Hydromechanics PDF

GABRIEL D WEYMOUTH HYDROMECHANICS 2 Preface Hydromechanics: This document introduces the fundamen- tals of maritime fluid dynamic analysis: from kinematic and dynamic equations, to applications in lift drag and waves. Each section has a summary at the end covering the key con- cepts, and example questions for further study. An appendix Please don’t print these notes! covers some mathematical background. The document is meant to be used as an intermediate refer- This document has been formatted to be read elec- ence, as well as in coursework and exam preparation. I hope tronically on a tablet or laptop so that you won’t you find it useful, have to print it. - Prof G. D. Weymouth Contents Introduction 5 Physical Equations of Motion 21 Incompressible, Irrotational Flow 45 Lifting Surfaces 67 Boundary Layers & Drag 87 4 Water waves 111 Mathematical Appendix 129 Bibliography 138 Introduction The Great Wave Off Kanagawa, Woodblock print by Hokusai. While a classic work of art, these fishermen are just hoping that their boat designers understood hydromechanics. 6 Why require hydromechanics? An understanding of fluid motion is prerequisite for the study of maritime engineering. Fluid forces and motions are critical design considerations for systems operating in the marine environment. Many of these forces will be familiar to you: Hydrostatic pressure on an object floating in water Fluid resistance against moving at a constant speed Figure 1: The container ship MOL Comfort moments Time-varying forces due to waves and maneuvers 1 before it split in two in 2013 due to fluid forces. Maritime systems are not fastened down, so these forces often 1 I will use American English spelling throughout. result in large scale motions such as the pitch and heave of I will work on a Dutch version of these notes in the a ship in waves and the strum of an underwater cable in a future. current. Without designers who understand fluid dynamics, these ships would fail (Figure 1) and the cables snap. Fluid forces and motions are critical marine design considerations. 7 What is a fluid? Figure 2: A mudslide in Washington State, USA, March 2014. Soil more than 200m thick flowed as a fluid, killing 43 people. Liquids and gasses are not the only fluids. Other states of matter such as plasma, amorphous solids, and super-fluids also exist. And under certain conditions, everything from snow, to rocks (as in Figure 2), to cars in traffic can be treated as fluids. A fluid, then, is not defined by a set of materials, A fluid, then, is not defined by a set of materials, but by a set but by a set of behaviors. A fluid flows. of behaviors. A fluid flows. 8 Figure 3 illustrates this concept for an elemental volume of material. A solid experiences a finite deformation for a finite tangential force.2 The deformation slope is called the shear strain ϵ, while the force per unit area is called the shear stress τ. In contrast to a solid, a fluid cannot resist a shear stress without continuously deforming. In experiments we see that a undeformed material element d ϵ. 3 fluid responds with a finite rate of strain dt Continuous deformation under shear is the defining charac- τ teristic of fluids. ϵ deformed solid element τ ∂ ∂t ϵ deforming fluid element Figure 3: Illustration of different stress vs. strain relationships. 2 This is true until the force is increased to the point that plastic deformation and failure is reached, at which point even a solid will ‘flow’. 3 The operator d denotes the time derivative. dt 9 Fluid forces How does a fluid react to force? Elements of fluid obey Newton’s second law, the same as rigid-solids like a pendulum. d dt (mu) = ∑F (1) where mu is the fluid momentum, the mass times velocity, and F are all the forces acting on the fluid. Applying this to an elemental fluid volume, such as sketched in Figure 3, we have d dt (ρu dV ) = ∑ dF (2) where dV is the elemental volume, ρ is the fluid density, and dF is the elemental forcing.4 4 Table 1 gives a few fluid density values. What forces act on a fluid element? In general, there are many types of forces that may be applied to a fluid element from surface tension to elasticity. However, there are three main forces which are fundamental to maritime engineering: gravity ϕ friction pressure mg Let’s take a first look at a few relevant examples which we will revisit in detail later. Figure 4: An ideal pendulum. 10 Water Waves and Gravity y η = a sin(kx − ωt) Before physicists stole the term 5 ocean waves were some- times called gravity waves by maritime engineers because they x are governed by a balance of gravitational potential energy and 2π k= λ kinetic energy, Figure 5. 2π In fact, water waves are strongly analogous to the simple ω= T pendulum of Figure 4. The weight of the wave crests pulls h them down and their inertia carries them past the undisturbed water height down to the trough, like a pendulum overshoot- ing the resting position beneath the pivot. Just as with a pen- dulum, we will find that the period of oscillation does not Figure 5: Sinusoidal waves with a flat seabed. We depend on the mass, only on gravity g and a length scale - in will define these terms when we revisit waves in the this case the wavelength λ. final chapter. Just as in solid mechanics, the force on a small volume dV of 5 Not to say that waves in the gravitational field the water due to gravity can be expressed as aren’t cool, they certainly are. dFg = ρg dV (3) Adding up all of these little contributions gives the total weight of the fluid. 11 Pipe Flow and Friction Perhaps the simplest flow fundamental to maritime engi- neering is that found in pipes. After an initial development region (Figure 6) the flow remains identical at every section (x-value) along the pipe. The friction between the pipe and the fluid slows down (and even stops) the elements of fluid very close to the wall. There is also friction between these slow moving elements of fluid and the fast ones closer to the center of the pipe, resulting in the smooth velocity profile shown in Figure 6. The force on a small piece of the surface area dA of the pipe due to friction stress τ (recall Fig 3) can be expressed as Figure 6: Flow near the inlet of a pipe. The arrows are the velocity vectors and the blue lines connecting dFτ = τ dA (4) their tips are the velocity profiles. and is in the direction tangent to the surface. Adding up all of these little contributions gives the total friction drag in the pipe. 12 fluid ρ (kg/m3 ) µ (kg/ms) air 1.204 18.13 × 10−6 Newtonian Fluid: A Newtonian fluid is one which does not fresh water 1000 1.00 × 10−3 elastically resist shear, and has a constant friction coefficient.6 sea water 1025 1.10 × 10−3 This can be expressed mathematically as Table 1: The density and viscosity of some common dϵ fluids at 20 0 C and atmospheric pressure. τ=µ (5) dt where µ is the viscosity of the fluid and recall ϵ is the strain 6 Newtonian fluids are so named because Isaac New- (Figure 3). A non-Newtonian fluid violates one of these con- ton was the first to derive the differential relation ditions, either by elastically resisting shear strain or having a describing them. viscosity which depends on the strain.7 7 A simple corn-starch slurry is an elastic fluid that can be made easily at home. It is sometimes called Water and air are nearly perfect Newtonian fluids. oobleck after the Dr. Seuss book. 13 Lifting surfaces and Pressure 36 MarineRuddersand ControlSurfaces 3.2.18 Cavitation Devices immersed in a liquid may have their performance degraded by the formation of voids or cavities in the liquid. For example, this may occur when the suction (negative pressure) becomes high on a lifting device. Cavitation occurs when the local fluid pressure drops to the vapour pressure of the liquid, that is the pressure at which the liquid vapourises. Vapour pressure depends on temperature and the quality and content of the liquid. Cavitation can occur on marine rudders, control surfaces such as stabiliser fins, support struts and propellers. This is discussed further in Chapter 5. L 3.2.19 Ventilation D Ventilation, or aeration, is a generally similar phenomenon to cavitation except that T the void is filled with air drawn down from the free surface. The pressure in the void is therefore usually atmospheric, although the void may close after formation and the pressure then drops below atmospheric as the volume of the void increases. Ventilation typically may occur on surface piercing rudders, such as transom-hung R rudders on yachts, and control surfaces on hydrofoil craft. Surface-piercing ruddersU Devices such as a rudder are called lifting surfaces because are discussed further in Chapter 5. they use lift forces to control a ship’s motion through the water (Figure 7). 3.3 Properties Figure 7: Topofview liftingsketch foils of the forces acting on a The pressure p on the surface is the main contributor to the ship in afluid A common turning device is manoeuvre. The lift a foil designed to produce a lifting force L force applied acting across the direction of the incoming flow. The lifting action of the foil arises from the lift force (Figure 8). In contrast to hydrostatic pressure, this by the rudder pushes the ship into the turn. difference in the average pressure of the fluid over the upper and lower surfaces of the lifting foil, Figure 3.19. Typical applications of such lifting foils are aircraft pressure is generated dynamically. As the body moves through wings, hydrofoils, rudders, stabiliser fins, propeller blades and sails. a fluid it induces the fluid to accelerate around it, which in / ~ - "~"',, Decreasein pressure turn induces a fluid reaction force on its surface. Netforce~ / ~ ~ ~ relativeto ambient The force on a small piece of the surface area dA due to ,'\ \ \',, pressure can be expressed as ,,i II \ \ \",, \~ dFp = − p dA (6) z / II I and is in the direction normal to the area. Adding up all of these little contributions gives the total lift on the surface. relativeto ambient Figure 3.19 Pressures around a lifting foil Figure 8: Pressure stress on a foil, from Molland and Turnock 2007. 14 Model testing The first scientific approach to predicting fluid forces was that of scale-model testing developed by William Froude and first documented in 1861. Before that time, as described by Nobel Laureate Cyril Hinshelwood, Fluid mechanics [suffered] an unfortunate split – between the field of hydraulics, observing phenomena which could not be explained, and theoretical fluid mechanics explaining phenomena which could not be observed. What made these experiments different than any before Figure 9: Two of the hulls originally tested by was that the nondimensional groups governing the test were Froude, the Swan (top) and Raven (bottom), from matched to full-scale. Not just the ratios of the ship length to the Science Museum, London. draft, or the entrance angle of the bow, but also the ratios of the fluid forces discussed above. The ratio of fluid forces should match in model- scale tests. 15 The Reynolds number8 characterizes the relative strength of 8 The parameter is named after the Irish engineer Os- d (ρuV ) and viscous force µ d ϵA the momentum rate dt bourne Reynolds who used it to quantify turbulent dt flow inside a pipe in 1883. ρU 2 L2 ρU L Re = = (7) µU L µ where U, L are the characteristic velocity and length-scales of the problem and we use the scaling relationships V ∼ L3 , A ∼ L2 , t ∼ L/U, and the strain ϵ is dimensionless. 9 9 Check the units to make sure these formulas are Typically maritime values might be a length of L ≈ 1 – 100m consistent. It’s an excellent habit, and will help you and speed of U ≈ 1 – 10m/s, and therefore Re ≈ 106 – 109. catch lots of mistakes! Unfortunately, it’s difficult to match the Reynolds number in scaled experiments since reducing the length of a scale-model requires increasing the speed proportionally. So a 1:100 scale- model would need to be towed at 100 times the service speed! 10 10 In air, you can greatly increase the density to viscosity ratio to reduce the velocity required, as in The Froude number characterizes the relative strength of the Princeton’s high pressure recirculating channel. momentum rate and gravity force ρgV s ρU 2 L2 U Fn = = p (8) ρgL3 gL Typical values for ships range from Fn ≈ 0.1 – 1, although most ships operate below Fn = 0.4.11 Unlike Reynolds number, 11 We’ll discuss why in the Waves chapter. matching Froude number is easily achievable since a model that is scaled-down by a length factor α2 requires reducing the model speed by a factor α. So now our 1:100 model requires testing at 1:10 speed, probably much less than 1 m/s.12 12 Matching both Re and Fn is impossible in water without modifying the gravitational constant g! 16 Force coefficients quantify the ratios of measured forces to the momentum rate. For example, the measured lift and drag forces L, D on the wing in Fig 8 can be quantified in terms of a lift coefficient L D CL = 1 , CD = 1 (9) 2A 2 2 ρU 2 ρU A where A is a characteristic area.13 If the wing is tested at the 13 We’ll discuss wing geometry and scaling in detail correct dimensionless conditions (Reynolds number, etc), then in the lifting surface chapter. the force coefficients measured on the scale-model will match the full-scale wing. This is the essence of scale-model tests. Froude’s hypothesis is required when testing a maritime model, since Re and Fn cannot be simultaneously matched. Therefore we must assume the drag can be decomposed as CD = CW (Fn) + CR (Re) (10) where CW is called the wave resistance and CR is called the residuary resistance. If Froude’s hypothesis were strictly true, the lines of con- stant Fn in Fig 10 would be parallel with spacing independent of Re. While the match is fairly good, there are clear deviations Figure 10: Total drag coefficient over a range of Re - especially at low speed. In practice, a single scale-model is (here denoted R) of a 58m vessel, along with 8 scale- models (around 1.2 - 9m), from Truust and Zakay tested and an empirical fit (such as Schoenherr’s line shown 1951. Note the gap in Re between the largest model in the figure) is used to estimate CR adding further predictive and the (midsize) ship is 17 × 106 versus 300 × 106. error. 17 Limits of model testing While the importance of model testing cannot be overstated, the process has limitations: 1. Expense of creating scale-models and running tests. 2. Unavoidable experimental measurement errors. 3. Difficulty (or impossibility) of matching Re. 14 Fully appended models are very expensive, and 4. Difficulty in matching geometric conditions.14 the scale effect are much more severe since the local Re on these structures is much smaller. 5. Difficulty (or impossibility) of matching environmental conditions.15 15 In the ocean, we have wind, waves, currents and a ship moving in (at least) 6 degrees of freedom! As such, analytic and numerical tools for the prediction of hydrodynamic loads are critical for the design, analysis, and operation of maritime vessels and structures. We begin the process of developing and understanding these tools in the following chapters. 18 Summary This section presents some of the fundamental concepts in hydromechanics. Fluid forcing and motion is an import design consideration in maritime engineering. A fluid flows, continuously deforming under shear stress. A fluid element’s change in momentum is due mostly to the contribution of: gravity, friction and pressure forces. A Newtonian fluid linearly resists the rate of shear, with viscosity coefficient µ. Force coefficients measured at one scale can be applied to any other scale if the dimensionless groups are matched. The Reynolds number Re = ρUL/µ and Froude number p Fn = U/ gL are the primary dimensionless conditions in hydromechanics. 19 Example questions Each chapter has a set of question to help you evaluate your understanding of the material. In some cases, these questions are more involved than a typical exam question. 1. A spherical bouy of volume 0.5 m3 and density half that of sea water is submerged on a cable 50 m long. Considering only hydrostatics, what is the tension T on the cable? 2. For developed laminar flow through a pipe with radius R (see Figure 6) the velocity in the pipe is given by ∆P 2 u x (r ) = B ( R − r2 ) µ where ∆P is the pressure drop along the pipe. What are the units of B? 3. Design an experiment where you can match the Re and Fn for two different sized models. Think outside the box, but back up your ideas with references and calculations. 4. List a few full-scale maritime systems with more than 6 Figure 11: Mooring buoy degrees of freedom. What is the simplest such system you can think of? 20 5. Consider a ship (or floating platform) with no forward motion but oscillating in roll as θ0 sin(2πt/T ). You are tasked with measuring the drag force generated by the bilge keels on a scale-model and predicting the full scale forces. (a) There are now two characteristic lengths L, the keel size h and the radius from the bilge to the center of rotation r, but no constant velocity U. What parameter must you use in Re, Fn, and CD instead? (b) What additional dimensionless conditions do you need Figure 12: Floating Production and Storage Offshore to match in the experiments? (FPSO) platform 1:40 scale-model tested at MARIN. (c) Figure 13 shows data from plates in an oscillating flow. Use these to estimate the bilge drag force per unit length of keel on a platform with h, r, θ0 , T = 0.35m, 30m, 0.25rad, 12s. Figure 13: Drag data for bilge keels versus the Keulegan-Carpenter number KC = UT/h. Physical Equations of Motion Dye visualization of the curved streamlines near the bow of a model-scale ship carried out at HSVA for Wartsilla. The path of the flow visualized here is governed by Newton’s second law applied to the curving, spinning, and deforming material elements. It all comes naturally to the fluid, but describing the pro- cess mathematically requires extensive use of vector calculus, which can be a bit less natural for students learning it for the first time. 22 Fluid kinematics Figure 14: Numerical solution of the flow around a thin lifting surface, such as a sail. This is an interac- tive application. Try it out! How can an element of fluid change over time? We need to quantify (and hopefully define limits on) the evolution of fluid elements. The description of fluid motion is called fluid kinematics. 23 streamline n̂ Position: Figure 14 visualizes the flow by tracing fluid ele- ŝ ments over time. Consider one of those fluid elements mov- ing along a general curved path, as in Figure 15. Define the r location of that element with the vector ⃗x.16 Depending on ⃗x y the situation we might express this in Cartesian coordinates θ ⃗x = ( x, y), in polar coordinates (r, θ ), or even in coordinates aligned with the motion of the element (s, n), Figure 15. x Figure 15: A vector may be described by its extent The velocity ⃗u is simply the rate of change of the position in each direction in a chosen coordinate system. In " # " # stream-aligned coordinates s is aligned with the d ⃗x (t + ∆t) − ⃗x (t) ẋ u ⃗x = lim = = x = ⃗u direction of the local velocity, and n is normal to it. dt ∆t→0 ∆t ẏ uy 16 You are already familiar with vectors and coordi- where the subscripts u x and uy indicate components of the nate systems, and the mathematical appendix has an overview. I highly recommend 3Blue1Brown’s video vector in the x and y directions respectively. I will often use the on vectors, if you need a bit of a review. shorthand u x , uy , uz = u, v, w since we use Cartesian compo- nents of velocity so often. Streamlines can be calculated from a given velocity field via integration; reversing the process above.17 From the def- 17 This is how the visualization in Figure 14 works. inition of velocity, the translation of a particle over a time ∆t is given by ∆⃗x = ⃗u∆t. Repeating this process and connecting these traces together results in a stream line which is therefore always tangent to the vector field. A streamline is always tangent to the local veloc- ity vectors. 24 A fluid element can do more than just translate, however. Figure 16 illustrates the three critical kinematic transformations to a fluid element: translation, rotation, and compression. dx dt ⃗ = ⃗u The rate of rotation is known as the vorticity ω ⃗. As in rigid-body dynamics, this vector points in the direction of the axis of rotation with the sign determined by the right- hand-rule. In figure 16 this means the vector points "out of" the Translation rate image. An element rotates when the velocity vectors around the element are unbalanced. ⃗ × ⃗u = ω ∇ ⃗ The rate of expansion/compression quantifies the change in fluid volume. Note that volume doesn’t have a di- rection so it is a scalar value, not a vector! An element compresses when the velocity vector into and Rotation rate out of the element are unbalanced. ⃗ · ⃗u ∇ Expansion rate Figure 16: Fundamental kinematic transformations on a fluid element. 25 Vector calculus We will need some new mathematical tools to quantify the vorticity and compression rate of a flow. In particular, both of these metrics depend the spacial derivatives of the velocity. For example, the compression depends on the change in x-velocity across an element in the x-direction ∂u u( x + ∆x, y) − u( x, y) = lim ∂x ∆x →0 ∆x and the vorticity depends on the change in x-velocity in the y-direction18 18 The shear-rate d ϵ also depends on a "cross" dt derivative. Generalizing Figure 3 to fluid elements ∂u u( x, y + ∆y) − u( x, y) moving on a general streamline, Figure 15, we have = lim ∂y ∆y→0 ∆y d ∂us τ=µ ϵ=µ (11) The partial derivative symbol ∂ is used instead of d to indicate dt ∂n that we are only taking the derivative with respect to one of for the stress of a Newtonian fluid. multiple possible variables. 26 The gradient operator is a shorthand for gathering all the spacial derivatives into a vector. In 2D Cartesian coordinates, the gradient is simply: " # ∂ ϕ ⃗ϕ= ∂ ∇ ∂x ∂y ϕ where ϕ can be any variable, such as the pressure p, and ∇ is the del operator. Physically, the gradient vector points in the direction of steepest ascent, Figure 17. Figure 17: Example field and it’s negative gradient using Python: # Create a grid of locations Example 1: Consider the field import numpy as np 1 x,y = np.meshgrid(np.linspace(-6,6,31), q ϕ = log( x2 + y2 ) = log( x2 + y2 ) np.linspace(-4,4,21)) 2 where we use the exponent property of logarithms to pull out # Create pressure field and gradient the square-root. It’s gradient is determined by taking deriva- p = (x**2+y-11)**2+(x+y**2-7)**2 tives with respect to x and y. Recalling that d( log u) = du u we p[p>160] = 160 # limit the values have dpdy,dpdx = np.gradient(p) # gradient x   ⃗ϕ= 2 y2 x + ⃗x ∇ y  = 2 # Plot it up x + y2 x + y2 2 import matplotlib.pyplot as plt plt.quiver(x,y,-dpdx,-dpdy,alpha=0.5) This vector field is plotted in Figure 32, and we’ll return to it plt.contour(x,y,p); plt.axis('equal') later. plt.xlabel('x'); plt.ylabel('y',rotation=0) 27 The divergence operator is the short-hand for the com- bination of partial derivatives needed to measure the rate of compression of a vector field, see Figure 16. In 2D Cartesian coordinates, it is ∂q ⃗ · ⃗q = ∂ , ∂ · (q x , qy ) = ∂q x + y ∇ ∂x ∂y ∂x ∂y which is the inner product of the gradient operator with the Example 2: Calculate the divergence and curl of components of a vector field. And as needed, it returns a scalar ⃗q = ( x, −y). field instead of a vector. which is called a "corner flow", and is plotted above. The divergence quantifies the net "flux" into or out of a ⃗ · ⃗q = 1 − 1 = 0, and so we The divergence is ∇ point of a vector field. call this field "divergence free". The curl is also zero: ⃗ × ⃗q = (0 − 0)ẑ. ∇ The curl operator is the shorthand for the combination of Example 3: Calculate the divergence and curl of partial derivatives needed to measure the rotation of a vector field, again see Figure 16. In 2D Cartesian coordinates, we have ⃗q = ⃗x /( x2 + y2 ) ∂qy ∂q x from our gradient example. ⃗ × ⃗q = ∇ − ẑ ∂x ∂y Using the quotient rule for derivatives d(u/v) = (v du −u dv)/v2 , the divergence is which is the cross-product of the gradient and vector com- ponents. As needed, it points into (or out of) the page for 2D 2 2 2 2 ⃗ · ⃗q = y − x + x − y = 0 ∇ flows. r 4 r 4 r4 The curl quantifies the rotation around a point of a vector which is divergence free everywhere except the field. origin, where it is singular. The curl is similar ⃗ × ⃗q = −2xy − −2xy ẑ = 0 ẑ ∇ r4 r4 r4 28 The Laplacian is the divergence of the gradient of a field. Again, in Cartesian coordinates we have ∂2 ϕ ∂2 ϕ ∂ ∂ ∂ϕ ∂ϕ ∇2 ϕ ⃗ ·∇ =∇ ⃗ϕ= , · , = 2+ 2 ∂x ∂y ∂x ∂y ∂x ∂y We will need this operator at the end of this chapter and it’ll be Example 4: Calculate the Laplacian of the star of the following chapter. q ϕ = log( x2 + y2 ) The Laplacian measures the total curvature of a field. We’ve already done this! We computed it’s gra- Coordinate systems: The examples above use 2D Cartesian dient in example 1, and then took it’s divergence in example 3. So ∇2 ϕ = 0 except at the singular point coordinates to illustrate the vector calculus operators. However, at the origin. just like vector’s themselves, the gradient curl divergence and Laplacian can be expressed in any coordinate system. Please the gradient curl divergence and Laplacian can be check the mathematical appendix for the 3D and cylindrical expressed in any coordinate system coordinate version of these operators. Depending on the appli- cation, switching to another coordinate system may simplify the analysis significantly. 29 Kinematic assumptions Now that we’ve clearly defined how to measure kinematic fluid properties like velocity, rotation, and compression, we can consider some constraints on these motions. Incompressible Flow: A mistaken statement often repeated by overzealous science teachers is that water is incompressible. Sound is a compression wave, so if that were actually true the ocean would be completely silent! 19 19 To the contrary, the ocean is full of amazing The speed of sound in a material gives an indication of how sounds. compressible that material is, and the Mach number Ma is the ratio of the fluid velocity ⃗u to the speed of sound. As long as Ma < 0.3, compression can be ignored in the kinematic description. As the speed of sound in water is 1.5 km/s we can safely assume incompressible flow20 20 The only common maritime violation of this as- sumption is cavitation - which we try to avoid! ⃗ · ⃗u = 0 ∇ (12) Again, this is an assumption about the flow, not the fluid. 30 Irrorational Flow: A surprising feature of flows around a body is that they are mostly irrotational, ie ⃗ × ⃗u = 0 ⃗ =∇ ω in large portions of the fluid domain. Moving a solid body through a fluid must generate some rotating flow, as illustrated in Figure 18. However, the low viscosity of air and water cause this region to be highly compressed into a thin boundary layer on the body surface and into vortices in the wake behind the body. Outside of these two regions, the flow is essentially irro- tational. Figure 18: The smoke from a fire on a wind turbine We will see that an irrotational flow enables a simplified in Palladam Indian July 2016 visualizes the regions mathematical description, which makes this an appealing as- of rotating flow: primarily in the wake of the turbine sumption. But we will also see that this simplification can lead housing and the blade tip helical vortices. Every- to overwhelming prediction errors in some cases.21 As such, where else is effectively irrotational. we will avoid this assumption for now, and revisit it in the following chapter. 21 This led to the schism mentioned in the quote in the previous chapter. 31 Conservation Laws We are now ready to write down a set of conservation laws for fluid mechanics. There are two complementary forms for these laws 1. Differential forms 2. Control volume forms The differential forms were developed first22 and come 22 Leonard Euler presented his fluid equations in directly from considering an element of fluid and applying the 1752, building on the Bernoulli family’s earlier work. vector calculus operators we’ve developed above. The control volume form came much later (the mid 1900s) and integrates the differential equations over some chosen vol- ume, often using integral identities to simplify things down.23 23 Consult these fluid mechanics textbooks for a full There are an infinite number of possible control volumes to treatment of control volumes: Frank M White. Fluid mechanics. McGraw-Hill, choose, but a clever choice can simplify the analysis of certain Boston, 1999; and John Anderson. Fundamentals of problems. Aerodynamics. McGraw hill, 2011 32 n̂ Stokes’ Theorem is very helpful in developing and simplify- ing differential and control-volume equations. 24 Physically, Stokes says that the sum of small changes hap- un pening inside a control volume can be measured by what’s happening on the boundary. There are many applications of this theorem in fluids, but ŝ the two more important are below and illustrated in Figure 19: us The flux Q is the total volume-rate that enters or leaves a control volume, and Stokes relates that to the total divergence I Z Q= ⃗u · n̂ d A = ⃗ · ⃗u dV ∇ Figure 19: If a flow is otherwise incompressible, imbalanced points of divergent flow cause a net H where d A is the integral over the whole boundary and n̂ is the normal unit vector to the boundary. flux out of the tank. Likewise, if a flow is otherwise irrotational, the sum of the vorticity induced by the stirring rods is the total circulation. The circulation Γ is the total vorticity within a (2D) control volume, and equals the velocity around the bounding curve 24 Stokes theorem generalizes the fact that integrals I Z ⃗ × ⃗u d A are anti-derivatives: Γ= ⃗u · ŝ ds = ∇ Z a df dx = f ( a) − f (b) b dx See the mathematical appendix. 33 Conservation of volume ⃗u n̂ If the flow is incompressible, then the volume of each ele- ment is constant. ∂Ω Ω They can’t compress or expand! We’ve already developed the differential form for the conservation of fluid volume, equa- tion 12. For the control volume version, consider the incompressible flow between two streamlines and draw a control volume Ω Figure 20: Control volume analysis of incompressible with the two lines as part of the boundary ∂Ω. Then Stokes flow through a contracting channel. Since the flux gives the flux as through the walls are zero, the the velocity must increase as the channel contracts. I Z Q= ⃗u · n̂ dA = ⃗ · ⃗u dV = 0 ∇ ∂Ω Ω since the flow is incompressible. As there is no flux through a wall or other streamline, we have the common engineering mantra: Flux in equals flux out This gives a simple way to determine average flow rates through confined regions like pipes and channels. 34 Conservation of mass All marine hydrodynamic systems conserve mass.25 If the 25 Violating this "assumption" would require rela- flow is incompressible the volume V of an element is constant, tivistic flow speeds! so we have d d d (m) = (ρV ) = V ρ = 0 dt dt dt Since the element volume is nonzero, the rate of change of density of an element is zero Dρ =0 (13) Dt where we use the capital D to remind us that this derivative is with respect to a material element, and call it the material derivative, after Bachelor.26 Figure 21: An incompressible and mass conserving ∂ρ Note that equation 13 does not imply that the density is a system. But ∂t ̸= 0!! 26 George Keith Batchelor. An introduction to fluid constant across the whole fluid, or even that it is constant at mechanics. Cambridge University Press, 1967 any one point in space! A water wave is incompressible and conserves mass, but the density at any point near the air/water interface fluctuates as the waves pass by (or crash down). 35 Since the element position varies in time, the density varies like ρ(t, x (t), y(t)) and we need to use the chain rule27 to ex- 27 See the mathematical appendix. pand the derivative in equation 13 D ∂ρ dx ∂ρ dy ∂ρ ∂ρ di⃗x ⃗ ρ= + + = + · ∇ρ Dt ∂t dt ∂x dt ∂y ∂t dt where we used the shorthand for gradient and dot product to write this in a coordinate-free form. Substituting the definition of the velocity, we have D ∂ρ ⃗ρ ρ= + ⃗u · ∇ (14) Dt ∂t and the differential form of mass conservation is ∂ρ ⃗ρ=0 + ⃗u · ∇ (15) Figure 22: A simulation of a wave crashing into a ∂t block that I made using equations 12, 15 and 17. The first term is the unsteady change in the density, and the second is the change due to convection. This equation governs how fluid density changes as the flow evolves over time, and is ∂ρ required to simulate flows like Figure 22. Any local rate of change in density ∂t is caused Integrating equation 15 and using the divergence theorem by a convective flux of density to a new location. gives the control volume version Z I ∂ρ dV + ρun dA = 0 (16) Ω ∂t ∂Ω where un = ⃗u · n̂. Again, the first term is the unsteady change in the density and the second ρun dA is the mass flux through the boundary. 36 Momentum balance ⃗f c · ŝ = µ ∂ us ∂n n̂ ⃗f c We have already discussed all of the elements needed for the momentum balance given by equation 2: ŝ ⃗f c · n̂=− p d D⃗u (ρ⃗u dV ) = ρ dV = ρ⃗g dV − pn̂ dA +⃗τ dA dt Dt where we’ve used mass conservation to isolate the velocity in the left-hand-side derivative, and substituted the gravity, S pressure, and viscous forces (equations 3-6) on the right. Figure 23: The contact force per unit area is Applying the material derivative expansion in equation 14 to ⃗u instead of ρ and integrating gives the control volume form of ⃗f c = − pn̂ + µ ∂us ŝ ∂n the momentum balance where p acts in surface normal direction n̂, and ∂⃗u Z I Z I I ρ dV + ρ⃗uun dA = ρ⃗g dV − ⃗pn̂ dA + ⃗τ dA the Newtonian fluid shear, equation 11, acts in the Ω ∂t ∂Ω Ω ∂Ω ∂Ω stream-aligned direction ŝ, Figure 15. The last two terms are the integrated pressure and viscous forces on the boundaries - exactly the forces we want to know in engineering applications. The left-hand side is also interest- ing; the momentum flux is the the velocity convection of the velocity itself.28 28 This is a nonlinear process and is the main con- tributor to the complexities of turbulence! 37 The Navier-Stokes equation The Navier-Stokes equation is the differential equation gov- erning the evolution of fluid momentum. 29 Using Stokes to The Navier-Stokes equation is the differential convert the pressure and viscous terms into volumetric form equation governing the evolution of fluid momen- we have tum. ∂⃗u ⃗ p + µ∇2⃗u ⃗ u = ρ⃗g − ∇ 29 The equation is named after Claude-Louise Navier, ρ + ρ⃗u · ∇⃗ (17) ∂t and Sir George Stokes. Navier was a French engineer where we’ve substituted the stress equation for Newtonian flu- who proposed a viscous extension of Euler’s equa- tions in 1822. Stokes was an English academic who ids, equation 11, and used the shorthand for the Laplacian.30 derived this viscous equation from first principles in Equation 17 can be written in 2D Cartesian coordinates as31 1845. ! 30 Note that the viscous term smooths out any sharp ∂u ∂u ∂u 1 ∂p ∂2 u ∂2 u changes in the velocity field, as discussed way back +u +v = gx − +ν + 2 ∂t ∂x ∂y ρ ∂x ∂x2 ∂y in Fig 6. 31 Imagine how they look in 3D spherical coordi- ! ∂v ∂v ∂v 1 ∂p ∂2 v ∂2 v nates! Vector notation has some advantages! +u +v = gy − +ν + 2 ∂t ∂x ∂y ρ ∂y ∂x2 ∂y where we’ve scaled by ρ and use ν = µ/ρ, the kinematic m2 viscosity.32 32 Typical values are ν w ≈ 10−6 s in water and m2 νa ≈ 10−5 s in air. 38 Because air and water are near perfect Newtonian fluids,33 33 And all maritime flows conserve mass. equation 17 accurately models the vast majority of marine and maritime flows. You may be asking yourself: If this equation is such a good model of fluid behavior, why not just solve it and end the book here? Why not indeed? First, the pressure is still unknown. For an incompressible flow, we add ∇ ⃗ · ⃗u = 0, but this is not a function of p. Therefore this condition and the Navier-Stoke equation must be solved simultaneously for the velocity and pressure. 34 34 In practice, this is the slowest part of finding nu- Second, the equation is nonlinear in the velocity ⃗u due to the merical solutions to the Navier-Stokes equation. convective term. Nonlinear PDEs are almost never analytically solvable, and in this case nonlinear convection leads to the 35 The situation is actually worse than that. Because seemingly unpredictable patterns of turbulent flow!35 of nonlinearity we can’t even prove that solutions Therefore, despite good agreement with experiments, the to the Navier-Stokes equation are physically reason- able. Perhaps solutions can blow up to infinity!?! full Navier-Stokes equation is difficult to use for predictions. Proving or disproving this is one of the Clay Institute However, we can simplify these equations in certain situations, Millennium Prizes. both numerically 36 and analytically! The analytic solutions are especially important since they can be readily understood. 36 Much progress towards approximate solutions has been made in recent years using Computational And maritime engineers who understand fluid fundamen- Fluid Dynamic (CFD). Consult references such as: tals are more valuable than ones who don’t. Joel H Ferziger and Milovan Perić. Computational methods for fluid dynamics, volume 3. Springer Berlin, 1996 39 As a geometric simplification, consider steady 2D flow through an extremely long channel narrow channel with per- fectly parallel walls, a 2D version of Figure 6. Defining the x-axis aligned with the walls and the walls at y = 0, h, we assume that there is no variation in the velocity along the length of the channel after the inlet transient. We express this mathematically as: ∂⃗u =0 ∂x This is a surprisingly restrictive assumption because the divergence free constraint simplifies to ∂v ∂y = 0 and so u is a constant which must be zero to avoid flow through the walls. This means ⃗u(t, x, y) simplifies to u(y) with all other contribu- tions zero. Figure 24: Velocity profile for a channel flow with variable wall velocities and pressure gradient. Substituting this into the x-component of the Navier-Stokes gives def channel(U_top,U_bottom,gradp): y = np.linspace(0,1,21) # 1D grid dp d2 u 0+0+0 = 0− +0+µ 2 zero = np.zeros_like(y) # x & u_y dx dy where every term other than the pressure derivative along # get profile array and plot it the pipe and the viscous stress across the pipe are zero! The u_x = gradp*y*(y-1)+U_top*y+U_bottom*(1-y) plt.quiver(zero,y,u_x,zero,scale=3) mighty Navier-Stokes has been demoted to a linear Ordinary plt.plot(u_x,y) Differential Equation (ODE) which can be solved by integrating plt.ylabel('$y/h$',rotation=0,size=14) twice and using the boundary conditions u(0) = Ubottom and plt.xlabel('$u_x$',size=14) u(h) = Utop plt.yticks([0,1]); plt.xlim(-1,2) channel(1,0,-3) h2 dp y y y y u= − 1 + Utop + Ubottom 1 − 2µ dx h h h h 40 Figure 24 shows this velocity profile. Try this code and play ds with the velocity boundary conditions and pressure forcing dz ds to get different flows - all of which are valid solutions to the Navier-Stokes equation. ŝ ẑ As a physical simplification, consider steady flow of an in- Figure 25: Since ⃗a · ⃗b = | a||b| cos θ, where θ is the viscid (µ = 0) fluid! In this case the stream-aligned component angle between the vectors, the inner-product of the of the momentum equation is37 two unit vectors is simply cos θ. Clearly, dz ds is the ∂|u| ∂p same as "adjacent over hypotenuse", and so equals ρ|u| = −ρgẑ · ŝ − cosθ as well. ∂s ∂s 37 Steady means ∂ = 0, stream-alinged means ∂t where we’ve assumed the direction of gravity is −ẑ. ⃗u = |u|ŝ + 0n̂ so no cross-stream convection, and Since ẑ · ŝ = dz , Figure 25, all three terms are defined by a inviscid means no viscous term. ds stream-wise derivative. Integrating along the streamline gives the steady Bernoulli equation 1 ρ|u|2 + p + ρgz = C (18) 2 which states that the sum of the fluid energy along an inviscid streamline is constant! the sum of the fluid energy along an inviscid streamline is constant! 41 Summary The fundamental fluid transformations include: 1. the rate of translation, the velocity ⃗u, 2. ⃗ × ⃗u = ω the rate of rotation, the vorticity ∇ ⃗ , and 3. ⃗ · ⃗u. the rate of compression ∇ Flow at low Mach number is essentially incompressible. Flow outside the boundary layer and wake is irrotational. Stokes theorem says the total changes within a control- volume can be measured on it’s boundaries. Conservation of volume, mass, and momentum equations govern fluid evolution and can be expressed in differential or control volume form. D determines the rate of change The material derivative Dt of a material element. Equation 14 expands this into it’s unsteady and convective terms. The Navier-Stokes equation (17) is Newton’s second law applied to viscous Newtonian fluids. 42 Example questions A prototype wind tunnel with a 50mm x 100mm test- ing cross section and a top flow speed of 40m/s. A wing sail is in the test section with span s ≈ 100mm and chord c ≈ 60mm. 0 mm 10 50mm 4% camber m m V=5m/s V=40m/s 00 R 60mm 1 1. (a) What is the flux Q through the wind tunnel and what size is the circular fan driving the tunnel flow if the flow speed directly after the fan is 5m/s. (b) Use the flow app in Figure 14 to roughly estimate the lift coefficient on the wing sail with zero angle of attack and maximum camber = 4% and use this to estimate the lift force measured in the wind tunnel. (c) Recklessly assuming that viscous forces in the tun- nel are negligible, use the Bernoulli equation to esti- mate the pressure pfan just after the fan if the pres- sure measured upstream of the foil in the test section is ptest = 102.10 kPa. 43 2. A fancier wind tunnel uses suction to minimize the bound- ary layer in a cylindrical test section. It achieves this via 1200 5-mm diameter holes per square meter of test section wall. (a) Compute the average suction flow rate when the suc- tion velocity into the holes is 8-m/s. (b) If the dimensions of the test section is 4-m long by 0.8- m diameter, what is the reduction in the average velocity from the inlet to the exit? State your assumptions. 3. Consider a velocity field ( given in cylindrical coordinates by Γr/R2 for r < R uθ = Γ/r else with ur = uz = 0, and where R is a fixed radius and Γ is the circulation of the flow with units of U L. (a) Sketch this velocity field and show that each flow region (r < R and r ≥ R) is incompressible. (b) Calculate the vorticity ω ⃗ × ⃗u of the velocity field ⃗ =∇ given above. Are both flow regions irrotational? 44 4. A 2D vortex in a square container of dimensions 2π by 2π is given by u = e−2νt sin( x ) cos(y) v = −e−2νt cos( x ) sin(y) where ν is the kinematic viscosity of the fluid.38 38 This is called the Taylor-Green vortex. (a) Use python to plot this vector field at time t = 0. Is this really the flow in a box? (b) Calculate the time-derivative ∂u ∂t and the viscous term 2 39 ν∇ u. Can you write down the solutions for ∂v and 39 Hint: Try to write this as a scaled version of the ∂t ν∇2 v without further calculation? original velocity field. (c) Is the field incompressible? Is it possible that this field satisfies the Navier-Stokes equations? 5. Consider a car undergoing sustained aquaplaning with speed U during hard breakage along a dock.40 Relative to 40 Now it’s a "maritime" example, right? the small gap distance h between the tires and the ground, the water flow may be considered 1-directional and uniform in the direction of motion. (a) This flow will obey a simplified version the Navier- Stokes equation. Which terms can be neglected and which will dominate? (b) Calculate the shear stress of the water on the tires assuming h = 1 mm and U = 15 m/s. Practically speaking, will this force slow down the car or is it going into the harbor? Incompressible, Irrotational Flow π/2 The Navier-Stokes equation is a dynamic law - based on forces and accelerations. However, this is not the only type of governing equation. Kinematic equations are determined by the geom- etry of the system constraints. This figure shows an R π example system of two interlocking gears. Given de- tails of the geometry and strong assumptions about the regularity of the system (both gears undergo R/2 rigid body rotation and the gears don’t slip or break) we can determine the relative displacement of every point in the system. An analogous process applies to develop kine- matic equations for fluid motions, avoiding the diffi- culties of the Navier-Stokes, but only under similarly strong assumptions... 46 Stream function ψ=c Figure 26: Streamline act as solid boundaries to a flow. The stream function ψ is an example of how kinematics can be used to simplify the velocity field description. Since the velocity is tangent to its streamline, the stream- lines set the direction of the flow. Moreover, an incompressible flow 41 slows down when streamlines diverge, and speeds up 41 This is the kinematic constraint. when they converge. This means the streamlines set the flow magnitude as well. Therefore, the streamlines determine the velocity vector uniquely in a 2D incompressible flow. 47 To make this quantitative, we assign a value to every streamline: the flux between the given streamline and some reference streamline.42 This defines the streamlines as the iso- 42 We set ψ = 0 on the body streamline by conven- contours of a continuous stream-function field ψ(⃗x ). tion. Stepping out a small distance d⃗n from a given streamline, Figure 27, the change in flux is dψ = ⃗u × d⃗n = u dy −v dx which means ∂ψ ∂ψ u= , v=− ∂y ∂x In polar coordinates this becomes, 1 ∂ψ ∂ψ ur = , uθ = − r ∂θ ∂r Applying the incompressible flow constraint results in a stream function description of the velocity field. Figure 27: The direction and magnitude of velocity is determined by the change in ψ. The stream-function description has a number of advan- tages: Contours of ψ are a natural way to visualize a flow since streamlines act like flow boundaries, Figure 26. Any flow defined by a stream function is guaranteed to be divergence free - no need to check! 43 43 Proof by substitution: We reduced the number of unknown fields: from u, v to ψ. ⃗ · ⃗u = ∂ ∂ψ + ∂ ∇ − ∂ψ =0 ∂x ∂y ∂y ∂x 48 It’s for these reasons that many analytic and numerical meth- ods for 2D flows are based on stream functions. Example 1: Show the stream function U ψ= xy L defines an incompressible velocity field. Taking the two partial derivatives gives " # U x ⃗u = L −y This is a scaled version of the incompressible "corner flow" velocity field we saw in the last chapter. Now we know it’s stream function. 49 Velocity potential ⃗x C2 C1 O One issue with the streamfunction is that it’s only defined in 2D. If we start by assuming the flow is irrotational we can Figure 28: The circulation going forward on C1 and instead define a function ϕ(⃗x ) using a path integral from the backward on C2 is ϕC1 − ϕC2 = Γ = 0 for any irrota- origin O to a point ⃗x tional flow. Therefore ϕC1 = ϕC2. Z ϕ(⃗x ) = ⃗u · d⃗x Example 2: A cylindrical version of the "corner" flow C where C is any path from O to ⃗x, as shown in Fig 28. Taking the U 2 ϕ= ( x + y2 − 2z2 ) gradient gives 2L ⃗ ϕ = ⃗u ∇ gives the irrotational velocity field recovering the velocity as the gradient of ϕ. Because of this U ⃗u = ( x, y, −2z) property the field ϕ is called the velocity potential, and de- L scribes any irrotational flow in 2D or 3D. 44 44 Again, proof by substitution ⃗ × ⃗u = 0 ⇔ ∇ ∇ ⃗ ϕ = ⃗u (19) ⃗ × ⃗u = ∂ ∂ϕ − ∂ ∂ϕ = 0 ∇ The velocity potential requires the risky irrotational flow ∂x ∂y ∂y ∂x assumption, but it defines the velocity field in any dimension! so any velocity defined by a potential is automati- cally irrotational. 50 Potential flow A flow which is both irrotational and incompressible is called a potential flow. In this case equations 19 and 12 are A flow which is both irrotational and incompress- combined to give45 ible is called a potential flow. 45 Set the divergence of the potential (from 19) to ∂2 ϕ ∂2 ϕ ⃗ ·∇ zero (from incompressible flow) and use ∇ ⃗ = ∇2. ∇2 ϕ = 2 + 2 =0 (20) ∂x ∂y Alternatively, in 2D, we can take the curl of the stream- function defined velocity ∂uy ∂u x ∂2 ψ ∂2 ψ ωz = − = − 2 − 2 = −∇2 ψ ∂x ∂y ∂x ∂y and so for 2D potential flow we also have ∇2 ψ = 0 (21) Once we solve for ϕ or ψ we know the velocity field. No need for the Navier-Stokes equation! No need for the Navier-Stokes equation! 51 Both of these are Laplace equations, and are extremely simple compare to the Navier-Stokes equation. 46 Specifically, it is 46 The Laplace equation and the Laplacian are named linear, the unknown ϕ is a scalar field, and there are no other for Pierre-Simon Laplace. Laplace made many major unknown coefficients or variables. As a bonus, there are many contributions to mathematics, from differential equa- tions to probability theory in late 1700s, but Equation specialized methods, such as separation of variables and super- 20 was first obtained by Euler in 1757. position, which can be used to solve this equation analytically. Example 3: Show that the 3D potential Example 4: Show that the stream function U 2 U ϕ= (r − 2z2 ) ψ= xy 2L 2L satisfies the potential flow equation. satisfies the potential flow equation. Using the cylindrical coordinate Laplacian defined in the Substitution gives appendix, we have ∂2 ψ ∂2 ψ U ∇2 ψ = + 2 = (0 + 0) = 0 1 ∂2 ϕ ∂2 ϕ 2 2L 1 ∂ ∂ϕ U ∂x ∂y ∇2 ϕ = r + + = (4 + 0 − 4) = 0 r ∂r ∂r r2 ∂θ 2 ∂z2 2L 52 Superposition Figure 29: Potential flow around a "half body" by superimposing a uniform flow and a source. Superposition is a simple approach for building potential flows. Because Equation 20 is linear, if we have a set of func- Superposition avoids "solving" the Laplace equa- tions {ϕ1 , ϕ2 ,...} that each satisfy tion at all: you only need to mix and match known characteristic solutions together. ∇2 ϕ1 = ∇2 ϕ2 =... = 0 then we may scale and add them together to give a new solu- tion ϕ = c1 ϕ1 + c2 ϕ2 +... which also satisfies Equation 20. 53 Visualizations are helpful when learning potential flows for the first time. Especially as you start trying superposition. I’ve put together two tools to help you get familiar with these equations: A superposition web-app that let’s you drag-and-drop flows together intuitively, Figure 29. A python function that visualizes streamlines and velocity vectors given a stream-function, Figure 30. Figure 30: Python function for visualizing stream- function flows: The first is fun and intuitive - start there. However, it’s quali- import numpy as np tative and you can only use the flows I’ve programmed for you import matplotlib.pyplot as plt (unless you hack the source). line = np.linspace(-4,4,21) The Python function is less intuitive, but it lets you plot any flow defined by a stream function. You can also augment and def curl_z(f): # velocity equation annotate the plot (like Figure 34) or customize the function dfdy,dfdx = np.gradient(f) return dfdy,-dfdx however you like. And once you have the function in Python, you can make any quantitative measurement you need. # Plot streamlines and velocity vectors def plot_flow(psi,mask=None,xl=line,yl=line): x,y = np.meshgrid(xl,yl) # make a 2D mesh if mask: # hide points where mask(x,y)=True x,y = (np.ma.masked_where(mask(x,y),x), np.ma.masked_where(mask(x,y),y)) plt.figure(figsize=(6,6),dpi=100) # make the plot plt.contour(x,y,psi(x,y),25) # streamlines plt.quiver(x,y,*curl_z(psi(x,y))) # velocity plt.axis('equal') # Test with corner flow plot_flow(psi = lambda x,y: x*y) 54 Characteristic solutions Laplace equation solutions are called harmonic functions, and there are LOTS of known solutions. Here are the three most fundamental 2D solutions: Uniform flow is defined as ϕ = Ux, ψ = Uy which, applying Equation 19, leads to the simple velocity field u x = U, uy = 0. Figure 31: Uniform potential flow showing the As the name suggests, the velocity field is everywhere uniform, streamlines (grey) and velocity vectors (blue). with magnitude U. This is often used as a background flow on which other solutions are added, as we shall see. 55 A source or sink in potential flow is defined as Q Qθ ϕ= ln r, ψ= 2π 2π where r2 = x2 + y2 is the radial distance from the origin and θ = arctan y/x. This gives the velocity field in polar coordi- nates as Q ur = , uθ = 0. 2πr The velocity is now only acting in the radial direction, with no azimuth component. The coefficient Q is the flux out of the origin (or flux into the origin if Q < 0). Note that the flow Figure 32: Source flow streamlines and vectors. has infinite velocity at the origin where r = 0. This is called a singularity. A vortex in potential flow is defined as Γθ Γ ϕ= , ψ=− ln r 2π 2π where Γ is the circulation defined above. The velocity field is given by Γ ur = 0, uθ = 2πr and we can see the components of velocity for the vortex are the reverse of the source; now only the azimuth component is Figure 33: Vortex flow streamlines and vectors. active. Depending on the sign of Γ the flow will spin clockwise, or anti-clockwise. Again, the velocity at the origin is singular. 56 Rankine bodies A Rankine half-body is a superposition of a source and uniform flow, Figure 34 47 Qθ Q ψ = Ur sin θ + = Uy + arctan y/x 2π 2π The velocity from the source and uniform flow cancel at a stagnation point on the x-axis. The radial velocity at this point is thus Q U+ =0 2πrstag Figure 34: Rankine half-body: So rstag = − Q/(2πU ). Using this relationship the equation for plot_flow(psi = lambda x,y: -y+source(x,y)) the streamline through this point is theta = np.linspace(-2.5,2.5,20) r = theta/np.sin(theta) ψbody = 0 = r sin θ − rstag θ plt.plot(r*np.cos(theta),r*np.sin(theta),'k'); Everything inside that streamline may be conceptually treated 47 William Rankine was a Scottish naval academy as a solid body, as in Figure 26; this is the so-called half body. graduate and Professor who did foundational work in thermo-fluid in the mid 1800s. 57 The Rankine oval adds a sink downstream of the source, closing the body streamline Q y y ψ = Uy − arctan − arctan 2π x−a x+a where the singularities are shifted to the left and right of the origin by an amount a. Using the difference of arctangents identity we have y y ! x−a − x+a y y arctan − arctan = arctan y y x−a x+a 1 + x−a x+a Figure 35: Rankine oval. The body streamline in 2ay black must be determined numerically. = arctan x 2 + y2 − a2 and so the equation for the body streamline is Q 2ar sin θ Ur sin θ − arctan =0 2π r 2 − a2 which I’ve solved numerically for r in the plot. 58 Dipole and cylinder flow A dipole is a superposition of a source and sink with vanish- ingly small distance between them. Since lima→0 arctan( a) = a the difference of arctangents is 2ar sin θ 2a sin θ lim arctan = a →0 r 2 − a2 r To avoid the singularities completely canceling, we set the dipole strength as κ = 2aQ giving κ sin θ κ cos θ ψ=− , ϕ= Figure 36: Dipole flow streamlines and vectors. 2π r 2π r The corresponding velocity field is κ cos θ κ sin θ ur = − 2 , uθ = − 2π r 2π r2 which now shows a second-order singularity. 48 48 A di-pole. 59 Flow around a circular cylinder can be obtained by superimposing a dipole and uniform flow, κ sin θ ψ = Ur sin θ − 2πr The definition of the "body" streamline on ψ = 0 gives the circle radius as R2 = κ/(2πU ), and substitution gives ! R2 ψ = Ur sin θ 1 − 2 r Figure 37: Potential flow around a circle. 60 Potential flow forces The Rankine body and circle flows are good "proto-types" for the kinds of engineering problems we are interested in. The half-body could be the flow into a bow, and circle flows are ubiquitous in engineering. Let’s use the circle flow to ask a few more important engineering questions: What is pressure far from the circle? What is pressure on the circle? What is the force on the circle? We will answer these questions with the known potential velocity field ⃗u and the Bernoulli equation: 1 ρ|u|2 + ρgz + p = C 2 61 Far from the circle we have r ≫ R and so the dipole contri- bution goes to zero. This means we quickly revert to a uniform flow with ⃗u = U x̂ and p = p∞ set by the boundary conditions. So the immersed circle has a fairly localized effect. On the circle, the radial velocity ur = 0 49 so we only need to determine uθ on r = R. U uθ |r= R = −2U sin θ Ω ∂Ω So we have stagnation points at θ = 0 and π, and maximum velocity |u| = 2U at θ = ±π/2; the top and bottom of the circle. And substitution into the Bernoulli equation 18 gives 1 p|r= R = p∞ + ρU 2 (1 − 4 sin2 θ ) 2 Figure 38: Normal stresses on the circle surface or 50 ∂Ω. The symmetry of the pressure ensures that the c p = 1 − 4 sin2 θ = 2 cos(2θ ) − 1 normal stresses cancel, such as the two blue arrows. 49 Can you see why? The max c p = 1 at the stagnation points and the min c p = −3 at the top and bottom. 50 The local pressure coefficient is defined as p − p∞ cp = 1 (22) 2 2 ρU 62 The fluid force is determined by p in a potential flow. 51

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