An Overview of Undergrad Fluids Notes Developed for MAE 222

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An Overview of Undergrad Fluids Notes developed for MAE 222

Written and compiled by:

Clayton P. Byers

Princeton University, 2016 Foreword

These notes are a compilation and edit of my precept notes I developed for MAE 222 in Spring 2016 at Princeton University. The goal here was to combine them into a sort of study guide that can be used by future students. They assume the reader is either familiar with, or concurrently studying, basic fluids mechanics. These notes will not stand alone as a way to learn fluids from scratch! Some sections go into fine detail about derivations, which aren’t necessary for the basic undergrad education. Other sections assume you have some base level of knowledge, and may be complicated if you haven’t seen the material before. They will serve well for an undergrad reviewing the material for your final, or even for a graduate student in the early preparation for your general examination.

The ﬂow and order of the material follows the general outline in which it was presented in the course. In general, each section should have a number of fully worked examples to demonstrate the core concepts. I go through each step in the math and logic to help develop your understanding of the application of the material.

I have tried to edit the content so that it is consistent in the internal equation and ﬁgure references, but there might be an equation or two that accidentally got cross-referenced. If you see any mistakes in notation (or more egregious errors in the content!) please let me know: claytonb (at) princeton (dot) edu.

This material was inspired by Prof. Marcus Hultmark’s class lectures, Prof. Alexander Smits’ book “A Physical Introduction to Fluid Mechanics,” and the book “Engineering Fluid Mechanics” by Crowe, Elger, Roberson, and Williams. You’ll find similarities to those texts or classes for good reason, but my goal was to present them in a new light or with a different approach. I hope you find this useful and enlightening!

Intro to Fluids Notes 1 C. P. Byers 2016 Contents

1 Introductory Concepts5 1.1 Dimensional homogeneity ...... 5 1.2 The unique characteristics of pressure in fluids ...... 5 1.3 Example: breaking a barrel ...... 7 1.4 Example: an airplane door ...... 8 1.5 Statics and moments ...... 10 1.5.1 Example: a dam wall with one fluid ...... 10 1.5.2 Example: a dam wall with two fluids ...... 13 1.6 Rigid body motion summary ...... 15 1.7 Specific weight and specific gravity ...... 18 1.8 Buoyancy ...... 18 1.8.1 Example: floating some aluminum ...... 19 1.8.2 Example: floating in a different fluid ...... 21 1.9 Control Volumes ...... 22 1.9.1 Example: accumulating mass ...... 22 1.9.2 Example: non-uniform flow ...... 23 1.10 Material Derivative ...... 25

2 Continuity and Momentum 27 2.1 Continuity equation ...... 27 2.1.1 Diﬀerential form of the continuity equation (bonus material) ...... 29 2.1.2 Example: continuity on a nozzle ...... 30 2.2 Momentum Balance ...... 32 2.2.1 Example: force on a nozzle ...... 32 2.2.2 Example: Balance with angled velocities ...... 34

3 The Momentum Equation 38 3.1 Integral & Differential momentum equation ...... 38 3.1.1 Example: momentum practice ...... 41 3.1.2 Example: differential momentum practice ...... 45 3.2 Bernoulli’s equation ...... 46 3.2.1 Rigorous derivation of Bernoulli’s equation (bonus material) ...... 47 3.2.2 Pressure and streamlines ...... 49 3.3 Example: flow over a bump ...... 49

Intro to Fluids Notes 2 C. P. Byers 2016 CONTENTS

4 Introducing Potential Flow 53 4.1 Some ﬂow deﬁnitions and characteristics ...... 53 4.1.1 Vorticity ...... 53 4.1.2 Velocity Potential ...... 54 4.1.3 Stream functions ...... 55 4.2 Example: a potential vortex ...... 56 4.3 The power of the velocity potential ...... 59 4.4 Example: superposition ...... 60

5 Non-Dimensionalization 63 5.1 Non-dimensionalization ...... 63 5.1.1 Similarity and non-dimensional parameters ...... 63 5.1.2 Non-dimensional equations ...... 66 5.1.3 Example: non-dimensionalizing the momentum equation ...... 66 5.1.4 Example: ﬂow in a channel ...... 68 5.2 Prandtl’s Boundary Layer ...... 69 5.2.1 x-momentum equation ...... 69 5.2.2 y-momentum equation ...... 72 5.2.3 Combining the x- and y-momentum equations ...... 73 5.2.4 Bonus: similarity solution ...... 74 5.3 Bonus: Energy Equation ...... 75

6 Analysis of the Boundary Layer 77 6.1 Summary of the boundary layer ...... 77 6.2 Zero pressure gradient ...... 78 6.3 Displacement thickness ...... 78 6.3.1 Displacement thickness and the zero pressure gradient assumption . . . . 80 6.4 Momentum thickness ...... 81 6.4.1 Momentum thickness and drag ...... 84 6.5 Example: approximate boundary layer ...... 84 6.6 Example: ﬁnding the skin friction ...... 85

7 Drag, Separation, and Shedding 89 7.1 Drag force from friction ...... 89 7.1.1 Example: drag on a plate ...... 90 7.2 Separation of the boundary layer ...... 91 7.2.1 Favorable pressure gradient ...... 93 7.2.2 Zero pressure gradient ...... 93 7.2.3 Adverse pressure gradient ...... 94 7.3 Resistance to separation ...... 94 7.4 Case: a curved surface ...... 94 7.5 Strouhal number ...... 95 7.5.1 Example: Tacoma Narrows Bridge ...... 96 7.5.2 Example: car antenna ...... 98

Intro to Fluids Notes 3 C. P. Byers 2016 CONTENTS

8 Internal Flows and Losses 99 8.1 Internal flows ...... 99 8.1.1 Continuity ...... 100 8.1.2 Inertial terms ...... 101 8.1.3 The x-momentum equation for a channel (and pipe) ...... 101 8.1.4 The pressure gradient ...... 102 8.2 The velocity profile in a pipe ...... 103 8.3 The friction factor in a laminar pipe ...... 105 8.4 Example: a vertical pipe ...... 106 8.5 Example: force to hold a pipe ...... 109 8.6 Losses in pipe flow ...... 110 8.7 Example: flow from a reservoir ...... 113

9 Supercritical and Supersonic Flows 115 9.1 Comparison of Mach number and Froude number ...... 115 9.2 Hydraulic jumps ...... 116 9.2.1 Example: hydraulic jumps on a dam ...... 118 9.2.2 Example: jump in a channel ...... 118 9.3 Compressible ﬂow relations ...... 120 9.3.1 Example: space shuttle re-entry ...... 121 9.4 Normal shocks ...... 122 9.4.1 Example: revisiting the shuttle ...... 123

Intro to Fluids Notes 4 C. P. Byers 2016 Chapter 1

Introductory Concepts

1.1 Dimensional homogeneity

This concept is one of the most important concepts in science and engineering, yet is often glazed over and not actively discussed as early as (I think) it should be. Whenever we write an equation, we aren’t just looking to plug in number, but are also dealing with dimensions or quantities of some sort. Something as simple as Newton’s Second Law, P F~ = m~a has no meaning or usefulness unless we use the proper quantities and dimensions.

m →[mass] a fundamental dimension

length a → [acceleration] time2

mass * length F →[force] time2 From this information, we can see that both sides of the equation balance in terms of fundamental dimensions - we have [mass ∗ length]/time2 on both sides. However, we must also make sure we use consistent units. It doesn’t make sense to have meters on one side and feet on the other. All equations we work with (which are based on some physical interaction in the world around us) will be dimensionally homogeneous. This seems simplistic, but keep it in mind and always take a look at it to make sure you’re approaching your equation correctly. In ﬂuids, we can sometimes get information in units like psi but need P a. The easiest way to make sure you’re doing things right is to write down the units when doing math. Not only will it help you keep track of what you’re doing, but it may clarify where a mistake could be.

1.2 The unique characteristics of pressure in ﬂuids

We like to make analogies of pressure with weight balancing on an area. However, this can be misleading, and requires a little more thought to grasp how pressure really works.

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Remember that we classify pressure as a “scalar,” which means it has a value, but no direction. This is analogous to temperature - does it have a direction? Heat transfer does, but temperature itself does not. It simply has a value at a location. Pressure is the same, while the force we feel (or calculate) due to pressure isn’t prescribed by the pressure field, but instead by the physical boundaries in which pressure acts. The force due to pressure, which is a vector, gets the magnitude from the pressure and area it acts on, but the direction comes from the prescribed boundaries on which the pressure is acting. One way to think about the force of pressure is to think of a person sitting on the bottom of the ocean. Assuming the depth this person is at stays constant, then we can say the pressure acting on him is constant. Recall the equation for hydrostatic balance: dP = −ρg (1.1) dz where the z-direction is defined as positive going up. Integration of the equation then will give us: P − P0 = −ρg(z − z0) Let’s think for a moment - what are the initial conditions, and what are the variables here? Since we’re dealing with the ocean, ρ will be the density of sea water, which we can wave our hands and say is 1000 kg/m3. Notice the units we’re using? Keep it consistent throughout! Let’s also assume that gravity isn’t going to change on us too much, which is usually a safe bet, and say g ∼ 10m/s2. Again, note the unit’s we’re using! Now think on the initial conditions. We can specify the pressure at the surface of the water as atmospheric pressure, Patm. In real life, this can change a bit day to day, but we usually are interested in departures from this pressure, not the actual value of it. So sometimes we’ll just leave it as a variable, Patm, while other times we will just set it to zero and note that we are looking for the departure from atmospheric pressure. As for the height, let’s set the surface of the ocean as zero. We could also set it as any other value, but why make it hard on ourselves? OK, here’s where we stand now with the equation: kg m P − P = −1000 ∗ 10 ∗ (z − 0) atm m3 s2 or N ∆P = −10000 ∗ (z) m3 All we’ve done is simply call the pressure difference ∆P since we’re talking about a departure from atmospheric. Now, let’s say our person is 33 feet below the surface. What is the pressure difference from atmospheric pressure? N ∆P = −10000 ∗ (−33ft) m3 Note that we’ve set the depth as a negative value, as we defined the positive z-direction as up. But look at the units! We shouldn’t be using feet! Recall dimensional homogeneity! We need to convert this to meters, which is about 3.3 feet to a meter. Therefore, we would find: N 1m N ∆P = −10000 ∗ (−33ft ∗ ) = 100000 m3 3.3ft m2

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The pressure at 33 ft below the surface is approximately 100 kP a higher than the atmospheric pressure! So even though our equation had the minus sign, we found the pressure increased with depth (which our intuition should agree with). Now why did I say that pressure can be tricky or unique? Let’s think about the water above the person. We specified them to be at the bottom of the ocean, 33 ft down (ok I’m tired of feet, lets use meters like real scientists). What if instead it was the bottom of a 10 m pool? Or a 10 m dunk tank? The person would still experience an additional 100 kP a of pressure. You could even shrink this down to a tube that encapsulates the person, but with 10 m of water above, and it would still be the same pressure. Even more crazy, you could use a 10 m long straw, place it on top of their head, and fill it with water, and the pressure at the top of their head would be 100 kP a higher than atmospheric pressure! So pressure isn’t so much about the weight of fluid above the point, because the straw full of water and the ocean exert the same pressure at a depth of 10 meters, even though their volumes are vastly different. Remember that the pressure is a scalar value that exerts itself on the surface, and the surface is what determines the direction of the resultant force.

Overall, just be careful when doing the integration, and deﬁne your zero point and positive direction clearly.

1.3 Example: breaking a barrel

This example highlights the “strangeness” of pressure while showing how powerful it can be. Recall how we just said a straw 10 meters tall and full of water exerts the same pressure as the ocean at 10 meters below the surface. This concept can be used to exert large forces through simple setups. Imagine a sealed barrel, expertly drawn in figure 1.1, with a pipe connected to the top. The pressure in the barrel, assuming it is completely sealed, is then dependent on the height of the fluid in the tube. Think: if the tube were disconnected, the barrel empty, and there was just the hole at the top, then that would be exposed to the atmosphere. Assuming the barrel isn’t absurdly large, the pressure within would be the same as atmospheric pressure. If it were filled with water to the hole, it would have increasing pressure towards the bottom, but again assuming it isn’t too big it wouldn’t be much different from Patm. Now, we put the tube on, but don’t fill it. What’s the pressure? Have we really changed anything? No, we haven’t! Sure, there is a column of air in the tube, but the pressure at the opening of the barrel will still be Patm, as that is the pressure of the air at the bottom of the tube. This air pressure then interfaces with the water in the barrel, so the pressure exerted is no different than if we didn’t have the tube. Now, let’s start filling the tube up with water. Again, the barrel is perfectly sealed. What happens as the water level in the tube increases? You can integrate equation 1.1 and set it all up, but your intuition should tell you that pressure will rise. What happens when the water level in the tube is 10m? (hint, we just talked about a straw that had 10m of water in it!) Since that pressure from the water column exerts itself on the water in the barrel, we would find that the pressure inside is suddenly 100 kPa higher than when there was no water column. Increase the water height to 20 meters and suddenly the water pressure in the barrel is 200 kPa above atmosphere! If you then integrated this pressure over the area inside the barrel, you would find

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Figure 1.1: a totally legit drawing of a barrel with some pipe contraption coming out of the top. a tremendous amount of force on the walls, which at some point would lead to it bursting. Of course that depends on the material properties and exact geometries.

The point from this thought exercise is to help you grasp how powerful but simple pressure can be. A very tall but thin column of water can end up exerting very large amounts of pressure, even though the actual weight of the water itself is small.

1.4 Example: an airplane door

Airplane doors open inward before swinging out. The inside cabin air is pressurized relative to the cruising altitude pressure, but still lower than atmospheric. This example will help illustrate why. Let’s say you’re cruising at 10,000m in a commercial airliner with your friend. You’re ﬂying

Intro to Fluids Notes 8 C. P. Byers 2016 CHAPTER 1. INTRODUCTORY CONCEPTS international, so they’ve fed you dinner. You decided to have the beef, while your friend had the fish. Now, 1 hour later, your friend is entering a hallucinogenic state brought about by the fish. He decides that the airplane is too confining, so he goes up to the cabin door to open it and leave. Should you try to stop him?

Let’s see what pressure says you should do! We know there is a pressure diﬀerence between the outside and inside of the plane, so we should calculate those values to determine the direction and magnitude of the resultant force. Lets start by assuming that the pressure inside the cabin is half that of regular atmospheric pressure, or 50kP a. Let’s also assume the cabin door is around 1 meter wide by 2 meters tall. Now what’s the outside pressure? You could look it up in a table, or online, or any number of sources, but let’s make it a little more work. The function describing the density in the troposphere (the ﬁrst 10,000m or so of the atmosphere) is: P kg ρ = (1.2) 286.9(288.1 − 0.0065 ∗ z) m3 This means we have density as a function of height in the atmosphere. This can then be plugged into our hydrostatic relation, equation 1.1, to get: dP −P ∗ g = dz 286.9(288.1 − 0.0065 ∗ z) or, by rearranging and plugging in 9.81m/s2 for g, we have:

dP −0.0342 = dz P (288.1 − 0.0065 ∗ z)

I hope that you’ve done some basic diﬀerential equations, or at least separation of variables. If not, then at this point just remember that you want all of one variable on one side, and the other variable on the other side. Here, you can see that all P variables are on the LHS, while all z variables are on the RHS. Then you can integrate both sides with respect to their own variables.

Z P (z=10,000) 1 Z z=10,000 −0.0342 dP = dz P (z=0) P z=0 (288.1 − 0.0065 ∗ z) P? 10,000 1 ln (P ) = − 0.0342 ln (288.1 − 0.0065 ∗ z) ∗ −0.0065 Patm 0 ln (P?) − ln (100kP a) = 5.26 ln (223.1) − ln (288.1) P ln ( ? ) = − 1.345 100kP a P? = 26, 000 P a

So the math was a little fun, but otherwise it was a straightforward plug and chug to get the pressure at 10,000m elevation. If you look up the value in a table, you’d see that we’re within a few percent, so all is well!

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Now, we have the outside pressure, and we know the inside pressure, so the force exerted on the cabin door will be: X F orces = Finside + Foutside = (−Pinside nˆi + −Poutside nˆo) ∗ Adoor

Remember how the definition of the force due to pressure has the minus sign in it. This means you also need to pay attention to the wall normal vectors. It’s pretty straightforward in this example, but for cases where the walls or surfaces are sloped, it becomes very important to pay attention ton ˆ. Assumingn î (inside wall normal direction) is directed in the positive direction, andn ô (the outside wall normal direction) is in the negative direction, then the above equation can be written to find: (−50kP a − −26kP a) ∗ 2m2 = −48kN This means the pressure difference across the two sides of the door leads to a net force of 48kN from the inside (pointing in the negative direction). In other words, your crazy friend would have to pull inwards with 48kN of force to overcome the pressure difference on the door. This is equivalent to lifting ∼5000kg off the ground. Don’t think it’s going to happen.

1.5 Statics and moments

This utilizes the concepts of hydrodynamics and a force/moment balance. These problems end up being a good exercise in your calculus skills. When a surface is horizontal (or equivalently at uniform depth) in a quiescent fluid, we can show that the pressure acting on it is uniform. Likewise, when the fluid is a gas you can usually assume the pressure acting on a non-horizontal wall as uniform. However, when we depart from this, like an angled wall in water, we find that the pressure acting on the surface is not uniform. We must then perform a more general type of analysis.

1.5.1 Example: a dam wall with one fluid Lets say we have a dam that is 100m high, 200m wide, holding up 100m of water, as seen in figure 1.2. What’s the total force exerted on the wall? Where would the equivalent point force occur to produce the same moment on the wall? This is a simple application of forces and moments, so let’s start by writing an expression for force. We know that the pressure varies with depth, since that’s exactly what the hydrostatic equation tells us! Water has a density of 1000kg/m3, so we start by writing: dP kg m N = −ρg = −1000 ∗ 9.8 = −9800 (1.3) dz m3 s2 m3 This is how fast the pressure changes with height. Note we have assumed a positive z-direction as upward, so the negative value would mean increasing pressure with decreasing height. We also define zero elevation at the surface, so going into the water yields negative height. By taking the reference pressure at the water surface as zero (in other words, looking at the departure from atmospheric pressure, or other-other words: gage pressure) we can calculate the force on the wall by decomposing into infinitesimally small portions:

dF = P dA

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Figure 1.2: schematic of a dam supporting a single ﬂuid. where dA is simply a small area, or the width (which is a constant 200m) times dz. Why dz? well, think on what is the variable here. Does the pressure vary with any other direction? No! Again, your intuition helps here! Now, P varies in z based on equation 1.3. We can integrate to ﬁnd the exact relationship: Z top Z top Z P (top) Z 0 dP = −9800 dz → dP 0 = −9800 dz0 bottom bottom P (z) z N P (top) − P (z) = −9800(0 − z) → P (z) = −9800z m2 So this tells us that the value of the pressure at a given elevation below the surface of the water. Note that you can see at 10m below the surface, the pressure is about 100kP a, one about 1 atmosphere! So we’re doing OK thus far. Now, the fractional area dA is simply 200 ∗ dz, so we can now write: Z top F = dF bottom Z top = P dA bottom Z top = −9800z ∗ 200 dz bottom Z 0 = −1960000z dz −100 0 2 = − 980000z −100 =0 − −980000(−100)2 F =9.8 GN That’s a lot of force. Notice that we got a positive value, which is good news considering we didn’t prescribe a direction the force is applied in! Even though it can be kind of annoying to

Intro to Fluids Notes 11 C. P. Byers 2016 CHAPTER 1. INTRODUCTORY CONCEPTS carry through all these minus signs, it will still get you to the correct spot if you are diligent in your definitions. How about the moment about the base of the wall? Instead of an infinitesimal force, let’s calculate the infinitesimal moment:

dM~ = ~r × dF~ → dM = −z − (−100) dF → dM = −z + 100 dF

I’m using the right-handed coordinate system, so the forces on the dam will produce a positive moment about the base of the dam. Also note that there is a z + 100 term, not just z. Thats due to the moment being about the base, so the greater the depth, the shorter the moment-arm. Therefore, we need to be better about our normal and deﬁne the force on the dam as acting in the negative direction. dF = −P dA This means the force will be acting towards the left on the page. Then, working with our cross product, we get positive moments about the base. Plugging this in gives:

Z top M = dM bottom Z 0 M = −(z + 100) dF −100 Z 0 = (z + 100)P dA −100 Z 0 = (z + 100) ∗ (−9800 ∗ z) ∗ 200 dz −100 Z 0 = −1960000(z2 + 100z) dz −100 3 0 z 2 = − 1960000( + 50z ) 3 −100 −1003 =0 − −1960000 + 50(−100)2 3 M =326 GN · m

Note that I was better with the directions of the forces and moments. The force from the fluid on the dam is applied in the negative direction, while the moment about the base is positive. The calculus worked how we had hoped it to! Next, we can find where the equivalent point force would be applied. By taking the moment, and dividing by the total force, we find: M z = cop F 326 GN · m = 9.8 GN zcop =33.3 m

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From this, we got a distance of 33.3m, which means the location we could apply the point force would be 1/3rd of the way from the moment axis, or 2/3rds of the distance below the surface. This is the location of our center of pressure, and since the pressure distribution is triangular, this makes sense! This simple setup is giving a result that matches what we should have seen in our mechanics classes.

1.5.2 Example: a dam wall with two fluids Now we will change the setup ever so slightly to see what happens. Imagine we have two layers of fluid, where fresh water is floating on top of salt water, seen in figure 1.3. Let’s calculate out the forces, moments, and center of pressure again.

Figure 1.3: schematic of a dam supporting two diﬀerent ﬂuids.

The pressure distribution will be slightly diﬀerent. Let’s ﬁrst look at the top 50 meters:

Z top Z top dP = −ρg dz z z Z 0 P (0) − P (z) = −9800 dz z N P (z) = − 9800z if 0 ≥ z ≥ −50 (1.4) m2 So for the first 50 meters, we wouldn’t see anything different than the previous example. How- ever, once we cross to the next fluid, we need to account for the different density as well as the pressure at the fluid interface.

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Z P (−50) Z −50 dP 0 = −ρg dz0 P (z) z Z −50 P (−50) − P (z) = −10094 dz0 z −50

490000 − P (z) = − 10094 ∗ z z N P (z) = − 10094z − 14700 if −50 > z ≥ −100 (1.5) m2 Notice that the second layer of a fluid has lead to a more complicated expression for our pressure. We have equation 1.4 for the first 50m in the fluid, then equation 1.5 for the second 50m in the fluid. That is how we found the value for P (−50). To calculate the total force, we perform the integration as was done before. We will note that the force is in the negative direction (putting the minus sign on it) and keep our expressions in mind for the different z-locations: Z 0 F = −P dA −100 Z 0 Z −50 = −(−9800 ∗ z) ∗ 200 dz + −(−10094z − 14700) ∗ 200 dz −50 −100 0 −50 2 2 =980000z + (1009400z + 2940000z) −50 −100 F = − 9.87 GN

Notice that this force is just a little bit higher than the previous example? That’s due to the denser ﬂuid on the bottom increasing the overall pressure on the bottom half! Also notice how the sign is negative? That’s due to us correctly accounting for the fact that the pressure acts on the wall to move it in the negative x-direction. The moment will be done in a similar manner, where the axis which we calculate the moment about is located on the bottom of the wall, and making sure to account for the change in the expression for pressure at the z = −50m point.

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Z top M = dM bottom Z 0 = −(z + 100) dF −100 Z 0 = (z + 100)P dA −100 Z 0 Z −50 = (z + 100) ∗ (−9800 ∗ z) ∗ 200 dz + (z + 100) ∗ (−10094z − 14700) ∗ 200 dz −50 −100 Z 0 Z −50 = −1960000(z2 + 100z) dz + −2018800z2 − 204820000z − 294000000 dz −50 −100 3 0 −50 z 2 3 2 = − 1960000( + 50z ) − (672933z + 102410000z + 294000000z) 3 −50 −100 −503 =0 − −1960000 + 50(−50)2 3 + −672933(−50)3 + 102410000(−50)2 + 294000000(−50) − −672933(−100)3 + 102410000(−100)2 + 294000000(−100) M =328 GN · m

Just like the force, the moment is ever so slightly higher than the previous example was. Again this intuitively make sense, as we have a larger force acting on the wall. If we then calculate out the center of pressure, we ﬁnd: M z = cop F 328 GN · m = 9.87 GN zcop =33.2 m

This is only 0.1m lower than the previous center of pressure. Not very significant, but then again the density difference isn’t very significant either. However, this isn’t the centroid of a triangle now! This goes to show that you need to take the time to calculate out the pressures/forces/moments in order to accurately find where the center of pressure is. Overall, this is to show you how being diligent with your signs and writing out your integrals is an important thing when calculating the center of pressure.

1.6 Rigid body motion summary

The previous examples illustrated the power of pressure, but we weren’t considering fluid flow. That is why we used the “Hydrostatic Equation” - the fluids were static! (Yes, the plane moves, but for the analysis we assume the outside pressure is constant for a given altitude). We can also use a similar analysis on fluid that is in motion, but only under a simplified case. Rigid body

Intro to Fluids Notes 15 C. P. Byers 2016 CHAPTER 1. INTRODUCTORY CONCEPTS motion is when the fluid particles do not move relative to one another. When this occurs, the forces on a fluid particle are only the normal and body forces, not shear stresses. So performing the same type of analysis in both the x- and y-directions as was done in the z-direction (for developing the hydrostatic equation) will give you expressions for pressure gradients in all three coordinate directions: ∂P = −ρa (1.6) ∂x x ∂P = −ρa (1.7) ∂y y ∂P = −ρ(g + a ) (1.8) ∂z z These equations of course depend on the way you define your coordinate system, but here we use the standard positive z-direction being up. Also note that the derivatives are now partials, because pressure is a function of multiple variables. The minus signs again can be confusing, but a little intuition helps us. If we were accelerating in the positive x-direction, then we would see a negative pressure gradient with respect to x. This signifies that the pressure decreases as you move in the positive x-direction. Think of a very long car, completely sealed. If it accelerated at 1g, then the fluid within would have the largest pressure at the back, and the smallest pressure up front. This is a negative pressure gradient, because the value drops as you move in the positive direction! Likewise with the z−direction, we can think about what the equation is telling us. If we were to be in an enclosed box with a fluid, then we accelerated in the negative z-direction at 1g, we would find there is no pressure gradient. That is because accelerating downward at 1g is simply freefall! If we didn’t accelerate in the z-direction, then we’re back to hydrostatics. If instead we accelerate in the positive z-direction, we would increase the pressure gradient, and the pressure at the bottom of the box would increase, proportional to the amount we accelerate.

If you need any clariﬁcation on the derivation of these equations, draw a ﬂuid particle and perform a force balance. Don’t ask me, I’m just a set of notes.

Additionally, under rigid body motion, the pressure from a sloped or curved surface can be found. Alternatively, the acceleration needed to produce certain surfaces can be found as well. For instance, an example taken from Prof, Smits’ book asks us to ﬁnd the acceleration required to make water spill out of a cart, as seen in ﬁgure 1.4. However, getting to the equation that determines this isn’t always the most straightforward. We start by writing the total derivative of pressure: ∂P ∂P ∂P dP (x, y, z) = dx + dy + dz ∂x ∂y ∂z This is a general expression for the total derivative of any function that depends on three variables. Now this example doesn’t have any y-dependence, so we can instead write: ∂P ∂P dP (x, z) = dx + dz ∂x ∂z

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Figure 1.4: schematic of an accelerated cart, taken from Prof. Smits’ book.

Next, we can plug in our expressions for the pressure gradients:

dP (x, z) = −ρax dx − ρ(g + az) dz Finally, we want to ﬁnd the slope of a surface (or specify the accelerations to achieve a particular slope). If we are on the surface of water in the atmosphere, we can say the pressure is constant. Think about a ﬂat body of water - the pressure all along the surface is Patm. If the body of water is then accelerated, it will slope somehow. However, the surface is still exposed to atmospheric pressure, so all along that surface the pressure is still constant. Therefore, for us, we can say surface of the water will have a constant pressure, even when under acceleration! Therefore we can write the following for the surface of water:

0 = −ρax dx − ρ(g + az) dz or

ρax dx = −ρ(g + az) dz or a dz − x = g + az dx In this example, there is no vertical acceleration besides gravity. Therefore, the slope is simply −ax/g. Based on the dimensions given, the water level at the back of the cart needs to increase in height by h/3, and the location in which the surface will “pivot” is at the halfway point, or an x-distance of 3h/2. This leads to a negative slope, so we can then write: a dz − x = g dx a − h − x = 3 g 3h 2 2 a = g x 9 Note that the height needed to increase and the pivot point of the free surface were straightforward to find here since the container is nearly full. If there were only a thin layer of water of the bottom, you’d have to get creative with defining the shape of the fluid in the container.

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1.7 Speciﬁc weight and speciﬁc gravity

Specific weight, γ, is simply the gravitational force of a fluid per unit volume. In other words, it is the weight of the fluid divided by the volume. It can be expressed simply as: weight ρV g γ = = = ρg (1.9) volume V This assumed a constant density throughout the fluid volume, which we can assume true for a fluid particle. We can then use the specific weight to re-write our hydrostatic relation: dP = −γ dz Specific gravity is simply a ratio of the specific weight of a fluid to that of water. This requires some reference state for water, but you can choose that. γ ρ g ρ SG = fluid = fluid = fluid (1.10) γwater ρwater g ρwater So a specific gravity greater than 1 means the fluid (or object) has a density greater than water, and less than 1 means the density is less than water.

1.8 Buoyancy

Buoyancy is intimately related to the hydrostatic relation. It also factors in speciﬁc gravity, since the ratio of densities will determine just how buoyant force the object can exert. A straightforward example is to look at a block of aluminum and a block of styrofoam.

SGAl = 2.7 and SGsty = 0.04

We can then show how each of these would behave in water. Your intuition will tell you that the aluminum should sink, and the styrofoam ﬂoat. We can mathematically show that! Utilizing Archimedes principle, we can solve for the amount of water that each would displace when at equilibrium. Starting with styrofoam: X F orces = 0 in equilibrium

Fb − Fweight = 0

ρwater ∗ g ∗ Vdisplaced − ρsty ∗ g ∗ Vblock = 0

ρwater ∗ g ∗ Vdisplaced = SGsty ∗ ρwater ∗ g ∗ Vblock

Vdisplaced = SGsty ∗ Vblock

= 0.04 ∗ Vblock

This tells us that the displaced volume of water will be less than the volume of the styrofoam block. In layman’s terms, that means the block will ﬂoat! In general, the expression Vdisp = SG ∗ Vobject will tell you how much water the material will want to displace. So long as the speciﬁc gravity (ratio of densities) of the material is less than 1, then the object will want to

Intro to Fluids Notes 18 C. P. Byers 2016 CHAPTER 1. INTRODUCTORY CONCEPTS displace less volume than it has. This means some portion of it will remain above the surface of the water. If the SG is greater than 1, then it will want to displace more water than it can. How does something displace more volume than it has itself? It can’t, which means it sinks! Going through the math for the amount of volume the aluminum block would want to displace, we would ﬁnd Vdisp = 2.7 ∗ Vblock. Now, just like we said, it can’t displace more volume than it has, so the block would sink. This methodology is how you can determine the size of a ship needed to move cargo. If you want to move 1000 metric tons of cargo across the ocean, you need to build a ship that can displace 1000 metric tons of water and not sink!

1.8.1 Example: floating some aluminum Lets imagine a container that is 2x2 meters square, and tall enough to contain the following problem (poorly drawn in figure 1.5). We want to float a 1x1x1 meter cube of aluminum in this container. What is the minimum amount of styrofoam that would be required to keep the block from sinking, and how much does the fluid depth increase in the container?

Figure 1.5: the basic schematic for ﬂoating an aluminum block in water.

In order to ﬂoat the block, we need to displace an amount of water equal to its weight.

FAl =ρAl ∗ g ∗ VAl kg m =2.7 ∗ 1000 ∗ 9.8 ∗ 1m3 m3 s2 =26460N

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We thus need to displace an equivalent amount of water, or:

Fwater =26460

ρwater ∗ g ∗ Vwater =26460N 26460N V = water m kg 9.8 s2 ∗ 1000 m3 =2.7m3

This seems familiar right? That’s because we’re putting the material in water, which is what a specific gravity is defined with! For every unit volume of aluminum, it requires 2.7 units of water to balance the force. That’s exactly what specific gravity is! Now we need to figure out how much additional buoyancy force we require to float. Think: the aluminum block will sink, but since it still displaces some water, it will have an upward buoyancy force acting on it. Ever moved rocks in a lake, or an anchor perhaps? It’s easier when it’s in the water than when it’s in air. That’s buoyancy working for you! So lets see what remaining force we need to overcome:

Fremaining =Fbuoyancy − Fweight

=ρwater ∗ g ∗ Vdisplaced − 26460N =9800N − 26460N = − 16660N

This means there is still a remaining 16660N of force that needs to be overcome for the aluminum block to be neutrally buoyant. Now we look into the amount of styrofoam needed to overcome this force. We know it ﬂoats, as was seen previously. We can then see how much force it will exert. From our previous calculations, we were able to show that the volume of water displaced by styrofoam will be 0.04 times the volume of the styrofoam itself. That was because there is some amount of weight to the object, which means there must be some displacement to result in a buoyancy force. Therefore, for every 1.00m3 of styrofoam, we get an eﬀective 0.96m3 of displacement. To obtain the 16660N of buoyancy force, we thus need:

16660N =ρwater ∗ g ∗ Vdisplaced

16660N =1000 ∗ 9.8 ∗ Vdisplaced 3 1.70m =Vdisplaced

This amount of displaced water would yield the appropriate buoyancy force. However, we know that there is only 96% of the volume of styrofoam that contributes to additional buoyancy beyond what is required for the weight of the material. Therefore, we need:

3 0.96 ∗ Vstyrofoam =1.70m 3 Vstyrofoam =1.77m

So to ﬂoat the 1m3 block of aluminum, we would need 1.77m3 of styrofoam to attach to it.

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Let’s do this in a diﬀerent manner. I had gone through and methodically laid out each and every step to see what volume of styrofoam was needed. Instead, we can start with a basic force balance! This is a hydrostatics problem, so all the forces should balance. Therefore:

X F~ =0

Fbuoyancy − Fweight =0

Fb,Al + Fb,st − Fw,Al − Fw,st =0

ρwater ∗ g ∗ VAl + ρwater ∗ g ∗ Vst − ρAl ∗ g ∗ VAl − ρst ∗ g ∗ Vst =0

ρwater ∗ g ∗ (VAl + Vst) − SGAl ∗ ρwater ∗ g ∗ VAl − SGst ∗ ρwater ∗ g ∗ Vst =0

ρwater ∗ g(VAl + Vst − SGAl ∗ VAl − SGst ∗ Vst) =0

(1 − SGAl) ∗ VAl + (1 − SGst) ∗ Vst =0 3 −1.7 ∗ 1m + 0.96 ∗ Vst =0 3 Vst =1.77m

We were able to get the exact same value using a diﬀerent method. Intuition should tell you that each block will have a force due to it’s weight, and a buoyancy force due to the displaced water. Summing the contributions of these forces will balance out to zero, since it’s hydrostatics. The only unknown is the size of the styrofoam block, so no additional equations are needed. Now, we know the volume of styrofoam needed to ﬂoat the aluminum, and we know the volume of aluminum, so let’s calculate how much the water level has increased in the tank. We know the tank has a 2m x 2m footprint, so the calculation is pretty straightforward: V h = A 1m3 + 1.77m3 = 4m2 =0.69m

The height of the water then increases 0.69m due to the additional material in the water.

1.8.2 Example: floating in a different fluid Let’s do the exact same setup, but now the fluid is honey! Assuming honey has SG = 1.42, what is the new volume of styrofoam needed? Doing the force balance again, we find:

Fb,Al + Fb,st − Fw,Al − Fw,st =0

ρhoney ∗ g ∗ VAl + ρhoney ∗ g ∗ Vst − ρAl ∗ g ∗ VAl − ρst ∗ g ∗ Vst =0

SGhoney ∗ ρwater ∗ VAl + SGhoney ∗ ρwater ∗ Vst − SGAl ∗ ρwater ∗ VAl − SGst ∗ ρwater ∗ Vst =0

(SGhoney − SGAl) ∗ VAl + (SGhoney − SGst) ∗ Vst =0 3 (1.42 − 2.7) ∗ 1m + (1.42 − 0.04) ∗ Vst =0 3 Vst =1.08m

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The volume of styrofoam needed to ﬂoat the aluminum in honey has decreased compared to that needed in water. That’s due to the honey being a more dense ﬂuid! Every unit volume of honey displaced has a higher buoyancy force it exerts compared to water.

1.9 Control Volumes

Now we’ll look into some basic applications of mass ﬂux and control volumes. The basic concept we will be sticking to is the conservation of mass: all mass must be accounted for in one way or another. This can be expressed as:

min − mout = ∆msys (1.11) In other words, the diﬀerence between the mass entering the system and the mass leaving the system is the amount of mass accumulated in the system.

1.9.1 Example: accumulating mass A quick demonstration of equation 1.11 can be shown through the following setup. A tank of water with cross-sectional area of 10m2 has an inflow of 10kg/s and outflow of 5kg/s, sketched in figure 1.6. Find the rate in which the water level in the tank changes.

Figure 1.6: a tank with diﬀering ﬂow rates on the inlet and outlet.

We start by looking to equation 1.11 and modifying it for a flow rate rather than a fixed mass. in other words, we differentiate the equation with respect to time. Plugging in values:

m˙ in − m˙ out =m ˙ sys kg kg 10 − 5 =m ˙ s s sys kg 5 =m ˙ s sys

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So the system accumulates 5kg/s of water. We can then look to how this aﬀects the height. The mass accumulation is simply a rate of change of the total mass of the system: d d d dh m˙ = m = ρV = ρAh = ρA sys dt sys dt dt dt Since the density isn’t changing, and the cross sectional area is a constant, the height is the only parameter that changes with time. Therefore, we can ﬁnd the change in height with time: kg kg dh 5 =1000 ∗ 10m2 ∗ s m3 dt m dh 0.0005 = s dt The water level in the tank will increase by 0.0005m/s, or 1.8m/hr.

1.9.2 Example: non-uniform flow The previous example only gave the total mass flux into and out of the tank. In general, many of these types of problems assume a uniform inflow and outflow condition. Let’s now look at a case of non-uniform flow in the inlet and outlet conditions. Assume a parabolic inlet flow with max velocity of 5m/s, and triangular outflow as shown in figure 1.7. If the inlet has an area of 1m x 1m, and the outlet has an area of 0.75m x 1m, what is the max outlet velocity, assuming an incompressible fluid?

Figure 1.7: nozzle with incompressible ﬂow (no mass accumulation).

Once again, this is an application of equation 1.11. We want the mass flux version, so differentiating gives: m˙ in − m˙ out =m ˙ sys But in this case, we are assuming there is no mass accumulating in the system (because it’s a steady flow problem with an incompressible fluid). All mass influx must be balanced by mass

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m˙ in − m˙ out = 0

m˙ in =m ˙ out

To get the max velocity at the outlet, we need to find the total mass flux into the system. We are given a parabolic velocity profile on the inlet, so we can integrate to findm ˙ in. ZZ m˙ in = ρUdA surface

First, let’s determine what the expression for U is. It is parabolic, reaching a maximum at the centerline, and being zero at the edges. We can therefore write it as:

y U = U ∗ 1 − 2 in max 0.5 ∗ h

In this, y is the vertical coordinate. We define the centerline as 0, so the top and bottom of the inlet are at 0.5h and −0.5h, respectively. Now, plugging in our known values, the velocity can be expressed as: m y U = 5 ∗ 1 − 2 in s 0.5 We next write dA based off it having a unit width (into the page). Since the only variation is in the y-direction, this gives dA = 1 ∗ dy. Then, integrating from −0.5m to 0.5m, since the height is 1 meter, we get: ZZ m˙ in = ρUdA surface Z 0.5 kg m y 2 =ρ 3 ∗ 1m ∗ 5 ∗ 1 − dy m −0.5 s 0.5 0.5 4 3 kg =5ρ y − y 3 −0.5 m · s 10 kg = ρ 3 s Now that we have the mass flow in, we know it has to be the mass flow out. Also, we know the area out, and the velocity profile shape. Therefore, we can put these together to find the maximum velocity at the outlet. Since it’s triangular in shape, we can write the profile as: y U =U 1 − for positive y out max 0.5 ∗ h y =U 1 + for negative y max 0.5 ∗ h

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Here, the zero point is on the centerline again, and the total height is 0.75m, so we can write this as: 8y U =U 1 − for positive y out max 3 8y =U 1 + for negative y max 3 Now, since the outflow is symmetric, we could always just integrate one half and double (which could have also been done on the inlet, but here since it’s discontinuous it makes more sense). Then, writing out the total mass outflux, which must be equal to its influx, we get: ZZ m˙ in =m ˙ out = ρUdA surface Z 0.375 10 kg kg 8y ρ =ρ 3 ∗ 1m ∗ 2 ∗ Umax ∗ 1 − dy 3 s m 0 3 2 0.375 5 m 4 2 =Umax y − (y) 3 s 3 0 5 m2 3 =U ∗ m 3 s max 16 80 m =U 9 s max So we find that the max velocity at the outlet is 8.89m/s.

Again, as with the buoyancy example, this was stepped through methodically. If instead you saw that the mass ﬂuxes must balance, and therefore equated the integrals, you would get the same answer in fewer steps:

m˙ in =m ˙ out Z 0.5 Z 0.375 y 2 8y ρ ∗ 1 ∗ 2 ∗ 5 ∗ 1 − dy =ρ ∗ 1 ∗ 2 ∗ Umax ∗ 1 − dy 0 0.5 0 3 0.5 0.375 4 3 4 2 5 ∗ y − y =Umax ∗ y − (y) 3 0 3 0 5 3 =U ∗ 3 max 16 80 =U 9 max 1.10 Material Derivative

One ﬁnal introductory topic that needs to be addressed is the material derivative. Hopefully you’ve already seen it, but the basic idea is that the material derivative is a way to relate changes in the Lagrangian reference frame to changes in the Eulerian reference frame. To refresh your memory, the Lagrangian reference frame is a way to look at a ﬂuid

Intro to Fluids Notes 25 C. P. Byers 2016 CHAPTER 1. INTRODUCTORY CONCEPTS particle where you follow along with it. Think about being a small fluid particle in the air that travels around, and everything you experience is because of the surroundings acting on you as you move through them. The Eulerian reference frame is a way to look at a single fixed point in the flow. Now you are an observer looking at one location, seeing many fluid particles pass through that point. The world of classical mechanics operates within the Lagrangian frame, but with fluid mechanics, we like to think in terms of the Eulerian frame. This stems from the fact that it becomes extremely difficult to track each and every fluid particle in a system - we’re talking orders of magnitude around 1020 particles just in a moderately sized room! Our current computational ca- pabilities cannot handle this, let alone people working it out by hand back in the early founding days of fluid mechanics. So instead, we desire to be able to look at a single point and evaluate what happens at that location. A simple example would be looking at a rocket nozzle. Do you care about the dynamics of each infinitesimally small fluid particle exiting the nozzle, or is it more productive to have an understanding of the flow at the exit of the nozzle? The Eulerian reference frame allows us to look at the nozzle exit and understand the flowfield without tracking each particle. Our problem is that mechanics works with Lagrangian concepts. We desire to take the Eulerian reference frame and apply the Lagrangian concepts we are more familiar with. In general, any property of a fluid particle can be described as a function of time and space when we utilize the Eulerian reference frame. We observe a location (x, y, z) and watch it change in time t. For example, the temperature, density, or any property Φ at that position can be described as: Φ = Φ(t, x, y, z) Since the Lagrangian frame follows a particle that can move, its position will change in time. Therefore: x = x(t), y = y(t), z = z(t) thus Φparticle = Φ t, x(t), y(t), z(t) Looking at the time derivative of the particle (and making sure we use the chain rule), we then see: d ∂Φ ∂Φ ∂x ∂Φ ∂y ∂Φ ∂z Φ = + + + dt ∂t ∂x ∂t ∂y ∂t ∂z ∂t ∂Φ ∂Φ ∂Φ ∂Φ = + U + V + W ∂t ∂x ∂y ∂z This can be compactly expressed as: d DΦ ∂Φ Φ = = + U~ · ∇Φ (1.12) dt Dt ∂t |{z} Lagrangian

This is our material derivative of the quantity Φ, which will be seen throughout the course. Note that we will reference it by utilizing D/Dt, since that is a more distinct nomenclature compared to the Lagrangian d/dt.

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Continuity and Momentum

2.1 Continuity equation

After utilizing the relatively simplistic equation for mass balance (equation 1.11), we want to develop a“fluid dynamics” version of it, which we call the continuity equation. Recall this applies in the Eulerian frame, where we fix a control volume and look at the flow through it. The simple way to explain the continuity equation is taking the concept of conservation of mass and re-wording it:

mass in − mass out = change in mass −or− rate of change of mass in C.V + total mass ﬂux out of C.V = 0

Mathematically, we expressed this as: d ZZZ ZZ ρ dV + ρU~ · dS~ = 0 dt volume surface or, using the unit normal, we can equivalently state: d ZZZ ZZ ρ dV + nˆ · ρU~ dS = 0 (2.1) dt volume surface

It is important to realize that both density and velocity can be a function of position and time, or ρ = ρ(x, y, z, t) and U~ = U~ (x, y, z, t). This makes things considerably more complicated in the most general sense of things, but lucky for us we usually get problems where one or both are constants. Just realize there is no restriction on their temporal or spacial dependence. Also note that the integral over the surface represents a net outﬂux of mass. This is because the dot-product ofn ˆ with U~ will be positive if they are in the same direction, andn ˆ is positive outwards. Therefore, a positive surface integral will have to be balanced with a negative volume integral, implying a net outﬂux of mass and decrease in system volume. Now, there is another way to obtain the continuity equation. We can utilize a wonderful mathematical tool called Reynolds Transport Theorem. If we have an extensive property

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B, and intensive property b, the RTT allows us to relate the change of the extensive property in a Lagrangian view to the Euerian view through: d ZZZ d d ZZZ ZZ b dV = B = b dV + nˆ · bU~ dS (2.2) dt dt sys dt V (t) Vs Ss | {z } | {z } Lagrangian Eulerian

You may have seen this equation before. It is sometimes re-cast in different forms, where instead of b, there is b ∗ ρ. I am defining b here as an intensive property (per unit volume) where the later case has b defined per unit mass. If this gets confusing, don’t fret. We won’t be focusing on RTT so much, mainly on the results that it can provide us. For our convenience, let’s review the Lagrangian and Eulerian viewpoints:

1. Lagrangian: Following the ﬂuid particle/system. Boundaries can deform so there is no ﬂux through them.

2. Eulerian: Fixed Control Volume that allows a flux through boundaries. You look at the whole flow field rather than individual fluid particles.

Therefore, looking to the LHS of equation 2.2, we can see that this is the classical mechanics viewpoint, where as the RHS is the continuum mechanics approach. The RTT is the bridge between ﬂuids and classical mechanics!

Now let’s look at an intensive property, such as density. if we set b = ρ, we plug it into the LHS of equation 2.2 and get: d ZZZ d b dV = B dt dt sys V (t) d ZZZ d ρ dV = B dt dt sys V (t) d d m = B dt dt sys

So, by deﬁning our intensive property b as density, we found the extensive property Bsys to be mass. Is mass an extensive property? Yes! Think about what happens if you cut a volume in half: the mass changes. Think about the intensive property density: it isn’t dependent on the volume of material (in the strictest of senses, of course internal density gradients change things, but let’s not get too complicated here). In the classical mechanics point of view (Lagrangian), what happens to the mass of a closed system? It is conserved! Therefore: d m = 0 dt

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Now let’s look to the RHS of equation 2.2 and plug in our intensive property: d ZZZ ZZ b dV + nˆ · bU~ dS dt Vs Ss d ZZZ ZZ ρ dV + nˆ · ρU~ dS dt Vs Ss and equating the RHS to the LHS, we get: d ZZZ ZZ ρ dV + nˆ · ρU~ dS = 0 dt Vs Ss This is simply equation 2.1! There are many ways to derive the continuity equation, but Reynolds Transport Theorem is just one way to quickly obtain it. There are long-winded proofs on how to obtain RTT, but let’s not worry about that.

2.1.1 Differential form of the continuity equation (bonus material) We can re-cast the continuity equation in differential form. The integral form allows us to look at a control volume and understand the whole system while “glossing over” the details within it. The differential form is a powerful tool for understanding the flow field itself. An important point to remember is that the volume/surface in which the integrals occur over in the Eulerian frame are fixed in space. They don’t move or deform (we can sometimes have a moving frame if needed, but lets not worry about that for now). Because they are fixed, the time derivative can be moved inside the limits of integration. That’s because the boundaries aren’t dependent on time! So the continuity equation can be re-written as: ZZZ ∂ ZZ ρ dV + nˆ · ρU~ dS = 0 (2.3) ∂t Vs Ss Note that when moving the time derivative within the integral, we change it to a partial derivative with respect to time. That’s because the integral “kills” the spacial dependence density may have when you integrate in space, so there is only a (possible) time dependency remaining. By moving into the integral, there still may be a spatial dependency on the density, so we must be clear in that it’s a partial derivative. Another tool we can use is the divergence theorem, also known as Gauss’s theorem. It relates the flux of a vector field through a surface to the change of the vector field within the volume bound by that surface. In mathematical terms, we can express this as: ZZZ ZZ ∇ · F~ dV = nˆ · F~ dS (2.4)

V S By applying equation 2.4 to the equation 2.3, we can now write it as: ZZZ d ZZZ ρ dV + ∇ · ρU~ dV = 0 dt Vs Vs

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Since both integrals apply over the same volume, they can be combined: ZZZ ∂ρ + ∇ · ρU~ dV = 0 ∂t Vs We see that this integral will be zero. Since the volume in which we integrate over is arbitrary (think: you can define a control volume however you want) then it will be zero no matter what boundary of integration is chosen. This in turn means the integrand itself must be zero, giving us: ∂ρ + ∇ · ρU~ = 0 ∂t Expanding out the dot product (a good exercise to practice your vector calculus skills) gives us: ∂ρ + U~ · ∇ρ + ρ∇ · U~ = 0 ∂t We can make one more simplification by utilizing the material derivative (which we introduced in equation 1.12) to get the final differential form of the continuity equation: Dρ + ρ∇ · U~ = 0 (2.5) Dt −with− D ∂ = + U~ · ∇ Dt ∂t 2.1.2 Example: continuity on a nozzle We know how to do a simple mass balance from the previous chapter. Let’s slightly modify a previous example (shown in figure 2.1) and apply the continuity equation to investigate the behavior of the system. We previously assumed a steady, constant density flow to solve for the max velocity at the outlet. Instead, we now have values prescribed at the inlet and outlet, then apply equation 2.1 to gain some insight into this system. First, let’s assume we know the velocity profiles at the inlet (parabolic) and exit (triangular), the depth is 1m into the page, and we’re given the density at the inlet and exit. What is happening to the mass of the system? d ZZZ ZZ ρ dV + nˆ · ρU~ dS = 0 dt volume surface The first term is unknown, while the second term is made of variables we have. The sum of the two should be zero, so it seems we’re good to go (one equation, one unknown). Plugging in the velocity profiles and densities: d ZZZ ZZ ρ dV = − nˆ · ρU~ dS dt volume surface ZZ ZZ = − nˆ · ρU~ dS − nˆ · ρU~ dS

inlet outlet Z 1 Z 0.5 Z 1 Z 0.375 2 8 = − 2 ∗ (−î) · 5ρin 1 − 4y î dy dz − 2 ∗ (î) · 10ρout 1 − y î dy dz 0 0 0 0 3

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Figure 2.1: modiﬁed example for applying continuity equation

Let’s note a few things at this point. First, I’ve performed the integration in y over half the area because it is a symmetric profile, thus there is an additional 2 in each integral. Second, I’ve put in the unit normals, −î in the inlet, and î in the outlet. Third, I’ve specified the unit vector direction of the velocity, which is î in both the inlet and outlet. Continuing the math:

d ZZZ Z 0.5 Z 0.375 8 ρ dV =1 ∗ 2 ∗ 5 ∗ 1.125 ∗ 1 − 4y2 dy − 1 ∗ 2 ∗ 10 ∗ 1 ∗ 1 − y dy dt 0 0 3 volume 0.5 0.375 4 3 4 2 =11.25 y − y − 20 y − y 3 0 3 0 =3.75 − 3.75 =0

So we find that the time rate of change in the overall mass in the system is zero. The system is not accumulating or losing mass! We refer to this type of situation as a steady flow, where the system isn’t changing in time. In this instance, we had a density that was higher at the inlet than the outlet, so there is obviously some sort of density gradient within the system. However, because we’re doing a control volume analysis, we don’t care about the exact details of the system within the boundaries. In fact we can’t even solve for the details within the boundaries - we only know the overall change within the system (in this example, 0). Using a control volume, the only detail we have (or could potentially solve for) is the fluxes at the boundaries!

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2.2 Momentum Balance

Just as we did with density, lets utilize another quantity, speciﬁc momentum b = ρU~ . Substi- tuting this for our intensive property in the RTT then gives the following: d ZZZ d ZZZ ZZ b dV = b dV + nˆ · bU~ dS dt dt V (t) Vs Ss d ZZZ d ZZZ ZZ ρU~ dV = ρU~ dV + nˆ · ρU~ U~ dS dt dt V (t) Vs Ss | {z } Lagrangian d d ZZZ ZZ mU~ = ρU~ dV + nˆ · ρU~ U~ dS dt dt | {z } V S Newton’s 2nd s s X d ZZZ ZZ F~ = ρU~ dV + nˆ · ρU~ U~ dS (2.6) dt Vs Ss This is a simpliﬁed form of the momentum equation. Some notes: 1. The LHS of equation 2.6 was obtained by noting that the integral of ρU~ over the volume in the Lagrangian view is the linear momentum of the entire system. That’s once again because no mass enters or exits the Lagrangian control volume. The rate of change of linear momentum is the true form of Newton’s Second Law, so we can re-write the LHS as the sum of the forces applied on the system.

2. The RHS of equation 2.6 has two parts: the ﬁrst being the time rate of change of momentum within the volume, the second being the net ﬂux of momentum out of the control volume. We will expand on equation 2.6 in later chapters, and express the LHS in terms of pressure, surface, and body forces. Either way, we can see that RTT is a powerful tool. Without going into the details of deriving the forces on the system, we’ll use an example to show how these forces arise.

2.2.1 Example: force on a nozzle Let’s explore equation 2.6 by applying it to the previous problem! We know the inlet and outlet conditions, and we also know that the system is steady in time. This allows us to simplify equation 2.6 even further: 0 ZZZ ¨¨* ZZ X ~ d ~¨ ~ ~ F = ¨ρ¨U dV + nˆ · ρU U dS dt ¨¨ ¨Vs Ss ¨ X ZZ ZZ F~ = nˆ · ρU~ U~ dS + nˆ · ρU~ U~ dS

Ain Aout

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Figure 2.2: it’s the previous problem again!

We split the surface integral to show there is a momentum ﬂux in and momentum ﬂux out, and they must balance with possible forces on the system. To determine what forces we might have, let’s draw the control volume and list out what could be acting on it. I’ll do this C.V. for convenience:

By defining the C.V. in this fashion, it allows us to not worry about any forces that could be acting on the surface of the top and bottom of the C.V. If it coincided with the inner walls of the nozzle, there could be frictional forces from the fluid! However, by “cutting through” the walls, we now have some sort of external force on our system. Think: if there was a net force from the flow, the walls would want to move. By having the C.V go through the wall, we must account for that force acting on the wall to hold it in place. Also, there could be pressure forces at the inlet and outlet, but let’s just assume there isn’t for now. Finally, we will assume gravity is down, so no gravitational forces will be affecting the flow.

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With all this info, we can ﬁnally write our momentum balance as: ZZ ZZ F~ext = nˆ · ρU~ U~ dA + nˆ · ρU~ U~ dA

Ain Aout Now we need to plug in what we know and solve: ZZ ZZ 8 8 F~ = − î · 1.125 ∗ 5 1 − 4y2 î 1 − 4y2 î dA + î · 1.0 ∗ 10 1 − y î 1 − y î dA ext 3 3 Ain | {z } | {z } Aout | {z } | {z } nˆ·ρU~i U~i nˆ·ρU~o U~o Z 1 Z 0.5 Z 1 Z 0.375 64 16 = − 2 1.125 ∗ 5 16y4 − 8y2 + 1 î dy dz + 2 1 ∗ 10 y2 − y + 1 î dy dz 0 0 0 0 9 3 Z 0.5 Z 0.375 64 16 = − 11.25 16y4 − 8y2 + 1 î dy + 20 y2 − y + 1 î dy 0 0 9 3 0.5 0.375 16 5 8 3 64 3 8 2 = − 11.25 y − y + y î + 20 y − y + y î 5 3 0 27 3 0 = − 3î + 2.5î F~ext = − 0.5î, 0 ˆj, 0 kˆ N It looks like the overall force balance was only in the x-direction, and has a negative value. Also note the units; they’re a force as expected. We didn’t carry it through each step, but you should see that they would work out from (kg/m3) ∗ (m/s) ∗ (m/s) ∗ (m2) to give you kg ∗ m/s2, or Newtons.

We should expect the force to only be in the x-direction, as that’s the only direction flow is in. What about the negative value? That signifies the external force on the control volume must be applied in the negative x-direction, or towards the left side of the page. Does that make sense? Let’s think about a nozzle on the end of a hose. The flow is slower in the body of the hose, then reaches the entrance of the nozzle, accelerates, and exits out the end of the nozzle. If there weren’t threading holding the nozzle into place, it would want to pop off and fly away from the hose. It therefore requires an external force in the opposite direction of the flow to hold it on. This is exactly what we found in this example!

The most complicated part (for me) when it comes to a momentum balance is remembering the dot products. This was a quasi-one dimensional problem (no balance needed for the y- or z-components), so it was pretty straightforward. Introducing additional directions can make things a little complicated.

2.2.2 Example: Balance with angled velocities Let’s do one last example with multiple directions of velocity. The previous example was a straightforward application of the momentum equations because there was no ambiguity over the direction of ﬂux or the component of momentum. This example will show how you need to think about what components of the vector apply to what terms in the equation.

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Suppose we have a simple square duct as shown in ﬁgure 2.3. We desire to determine the

Figure 2.3: flow where the velocity has two components. outflow velocity and balance of forces on the system. We will assume that the flow is steady, has uniform inflow and outflow conditions, and has no contributions from pressure.

Start by drawing a control volume:

Once again, I’ve drawn the C.V. in a way that allows us to not worry about possible surface forces. If the C.V. had coincided with the inner walls, we would need to think about what is happening on the surface of the C.V. where the fluid is interacting with it. Starting with the continuity equation, we should apply the integral form, since this is a C.V. application. Recall, the integral approach allows us to look at inflow and outflow conditions while “glossing over” the details of the flow. The differential form would require us to know the flow at all locations in the problem.

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Since we are steady, we can simplify equation 2.1 to get:

ZZZ ¨* 0 ZZ d ¨¨ ¨ ρ dV + nˆ · ρU~ dS =0 dt¨¨ ¨ volume surface ZZ ZZ nˆ · ρU~ dA + nˆ · ρU~ dA =0

Ain Aout Fortunately, we have constant areas, constant density, and velocities that are constant. This allows considerable simplification. However, we have to be careful about the unit normal for the surface and the direction of flow. For the outlet: m U~ = U î, 0 ˆj, 0 kˆ andn ˆ = î o o s

Therefore, at the outlet,n ˆ · U~o = Uo. The inlet is a little more complicated. However, we just need to break the flow out into its components: m U~ = |U | cos 60î, |U | sin 60 ˆj, 0 kˆ andn ˆ = −î i i i s

So at the inlet,n ˆ · U~i = −Ui cos 60. We then can solve:

−ρ ∗ 10 ∗ cos 60 ∗ 1 + ρ ∗ Uo ∗ 1 =0 m U =5 o s Continuity tells us that the mass flow in must balance the mass flow out (because it is steady), and only the x-component of velocity contributed to this mass flux. That’s because there isn’t a surface in which the y-component of velocity could enter the system. Now to momentum, we simply apply the same treatment to the steady momentum equation. We know the inflow and outflow velocities, so let’s plug them in: 0 ZZZ ¨¨* ZZ X ~ d ¨~¨ ~ ~ F = ¨ρU dV + nˆ · ρU U dS dt ¨¨ ¨Vs Ss ¨ZZ ZZ F~ext = nˆ · ρU~i U~i dA + nˆ · ρU~o U~o dA

Ai Ao √ √ F~ext =ρ ∗ − î · 5î, 5 3 ˆj, 0 kˆ 5î, 5 3 ˆj, 0 kˆ ∗1 + ρ ∗ î · 5î, 0 ˆj, 0 kˆ 5î, 0 ˆj, 0 kˆ ∗1 | {z } | {z } | {z } U~i | {z } U~o nˆ·U~i nˆ·U~o √ F~ext =ρ ∗ −5 ∗ 5î, 5 3 ˆj, 0 kˆ + ρ ∗ 5 ∗ 5î, 0 ˆj, 0 kˆ √ F~ext = 0î, −25 3ρ ˆj, 0 kˆ

All we have is a force in the y-direction! The momentum ﬂux in the x-direction balances out so there is no resultant force (think - same area, density, and velocity at the inlet and outlet in the

Intro to Fluids Notes 36 C. P. Byers 2016 CHAPTER 2. CONTINUITY AND MOMENTUM x-direction). However, the y-direction has a force holding the duct in place. That is because there is a net ﬂux into the C.V. that carries y-momentum. Being diligent with your dot product is very important when dealing with momentum ﬂuxes.

Here’s another way to think about momentum vs. mass ﬂux.

• MASS: The velocity dotted with the surface normal provides the ﬂux itself. It carries with it certain properties. In the case of continuity, we’re looking at a mass per unit volume, which is a scalar. That means it has a value at a point. Therefore, for continuity, it’s like we’re carrying an amount of mass through the surface, and it’s dependent on the velocity going into the surface.

• MOMENTUM: We use the same velocity dotted with the surface normal to provide the flux. No matter what it is we are “fluxing,” it is given a magnitude ofn ˆ · U~ . For momentum, we’re fluxing some momentum per unit volume (ρU~ now), which is a vector quantity. If there was no velocity normal to the surface which we pass through, then there would be no flux, thus no momentum being carried into the C.V. However, with a flux into the surface, it carries the momentum of the fluid into it. Think: a small particle just on the outside of the C.V can have x-, y-, and z-momentum. An infinitesimally small amount of time later, it will have crossed the control surface (thanks to the flux into it), but still carry its momentum.

Big wall of text, I know, but I hope there is a bit of insight in it for you. The point I’m trying to convey is that for our example, only the x-component of velocity carried mass into the system. But that same component is responsible for the ﬂux of any parameter we want to look at. This means that, for momentum, a y-component of velocity will contribute to the force balance even though it doesn’t contribute to the ﬂux.

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The Momentum Equation

3.1 Integral & Diﬀerential momentum equation

Our work with the momentum equation thus far has been limited to the integral form, and a simplified one at that. There are a number of forces that can be applied to a fluid particle, so lets break down the few that are common in fluids. We know the pressure acting on a face is the magnitude of the pressure integrated over the area, and acts in the opposite direction of the surface normal. We represent this as: ZZ F~pressure = − nPˆ dS (3.1) S Note that the pressure is a scalar, but when acting on the normaln ˆ, we get a vector. Additionally, the minus sign indicates that a pressure is pushing on a surface, since it’s direction is opposite the surface normal. Another force to consider is the gravitational force, which is just a body force. This means it acts throughout the body (not just on the surface like pressure did). We know it is proportional to the weight, so we take an infinitesimal volume, calculate the weight, and sum them up to get the total gravitational force: ZZZ F~gravity = ρ~gdV (3.2) V Note that the direction of the force is set by the gravitational vector. This is why we ignore gravitational effects when considering flow in the x-direction: gravity often applies in the y- or z-direction (under traditional coordinate systems). The last force we should consider (for now) is the viscous force. We’ve been neglecting it thus far, and for good reason. It likes to make things a bit more complex. However, it is a very important component in most fluid flow applications. For now, we can incorporate equations 3.1 and 3.2 into the simplified momentum equation (equation 2.6) to get: d ZZZ ZZ ZZ ZZZ ρU~ dV + (ˆn · ρU~ )U~ dS = − nPˆ dS + ρ~gdV + F~ (3.3) dt viscous V S S V I’ve neglected the external force that acts through control volumes. That’s because this analysis is only going to pertain to small fluid particles, and not through solid objects in the system.

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The viscous component is one that we could spend weeks discussing at a graduate level, going into the form of the viscous stress tensor, whether the fluid is isotropic or not, and considering compressibility issues. That’s not productive at the introductory level, and only serves to confuse. For the most part, we will allow for a simple treatment of viscous effects. If we consider that viscous effects are shears and stresses on a surface, we can conceptually think of the viscous component of force as being the sum of these forces over the surface of the control volume. In other words, we would have: ZZ F~viscous = nˆ · T v dS (3.4) S

I’ve not said anything about the form of T v, but lets just call it some viscous force tensor that acts on the surface. Now we substitute equation 3.4 into 3.3 to get: d ZZZ ZZ ZZ ZZZ ZZ ρU~ dV + (ˆn · ρU~ )U~ dS = − nPˆ dS + ρ~gdV + (ˆn · T ) dS (3.5) dt v V S S V S Equation 3.5 is the integral form of the momentum equation in a fluid flow! In the previous chapter there was some bonus material that took the integral form of the continuity equation to the differential form, and we will do the same here. We apply a lovely math trick called the divergence theorem, which relates surface fluxes of a vector to a change in the volume of that vector. We start by re-writing the pressure term in the form of a dot product: ZZ ZZ − nPˆ dS = − (ˆn · P I) dS

S S where I is the identity matrix. I’ve written it in this fashion to help show that it too is a tensor, but a simple one at that. Think about what happens if you take the unit normaln ˆ and dot it with the identity matrix: you get a vector of each component of the normal, or simply just get nˆ back! Do note that the magnitude ofn ˆ below is not unity, but the idea is all the same.   P 0 0 nˆ · P I = î, ˆj, kˆ ·  0 P 0  = Pî, Pˆj, P kˆ =nP ˆ 0 0 P By recasting the pressure term in this form, we can now apply the divergence theorem to the second, third, and fifth terms in our momentum equation. The divergence theorem for a unit normaln ˆ and vector F~ can be expressed as: ZZ ZZZ nˆ · F~ dS = ∇ · F~ dV

S V so equation 3.5 can be written: ZZZ ∂ ZZZ ZZZ ZZZ ZZZ ρU~ dV + ∇ · (ρU~ )U~ dV = − ∇ · P I dV + ρ~gdV + ∇ · T dV ∂t v V V V V V

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Because we’re using a ﬁxed control volume, the derivative on the ﬁrst term was able to be brought inside (the limits of integration don’t depend on time, so there are no additional terms introduced by moving it in). Now, since these integrals are all evaluated over the same control volume, we can group them together:

ZZZ ∂ ρU~ + ∇ · (ρU~ )U~ + ∇ · P I − ρ~g − ∇ · T dV = 0 ∂t v V Just like we saw with the continuity equation, we see that the integral is zero. Since the volume of integration is arbitrary, the integrand must always be zero no matter what CV is chosen. This means: ∂ ρU~ + ∇ · (ρU~ )U~ + ∇ · P I − ρ~g − ∇ · T = 0 ∂t v We can simplify the pressure term much in the same way we had made it a tensor:

 P 0 0  ∂ ∂ ∂ ∂P ∂P ∂P ∇ · P I = , , · 0 P 0 = , , = ∇P ∂x ∂y ∂z   ∂x ∂y ∂z 0 0 P

So rearranging the terms, we get: ∂ ρU~ + ∇ · (ρU~ )U~ = −∇P + ρ~g + ∇ · T (3.6) ∂t v Equation 3.6 is the differential form of the momentum equation in a fluid flow! This is almost the same as the Navier-Stokes equations, but we need to do a little bit of simplifying first. Starting by expanding out the LHS derivatives, we get: ∂ ∂ U~ ρ + ρ U~ + (ρU~ · ∇)U~ + (∇ · ρU~ )U~ = −∇P + ρ~g + ∇ · T ∂t ∂t v This expansion may seem confusing, but if you write it all out in terms of each derivative and each component of velocity, you’ll find this is indeed the correct form. Do it yourself to practice your vector calculus skills! Then, grouping together some terms:

∂ ∂ U~ ρ + (∇ · ρU~ ) + ρ U~ + (ρU~ · ∇)U~ = −∇P + ρ~g + ∇ · T ∂t ∂t v

Do you see something unique about the term in brackets? This is just the diﬀerential form of the continuity equation! Since the continuity equation tells us the bracketed term is always zero, we can drop it to achieve

∂U~ ρ + ρ(U~ · ∇)U~ = −∇P + ρ~g + ∇ · T (3.7) ∂t v D ∂ ~ We can further simplify this by utilizing the material derivative, Dt = ∂t +U ·∇, and by assuming a form for the viscous stresses. By assuming the ﬂuid is incompressible, we can represent the

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T viscous forces as T v = µ ∇U~ + (∇U~ ) , and by further assuming viscosity µ is constant, we get the viscous term as:

T 2 ∇ · T v = ∇ · µ ∇U~ + (∇U~ ) = µ∇ U~ with ∂2U ∂2U ∂2U ∂2V ∂2V ∂2V ∂2W ∂2W ∂2W ∇2U~ = + + ˆi + + + ˆj + + + kˆ ∂x2 ∂y2 ∂z2 ∂x2 ∂y2 ∂z2 ∂x2 ∂y2 ∂z2

This is another great exercise in your vector calculus skills. First, note that ∇U~ is a vector operation on each component of another vector, which results in a tensor. Then we add the transpose of that tensor to itself. When we take the divergence of this tensor, we are applying a dot product across each row, resulting in a vector, as shown above. Finally, recall that we assumed the ﬂuid was incompressible (∇ · U~ = 0), which means that there will be a number of terms that cancel out, leaving us with the ﬁnal form of µ∇2U~ . While there have been a number of simplifying assumptions to get to this point, we don’t need to worry about it at this level. The point is, we have a nice concise expression now for the viscous term. Putting this and the material derivative into equation 3.7 then gives the the incompressible Navier-Stokes Equations:

DU~ ρ = −∇P + ρ~g + µ∇2U~ (3.8) Dt

3.1.1 Example: momentum practice Before we get into what the Navier-Stokes Equations can tell us about a flow when we know the velocity profile, lets do a classic CV analysis. Assume for a rectangular channel that the velocity profile can be described by the equation: m U(y) = 51 − y2 s where the geometry is defined in figure 3.1. Note that we will assume it is infinitely long in the

Figure 3.1: rectangular channel with a parabolic velocity proﬁle z-direction, so we can neglect any velocities or gradients in z. This is equivalent to calling it a

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2D flow. Let’s also assume gravity is acting downwards in the y-direction. What would happen if you did a CV approach to this problem? Let’s start with analyzing CV1 in figure 3.2. This problem is steady, and there are no fluxes in the y-direction. We can then simplify equation 3.5 to get our integral form: ZZ ZZ ZZ ZZZ ZZ 0 + (−ρUx)U~ dA + (ρUx)U~ dA = − nPˆ dS + ρ~gdV + (ˆn · T v) dS + F~ext

Ain Aout S V S

Figure 3.2: two diﬀerent control volumes applied to the problem

At this point, we can simplify the LHS by plugging in our velocity proﬁle. If we assume a unit depth into the page for simplicity, we get:

Z 1 Z 1 −ρ U 2 dy + ρ U 2 dy = RHS −1 −1 You can see that the LHS is now zero! This is because the velocity profile doesn’t change from the inlet to the outlet - the flux in matches the flux out, so the sum of the two is zero. Then, we’re left with the RHS terms: Z 1 Z 1 ZZ x-direction: 0 = Pin dy − Pout dy + (ˆn · T v) dA +Fext,x −1 −1 A | {z } just x-component ZZ y-direction: 0 = −Wfluid + (ˆn · T v) dA +Fext,y A | {z } just y-component

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These simplifications take place because gravity doesn’t act in the x-direction, and the gravitational force integrated with density over the volume gives the weight of the fluid in the y-direction. The pressures in the y-direction balance over the control surface, so they can be neglected. We then need to know what the viscous forces are doing. By writing out the 2D viscous force tensor, we can find: ∂U ∂U ∂U ∂V ! ∂x + ∂x ∂y + ∂x 0 −10µy T v = µ ∗ ∂V ∂U ∂V ∂V = ∂x + ∂y ∂y + ∂y −10µy 0 Now keep in mind that this tensor is only valid in the flow. Therefore, we can immediately see that the top and bottom surfaces of the CV will not have any viscous contributions, as those faces lie outside the flow. The inlet and outlet surfaces have the unit normal as ±(1î, 0ˆj), so when we dot this with the viscous tensor we get: 0 −10µy 0 nˆ · T = (±1 0) · = v −10µy 0 ±10µy Looking in the x-direction, we end up getting zero! In the y-direction, we have an antisymmetric function (positive on the inlet, negative on the outlet) that is integrated from −1 to 1 at both the inlet and outlet, giving us zero contribution from the viscous forces! Therefore, for this control volume, there are no viscous forces! Overall, we find for CV1:

Z 1 x-direction: ∆Px dy = Fext,x −1

y-direction: Wfluid = Fext,y

Now let’s look at CV2. It should be apparent that the influx and outflux of momentum doesn’t change, so the LHS of the equation is still zero. However, there is no external force now because we’re not cutting through a solid object! Then, our expressions in each direction becomes: Z 1 Z 1 ZZ x-direction: 0 = Pin dy − Pout dy + (ˆn · T v) dA −1 −1 A | {z } just x-component Z L Z L ZZ y-direction: 0 = −Wfluid − Pbottom dy + Ptop dy + (ˆn · T v) dA 0 0 A | {z } just y-component Notice how we need to account for the pressure along the top and bottom of the CV, because the surface is in the flow and may not have the same pressure on them anymore! Additionally, because the new top and bottom surfaces are in the flow, we need to include these normals in our viscous integrals. Lets look: ZZ ZZ ZZ ZZ ZZ (ˆn · T v) dA = (ˆn · T v) dA + (ˆn · T v) dA + (ˆn · T v) dA + (ˆn · T v) dA A inlet outlet top bottom

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With our viscous stress tensor being the same as it was before, we can now apply the following normals: nˆinlet = (−1, 0)n ˆoutlet = (1, 0)n ˆtop = (0, −1)n ˆbottom = (0, 1) and then the viscous integral becomes:

ZZ ZZ 0 ZZ 0 ZZ 10µy ZZ −10µy (ˆn · T ) dA = dA + dA + dA + dA v 10µy −10µy 0 0 A inlet outlet top bottom

Once again, the integration over the inlet and outlet goes to zero, since it is an antisymmetric function that is integrated from −1 to 1. You should hopefully then see that there is only an x-contribution to the force balance from viscous forces! This should have been your initial thought as well, since we know viscous stresses are caused by shearing forces on a surface. Since there are only going to be shears on y-normal surfaces, and they are acting in the x-direction, we would expect there to only be a viscous force in the x-direction. We can then write our integrals out along the top and bottom, and simplify our force balances: Z 1 Z L Z L x-direction: ∆Px dy = 10µ(1) dx + −10µ(−1) dx −1 0 0 Z L Z L y-direction: 0 = −Wfluid − Pbottom dy + Ptop dy 0 0 Note that we’ve plugged in the y-coordinate in which the force takes place for each of the integrals, and then we integrate along the length of the wall in which the CV applies. Simplifying this gives us: Z 1 x-direction: ∆Px dy = 20µL −1 Z L y-direction: Wfluid = ∆Py dy 0 So in the x-direction, we now see that the pressure at the inlet will be higher than the pressure at the outlet, proportional to the length of the control volume. Using a different control volume lead us to getting a different force balance. If we look to the two different results, we find that the viscous force is that external force we initially found in the x-direction, and the pressure difference is the external force we found in the y- direction! However, we still don’t know the fine details, just that some sort of pressure drop in the x-direction will have to balance the viscous forces on the wall, and that some pressure drop change in the y-direction will have to balance the weight of the fluid.

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3.1.2 Example: differential momentum practice If we were to plug the velocity profile into equation 3.8 rather than doing the integral approach, we’d find the following:

∂U~ ρ + ρU~ · ∇U~ = −∇P + ρ~g + µ∇2U~ ∂t ∂U ∂U ∂U ∂P ∂2U ∂2U x-direction: ρ + ρ U + V = − + µ + ∂t ∂x ∂y ∂x ∂x2 ∂y2 ∂V ∂V ∂V ∂P ∂2V ∂2V y-direction: ρ + ρ U + V = − − ρg + µ + ∂t ∂x ∂y ∂y ∂x2 ∂y2 Take a moment to look at how we broke the equation into the two vector components. The most diﬃcult part is the “advective” term, or the non-linear term. The best way to think of how this works is to treat the term U~ · ∇ as an operator. This operator acts the same on all three components of velocity. It’s much the same as the ∇2 operator, in that the form looks the same for all three components of velocity: ∂ ∂ ∂ U~ · ∇ = U + V + W ∂x ∂y ∂z and ∂2 ∂2 ∂2 ∇2 = + + ∂x2 ∂y2 ∂z2 So hopefully you can see that the 2D form of these operators have been applied in both the x- and y-directions, and that they are identical in their form, regardless of which component it acts on!

Since U = U(y), and V = 0, these equations reduce to:

∂P x-direction: 0 + 0 + 0 = − + µ 0 − 10 ∂x ∂P y-direction: 0 + 0 + 0 = − − ρg + µ 0 + 0 ∂y which we then can rearrange to show: ∂P P a = −10µ ∂x m ∂P P a = −ρg ∂y m The ﬁrst equation is telling us the magnitude of the pressure change in the x-direction. We see that it is negative in value, implying that pressure is higher at the inlet and lower at the outlet, exactly as we found in the CV analysis! There’s actually a lot of physical meaning in this: 1. The Navier-Stokes Equation has a viscous component, which we found was non-zero in the x-direction.

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2. The LHS was zero, implying there was no net change of momentum due to the flow velocity itself. Recall this side of the equation was from the fluxes in the integral equation. It’s d nd also equivalent to the dt (m~v) term of Newton’s 2 law. 3. Since there wasn’t a net force from a momentum change in the flow, the viscous forces must be balanced by another force, which we find is the pressure forces!

Therefore, we can see that this expression in the x-direction means that the pressure within the channel must drive the ﬂow to oppose the viscous stresses, and this results in a pressure drop as you move towards the right. This is quite literally what a pump is doing - increasing the pressure of the system to drive the ﬂow and overcome viscous forces.

The second equation is simply the hydrostatic equation but in terms of y, not z. But how is this possible? The ﬂuid isn’t static, right? That’s indeed right, the ﬂuid is in motion! However, this is something important to notice and realize:

The hydrostatic relation is always aﬀect- ing the ﬂuid, even when it is in motion!

This means that even with the flow, the pressure will increase from the top of the channel to the bottom of the channel. Now, in a fluid like air, and with relatively small distances, this pressure increase can be extremely small. Even in water, with small distances we can neglect this. However, over the atmosphere, or into a lake or ocean, the pressure change with elevation is still governed by the hydrostatic equation. However, any additional flow in the vertical direction may change this pressure gradient.

Overall, we see that the Navier-Stokes Equations can give us a lot of information about a flow field. In our case, since the parabolic velocity profile is assumed to be all throughout this flow, we can say that the pressure gradient relations we calculated apply everywhere in the flow. That’s the power of the differential fluid equations - they apply to the flow field itself. A control volume analysis would lose all these fine details.

3.2 Bernoulli’s equation

1 ρU 2 + P + ρgz = Const. (3.9) 2 This is a favorite equation for most undergrad ﬂuids courses. It’s quite powerful in it’s use, but only for a certain class of ﬂows. It can only be applied if one of the the following set of conditions are true:

1. steady, inviscid, constant density, irrotational ﬂow

2. steady, inviscid, constant density ﬂow along a streamline

If either of these sets of rules isn’t true, then equation 3.9 cannot be used. However, we will usually apply some assumptions or simplifications to many of our fluids problems to allow us the use of Bernoulli’s equation. You’ll even find out that there are situations that we use it where

Intro to Fluids Notes 46 C. P. Byers 2016 CHAPTER 3. THE MOMENTUM EQUATION the assumptions aren’t strictly true, but give us a close enough approximation. That’s the heart of engineering for you. Since the LHS of equation 3.9 is equal to a constant, then it should hold for anywhere in a flow that Bernoulli’s equation applies. That means, for any two points in a flow: 1 1 ρU 2 + P + ρgz = ρU 2 + P + ρgz 2 1 1 1 2 2 2 2 This becomes useful when you can pick points in a flow where you know some reference values. For example, a good point to pick would be at the surface of a fluid, where P = Patm, or perhaps at the top of a large body of water where V = 0. Also, we can see that rearranging the gravitational terms gives us a difference in height. Therefore, it is advantageous to define z = 0 at one of the locations, since we’re only concerned about the relative distance between the two.

3.2.1 Rigorous derivation of Bernoulli’s equation (bonus material) This is a nice look into how you can derive equation 3.9 from the momentum equation.

First, we start with the Navier-Stokes equation, which we were able to get earlier: ∂U~ 1 1 + U~ · ∇U~ = − ∇P + ~g + ∇ · T ∂t ρ ρ v This is the most general form, where we haven’t assumed anything about the variables involved in the equation. The continuity equation has been applied, allowing us to write the LHS as it is. Also, we have divided through by ρ, but we haven’t assumed anything about ρ. Let’s start by re-writing some terms by using some vector identities: 1 U~ · ∇U~ = ∇ U~ 2 − U~ × ∇ × U~ 2 Plugging this in and rearranging gives: ∂U~ 1 1 1 + ∇ U~ 2 + ∇P − ~g = U~ × ∇ × U~ + ∇ · T ∂t 2 ρ ρ v Next we look to the gravity vector. It is a conservative body force because the net work done by it is independent of the path taken. From this property, we can write any conserved force as a gradient of a potential: ~g = −g∇z Here, g is a constant, and z is the elevation. Additionally, we can re-write U~ 2 as U 2, because the square of a vector is simply the dot product of that vector with itself. Therefore, we can write it as the scalar U 2. Now, plugging in these expressions, we get: ∂U~ 1 1 1 + ∇ U 2 + ∇P + g∇z = U~ × ∇ × U~ + ∇ · T (3.10) ∂t 2 ρ ρ v This equation is simply the momentum equation with a few re-arrangements. Again, this is all rigorous without any assumptions to this point. However, now we can simplify equation 3.10 if we assume:

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1. Inviscid: ∇ · T v = 0 ∂ 2. Steady flow: ∂t → 0 1 P 3. Constant density: ρ ∇P = ∇ ρ The first assumption allows us to neglect the stress tensor. This term is where viscosity lies, so if we assume an inviscid solution, then we don’t have it! The second assumption allows us to get rid of the first term in equation 3.10, and the third assumption allows us to move density inside the gradient operator on the pressure. We then have: 1 P ∇ U 2 + ∇ + g∇z =U~ × ∇ × U~ 2 ρ 1 P ∇ U 2 + + gz =U~ × ∇ × U~ 2 ρ At this point, you should notice that the three assumptions we’ve made all appear in the two sets of assumptions we made at the beginning of section 3.2. However, there is still two more simplifications we can make: either irrotational flow, or flow along a streamline. 1. Irrotational Flow: In irrotational flow, we can represent the velocity as a gradient of a potential: U~ = ∇Φ. If this is then plugged into the RHS, we get ∇ × ∇Φ, which is the curl of a gradient. This is always zero! The expression is now: 1 P ∇ U 2 + + gz = 0 2 ρ Since the terms in the brackets are scalar values, and the gradient of their sum is zero, that implies that everything in the brackets is constant. That returns us to equation 3.9! 2. Flow Along a Streamline: When moving along a streamline, we follow the trajectory of the velocity at a point. In other words, an infinitesimal distance along a streamline d~` is parallel with the velocity field. To represent movement along the streamline, you take the dot product of our equation with d~`. However, the expression U~ × ∇ × U~ gives a vector perpendicular to the velocity (think about what a cross product does). Therefore, U~ × ∇ × U~ · d~` = 0. So now we are at: 1 P ∇ U 2 + + gz · d~` = 0 2 ρ The gradient of any scalar, let’s say B, dotted with a streamline can be represented as: ∂B ∂B ∂B ∇B · d~` = , , · d~` ∂x ∂y ∂z ∂B ∂B ∂B = , , · dx, dy, dz ∂x ∂y ∂z ∂B ∂B ∂B = dx + dy + dz ∂x ∂y ∂z =dB

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This tells us that the LHS dotted with a streamline gives us the perfect differential of the quantity in the brackets: 1 P d U 2 + + gz = 0 2 ρ Since that equals zero, integration once will give a constant, once again bringing us back to equation 3.9. So we can see that there are two different sets of assumptions that both give us Bernoulli’s equation. However, there are still more ways to go about deriving equation 3.9 that utilize different approaches! We don’t need to worry about that here, and in all reality this derivation is probably more rigorous than necessary for an undergrad course. However, it’s always a good thing to see how and why we use the equations we use. It also helps reinforce the conditions in which we can apply Bernoulli’s equation and the limitations involved.

3.2.2 Pressure and streamlines I just want to give a quick summary of an important concept in ﬂuids: pressure and streamlines. From the rigorous derivation of Bernoulli’s equation along a streamline (case 2 above), we know that the force balance along a streamline is simply Bernoulli’s equation! That’s because we took the momentum equation (a continuum representation of force balance), dotted it with an inﬁnitesimal streamline d~`, and found equation 3.9. A similar process can be done to look at what happens across streamlines, which is outlined in a number of undergrad books. I won’t step through the derivation here, but I do wish to reiterate the summary we get from that math: The pressure across straight streamlines is a constant if you can neglect gravitational forces!

3.3 Example: ﬂow over a bump

Determine the magnitude and direction of the horizontal component of force exerted by the fluid on the bump (figure 3.3) in terms of ρ, V1, H1, and width w. Assume uniform inflow and outflow over the areas, and that H1/H2 = 2. As with most fluids problems, the first step we can take is to utilize the continuity equation. Assuming the flow is steady, let’s determine the value of V2 in terms of our inflow parameters: 0 ∂ ZZZ ¨¨* ZZ ¨ρ dV + nˆ · ρU~ dS =0 ∂t ¨ ¨¨ ¨ Vs Ss ZZ ZZ nˆ · ρU~ dA + nˆ · ρU~ dS =0

Ai Ao

−ρV1wH1 + ρV2wH2 =0 H ρV w 1 =ρV wH 2 2 1 1 V2 =2V1

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Figure 3.3: Image credit thanks to A. Smits, “A Physical Introduction to Fluid Mechanics.” John Wiley & Sons Inc. 2000

Therefore, the velocity out is twice the velocity in. This could be intuited since all fluid properties are constant, and the area out is half the area in, but it’s nice to see the math prove it. Next, we need to look into what the pressures in the system are. There are a few tools we can utilize to figure this out. First, note at the exit how the streamlines look. The flow is uniform out, which implies straight streamlines. That means there are no pressure gradients across the streamlines (assuming we neglect gravity). Therefore, the pressure across the outlet is the same as the pressure below the outlet. We also see the same case on the inlet, so the pressure is uniform at that location as well. Let’s investigate what the system would look like with no flow:

We can see that without any flow, we just have a constant atmospheric pressure everywhere. Then, when we move back to the case of flow, since the outlet streamlines imply the pressure doesn’t change across the outlet, we can assume the pressure in that region is just Patm. Additionally, the pressure at region 1 is just Patm plus whatever additional flow pressure may exist. Since we already know the flow is steady and has constant density, let’s further assume it’s

Intro to Fluids Notes 50 C. P. Byers 2016 CHAPTER 3. THE MOMENTUM EQUATION inviscid (a safe bet in this case) and irrotational. That allows us to apply Bernoulli’s equation! Utilizing the locations below, we know the velocities, and now the pressures. Additionally, we’ve

neglected gravitational effects in our assumption that straight streamlines mean no pressure gradient, so we can neglect the gravitational contribution in the Bernoulli equation. Therefore, we have: 1 0 ρU 2 + P + ρgz¨¨* = const. 2 ¨ 1 1 ρV 2 + P + P = ρV 2 + P 2 1 1 atm 2 2 atm 1 1 ρV 2 + P = ρ(2V )2 2 1 1 2 1 3 P = ρV 2 1 2 1 So now we have the pressure in terms of the velocity and density that we’re given. The final task is to find the force on the step, so we can perform a CV analysis using the momentum equation. There are a number of ways to draw the CV, but I’ll just choose one that cuts through the bump:

Looking to what is acting on the CV, we have ﬂow into it on face 1, ﬂow out of it on face 2, pressure acting on both the left and right faces (and that’s all of the right face, not just the

Intro to Fluids Notes 51 C. P. Byers 2016 CHAPTER 3. THE MOMENTUM EQUATION portion where the flow exits), and a cut through a solid body on the bottom. The savvy student would ask: “what about the fluid interacting with the walls on the top and bottom?” The even more savvy student will say: “we are treating it as inviscid, so there aren’t any frictional forces!” Hooray for being savvy! Then, applying the momentum equation, we only have an external force and pressure forces. This gives us: 0 ZZZ ¨¨* ZZ X ~ d ~¨ ~ ~ F = ¨ρÜ dV + nˆ · ρU U dS dt ¨¨ ¨Vs Ss ZZ ¨ZZ ZZ F~ext + − nPˆ dS = nˆ · ρV~1 V~1 dA + nˆ · ρV~2 V~2 dA

Ss Ai Ao ZZ ZZ ZZ ZZ F~ext + − nˆ(P1 + Patm) dA + − nPˆ atm dA = nˆ · ρV~1 V~1 dA + nˆ · ρV~2 V~2 dA

Aleft Aright Ai Ao 2 2 Fext,x + P1wH1 + PatmwH1 − PatmwH1 = − ρV1 wH1 + ρV2 wH2 (in x-direction) | {z } | {z } left side right side H F + P wH = − ρV 2wH + ρ(2V )2w 1 ext,x 1 1 1 1 1 2 3 F + ρV 2wH =ρV 2wH ext,x 2 1 1 1 1 1 F = − ρV 2wH ext,x 2 1 1 So we ﬁnd that the external force on the control volume is in the negative x-direction. Since this force required to hold the CV in place is negative, the force exerted by the ﬂow is thus in the positive x-direction. Therefore, the overall force on the step (which is what the external force 1 must be acting through) is ρV 2wH . 2 1 1

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Introducing Potential Flow

4.1 Some ﬂow deﬁnitions and characteristics

This is just a review of a few key definitions for fluid flows. We want to somehow combine these to solve for different types of flow fields under different assumptions. The idea is to see when we can combine these, and how they work together.

4.1.1 Vorticity Vorticity, ~ω, is an important concept in fluid mechanics, and we use it to measure rotation. It arises in many flows, and can make a headache of some problems. In fact, vorticity is one of the defining characteristics of turbulent flows, and we can build transport equations for vorticity, much in the same way we build transport equations for mass (continuity) and momentum (Navier-Stokes). We’re not so interested in those details at this level, so for now we simply desire to define ~ω as the curl of the velocity field: ∇ × U~ = ~ω (4.1) We can evaluate the curl by taking the determinant of the following matrix: ˆ î ˆj k ∂ ∂ ∂ ∂w ∂v ∂w ∂u ∂v ∂u ∇ × U~ = = î − − ˆj − + kˆ − ∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y u v w So then we have vorticity respresented as:

∂w ∂v ∂u ∂w ∂v ∂u ~ω = − , − , − (4.2) ∂y ∂z ∂z ∂x ∂x ∂y

Notice that each component of vorticity contains the other components of velocities and gradients. For example, the x-component of vorticity has the y- and z-components of velocities, and y-and z-derivatives. The same holds for the other components of vorticity. This means that a 2D ﬂow that has vorticity will have it in the neglected dimension. For example, consider the parabolic velocity proﬁle: y U(y) = U 1 − 2 cl D

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We we say this velocity profile represents a 2D flow. That’s because we have a velocity in the x-direction, and it varies in y. There is no velocity or variation in z, thus only 2 dimensions are relevant! Now, when we compute the vorticity we get: 2U y ~ω = 0, 0, cl D So even though the flow is 2D and in the x-y plane, the vorticity is in the z-direction. This is akin to a moment in statics. That’s just a little bonus information to keep in mind.

Why do we care about vorticity? When we can neglect it, we are dealing with considerably simplified flows! In the parabolic velocity profile above, we found there was a component of vorticity due to the velocity gradient. This gradient will also give rise to viscous forces in the flow. In general, a viscous shearing stress requires a gradient in the flow. Another property of a viscous fluid flow is that any contact with a surface results in rotational effects. Since there are rotational effects due to viscosity, which in turn means there is vorticity, we can say that an irrotational flow can be treated as inviscid. Let’s see how this actually works. We look to the vector curl identity:

∇ × (∇ × U~ ) = ∇(∇ · U~ ) − ∇2U~

This is a vector identity, and is always true for any vector U~ , not just our velocity field. Now, if we have an incompressible flow (or like we said for Bernoulli, ρ is constant), then the differential form of the continuity equation takes the form ∇ · U~ = 0. Then, if we say that the flow is irrotational, we have ~ω = ∇ × U~ = 0. These two properties combined gives us:

∇ × (0) = ∇(0) − ∇2U~ → ∇2U~ = 0

The viscous term in the Navier-Stokes Equations is of the form µ∇2U~ , so we can see that incompressible, irrotational ﬂow will imply an inviscid ﬂow as well!

4.1.2 Velocity Potential If we’ve assumed a ﬂow is irrotational, we’ve decided to neglect viscous eﬀects. Mathematically, we can represent that as: ∇ × U~ = 0 (4.3) Since this is a vector, each component of vorticity must be zero: ∂w ∂v ∂w ∂v − = 0 → = ∂y ∂z ∂y ∂z ∂u ∂w ∂u ∂w − = 0 → = ∂z ∂x ∂z ∂x ∂v ∂u ∂v ∂u − = 0 → = ∂x ∂y ∂x ∂y one possible solution to these three equations would be a function φ such that: ∂φ ∂φ ∂φ u = , v = , w = (4.4) ∂x ∂y ∂z

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We can then represent this as: U~ = ∇φ (4.5) Plugging equation 4.5 into 4.3 then gives us:

∇ × ∇φ = 0

Which is always true, since this is a vector identity. This confirms that equation 4.5 is indeed a valid solution to the irrotational flow field. We call this a potential flow, since the velocity is represented as the gradient of a potential. Keep in mind this is only valid if the flow field is assumed irrotational. We also can express this in cylindrical coordinates, which is given in a number of books.

The way I like to think about velocity potentials is to compare it to other “potential-type” quantities. One that was mentioned in a previous chapter was gravity. It’s a conservative force, which means we can express it as a gradient of some potential function. This lead to the expression: ~g = ∇(−gz) = −g∇z = (0, 0, −g) This means that the gravity vector is dependent on some gravitational constant g and the change in height z, with the negative sign meaning it acts in the minus z direction. We think of the gravitational potential energy as the height in which a weight is elevated, or equivalently, the change in potential energy as the change in height:

∆PE = mg∆z

Therefore, with gravity, we think of equal heights as having equal potential. Equivalently, we can think of gravity as having the same action at all elevations, as ∇z = (0, 0, 1), which means gravity is a constant −g and acts only in the z-direction. This same reasoning can be applied to a velocity potential. If we’re given a velocity potential of the form φ = Ax, then we can ﬁnd the velocity ﬁeld based on this potential:

U~ = ∇φ → (u, v, w) = ∇(Ax) = (A, 0, 0)

Thus, for the given potential, we found that the velocity ﬁeld has a constant action in the x- direction only. This is quite analogous to the gravity potential!

Potential functions become really useful when constructing different types of flow fields, and we will see that in an example later.

4.1.3 Stream functions Just as we defined a function that satisfies the requirement of irrotational flow, we can define a function to satisfy the continuity equation. For a 2D, incompressible flow, our continuity equation can be expressed as: ∂u ∂v + = 0 ∂x ∂y

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Just as we previously introduced a function to satisfy the irrotational requirement for potential flow, we can introduce a function to satisfy the incompressible requirement here. We define a function ψ such that: ∂ψ ∂ψ u = and v = − (4.6) ∂y ∂x By substituting this into the 2D incompressible continuity equation, we find: ∂u ∂v ∂ ∂ψ ∂ ∂ψ + = + − ∂x ∂y ∂x ∂y ∂y ∂x ∂ ∂ψ ∂ ∂ψ = − ∂x ∂y ∂x ∂y =0

Therefore, our stream function ψ satisfies the continuity equation. Again, there is also a cylindrical coordinate representation for the stream function. One thing to keep an eye out for is the way we defined the stream function. For this course, and hopefully for the rest of your undergrad work, we put the minus sign on the v differential expression. Some authors and courses like to put the minus sign on the u differential. So long as only one of the derivatives has a negative sign, then the stream function will satisfy the 2D continuity equation. We will be consistent with our definition of equation 4.6.

An important property of stream functions is that lines of constant ψ are streamlines (it’s almost as if it were named this way for a reason). This can be seen by writing the total differential of a streamline: ∂ψ ∂ψ dψ = dx + dy = −v dx + u dy ∂x ∂y For a constant ψ, dψ = 0, so we can rearrange the equation to get: dx u v dx = u dy → = dy v This expression is saying that a differential change in position is in the direction of the velocity. In other words, lines of constant ψ have a direction that is tangential to the velocity field. Therefore we can say streamlines are represented by a stream function of constant value.

4.2 Example: a potential vortex

One type of flow we can analyze with stream functions and velocity potentials is called the potential vortex. A name like “vortex” seems to imply some sort of vorticity, doesn’t it? Wouldn’t that violate the irrotational flow requirement for potential flow? We’ll find out if that is indeed true in this example. Let’s look at the stream function we can define for a potential vortex: Γ ψ = − ln(r) 2π Note that we’ve use the coordinate r, which implies this is in cylindrical coordinates. Using this stream function, let’s draw the streamlines and define a corresponding potential function,

Intro to Fluids Notes 56 C. P. Byers 2016 CHAPTER 4. INTRODUCING POTENTIAL FLOW if one exists. Then, let’s explain the relationship between the streamlines and lines of constant potential. We can start by drawing out a few lines of constant ψ: The exact value of the stream function

Figure 4.1: coordinate system and three lines of constant ψ itself isn’t so important, but what we see is that for a constant r, the stream function has a constant value, meaning we get streamlines that are circles. The velocity is tangential to these lines, so we see that the ﬂow is in a circular direction. We can then look to ﬁnd the exact magnitude of the velocity components:

1 ∂ψ 1 ∂ Γ u = = − ln(r) = 0 r r ∂θ r ∂θ 2π ∂ψ ∂ Γ Γ u = − = − − ln(r) = θ ∂r ∂r 2π 2πr

This says that there is no radial component of velocity, only an angular component, which decays in magnitude with increasing radius. This matches the streamlines we had drawn! Since this was a stream function, we know that it is a 2D incompressible ﬂow. Incompress- ibility can be conﬁrmed however:

1 ∂(ru ) 1 ∂u 1 ∂ Γ ∇ · U~ = r + θ = 0 + = 0 r ∂r r ∂θ r ∂θ 2πr So indeed the incompressible continuity equation (in cylindrical coordinates) has been satisfied. Can we write a velocity potential for this flow? That requires the flow to be irrotational. We can check by taking the curl of the velocity field. In cylindrical coordinates, this can be written as: 1 ∂u ∂u ∂u ∂u 1 ∂(ru ) 1 ∂u ∇ × U~ = z − θ eˆ + r − z eˆ + θ − r eˆ r ∂θ ∂z r ∂z ∂r θ r ∂r r ∂θ z

We don’t have any uz component, and no gradients in z, therefore the ﬁrst two vorticity components will be zero. This should make sense, as we are only a 2D problem, and the curl gives you a vector perpendicular to the two components in the operation. Since we have r- and θ- components, the curl should give a z-component of vorticity, if any. Therefore, plugging in our expressions for ur and uθ gives us: 1 ∂ Γ 1 ∂ 1 ∂ Γ ∇ × U~ = r − (0) eˆ = − 0 eˆ = 0e ˆ r ∂r 2πr r ∂θ z r ∂r 2π z z

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The curl of the velocity field of a potential vortex is zero, meaning we can define a velocity potential for the flow field! Perhaps that’s why it’s called a potential vortex! We then use the cylindrical coordinate version of a potential flow to find the function: ∂φ u = = 0 → φ = const. + f(θ) r ∂r 1 ∂φ Γ Γ u = = → φ = θ + const. + f(r) θ r ∂θ 2πr 2π By inspection, we can see the potential function has no r-dependence. Also, since we define the potential function so that we can extract the velocity through the gradient of it, it stands to reason that we don’t care about the possible additive constant. Therefore, we can set it to zero to get: Γ φ = θ 2π We can then plot lines of constant potential on our graph with the streamlines, as shown in figure 4.2. Once again, the actual values of the lines of constant potential aren’t so important

Figure 4.2: Potential vortex with lines of constant ψ and φ here. What we see is that φ is constant for a constant angle, giving these radial lines. Also of interest is the fact that the lines of constant velocity potential are perpendicular to the lines of constant stream function.

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In general, if we can define both a stream function and velocity potential, we find that lines of constant φ and lines of constant ψ are orthogonal. This can be shown by taking the total differential of a velocity potential (now back in Cartesian coordinates): ∂φ ∂φ dφ = dx + dy = u dx + v dy ∂x ∂y For a constant φ, dφ = 0, so rearranging gives: dx v −v dx = u dy → = − dy u This most certainly isn’t the same case as the streamlines, where the instantaneous slope coincided with the velocity at that point. We can see the two different slopes are orthogonal because their product, u/v × −v/u, is −1 indicating perpendicular lines. This is a general result for any field with both a potential function and stream function!

4.3 The power of the velocity potential

We’ve seen that we can define stream functions if the flow is 2D and incompressible, and velocity potentials if the flow is irrotational, but what benefit does this provide us? Let’s focus on the velocity potential for now to see how we can use it to our advantage. If, in addition to irrotational flow, we say a flow is incompressible, we can then apply the simplified continuity equation to our velocity potential:

∇ · U~ = 0 → ∇ · ∇φ = 0

However, just like we did with the viscous term in the Navier-Stokes equations, we can write this using the Laplacian operator, ∇2:

∇2φ = 0 (4.7)

Likewise, since we have an incompressible, irrotational flow, if we say it is 2D, we can do the same for the stream function. With the velocity components from a stream function defined as: ∂ψ ∂ψ u = and v = − ∂y ∂x we then plug this into our irrotational definition: ∂v ∂u ∇ × U~ = − kˆ =0 ∂x ∂y ∂ ∂ψ ∂ ∂ψ − − kˆ =0 ∂x ∂x ∂y ∂y ∂2ψ ∂2ψ − + kˆ =0 ∂x2 ∂y2 This too is simply represented as: ∇2ψ = 0 (4.8)

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What equations 4.7 and 4.8 represent is something we call Laplace’s Equation. This is a very compact and powerful equation, and is a nice simplification for our fluids equations. Before, we had to solve the Navier-Stokes equations to know the flow field. This was a tough thing to do, as it’s a non-linear partial differential equation. Even if we say the flow field is inviscid, the equation is still difficult. We called this form Euler’s Equation:

∂U~ ρ + ρU~ · ∇U~ = −∇P + ρ~g ∂t Note how it is still a non-linear partial differential equation with 4 dimensions (x, y, z, t), possibly non-constant coefficient ρ, and 4 variables (u, v, w, P ). Solving an equation like this is no easy task, and in fact we only have a small number of exact solutions. However, Laplace’s Equation is a different story. It is much more simple to solve, and more importantly, it is linear! That means that for any two solutions to the equation, their sum (or difference) is also a solution.

2 2 2 ∇ φ1 = 0 and ∇ φ2 = 0 → ∇ (φ1 + φ2) = 0

This means we can build ﬂow ﬁelds using multiple velocity potential solutions! This methodology is called superposition.

4.4 Example: superposition

Let’s look at a flow field where we have a source in a uniform flow. Where is the stagnation point? What is the pressure at the stagnation point? If we have a uniform flow described by the potential φ = Ur cos(θ), and the source described q 2 by φ = 2π ln(r), with U = 2m/s and q = 4m /s, we can assemble the total potential function of the flow: 4 φ = φ + φ = 2r cos(θ) + ln(r) 1 2 2π We could express this in Cartesian coordinates, but that would just give complicated expressions with ln px2 + y2 that we don’t want to deal with. If we plot this out (preferably with a computer) we would find the find the lines of constant velocity potential. However, velocity acts across the gradient of the potentials, which means the plot wouldn’t look so meaningful! Instead, let’s look to the stream function, since the velocity is tangential to streamlines. First, let’s start by decomposing the potential into the radial and angular velocity components: ∂φ 2 u = = 2 cos θ + r ∂r πr 1 ∂φ u = = −2 sin θ θ r ∂θ So we have the velocity field, and we know the flow is 2D and irrotational. To use a stream

Intro to Fluids Notes 60 C. P. Byers 2016 CHAPTER 4. INTRODUCING POTENTIAL FLOW function, we need to know if the flow is incompressible. Apply continuity and check to see: ∇ · U~ =0 1 ∂ 1 ∂ (ru ) + (u ) =0 r ∂r r r ∂θ θ 1 ∂ 2 1 ∂ r(2 cos θ + ) + (−2 sin θ) =0 r ∂r πr r ∂θ 1 1 2 cos θ + 0) − (2 cos θ) =0 r r 2 cos θ − 2 cos θ =0 r The incompressible continuity equation is indeed satisfied, so we can define a stream function! 1 ∂ψ 2 u = = 2 cos θ + r r ∂θ πr ∂ψ u = − = −2 sin θ θ ∂r Integration of each equation then gives us: ∂ψ 2 2θ = 2r cos θ + → ψ = 2r sin θ + + f(r) + C ∂θ π π 1 ∂ψ = 2 sin θ → ψ = 2r sin θ + f(θ) + C ∂r 2 Again, the additive constant isn’t important in this situation. We can see from the two equations that our stream function ψ is defined as: 2θ ψ = 2r sin θ + (4.9) π Plotting out equation 4.9 through the aid of a computer gives us the following plot:

Figure 4.3: streamlines for a point source in a uniform ﬂow

To get the stagnation point, we look for when the velocity has stagnated, meaning we have 0 velocity in both the radial and angular directions. Therefore, setting ur = 0 and uθ = 0 gives: 2 0 = 2 cos θ + πr 0 = −2 sin θ

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The second expression constrains our θ value to 0 or π. If we chose θ = 0, then we would have to have a negative r value, which doesn’t make sense. Therefore, we find stagnation occurs at the point: 1 (r, θ) = ( , π) π To find the pressure at the stagnation point, we need some sort of reference pressure to start with. Let’s assume the pressure far upstream is just atmospheric pressure, Patm. The assumptions we’ve already used for this problem (incompressible, irrotational, inviscid, steady) allows us to use Bernoulli’s equation. Therefore, picking one point far upstream (r = ∞ and θ = π), and the second point at our stagnation point, we find:

1 2 : 0 1 2 :0 ρU + P1 +ρgz1 = ρU + P2 +ρgz2 2 1 2 2 1 2 2 1 2 2 ρ (2 cos(π) + )êr + (−2 sin π)êθ + Patm = ρ (2 cos(π) + 1 )êr + (−2 sin π)êθ + P2 2 π ∗ ∞ 2 π ∗ π 1 2 1 2 ρ 2ê + 0ê + P = ρ 0ê + 0ê + P 2 r θ atm 2 r θ 2

2ρ + Patm = P2

This simply says the pressure at the stagnation point is 2ρ Pascals higher than the free stream pressure. This matches up with the expression we have from a pitot tube: s 2∆P U = ρ

If you plug in the freestream velocity of 2m/s, you ﬁnd that ∆P is 2ρ!

Intro to Fluids Notes 62 C. P. Byers 2016 Chapter 5

Non-Dimensionalization

5.1 Non-dimensionalization

In fluids, we love to use non-dimensional numbers. One primary reason for this is the simplification of data analysis that is possible due to non-dimensionalization. Another reason is to find the leading order terms in the equations of motion. I’ll go through an example of both of these to show just how they can be used and their implications.

5.1.1 Similarity and non-dimensional parameters Data analysis can be simplified through the use of non-dimensional parameters by a concept we call similarity. By matching all relevant non-dimensional parameters, we achieve flow similarity. This is especially relevant when we want to do scale modeling, which is the engineers playground. We can construct non-dimensional parameters by investigating what sort of terms show up in the problem itself. Let’s start by looking at a simple cylinder in a flow (figure 5.1) and determine what terms are important. The flow itself has a velocity, U∞, which we would expect to influence the drag. The density ρ and viscosity µ will also probably have an effect on the drag. We could also surmise that the size of the object, or diameter D, will influence the amount of drag. Since we want to know the drag FD, it’s also an important parameter. Now, we could also include other things that may influence the flow, such as surface roughness ks, or maybe freestream temperature T∞, or even perhaps what phase the moon is in, φmoon. However, we have to decide we have sufficient parameters at some point. In our case, we’ll just limit ourselves to U∞, ρ, µ, D, and FD. With this, we can then say that the drag is a function of these parameters:

FD = f(U∞, ρ, µ, D) (5.1)

The next step is to see what fundamental units are involved in this expression. These are dimensions such as length, time, mass, temperature, etc. So for us, we have: ML L M M F = U = ρ = µ = D = L D T 2 ∞ T L3 LT

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Figure 5.1: Cylender in a ﬂow with relevant parameters

So in our example, we have units of mass, length, and time, or three fundamental units. We decided that there were ﬁve relevant parameters. Therefore, we will have 5 − 3 = 2 non- dimensional groupings. The power of non-dimensional numbers is that we can deﬁne the problem with these 2 non-dimensional parameters! This means we’ll have one non-dimensional parameter, Π1 that is a function of the second, Π2, giving us a pretty simple relationship:

Π1 = f(Π2) (5.2) This is the basis of Buckingham-Pi theory, where we build up a number of non-dimensional “Pi groups”. Notice how equation 5.2 is much more simplistic than equation 5.1. I’d rather deal with a function of one variable than a function of four! I encourage you to read up on Buckingham-Pi a bit, but hopefully the basics here get you on your way.

There are a few ways to build our non-dimensional parameters. One way is to multiply terms together and give them unknown powers, then solving the powers so that the grouping is dimensionless. This is a classic way to find our “Pi groups,” but I like to do it in a different fashion. Let’s instead pick a parameter and work on getting rid of its dimensions. We know the drag is what we’re solving for, so it should make up one of the dimensionless parameters. Starting with it, we look at what it’s fundamental dimensions are: ML F = D T 2 So we need to somehow eliminate the M, L, and two T ’s. Let’s start by dividing by ρ, which will kill off the M dimension: F L4 D = ρ T 2 2 So mass is no longer in this grouping. Let’s next eliminate T by dividing by U∞: FD 2 2 = L ρU∞

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Finally, we can divide by D2 to get rid of the length dependence: FD 2 2 = dimensionless (5.3) ρU∞D

FD So we’ve found one non-dimensional parameter, CD = ρU 2D2 , which is similar to the classic deﬁnition of the drag coeﬃcient! However, what would have happened if we instead used µ to get rid of the mass term rather than ρ? Let’s calculate it through: F L2 D = µ T We can then eliminate time by dividing by velocity:

F D = L µU∞ And ﬁnally, to get rid of length we divide by D:

F D = dimensionless (5.4) µU∞D So there is a completely diﬀerent non-dimensional force grouping we could have come up with. Either grouping from equations 5.3 or 5.4 are valid, it’s just our experience and actual data collection that tells us which is a better one to use. So let’s select equation 5.3 so we have one non-dimensional parameter. We still need to build our second parameter. Let’s start with density and get rid of the mass component by dividing with viscosity: ρ T = µ L2 Now, we eliminate time by multiplying with velocity:

ρU 1 ∞ = µ L

Finally we can get rid of the length by multiplying with D:

ρU D ∞ = dimensionless (5.5) µ

Equation 5.5 is once again the Reynold’s number, which we’ve seen before. Then, using our two non-dimensional parameters, we can simplify equation 5.1 to: FD ρU∞D 2 2 = f ρU∞D µ With non-dimensionalization, you’ll often ﬁnd it’s your intuition and experience that drives the selection of parameters. It’s deﬁnitely more of an art, but yields some powerful science!

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5.1.2 Non-dimensional equations Rather than list out terms like we did in the last section, we can instead look to the governing equations to tell us what the ﬂow is doing. We select each variable and assign a scaling parameter and scaling function. This sounds complicated, but in practice it’s quite simple. For example, let’s say that the U~ velocity is in our equation, then we can re-deﬁne it as:

∗ U~ = UscaleU~

∗ The idea here is we pick an appropriate value for Uscale based on our ﬂow, and U~ is the scaling function. Another way to think of this is to take an appropriate scale, like Uscale, and divide our relevant variable by it. This will then give us a non-dimensional variable. So we see:

U~ = U~ ∗ Uscale This is just a re-arrangement of our previous relation. Likewise, we can re-deﬁne derivatives in a similar fashion: d d = ∗ dx d(Xscalex ) Where our x-coordinate is a non-dimensional x∗ scaling function times the scaling parameter Xscale. Again, we must pick Xscale to be appropriate to our ﬂow. So, let’s look to an example to see what these terms mean.

5.1.3 Example: non-dimensionalizing the momentum equation If we look to the steady, incompressible momentum equation neglecting gravity, we can non- dimensionalize it using the scaling parameters and functions. This means our momentum equation can be written as: ρ(U~ · ∇)U~ = ∇P + µ∇2U~

Now, if we say we have some velocity scale U, a length scale L, and pressure scale Ps, we can deﬁne: 1 U~ = UU~ ∗ and ∇ = ∇∗ and P = P P ∗ L s Note that derivatives deal with changes over a direction, so that’s why the L is on the bottom of the fraction. We don’t know what the pressure scale Ps is, but we can ﬁgure that out later. Plugging in our parameters then yields: 1 1 1 ρ(UU~ ∗ · ∇∗)UU~ ∗ = ∇∗(P P ∗) + µ( ∇∗)2UU~ ∗ L L s L ρU 2 P µU (U~ ∗ · ∇∗)U~ ∗ = s ∇∗P ∗ + ∇∗2U~ ∗ L L L2

The second line has all the scaling parameters (which are just constants!) pulled out front. We then have to choose what parameter we feel is the most important. For instance, in many ﬂows

Intro to Fluids Notes 66 C. P. Byers 2016 CHAPTER 5. NON-DIMENSIONALIZATION we know the acceleration term (or advective term) is important in our flow. Therefore, we can then divide the whole equation by the constants on the first term (ρU 2/L) to find:

P µ (U~ ∗ · ∇∗)U~ ∗ = s ∇∗P ∗ + ∇∗2U~ ∗ (5.6) ρU 2 ρUL Look at what we found - two non-dimensional numbers we’ve dealt with before! The ﬁrst one looks very familiar to our coeﬃcient of pressure, and the second one is the reciprocal of the Reynolds number. Recall: P − P ρUL C = 0 and Re = p 1 2 L 2 ρU µ

The diﬀerence with Cp is a multiplicative constant of 1/2, and since Ps is just some scale pressure, setting it to P − P0 is perfectly valid. As for the Reynolds number, a body of 1m length in air at 1m/s will have a Reynolds number of:

ρUL Re = L µ 1.2 kg ∗ 1 m ∗ 1m = m3 s − kg 1.846 ∗ 10 5 ms ∼66000

So if we look to equation 5.6, we see that the last term has a factor of 1/65000 in front of it. That means that term is negligible, leaving us with: P (U~ ∗ · ∇∗)U~ ∗ = s ∇∗P ∗ (5.7) ρU 2 This equation is simply a non-dimensional form of Euler’s Equation without the gravity term! Since the LHS is all non-dimensional scaling functions, then it should have an order of magnitude around 1. This means the non-dimensional parameter on the RHS, Cp, should have an order of magnitude around 1 as well. If you look at a plot of Cp over a cylinder, you’ll find that it varies from -3 to 1, which is what we expected! If it were to vary up to 100, then we’d have done something wrong. Therefore, from equation 5.7 we can see that the appropriate scaling for pressure is 2 Ps = ρUs . However, this is under the assumption that the advective terms were the important parameters, (or equivalently stated, the inertial effects are large), and viscous effects small. This process is exactly how we had issues with viscosity and the equations of motions up to the early 1900s. Since the Reynolds number was so high, the viscous term was negligible, so people would argue that Euler’s equation is the way to go. However, we know viscosity is important, so we can’t just neglect it! That second derivative allows us to utilize a second boundary condition in our PDE (because it was a second order equation), which is the no-slip condition. It took a trick with how we think about scaling to finally solve our problem. We will investigate that in a later example.

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5.1.4 Example: ﬂow in a channel Imagine a 2-D rectangular channel, length 100m and height of 1m, with a centerline velocity of m U = 5 s . Assuming the ﬂow is incompressible, let’s non-dimensionalize the continuity equation and see what it tells us: ∂U ∂V Continuity: + = 0 ∂x ∂y If we non-dimensionalize the x- and y- directions with the lengths we’re give, we can write: ∂ ∂ 1 ∂ ∂ ∂ ∂ = = and = = ∂x ∂(100x∗) 100 ∂x∗ ∂y ∂(1y∗) ∂y∗ Likewise, we can non-dimensionalize the velocities with what we have:

∗ ∗ U = 5U and V = VsV

Since we don’t have a V velocity scale, we’ll just use some constant Vs as a placeholder. Plugging everything in, we then find our continuity equation: 5 ∂U ∗ ∂V ∗ + V = 0 100 ∂x∗ s ∂y∗ Here is where non-dimensionalization become powerful and important. We’ve chosen scaling values for each variable, and we determined those to be important based on the flow. The non-dimensional functions are now the same order of magnitude, that being order 1. Think about it this way, we took changes in x and divided by 100. Since our geometry in this example is 100 m long in the x-direction, then the change in x∗ would be at most 1. This is how we non-dimensionalized all the variables! Since the continuity equation is only made of 2 terms, we must always have a balance between ∂U ∗ ∂V ∗ those two terms. Therefore, with ∂x∗ and ∂y∗ both being order 1, their coefficients must also balance. The coefficient of the first term is 1/20, so that tells us that:

1 m V = s 20 s This means that in our channel, the V -velocity will be small compared to the U-velocity. This doesn’t tell us the actual value of V -velocity, just its relative magnitude. Therefore, in this pipe, we see that velocities in the y-direction are very small compared to velocities in the x-direction.

Imagine next the balance of momentum for our equation. For a 2-D, steady, incompressible ﬂow, neglecting gravity, the x-momentum equation can be written as: ∂U ∂U 1 ∂P ∂2U ∂2U U + V = − + µ + µ ∂x ∂y ρ ∂x ∂x2 ∂y2 We can non-dimensionalize this with the scaling parameters we previously used for both velocity and length. We don’t have a parameter for pressure, so let’s use Ps again. Plugging in the values will yield: ∂(5U ∗) 1 ∂(5U ∗) 1 ∂(P P ∗) µ ∂2(5U ∗) µ ∂2(5U ∗) (5U ∗) + V ∗ = − s + + ∂(100x∗) 20 ∂(1y∗) ρ ∂(100x∗) ρ ∂(100x∗)2 ρ ∂(1y∗)2

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Pulling out all constants from their derivatives and re-arranging yields: 5 ∗ 5 ∂U ∗ 5 ∂U ∗ P ∂P ∗ 5µ ∂2U ∗ 5µ∂2U ∗ U ∗ + V ∗ = − s + + 100 ∂x∗ 20 ∂y∗ 100ρ ∂x∗ 1002ρ ∂x∗2 ρ ∂y∗2 ∂U ∗ ∂U ∗ P ∂P ∗ µ ∂2U ∗ 20µ∂2U ∗ U ∗ + V ∗ = − s + + (5.8) ∂x∗ ∂y∗ 25ρ ∂x∗ 500ρ ∂x∗2 ρ ∂y∗2 Now we can see a little more about what terms are important in our x-momentum equation. Recall how the non-dimensional terms have an order of magnitude of 1. This means the LHS of this equation is order 1. Then, looking to the RHS, we can conclude that the second term (first viscous term) is smaller than the third term. This conclusion is similar to what we found with equation 5.6, where the whole viscous term was deemed negligible. However, now we see that the second viscous term in equation 5.8 is more important than the first, so it isn’t strictly correct to neglect the whole µ∇2U~ term when non-dimensionalizing. In a future chapter, we will investigate internal flow (pipe and channel) equations in further detail. For now, just focus on how applying physical dimensions from an example resulted in a change in the balance in our non-dimensional relation. We see that different components of viscosity are more important than others. (In this case by four orders of magnitude!)

5.2 Prandtl’s Boundary Layer

Hopefully the last few sections were able to show you how non-dimensionalization can be performed, and how it can tell you what terms in an equation are important. However, as was shown with equation 5.6, a traditional scaling analysis led us to believe that the viscous term was negligible, and thus Euler’s equation is the valid representation of a flow, even around or through a body with solid walls. Granted, we can get decent approximations in some cases, but we’ve lost the physics of what’s really going on - we violate the no-slip condition. We then saw in equation 5.8 that one of the two viscous terms was smaller than the other (by a factor of 104 in that example). This should lead you to believe that we need to think ∂2U about how we non-dimensionalize the equations. We saw the µ ∂y2 term ended up being more ∂2U significant than the µ ∂x2 term. This can be explained by thinking about the distance over which these gradients were occurring. The x-direction had a characteristic length scale of 100m, where the y-direction had a characteristic length scale of 1m. So gradients would be much stronger in the y-direction compared to the x-direction. Prandtl had made this same conclusion when thinking about viscous effects on solid boundaries.

5.2.1 x-momentum equation To re-create Prandtl’s analysis, let’s look at a flat plate, 2-D, incompressible, steady flow with no gravitational effects. The physical setup can be seen in figure 5.2. This is actually a popular setup to have in a lab, so these simplifications aren’t that far- fetched. The x-momentum equation will then look like: ∂U ∂U 1 ∂P µ ∂2U µ ∂2U U + V = − + + ∂x ∂y ρ ∂x ρ ∂x2 ρ ∂y2

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Figure 5.2: Flow over a ﬂat plate, vertical and horizontal distances are not to scale.

We then say that the U velocity has a characteristic velocity of U∞, which is the free-stream velocity. We’ll set the pressure scale as Ps, then refine it further when have non-dimensionalized. 2 Based off what we’ve seen before, it’s likely that it will be ρU∞. The changes in the x-direction will occur over some length, let’s say L. The changes in the y-direction are then argued to occur over a very small distance, δ, where δ < L. We constrain the vertical length scale in this manner to make the viscous term important, as we know that there must be viscous effects in the near-wall region. We don’t worry about scaling ρ or µ because they are constant in this problem. If they were varying throughout the flow, then yes, we would need to use a scaling parameter for them. Next, to find the appropriate Vs scale, we turn to the continuity equation: ∂U ∂V + =0 ∂x ∂y ∂(U U ∗) ∂(V V ∗) ∞ + s =0 ∂(Lx∗) ∂(δy∗) U ∂U ∗ V ∂V ∗ ∞ + s =0 L ∂x∗ δ ∂y∗

We then argue that the terms in brackets must be the same order of magnitude, since they are the only two terms in this equation. The non-dimensional terms again have an order of magnitude of 1, so the order of magnitude of the bracketed terms have to match. Therefore, our appropriate velocity scale in y, or Vs, is: U δ V = ∞ s L Look at what this is saying - the ratio of δ/L is small, so our characteristic velocity in the y-direction is small compared to the free-stream velocity. Now, using this with our other scaling

Intro to Fluids Notes 70 C. P. Byers 2016 CHAPTER 5. NON-DIMENSIONALIZATION parameters, we turn to the momentum equation:

∂U ∂U 1 ∂P µ ∂2U µ ∂2U U + V = − + + ∂x ∂y ρ ∂x ρ ∂x2 ρ ∂y2 ∂(U U ∗) U δ ∂(U U ∗) 1 ∂(P P ∗) µ ∂2(U U ∗) µ ∂2(U U ∗) (U U ∗) ∞ + ∞ V ∗ ∞ = − s + ∞ + ∞ ∞ ∂(Lx∗) L ∂(δy∗) ρ ∂(Lx∗) ρ ∂(Lx∗)2 ρ ∂(δy∗)2 U 2 ∂U ∗ U 2 ∂U ∗ P ∂P ∗ µU ∂2U ∗ µU ∂2U ∗ ∞ U ∗ + ∞ V ∗ = − s + ∞ + ∞ L ∂x∗ L ∂y∗ ρL ∂x∗ ρL2 ∂x∗2 ρδ2 ∂y∗2 ∗ ∗ ∗ 2 ∗ 2 ∗ ∗ ∂U ∗ ∂U Ps ∂P µ ∂ U µL ∂ U U ∗ + V ∗ = − 2 ∗ + ∗2 + 2 ∗2 ∂x ∂y ρU∞ ∂x ρU∞L ∂x ρU∞δ ∂y Once again, we’re able to look at the bracketed terms to determine what it important in this equation. The LHS should all have an order of magnitude around 1. The first bracketed term 2 on the RHS is like the Cp term we’ve seen before. We can then scale pressure as Ps = ρU∞, or 1 the dynamic pressure. The factor of 2 that classically comes from Bernoulli’s equation doesn’t really affect the order of magnitude, so we don’t worry about including it. The second bracket on the RHS is 1/Re, and we know Reynolds number tends to be large in a boundary layer (1m/s velocity with a 1m long plate in air gives us Re ∼ 66000). So we can conclude this first viscous term is negligible. What about the second viscous term on the LHS - is it negligible? If it is, then we’ve once again lost viscosity! Prandtl argued that cannot be true, especially since we’re saying viscous effects need to be important in order to satisfy the no-slip condition. Therefore, we need the bracketed parameter on our second viscous term to also have an order of magnitude of 1: µL 2 ∼1 ρU∞δ νL δ2 ∼ U∞ r νL δ ∼ U∞

Note that we’ve used ν = µ/ρ. We can divide both sides by the L scale to re-cast in a different light: δ r ν 1 ∼ = √ (5.9) L U∞L ReL This tells us that the length scale in the y-direction, which is our boundary layer thickness, is very small compared to the x-direction length scale. We assumed this to be the case by saying δ < L, and of course we constrained it by solving for δ assuming it is small enough to make the last viscous term important. Nonetheless, by using some dimensional analysis, we’ve been able to understand how the boundary layer thickness is related to flow parameters! Lastly, we can then go back to our non-dimensional x-momentum equation to find the final

Intro to Fluids Notes 71 C. P. Byers 2016 CHAPTER 5. NON-DIMENSIONALIZATION form. Plugging in our expression for the boundary layer thickness, we get: 0 ∗ ∗ ∗ > 2 ∗ 2 ∗ ∗ ∂U ∗ ∂U Ps ∂P 1 ∂ U µL ∂ U U ∗ + V ∗ = − 2 ∗ + ∗2 + 2 ∗2 ∂x ∂y ρU∞ ∂x ReL ∂x ρU∞δ ∂y ∗ ∗ 2 ∗ 2 ∗ ∗ ∂U ∗ ∂U ρU∞ ∂P µL ∂ U U ∗ + V ∗ = − 2 ∗ + 2 ∗2 ∂x ∂y ρU∞ ∂x q ∂y ρU νL ∞ U∞ ∂U ∗ ∂U ∗ ∂P ∗ µ ∂2U ∗ U ∗ + V ∗ = − 1 + ∂x∗ ∂y∗ ∂x∗ ρν ∂y∗2 ∂U ∗ ∂U ∗ ∂P ∗ ∂2U ∗ U ∗ + V ∗ = − + ∂x∗ ∂y∗ ∂x∗ ∂y∗2

We see that the non-dimensional form of the x-momentum equation has four terms that are all of order 1. This means the four original terms (not non-dimensional form) can be retained in our equation, while we then neglect all other terms. Therefore, the x-momentum equation for our 2D, steady, incompressible, negligible gravity boundary layer is: ∂U ∂U 1 ∂P ∂2U U + V = − + ν (5.10) ∂x ∂y ρ ∂x ∂y2 Even though viscosity is small, the last term remains because the gradients in the vertical direction are extremely strong in the near wall region.

5.2.2 y-momentum equation The next step in this analysis is to scale the y-momentum equation. There is one important factor when doing the scaling analysis on this equation - it is just one of the components of a vector equation! This means that it is directly tied to the other components of the momentum equation. Think on this: if you scale one component of a vector in a certain way, you have to scale the other components of that vector in the same manner in order to not “skew” the vector. The magnitude may have changed, but the direction should be the same. Since we have to scale in the same fashion, we re-use all the previous scaling parameters from the x-momentum equation on the y-momentum equation, which is written as: ∂V ∂V 1 ∂P µ ∂2V µ ∂2V U + V = − + + ∂x ∂y ρ ∂y ρ ∂x2 ρ ∂y2 Notice that this looks extremely similar to the x-momentum equation, but instead is acting in the y-direction. The same assumptions apply as before, but now we’re concerned about the vertical components. Now, applying the same scalings as before, we ﬁnd:

∂ U∞δ V ∗ U δ ∂ U∞δ V ∗ 1 ∂(ρU 2 P ∗) µ ∂2 U∞δ V ∗ µ ∂2 U∞δ V ∗ (U U ∗) L + ∞ V ∗ L = − ∞ + L + L ∞ ∂(Lx∗) L ∂(δy∗) ρ ∂(δy∗) ρ ∂(Lx∗)2 ρ ∂(δy∗)2 U 2 δ ∂V ∗ U 2 δ ∂V ∗ U 2 ∂P ∗ µU δ ∂2V ∗ µU ∂2V ∗ ∞ U ∗ + ∞ V ∗ = − ∞ + ∞ + ∞ L2 ∂x∗ L2 ∂y∗ δ ∂y∗ ρL3 ∂x∗2 ρδL ∂y∗2

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At this point, it would be tempting to divide by the bracketed term from the LHS of the equation. However, we had argued that we must scale both equations in the same manner! Since the x- 2 U∞ momentum equation was re-scaled by dividing it by L , we must do the same here so our vector equation isn’t “skewed.” This will result in:

∗ ∗ ∗ 2 ∗ 2 ∗ δ ∗ ∂V δ ∗ ∂V L ∂P µ δ ∂ V µ L ∂ V U ∗ + V ∗ = − ∗ + ∗2 + ∗2 L ∂x L ∂y δ ∂y ρU∞L L ∂x ρU∞L δ ∂y

The grouping here is for us to see how each of the√ bracketed terms evolve. Recall that ReL = ρU∞L δ µ , and from equation 5.9 we know that L ∼ 1/ ReL. Therefore, we can plug these into our above relation to get:

1 ∂V ∗ 1 ∂V ∗ p ∂P ∗ 1 ∂2V ∗ 1 ∂2V ∗ √ U ∗ + √ V ∗ = − Re + + √ ∗ ∗ L ∗ 3/2 ∗2 ∗2 ReL ∂x ReL ∂y ∂y (ReL) ∂x ReL ∂y As was argued with the x-momentum equation, certain terms become less and less important as the Reynolds number grows. In this case, we see only one term remains important with increasing ReL! Therefore, all other terms are negligible and we can write our y-momentum equation as: ∂P 0 = (5.11) ∂y This is a very important result for us that study ﬂuid mechanics. It says that through a boundary layer, there is no gradient in pressure in the y-direction. That means that whatever the pressure is outside the boundary layer, it will be the same all the way through to the wall. We can show this through integrating this equation:

freestream freestream Z Z ∂P 0 dy = dy ∂y wall wall

0 =P∞ − Pwall

In other words, there is no diﬀerence between the freestream pressure and the wall pressure. Therefore, for a given x-location, we can treat the static pressure as a constant!

5.2.3 Combining the x- and y-momentum equations Since we’ve seen that the pressure isn’t changing in the wall-normal direction, then P inside the boundary layer is the same as P∞, which is defined as the “inviscid” region that lies outside the boundary layer. Since this outer flow is inviscid, incompressible, steady, and irrotational, Bernoulli’s equation applies! Therefore, we can say for the outer flow: 1 ρU 2 + P = const. 2 ∞ ∞

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Diﬀerentiating this with respect to the x-direction gives us:

d 1 ρU 2 + P =const. dx 2 ∞ ∞ dU dP ρU ∞ + ∞ =0 ∞ dx dx dU 1 dP U ∞ = − ∞ ∞ dx ρ dx From the y-momentum equation, we know P isn’t a function of y, so the x-momentum equation’s pressure gradient can be a total derivative with respect to x rather than a partial derivative. This allows us to substitute the above expression into equation 5.10 to get:

∂U ∂U dU ∂2U U + V = U ∞ + ν (5.12) ∂x ∂y ∞ dx ∂y2

This is our final expression for the x-momentum equation in a 2-D, incompressible boundary layer. If the flow outside the boundary layer weren’t changing, then U∞ is a constant (this is another way of saying there is no imposed pressure gradient on the flow!), and we get:

∂U ∂U ∂2U U + V = ν (5.13) ∂x ∂y ∂y2

5.2.4 Bonus: similarity solution One of Prandtl’s students, Blasius, was able to perform a coordinate transformation on the zero pressure gradient equation. By utilizing a streamfunction to simplify the non-linear partial diﬀerential equation above into a non-linear ordinary diﬀerential equation, he found:

f 000 + f 00f = 0

The primes indicate diﬀerentiation with respect to the new coordinate. The math is very involved, but it nonetheless reduced the equation to a form that can be solved numerically. Of course in 1908 we didn’t have computers, so the numerical solution was done by hand. Com- pletely doable, but time intensive. From his analysis, Blasius was able to show: δ 5 = √ x Rex

Here, x is the downstream position.√ This is just like we found in equation 5.9! We said the ratio of δ/L was proportional to 1/ ReL, and the “exact” solution of Blasius found the constant of proportionality is 5! This goes to show that a scaling analysis and non-dimensionalization are extremely powerful and useful for us.

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5.3 Bonus: Energy Equation

This section is a nice exercise of non-dimensionalization. Additionally, it shows a few more non-dimensional numbers that come up in ﬂuids. Let’s take the diﬀerential form of the energy equation: DT DP ρc = k∇2T + + τ · ∇U~ (5.14) p Dt Dt v

In this equation, k is the thermal conductivity of the fluid, cp is the specific heat of the fluid, D Dt is the material derivative, and τv is our viscous stress tensor, defined as:

T τv = µ∇ · U~ + µ(∇ · U~ ) (5.15)

We’ve assumed an incompressible ﬂow and used continuity and a number of other relations to simplify this expression. It can be found in a similar manner to how you found the Navier- Stokes equation, but the math isn’t important here. Let’s just look to non-dimensionalize this ∂ equation and see what pops out! First, let’s assume a steady ﬂow, so all ∂t terms go to zero. Then we write our equation as:

2 ρcp(U~ · ∇)T = k∇ T + (U~ · ∇)P + τv · ∇ U~

∗ Next we introduce appropriate scaling parameters. We start by using T = TsT for the temperature scaling. Next, for τv, we look to equation 5.15. Note that it’s a gradient of velocity times µUs ∗ µ, so we can scale it with our velocity and length scales. Therefore, we’ll write τv = L τv . Plugging these into our energy equation gives: 1 1 1 µU 1 ρc (U U~ ∗ · ∇∗)(T T ∗) = k( ∇∗)2(T T ∗) + (U U~ ∗ · ∇∗)(P P ∗) + s τ ∗ · ∇∗(U U~ ∗) p s L s L s s L s L v L s Moving all constants out of the derivatives gives:

ρc U T kT U P µU 2 p s s (U~ ∗ · ∇∗)T ∗ = s ∇∗2T ∗ + s s (U~ ∗ · ∇∗)P ∗ + s (τ ∗ · ∇∗U~ ∗ L L2 L L2 v

Note here that the bracketed terms are our scaling parameters and physical constants. These are all constants and set by the flow we’re looking at. If we determine that the inertial terms dominate (which is the LHS, and is usually the case), then we can divide by that parameter grouping. This gives us: ~ ∗ ∗ ∗ k ∗2 ∗ Ps ~ ∗ ∗ ∗ µUs ∗ ∗~ ∗ (U · ∇ )T = ∇ T + (U · ∇ )P + (τv · ∇ U ρcpUsL ρcpTs ρcpTsL Now we can start defining some new non-dimensional parameters. Starting with the first term on the RHS, we have: k µ k 1 1 = = ρcpUsL ρUsL cpµ ReL P r So we’ve once again found the Reynolds number, but we’ve also discovered the Prandtl number, cpµ defined as P r = k , which is a measure of viscous diffusion to thermal diffusion. Next, the second term on the RHS has Ps. From our previous scaling of the momentum equation, we

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2 determined an appropriate scale for Ps is ρUs . This comes from the fact that pressure and inertial effects tend to balance. If we had a different scenario, we would have to scale differently. 2 However, setting Ps = ρUs , we get: P ρU 2 U 2 s = s = s = Ec ρcpTs ρcpTs cpTs Here, we’ve defined a new non-dimensional number called the Eckert number. This is can be shown to be proportional to the Mach number squared in a perfect gas. The last term is then:

µU µ U 2 Ec s = s = ρcpTsL ρUsL cpTs ReL So by just re-arranging the terms, we came up with two non-dimensional numbers we’ve seen before. All together now, we ﬁnally get:

~ ∗ ∗ ∗ 1 ∗2 ∗ ~ ∗ ∗ ∗ Ec ∗ ∗~ ∗ (U · ∇ )T = ∇ T + Ec (U · ∇ )P + (τv · ∇ U ReL · P r ReL

While you likely won’t be using the full energy equation in an undergrad fluids class (at least not in this form), we can see that we are able to get a lot of information through non- dimensionalizing. We find non-dimensional parameters that are relevant to the flow without having to make a list. The equations hold a lot more information than they first seem to! Also recall that we can analyze and compare flows that are similar, so long as we match non-dimensional parameters. The energy equation (eq 5.14) is telling us how the temperature field in a flow changes due to heat, pressure, and viscous energy contributions. So if we want to analyze a flow where we have changes in temperature, we should be looking to match not just the Reynolds number, but also the Prandtl and Eckart numbers too.

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Analysis of the Boundary Layer

6.1 Summary of the boundary layer

After discovering the concept of the boundary layer, it is important to try and understand some parameters that we use to describe and characterize these types of flow. We previously learned that the boundary layer itself is a thin region near the wall where viscosity plays an important role in the flow. Outside the boundary layer, the fluid can be described as inviscid, and thus the flow is governed by Euler’s Equation. Within the boundary layer, we cannot neglect viscosity (which is the whole reason we have defined a boundary layer). We then utilized an order of magnitude analysis to reduce the x-momentum equation to:

∂U ∂U 1 dP ∂2U U + V = − ∞ + ν (6.1) ∂x ∂y ρ dx ∂y2 Note that the pressure gradient is not a partial derivative. This was found from the y-momentum equation, where the gradient in the y-direction is zero, thus pressure only varies in x and is set by the outer ﬂow. Note that this is the static pressure we speak of, and the velocity still contributes a dynamic pressure to the system. In order to ﬁnd this equation, we had to set the y-direction scale as δ, which was assumed to be much smaller than the x-direction scale, L. From this assumption, we found: δ 1 ∼ √ (6.2) L ReL The “exact” Blasius solution (which assumed the pressure gradient was zero) found the unknown pre-factor, giving the relation as: δ 5 = √ (6.3) x Rex One quick note: we have switched L and x between the two expressions above. This isn’t an issue, since L represents some downstream length scale, which is occurring in the x-direction. Therefore, we can just use x as a variable instead.

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6.2 Zero pressure gradient

The laminar boundary layer analysis is performed under the assumption that the pressure gradient, dP/dx, is zero. This allows a considerable simplification to the equations of motion, and was what allowed Blasius to find a similarity solution. How is it that a flow can exist without a pressure gradient? It does seem counter intuitive, doesn’t it? We can consider it from a few perspectives. One way is to imagine a flat plate being pushed or moved through still air. There is no external pressure gradient on the flow, since it is all still and quiescent except for the immediate region around the plate. Another way to imagine this is to appeal to the fact that there is inertia to the flow, and the inertia carries this flow. Sure, some pressure gradient likely had to get the flow moving, but then we argue that the flow is at its set speed, and we no longer are forcing it, but letting it continue. Since the outer portion of the flow (outside the boundary layer) is inviscid, there is no resistance to its inertia, so it will continue to move at the free stream velocity. Here’s a mathematical way to think of it. Outside the boundary layer, the viscous effects are negligible. This allows us to use Euler’s equation:

DU~ ρ = −∇P − ρg∇h Dt Now let’s think about what assumptions go into analyzing the flow in and around the boundary layer. We neglect gravity, say it’s steady, 2-D, and constant density. This means our x-momentum equation for the outer flow simplifies down to: ∂U ∂U dP ρU ∞ + ρV ∞ = − ∞ ∞ ∂x ∞ ∂y dx Recall how in the last chapter we found the pressure to not change in the y-direction. This allows us to write the pressure derivative as a total derivative, not a partial. Now let’s think on what’s happening in this outer flow - the velocity is constant! Of course, there is some small V∞ that changes with downstream position (think about continuity for that, or look ahead to our integral analysis in the next section). However, there isn’t a change in our external U∞, the x-component of the free stream velocity. This means: 0 0 ∂U∞ ∂U∞ dP∞ ρU∞ + ρV∞ = − ∂x ∂y dx dP 0 = ∞ dx The pressure gradient in the external flow is zero! We’ve shown that a constant external flow velocity means the problem can be considered a ZPG boundary layer. Let’s discuss this a little more in the next section.

6.3 Displacement thickness

One parameter we use to describe a boundary layer is the displacement thickness. This can be thought of as a measurement of the displacement of an ideal (inviscid) ﬂow due to the decrease

Intro to Fluids Notes 78 C. P. Byers 2016 CHAPTER 6. ANALYSIS OF THE BOUNDARY LAYER in velocity in the boundary layer. If we account for the “lost mass” due to the no-slip condition slowing the ﬂow down, we can then equate it to an idealized ﬂow that is displaced from the wall.

Figure 6.1: concept of displacement thickness

Figure 6.1 shows the conceptual way to think about this. We can see that there is a velocity deficit from the free-stream, which represents a departure from the idealized case of perfect slip at the wall. Assuming a width w into the page, the left side of figure 6.1 shows this lost mass flow, and can be expressed as:

lost mass flow = total mass flow with slip − total mass flow without slip ∞ ∞ Z Z lost mass flow = w ρU∞ dy − w ρu(y) dy 0 0 ∞ Z lost mass flow = ρw U∞ − u(y) dy 0 The next step is to define this lost mass flow as a “chunk” removed from the ideal slip case, ∗ which we define on the right side of figure 6.1. This lost mass flow is simply ρU∞wδ . In other words, the ideal no-slip velocity is displaced a distance δ∗ from the wall. If we equate these two “lost mass flows” we then get:

∞ Z ∗ ρU∞wδ =ρw U∞ − u(y) dy 0 ∞ Z ∗ 1 δ = U∞ − u(y) dy U∞ 0

Since the free stream velocity U∞ is a constant, we can move this inside the integral to get:

∞ Z u(y) δ∗ = 1 − dy (6.4) U∞ 0

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The expression in equation 6.4 gives the displacement from the wall that an ideal flow would be to give an equivalent loss of mass flow that the viscous case has. This is another parameter to characterize a boundary layer, but is distinctly different from the boundary layer thickness. Will δ∗ be larger or smaller than δ? Think about figure 6.1: we will always find that δ∗ < δ.

6.3.1 Displacement thickness and the zero pressure gradient assumption Remember the whole ZPG boundary layer assumption? What happens when we enclose our flow in a wind tunnel? The mass flow into the entrance of the tunnel must equal the mass flow at the exit of a tunnel. However, we just came up with this concept of a displacement thickness to account for the “lost mass flow” in the near wall region. At the exit of the tunnel, we can approximate the flow as an inviscid external flow displaced from the wall by a distance δ∗. This means we’ve “modified” the exit area, as there won’t be any flow in the displaced area. Since mass must be conserved, this then means we’ve modified the free-stream velocity! Let’s take a quick look at why this is happening.

Figure 6.2: Wind tunnel - viscous eﬀects at wall give a displacement thickness at the exit

Figure 6.2 shows the schematic setup of a wind tunnel. We have an inviscid ﬂow into the tunnel, then the boundary layer at the top and bottom walls results in an equivalent displacement thickness at the exit. With boundary layers, we assume a 2-D, steady, incompressible ﬂow, which means we can write our continuity equation as:

ZZ nˆ · ρU~ dA = 0

A ∗ −ρU∞,inwh + ρU∞,outw(h − 2δ ) = 0 h U = U ∞,out ∞,in h − 2δ∗ This is saying the free stream velocity at the exit is larger than the free stream velocity at the inlet. In other words, we have some dU∞/dx. Based oﬀ our analysis for the inviscid outer ﬂow, this means we don’t really have zero pressure gradient, as we end up with: dU dP ρU ∞ = − ∞ ∞ dx dx

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This is non-zero in the wind tunnel! That means there’s some sort of pressure gradient due to the change in the free stream velocity. In general, boundary layers are small, and thus δ∗ is very small. So long as the height h >> δ∗, this should have little eﬀect on the free stream, and the assumption of ZPG will hold.

6.4 Momentum thickness

Utilizing the same methodology we used to calculate the displacement thickness, we can find another length-based parameter associated with the boundary layer. We will look to find the lost momentum flux from the flow, just as we found the lost mass flow. This lost momentum will be related to a momentum thickness that the inviscid flow is moved from the wall. Once again, we look to see how much momentum is lost from the ideal no-slip case. Why do we care about lost momentum from the flow? Think about Newton’s second law - rate of change of momentum is equal to the total force on the system. Therefore, a loss of momentum means some force was imparted on the flow (we will see this is the viscous drag)! The math involved is a little more complicated compared to the displacement thickness. Let’s look to a control volume analysis of the boundary layer. We want to find the lost momentum

Figure 6.3: control volume used to find lost momentum from ideal case from the ideal no-slip case, so we can just do a CV analysis on the red box in figure 6.3. We will compare the fluxes of momentum in the x-direction, and find their overall balance. This value will be the difference in momentum from the ideal no-slip case. Since this is a steady, 2-D, constant density problem, we will find:

Net x-momentum change = sum of x-momentum into and out of the control volume ZZ Net x-momentum change = (ˆn · ρU~ )U~ dA

A We need to determine what is being carried in and out through the surfaces. The inﬂow of momentum is simply the left surface, which has a velocity equal to the free stream U∞, and in this 2-D problem occurs over a height of H and some width w. This corresponds to the ideal case, where if there were slip at the wall we would have this momentum across the entire inlet. The outﬂow of momentum occurs through the top and right surface. For the top, we have a horizontal freestream velocity U∞ and a vertical component v over a length of L and width w.

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However, for the x-momentum balance, we see that then ˆ · U~ component is occurring through a surface where the normal is vertical. Therefore,n ˆ · U~ = ˆj · U~ = v, the vertical velocity out. The right outflow is like the left inflow, except the velocity is u(y), not the free stream. Now, the velocity profile u(y) will reach the free stream value before the end of the control volume, but that’s just accounted for in the function (which we don’t need to know the exact value at this time). Therefore, using these values, we can write our momentum loss as: ZZ ZZ ZZ Net x-momentum change = ρ (−î · U~ )Ux dA + ρ (ˆj · U~ )Ux dA + ρ (î · U~ )Ux dA

Aleft Atop Aright H L H Z Z Z 2 2 Net x-momentum change = ρw −U∞ dy + ρw vU∞ dx + ρw u dy 0 0 0 L H Z Z 2 2 Net x-momentum change = −ρwHU∞ + ρwU∞ v dx + ρw u dy 0 0 The next step is to replace the currently unknown v expression (which is bracketed for our convenience) with known parameters. Utilizing continuity, we simply balance the mass flow in with the mass flow out to find v as:

ZZ ZZ ZZ nˆ · ρU~ dA + nˆ · ρU~ dA + nˆ · ρU~ dA = 0

Aleft Atop Aright L H Z Z −ρwHU∞ + ρw v dx + ρw u dy = 0 0 0 L H Z Z v dx = HU∞ − u dy 0 0 This expression for the integral of v is exactly the bracketed parameter in the momentum expression above! We then substitute it back into the momentum equation:

H H Z Z 2 2 Net x-momentum change = −ρwHU∞ + ρwU∞ HU∞ − u dy + ρw u dy 0 0 H H Z Z 2 Net x-momentum change = −ρwU∞ u dy + ρw u dy 0 0 H Z 2 Net x-momentum change = −ρw uU∞ − u dy 0

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A little re-arranging was done for the last line. Now, we know that U∞ ≥ u(y), so the integral will always be positive. This means our net x-momentum change is negative, implying a loss of momentum! We expected that, since we can think of this value as the momentum lost from the ideal wall-slip case. Now, we will deﬁne a new thickness θ that represents the displaced

Figure 6.4: concept of the momentum thickness, similar to the displacement thickness momentum. This can be visualized in figure 6.4. The missing momentum from the ideal con- 2 dition would simply be ρU∞wθ. Equating this “idealized lost momentum” with the actual lost momentum from the flow, we find:

H Z 2 2 ρU∞wθ =ρw uU∞ − u dy 0 H Z 1 2 θ = 2 uU∞ − u dy U∞ 0

Once again, U∞ is a constant, and can be brought into the integral. Additionally, the limit of integration can be set to ∞ rather than some arbitrary H. Think about why that works: u/U∞ goes to 1 as y → δ, so the integrand goes to zero as the limit goes to δ. This means there will be no additional contribution to our equation for an increase in the upper integrating limit beyond y = δ. Some rearranging gives us the following result:

∞ Z u u θ = 1 − dy (6.5) U∞ U∞ 0

The expression in equation 6.5 gives us the equivalent distance from the wall that an ideal inviscid flow would be moved to account for the lost momentum due to viscous effects. Like δ∗, this is a new parameter to characterize a boundary layer. Also, we can see that the momentum thickness will be smaller than the displacement thickness (compare the integrands of the two definitions to see).

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6.4.1 Momentum thickness and drag We calculated the momentum thickness by doing a control volume analysis and finding the change in x-momentum. We saw this was a net loss, and equated it to the “loss” in x-momentum from displacing an inviscid flow from the wall. However, you can also see that this net loss of P x-momentum will just be Fx from the Navier-Stokes Equations: d ZZZ ZZ X ρU~ dV + (ˆn · ρU~ )U~ dA = F~ dt V A When we look in the x-direction for our flow, like in figure 6.4, this simplifies down to: 0 ZZZ ¨¨* ZZ d ¨¨ X ¨ρUx dV + (ˆn · ρU~ )U~ dA = Fx dt ¨¨ ¨¨ V A | {z } our momentum deficit

2 This simpliﬁes to ρU∞wθ = Fvisc. The forces reduced to only viscous forces because we like to deal with a zero pressure gradient ﬂow, which means forces from pressure are 0. Additionally, there are no x-direction gravity forces, and the CV doesn’t cut through a surface. Therefore, we can re-express our momentum thickness (for some given width w into the page) in terms of surface drag: ∞ Z Fvisc u u θ = 2 = 1 − dy (6.6) ρU∞w U∞ U∞ 0

6.5 Example: approximate boundary layer

We know from class that the Blasius solution isn’t an analytical solution, so we don’t have a direct functional relationship between velocity and position. However, we can approximate the velocity profile in a boundary layer with different functional forms. One way to do it is to specify the boundary conditions and pick a function that satisfies them (and at least looks like a velocity profile). We will assume a set x-position, so there isn’t any functional dependence on x for this case. We have the following boundary conditions:

∂u u(0) = 0 and u(δ) = U∞ and = 0 ∂y y=δ These three BCs are the no-slip condition, the deﬁnition of the boundary layer height, and no- shear at the outer boundary layer edge. You can think of any number of functions that satisﬁes these conditions, but let’s just use a simple parabolic relation:

2y y2 u = U − (6.7) ∞ δ δ2 You can see that this function satisﬁes all three of our necessary boundary layer BCs. Given this function, let’s calculate out the displacement thickness and the momentum thickness.

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∞ Z u(y) δ∗ = 1 − dy U∞ 0 δ Z 2y y2 = 1 − − dy δ δ2 0 2 3 δ y y =y − + 2 δ 3δ 0 Therefore, we see our displacement thickness from this velocity proﬁle is:

δ δ∗ = 3

This is saying the displacement thickness is a third the height of the boundary layer thickness, which lines up with our assumption that δ∗ < δ. Next, the momentum thickness can be found:

∞ Z u u θ = 1 − dy U∞ U∞ 0 δ Z 2y y2 2y y2 = − 1 − − dy δ δ2 δ δ2 0 5 4 3 2 δ y y 5y y = − 4 + 3 − 2 + 5δ δ 3δ δ 0 And our momentum thickness comes out to be:

2δ θ = 15

Again, this matches our intuition that θ < δ∗ < δ. These approximations actually turn out to ∗ not be too terrible compared to the Blasius solution, which gives δB = 0.34δ and θB = 0.132δ.

6.6 Example: ﬁnding the skin friction

A laminar boundary layer velocity proﬁle may be described approximately as u = U∞ sin(πy/2δ). Find the shear stress at the wall, τw, and express both the total and local skin friction coeﬃ- cients, CF and Cf in terms of the Reynolds number. (Problem adapted from Lex’s book.)

Let’s start by listing our deﬁnitions of the total and local skin friction coeﬃcients: F τ C = visc and C = w F 1 2 f 1 2 2 ρU∞A 2 ρU∞

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Note that CF is based on the total drag force Fvisc that occurs over an area A, and Cf is based the wall shear τw at a single location. In order to calculate these, we turn to the velocity proﬁle: πy u = U sin ∞ 2δ

We can start by ﬁnding τw through the velocity gradient:

∂u τw =µ ∂y y=0 µπU π ∗ 0 = ∞ cos 2δ 2δ µπU = ∞ 2δ So we have the local shear stress as a function of delta. Next, we need to ﬁnd the total drag force, Fvisc, which we know is related to the momentum thickness through equation 6.6:

∞ Z Fvisc u u 2 = 1 − dy ρU∞w U∞ U∞ 0 We then plug in our velocity proﬁle to get the total viscous drag:

δ Z πy πy F =ρU 2 w sin 1 − sin dy visc ∞ 2δ 2δ 0 δ 2 δ πy 2δ πy y =ρU∞w sin − cos − 2π δ π 2δ 2 0 2 1 =ρU 2 wδ − ∞ π 2

Now we want the non-dimensional parameters, CF and Cf , as a function of the Reynolds number, ρU∞x/µ. This means we need to find some sort of x-dependence that is in this system. One parameter here that would have an x-dependence is the boundary layer thickness, δ. Therefore, we should hope to find a way to get that functional dependence! Lucky for us, there are other terms that depend on x, namely the wall shear τw and the total drag Fdrag. Think about it this way - since the boundary layer thickness grows, the velocity profile will change, thus the shear will change downstream. Additionally, the total drag is dependent on how far down you integrate over the surface!

So let’s look at the total drag and relate it to the local friction. We can get the total drag (due to friction) by integrating the local friction over the area of the surface exposed to ﬂow: Z x Fdrag(x) = w τw dx 0

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Here we assume the shear isn’t changing over the width w. Then, we can diﬀerentiate this expression to get: 1 d F = τ w dx drag w

We have expressions for both Fdrag and τw that have the boundary layer thickness δ in them! We can then plug those into the above relation and solve:

1 d 2 1 µπU ρU 2 wδ − = ∞ w dx ∞ π 2 2δ 4 − π dδ µπU ρU 2 = ∞ ∞ 2π dx 2δ µπ2 δ dδ = dx ρU∞(4 − π) 1 µπ2x δ2 = 2 ρU∞(4 − π) s 2π2 r µx δ = · (4 − π) ρU∞ r µx δ =4.8 ρU∞ This relation for δ should look pretty familiar. If we were to divide by x, we would get: δ 4.8 = √ x Rex This is nearly the same as the result from the Blasius solution, which instead had a constant of 5! That’s pretty remarkable that we can just use an approximate shape and get nearly the same result as the “exact” solution. So, using our expression for δ, we can now ﬁnd expressions for our CF and Cf :

τ C = w f 1 2 2 ρU∞ 1 µπU = ∗ ∞ 1 2 2 ρU∞ 2δ µπ 1 = ∗ q ρU∞ 4.8 µx ρU∞ √ 0.655 µ = √ ρU∞x 0.655 Cf = √ Rex

Our skin friction coefficient, Cf , has a pre-factor of 0.655 from this velocity profile estimate. The Blasius skin friction coefficient has a prefactor of 0.664 for comparison.

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Performing the same math with the total drag coeﬃcient, we ﬁnd: F C = visc F 1 2 2 ρU∞A 1 2 1 = ∗ ρU 2 wδ − 1 2 ∞ 2 ρU∞wL π 2 δ 2 1 = ∗ 2 − L π 2 q µL 4.8 ρU = ∞ ∗ 0.273 L 1.311 CF = √ ReL

Note that when plugging in the expression for δ, we assumed the x-location was the length L in which the area was defined. Our drag coefficient CF has a pre-factor of 1.311, which compared to the Blasius prefactor of 1.328 is again a good estimate. So while there was a bit of math, we were able to find skin friction and drag coefficients for a given boundary layer velocity profile! We had utilized the concept of the momentum thickness to provide a relationship between drag force, boundary layer thickness, and downstream location. We see that this seemingly strange concept ended up providing a powerful tool for our analysis.

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Drag, Separation, and Shedding

7.1 Drag force from friction

In the previous chapter, we were able to relate the boundary layer thickness, drag force, and skin friction through the momentum thickness. We were able to take different approximations to the velocity profile in a boundary layer (such as a quadratic function, or a sine function) and build these relations. In general, for a flat plate, zero pressure gradient boundary layer, the Blasius solution results in the following: δ 5 = √ x Rex τ 0.664 C = w = √ f 1 2 2 ρU∞ Rex F 1.328 C = friction = √ F 1 2 2 ρU∞A ReL

And with respect to the ﬂat plate boundary layer, Ffriction was simply the total frictional force due to wall shear stress, which for some downstream length L and uniform width w means:

Z L Ffriction = w τw dx 0

Note that CF is not the same as CD, which is the drag coeﬃcient. CD comprises of both the frictional and pressure forces that make up the total drag force:

Fdrag = Ffriction + Fpressure

So make sure you pay attention to what it is you’re calculating!

Another important point to note is the A in CF (as well as CD). This is the characteristic area of the object, which can be different depending on what you’re looking at. In general, it’s the area that is responsible for the main contribution of the force. It’s likely to be the frontal area, or projected area to the flow. However, depending on the setup, it may be defined differently, such as using the surface area. Just be mindful of how it’s being use in your specific

Intro to Fluids Notes 89 C. P. Byers 2016 CHAPTER 7. DRAG, SEPARATION, AND SHEDDING application.

These differences in non-dimensional force coefficients help us to understand why we call a body either “streamlined” or “bluff.” If the primary contribution to the drag is due to pressure forces, then we call the object a bluff body. If the primary contribution is instead from frictional forces, we call it a streamlined body. Think of an airfoil: this is the classic streamlined body. We try not to have it at too high of an angle of attack, so it is primarily subjected to frictional forces. Conversely, a ball has a majority of its drag come from the pressure differences between the front and back, so it would be a bluff body.

7.1.1 Example: drag on a plate Find the drag force on a thin, 1m wide by 5m long plate in a ﬂow of air at 1 m/s.

1∗5 First, checking the Reynolds number, we ﬁnd that for this plate, ReL = 0.000015 ≈ 300, 000, which is less than the critical reynolds number. Therefore, we can assume this ﬂow remains laminar, and thus can use the Blasius relations. For the total drag force, we can assume it’s only due to frictional forces, as this body is assumed thin. Therefore, there are little to no pressure forces acting on the plate.

: 0 Fdrag = Ffriction +Fpressure

Since we can assume that the total drag force is from the viscous friction, we say CD ≈ CF . Then, we can use our ﬂat plate relations to calculate the drag force: F 1.328 C ≈ C = friction = √ D F 1 2 2 ρU∞A ReL where the last equality came from the Blasius ﬂat plate drag relation. So we plug in our Reynolds number to the relation: 1.328 CF = √ = 0.00242 300000 Then our total drag force can be found: 1 F ≈ F =C ρU 2 A drag friction F 2 ∞ 1 kg m2 =0.00242 ∗ ∗ 1.2 ∗ 1 ∗ (2 ∗ 1m ∗ 5m) 2 m3 s

Fdrag = 0.015N

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Note that we used the total surface area of the plate, which is the two sides. Overall, the drag from the ﬂow is very small!

What would happen if this was oriented vertically so the flow was perpendicular against it? Then we’d have a bluff body, and pressure forces would dominate. To get the total drag here, we turn to tables or charts that list out the drag coefficient as a function of Reynolds number. In this instance, a flat plate perpendicular to flow has a CD = 2.0. Then, we find: 1 F =C ρU 2 A drag D 2 ∞ 1 kg m2 =2 ∗ ∗ 1.2 ∗ 1 ∗ (1m ∗ 5m) 2 m3 s

Fdrag = 6N

Note here that the area was now the frontal area, which is just 1m by 5m rather than double that for the streamlined body scenario. Overall, we see that bluﬀ bodies tend to have more drag force!

7.2 Separation of the boundary layer

We performed a variety of analyses on the boundary layer under the assumption of zero pressure gradient. However, we were able to see that the concept of the displacement thickness lead to a change in the free stream velocity in a wind tunnel. That was due to the streamlines being displaced, and continuity leading to an increase in free stream velocity at the exit. Looking to the momentum equation in the free stream, we saw it was Euler’s Equation that governed it since there weren’t viscous eﬀects. ∂U ∂U 1 dP U ∞ + V ∞ = − ∞ (7.1) ∞ ∂x ∞ ∂y ρ dx

We then argued that U∞ doesn’t change with y, but continuity showed that U∞ was larger at the exit than the entrance. This means we had: ∂U 1 dP U ∞ = − ∞ (7.2) ∞ ∂x ρ dx Since the LHS is a positive value, it means the RHS is also positive, indicating the pressure gradient must be negative: dP ∞ < 0 dx This negative pressure gradient was associated with an acceleration of the flow. We call this a favorable pressure gradient. If the velocity had slowed down, we would have the LHS be negative, thus our pressure gradient would be positive, indicating an adverse pressure gradient. The naming convention comes from the effect this pressure gradient has on the flow field. Recall that we found ∂P/∂y = 0 in the boundary layer, so the static pressure within the boundary layer is set by the external pressure. If there is a favorable pressure gradient in the external

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Figure 7.1: external pressure gradient acting within the boundary layer

flow, then this same pressure gradient must apply through the boundary layer as well. Since this static pressure in the outer flow is the same at the wall, the negative pressure gradient is felt as the wall as well, where we find Pw,1 > Pw,2. Note in figure 7.1 that it’s the displacement thickness δ∗ that’s shown, since it was what led us to conclude there could be an increase in free stream velocity.

Let’s then look at the momentum equation at the wall to try and understand what this “imposed pressure gradient” might be doing to our ﬂow. Including the pressure gradient, our x-momentum equation for the boundary layer is:

∂U ∂U 1 dP ∂2U U + V = − ∞ + ν (7.3) ∂x ∂y ρ dx ∂y2

Note that we can still use P∞ since it is imposed throughout the boundary layer. Now, at the wall both the U and V velocities are zero, so the momentum equation reduces down to:

2 1 dP∞ ∂ U 0 = − + ν 2 ρ dx ∂y y=0 2 1 dP∞ ∂ U = 2 µ dx ∂y y=0 Recall from calculus what a second derivative is telling you. It’s an indication of curvature. So, with µ always positive, the curvature of the velocity profile at the wall is determined by the value of the pressure gradient. This is illustrated in figure 7.2, where three values of a pressure gradient are shown. Note that this is the inflection at the wall, while near the boundary layer edge the inflection must be negative.

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Figure 7.2: Inﬂection (at the wall) of the velocity proﬁles due to the pressure gradient

7.2.1 Favorable pressure gradient Imagine we are back in the wind tunnel, and the displacement of the streamlines has lead to an increase of freestream velocity. This in turn causes the free stream pressure to drop. Since the external pressure is imposed on the wall, the pressure gradient along the wall is negative. This corresponds to case (3) in ﬁgure 7.2, in which we can ﬁnd:

2 1 dP∞ ∂ U < 0 → 2 < 0 µ dx ∂y y=0 A negative value for curvature at the wall then indicates that the curvature is negative throughout the boundary layer. This results in no ﬂow separation and all the world is happy.

7.2.2 Zero pressure gradient This is the classic laminar boundary layer case, which is shown in (2) in figure 7.2. The curvature of the velocity profile at the wall is zero, then it goes negative as it moves towards the free stream. Take a look at the actual Blasius boundary layer velocity profile, and you’ll see that the shape does indeed have zero curvature at the wall! Like the favorable pressure gradient, this flow will not separate from the wall.

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7.2.3 Adverse pressure gradient If we have a situation where the velocity in the free stream is decreasing, then the pressure gradient must be positive. This is what is shown in case (1) in figure 7.2. Think about when this situation might happen - it would have to be the opposite of the favorable case. We saw that a favorable pressure gradient resulted from the flow “necking down”, thus increasing free stream velocity. Therefore, if we have the flow “opening up,” like in a diffuser, then we could expect some free stream deceleration. With a positive pressure gradient, we find the curvature of the velocity profile at the wall is positive. Since the velocity profile has negative curvature towards the boundary layer edge, we must have an inflection point somewhere in the flow field. This inflection point is how we characterize if flow separation will occur. If the adverse pressure gradient is sufficient enough to drive the wall shear to zero (∂U/∂y = 0), then we would have separation. This is characterized by the flow becoming detached from the surface and mixing, increasing system pressure, and being more susceptible to turning turbulent. This region of flow is far more difficult to characterize and measure. In addition to this increased mixing, we usually see a significant increase in the pressure field, which can lead to an increase in drag. This is problematic for the engineer who is trying to overcome drag forces, and thus is why we care about pressure gradients in a flow.

Note that we can only separate if there is an inﬂection point in the velocity proﬁle, and even further, we must have zero shear at the wall. That is why the favorable pressure gradient and zero pressure gradient boundary layers do not undergo separation.

7.3 Resistance to separation

From the three different types of flows (favorable, zero, and adverse pressure gradient), we saw that it’s only under certain circumstances in the adverse pressure gradient that we see separation. Once exposed to this adverse PG, the flow near the wall will be affected by the pressure acting against the motion. We can visualize this effect on two different flows: A turbulent boundary layer has more momentum in the near wall region compared to the laminar boundary layer. This results in the pressure having to apply more force to overcome this momentum. Additionally, turbulence is characterized by its ability to mix the flow, bringing low momentum fluid from the wall to the outer region, and bringing high momentum fluid from the outer region towards the wall. This means a turbulent profile has a way of “healing” the adverse pressure gradients effect, and thus delaying when separation may occur. This concept is illustrated in figures 7.3 and 7.4

7.4 Case: a curved surface

To demonstrate the importance of separation, let’s look to an airfoil. Imagine a nice, slender airfoil in flow at a low angle of attack, as pictured in figure 7.5. As the streamlines diverge, the flow rate between them must remain constant, so their velocity decreases. Bernoulli’s equation tells us this will lead to an increase in pressure, thus we have a positive pressure gradient in the flow. This was defined as an adverse pressure gradient, meaning it can induce separation!

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Figure 7.3: adverse PG acting against laminar and turbulent velocity proﬁles

Figure 7.4: turbulence moving momentum towards the surface

However, just because we have an adverse pressure gradient, it doesn’t mean we have to separate. Now, let’s look at a higher angle of attack: Look how the velocity profile in figure 7.6 not only has a positive inflection at the wall, but it also has backwards flow (relative to the incoming flow). This is exactly what flow separation is! The pressure gradient is very positive, and this high pressure ends up “pushing” the boundary layer away from the wall, causing separation. This results in decreased lift, and many complications for anyone trying to fly.

7.5 Strouhal number

Once separation occurs, a region of recirculation (which may or may not be turbulent) builds up. If the recirculation is behind a bluﬀ body, then these recirculation zones can shed in

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Figure 7.5: airfoil at slight angle of attack - note increase in pressure in direction of ﬂow from streamline divergence, giving an adverse pressure gradient (yes I drew this in paint, don’t ask how long it took)

Figure 7.6: airfoil at large angle of attack - note that there is separation now alternating patterns. There will be a frequency f associated with this shedding. If you perform a dimensional analysis, you can find a non-dimensional number based around this frequency, which we call the Strouhal number: fD St = (7.4) U∞ In figure 7.7, you see nice groupings of vorticity. However, even if the wake were to be turbulent, a fundamental frequency would still exist for the body. There are charts of Strouhal number vs. Reynolds number for a cylinder in cross flow. For a wide range of Reynolds number, the Strouhal number remains constant at around 0.21!

7.5.1 Example: Tacoma Narrows Bridge It’s likely that you’ve seen a video of the Tacoma Narrows bridge oscillating in the wind (if you haven’t take a moment to look it up on YouTube). On the day it collapsed, the winds were up to 20m/s, and the height of the roadway is approximately 3m. Was vortex shedding the cause

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Figure 7.7: von K´arm´anvortex street behind a cylinder at ReD = 100 (top) and 300 (bottom) of the collapse?

We start by calculating the Reynolds number of the ﬂow. We have the velocity and “diameter,” and because it’s in air, we know the kinematic viscosity is 0.000015m2/s. 20 ∗ 3 Re = = 4 × 106 0.000015 For the most part, the Strouhal number can be considered to be a near constant St = 0.21 once Reynolds number is greater than 100. Of course it is more complicated, especially with diﬀering geometries, but let’s stick to this for now. Using this estimation, we then calculate the frequency of shedding as: f ∗ 3 St = 0.21 = → f = 1Hz 20 If you watch the footage, the bridge is oscillating at around 0.25Hz, which isn’t the frequency from the shedding. This means a more complex interaction occurred to cause the collapse, but I’d bet the wind didn’t help.

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7.5.2 Example: car antenna You’re gifted an older car with an antenna on it. When traveling at highway speeds, you hear an low, throbbing noise. Should you attribute the noise to the antenna, or get your car checked up?

Let’s do a quick check on the Reynolds number, and thus Strouhal number, for the antenna. It’s likely around 1/8in in diameter, and you follow the speed limit in NJ (right?) and do 65mph on the highway. We then calculate the Reynolds number:

UD 65 mi ∗ 1600 m ∗ 1 hr ∗ 1 in ∗ 0.0254 m Re = = hr mi 3600 sec 8 in ≈ 6100 d ν m2 0.000015 s

From a chart relating St to ReD for a cylinder, we ﬁnd St = 0.21, which matches what we’ve previously said! Then, we can calculate the frequency associated with this antenna: f ∗ 0.0032m St = 0.21 = → f ≈ 1900Hz 29m/s

A frequency of 1900Hz is not a low throbbing noise, but more of a high pitched whine. In fact, if you’ve ever been to a very loud concert, you’ve probably left with your ears ringing. That ringing sound is somewhat similar in pitch to this frequency! So a low throbbing noise in your car isn’t from the antenna, and you should probably get it looked at.

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Internal Flows and Losses

8.1 Internal ﬂows

The study of boundary layers is classified under external flows, where the flow goes over the outside of a body. Other examples are flow over a cylinder or an airfoil, or the flow in the atmosphere. We are going to now look at internal flows, where the flow is surrounded by the body. In other words, we’ll be looking to channel and pipe flows.

We like to simplify the analysis by considering flows that are fully developed. This means that the velocity profile within the pipe has developed to it’s fullest extent, and will not change as it moves downstream. But what do we mean by “developed?” If we look to the entrance of fluid in a 2-D channel, we can see:

Figure 8.1: development of the velocity proﬁle in a channel entrance

As the fluid flows in the channel, a boundary layer starts to develop and grow. Note that there will be a boundary layer on the top of the channel as well as the bottom. These will grow towards each other, until at some downstream location they merge. Since the velocity profile changes with x in a boundary layer, we expect the velocity profiles to change up to this merging. At this point, the flow has developed, and there is no more growth of these boundary layers or change to the velocity profile with further movement in x. We now call the flow fully developed.

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Imagine that you were to treat the two walls independent from one another. You could then solve for the distance into the channel that the boundary layer develops. In other words, how far downstream must you go until the boundary layer thickness is half the height of the channel? If the ﬂow is laminar, we can just use the Blasius relation: δ 5 = √ x Rex If the height of the channel is H, then we want to see when the boundary layers develop to half that height (they both grow from opposite sides, so they only need to meet in the middle). Therefore, we can estimate the “development length” from plugging this into the Blasius relation:

H H 5 δ = → 2 = √ 2 x Rex H x = q 10 ρU∞x µ H2U x = ∞ 100ν x = 0.01 ∗ Re (8.1) H H Equation 8.1 is an approximate expression for the entrance length for a 2-D channel, assuming the flow is laminar. This same expression can be found for a pipe as well. The exact solutions for the two cases wouldn’t be this expression, but they are very similar. Why do we say approximate solution? Think on the laminar flow equations - we assumed zero pressure gradient to get the Blasius solution. We know there is some sort of acceleration to the outer flow here, as we discussed with respect to displacement thickness. This acceleration means there is a pressure gradient, thus our Blasius solution is no longer valid. However, this at least gives us a decent approximation to understand how long it takes to turn fully developed.

Stating that a flow is fully developed has two primary implications. • Viscosity is important everywhere. This is because the two boundary layers have merged, and the boundary layer is the region in a flow where viscous effects are important. Therefore, viscosity will play a roll all throughout the flow in a channel (or pipe) • There are no velocity gradients in the x-direction. Since it is fully developed, there is no more change in the shape of the velocity profile that needs to occur. This means U becomes a function of y only, not x. Let’s investigate the result of these effects.

8.1.1 Continuity By turning to the incompressible, 2-D continuity equation, we can immediately ﬁnd a fascinating result for fully developed ﬂows: ∂U ∂V + = 0 ∂x ∂y

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We just said that a fully developed ﬂow has no changes to U in the x-direction, which means the ﬁrst term is zero! Therefore, we can show: 0 ∂U7 ∂V + = 0 ∂x ∂y Z ∂V dy = 0 ∂y V (y) + const = 0

This says that our vertical velocity V (y) is actually just a constant. However, applying the kinematic boundary condition (no ﬂow through the wall) means V = 0 at the wall. Since V is just a constant, then:

V = 0 everywhere in fully developed pipe/channel ﬂow

8.1.2 Inertial terms Using this result, let’s look at the inertial terms in the x-momentum equation: ∂U ∂U ρU + ρV =··· ∂x ∂y We just saw that V is zero, so the second term is gone. However, we’ve also said that the U velocity doesn’t change in the x-direction, so the ﬁrst term is also gone! This means that the inertial terms do not have a contribution to the x-momentum equation! If you were to look at the y-momentum, you could probably guess that the inertial terms there are also zero!

8.1.3 The x-momentum equation for a channel (and pipe) Due to there not being any inertial eﬀects, we can write the x-momentum equation as:

dP d2U 0 = − + µ dx dy2 Or equivalently, we can write: dP d2U = µ (8.2) dx dy2 Equation 8.2 is saying that in a channel, the pressure gradient is balanced by viscous eﬀects. In other words, the viscous force balances the pressure force. The inertial terms are negligible compared to these two forces! If we were to do the same analysis in cylindrical coordinates, we would ﬁnd the x-momentum equation in a pipe is:

dP µ d dU = r (8.3) dx r dr dr

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Overall, the flow is viscously dominated throughout, and inertia doesn’t play an effect to the dynamics of the fluid. So a pressure gradient is the only thing left to overcome the effect of viscosity slowing the fluid down. This is what equations 8.2 and 8.3 are telling us, since it’s the pressure term balancing the viscous term.

8.1.4 The pressure gradient One aspect of pipe and channel flow that can be confusing stems back to the “fully developed” description. We said that the boundary layers from the entrance of the pipe converge, thus the flow has developed to its final profile. After that point, the velocity profile will not change with increasing downstream distance. Put another way, there are no changes in the x-direction. How- ever, we see there is a pressure gradient, where we have an x-derivative. How do we explain that?

Let’s step away from the equations and try to use intuition for a moment. If you want to push fluid through a pipe, we do that using a pump. This is just a way to increase the pressure at one end of the pipe. The increased pressure at one end drives the fluid out the other. However, viscous effects will work to resist the flow down and induce “energy losses” in the system. Therefore, this pressure must be decreasing throughout the pipe. That’s exactly what a pressure gradient is - a change in pressure over a distance. Now, this explains why dP/dx can exist, but we said there aren’t any changes to the flow velocity in the x-direction. This means that the gradient can’t be changing in the x-direction either. So while the pressure changes in x, the gradient itself is constant!

Let’s do a little math to show this has to be true. Start by taking a derivative of our momentum equation to show the result:

d dP d2U = µ dx dx dy2

We can interchange x- and y-derivatives (this is a calculus thing, hopefully you are convinced this is true), so then we can write:

d dP d2 dU = µ dx dx dy2 dx

Since U is not changing in the x-direction, the RHS of this equation is zero.

d dP = 0 dx dx dP = const. dx So in addition to our “gut check” for the pressure gradient, the math also showed that the pressure gradient must be constant. The value of the constant will set the magnitude of the velocity, with increasingly large magnitudes corresponding to increasing velocity. What about the sign of the constant - is the pressure gradient positive or negative? Think about the discussion we had with separation. The second derivative of velocity indicates the curvature.

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We should have negative curvature throughout the velocity proﬁle for an internal ﬂow, so we would expect the value of the pressure gradient to be negative. This means the pressure is falling with increasing distance, which again falls in line with the gut check!

8.2 The velocity proﬁle in a pipe

This problem is worked out in any undergrad textbook, but perhaps this will be another way to think about it. We want to ﬁnd the laminar velocity proﬁle in a pipe.

We start by looking to the momentum equation for a pipe and applying some boundary conditions: dP µ d dU = r dx r dr dr We know there is no slip at the wall, so the velocity at the wall is U = 0 at radial position r = R. Additionally, the velocity proﬁle will be symmetric (or equivalently, there is no shear on the centerline), so the gradient should be zero at r = 0, or dU/dy|r=0 = 0. Using the equation and boundary conditions, we can start rearranging:

µ d dU dP r = r dr dr dx d dU r dP r = dr dr µ dx

Now, we had just done some math and shown that the pressure gradient itself is a constant, so we can integrate both sides with respect to r:

Z d dU Z r dP r dr = dr dr dr µ dx dU 1 dP Z r = r dr dr µ dx dU 1 dP r2 r = + C dr µ dx 2 1 dU 1 dP r C = + 1 dr µ dx 2 r Applying the no-shear boundary condition, we find: C 0 = 0 + 1 0 We can’t have a singularity in the flow! We know that the flow exists, so this means we have to

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set C1 = 0 to keep the ﬂow physical. Then, we can integrate the expression again: 0 dU 1 dP r C¡ = + ¡1 dr µ dx 2 ¡r Z dU 1 dP Z r dr = dr dr µ dx 2 1 dP r2 U = + C µ dx 4 2

Now, setting U = 0 at r = R as our second boundary condition, we can find C2: 1 dP R2 0 = + C µ dx 4 2 1 dP R2 − =C µ dx 4 2 With a little rearranging, we now have an expression for our velocity profile in laminar pipe flow: 1 dP U(r) = r2 − R2 (8.4) 4µ dx If this expression doesn’t quite look like what you have in your book, that’s OK: it’s just based off how we defined our boundary conditions. Also, it looks like the velocity will be negative, since we have r2 − R2 for all r ≤ R. But remember what we said about the sign of the pressure gradient. The pressure falls as we move down the pipe, as it’s the pressure force overcoming the viscous force. Therefore, in addition to knowing that the pressure gradient is a constant, we can write: dP = −K dx We do this simply because there will be a given pressure gradient. We know it’s negative, and we know it’s constant, so defining it as −K just makes the following expression easy to look at. Then we write the velocity profile as: K U(r) = R2 − r2 4µ Notice how we can see this is a parabola that has a maximum at r = 0 and is zero at r = R. We can re-cast this in terms of the pipe diameter by substituting R = D/2 to get: K U(r) = R2 − r2 4µ K D U(r) = 2 − r2 4µ 2 KD2 2r 2 U(r) = 1 − 16µ D Overall, we see that the velocity profile in a laminar pipe flow is parabolic!

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8.3 The friction factor in a laminar pipe

The skin friction is a non-dimensional parameter that we had deﬁned in the boundary layer: τ c = w f 1 2 2 ρUs

Where τw is the local wall shear stress, and Us is a scaling velocity. In the boundary layer, we use U∞, but in a pipe we will have to think about what the appropriate velocity scale is. Now, this isn’t exactly like the friction factor f, which is deﬁned as:

∆P D f = L 1 2 2 ρUs

Let’s turn to the laminar pipe ﬂow equation and see why f and cf are a bit diﬀerent:

dP µ d dU = r dx r dr dr integrating, rearranging, evaluating at wall

R dP dU = µ = τw 2 dx dr r=R We see that the wall shear stress is balanced by the pressure gradient times a length scale in the radial direction. If we plug this relation for τw into our expression for cf , we ﬁnd:

τ R dP c = w = 2 dx f 1 2 1 2 2 ρUs 2 ρUs Now, we know that the pressure gradient is a constant, and is negative in magnitude. Therefore, we can deﬁne this pressure gradient as some pressure drop over a length: dP ∆P = dx L Plugging in, we ﬁnd: R ∆P 1 ∆P D f c = 2 L = L = f 1 2 1 2 2 ρUs 4 2 ρUs 4

So the friction factor f is just four times the skin friction coefficient cf . It doesn’t matter that we have this constant 4 involved, it’s just scaling the results up by a factor of 4. If we plotted cf and f on the same chart, we’d find they are the exact same shape. For pipes, we just use f for historical purposes, so we’ll stick to that Now, let’s see if we can find an actual relationship for the friciton factor f based off of flow parameters. We need to chose an appropriate velocity scale Us. Let’s pick an easy velocity to

Intro to Fluids Notes 105 C. P. Byers 2016 CHAPTER 8. INTERNAL FLOWS AND LOSSES measure, which would be the bulk velocity U¯, or area averaged velocity: 1 Z U¯ = U(r) dA A 4 Z D/2 KD2 2r 2 = 2 1 − 2πr dr πD 0 16µ D K Z D/2 2r 2 = 1 − r dr 2µ 0 D K r2 r4 D/2 = − 2 2µ 2 D 0 KD2 = 32µ We can then plug this expression for our bulk velocity in to the friction factor expression: ∆P D f = L 1 ¯ 2 2 ρU We know that ∆P/L is just the pressure gradient, and since the pressure drop is negative but we want a positive value, we plug in K for it. Also, rather than plug in U¯ 2, let’s plug in one U¯ and leave the second as a variable. We’ll see why: KD f = 1 ¯ KD2 2 ρU 32µ 1 = ρUD¯ 64µ You can see that leaving one of the U¯ has allowed us to get a familiar non-dimensional number! So for laminar pipe ﬂow, we ﬁnd the friction factor f is: 64 f = (8.5) ReD

The friction factor is inversely related to the Reynolds number. That means for increasing ReD, the friction factor decreases! Does that make sense? Let’s think - friction factor falls as f ∼ 1/U¯, 2 but it is related to drag by FD ∼ f ∗ U¯ . Therefore, even though the friction factor is falling, the total drag is increasing. That makes more sense, so that’s good to see! Remember that this relationship is only good for laminar pipe ﬂow, so ReD < 2300 or so. If you are in the turbulent regime, you’ll have to use a table, chart, or empirical formula instead.

8.4 Example: a vertical pipe

Water of density ρ and viscosity µ flows steadily down a vertical circular pipe of diameter D. The flow is fully developed and has a parabolic velocity profile described by: U 2r 2 = 1 − Ucl D

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where Ucl is the centerline velocity. Find the wall shear stress and express the kinematic viscosity ν in terms of ﬂow parameters.

Figure 8.2: fully developed gravity-driven ﬂow

The flow here is a fully developed pipe flow, but we’re driven by gravity. Rather than have a pressure gradient in the z, we will have the gravitational force. Let’s think on why - if the flow were stationary, we’d have a pressure gradient balanced by gravitational forces. This is exactly the hydrostatic equation! dP no flow: = ρg dz Notice that there isn’t a minus sign. Look at figure 8.2 and think about what direction the z is defined as positive, and what direction g is pointing. Therefore, with the pressure increasing in the direction of gravity, the pressure gradient is positive. Now, by allowing flow to occur, the frictional forces will oppose the motion. However, we see that the pressure gradient is just set by the gravitational forces, so we can instead say our z-momentum equation is:

µ d dU 0 = r + ρg r dr dr

So the gravitational term has replaced our pressure gradient as the “ﬂow driver.” Also, we have U in the z-direction. Yeah, that’s a pretty sloppy move on our part since it should really be W

Intro to Fluids Notes 107 C. P. Byers 2016 CHAPTER 8. INTERNAL FLOWS AND LOSSES in the z-direction, but let’s just go with it for now. Now, integrate and solve for the shear stress:

d dU −ρgr = µ r dr dr 2 r dU −ρg = µr 2 r=D/2 dr r=D/2 2 D D dU −ρg = µ 8 2 dr w

D dU −ρg = µ 4 dr w

We then ﬁnd that the wall shear stress τw is:

D τ = −ρg w 4

Note that the wall shear is a negative value. That’s because it is acting in the negative z- direction to oppose the ﬂow in the positive z-direction.

Now we want to solve for the kinematic viscosity, ν = µ/ρ. We’re given the shape of the velocity proﬁle, so we can plug that in for our expression of the wall shear stress:

D τ = −ρg w 4

dU D µ = −ρg dr w 4 2 d 2r D µ Ucl ∗ 1 − = −ρg dr D w 4 8 D µUcl − 2 r = −ρg D r=D/2 4 4 D µU − = −ρg cl D 4

Now, we can simply re-arrange to get the kinematic velocity:

gD2 ν = 16Ucl

Therefore, if we have a vertical pipe that we know is in steady laminar flow, but we don’t know the fluid viscosity, we can extract its value by measuring the bulk velocity! This is just one way to measure the viscosity of a fluid.

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8.5 Example: force to hold a pipe

On a very long pipeline, large sections are held in place by a series of flanges. If we assume these flanges act independently from one another, what would be the force a flange exerts on a section of pipe that is 0.5m in diameter, 100m long, carrying crude oil at a velocity of 1m/s?

Crude oil has a wide range of viscosity and density values, so let’s just use an average value −6 m2 kg of ν = 5 × 10 s and ρ = 900 m3 . We then calculate the Reynolds number for this flow: UD 1 ∗ 0.5 Re = = = 105 D ν 0.000005 So we are most definitely in the turbulent regime. To figure out the force exerted, we turn to the friction factor, f, which we can look up in a chart. Assuming the pipe is smooth, we find 5 for ReD = 10 that f = 0.018. We then use the definition of the friction factor to extract the pressure drop: ∆P D dP D f = L = dx 1 2 1 2 2 ρUs 2 ρUs We also know from our previous math with the momentum equation that the pressure drop is related to the shear stress by: D dP τ = w 4 dx Therefore, we have: 4τ f = w 1 2 2 ρUs 8τ 0.018 = w 900 ∗ 12 τw = 2.025P a To find the total force, we need to integrate over the area of the pipe this acts on. That’s the inner cylindrical surface area A = πDL. Since the flow is fully developed, the velocity profile isn’t changing, and thus the shear force is a constant! So we simply multiply the shear over the area to find: ZZ FD = τw dA

FD =τw ∗ A

FD =2.025P a ∗ π ∗ 0.5m ∗ 100m

FD = 318N This doesn’t seem to be too large of a value for the large amount of oil the pipe is carrying. However, let’s think on what would happen if we tried to have the same volume ﬂow rate through a smaller pipe. Let’s start by calculating the volume ﬂux (since density is constant, we don’t worry about it): m 0.52m2 m3 V˙ = UA¯ = 1 ∗ π = 0.196 s 4 s

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If you were to try and pass this much ﬂuid through a pipe with half the diameter, we would have a higher average velocity:

0.252m2 m V˙ = 0.196 = U¯ ∗ A = U¯ ∗ π → U¯ = 4 2 2 2 4 2 s With four times the velocity but half the diameter, the Reynolds number will be twice as high. 5 The friction factor for ReD = 2 × 10 is f = 0.016. Then, we can ﬁnd the force needed to exert on 100m of pipe at this velocity: 4τ f = w 1 2 2 ρUs 8τ 0.016 = w 900 ∗ 42 τw = 28.8P a

And once again, multiplying by the internal area over which this shear occurs, we ﬁnd:

FD =28.8P a ∗ π ∗ 0.25m ∗ 100m

FD = 2262N

This is why pipelines use very large pipe diameters - it considerably reduces the total force they must overcome for a given ﬂow rate for a nominal increase in construction costs!

8.6 Losses in pipe ﬂow

We’ve used Bernoulli’s equation to solve many flow problems before. It’s a powerful tool, so long as you can satisfy the assumptions necessary for it to be valid. However, we can still use a type of Bernoulli’s equation that is modified to account for losses in the flow that occur due to a departure from our initial assumptions. Let’s start by looking at Bernoulli’s equation in a horizontal pipe: P 1 + gz + V 2 = H ρ 2 Note that we’ve divided by ρ to simply re-cast the equation in a slightly different form, and H is our constant. If we were to apply this at two points in a horizontal pipe flow that is fully developed, we’d find: P 1 P 1 1 + gz¨ + V 2 = 2 + gz¨ + V 2 ρ ¨ 1 2 1 ρ ¨ 2 2 2 P − P 1 1 2 = V 2 − V 2 ρ 2 2 1

In fully developed pipe ﬂow, the velocity isn’t changing downstream, so we would get: P − P 1 2 = 0 ρ

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This indicates that the pressure in the pipe is constant, but we know better! This is an example of why Bernoulli is restrictive in its use - it cannot account for the viscous eﬀects.

We can instead modify Bernoulli’s equation to account for these losses. If you recall from Chapter3, we did a rigorous derivation for Bernoulli’s equation starting from the Navier-Stokes equation. Through some vector identities and assuming a steady flow and constant density, we got: 1 P 1 ∇ U 2 + + gz = U~ × ∇ × U~ + ∇ · T 2 ρ ρ v The next step for Bernoulli’s equation is to assume irrotational and inviscid flow, then the RHS is zero, implying the bracketed term is a constant. Instead, we know there are viscous effects and geometric losses (like valves, orifices, elbows, etc.) that can affect the flow. We want to try and account for those losses. Now, we usually utilize this as a quasi-1-D type equation. In a pipe, the velocity is most certainly not constant. However, we usually use the area averaged velocity, such as V¯ . When using this averaged value, we utilize what is called a kinetic energy coefficient, α, which is defined as: 1 Z α = V 3 dA AV¯ 3 This comes from the integrated flux of kinetic energy. Overall, this coefficient α is a way to account for the different type of flow. In laminar flow, α = 2, while in turbulent pipe flow we can approximate α ≈ 1. Now, using this coefficient and the bulk velocity (or area averaged velocity), we can write our modified Bernoulli equation as:

1 P 1 ∇ αV¯ 2 + + gz = U~ × ∇ × U~ + ∇ · T 2 ρ ρ v

Looking at the first term on the RHS, that has a rotational component. If we were to look at this function along a streamline, then we would see the first term on the RHS is zero (because U~ × ∇ × U~ is perpendicular to the flow).

: 0 1 ¯ 2 P ˆ ~ ~ ˆ 1 ˆ ∇ αV + + gz · dl = U ×∇ × U · dl + ∇ · T v · dl 2 ρ ρ

The remaining term on the RHS is the viscous contribution. We will say this can be represented by some sort of parameter to characterize the losses due to viscous eﬀects. For now, let’s not try to simplify it. Instead, we integrate both sides. Recall from chapter3 that the LHS turns into a perfect diﬀerential when you take the dot product along a streamline. Therefore, integration between two points along a streamline will give us:

Z 2 Z 2 1 2 P 1 d αV¯ + + gz = ∇ · T v · dˆl 1 2 ρ 1 ρ 2 Z 2 1 2 P 1 αV¯ + + gz = ∇ · T v · dˆl 2 ρ 1 1 ρ

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Now our LHS is back to the original Bernoulli expression: the diﬀerence in the Bernoulli constant between two points! However, we have a RHS that is not zero, so long as there are viscous contributions. We can expect that this value is negative, as it is a viscous force that is working to reduce the momentum in the ﬂow. Let’s simplify the expression by using the following:

Z 2 1 ˆ ∇ · T v · dl = −g ∗ hl 1 ρ

Why do we chose to represent this loss as −g ∗hl? Look at the units on the LHS of our Bernoulli expression - they are all units of m2/s2. We could divide this equation through by g to get all terms in the dimension of some length scale. This is often referred to some sort of head, such as pressure head, or velocity head. Think about a pipe ﬂow with liquid manometers on top to show the pressure as a column of water - that’s the “head” at that point. Now (taking notice of the minus sign) we have the following expression:

1 P 1 P αV¯ 2 + 1 + gz − αV¯ 2 + 2 + gz = g ∗ h (8.6) 2 1 ρ 1 2 2 ρ 2 l The two bracketed terms on the LHS are just like the original Bernoulli equation, and the right hand side is our losses in the flow. Now let’s think about what happens in a horizontal pipe flow. Before, the classic Bernoulli equation found that the pressure would remain the same, but here we see something different. The elevation is the same, so gz cancels in both brackets. The flow is fully developed, so the velocity terms cancel. You are then left with:

P1 − P2 = ρ ∗ g ∗ hl

This is saying that there is a pressure difference due to losses in the system! That’s what we know happens! The actual values of the head loss coefficients can be found from tables. There are different expressions for different types of losses. Major losses are characterized by the viscous drag. Minor losses are characterized by fittings, valves, changes in diameter, etc. The equations we use are defined as:

L V¯ 2 major losses: h = f l D 2g V¯ 2 minor losses (a): h = K l 2g L V¯ 2 minor losses (b): h = f e l D 2g

The values of K and Le can be looked up in the a table for specific flow situations, and f is our friction factor. We can find that in a table, or in laminar flow we can calculate it. We’ll use these in an example to get an idea of what is going on.

This equation still holds for turbulent flow! That’s because we derived this from the Navier- Stokes equations, which also hold for turbulent flows. The factors that change will be α, which you set as 1 in turbulent flows, and f, which you can look up for different ReD.

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8.7 Example: ﬂow from a reservoir

Lets imagine we have a large reservoir of ﬂuid that empties through a pipe, as pictured in the ﬁgure below. The entrance can be approximated as a square-edged entrance. The pipe material

m3 is galvanized iron, and the ﬂuid is water. If the ﬂow rate is 0.05 s , what is the height of the reservoir?

We start by first recognizing the types of losses that can occur. First, the entrance to the pipe has an abrupt square edge, so we have a minor loss there. Then, we have frictional losses in the flow, which is a major loss. We need the friction factor, and have to account for the surface roughness. The exit is an abrupt opening, so that also has a minor loss associated with it. Turning to Bernoulli, let’s start with the balance from the top of the reservoir to the entrance of the pipe: P P 1 atm + gH = enter + αV¯ 2 ρ ρ 2 Here we’ve assumed the pipe entrance is the zero height reference point. We don’t have any losses up to this point, so we’re set for this expression. Next, let’s balance from the pipe entrance to the pipe exit: P 1 P 1 enter + αV¯ 2 = atm + αV¯ 2 − g(200) + g ∗ h ρ 2 ρ 2 l We have the losses we need to account for, so we include the g ∗ hl term. Note that the velocity is assumed the same at the entrance and exit, since it’s some sort of average velocity. That means the balance on this second equation can cancel out velocities, but we won’t do that yet. Instead, let’s substitute this into our first expression to get: P P 1 atm + gH = atm + αV¯ 2 − g(200) + g ∗ h ρ ρ 2 l 1 αV¯ 2 = g(200 + H − h ) (8.8) 2 l

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If we would have done the balance between the surface of the reservoir and the exit of the pipe, we would have obtained this expression from the start! Bernoulli is still working for us, so long as we account for the losses. Before we even do any math, we can see something intriguing by this expression. Just solving for velocity (and ignoring α for now) gives the expression: p V¯ = 2g(H + 200 − hl) When ignoring the head loss term, this is exactly the same expression Bernoulli’s equation gives for the velocity at the exit of a tank of height H + 200. The losses in the system, which are accounted for in the hl term, will work to reduce this exit velocity. That should hopefully make sense, so we are still on the right track!

Now, let’s ﬁgure out the velocity. We have a ﬂow rate and pipe diameter, so we can get V¯ :

3 Q˙ 0.05 m m V¯ = = s = 6.4 A 0.12 2 s π 4 m Then with this, we find the Reynolds number in the pipe is: m 6.4 s ∗ 0.1m ReD = = 640000 −6 m2 10 s This is most certainly turbulent! First, this means the kinetic energy coefficient α is just one. Then, using this Reynolds number, we can find the friction factor for the pipe. Recall that it’s galvanized iron, which means it has a surface roughness of k = 0.15mm, meaning k/D = 0.00075. Then, looking up on a Moody plot for this roughness and Reynolds number, we can find the friction factor f ≈ 0.019. We next look to the entrance condition, which we said was square edged. This gives a minor loss coefficient of Kentry = 0.5. Finally, there is an abrupt exit, which gives a minor loss coefficient of Kexit = 1. We can now put together our relationship for equation 8.8: 1 αV¯ 2 = g(200 + H − h ) 2 l 1 αV¯ 2 = g(200 + H − h − h − h ) 2 major entry exit 1 L V¯ 2 V¯ 2 V¯ 2 (α)V¯ 2 = g 200 + H − f − K − K 2 D 2g entry 2g exit 2g 1 2000 (6.4)2 (6.4)2 (6.4)2 (1)(6.4)2 = (9.8) 200 + H − 0.019 − 0.5 − 1 2 0.1 2 ∗ 9.8 2 ∗ 9.8 2 ∗ 9.8

Solving for H, we then ﬁnd: H = 599m This is telling us that the losses in the system require the tank height to be 599m to get the given ﬂow rate through the pipe. If it were ideal, this height would result in a bulk pipe velocity of 125m/s, which most certainly higher than the 6.4m/s we have here.

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Supercritical and Supersonic Flows

9.1 Comparison of Mach number and Froude number

We group the discussion of water waves and compressible flow together due to their similarities. The striking analogy that can be made deals with the two non-dimensional numbers we form for the respective flows. These two numbers are the Mach number M and Froude number F r: U U M = and F r = √ a gy These two numbers are a ratio of the flow velocity U and a characteristic velocity. In compressible √ flow, we use the speed of sound a, and in shallow water waves we use the wave speed gy, where g is gravity, and y is the water depth. Another way to think about the Froude number is to square it and re-arrange: U 2 1 ρU 2 F r2 = = 2 · 2 gy ρgy This shows that the Froude number is a ratio of kinetic energy to potential energy. The similarity between the two non-dimensional numbers can be shown in a simple schematic. In figure 9.1, we illustrate the propagation of a disturbance in a flow field. In this case, there is some disturbance (the black dot) which is subjected to a flow of speed U, resulting in the respective Froude number. From the point of the disturbance, waves will propagate outward √ at the wave speed, gy. However, they are also convected along at the flow speed. When there is no flow we have F r = 0, and waves propagate outward equally. When 0 < F r < 1, then the waves are convected downstream at a velocity which is slower than their wave speed. The wavefront then moves more quickly downstream, but more slowly upstream. Think on the doppler effect for this one. When F r > 1, then the waves cannot move upstream, since they are convected away at a speed greater than their traveling speed, so a “cone” is formed that “knows” about the disturbance. Upstream of the cone, no information is known about the disturbance. Downstream, the wake is the only region that has experienced or been influenced by the disturbance. The angle of this cone can be calculated by relating the distance traveled by the flow vs. the distance the disturbance travels. This simplifies down to: 1 sin α = F r

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Figure 9.1: shallow water waves from a disturbance in diﬀerent ﬂow speeds

Now, if we were to do the same analysis with sound waves, we would have the schematic as seen in ﬁgure 9.2. Notice that this is exactly the same setup as was with the shallow water waves!

Figure 9.2: sound waves from a disturbance in diﬀerent ﬂow speeds

The primary diﬀerence here is the wavefront buildup in the sound wave case leads to shocks, where in the shallow water waves we get a breaking wave.

9.2 Hydraulic jumps

A hydraulic jump occurs when a high velocity ﬂuid undergoes a rapid deceleration and sudden rise in depth. This often occurs due to a downstream condition that requires the ﬂow to slow down. One way you can see this in every day life is in a sink. When you turn the faucet on, the water comes out the spout and hits the bottom of the sink. From there, it rapidly spreads in a

Intro to Fluids Notes 116 C. P. Byers 2016 CHAPTER 9. SUPERCRITICAL AND SUPERSONIC FLOWS radial fashion, but at some distance it will undergo a jump. This is exactly a hydraulic jump! The downstream condition in this case is likely an interaction of unstable flow with surface roughness. Utilizing the continuity and momentum equations, you can extract an expression for a hydraulic jump. Figure 9.3 illustrates the simplified flow scenario. The math to find the ratio of heights isn’t too terribly complicated, but the derivation is a little tedious so we’ll just utilize the result: h 1p 1 = 1 + 8F r2 − 1 h0 2 Note here that F r is the Froude number of the incoming flow. We have three options for F r:

1. F r = 1 → h1 = h0, no change in height

2. F r > 1 → h1 > h0, a jump in height

3. F r < 1 → h1 < h0, a drop in height

Figure 9.3: schematic for a hydraulic jump (thanks to wikipedia for the image)

The third option of a hydraulic drop turns out to be unphysical. To understand why, compare the Bernoulli constants for the two flows. The jump is a turbulent mess of flow, so we would expect losses to occur across it. Along the surface, we would have the same pressure upstream 1 2 and downstream, so it’s just a comparison of 2 ρU and ρgh on each side of the jump. Once again, an expression can be built, and it is once again a tedious process. The math is done in Lex Smits’ book, but the result shows that if a hydraulic drop were to occur, there would be more energy after the drop than before. Since losses occur through the drop, this turns out to be non-physical. Therefore, we can only have hydraulic jumps, which occur for F r > 1. This once again turns out to be analogous to compressible flow, where a hydraulic jump is like a shock. Shocks are regions of intense change in fluid properties that occur when M > 1. Across a shock, the flow will go from supersonic to subsonic, and there are losses that can be accounted for by an increase in entropy. Do you think there could be shocks where a flow goes from subsonic to supersonic? No! that would be an incredible phenomenon to see in the first place, but more importantly that process would result in a net decrease in entropy. Since this cannot happen, then subsonic shocks do not occur. This is quite similar to sub-critical flow (F r < 1) being unable to experience a hydraulic drop.

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9.2.1 Example: hydraulic jumps on a dam One example of the importance of hydraulic jumps lies in the creation of dams. Many dams have some sort of spillway for when the water level is too high to maintain. An example can be seen in figure 9.4. As the water flows down the spillway, it’s velocity increases, and the height of the water level narrows due to continuity. This means the Froude number is increasing from zero at the top, until it goes supercritical (F r > 1). At some point, a hydraulic jump will occur. The issue with jumps is the intense mixing that occur in the jump itself. This region is highly turbulent, promoting mixing and large gradients in the velocity field. Imagine if this mixing were to occur over the riverbed - it would work to quickly significantly erode the material it is over. This would end up compromising the structural stability of the dam/riverbed system.

Figure 9.4: picture of a dam spillway with a hydraulic jump (thanks wikipedia)

Due to these concerns, engineers go out of their way to assure the hydraulic jump occurs on the spillway itself. Since it is made of concrete, it will be more structurally sound than the earthen floor. There are a few ways to assure this jump occurs early enough - either make the spillway long enough, or putting flow disturbance devices (like baffles, sills, blocks, etc.) towards the end of the spillway to assure a jump occurs.

9.2.2 Example: jump in a channel Water is ﬂowing in a rectangular channel at a depth of 30cm with a velocity of 15m/s. If downstream conditions force a hydraulic jump to occur, what will be the depth and velocity

Intro to Fluids Notes 118 C. P. Byers 2016 CHAPTER 9. SUPERCRITICAL AND SUPERSONIC FLOWS downstream? What is the head loss across the jump?

We start by using the relation for a hydraulic jump. For this, we need the Froude number:

U 15m/s F r = √ = = 8.75 gy p0.3m ∗ 9.8m/s2

This ﬂow is most certainly supercritical. Therefore, we can use the Froude number and the incoming height to calculate the height of the ﬂow after the hydraulic jump: h1 1 p 2 0.3m p 2 = 1 + 8F r − 1 → h1 = 1 + 8 ∗ 8.75 − 1 h0 2 2

h1 = 3.56m

The height of the water increased by (around) tenfold! We then can calculate the downstream velocity by utilizing continuity. We know water is pretty much incompressible, so we have:

m˙ out =m ˙ in

ρUoutAout =ρUinAin

Uout ∗ 3.56m ∗ w =15m/s ∗ 0.3m ∗ w

Uout =1.26m/s

The velocity dropped by (around) tenfold. That should have been expected√ from the result with the height increase. Checking the Froude number, we see that F r2 = 1.26/ 9.81 ∗ 3.56 = 0.21, or subcritical. Now, to calculate the head loss, we can use the modiﬁed form of Bernoulli’s equation from the previous chapter:

1 P 1 P αV¯ 2 + 1 + gz − αV¯ 2 + 2 + gz = g ∗ h 2 1 ρ 1 2 2 ρ 2 l

If we look along the surface of the water, we can neglect the pressure contribution since it’s just atmospheric pressure at both locations. Then, assuming the coeﬃcient α = 1, we can plug in the velocity and heights to get:

1 1 h = αV¯ 2 + z − αV¯ 2 + z l 2g 1 1 2g 2 2 152 1.262 h = + 0.3 − − 3.56 l 2 ∗ 9.8 2 ∗ 9.8 hl = 8.13m

This is stating that there was an “energy loss” of 8.13m of the working ﬂuid, which in this case is water. Think about the pressure exerted by water - we know that 10m is approximately one atmosphere of pressure. That’s a lot of lost potential working energy in the jump!

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9.3 Compressible ﬂow relations

The behavior of compressible flow is unique and quite unlike what we’ve seen in undergrad fluids. We’ve for the most part treated any flow as incompressible, and even further we’ve said that density, ρ, is constant. We’re now entering a regime where density is most certainly not constant. We’ll find a number of relations for the flow, but first let’s tie this back to the shallow water waves. Recall figure 9.2, where we saw that flows with M > 1 result in a shockwave. We can find a relationship for the angle α of the wave by relating the distance traveled for the different regimes: distance traveled by wave a · ∆t a 1 sin α = = = → = sin α distance traveled by flow U · ∆t U M Note that this is exactly analogous to the angle made with the shallow water waves. Also note what happens with M < 1: you cannot get a solution. This means that the soundwaves do not build up to a shock unless M > 1, as you probably would have expected.

There are many relations for compressible flow, and it involves a lot of thermo with fluids. We will once again skip over a ton of these derivations and instead just start with the results themselves. If you wish, you can find the actual math in any compressible fluids book, including Smits’ book. We’ll start with a few basic relations. For a perfect gas (which is an ideal gas with constant specific heats cp and cv) with a temperature T , density ρ, pressure P , and heat capacity ratio γ = cp/cv, we have:

1 γ ρ T γ−1 P T γ−1 P = c · ργ and 0 = 0 and 0 = 0 ρ T P T

In these expressions, c is a constant, and the subscript 0 refers to a reference condition, which we will choose as the stagnation quantities. In other words these expressions show how the density or pressure in a compressible flow relate to the stagnation quantities of that flow. For example, if you were to smoothly decelerate a flow to rest (and thus stagnate it), then you would recover ρ0, P0, and T0. Additionally, the speed of sound is defined as the square root of the change in pressure with respect to density at constant entropy:

2 ∂P a = ∂ρ s=const We can then use our above relationship for the perfect gas (and ideal gas) to get an expression

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2 ∂P a = ∂ρ s=const ∂ a2 = c · ργ ∂ρ ργ a2 =cγργ−1 = cγ ρ P/c P a2 =cγ = γ ρ ρ a2 =γRT

a = pγRT

We can then plug in some values to see what the speed of sound is! At sea level, T = 293K, J γair = 1.4, and Rair = 287 kgK , giving us a = 343m/s. If you’re in the upper atmosphere, things can be quite cold, so T = 100K then gives a = 200m/s. Does this make sense? As you get closer to space, the temperature drops, but we also have less and less atmosphere. Sound is simply pressure waves, which is a molecular transmission of kinetic energy. If the temperature is very low, then there is less kinetic energy in the molecules, and thus pressure waves move more slowly! Another important relationship that can be found for compressible ﬂow relates the temperature to the Mach number: T γ − 1 0 = 1 + M 2 T 2

9.3.1 Example: space shuttle re-entry This is a simplistic way to see why the space shuttle (and other re-entry vehicles) have heat shielding on them. If the shuttle enters the atmosphere at M = 15, where the temperature is T = 100K, approximately how hot does the shuttle get?

If we assume γ = 1.4 for the upper most reaches of the atmosphere (which it probably doesn’t but let’s not worry about it), then we can calculate: T γ − 1 0 = 1 + M 2 T 2 We pick the temperature of the flow to be T = 100K, and we solve for the stagnation temperature. Why isn’t the stagnation temperature 100K if the atmosphere is at rest? We must be mindful of the reference frame! The shuttle sees this fluid coming towards it at M = 15, which then will stagnate on the surface of the shuttle. Of course there is a shock, but let’s forget about that at this time. We then find: T 1.4 − 1 0 = 1 + 152 → T ≈ 4500K 100K 2 That’s pretty hot. It’s an over-approximation since the sparse gas turns into plasma, but it helps explain why there are special ceramic tiles on the shuttle. This also illustrates the explanation

Intro to Fluids Notes 121 C. P. Byers 2016 CHAPTER 9. SUPERCRITICAL AND SUPERSONIC FLOWS of a common misconception about the shuttle - it’s the compressibility of the air that causes the heating, not friction!

9.4 Normal shocks

Much like hydraulic jumps, shocks can occur when a supersonic flow encounters a downstream condition or obstruction that requires it to decelerate. There are different types of shocks, but we will concern ourselves with normal shocks for now. The name implies the shock will be oriented normal to the flow direction. This means shockwaves from a plane moving at M > 1, which are usually angled, are not normal shocks. Once again, there are many different relations

Figure 9.5: bow shock oﬀ a blunt object. The area near the centerline can be thought of as a normal shock, while the outer edges are oblique shocks (thanks NASA) and derivations for normal shocks. It involves more thermo, usually using the energy equation and the second law. We don’t want to concern ourselves with the derivations, so instead we will look into the results! One important relation is the temperature across the shock:

T 1 + γ−1 M 2 2 = 2 1 T γ−1 2 1 1 + 2 M2

Notice that this is just the ratio of the previous temperature relation we had used? That’s

Intro to Fluids Notes 122 C. P. Byers 2016 CHAPTER 9. SUPERCRITICAL AND SUPERSONIC FLOWS because the stagnation temperature across a shock is constant. This again is a result of some fancy work with equations. Another important relation would be the relation of the Mach numbers across a shock. For some upstream Mach number M1, the downstream Mach number M2 is found by:

2 2 M1 + M 2 = γ−1 2 2γ 2 γ−1 M1 − 1

This relationship shows that for any given M1 > 1, the downstream M2 is going to be less than 1. If M1 < 1, then we would get M2 > 1. Does that make sense? No! This would violate the second law of thermodynamics, and can be shown mathematically. Another relationship we’ll use is the strength of the shock, which is a ratio of pressures across the shock:

P2 2γ 2 = 1 + M1 − 1 P1 γ + 1 We can see that the higher the Mach number of the incoming ﬂow, the stronger the shock. At M1 = 1, the pressure ratio is zero, indicating no shock occurs.

We can utilize these diﬀerent relations to calculate other properties, such as density ratios, by using the ideal gas law.

9.4.1 Example: revisiting the shuttle Because of the high Mach number of the shuttle at re-entry, we should expect a shock to form. We see that a blunt object, such as shown in ﬁgure 9.5, will create a normal shock in front of it. Let’s calculate the change in air properties across this shock. From the previous problem, we assumed T1 = 100K, M1 = 15. Let’s also assume the pressure in this upper atmosphere is 100P a, and the adiabatic gas constant is γ = 1.4. What’s the mach number, temperature, and pressure after the shock (but before it stagnates on the surface)?

We can start by calculating the Mach number after the normal shock:

2 2 2 2 M1 + 15 + M 2 = γ−1 = 1.4−1 2 2γ 2 2∗1.4 2 γ−1 M1 − 1 1.4−1 ∗ 15 − 1

M2 = 0.38 The Mach number after the shock is less than 1. This is much like the hydraulic jump, where the Froude number after a jump is less than 1. Then, using this Mach number, we can calculate the temperature of the ﬂow after the shock: T 1 + γ−1 M 2 1 + 1.4−1 ∗ 152 2 = 2 1 → T = 100K ∗ 2 T γ−1 2 2 1.4−1 2 1 1 + 2 M2 1 + 2 ∗ 0.38

T2 ≈ 4500K

Intro to Fluids Notes 123 C. P. Byers 2016 CHAPTER 9. SUPERCRITICAL AND SUPERSONIC FLOWS

This is the same result we had found in the stagnation case! Think about why - the Mach number of the ﬂow after the shock is 0.38, which isn’t all too fast. The temperature ratio compared to stagnation temperature due to that mach number is T0/T = 1.02, indicating that it will only heat up by 2% from post-shock to stagnation. Finally, we can calculate the pressure after the shock: P2 2γ 2 2 ∗ 1.4 2 = 1 + M1 − 1 → P2 = 100P a ∗ 1 + 15 − 1 P1 γ + 1 1.4 + 1

P2 = 26kP a

The pressure rises an absurd amount! This is well over a 100 fold increase. Imagine then if you were to try and have a hypersonic vehicle near sea level, where Patm = 100kP a. Even moving at M = 5 gives a pressure after a normal shock of P ≈ 2.9MP a, which would result in a crazy amount of drag. This is why supersonic vehicles are slender and ﬂy in the upper atmosphere.

Hopefully this gives you an idea of some of the things that need to be taken into consideration when dealing with compressible ﬂow!

Intro to Fluids Notes 124 C. P. Byers 2016