, due to inertia the mass will continue to move even if the spring is at its equilibrium position and thus an oscillation sets in. The spring is represented by the pressure of the gas; the inertia is the water surrounding the bubble. Then what is the velocity? It is not the center of gravity of the bubble that is oscillating. It is the size of the bubble that does so. Minneart has in 1933 studied this effect and tried experimentally to find the frequency of the oscillating bubbles by comparing with the sound of a tuning fork (see [3]). Figure 4.1: Oscillation of a mass m coupled to a spring. The mass-spring system (fig.(4.1)) can be analysed on the basis of Newton s law: F = m a or rephrased in terms of momentum: dp = F. The force due to the spring dt is: F = k(x l 0 ) with l 0 the equilibrium length of the spring. Thus, using Newton s law we have with p = mv = m dx: dt m dv dt = k(x l 0) m d2 x dt 2 + k(x l 0) = 0 (4.1) 39

42 40 CHAPTER 4. BUBBLES As we are interested in the motion of the mass around its equilibrium position, it is natural to introduce x x l 0, the displacement from the equilibrium position. Thus, we can simplify eq.(4.1): d 2 x + k x = 0 (4.2) dt 2 m This is the well-known wave equation with as general solution: x(t) = A sin ωt+ B cosωt. The oscillation frequency ω is given by: ω 2 = k m. What we need to do now, is derive an equation for the oscillating bubble that looks the same. If we have done so, we will have the oscillation frequency as well. Let s first think about the momentum in our bubble-water system. We put the center of the bubble in the origin of our coordinate system. Further, we assume that the bubble radius, R, oscillates around an equilibrium value R 0. Due to this oscillation both the gas inside the bubble and the water surrounding the bubble will oscillate. It is easily seen that the momentum of the gas is negligible in comparison with that of the water as ρ air ρ water The momentum of the liquid can be calculated if we know the liquid velocity. This is, however, changing from place to place, since we can safely assume, that the liquid at infinity will not notice the presence of a tiny bubble. The velocity field is obtained by realising that the problem is spherical symmetric: in spherical coordinates, the velocity has only a radial component and no quantity can depend on the coordinates φ and θ. Next, we can use that water is incompressible. Consider the water present between two spheres, one of radius R 1, the other of R 2 (see fig.(4.2)). Figure 4.2: The amount of water between R 1 < r < R 2 is constant as water is incompressible. As water is incompressible, we have that the total amount, or better total mass, of water enclosed by the two spheres is constant. Thus what flows in at r = R 1 must flow out at r = R 2. But we can use the symmetry: the velocity of the water has only a radial component, which may only depend on the radial coordinate: v r (r). Into our volume R 1 < r < R 2 flows water with a velocity v r (R 1 ), the area through which it flows is 4πR 2 1. Thus the inflow is ρ water 4πR 2 1v r (R 1 ). Similar we find that at

43 4.1. THE SOUND OF MOUNTAIN STREAMS 41 r = R 2 the outflow of water is: ρ water 4πR 2 2v r (R 2 ). Since the inflow must be equal to the outflow we have: ρ water 4πR 2 1 v r (R 1 ) = ρ water 4πR 2 2 v r (R 2 ) = const (4.3) As the choice of R 1 and R 2 is arbitrary, we find that the total flow through any surface of a sphere with radius r must be the same constant. So, we found that v r (r) 1. Furthermore, the motion of the water is coupled to the oscillation of the r 2 bubble radius. We will call the velocity at the bubble surface (r = R): v R. Thus we have for the velocity at any point in the water that it has only a radial component which is equal to: v r (r) = R2 r 2 v R (4.4) Now we could calculate the total momentum of the liquid. Take the mass of water between two concentric spheres of radius r and r + dr, respectively. This mass is dm = ρ water 4πr 2 dr and it has a velocity of v r (r) in the radial direction. So we compute the momentum of the entire water mass as P = = R R ρ water 4πr 2 v r (r)dr ρ water 4πr 2 R2 r 2 v Rdr = ρ water 4πR 2 v R dr (4.5) Now, we have a problem: the integral of the last equation is divergent! What did we do wrong and how can we cure it? Well, we have been a bit sloppy with the momentum. It is actually a vector quantity. But in eq.(4.5) we have been summing the contribution of all water elements without paying attention to the direction of the momentum of each small piece of water mass. And as the velocity has only a radial component, so does the momentum. This means that from symmetry we know that the total momentum is zero! Of course, the same will hold for the force exerted by the pressure inside the bubble on the water. A way out of this is to make use of the total kinetic energy of the water and than find the momentum via E kin = 1 2 mv2. Thus,we calculate the kinetic energy (which is a scalar, so no problems expected): R E kin = R 4πr ρ waterv 2 r(r)dr = 2πρ water v 2 R 4 R R r dr 2 = 2πρ water vrr 2 3 (4.6) So, if we now take P = mv = de kin dv we find for the momentum of the water: P = 4πρ water R 3 v R (4.7)

44 42 CHAPTER 4. BUBBLES The force exerted by the bubble on the liquid is: F = (p p 0 ) 4πR 2 (4.8) in which p is the pressure in the bubble and p 0 the equilibrium pressure when the bubble has radius R 0. Applying Newton s law thus gives: d 4πρ water dt R3 v R = (p p 0 )4πR 2 (4.9) What we need to do still, is to couple the velocity of the bubble-water interface, v R, to the bubble radius. That is trivial as v R dr. Furthermore, we need to do dt the same for the pressure in the bubble. Obviously, the pressure is related to the bubble volume by the ideal gas law. However, we need to make an assumption for the temperature. Is the process isothermal or adiabatic or what else? We will assume that the oscillation is fast enough to assume an adiabatic process. Then, we can use (as we have done before): pv γ = p 0 V γ 0. This couples the pressure to the radius of the bubble. The final step is assuming that the amplitude of the oscillation is small, so we may neglect higher order terms. We write formally: R(t) = R 0 + R(t), with R/R 0 1. Now we can express the pressure as a function of the radius: p p 0 = = ( ) γ V0 V ( R0 3 (R 0 + R) 3 ) γ 1 3γ R R 0 (4.10) R This means: p p 0 3γp 0 R 0 in other words the pressure difference is linear in the small radius-variation. So for evaluating the force at the right hand side of eq.(4.9), we need to take for the surface area 4πR0. 2 The higher order terms may safely be neglected. At the left hand side of eq.(4.9) we have to take the time derivative of R 3 v R. This gives two contributions, namely 3R 2 dr v R and R 3 dv R dt dt. As we are looking for oscillations with small amplitude, we can safely assume that R will be of the form: R = A sin ωt (4.11) where the amplitude A is small with respect to R 0. So, the velocity v R, which is equal to dr will be proportional to A and thus vary linearly with R. Consequently, dt the term 3R 2 dr v R is quadratic in R and can be neglected with respect to R 3 dv R dt dt which is linear in R: R 3 dv R dt R0 3 d2 R. If we put this all together, we find for dt 2 eq.(4.9): d 2 R dt + 3γp 0 R = 0 (4.12) ρ water R0 2 Thus, we have the type of equation similar to the mass-spring we were looking for. The solution of the variation of the bubble radius is:

45 4.2. THE SOUND OF BOILING WATER 43 ( ) 3γp 0 R = A sin t ρ water R0 2 The oscillation frequency is equal to f = ω 2π = 1 3γp 0 2π ρ water R0 2 (4.13) (4.14) For a bubble with a radius of 1mm, we find: f = 3300Hz, which is in the range of what we here from the streams. 4.2 The sound of boiling water Water in a whistling kettle, heated on a stove has its own characteristic sounds. Before it really boils the first sound can be heard. The kettle seems to click and hiss. Hardly any steam is seen to escape from the kettle. If we wait longer the hiss gets louder. After a while, the water in the kettle is fully boiling and the sound made seems to be softer, less intense. What all is happening here? Well, the process of getting water to boil is perhaps more complicated then you might think. Of course, the actions taken to put water in the kettle and subsequently that one the stove are simple enough. But what is happening to the water inside the kettle? The bottom of the kettle will transfer heat to the water, which, obviously, will get warmer. But the temperature of the outside of the bottom of the kettle will be pretty high, in any case much higher than the water temperature itself. After a while the mean temperature of the water will be close to 100. The inside of the kettle bottom is then above 100. So, locally the temperature of the liquid will also rise above 100. As this is the boiling temperature (at a pressure of 1 bar) the first small vapor bubbles will be formed. These are created at specific spots at the bottom plate, where a small roughness acts as a nucleus where bubbles can grow on. Similar creation of bubbles from certain spots can be seen in a fresh glass of mineral water or beer. Here, usually a train of bubbles is formed. The creation of the bubbles is accompanied with the clicking sound Surface tension The bubbles in the water are super saturated, that is they are formed in a very thin layer of water that has a water temperature well above 100. Why are the bubbles formed first in that layer only? The reason for this super saturation is the higher pressure in a bubble in comparison with its surrounding due to the surface tension. The surface tension can be seen as a kind of film around a bubble or droplet that has elastic properties. If one tries to stretch it, it will resist. More precisely, if one tries to increase the area of the film one has to supply extra energy to the system, proportional to the increase in surface area. Moreover, if the surface is bend, it will exert a force that will try to straighten it. So, for a soap bubble, the soap film will act like the rubber of a balloon. The pressure inside the soap bubble is, like that in the balloon, somewhat higher than the pressure in the surrounding air. The extra pressure due to surface tension is given (for a spherical object) by Laplace s equation:

46 44 CHAPTER 4. BUBBLES p = 2σ R (4.15) with p the pressure difference between inside and outside of the object of radius R; σ is the coefficient of surface tension, or short: surface tension. It is a property of the two materials forming the surface. How thick is the film of a soap bubble? For the soap bubble, the surface tension is on the order of Pa.s. Thus, the overpressure in a soap bubble of 10cm is only 2Pa! Compared to the pressure of the surrounding air, this is negligible. The pressure inside and outside the soap bubble are thus virtually the same. Why do most soap bubbles first rise and then come down again (if they live long enough)? Obviously, the soap film around the bubble adds to the weight of the bubble. Nevertheless, the buoyancy force is initially greater than the force of gravity that tries to pull the bubble towards the ground. This is caused by the higher temperature of the air inside the soap bubble. This air is warmed by our longs. Let s use this to estimate the thickness of the soap film. We take a bubble with a diameter of 5cm. Further, we assume that the temperature of the air inside the bubble is half way between our body temperature and the surrounding air temperature. It is partly cooled down already and partly mixed with surrounding air when we blew it out of our mouth into the soap bubble. As we have seen, we can neglect safely any overpressure due to surface tension. Thus the buoyancy force is ρ air (T)g π 6 D3, gravity s force is mg. The mass of the bubble is ρ air (T + T) π 6 D3 + ρ soap πd 2 δ, in which δ is the thickness of the soap film. Thus, for a floating bubble we find: ρ air (T)g π 6 D3 ( ρ air (T + T) π 6 D3 + ρ soap πd 2 δ ) = 0 δ = 1 ρ soap ρ T T D 6 (4.16) In the above equation we have used that T is much smaller than T. For air we can use the ideal gas law to calculate ρ = ρ. Thus, for a soap bubble T T of D = 5cm, with T = 300K, T = 10K, ρ air = 1.2kg/m 3 and ρ soap = 10 3 kg/m 3, we find: δ = m. Let s return to the boiling water. Here the effect of surface tension is much more important. The bubbles formed are very small, so the overpressure will be relatively large. If we take the initial size as 0.01mm, we can calculate p = Pa (taking σ = 70mPas). From the steam tables we can find, that at a pressure of 1.15bar the saturation temperature of liquid water and water vapor is equal to T sat = 103 C. Thus, the bubbles formed are hotter than most of the water in the kettle. If they detach from the kettle bottom and rise upwards they will be surrounded by water that is colder. Thus, the bubbles will lose heat and collapse as they are now in an environment that is below the saturation temperature, even at the local pressure. This collapse is heard as the hiss. Of course, due to the collapse of the bubbles no bubbles reach the free liquid surface in the kettle and thus hardly any steam is produced.

47 4.3. THE SOUND OF A FRESH, HOT CUP OF INSTANT COFFEE 45 As the stove heats the water further, a point is passed at which the water temperature every where is at 100 C or above. Now boiling can be sustained everywhere and bubbles are formed continuously. Furthermore, they no longer collapse, but break through the free liquid surface. The sound becomes more continuous and softer. A splashing sound is heard that is due to the erupting bubbles throwing some liquid water up. Besides, steam is produced that leaves the kettle and might cause the kettle to whistle. The above also provides an explanation of the explosive situation that can arise when heating a cup of water in a micro wave. Here the water is heated from the inside due to the absorption of electro-magnetic waves by the water. So, the heating is rather uniform: no hot spots as compared to the heating in the kettle. Furthermore, the cup itself is heated indirectly and thus lags behind a bit in temperature. If now the wall of the cup is smooth, no nucleation spots for the generation of small vapor bubbles may be present. Thus, the water can reach a temperature above the saturation temperature at the given pressure: the entire liquid is super saturated. This is rather dangerous as now the introduction of nuclei (e.g. by adding sugar, salt or any other powder) will set the entire liquid into a vivid boiling. Due to the formation of vapor the mixture level will swell with the danger of overflowing. Furthermore, the sudden expansion will blow water out of the cup with the possibility of serious consequences. 4.3 The sound of a fresh, hot cup of instant coffee A very peculiar sound effect can be found in a fresh cup of instant coffee. If the bottom of the cup is tapped by a spoon a clear sound is heard. However, if the coffee is first stirred and then the bottom tapped, you will hear a lower pitch which is getting higher with time. After a while the pitch stays constant and of course the stirred coffee stops rotating. What is the reason of this changing pitch? Does warm water do the same? The answer to the last question is: no. So, changes in the distribution of the temperature in the liquid are irrelevant. One could have thought that perhaps in the unstirred case the temperature of the liquid is not uniform, whereas in the stirred case it is. But the warm water experiment shows that this aspect is not responsible for what we hear. Then, what is so special about the coffee? The answer is, that small air bubbles have been trapped into the powder mass. They are so small that they can not escape the coffee easily. The bubbles, over time, will collect close to the free surface of the coffee. When stirred these bubbles will be dispersed again over the entire coffee. If the stirring stops, the velocity of the coffee will decrease and the bubbles will again move upwards to the free surface. This suggest that the sound we hear is connected to the presence of the bubbles in the liquid. And indeed, it is. To understand why we have to consider waves in a liquid, as after all this is what the sound is. The tapping will send waves into the liquid. Due to resonance we will hear dominantly those waves that fit into the liquid. If the height of the liquid is H, we can estimate the frequency from: H = (k ) λ 2 λ = 4H 1 + 2k (4.17)

48 46 CHAPTER 4. BUBBLES with λ the wave length and k N. Furthermore, the frequency and wave length are coupled: f λ = c w (c w is the speed of sound in water). Thus we have that the sound we hear has a frequency f = c w λ = 1 + 2k c w 4 H f 0 = c w = 3750Hz (4.18) 4H where we have taken: c w = 1500m/s and H 10cm. This reasoning also holds for the coffee with the bubbles. Here we will also hear the standing waves. Thus, we conclude that the velocity of sound in a liquid with bubbles is different from that without! But why? To understand this we will have to look deeper into sound and what determines the speed of sound in a given material Speed of sound We will start our analysis with a vibrating string. A string is suspended between two fixed points and brought out of its equilibrium position (which is a straight line between the two fixed points, that we will call the x-axis). By doing so, the tension in the string is increased, resulting in a force that tries to restore the situation. Thus the string starts moving back to its equilibrium position. However, due to its inertia it will keep moving even if the equilibrium position is reached and the restoring force has vanished. Just like the pendulum, an oscillation has set in. And we will see, that for small amplitudes, the oscillation time is independent of the amplitude, again just like with the pendulum. To analyse the motion we will treat the string as consisting of N beads, each with a mass m, that are couple to one another by N + 1 identical, massless pieces of string, see fig.(4.3). Figure 4.3: Discrete representation of an oscillating string. Each bead feels the force of two string-parts. We need to set up Newton s law for each of the beads, so let s concentrate on bead i. Each bead is displaced vertically by a distance y i from its equilibrium position. The initial tension in the string is F. Due to the displacement, the strength might have changed a bit and the direction of the force on bead i is no longer parallel to the x-axis. The y-component of the force on bead i by string i 1 (see fig.(4.4)) is given by: F y,i 1 = (F + F)sinα i 1 ( = F + F ) l l yi y i 1 i 1 l + l i 1 = F l (y i y i 1 ) + h.o.t. (4.19)

49 4.3. THE SOUND OF A FRESH, HOT CUP OF INSTANT COFFEE 47 Figure 4.4: Three consecutive beads on the string. Similarly, for the y-component of the force by string i we find: F y,i = F l (y i+1 y i ) + h.o.t. (4.20) Thus for the motion of bead i in the y-direction we find: m d2 y i dt 2 = F l (y i 1 2y i + y i+1 ) (4.21) We have this equation for all beads i, with the restriction that y 0 = 0 and y N+1 = 0 as these are the fixed points of the bead string. The set of N equations for the N beads can be solved by realising that we are looking for small-amplitude oscillations around an equilibrium, thus the form y i (t) = A i sin ωt can be introduced. A complete analysis can be found in e.g. ( [4]). For us, it is more important to see what happens if the number of beads goes up and the length of each piece of string goes down: we turn the bead-string analysis into a continuous description that deals with the original string and its oscillations around an equilibrium position. More precisely, we take N and thus l 0 such that: lim N lim N N i=1 m i = M N l i = L (4.22) i=1 As a result, the set {y i } changes to a continuous function, y(t). The differences that appear in the right hand side of eq.(4.21) will change to derivatives with respect to the horizontal coordinate. This can be seen as follows. Consider the position of the consecutive beads, like in fig.(4.5). If we first consider y 2 y 1 in the limit l 0. Then we replace the discrete variables y 1,y 2 etc. by the continuous variable y(x 1 ),y(x 2 ) etc.:

50 48 CHAPTER 4. BUBBLES Figure 4.5: Three consecutive beads on the string. y 2 y 1 l = y(x 2) y(x 1 ) l = y(x 12 + l/2) y(x 12 l/2) l y x (x 12) for lim l 0 (4.23) where x 12 = x 1+x 2 2. Similar we have: y 3 y 2 l = y(x 3) y(x 2 ) l = y(x 32 + l/2) y(x 32 l/2) l y x (x 32) Finally, if we combine the last two equations we get: (4.24) y (x x 32) y (x x 12) l = y (x x 2 + l/2) y (x x 2 l/2) l 2 y x 2(x 2) (4.25) Summing up all these mathematical tricks and defining µ m l the equation for the y-component of the string as a function of {x,t} is: 2 y t = F 2 y (4.26) 2 µ x 2 This is the well-known wave equation, that describes oscillations or waves of a continuous medium, like our string. Solutions of the form y(x,t) = A sin (ωt kx) obey this equation, as is seen by inserting this into the wave equation. Notice, that ) 2 = F µ. The coefficient k is called. From the solution we see, that at every point x the string oscillates with the same frequency (and the same amplitude) but that the phase of the oscillation is different from point to point. the angular frequency ω and k are related via: ( ω k the wave number and is coupled to the wave length λ: k = 2π λ

51 4.3. THE SOUND OF A FRESH, HOT CUP OF INSTANT COFFEE 49 Or alternatively, at every time t the shape of the string is sinusoidal. These wave move along the string and their speed is given by c = ω as is easily seen by writing k the solutions as y(x,t) = A sin ( k [ x ωt]) = A sin (k [x ct]). Notice that thus the k speed c is given by: c 2 = F, an equation that has obviously two solutions c = ± F/µ µ showing that the waves can run to the left or to the right sound in air Sound in solids, liquids or gases is governed by a similar type of equation. Rather then the transversal waves we have analysed for the string, here we have to deal with longitudinal waves, i.e. the wavy motion around the equilibrium position is parallel to the direction of the wave propagation and no longer perpendicular. The analysis we will follow is given in ([2]). Consider a gas of given, constant temperature and pressure. Then also the density is constant. We focus on a small portion of the gas between {x,x + x}. Due to a small perturbation all gas particles move a distance χ(x, t) along the x-direction. This function χ is the displacement from the equilibrium and is similar to y(x,t) for the string. Thus the gas originally at x will move to x + χ(x,t) and that at x + x to x + x + χ(x + x,t). This is illustrated in fig.(4.6). Figure 4.6: Gas between x + x gets displaced due to a small perturbation. Due to the motion of the gas, the density is no longer the same. Originally the mass was between x,x + x, now it is between x + χ(x,t),x + x + χ(x + x,t). But since mass is a conserved quantity, the amount stays the same and we thus have, with ρ 0 the original, unperturbed density of the gas: ρ 0 x = (ρ 0 + ρ) [ x + χ(x + x,t) χ(x,t)] (4.27) If we simplify this equation, using χ(x + x,t) χ(x,t) χ x we find that the x change in density is given by (neglecting higher order terms, i.e. products in x and ρ): χ ρ = ρ 0 (4.28) x This is step 1. The next step is coupling this density change to a change in pressure. This is simple as the pressure is a function of the density and vice versa. Of course, the

52 50 CHAPTER 4. BUBBLES pressure is also a function of the temperature, as we( know ) from e.g. the ideal gas law. So, p = f(ρ, ) and we can formally write: p = ρ, without specifying what the other variables are that we have to keep constant when taking the derivative. This is step 2. The last step is setting up Newton s law for the motion of the gas that was between x,x + x. This gas moves because from the left side a pressure is pushing it to the right, whereas from the right side the pressure is trying to push it back. Sound is a manifestation of the unbalance in these two pressure forces. The situation is depicted in fig.(4.7). p ρ Figure 4.7: Unbalance in pressure on left and right side leads to motion. Here, we call the area perpendicular to the x-axis A. Then the total mass of the gas between {x,x+ x} is ρ 0 A x. The pressure on the left exerts a force +Ap(x,t) and that on the right Ap(x + x,t). Thus, when we use that the acceleration a is the second derivative of the displacement χ: a = 2 χ, Newton s equation of motion t 2 reads as: ρ 0 A x 2 χ t 2 If we simplify this equation we may write: ρ 0 2 χ t 2 = Ap(x,t) Ap(x + x,t) = A p x (4.29) x = p x = p ρ ρ x = ρ 0 p ρ 2 χ x 2 (4.30) Where, for the last equality we have used the outcome of step 1. So, we arrive at 2 χ t = p 2 ρ 2 χ (4.31) x 2 This is the form of the wave-equation we had before. Thus, we have the same type of solution: the density variations are like sin(ωt kx) and the speed of this wave, which is the speed of sound, is given by:

53 4.3. THE SOUND OF A FRESH, HOT CUP OF INSTANT COFFEE 51 c 2 s = ω k = p ρ (4.32) Speed of sound in air Now we can calculate the speed of sound in air, which is 341m/s at 1bar and room temperature. If we assume that air is an ideal gas and that sound is an isothermal phenomena, we have: with M the molar mass of air. Thus, we would find that c s = p = RT M ρ RT M ( ) p ρ T = RT M (4.33). Conclusion (1), the speed of sound in air is only depending on the absolute temperature! Conclusion (2) its value at 300K is 294m/s, which is obviously much too low! The reason is that sound is not an isothermal process, but adiabatic. The expansion and compression (i.e. changes in density!) are to fast for thermal processes. Thus, like we have done many times, the relation pv γ = const should be used, or written as a relation between the pressure, p, and the density, ρ, of the gas: p = cρ γ. If we use this relation in eq.(4.32) we get: c s = γ RT M For air γ = 1.4, thus at T = 20 C we calculate: c s = 340m/s. (4.34) Donald duck like voice After inhaling helium gas ones voice start to sound like that of Donald Duck. Why? Well, first of all the speed of sound in Helium gas is different from that in air. According to the equation above, we have for He-gas (γ = 5/3) at 20 C: c s = 1007 m/s! Much higher than in air due to the rather low molar mass M. But why does this influence the sound we make? This is because we form sounds by using our mouth as a kind of resonance-box. This means that we fix the wave length, λ, of the sound we make. The frequency we hear follows from: f = c s /λ. Consequently, when we inhale helium, the frequency is about three times as high if we speak in the same way. Furthermore, the amplitude of the overtones is different for the different gases. Therefore, the sound is Donald Duck like. After the tragedy of the Russian submarine Koersk, the eight o clock news showed Norwegian and British divers that had to stay in a pressurised cabin for several weeks in order to avoid dangerous complications due to decompression effect. When spoken to, the divers answered with a Donald Duck like voice. Most likely, their cabin is not only pressurised but the gas inside is no longer standard air, as the speed of sound and therefore the sound of the divers voices does not depend on the absolute value of the pressure! Return to the tapping of a cup of instant coffee We left the tapping of the coffee cup by realising that the sound we hear is caused by standing waves whose frequency is determined by the height of the liquid level in the cup and the speed of sound in the coffee. The presence of small bubbles in the

54 52 CHAPTER 4. BUBBLES coffee has an important influence on the speed of sound in the coffee. We can now understand why. When we were dealing with the speed of sound in air we found that this is given if we know the relation between the pressure and the density. That still holds for water with bubbles. However, nature has a surprise. Most people would guess that the speed of sound in a mixture of air bubbles and water is somewhere between the values of the speed of sound in air (340m/s) and in water (1500m/s). The reasoning would be something like: if the sound goes through the bubbles it will do so with a velocity of 340m/s. When it has passed through the bubble, it will continue its way through water at a speed of 1500m/s. Then it will propagate through a bubble and so on. Conclusion: the speed of sound is reduced with respect to that of water, but increased above that of air. However, this reasoning is in many cases wrong. It does not take into account, that the wave is a coherent motion of the system. So if the wave length is large, i.e. much larger than the bubble size and the bubble-bubble distance, the sound sees the bubbly mixture as a homogeneous liquid. Think of the sound waves as the stringbead system. The strings were responsible for the restoring force, the beads had the inertia to keep the oscillation going. The same holds here: water brings the inertia, but the bubbles can be compressed easily until finally their increasing pressure stops the water motion and tries to restore the equilibrium situation. We have to find the relation between a change in pressure and resulting change in the density of the bubbly mixture. Our starting point is fig.(4.8). What we need to calculate is c 2 = p ρ. But we may equally well calculate the inverse: ρ p Figure 4.8: Compression of a bubble/water mixture. We shall ignore the relative velocity between the gas bubbles and the liquid (or do the experiment in the space shuttle, were the influence of gravity is absent). The amount of gas present is quantified as the volume fraction of gas. That is the total volume of the gas bubbles divided by the total volume: α = Vgas. If we increase the V pressure by p, each of the phases will be compressed. However, they will not do so at the same rate: we may expect the gas phase to be much more compressed than the liquid phase. Thus we can expect that also α will change due to the compression. The changes in volume of each of the phases is given via the compressibility of each phase: V gas = V gas,0 + V gas p p V liq = V liq,0 + V liq p (4.35) p

55 4.3. THE SOUND OF A FRESH, HOT CUP OF INSTANT COFFEE 53 and thus for the total volume we have after the compression: [ Vgas V = V gas + V liq = V 0 + p + V ] liq p (4.36) p in which subscript 0 indicates the situation before the compression. Now we are able to calculate the change in the volume fraction of gas when the bubbly mixture is compressed: α V gas ( V gas,0 V gas V gas,0 V = ( [ V Vgas V 0 p V ( gas, V gas V 0 V gas,0 p p 1 p p ) + V liq p V gas = α 0 (1 + (1 α 0 ) V gas,0 ) ( ] p ) 1 1 [ Vgas V 0 p + V liq p p p (1 α 1 V liq 0) V liq,0 ] p p ) p ) (4.37) where we have neglected higher order terms in p. From the above equation we obtain dα dp : dα dp ( 1 ρ liq = α(1 α) ρ liq p 1 ) ρ gas ρ gas p (4.38) In this equation we have dropped the subscript 0 and we have used 1 V 1 V p ρ Now we are almost ready to calculate the speed of sound in a bubbly mixture. First we write down an expression for the mixture density, ρ, in terms of the densities of the gas and liquid: ρ. p ρ = M V From this equation we calculate ρ p : = M gas + M liq = V gasρ gas + V liq ρ liq V V = αρ gas + (1 α)ρ liq (4.39) ρ p = (1 α) ρ liq p + α ρ gas p + (ρ gas ρ liq ) α p (4.40) Finally, we plug into this equation the result we had for α, simplify and use the p definition of the speed of sound: c 2 = p or 1 = ρ. This gives us the speed of sound ρ c 2 p in our bubbly mixture: 1 (1 α) c = 2 [ 1 α c 2 liq ( 1 ρgas ρ liq )] + [ ( )] ρliq α 1 + (1 α) ρ gas 1 c 2 gas (4.41) In fig.(4.9) the mixture velocity is plotted as a function of the volume fraction of the gas bubbles (notice the log-scale on the vertical axis). It is clear from the graph,

56 54 CHAPTER 4. BUBBLES Figure 4.9: Speed of sound in an air-bubble/water mixture. Notice the log scale on the vertical axis. On the left vertical axis the speed of sound in water is given; on the right one that of air. that the speed of sound in a bubbly mixture drops dramatically below the speed of sound in air. Now, we have to keep in mind that our result is only valid for long wave as we discussed above. The volume fraction where the speed of sound of the mixture is equal to that of air is α 0.11%. For volume fractions larger than this value the velocity of sound is lower than in air. So, if we go back to the coffee problem, we can now understand that when bubbles are stirred into the coffee, the frequency we hear from our tapping must be lower than in the clear case as f 1. Furthermore, after the stirring has c stopped, the bubbles will gradually move to the surface of the coffee and thus the frequency we hear will rise gradually as well. This experiment is simple enough to do at home! 4.4 A bubble bath In many modern swimming pools a bubble pool or a bubbly section of the swimming pool can be found. The well-to-do might have a bubble bath at home. Anyone who has been sitting in a bubble bath knows the sensation : the bubble tickle and it feels cold. What is the reason of the latter? Usually, the water is warm (at least 30 C). Do they use cold air? And if so, why not preheat the air to the water temperature? Well, the answers are: the air is probably not very cold and heating up the air to the water temperature does not help. Why? Because the temperature of the bubbles is only partially controlled by the temperature at which they are injected into the water. If a bubble is formed and moves upwards through a pool of water two things happen that might influence the bubble temperature. On the one hand, the bubble expands when moving upwards as the hydrostatic pressure decreases. Thus, the bubble has to perform work against the hydrostatic pressure. This goes at the expense of the internal energy of the bubble, hence it cools. Secondly, the bubble is now surrounded by water and, if the bubble is not saturated with water vapor, water

57 4.4. A BUBBLE BATH 55 will evaporate into the bubble. The heat necessary for this will be supplied by the bubble and the water surrounding it Effect of bubble expansion Let s first make the cooling by expansion quantitative. When the bubble rises upwards a little, the hydrostatic pressure of its surroundings will change by an amount p. Thus, the bubble will expand by V. The effect this has on the energy of the bubble can be found via the first law of thermodynamics: du = dq pdv (4.42) If we assume, that the process is adiabatic (that gives us the biggest change of the internal energy of the bubble), we have dq = 0. The work done against the pressure is for a small change in volume obviously p 0 V, with p 0 the pressure at the starting point of the expansion. For an adiabatic process, we use, like we have done before: pv γ = const. Thus, the change in V is coupled to that in p as: p 0 V γ 0 = (p 0 p)(v 0 + V ) γ V = 1 γ V p 0 (4.43) p 0 where we have ignored higher order terms in both p p 0 and V V 0. For the internal energy we can write: U = nc V T, with n the number of moles and c V the specific heat, which for air is equal to 5 RT. If we combine the last two 2 equations we get (with γ = cp c V = 7): 5 T = 2 p T T 7 p 0 L = 2T p 7p 0 L (4.44) For a bubble in water, p is equal to ρ L wg: just the hydrostatic effect. So, we find for the temperature change when rising through 1 meter of water of 20 C: T = 8 C. This is substantial, but now we may not be so sure that the process is really adiabatic. A temperature difference of, on average, some 4 C between the water and the air in the bubble will result in a heat flow towards the bubble. How much heat will, given this temperature difference, go from the water to the bubble? Is this little or much compared to the change in internal energy we have calculated? The heat flow from the water to the bubble is proportional to the temperature difference between water and air. Furthermore, the heat flow is proportional to the surface of the bubble. This is logical: the larger the surface across which the driving temperature difference is found, the more heat will flow across it. The proportionality constant is called the heat transfer coefficient, h. For a bubble with a diameter of 3mm, rising at its terminal velocity of 25cm/s upwards through the water, h has a value of W. It takes the bubble 4 seconds to flow 1 meter up through the m 2 K water. Thus for the heat flow rate, φ q, into the bubble we have as an estimate: φ q = ha T 0.8W (4.45)

58 56 CHAPTER 4. BUBBLES Hence, in 4 seconds an amount of heat equal to 3.2J will be put into the bubble. Of course, this is only so if the bubble would lose at least this amount due to the expansion. The change in internal energy is: U = nc V T = J! So, most likely the process of expansion is far from adiabatic. A much better assumption is that it is isothermal. Conclusion: expansion can not be responsible for the cold sensation in the bubble bath Effect of evaporation The air used is most likely not saturated with water. Thus when the air bubble is formed and starts moving upwards, water will evaporate into the bubble until saturation is reached. The evaporation requires heat, which is initially supplied by the bubble. Thus, the internal energy of the bubble will decrease: the air cools. As we have seen above, this cooling is counteracted by a heat flow from the water that will reduce the cooling. So, in order to see whether of not the bubble will cool substantially, we will have to keep track of the amount of evaporation and the accompanying withdrawal of heat, of the heat flow from water to bubble and of the change in internal energy of the bubble. A sketch of the situation is given in fig.(4.10). Figure 4.10: Flow of mass and heat into the bubble as a consequence of evaporation. The evaporation is controlled by the difference between the water vapor concentration present in the bubble and the saturation concentration. The latter is a function of the (yet unknown) temperature of the air in the bubble and the pressure. The effect of the pressure can be ignored as the change in pressure for the bubble is small. We model the mass flow of water that is evaporating into the bubble similar to what we did above for the heat flow: the mass flow is proportional to the difference in the saturation concentration of water vapor, c, and the actual water concentration, c, as well as to the surface of the bubble. We call the proportionality constant k. For every gram of water evaporating, an amount of heat equal to the latent heat of that gram will be necessary. In formula: φ m = k πd 2 (c c) = φ q = H ev φ m (4.46) with D the bubble diameter, φ q the necessary heat flow and H ev the latent heat of water. As we have seen in the previous section, a difference in temperature between water and the bubble will cause a heat flow that can be modeled as: φ q2 = h πd 2 (T w T b ) (4.47)

59 4.4. A BUBBLE BATH 57 with T w the temperature of the water and T b the (mean) temperature of the air in the bubble. The next step in our modeling is keeping track of the increasing water vapor concentration in the bubble, as this is important for knowing how much water will evaporate. At the same time, we need to calculate the temperature of the air in the bubble at each time as this in turn influences the value of the saturation concentration, c, that is also influencing the rate of evaporation. So we set up a mass balance for the water vapor inside the bubble and an energy balance as well: dm w dt du dt = φ m = k πd 2 (c (T b ) c) = φ m H ev + h πd 2 (T w T b ) (4.48) These have to be solve simultaneously. Furthermore, as mentioned the saturation concentration, c (T b ), depends on the actual temperature, so we have to update its value as well in the calculations. The system of equations we now have is easiest solved numerically. The result is show in fig.(4.11). Figure 4.11: Temperature inside the bubble versus time: (a) different initial vapor concentration, (b) different initial air temperature. In fig.(4.11a) the temperature is shown of a bubble of 5mm. The initial air temperature is equal to the water temperature: 33 C. A steep drop in temperature is observed. The three curves represent three different initial water vapor concentrations of the air: 0%, 25% and 50% of the saturation concentration at T b = 33 C. As can be seen, even a 50% saturation at the beginning does not prevent the bubble from cooling significantly. In fig.(4.11b) a similar plot is shown, now for the initial air temperatures of 33 C and 45 C, resp. From this plot it is clear, that the cold sensation can be reduced by using pre-heated air. But at the same time it is clear that this is not a practical option. The required initial temperature will be so high, that personal injuries will almost certainly be the result: bath guest will also be in contact with air that is just released and might not have been cooled. The alternative option of saturating the air used with water vapor is much saver, although from a hygienic point of view not trivial (think of humid warm air, a perfect place for micro-organism).

60 58 CHAPTER 4. BUBBLES

61 Chapter 5 Bernoulli 5.1 Bernoulli The Swiss physicist Daniel Bernoulli ( ) came to the conclusion that energy conservation applied to the flow of liquids and gases couples pressure, velocity and gravity as these are in essence all forms of energy. For a steady state, and when friction can be ignored, the three forms of energy mentioned above can only changes from one into the other. The total sum can not change. This is very useful, as in many cases we will be able to treat a situation at hand as (quasi-) steady. Furthermore, the amount of mechanical energy, as the three form together, changes very little due to friction. We will use Bernoulli s law, as this form of energy conservation is called, to deal with a couple of situations. Therefore, it is handy to put the law in a mathematical form: 1 2 v2 + p + gz = const (5.1) ρ with v the velocity of the liquid or gas, p and ρ its pressure and density, respectively. Finally, g is the gravitational acceleration and z the vertical position. Note, that actually these quantities (except for g) may change from place to place in the liquid. The constant is not universal but may very from stream line to stream line. We will not go into the details here. 5.2 Shower curtain Ever noticed that the curtain of the shower does move towards you when taking a shower? What is the cause of this? Gravity will certainly try to keep it as vertical as possible. So, apparently a second force with a net component in the horizontal direction acts on the curtain. We could think of two possible candidates. On the one hand, the temperature behind the curtain is certainly higher than in the rest of the shower due to the warm water that is flowing. This influences the density and pressure distribution on both sides of the curtain. Thus, a pressure difference due to temperature differences might be the cause of the horizontal force. On the other hand, the water droplets fall through air. Hence, the air exerts a drag force on the droplets. But as we know from Newton s law, action = reaction, the droplets will also try to drag the air with them. This causes air to be sucked into the droplet spray 59

62 60 CHAPTER 5. BERNOULLI that comes out of the shower head. As a result the air in the shower behind the curtain is flowing. Due to the (usually) confined area the air flow inside the shower (i.e. behind the curtain) is much more pronounced than the one in the rest of the room. Or rephrased, the velocity of the air in front of the curtain is negligible compared to the one behind the curtain. Now, we can use Bernoulli s law (ignoring frictional effects) and conclude that there will be a pressure difference across the curtain, as there is a velocity difference: v 2 p. The pressure will be highest where the velocity is lowest and vice versa. Thus: the pressure in front of the curtain is higher than behind it in agreement with our observation on the curtain hanging inwards. Which of the two possible explanations can explain the effect? For the temperature effect, a simple test would be: does the curtain still move inwards if we take a cold shower? In fact the water temperature is probably lower than the air temperature, so in this case the curtain should move outwards. Try it your self and see. Furthermore, after a while we will have a steady state in the shower room. It is then not realistic to assume that there is still a large enough temperature difference across the curtain to account for the effect. Moreover, it does not seem logical that in a steady state a pressure difference across the curtain will be found. A relatively small air flow for a short time will even out this pressure difference and there is no driving force to maintain it. The Bernoulli effect, on the other hand, will be present as long as the shower is on. The driving force for the air velocity is the water spray. Thus, even in a steady state we may expect the air to circulate around the curtain. As the air velocity before the curtain will be small in comparison to the one behind, we will have a steady pressure difference. So, this is a good candidate for explaining the sucking inwards of the curtain. Bernoulli in action It is rather easy to see Bernoulli in action. Take two sheets of paper (A4 will do). Hold them parallel in front of your mouth, with a gap of a few centimeters between them (see fig.(5.1)). Then blow air through the slit formed by the two pieces of paper and observe. 5.3 Boat trip through the channels of Delft Most of the old cities with an extensive channel system offer boat trips for tourists. From the channels the cities look different. Many of these channels are rather narrow. The boats just fit. If a boat passes, water will have to move in the opposite direction. Now the question is: what will the water surface between the quay and the boat do? We could provide a multiple choice answer: The water surface stays flat; it always tries to do so. The water level rises as the water that is displace by the boat will have to show up somewhere. The water surface will drop, as this completes the possible choices.

63 5.3. BOAT TRIP THROUGH THE CHANNELS OF DELFT 61 Figure 5.1: Bernoulli in action. The right answer is the third one: the water level drops. Why? Bernoulli s law provides the answer. The situation is sketched in fig.(5.2). Figure 5.2: A boat sailing in a narrow channel causes the water level to drop. Consider point 1, on the free surface of the water, in fig.(5.2): here the velocity is still zero, the pressure is the atmospheric pressure and the z-coordinate is equal to h 0, the undisturbed level of the water in the channel. The same holds for point 3. Thus, from Bernoulli s law we see that indeed the level of the water upstream and downstream of the boat are the same. No surprise here. But for point 2 (also on the free surface) the situation is different. Here the water is flowing at velocity v, which is proportional to the velocity of the boat: the faster the boat sails, the more water has to flow per unit time passed point 2. As point 2 is also at the surface of the water and thus in direct contact with the atmosphere, its pressure is the same as that of the points 1 and 3. Thus we have from Bernoulli s law (friction is negligible): 1 2 v2 2 + gz 2 + p atm ρ = gh 0 + p atm ρ So, we conclude that the water level will be lower. z 2 = h 0 v2 2 2g < h 0 (5.2)

64 62 CHAPTER 5. BERNOULLI 5.4 Water flowing over an elevation in the river bottom A similar phenomena can be observed in flowing brooks or rivers. If in the bed of the river a bump is present, the water level above the bump does not have to rise. Actually it may drop. Figure 5.3: Water level of a brook when flowing over a bump in the brook bed. Suppose that the water level would stay flat, i.e. no sign of the bump would be visible. Then, the area of the cross-section above the bump where the water could flow through would be smaller than in front of or behind the bump. Thus, the velocity of the water above the bump will be higher than on either side of it. Hence, Bernoulli s law dictates that the gain in kinetic energy goes at the expense of the pressure and/or potential energy of gravity. So, if we focus on points of the water surface, we can use that the pressure at the free surface is constant, namely equal to the atmospheric pressure. Thus, if we compare a point (1) on the free surface far up (or down) stream of the bump with a point (2) on the free surface above the bump (see fig.(5.4)) we have 1 2 v2 1 + gh = 1 2 v2 2 + gh (5.3) For simplicity we assume that the river has a rectangular cross-section with width W. The height of the bump we will call δ and the height of the free surface above the bump h. Then, besides the Bernoulli equation given above, the water will also have to obey a mass balance, stating that the same amount of water has to flow through the river at the position of point 1 as above the bump at 2. If we express this mathematically, we have: WHv 1 = W (h δ)v 2 v 2 = So, we can eliminate v 2 from Bernoulli s equation and write it as: H h δ v 1 (5.4) Figure 5.4: Schematic of bump in a river.

65 5.5. WHY CAN AIR PLANES FLY? 63 ( ) 2 H + 2gh h δ v1 2 = 1 + 2gH v 2 1 (5.5) Obviously, this is a cubic equation for the unknown height h. In principle, the solution for h can be larger, equal or smaller than H, depending on the two parameters δ and v2 1. This equation can be solved analytically. An example is given of the H 2gH dependence of h on v 1 H 2gH for δ = 0.1 in fig.(5.5). H Figure 5.5: Dimensionless height of the free surface above the bump as a function of the dimensionless water velocity upstream of the bump. The equation has either 3 or 1 real roots. As can be seen in fig.(5.5) for low velocities the height of the free surface will drop, with for some values of the velocity several possible solutions. Which one will be chosen by the river needs more physics, like a stability analysis that will show which of the solution is stable against perturbations and which one is not. Notice further that for a certain range of velocities there seems to be no solution, as the only one we have calculated has a negative value for the height. Obviously, something else must be going on, as the water will definitely have to flow over the bump. 5.5 Why can air planes fly? A flying air plane is able to control its height above the earth surface reasonably well. Why does an air plane anyhow not drop to the earth like most objects thrown in the air? Of course, we know the answer: an upward pointing force is compensating the gravitational pull on the plane. Again this is a force with a hydrodynamic (or aerodynamic) background. It is present because of the air flowing around the air plane wings. To understand the basics, we employ Bernoulli law. A wing on an air plane has a special shape. In principle it has a flat bottom and a curved upper surface. A cross-section is shown in fig.(5.6).

66 64 CHAPTER 5. BERNOULLI Figure 5.6: Cross section of the profile of a wing and the stream lines around it. The velocity of the air on stream lines at the upper side of the wing is due to the shape of the wing higher than at the bottom part. Consequently, Bernoulli will dictate a higher pressure at the bottom than at the upper part of the wing. Thus, a net force that is proportional to this pressure difference and to the area A of the wings will try to push the air plane upwards. This force is, not surprisingly, commonly referred to as lift. In a steady state, it will be balanced by the weight of the plane. For control of the height, the air plane uses the area. Note also, that Bernoulli s law also incorporates the effect of the density. If there is no friction of any importance, the air changes its kinetic energy, 1 2 ρv2, for pressure energy, p. So we see that the pressure is also proportional to the air density. If we summarize this, we have that the upward force is proportional to: the air density, the wing area and the velocity of the plane squared: F lift A wing ρ air v 2 plane (5.6) A few observations can be made now. For instance, an air plane that will cruise at a high altitude will be surrounded by low density air and therefore will have to have larger wings or cruise at a high speed. Secondly, the weight of an air plane is to a large extend determined by its size. This scales like the third power of the linear dimensions of the plane. The lift force via the wings, however, scales with the square of the linear dimensions as it is proportional to a surface and not to a volume. This has as a consequence that a Boeing 747 must have a much larger wing area to volume ratio than e.g. a Chesna. This also poses problems to the engineer: how to design and make bigger and bigger wings that are not excessive in weight as that would require even larger wings, etc. Is Bernouli s law together with the typical shape of a wing the only way to understand flying? Certainly not! Think of paper folded airplanes, or of kites. They do not have the special shape and still they fly. So, there is at least an alternative. What keeps a kite airborn? Obviously the wind. But there is an other important point: the orientation of the kite. Kites make an angle with the vertical. Actually, this angle is a tuning parameter for kites, so they can deal with varies wind speeds. A schematic representation is shown in fig.(5.7). Let s assume that the kite makes an angle of 45 with the vertical and that the wind is coming in horiztally. As the kite is kept in its place by its rope, the incoming air stream has to move around the kite. Part of the inflowing air will move downwards. This is schematically shown in fig.(5.7). The kite has to exert a force on the air as the incoming air has only horizontal momentum, whereas the outflowing air has

67 5.6. TOP AND BACK SPIN 65 Figure 5.7: A kite floating in the air. only vertical momentum. This means that the force from the kite on the air has a horizontal component that is pointing in the opposite direction of the incoming wind velocity. This force is such, that it takes out exactly the horizontal momentum. Furthermore, the kite s force has a vertical component that is pointing downwards. This force is needed to accelerate the air in the downward direction. So, the kite s force points downwards at, roughly speaking, an angle of 45 with the vertical. Using Newton s principle of action=-reaction, the air in its turn pushes the kite backwards and upwards. Gravity and the rope do the rest, keeping the kite flying. Obviously, the same principle holds for an airplane. So, if the wing makes the right angle with the horizontal, the air is pushed downwards resulting in a reacting force from the air that has an upwards component. This mechanism also makes it possible for airplanes to fly upside down, something that with Bernoulli s law and the shaped wing would be impossible. Some interesting thoughts on lift and related can be found in: De Wetten van de Vliegkunst - over stijgen, dalen, vliegen en zweven by H. Tennekes ( [5]). 5.6 Top and back spin In many sports, the principles of aerodynamics are an essential part of the game. For instance, soccer, tennis, table-tennis or golf all employ them. Drag and lift are of considerable influence on the curve of a flying ball. Golf players need back spin to prevent the ball from rolling of the green after a strike. Tennis and table-tennis player can hit only hard and in at the same time if they give the ball enough top spin. A soccer player, attempting to score its free kick with a wall of players at about 9 meter in front of him, will try to curve the ball over the wall into the far corner out of reach of the keeper. In all these examples, the same principle is used: let the ball spin and thus induce a lift force that curves the trajectory of the ball. Again the lift force can be understood by looking at the local flow around the spinning ball. In fig.(5.8) a ball flying through air at velocity v is sketched. The ball is rotating with an angular velocity, ω, around an axis through its center. At the top part, the direction of the rotation is parallel to the speed of the ball, at the bottom it is opposite. For the air stream around the ball, this means that the velocity of the air moving around the ball at the top side will be smaller than at the bottom

68 66 CHAPTER 5. BERNOULLI Figure 5.8: Lift on a spinning ball. side as a consequence of friction. The friction has, however, rather little effect on the energy: not to much mechanical energy will be dissipated. Thus, Bernoulli s law will still predict a higher pressure at the top part of the ball, where the velocity is lowest. Thus, the ball will experience a net force from the air in the downward direction. This lift force works perpendicular to the velocity direction and the trajectory of the ball will be curved more than without the spinning motion. How much does lift affect the trajectory of the flying ball? Let s take as an example a ball (diameter 10cm, hollow, shell thickness 5mm, material density kg/m 3 ) that starts from the origin with a velocity of 30m/s at an angle of 10 with the horizontal. If flying through vacuum, the ball follows a parabolic trajectory, described by: x(t) = v x0 t y(t) = v y0 t 1 2 gt2 (5.7) Hence, for the initial conditions given, the ball will travel a horizontal distance of 31.4m, which takes 1.06 seconds. As we have seen before, the air will exert a drag on the flying ball, thereby dissipating some of the balls mechanical (= kinetic + potential) energy into heat. So, if we want the ball to reach the ground at the same point if we give it the same initial velocity, we have to change the angle at which it leaves the horizon. Instead of the 10, we have to set the initial angle at The travelling time is now increased to 1.37 seconds. Both trajectories are shown in fig.(5.9). The magnitude of the velocity at the point of hitting the ground is for the free flight equal to the initial one, i.e. 30m/s. If the drag force is taken into account this magnitude is reduced to 19.1m/s. If the ball is made to spin, we can have top spin or back spin. Now, if we want to calculate the ball trajectory, we have to expand the equation of motion with the lift force. This does affect both the x-component and the y-component of the equation of motion: the trajectory is a curved path, so the direction of drag and lift changes. Mathematically, we get: m dv x dt = F drag,x + F lift,x m dv y dt = F drag,y + F lift,y (5.8)

69 5.7. EMPTYING CONTAINER 67 Figure 5.9: Trajectories of a ball: influence of drag and spin. The lift force is a function of the rotational velocity of the ball. Let s give the ball an initial angular frequency of 10rad/s. Due to drag, also the rotation of the ball will decrease. So, we need to calculate also the change in time of the angular velocity. This has a similar form as the equation of motion: I dω dt = G = C R ( ) 5 D ρ airω 2 (5.9) Indeed this equation looks familiar: we see that instead of the mass, m, we now have a measure of the resistance against changes in the rotational speed, I (called the moment of inertia). The left hand side is not the momentum, but instead the angular momentum. Similar, the right hand side is not a force, but a moment. But for the rest, the equation does the same: it describes the change in time of the angular velocity of a body due to a moment (=force*arm) acting on it. For top and back spin, the trajectories are given in fig.(5.9) as well. Again, the initial angle is adjusted such that the ball hits the ground at the same point as the free flight. Obviously, the top spin ball follows a much more curved trajectory. Everybody who has been playing e.g. tennis or table-tennis knows that. The back spin trajectory is much flatter. Some relevant data are given in the table below. initial angle travelling time, τ v x (τ) v y (τ) ( ) (s) (m/s) (m/s) free flight drag drag & top drag & back Emptying container Consider a tin can (or any container with a cylindrical shape). The can is filled with water up to a level H 0. A small hole is made in the bottom of the can through which the water flows out. After a while, the can will be empty. So far, no surprise. But now, a straw is connected to the hole so that the water flows out of the can through

70 68 CHAPTER 5. BERNOULLI the straw. The question to be answered: What will happen to the time needed to empty the can? Possible answers: it stays the same; the emptying goes faster; it takes longer to empty the can. What is the right answer and does it matter whether we use water of thick honey? For the analysis we take a look at fig.(5.10). Figure 5.10: Which of the two is emptied faster: with or without the straw? Obviously, the can with the highest liquid velocity in the exit will be emptied quickest. If friction is negligible and the volume of the can is large we can use Bernoulli s law to judge which of the two has the highest exit velocity. The second requirement assures that the situation we deal with is quasi-steady. Actually, a large volume here means that the time needed to change the volume is so large that the flow adjusts itself in a quasi-steady way. If the two restriction apply, we have that a liquid element that flows out of the can does so at constant mechanical energy. Thus the sum 1 2 v2 +gz+ p is constant. Then compare for the two cases a fluid element that ρ flows from the free surface at the top through the exit. For both cases it holds that the fluid element starts with the following conditions: v 0, z = H and p = p atm. However, in the exit the situation is different for the two. In case of the hole we have: v = v h, z = 0 and p = p atm. Thus, we find v h = 2gH (5.10) For the straw we find: v = v s, z = L and p = p atm and we get v s = 2g(H + L) (5.11) So, the straw will cause the can to be empty quicker! Of course, this result brakes down when friction becomes too large. Then the mechanical energy is no longer conserved, but the fluid flowing through the straw will dissipate energy, i.e. mechanical energy will be transformed into heat and the last equation is no longer

71 5.8. TSUNAMI 69 valid. So, if we use honey instead of water we will most likely find, that the can with the hole will be empty quicker than the one with the straw. But even for water one has to be careful. Take for instance a straw with an inner diameter of 2mm. Then for H = 5cm, the velocity in case of a 10cm long straw will be, if we include friction, 1.09m/s, which is higher than the frictionless guess for the no-straw case, as 2 g H = 0.99m/s. If, however the water height is H = 10cm, we would have v with straw = 1.34m/s and v without = 1.40m/s (again for a straw of 10cm length). 5.8 Tsunami At the end of December 2004 an earthquake of the cost of the Indonesian island Sumatra shook the floor of the Indian Ocean, causing a devastating tsunami to develop. What is a tsunami and why can it turn into such an enormous disaster as it did? A tsunami (Japanese for harbor wave) is a disturbance of the level of the ocean surface with a long wave length. This makes the tsunamis member of the family of shallow water waves, for which we will below derive the propagation speed. In most cases they are generated by (sub-sea) earthquakes or volcanic activity. Tsunamis run at enormous speeds and can cross an entire ocean, such as the Pacific without losing too much energy to become harmless. At mid-sea, the tsunami may have only a small amplitude of several tens of centimeters, hardly noticable due to the long wave length of easily 100km. However, when approaching the shore, a tsunami can build up its amplitude to several meters or more. Morever, its wave shape may change into bore: a wall of water that rushes onto land. Obviously, the combination of velocity, wave or bore height and the high density of water can be truely destructive. Wave speed For all members of the family of shallow water waves, or more precise gravity waves on shallow water, a simple relation between the propgation speed and the water depth holds: c = gh (5.12) This relation can be derived using Bernoulli s law and the geometry of the wave: a long wave length compared to the water depth. Figure 5.11: Long wave on shallow water: reference frame is moving with the wave Consider a long wave on shallow water as dipicted in fig.(5.11). In the drawing, the wave amplitude is exaggerated for clarity. The wave is long: its wave length is much longer than centimeters. Then, surface tension does not play any role and the

72 70 CHAPTER 5. BERNOULLI waves are completly governed by gravity. We take a frame of reference that moves with the velocity of the wave. Hence, the shape of the water surface is steady and the wave seems stationary. As we have assumed that the water depth, H, is much smaller than the wave length λ, and the wave amplitude is very small compared to the water depth, we can safely ignore the vertical component of the water velocity and treat the horizontal velocity, v x, as being independent of the vertical coordinate z. But, v x has to vary slightly with the horizontal coordinate x, as the area for the water to flow through is a function of x. Water that flows in our frame of reference from right to left will have to flow in the same amount everywhere, as water is incompressible. Thus we deduce, that the product of the velocity v x and the local water depth at x, H + ξ must be constant: v x (H + ξ) = (v x + v x ) (H + ξ) = constant = v x H (5.13) with v x the mean liquid velocity (i.e. at water depth H) and ξ the vertical displacement of the water level with respect to the undisturbed level. As ξ << H we can approxiamete the above equation to first order in ξ and v x : H v x + v x ξ = 0 (5.14) Next, we use Bernoulli s law and apply that to points on the free surface. As these points have all the same pressure (i.e. the atmospheric) we find: 1 2 v2 x + g(h + ξ) = const. (5.15) For points on the surface at ξ = 0, this gives: 1 2 v x 2 + gh = const. (5.16) However, both last equations have the same constant and thsu, again to first approxiamation in x and ξ we now find: If we combine eq.(5.14) with eq.(5.17) we find: v x v x + gξ = 0 (5.17) v x = gh (5.18) Thus in a frame of reference in which the water is not drifting, we will see the wave passing by from left to right at a velocity: c = gh (5.19) Long waves at the ocean can travel at high speeds. Take for instance a wave with a wave length of 100km, traveling at the ocean with a depth of 4km. Indeed, that wave length is much longer than the water depth and thus the shallow water wave theory applies. Thus we find for the speed some 200m/s or 713 km/h! These waves will traveling hardly loose their energy. So, apart from a reduction in wave amplitude due to spreading of the wave, the wave arrives at the shore with its full energy. As it approaches the beach, it will find that the water depth is decreasing.

73 5.8. TSUNAMI 71 Consequently, the wave front slows down. Its tail still moves at the original velocity and thus the wave length shortens. However, still dissipation of energy is small and the energy of the wave per wave length is conserved. Thus the same energy is now found in a shorter wave, which implies that the wave amplitude increases, with devastating consequences. Like almost all waves, the tsunami will come in parallel to the coast. This is also a consequence of the slowing down of the wave speed. It is completely analogous to the breaking of light waves when they move from medium 1 with refraction index n 1 to medium 2 with index n 2. If n 1 < n 2, the waves will break towards the normal and move more parallel to the interface between the two media. As the refraction index is inversely proportional to the propagation velocity of the wave in medium, we can apply similar arguments as for breaking of light waves to the water waves: Snellius law holds equally well for the water waves.

74 72 CHAPTER 5. BERNOULLI

75 Chapter 6 Heat 6.1 Chinese egg roll In the NRC Handelsblad of May , a problem concerning a Chinese egg roll is discussed (see e.g. [6]). The question is: when a hot egg roll is purchased and needs to be taken home for a small diner, how can we get it warm home when being on a bicycle? What is wise, biking fast to shorten time at the expense of a larger cooling do to the induced air flow, or biking slow and so minimizing heat losses? The problem was brought into connection with the problem of walking in the rain. For the latter the question is raised: when does one get least wet, walking slowly or running. The egg-roll-problem can be solved by setting up a heat balance for the egg roll. Crucial in this is of course an adequate description of the heat losses. First we will start to model the egg roll as a cylinder of diameter D and length L. We will focus on the mean temperature, < T >, of the roll, that is averaged over the volume. The energy of the egg roll is ρc p V < T >. As the temperature of the egg roll is high compared to the surrounding air the heat flow will be from the roll to the surrounding air. Let us ignore for simplicity effects of evaporation of water, i.e. we assume that the roll is wrapped in some kind of paper. So, we have the situation as sketched in fig.(6.1). Figure 6.1: Schematic egg roll: heat flow from roll to air. We have dealt with similar situations. The heat flow is modeled as h A (< T > T air ). This gives for the energy balance of the roll: d dt ρc π p 4 D2 L < T > = h πdl (< T > T air ) d dt < T > = 4h ρc p D (< T > T air) (6.1) 73

76 74 CHAPTER 6. HEAT The above equation can be integrated if we know the dependence of h on < T > and t. Well, this is simple. For not too short time intervals h does not depend on t. If the velocity of the air flowing over the roll is not too small, h does not depend on < T > either. For our biker, trying to get home with the roll both for and if are valid. So, h is a constant with respect to the roll temperature and time. Thus the above equation can be integrated to: [ < T > T air = exp 4h ] < T > 0 T air ρc p D t (6.2) with < T > 0 the initial (mean) temperature of the egg roll. All we need to know for answering the question is how h depends on the biking velocity v, as this is the induced air velocity over the egg roll. We can distinguish two extreme cases: the resistance to the heat flow is in the wrapping material and the crust of the egg roll; the resistance to the heat flow is in the air layer around the roll. In the first case the resistance is constant, only depending on the material properties of the wrapping paper and crust. Hence also h is a constant as this is the reciprocal of the resistance. Now it is clear, that the faster one bikes the higher the temperature of the roll will be as the total cooling time, τ, is inversely proportional to the biking speed: τ 1/v. For the second case, things are a bit more complicated. Now the heat flow is directly governed by the velocity of the air flowing over the roll. For any reasonable biking velocity, the coefficient h will vary roughly like v. Thus, the exponent in the solution varies as 1/ v. So, again biking fast is better. 6.2 Frozen slice of bread Imagine you forgotten to take your bread out of the freezer in time. Nevertheless, it is lunch time and you want to eat. As you don t want to waist energy, the micro wave will not be used, but instead you spread out a few slices across the table. How should you do that: place the slices horizontal or vertical, or doesn t that matter? First we ask the question: what determines the defrosting time? Obviously, this is dictated by the heat flow to the slice. If we start analyzing, we see that is important to realize that the heat flow to the slice is proportional to the slice area and the driving temperature difference between the slice and the surrounding air: φ q = Ah T. The proportionality constant, h, will be different for different cases. It is convenient to try and locate the resistance against the flow of heat. Or more precise: is the resistance to be found in the slice or in the air surrounding the slice? Obviously, this is important to know, as in the case the resistance is in the slice it doesn t matter whether the slice is horizontal or vertical. Actually, we have a picture in mind that is quite comparable to Ohm s law. V = I R T = 1 h φ q (6.3)

77 6.2. FROZEN SLICE OF BREAD 75 Instead of an electric current, I, we use here a heat flow per unit area, φ q. The electric potential difference, V, is replaced by the driving temperature difference T. If we think about the heat flow to the slice of bread, it is easy to split the problem into two parts: the flow of heat from the bulk of the air to the bread surface and the flow of heat from the surface to the interior of the slice. If we go back to the question, we need to find the values for 1/h for in and outside. The internal value of the transfer coefficient, h i, can be estimated from h i = 4.93 λ b (6.4) δ with λ b the thermal conductivity of bread and δ the thickness of the slice, which is some 5mm. For λ b we estimate it to be that of ice, i.e. on the order of 0.3W/mK. 1 Thus, the internal resistance is: h i m 2 K/W. Note, that this is true for any orientation of the slice. For the outside, it is more complicated. Here, we need to realise that the heat flow in the air can be due to conduction but also to convection. The latter is a rather powerfull way of transport. Simplified we could say that with conduction in air (or any fluid) the molecules carry the heat individualy in an unorganized way, whereas with convection they move collectively. Thus, we need to think about the possibility of the air flowing as well. Now, the orientation comes in! If the slice is horizontal, the temperature of the upper surface is low in comparison with that of the air. Hence, a stable situation is created: a cold, dense air layer below a warm one. At the bottom surface of the slice a thin layer of air is trapped between the bread and the table. This acts like an insulator, just as the air gap between to layers of glass in the window. So, we conclude that in the horizontal position, the heat flow towards the slice is dictated by conduction. We can then estimate the transfer coefficient, h hor, as h hor λair D 0.12W/m2 K (6.5) So, we have that the resistance for heat flow in the air in the horizontal case is: 1 h hor 7.8m 2 K/W. Thus, it is concluded that in the horizontal case the resistance is in the air layer. Finally, we examine the vertical case. Here, the situation is unstable: the cold air layer along the slice surface is dense. Thus gravity will force it to flow downwards. Obviously, other, warmer air will replace the air in the layer. Thus, a convective motion is set up. We can anticipate that the vertical position is better for defrosting. The convective flow can still have two states: it can be laminar or turbulent. For the heat flow this matters. In laminar flow, the air flows in parallel layer along the slice. This means that perpendicular to the slice the heat still has to be transported by conduction. If, however, the flow is turbulent, the eddies will transport heat also perpendicular to the slice. Which of the two will occur depends on the following D 3 g T aν <T> (dimensionless) combination of parameters:. In this D is the characteristic length scale of the problem, which is in this case the length scale of a slice of bread, say 20cm; g is the acceleration of gravity, ν the kinematic viscosity, a the thermal diffusivity, T the difference in temperature between the bulk of the air and the slice, < T > the mean of these two temperatures (in Kelvin). It is quite logical that g should be present, as the flow is induced via density difference on which gravity acts. For the same reason, the ratio T <T> should be present as this is a direct measure

78 76 CHAPTER 6. HEAT for the density difference between the cold air close to the slice and the warm air further away from the bread. If the value of the dimensionless group is smaller than 10 8 the flow is laminar, when it is larger the flow is turbulent. If we take the properties of air at T = 20 C (i.e. a = m 2 /s,ν = m 2 /s) and assume that the temperature difference between bread and air is 30, we find a value of Thus we conclude that the induced air flow will be laminar. In that case we have that the transfer coeffcient is equal to: ( ) D 3 1/4 g T λ h vert = 0.55 aν < T > D 5.2W/m2 K (6.6) 1 Thus, h vert = 0.2m 2 K/W. So, we see that the resistance against heat is in the air for both the horizontal and vertical orientation of the slice. Furthermore, the resistance in the vertical case is about 40 times smaller than for the horizontal case. Thus, we can safely conclude that a vertical positioned slice of bread will defrost much faster than a horizontal one. 6.3 Black or white radiator? Every one knows that on a bright, sunny summer day a black car gets much warmer than a white one. The obvious reason being: a black surface absorbs much of the sun light that hits the surface, while a white one reflects much more and thus absorbs less energy. What holds for the absorption is also true for emission. Therefor, the reverse also holds: the black car will emit more than the white one. Then why do we all paint our central heaters white and not black? Is the latter not much efficient for the heating. Does radiation play any role in heating of a room? If we want an answer, we should know how much a body of a given temperature radiates. As we all know, a red hot piece of metal can be felt from quite a distance via its radiation. The higher the temperature, the more its color changes from red to white and the better we feel it radiating. The law connecting the body s temperature and the radiation is called Planck s law. It couples the spectrum of the radiation to the temperature of a black body. It is shown in fig.(6.2). Figure 6.2: Intensity of radiation of a black body as function of the wave length. The plot shows two spectra: one at relatively high temperature and one at low. We can see that the maximum in the spectrum shifts to shorter wave length as the

79 6.3. BLACK OR WHITE RADIATOR? 77 temperature increases and that the contribution of each wave length increases as the temperature goes up. The shift of the maximum is a simple function of temperature. It is known as Wien s displacement law: λ max T = (mk) (6.7) with λ max the wave length at which the spectrum has its maximum. Wien s law is shown in fig.(6.3). It is clear from Wien s law, that for temperatures below 4000K the maximum of the spectra is found outside the visible range. So, if we see iron getting red hot, say at a temperature of 1500K, λ max = 1.93µm and we are only seeing the low wave length part of the spectrum. Figure 6.3: Shift of the maximum in the black body spectrum with temperature. If on the other hand we take the sun as our black body, we find that λ max 500nm. Thus the surface temperature of the sun is about 6000K. The amount of energy radiated by a black body is governed by its temperature as well. From the Planck-spectrum it is easily shown, that this is proportional to the fourth power of the temperature: φ rad = σt 4 (6.8) with φ rad the energy per unit area radiated per second. This law is known as the Stefan-Boltzmann law; the constant σ = W/m 2 K 4. Let s return to the color of the radiator. A radiator will have a very modest temperature. Probably not much higher than 60 C. So, it radiates primarily at wave lengths around 9 µm. This is in the infra-red, no wonder that we can not see it. This has an unexpected consequence. What we can see is the color of the radiator: white or black. This is obviously an observation using light in the visible range. From a perspective of infra-red wave lengths, both paints will be almost black in the sense of radiating. What do we mean with that. Well, almost all objects are not black bodies. So, the Stefan-Boltzmann needs some modification. As a first approximation we introduce an emission coefficient, e, which has a value ranging from 0 to 1. If e = 1 the body is black, otherwise we call it gray. We write for the radiated energy: φ rad = e σt 4 (6.9)

80 78 CHAPTER 6. HEAT In general, the emission coefficient is wave length dependent. For the radiator paint, the coefficient in the visible range of wave lengths are: 0.1 for white paint and 0.9 for black. However, in the infra-red both emission coefficients are about 0.9. Thus, as far as the radiation is concerned, it doesn t matter whether or not one paints the radiator black. By the way, the same holds for an ice-bear: its white fur doesn t help it from radiating energy. Much more important is for all polar animal not to stick out too much! 6.4 Sipping Bird Perpetuum mobile are impossible, isn t it? Then, what is the explanation for the bird-toy, who dips its beak into a glass of water, takes a sip and moves back to stand upright. Then, when it is thirsty again it leans over for a new drink. What is its motor? Figure 6.4: The sipping bird takes another drink. In fig.(6.4) the bird is shown, together with its digestive tract. The bulb (with its stomach ) are at room temperature. However, its head might not be! The outside of the beak and part of the head is covered with felt. Take the bird in the upright position, just after it has dipped its beak into the water. The sip has made this wet and due to evaporation the head will cool down. The body of the bird is partially filled with a liquid (e.g. methylene chloride). But then is the vapor pressure of this liquid in the head lower than in the body. So, part of the liquid is pushed into the head of the bird. This shifts the center of gravity of the bird upwards, the bird becomes unstable and tips over. The beak takes another sip of water. In this horizontal position, the tube connecting head and body is only partially filled with liquid. Thus, the vapor can move freely through it and the driving pressure difference disappears. The liquid flows back into the body of the bird and the center of gravity is lowered again: the bird moves upright and the cycle is closed.

81 6.5. MILK IN THE COFFEE AND LEAVES IN THE TEA Milk in the coffee and leaves in the tea Suppose you have a hot cup of black coffee. However, you always drink your coffee with milk in it. As you are busy, you need to wait 5 minutes before you can drink it. What should you do, if you want your coffee as hot as possible: poor in the (cold) milk first and wait or wait first and poor in the cold milk just before you drink it? The answer is not difficult to give, if we think for a moment about the temperature of the coffee. Or, rephrased in terms of energy: let s concentrate on the internal energy of the coffee. Why and how will it change? The initial internal energy of the coffee decreases because of a heat flow to the surroundings. This heat flow is driven as we have seen before by the temperature difference between the coffee and the surroundings. As the temperature of the surroundings does not change during the waiting, the driving force is high if the coffee is hot. Furthermore, the coffee looses heat due to evaporation. The heat flow associated with the evaporation is proportional to the mass flow (in kg/s), the proportionality constant being the heat of evaporation. The evaporation is controlled by the vapor pressure of water above the coffee, which in its turn is a function of the temperature of the free surface of the coffee: the higher the temperature, the higher the vapor pressure and thus the higher the mass flow associated with the evaporation. Finally, we have to keep in mind that the energy contents of the coffee is proportional to its volume and that the heat loss is proportional to the heat exchanging surface. So, if we summarize our findings, we have that to a first approximation the internal energy (which is proportional to the temperature of the coffee) will decrease as: du dt = Aα 1 T dt dt = A V α 2(T T sur ) (6.10) Thus, the hotter the coffee, the faster it cools down. The rate of cooling is not only a function of the temperature difference between coffee and surroundings, but is also proportional to the cooling area (split in cooling via the free surface and via the cup-wall) and inversely proportional to the amount of coffee. However, we have to add cold milk. Obviously, the coffee & milk will have a temperature that is lower than the coffee temperature before mixing. Actually, we combine a volume V c of coffee with V m of milk, each with its own internal energy per unit volume. As coffee is for almost 100% water, we essentially mix hot water with cold water. So, if the hot water has initially temperature T c and the cold water T m after mixing we have: T mixed = V ct c + V m T m V c + V m (6.11) So, there are two scenarios: first mix or last mix. Let s make life simple and ignore the heat loss via the cup wall (after all, evaporation is a strong cooling mechanism). Furthermore, we will ignore the heat capacitance of the cup material. Finally, our cup is cylindrical in shape, so adding milk does not change the area of the free surface. First mix, then wait After mixing we have that the temperature, T mixed, is as given above. This tempera-

82 80 CHAPTER 6. HEAT ture serves as the initial condition of the cooling differential equation. The latter has as general solution: T(t) T sur = (T 0 T sur )e A V αt (6.12) with T 0 the temperature at t = 0. So, we have to substitute as initial temperature T mixed, given above. Further, we have to put for the volume now V c + V m as we have added the milk. This gives us for the temperature after waiting: ( ) Vc T c + V m T m T(τ) = T sur + T sur e A Vc+Vm ατ (6.13) V c + V m First wait, then mix Now we have that the cooling goes as: T(t) = T sur + (T c T sur )e A Vc αt (6.14) since here the initial condition is the starting temperature of the coffee, T c, and the volume has not changed. Now, we mix this at t = τ with the milk, so we get according to eq.(6.11): V c (T sur + (T c T sur )e A ατ) Vc + V m T m T(τ) = (6.15) V c + V m Both expressions look complicated. But the main point is that in the mix first case we cool slowly as the volume has increased but the free surface stays constant. So the exponent that governs the cooling rate is smaller. On top of that is that the temperature difference is smaller as we have mixed cold milk with the coffee. A numerical example is given in fig.(6.5). It is clear from the figure that indeed mixing first is better if one wants its coffee hot. Leaves in a cup of tea Imagine you have a cup of tea, made the old fashioned way: with real leaves. If you stir your tea hard enough, all leaves will be suspended in the liquid. But, if you wait for the leaves to settle, you will find that they collect at the central part of the bottom. What is causing them to do so? Let s analyse what is happening. By the stirring action, we have set the liquid into a rotating motion, which looks like a solid body rotation. That is the obvious part. But we need to be concerned with the motion of the leaves. Question (1): why are the leaves suspended anyhow? If the liquid motion would be like a solid body rotation nothing would probably happen, as there is no net force exerted by the liquid on the leaves in the vertical direction. So, all the leaves would do, is due to a drag force start rotating as well. Apparently, more is going on. For the liquid to move in a circular motion, a centripetal force must act on it. This force is due to a pressure distribution in the liquid: the pressure increases with increasing radial distance from the center. The leaves have a higher density than the

83 6.5. MILK IN THE COFFEE AND LEAVES IN THE TEA 81 Figure 6.5: Cooling curves of coffee with milk. liquid (as they all sink to the bottom if no motion is present in the liquid). Thus a higher force is required to keep them in circular orbits. This force is not available. Thus the rotation will cause the leaves to move outwards, away from the center. However, fluid flow is a complicated subject. The liquid moves coarsely speaking in a solid body rotation. But a closer look will reveal all kind of swirling motion: the liquid is full of eddies. The liquid is flowing under turbulent conditions! These eddies have more or less random orientation. And these eddies are responsible for lifting the leaves. Of course, the eddies will also transport leaves back to the bottom, but as there are at least initially much more leaves on the bottom than in the liquid the net effect is dispersion of leaves all over the tea. But we miss still a force on the leaves that finally will move them back to the center of the bottom. For this again we need to refine our view of the motion of the liquid. Due to friction between the liquid and the solid walls of the tea cup, the liquid velocity close to the walls is lowered. This is especially felt at the bottom. But if the rotational velocity is decreased close to the bottom, than also the pressure does not vary as much! Thus, if we now look at the region close to the vertical wall, we see that the pressure high in the tea cup due to the rotation is higher than that one close to the bottom/vertical wall region. This causes the liquid to flow downward along the vertical wall. And, of course, the liquid in the central region of the cup has to flow upwards: a net circulation has been established. The leaves have to respond to this, as they feel the drag of the flowing liquid. Thus, if the motion of the liquid becomes weaker, due to friction, the eddies loose there strength. They are no longer able to lift the leaves. The induced circulation is still powerful enough to drag the sinking leaves to the central part of the bottom!

84 82 CHAPTER 6. HEAT 6.6 Freezing lakes and channels As soon as the temperatures in the Netherlands drop below zero (Celsius), the people start to speculate how long it will take before the lakes and channels are covered with a solid layer of ice: can the skates be taken out of the grease? Let us analyze that problem. We will first make some assumptions and simplifactions, but leaving the heart of the matter untouched. The first assumption is, that the water in the lake or channel is at 0 C. Of course, we make here an error and we will have to estimate later its consequences. Secondly, we assume that the temperature of the surface of the ice (at the air side) is constant, say T 0 i.e. below zero. This is not true, as the temperature changes during the day and night. Furthermore, we usually only know the temperature of the air above the water. We can refine this later. Finally, we will neglect evaporation. Our next step is, that we will set up a heat balance in which we consider the freezing of an extra layer of ice with thickness dx on to the allready formed ice layer of thickness x. The situation is sketched in fig.(6.6). T = - T air heat flow x dx T = 0 o C ice water new layer Figure 6.6: Freezing of an extra layer of thickness dx to the ice sheet that has thickness x. The freezing water will release its heat of solidification upon turning into ice. Hence, if this heat can not be removed, the new layer cannot freeze. So, we conclude that the rate at which this heat can be tranported from the freezing water to the cold air controls the pace at which the water freezes. Let us think a moment about why there is transport of heat anyway. The answer is, that there is a temperature difference between the water and the cold air. Hence, a flow of heat will automatically start. Consequently, the water looses heat and has no other way to respond then solidifying as it can not cool down any further before it has gone through a phase transition. We assume that the process of ice fromation is slow. Experience tells that is a matter of days rather than minutes. So, it is reasonable to for the temperature profile in the allready existing ice layer the solution of a steady state situation in which the lower side is kept at 0 C while the upper side is fixed at T 0. Obviously, also here heat is transported from the lower to the upper side. This heat flow is due to conduction of heat and can be modeled using Fourier s Law: φ q = Aλ dt dx (6.16)

85 6.6. FREEZING LAKES AND CHANNELS 83 in which φ q represents the heat flow (in J/s), A the surface through which the heat flows, λ the heat conduction coefficient (for ice: λ = 0.55W/mK) and dt the dx temperature gradient. For the steady state problem of the ice sheet one finds that the temperature distribution in a straight line. The corresponding heat flow is then given by: φ q = Aλ T (6.17) L with T the temperature difference over the layer and L its thickness. Note that the direction of the heat flow is from the warmer to the colder side. This linear temperature profile is allready sketched in fig.(6.6). The new layer of thickness dx, freezes in a time interval of dt and releases an amount of heat equal to H sol ρ ice Adx with H sol the heat of solidification of water (i.e. 335kJ/kg) and ρ ice the density of ice (915kg/m 3 ). Hence, the amount of heat released per unit of time equals H solρ ice Adx. As we concluded this must be transported to the dt cold air and this is done through conduction through the ice layer with thickness x. So balancing the heat released and transported we have: H sol ρ ice Adx = Aλ T (6.18) dt x This is a simple differential equation in x, the ice thickness. It can be solved as follows: x dx = λ T dt 1 H sol ρ ice 2 x2 = λ T t + C (6.19) H sol ρ ice with C an integration constant. As we have that at t = 0 no ice was formed, i.e. x = 0 we find that C = 0. Thus, if we need a layer of ice with a thickness of 10cm, it takes for an air temperature of 7 C some 4.6 days. In fig.(6.7) is shown how the required time depends on the temperature difference and on the ice thickness. 30 t (days) 20 10cm 8 t (days) o C -7 o C 10 4cm 7cm 2-10 o C T ( oc) ice thickness (cm) Figure 6.7: Freezing time as a function of the temperature difference (left) and of the thickness of the ice layer (right). Let us now examine the consequence of assumption 2: we have taken tht the surface temperature of the ice at the air side is a given constant. Actually, it is probably better to have it fluctuate from day to night. So our second guess could be:

86 84 CHAPTER 6. HEAT T(x = 0) = T 0 + δt 0 sin(2πt/τ) (6.20) with τ the period of the oscillation, being 24 houres. Like before, we still assume that the changes with time are slow enough that the temperature profile in the ice layer is very close to the steady state solution. So, our analysis above does not change up to eq.(6.18). However, in solving we need to realize that the temperature difference T is now a function of time. Thus our solution reads as: 1 2 x2 = The integral is easy to evaluate: λ H sol ρ ice Tdt + C (6.21) τ Tdt = T 0 t δt 0 cos(2πt/τ) (6.22) 2π So we see, that if the time required is long compared to τ, i.e. several days, the second term with the cosine is negligible compared to the first, linear term provided that the amplitude of the oscillation is small with respect to mean temperature T 0. Thus we conclude that our assumption of constant surface temperature gives reasonable estimates. What about the asumption that the surface temperature is constant? Actually, the air temperature above the water/ice is known. If the heat is transported from the freezing layer to the bulk of the air, it not only has to pass the existing ice layer that acts as a resistance but also it has to flow from the surface to the bulk of the air. So, a second resistance is present. The value of the latter depends on the weather conditions. If there is no wind blowing over the lake, the heat transport is governed by natural convection: the air just above the ice is relatively warm with respect to the colder air higher above the surface and will have the tendency to rise due to its lower density. If, on the other hand, the wind is blowing a forced convective heat transport dominates the heat transport from the ice to the bulk of the air. In both cases an extra resistance against heat transport is present. The temperature profile is sketched in fig.(6.8). R 2{ air x R 1{ T = - T heat flow ice dx new layer T = 0 o C water Figure 6.8: Freezing of an extra layer of thickness dx to the ice sheet that has thickness x when a resistance against heat transport is present in the air as well.

87 6.6. FREEZING LAKES AND CHANNELS 85 In this figure, now two resistances (R 1,R 2 ) in series can be seen. The heat flow from the surface to the bulk of the air is modeled in the same way as done in the case of the chinese egg roll: φ q = h A (T surf T air ). If we combine this expression for φ q with the one that deals with the ice layer (note that this now has as T the temperature difference over the ice layer which contains the unknown T surf ), we get: ( x φ q = A λ + 1 ) 1 (T ice T air) (6.23) h As this equation is similar to eq.(6.17), we can follow the analysis of the first case and find thus (for constant T air. Hence, the differential equation to be solved for the ice thickness, x, reads as: ( x λ + 1 ) dx = T ice T air dt (6.24) h H sol ρ ice Thus we find: ( 1 x 2 2 λ + x ) = T ice T air t (6.25) h H sol ρ ice A good value for h is 20W/mK. If we use this value and calculate how long it takes to form a layer of 10cm at a temperature difference of 7 C, we find for the required time 7.1 days. Obviously, an increase in time as there is more resistance in the system to the transport of heat from the freezing layer to the cold air.

88 86 CHAPTER 6. HEAT

89 Chapter 7 Mechanics at work 7.1 Catastrophe Toy A nice example of interesting physics can be find in an old children s toy. This toy consists of a small clown that is surrounded by four leaves of a waterlily. The leaves at their base fixed on a disk that can be made to spin. The toy is shown in fig.(7.1). Figure 7.1: Photo of the catastrophe toy: leaves closed (left), leaves open revealing the clown (right). On the inside of the four leaves a rubber band pulls on the leaves, so that they are closed when the disk is not spinning. This toy is called to catastrophe toy, a name that will become apparent when we analyze the physics of the toy. We will start by bringing the complicated form of the toy back to a more basic form. In rest the leaves are with their tips resting against one another. Both the rubber band and gravity try to move to leaves as far inwards as possible: this is sketched in fig.(7.2). We can schematize this by replacing each leave by a point mass m half way a massless rod. The rod is with its free end fixed to the rotating disk. As now the center of gravity of a leave is moved to the end of the rod, the action of gravity is even more pronounced and we can neglect the rubber band. If the disk is in rest, i.e. not rotating, the masses will rest against one another. Due to symmetry the masses will be on the rotation axis of the disk. If the disk is brought into rotation, 87

90 88 CHAPTER 7. MECHANICS AT WORK Figure 7.2: Schematic representation of the catastrophe toy and the forces acting on it. the centrifugal force will try to move the masses outwards, whereas gravity will try to keep them at the rotation axis. Obviously, if the masses are real point masses, their starting position (when the disk is in rest) is at the rotation axis. Consequently when then the disk is brought into rotation no centrifugal force is felt. In the following analysis we will assume, that the rest position of the masses is just outside of the rotation axis. To understand the physics of the catastrophe toy, it is sufficient to concentrate on a single rod with mass. The situation we would like to analyze is shown in fig.(7.3). Figure 7.3: Simplified catastrophe toy. It is convenient to analyze this system in a rotating frame of reference, that rotates with the constant angular velocity, ω. Two forces are acting on the mass: gravity: and the centrifugal force: F g = mgẑ (7.1) F cf = mω 2 rˆr (7.2) In the above, ẑ and ˆr are a unit vector in the z-direction and the r-direction respectively. Obviously, a third force, exerted by the rod, acts on the mass. But as its working line goes through the point around which the rod can rotate, its angular moment is always zero. Consequently it does not influence the motion of the mass. It merely is there to keep the distance from the mass to the rotation point of the rod constant.

91 7.1. CATASTROPHE TOY 89 For each of the two forces we can calculate their contribution to the potential energy of the mass. This will turn out to be an elegant way to look at the toy. First we obtain the potential for the centrifugal force: V cf = F cf d r mω 2 rdr = 1 2 mω2 r 2 + const (7.3) Similar for gravity: V g F g d r = mgdz Hence for the total potential energy we find: = mgz + const (7.4) V = 1 2 mω2 r 2 + mgz (7.5) Here we have chosen the reference point for the potential energy such, that the potential energy is zero when the mass would rest at the center of the rotating disk. Now we will replace the coordinates {r,z} by a single coordinate, θ, as this coordinate is sufficient to describe to position of the m in the rotating frame: Thus for the potential energy we get: z = 1 Lsinθ (7.6) 2 r = R 1 Lcosθ (7.7) 2 V (θ) = 1 2 mglsinθ 1 2 mω2 ( R 1 2 Lcosθ ) 2 = 1 2 mglsinθ 1 ( 2 mω2 L cosθ min 1 ) 2 2 cosθ (7.8) In the last equality R has been replaced by Lcosθ min, with θ min the (minimum) angle when the mass is at rest at the rotation axis. It is sufficient to inspect the potential divided by 1 2 mgl: ( V (θ) = sinθ ω2 cosθ min 1 ) 2 gl 2 cosθ (7.9)

92 90 CHAPTER 7. MECHANICS AT WORK Figure 7.4: Potential and its derivative: left p = 0.5, right p = 4. Notice the different values of θ where dv dθ = 0. The factor ω2 acts here as a single parameter, p. In the figures below the potential gl V and its derivative are shown for θ min = 1rad 57 for two different values of p: p=0.5 and p=4. There is an important difference between the two figures: for small p the potential has a maximum with a value larger than θ min. As p is determined by the angular velocity, ω, we have in this situation that the initial state of the mass and thus the leaves of the toy (i.e. no rotation and the leaves are closed) is stable. In other words the leaves will stay closed and the clown is not visible. If on the other hand the value of p is sufficiently large, the maximum of the potential is found for θ smaller than θ min. The potential is now a decreasing function of θ. The mass will move to the position corresponding to the minimum in the potential, that is now the new stable point. For the toy it means that the leaves open and the clown is visible. It is easy to find the value of p (and thus of ω) at which the initial position of the leaves becomes unstable. We need to find that value of p, say p, for which the extremum of the potential is found at θ min : dv = cos(θ) p (cos(θ min 12 ) dθ cos(θ) sin(θ) 0 = cos(θ min ) p (cos(θ min 1 ) 2 cos(θ min) sin(θ min ) p = 2 sin(θ min ) (7.10) For the value of θ min used (1 rad), we find: p = Estimating the length of the leaves as 5cm, we have for the critical value of the angular velocity: ω = 0.35, or 2.20 rps. So, we can now understand the name catastrophe toy. If the angular velocity is below the critical one, the leaves stay closed. Once the critical value is surpassed, the leaves suddenly open widely to reveal the small puppet. The role of the rubber band

93 7.2. BUNGEE JUMPING 91 is twofold. On the one hand it will close the leaves if the rotation is stopped. On the other hand it will increase the necessary critical rotation speed, hence allowing the disk to spin so fast that when the leaves open, the eye hardly see the leaves anymore and the puppet will be continuously visible. 7.2 Bungee jumping Bungee jumping is a relatively new thrill. It is simple enough: dive from a height with a strong, long rubber band attached to e.g. your legs. The first part of the flight is a free fall, until the rubber band is stretched at its total length. Then the band starts acting as a spring, will stop the fall and pull up the jumper again into the air. The band will be unstretched and the jumper will fly free into the air, fall back en the cycle repeats itself. If there would be no dissipation of energy, the bungee jumper would continue its oscillatory motion for ever. However, there is quite some energy dissipation, so after a few oscillations the amplitude of the oscillatory motion has become small and the trhill is over. Massless Cord In a first analysis we will assume that the elastic cord is massless. The model of the physics is simple, and allows to estimate the various parameters that govern the bungee jumping. As mentioned there are two stages: a free fall and a mass-spring stage. The former is described by (regarding the jumper as a point particle of mass m): m d2 z dt 2 = mg z(t) = 1 2 gt2 + v 0 t + z 0 (7.11) with v 0 and z 0 the initial velocity and position at t = 0, respectively. We will denote the starting position of the jumper by z = 0. Notice that the positive direction of our z-axis points in the direction of gravity. This will be convenient in the description of the second stage. When the jumper has fallen over a distance of L 0, the rubber band is just stretched and the jumper will start to experience a force from the band. This acts just like a spring, i.e. proportional to the extra stretching, the proportionality constant being k. If we neglect energy dissipation for the moment, we get for the motion of the mass: m d2 z dt 2 = mg k (z L 0) (7.12) This equation can be easily solved, e.g. by denoting s = z L 0 mg k : m d2 s + ks = 0 (7.13) dt2 This gives sinusoidal motion with an angular frequency given by: ω 0 = k m. To make a more realistic description, we will assume that energy gets dissipated when the string is being stretched. A good attempt to try and incorporate this effect is by coupling this dissipation to the speed at which the rubberband is stretched. The reasoning behind this is, that if the rubber band is stretched by kept at a fixed length,

94 92 CHAPTER 7. MECHANICS AT WORK we do have to ably a force, but since there is no path along which this force has to act, no energy is involved. If, however, the forces is able to stretch the band further there is work involved (F x). Part of this is turned into spring or potential energy, part is turned into heat. Mathematically we can express this in the equation of motion via a term µ v, µ denoting a kind of friction coefficient: m d2 z dt = mg k (z L 0) µ dz 2 dt Again, using s = z L 0 mg we can write this as: k (7.14) d 2 s dt + µ ds 2 m dt + k m s = 0 (7.15) This describes a damped sinusoidal motion. The attenuation and the frequency can be found from the characteristic equation: This has as solutions: λ 2 + µ m λ + k m = 0 (7.16) λ 1,2 = µ m ± = µ 2m ± i ( µ ) 2 m 4 k m 2 k m ( µ 2m ) 2 (7.17) If ω 2 0 > µ then the motion is sub-critical. The oscillation frequency is given by 2m k ω = ( µ 2, m 2m) which is smaller than ω0. The general solution is of the form z(t) = L 0 + mg k + µ t) e( 2m [Asinωt + Bcosωt] (7.18) The amplitudes A and B follow from the given initial position and velocity that are obtained from the previous stage. Now we can make some estimates of the unknown quantities k and µ. Roughly speaking, there are two kinds of bungee-cords: so-called Mill-Spec cords and New Zealand cords. The former give a relatively short oscillation time, whereas the latter provide a more gentle ride with more rebounds. Mills-Spec cord Assume that the length of the rubber band, L 0, is equal to 10m. This cords gets elongated by about % (notice that the cords are adjusted to the weight of the jumper). The number of oscillations is limited, usually no more than 5. By setting up a simulation model based on the above, we can try and find the parameters k and mu that match the characteristics of the cord. Take for example, a jumper with a mass of 70kg. If we take k = 250kg/s 2 and µ = 10kg/s, we find an elongation of 100%, an equilibrium of the cord length of 12.7m and an oscillation time of 3.3s; the number of oscillations that has a free fall part is 5.

95 7.2. BUNGEE JUMPING 93 New Zealand cord Let us again take a cord of L 0 = 10m. Now the maximum elongation will %. The jumper is again 70kg. If we choose k = 40kg/s 2 and µ = 2kg/s, we get an elongation of 400%, an equilibrium length of 51.3m and an oscillation period of 8.3s. Only 3 oscillations have a free fall part, but the number of oscillations with a large amplitude is much higher than with the other cord. The amplitude of the oscillation after 20 oscillations is still 3m. In fig.(7.5), the position of the jumper is given as a function of time for both cases discussed above. Figure 7.5: Position of the jumper is given as a function of time: black line represents the Mills-Spec cord, the grey line the New Zealand cord. Cord with mass In practise, of course, the cord has mass. This has a nice consequence, which makes the thrill of the jump even higher. This has been analyzed by [7]. We will analyze only the first part of the jump: the free fall with unstretched cord. A schematic representation just before the start is given in fig.(7.6). m z M L/2 Figure 7.6: Situation just before the jump. We will use here the coordinate system as given in the figure: the z-axis points

96 94 CHAPTER 7. MECHANICS AT WORK upwards. At the moment of the jump, the kinetic energy of the system is zero. The potential energy of the jumper, which has mass m, is taken as zero (we define z = 0 as the height of the platform). As is seen in fig.)7.6), the cord is hanging from the platform. We take for the cord a total mass of M and assume that the density of the cord is homogeneous. The total cord length is L, hence in this position its center of gravity is located at z = L/4. Thus, the potential energy of the system is E pot = MgL (7.19) 4 During the jump obviously the jumper gains kinetic energy and so does part of the cord. If the jumper has fallen to position z, the velocity of the jumper is v. So is the velocity of the part of the cord that is moving. This has length L/2 + z/2: the top is lowered by z, but the point where the cord bend upwards is also lowered by z/2 (remember that z is negative below the platform). The kinetic energy is given by E kin = 1 2 mv M L + z 2L v2 (7.20) The potential energy is then given by: E pot = mgz + 0 L/2+z/2 gz dm + z L/2+z/2 gz dm (7.21) The first part is obviously the potential energy of the jumper. The second part is the potential energy of the part of the cord that is not moving, wheras the third part is the part of the cord that is moving. The integrals can be solved by using dm = ρdz. The total mass is then M = ρl. Calculating the integrals gives: E pot = mgz MgL 4 ) zl z2 (1 2 L 2 If we ignore any friction, the mechanical energy is conserved and we find: (7.22) MgL 4 = mgz MgL 4 ) zl z2 ( L 2 2 mv M L + z 2L v2 (7.23) From the above equation, we can solve v 2 : v 2 4mL + M(2L + z) = 2gz 4mL + M(2L + 2z) (7.24) We can compare this to a free fall of the jumper only, which is obviously the same as taking M = 0: 0 = mgz mv2 v 2 = 2gz (7.25) In fig.(7.7) the velocity of the jumper is compared for the case of m = M to that of the free fall (M = 0, dashed line). So, we see that the mass of the cord gives the the jumper a higher velocity which of course means that the acceleration of the jumper must be higher than g. To calculate

97 7.2. BUNGEE JUMPING v (m/s) 10 m=m M= z (m) -20 Figure 7.7: Velocity of a jumper with a cord of the same mass as the jumper and of a massless cord. this, we differentiate the velocity with respect to time. First we will shorten the notation for the velocity: v 2 4mL + M(2L + z) = 2gz 4mL + M(2L + 2z) v2 = 2gz N D (7.26) where we have shortened the numerator as N = 4mL + M(2L + z) and the denominator as D = 4mL + M(2L + 2z). Note that we may use N = D Mz. Using this notation we can differentiate the velocity and derive an expression for the acceleration: 2v dv dt a dv dt = 2g dz dt = g N D ( 1 Mz 2 dz dz D 2M dt dt 2gzM N D 2 4mL + M(2L + z) (2mL + M(L + z)) 2 ) (7.27) If we define µ = M, the mass of the cord relative to that of the jumper ans m introduce the relative coordinate s z, we can rewrite the previous equation as: L ( a = g 1 µs 2 ) 4 + µ(2 + s) [2 + µ(1 + s)] 2 (7.28) For any µ 0 the acceleration of the jumper is highest at s = 1, i.e. when the cord is just streched. We have at this position: ( a(s = 1) = g 1 + ) µ(4 + µ) 8 (7.29) Obviously, this is a monotoneously increasing function of µ and we find the heavier the cord the bigger the thrill for the jumper. The maximum acceleration for, for instance a cord of the same weight as the jumper is g, for µ = is this value 2g.

98 96 CHAPTER 7. MECHANICS AT WORK 7.3 What is the shape of a suspended cable? Suspended ropes or chains are found seen frequently in our daily life. Take for instance, the cables of electric power lines (see fig.(7.9)), or washing lines (fig.(7.8)). But also the shape of a hammock, or the cables of the San Fransisco Golden Gate Bridge show the peculiar form of suspended ropes. (a) (b) Figure 7.9: Suspended power lines (left) and the Golden Gate Bridge (right). The shape of a rope (or chain) suspended from its two ends, is an old problem. Galilei thought is was a parabolo, but a prove did not exist. The Dutch scientist Christiaan Huygens ( ) proved (at the age of 17!) that this assumption was wrong: the rope shape is not a parabola. This achievement of Huygens becomes more impressive, if we realize that at that time the differential calculus did not exist. With the development of differentiating and integrating, the analysis of the chain-curve (or catenary) become macu easier. The problem was solved in 1691 by Johan Bernoulli ( ) using differential- and integral calculus Problem formulation The differential equation describing the shape of the rope, can be found from a force balance. In the figure below, a rop is sketched. The x- and y-axes are chosen such that the lowest point of the rope is on they-axis. The yellow part of the rope (with length s) hangs in equilibrium. Hence, the sum of the forces is zero. Three Figure 7.8: Wahing lines suspended between houses. forces are acting on this piece. Gravity (F g ) in the negative y-direction. This force equals F g = m sg, with m the constant mass of the rope per unit length (thus m s is the mass of the yellow part). At the left end, the left half of the rope puls with a tension F 0. Tension forces are tangent to the rope. Due to symmetry, F 0 is horizontal. At the right end also a tension, F s, is exerted on the yellow piece. This force makes an