Fluid Mechanics
 The work of a curious fellow

Its hard to get a grip on this stuff..
 Somewhere along the line you have probably been introduced to the idea of three states of matter, solid, liquid and gas. Every living person has experience with all three states. The objects that we find in our surroundings include solids like rocks, chairs and houses, the water we drink is liquid and the air we breathe is gas. From this experience we have some intuition about the distinction among these three. Solids have a definite volume and shape, liquids have a definite volume but not shape and gasses have neither a definite volume nor shape. That intuitive distinction will be precise enough for now. There is another characteristic that separates solids from the other two states of matter. Solids can support shear stress, as we discussed in Equilibrium and Elasticity. Neither liquids nor gasses can. Those materials that do not support shear stress are called fluids. Fluid mechanics is the study of such materials and includes those materials that we intuitively regard as liquids and gasses. There are a few oddball materials where the ability to support shear stress depends on how rapidly the stress is applied. Glass for instance is fluid over very long time frames. Silly putty is solid with regard to very short time application of shear stress. We will not be treating these pathological cases in this course. Fluid mechanics is divided into statics and dynamics. Fluid statics deals with the behavior of fluid essentially at rest in our reference frame. Fluid dynamics deals with fluids in motion or objects moving through stationary fluids. We will begin with some ideas from fluid statics. Notice that in fluid mechanics we begin talking about materials rather than objects. Up to this point in this course we have dealt with particles either individually or in relatively closely coupled groups. Now we are going to back away from the behavior of individual particles and tight groups of particles and begin to consider fluid materials as an environment in which particles and other solids may find themselves. The interaction between the materials as a whole with other objects is the focus of fluid mechanics. Of course the behavior of fluids is determined by the behavior of the particles that make them up, but we will in general not pry into the private lives of individual particle, just consider how the whole society of particles behaves in fluids.
 One of the properties of substances that influences fluid mechanics is density, the mass per unit volume. Density in SI units is kg/m3. Solids and liquids have densities in the range of about 5e2 to about 2e4 kg/m3. The density of a solid or liquid depends slightly on temperature, generally decreasing as temperature increases, but remain in the stated range under any reasonable temperature. The density of a gas of course is not as fixed as for solids and liquids since we defined a gas as a substance without a fixed volume and density involves the volume of the material. The image at the left is a computer generated map of the density of inter-stellar gas in a binary star system. So that we can still talk about the density of a gas, we refer it to a set of standard conditions called STP for standard temperature and pressure. Those conditions are a pressure of one atmosphere and a temperature of zero degrees C. Under these conditions gas densities range from about 1e-1 to about 2 kg/m3. The fact that gas densities at STP are about 1/1000 the density of liquids and solids implies that the particles in the gas are separated by about 10 times the separation in the liquid and solid case. The density data also indicates that the difference between solids and liquids may depend on something other than simple the separation between the atoms or molecules making up the substance.
 There are two important consequences of the fact that fluids do not support shear stress. Firstly the only force that a fluid can exert on an object submerged in it is a force everywhere perpendicular to the surface of the submerged object. That is the reason that spherical vessels are favored for deep submergence. The compressive force is uniformly distributed over the whole sphere. Secondly, the force exerted by a fluid at a particular depth is felt uniformly in all directions. This is really the first situation looked at from the point of view of the fluid rather than the object. The pressure exerted by a fluid at a particular depth, h, is defined as the force per unit area that the fluid exerts on an object submerged to that depth. As we will show shortly, the pressure depends on depth, P=P(h) so to get the total force on any finite surface other than a horizontal one, we must define the pressure on an infinitesimal surface area, P(h)=Df/DA, where Df is the little increment of force on the area DA and DA is made as small as necessary so that the variation in depth can be ignored. Then the force on the finite surface area is found by adding up all the little Df's. This should begin to remind you of the rate of change discussion. The rate of change of force with respect to area is P(h) so Df = P(h)*DA. To add up all the little Df's we can slice the depth of the finite surface up into horizontal slices of constant depth, multiply the area of each slice times the pressure at that depth and add up all the products. The units on pressure being force divided by area are newton/m2. The newton/m2 is given the name pascal(Pa) 1Pa=1N/m2. The standard atmospheric pressure, mentioned above, is 1.013e5 pascals.
 To get at the variation of pressure with depth, the P(h) mentioned above, we are going to introduce a trick that your will see again in different contexts. Suppose there is a container of liquid and imagine a cylinder of cross sectional area A extending from the surface of the liquid to a depth h. This cylinder is an imaginary construct. It is there only in our minds, to form a boundary between the sample of liquid we wish to consider and the remaining liquid. This situation is shown in the pressure depth dependence model. The weight of the fluid in the imaginary cylinder exerts a force on the bottom face. Let's call the pressure exerted by the weight of our sample of fluid on the bottom face of this imaginary cylinder, P. There is also a pressure, P0, exerted on the top of our imaginary cylinder by the weight of all the air above it. Because the fluid is not flowing, we know that it is in equilibrium. This means that the sum of the forces on any cylinder of fluid must be zero. The upward pressure on the bottom of the imaginary cylinder therefore must also be P + P0. The weight of the fluid in the cylinder is the density, r, of the fluid times the acceleration due to gravity times volume, w=r*g*V. But the volume is A*h so the downward force on the bottom face of the cylinder due to the weight of the fluid is r*g*A*h. The total force at the bottom of the imaginary cylinder is P0*A + r*g*A*h. Dividing through by A, we find P = P0+r*g*h. indicating that pressure is indeed a function of depth.
 Suppose we want to find the total force on the vertical face of a dam of height H and width W. Since the pressure depends on depth, one approach would be to imagine the dam as a series of horizontal strips of height Dh and width W, where Dh was small enough that the pressure variation from top to bottom could be neglected. Then we could calculate the force on each strip and add them all up. This is a good job for a computer. In the case of our dam, we can use the P(h) to calculate the total force. The force on each little strip is Df = P(h)*W*Dh = (P0+r*g*h)*W*Dh. If Dh is small enough, the total can be found by finding the anti-derivative or integral of Df. Remembering the trick for getting the anti-derivative by increasing the exponent of the variable by 1 and dividing by the new exponent, we find f = P0*h*W+1/2*r*g*h2*W where we have made use of the fact that the sum of all Dh is h. This technique works even if the W is not constant but depends on depth as long as W(h) can be written in a mathematical expression. Because the dam has such a simple geometry and the pressure varies linearly with depth, there is another way to get the total force. We know the average pressure on the dam is 1/2*(P0+r*g*h). If we multiply this average pressure by the area of the dam we will get the total force f = P0*h*W+1/2*r*g*h2*W These three approaches to the same problem illustrate a very general principle in the relation of physics to mathematics. The greater the symmetry of a problem, the more likely we will be able to solve it by elementary means. For a rectangular dam we can just use algebra. If the dam had a face shaped like a triangle, we could use calculus. If the dam had a face shaped like the walls of a canyon the computer approach would be best. In fact we could always use the computer, but if we did we would soon loose our understanding of the situation and only keep the knowledge of what buttons to push. Replacing understanding with knowledge is never a good idea. Then you would not recognize the similarity between a dam and a hot air balloon and have to start all over again to figure out how tough the balloon's fabric would have to be to support the total force. Besides, you want to be the person instructing the computer how to solve the problem.
 Notice that in the dam problem we carried the P0 term throughout. In fact since the dam will have atmospheric pressure on its dry side as well as its wet, the dam design does not have to include this term. It must only be strong enough to support f = 1/2*r*g*H2*W. The pressure including the atmospheric pressure is called the absolute pressure. The pressure without the atmospheric pressure term is called "gage pressure". It is the pressure that would be measured on a normal pressure gage that detects the difference between the pressure measured and atmospheric pressure. The way we handled atmospheric pressure in our treatment of the dam problem has hidden in it an assumption that pressure applied at the surface of a liquid is felt throughout the liquid, undiminished by distance, and the pressure is transmitted to the walls of the liquid container. This may seem obvious but in 1650 AD it was a novel idea, first formalized by Blaize Pascal and hence known as Pascal's law. The whole physics of hydraulics derives from that idea. Consider a liquid container with two movable pistons forming part of the pressure boundary, one large and one small. Applying a force to the smaller piston will increase the pressure in the liquid by an amount equal to the applied force divided by the area of the smaller piston. This increased pressure will be felt at the larger piston where the resulting force will be the applied pressure multiplied by the area of the larger piston. The ratio of the applied force to resulting force will be the ratio of the large piston area to the small piston area. Run the hydraulic jack model to see how this works. Of course another consequence of Pascal's law is that in a container of liquid that is open to atmosphere at several locations we will find the liquid surface at the same height in each opening regardless of the size of the opening or the shape of the container in the vicinity. In the absence of differences in applied pressure liquid seeks it own level.
 Now let's consider a cube of solid stuff placed in a container of liquid so that it is completely submerged. Since the cube and the liquid cannot occupy the same space at the same time, the volume of the cube displaces some liquid. Since liquids are essentially incompressible the total liquid volume remains the same. Since the volume occupied by the cube is no longer available the displaced liquid must go elsewhere. The walls of the container prevent it from going out to the side so it must go up. In effect pushing the cube into the liquid lifts the displaced liquid. The force required lift the displaced liquid must come from the pushing of the cube into the liquid. The weight of the displaced liquid, wl is its density, rl, times the displaced volume, Vl, times the acceleration due to gravity, g. wl = rl*Vl*g The force wl must be provided by the force on the cube of stuff. This force is a combination of the weight of the cube and any applied force either up or down. The weight of the cube is its mass times the acceleration due to gravity. The mass is the density of the stuff, rc times its volume, Vc, so the weight comes out to be wc = rc*Vc*g. If the weight of an object is greater than the weight of the displaced liquid the net force on the object will be downward and the it will sink. If the weight of an object is less than the weight of the displaced liquid, the net force on the object will be upward and it will rise. The weight of the object is fixed but the weight of the displaced liquid depends on how much of the object is submerged. If the entire object is submerged and still the weight of the displaced liquid is less than the weight of the object, it will accelerate downward until the excess object weight is supported either by the force of friction as the object moves through the liquid or by the bottom of the liquid container. If the weight of the object is less than the weight of an equal volume of liquid, the object will float with a portion of its volume not submerged, such that the submerged volume displaces just enough liquid to support the weight of the thing.
 Run the buoyancy model to play around with various densities of liquid and solid stuff. In the buoyancy model we find a block of solid material held at the surface of a pool of liquid. A click of the Action button releases the block. As it settles into the water the buoyant force, equal to the weight of the displaced fluid, opposes the weight of the object. In the example given the object weight is a bit less than weight of an equal volume of the liquid so the object will ultimately float. Because the cube is dropped from an elevated position with its lower surface just touching the liquid, it initially submerges due to its downward momentum. It takes a while for the slight positive buoyancy to turn it around. The mass of this 1 cubic meter object is just the density in kg/m3. By changing the mass for a fixed volume we in effect control the object's density. Enter 500 in the box mass field and re-run the model. You should see the sort of bobbing behavior that a massive, somewhat buoyant object exhibits when slid into the water. Use the arrow keys to run the model in stop action and observe the flow resistance vector label. Click on the drawing area to give the keyboard focus to the model window, making the arrow keys active. I could not bring myself to ignore the effect of fluid friction. Frequently we ignore air resistance for slow moving objects but ignoring water resistance throws out a major contributor to the motion of the object. You may remove the effect of friction by setting the flow resistance coefficient, frc=0. You may change the density of the fluid by entering a new value in the fluid density field. This model offers another view of the buoyant object's behavior. Select the graph view for a plot of depth vs. time. By clicking on the plotted variable labels in the top-right quadrant of the graph, you may toggle on and off the individual plots. The model automatically re-scales the plots to accommodate the selected variables. Because the flow resistance is so out of scale with the other plotted parameters, the model scales it back by a factor of 0.001 for plotting.
 Next we take a brief look at fluid dynamics. Fluids in motion relative to solid objects, like rivers flowing past rocks and airplane wings passing through air, is a topic not yet completely understood by anyone. Advances in chaos theory may shed some light on exactly what is going on in these situations. Even what is understood of the subject tends to lead us into mathematical difficulties that are just too hard. In introductory physics courses like this we need to introduce some major simplifications. Even so there is a lot we can do in this area. In this course we will consider the flow of an ideal fluid with the following properties. 1. An ideal fluid in non-viscous. This means that internal friction in the fluid may be neglected. 2. An ideal fluid exhibits steady flow. This means that the velocity of the fluid at each point remains the same as time passes. No starting and stopping transients are considered. 3. An ideal fluid is incompressible. This means that the density is not a function of time or space. 4. An ideal fluid has zero angular momentum everywhere. This means that the flow doesn't swirl around in eddies or whirlpools. Properties 2. and 4. limit the velocity of an ideal fluid to less than the critical velocity where turbulence begins. Flow without turbulence is called laminar flow because the fluid layers sort of slide over one another. The path taken by a fluid particle in an ideal fluid is a smooth line called a streamline. The particle velocity at any instant is tangent to this streamline. No two streamlines can cross each other because if they did, a particle could move either way at the crossover point and the flow would not be steady. The entire bundle of streamlines in a flow is called the flow tube. No fluid may enter of leave this flow tube; otherwise the streamlines would be crossing each other. The image at the right shows laminar flow except near trailing edge of the object in the flow path.
 Now let's consider an ideal fluid flowing through a pipe with non-uniform cross-section. We will think of the region of large cross-section region 1 and that of smaller cross-section, region 2. The continuity equation model illustrates the following discussion. The particles of fluid move along the streamlines in steady flow. Imagine a surface perpendicular to the flow in region 1. If the velocity of the fluid flowing through this surface is v1, then in time Dt an imaginary cylinder of fluid, of length Dx1=v1*Dt, will pass through the surface. If A1 is the cross-sectional area of the pipe in this region then the volume of the imaginary cylinder is DV1 = A1*Dx1 = A1*v1*Dt. The mass of the fluid in this imaginary cylinder is Dm1 = r*A1*v1*Dt Now let's place another imaginary surface perpendicular to the flow in region 2 where there is a smaller cross-sectional area, A2. By exactly the same reasoning the mass that flows through this surface in time Dt is Dm2 = r*A2*v2*Dt. Now because the mass is conserved and the fluid is incompressible we know that Dm1 = Dm2 and A1*v1 = A2*v2 = constant This is called the continuity equation. The continuity equation tells us that where the cross-section is small, the velocity is large and where the cross-section is large the velocity is small. The product of velocity and cross-sectional area is called the volume flow rate and has units of volume/time. The constant volume flow rate tells us that the amount of fluid put in at one end of a pipe is the amount coming out at the other end, regardless of the shape or length of the pipe.
 Next we will look at the work done as fluid flows through our pipe of varying cross-section and this time we will include variations in elevation between region 1 and region 2 as well. The Bernoulli equation model illustrates this situation. Consider the flow of an ideal fluid through a non-uniform pipe in time Dt. The force, f1, pushing fluid into region 1 of the pipe is P1*A1 where P1 is the pressure in region 1 of the pipe and A1 is the cross-sectional area of that region. The work, w1, done by this force is f1*Dx1. w1 = P1*A1*Dx1 = P1*DV1 Likewise the work done on the fluid in region 2 of the pipe is w2 = -P2*DV2. The minus sign on the work is because the force and displacement are in opposite directions in region 2, since the fluid flows from region 1 to region 2. The continuity equation tells us that DV1=DV2 so the net work done, w1+w2 is Dw = (P1-P2)*DV Part of this work goes into changing the kinetic energy of the fluid and part goes into changing the gravitational potential energy. The change in kinetic energy is Dke = 1/2*Dm*v22 - 1/2*Dm*v12 where Dm is the mass moved through the pipe in time Dt. The change in potential energy is Du = Dm*g*y2 - Dm*g*y1 where y1 and y2 are the elevations of regions 1 and 2 respectively. The work-energy theorem says that for this volume of fluid Dw = Dke + Du or (P1-P2)*DV = 1/2*Dm*v22 - 1/2*Dm*v12 + Dm*g*y2 - Dm*g*y1 Dividing by DV and replacing Dm/DV by the density r we get P1+1/2*r*v12+r*g*y1 = P2+1/2*r*v22+r*g*y2 This last is Bernoulli's equation for an ideal fluid. It says that the sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume has the same value at all points along a streamline. This is an extension on the simple conservation of energy because it brings the pressure into the equation. If there is no change in elevation the potential energy terms drop out and we see that P2=P1*v12/v22 is, indicating that pressure is lower in the high velocity areas in a system.
 You can apply Bernoulli's equation to any number of fluid flow problems, as you will see from your physics text. This equation is so generally applicable that people sometimes stretch it to cover situations where it does not really fit. One classic is the explanation of the lift from an airplane wing. The story goes that the curved shape of the top of the wing forces the air passing over the top to take a longer path than the air passing beneath the wing. Therefore the velocity of the air above must be faster than the velocity of the air underneath; resulting in reduced pressure above and lift from the wing. If that were the mechanism for producing lift, inverted flight would be impossible and the wing of a typical light aircraft wound have to have a cross-section like this.
 Alternatively a four-seater Cessna 172 would have a stall speed of 400 miles per hour with its existing wing if Bernoulli was solely responsible for lifting the weight of the plane. The underlying assumption is that it takes the over-wing air the same length of time to reach the trailing edge of the wing as the under-wing air. There is no guarantee of course that this happens. In fact the over-wing air gets there considerably faster. If we think of the airplane wing in terms of Newton's third law, the lifting force must have an equal an opposite downward thrust. Actually the air passing over the wing has a significant downward velocity when it leaves the trailing edge. The details of the actual airflow are nowhere near the conditions for an ideal fluid. We should not be surprised that Bernoulli comes up short as an explanation. It is true that the air pressure on top of the wing is reduced but the Bernoulli explanation is not even close. Are there any questions?
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