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Home , Flow velocity, Inviscid flow, Potential flow, Velocity potential

Inviscid (Frictionless) Flow ME 306 Fluid Mechanics II • Continuity and Navier-Stokes equations govern the flow of fluids. • For incompressible flows of Newtonian fluids they are Part 1 훻 ⋅ 푉 = 0 퐷푉 휕푉 휌 = 휌 + 푉 ⋅ 훻 푉 = 휌푔 − 훻푝 + 휇훻2푉 퐷푡 휕푡 • These equations can be solved analytically only for a few problems. • They can be simplified in various ways. These presentations are prepared by −3 Dr. Cüneyt Sert 휇푤푎푡푒푟 = 10 Pa s • Common fluids such as water and air have small . 휇 = 2 × 10−5 Pa s Department of Mechanical Engineering 푎푖푟 Middle East Technical University • Neglecting the viscous term (zero shear force) gives the Euler’s equation. Ankara, Turkey 퐷푉 [email protected] 휌 = 휌푔 − 훻푝 퐷푡 You can get the most recent version of this document from Dr. Sert’s web site. • Still difficult get a general analytical solution for the unknowns 푝 and 푉. Please ask for permission before using them to teach. You are not allowed to modify them.

1-1 1-2

Inviscid Flow (cont’d) Inviscid and Irrotational Flow

• Bernoulli Equation (BE) is Euler’s equation written along a streamline. • To simplify further we can assume the flow to be irrotational. 푝 푉2 휉 = 2휔 = 훻 × 푉 = 0 (irrotational flow) + + 푧 = constant along a streamline 휌푔 2푔 (ksi) Angular velocity Exercise: Starting from the Euler’s equation derive the BE.

• Question: What’s the logic behind irrotationality assumption? Inviscid flow away from the object (negligible shear forces) • Irrotationality is about velocity . 휕푤 휕푣 휕푢 휕푤 휕푣 휕푢 휉 = − 푖 + − 푗 + − 푘 = 0 휕푦 휕푧 휕푧 휕푥 휕푥 휕푦

Airfoil or 1 휕푉 휕푉 휕푉 휕푉 1 휕 푟푉 1 휕푉 휉 = 푧 − 휃 푖 + 푟 − 푧 푖 + 휃 − 푟 푖 푟 휕휃 휕푧 푟 휕푧 휕푟 휃 푟 휕푟 푟 휕휃 푧 Adapted from http://web.cecs.pdx.edu/~gerry/class/ME322 Viscous flow close to the object and in the wake of it (non negligible shear forces)

1-3 1-4 Inviscid and Irrotational Flow (cont’d) Inviscid and Irrotational Flow (cont’d)

• One special irrotational flow is when all velocity gradients are zero. • In an inviscid flow net shear force acting on a fluid element is zero. • An example is uniform flow such as 푢 = 푈, 푣 = 0, 푤 = 0. 푦 푢 = 푈 • Only and body forces act on the fluid element. But they cannot cause 푣 = 0 푥 rotation because • In many flow fields there will be uniform-like flow regions. 푧 푤 = 0 • pressure forces act perpendicular to the element’s surface. Away from the body, flow has small velocity • body forces act through element’s center of gravity. gradients (uniform-like flow), small shear forces and can remain irrotational. In an inviscid flow, a fluid element Uniform that originates from an irrotational approach flow region will remain irrotational. velocity (irrotational) Uniform Close to the body velocity gradients upstream are high, shear forces are high and flow Airfoil flow becomes rotational. (irrotational)

Exercise: Sketch the developing flow inside a pipe with uniform entrance and Exercise: Show how a fluid element will rotate inside the developing flow region show the uniform and non-uniform flow regions. of a pipe with uniform entrance. 1-5 1-6

Inviscid and Irrotational Flow (cont’d) BE for Irrotational Flow

• In general, flow fields are composed of both Note: There are other Exercise: Repeat the exercise of Slide 1-3 (derive BE) for irrotational flow. factors that can cause • irrotational regions with negligible shear forces rotation, but they are • In an irrotational flow BE is valid between any two points of the flow , not • and rotational regions with considerable shear forces not as common as viscous effects. necessarily two points on the same streamline.

• Sometimes rotational regions will be very thin such as high speed external flow over an airfoil. 1 2

• But still neglecting them and assuming the flow to be totally irrotational would 3 Inviscid, irrotational yield unrealistic results. flow over an object

Assuming external flow over a body to be inviscid and irrotational everywhere 퐹drag = ? will result in zero air , which is not 푝 푉2 푝 푉2 푝 푉2 correct. This is known as d’Alambert’s + + 푧 = + + 푧 = + + 푧 휌푔 2푔 휌푔 2푔 휌푔 2푔 paradox. 1 2 3

Exercise: What will happen if we assume pipe flow with uniform entrance to be inviscid and irrotational? 1-7 1-8 (휙) Potential Flow

휕휙 휕휙 • For an irrotational flow : 훻 × 푉 = 0 • For a 2D flow in the 푥푦 plane : 푉 = 훻휙 → 푢 = , 푣 = 휕푥 휕푦 • As studied in ME 210, of the of any is zero. 휕휙 1 휕휙 • For a 2D flow in the 푟휃 plane : 푉 = 훻휙 → 푉푟 = , 푉휃 = 훻 × (훻휙) = 0 휕푟 푟 휕휃

• Using these two equations we can define a velocity potential function (휙) as • If the irrotational flow is also incompressible (In ME 306 we’ll NOT study compressible irrotational flows) 푉 = +훻휙 : 훻 ⋅ 푉 = 0 Some books use a minus sign so that 휙 decreases in the flow direction, similar to Phi: A scalar function 훻 ⋅ 훻휙 = 0 temperature decreasing in the heat flow called velocity potential 2 direction. But we use plus in this course. 훻 휙 = 0 Laplace’s equation

훻2 = 훻 ⋅ 훻 : • In an irrotational flow field, velocity vector can be expressed as the gradient of a scalar function called the velocity potential. • For an incompressible and irrotational flow, velocity potential satisfies the Laplace’s equation. These flows are called potential flows.

1-9 1-10

Velocity Potential (cont’d) Potential Flow Exercises

휕2휙 휕2휙 Exercise : Using Cauchy Riemann equations show that streamfunction also satisfies In the 푥푦 plane : + = 0 휕푥2 휕푦2 the Laplace’s equation for incompressible, potential flows. • 훻2휙 = 0 2 1 휕 휕휙 1 휕 휙 Exercise : Show that constant streamfunction lines (streamlines) are always In the 푟휃 plane : 푟 + 2 2 = 0 푟 휕푟 휕푟 푟 휕휃 perpendicular to constant velocity potential lines for incompressible, potential flows. Streamfunction

• Note that the relation between 푉 and 휙 is similar to that of 푉 and 휓. Exercise : Draw constant velocity potential lines of the following flow fields for which streamlines are shown. Constant velocity potential lines and streamlines drawn together form a flow net. What’s the ‘‘heat transfer’’ analogue of a flow net? Cauchy Riemann Equations Flow near a corner Flow over a cylinder In the 푥푦 plane In the 푟휃 plane 휕휙 휕휙 휕휙 1 휕휙 푢 = 푣 = 푉 = 푉 = 휕푥 휕푦 푟 휕푟 휃 푟 휕휃 휕휓 휕휓 1 휕휓 휕휓 푢 = 푣 = − 푉 = 푉 = − 휕푦 휕푥 푟 푟 휕휃 휃 휕푟 1-11 1-12 Potential Flow Exercises (cont’d) Potential Flow Exercises (cont’d) Outer region Exercise : The two-dimensional flow of a nonviscous, incompressible fluid in the Exercise : A horizontal slice through a tornado is 푦 vicinity of a corner is described by the modeled by two distinct regions. The inner or core Inner 푟 region (0 < 푟 < 푅) is modeled by solid body rotation. region 휓 = 2푟2 sin(2휃) 휃 The outer region (푟 > 푅) is modeled as an irrotational where 휓 has units of m2/s when 푟 is in meters. Assume the fluid density is region of flow. The flow is 2D in the 푟휃-plane, and the 푥 1000 kg/m3 and the 푥푦 plane is horizontal. components of the velocity field are given by

a) Determine, if possible, the corresponding velocity potential. 휔푟 0 < 푟 < 푅 Inner b) If the pressure at point 1 on the wall is 30 kPa, what is the pressure at point 2? 푉푟 = 0 , 푉휃 = 휔푅2 푟 > 푅 region Outer region Reference: Munson’s book. 푟 0 푦 where 휔 is the magnitude of the angular -0.2 velocity in the inner region. The ambient -0.4 pressure (far away from the tornado) is 푝 − 푝∞ 2 2 1 equal to 푝∞. Obtain the shown nondimen- 휌휔 푅 -0.6 푟 sional pressure distribution. 0.5 m -0.8 휃 푥 Reference: Çengel’s book. -1 2 0 1 2 3 4 5 1 m 1-13 푟/푅 1-14

Superposition of Elementary Potential Flows 1. Uniform Flow

• Laplace’s equation is a linear PDE. • Consider uniform flow in the 푥푦 plane in +푥 direction. • Superposition can be applied to both velocity potential and streamfunction. 푢 = 푈 , 푣 = 0 • Let’s find the equation for velocity potential. Potential flow 1 Potential flow 2 Potential flow 3 휕휙 휕휙 푢 = → 푈 = → 휙 = 푈푥 + 푓(푦) + = 휕푥 휕푥 휕휙 휕휙 푑푓 푣 = → 0 = → = 0 → 푓 = constant 휕푦 휕푦 푑푦

• 휙1 + 휙2 = 휙3 , 휓1 + 휓2 = 휓3 , 푉1 + 푉2 = 푉3 • Taking 푓 = 0 for simplicity • To obtain complicated flow fields we can combine elementary ones such as 휙 = 푈푥 • Uniform flow • Constant 휙 lines correspond to constant 푥 lines, i.e. lines parallel to the 푦 axis. • Line source/sink

• Exercise : Show that streamfunction equation is 휓 = 푈푦

1-15 1-16 1. Uniform Flow (cont’d) 2. Line Source at the Origin

• Constant 휙 and constant 휓 lines are shown below. • Consider the 2D flow emerging at the origin of the 푥푦 plane and going radially outward in all directions with a total flow rate per depth of 푞.

푦 View from the top 푦 푥 푥 푦 Streamlines 휓 = 휓1

휓 = 휓0 푥

1

0 휙

휙 푞 : Flow rate per depth = = Constant

(Strength of source [m2/s]) 휙 휙 휙 lines

푈 푞 푉 = , 푉 = 0 Exercise : Determine the equations of 휙 and 휓 푦 • Conservation of mass: 푞 = 2휋푟 푉푟 → 푟 2휋푟 휃 for uniform flow in a direction making an angle 푥 of 훽 with the 푥 axis. 훽 • 푉푟 decreases with 푟, i.e. effect of source diminishes with 푟.

• Origin is a singular point with 푉푟 → ∞, which is not physical, so don’t get too close. 1-17 1-18

2. Line Source (cont’d) 2. Line Source (cont’d)

• Let’s find the equation for velocity potential. • Consider a line source that is located NOT at the origin. 휕휙 푞 휕휙 푞 • Equations for 휙 and 휓 change as follows 푉푟 = → = → 휙 = ln(푟) + 푓(휃) 푦 휕푟 2휋푟 휕푟 2휋 푞 (푥, 푦) 휙 = ln 푟 2휋 1 푟 1 휕휙 1 푑푓 푑푓 푟 1 푉휃 = → 0 = → = 0 → 푓 = constant 푞 푟 휕휃 푟 푑휃 푑휃 휓 = 휃1 휃1 2휋 푏 휃 푥 • Taking 푓 = 0 for simplicity or equivalently using 푥 and 푦 coordinates 푎 푞 푞 휙 = ln(푟) 휙 = ln 푥 − 푎 2 + 푦 − 푏 2 2휋 2휋 • Constant 휙 lines correspond to constant 푟 lines as shown in the previous slide. 푞 푦 − 푏 Some useful relations 휓 = arctan 2휋 푥 − 푎 푥 = 푟 cos(휃) 푞 푦 = 푟 sin(휃) Exercise : Show that the streamfunction equation is 휓 = 휃 2휋 푟 = 푥2 + 푦2 푦 • 휃 = arctan To study a line sink for which the flow is radially inward towards a point, simply 푥 use a negative 푞 value. 1-19 1-20 3. Irrotational Vortex 3. Irrotational Vortex (cont’d)

• Studied in ME 305 as free vortex. Its velocity components are • Strength of a vortex is not given by 퐾. Instead we use its Γ. 푉휃 = 퐾/푟 퐾 • Circulation is the line integral of the tangential component of the velocity vector 푉 = , 푉 = 0 휃 푟 푟 around a closed curve. It is related to the rotationality of the flow.

• Using Cauchy Riemann relations we get 푦 Closed 2 휙 = 퐾휃 , 휓 = −퐾 ln(푟) curve 퐶 Γ = 푉 ⋅ 푑푠 [m /s] 퐶 푑푠 푉 • Compared to a line source, streamlines and constant potential lines are interchanged. Differential vector along the path of Closed path of integration integration 푦 Streamlines Similar to a line source/sink, 푥 origin is a singular point, 푥 where the velocity shoots to Constant 푉 = 푢푖 + 푣푗 infinity. • For the 2D flow in the 푥푦 plane 휙 lines Γ = (푢푑푥 + 푣푑푦) shown above 푑푠 = 푑푥푖 + 푑푦푗 퐶 1-21 1-22

3. Irrotational Vortex (cont’d) 3. Irrotational Vortex (cont’d)

Exercise: Calculate the circulation of an irrotational vortex for the following curve 퐶 Exercise: A drains from a large tank through a small opening as illustrated. A vortex forms, whose velocity distribution away from the opening can be 푦 approximated as that of a free vortex. Determine an expression relating the surface Curve 퐶 shape to the strength of the vortex Γ. 푉휃 = 퐾/푟 Reference: Munson’s book 푥 푅

푧 푝푎푡푚 푟 • Irrotational vortex is irrotational everywhere except the origin. All the circulation is 푦 squeezed into the origin, which is a singular point. 푥 • Circulation Γ = 2휋퐾 can be understood as the strength of the vortex. It’s in m2/s. 푧 푟 =? Γ Γ 휙 = 휃 , 휓 = − ln(푟) 2휋 2휋

• Direction of the vortex is determined as Γ > 0 : CCW (+푧) rotating vortex Γ < 0 : CW (−푧) rotating vortex 1-23 1-24 Exercises for Elementary Potential Flows Source in a Uniform Flow

Exercise : Elementary components of a potential flow of water is shown below. (Flow Past a Half Body) Find the velocity and pressure at point A if the pressure at infinity is 100 kPa. Exercise: Study the flow obtained by the combination of uniform flow in 푥 direction and a source at the origin. Obtain the location of the (s) and draw the stagnation streamline. 푦 푈 푈 = 3 m/s 푦 훼 푈 훼 = 30° A 2 푞 = 10 m /s 푥 푞 푎 푏 푎 = 0.8 m 푥 푏 = 0.6 m Source (푞) Sink (−푞) Uniform flow: 푢 = 푈 , 푣 = 0 , 휙 = 푈푥 , 휓 = 푈푦

푞 푞 푞 Source: 푉 = , 푉 = 0 , 휙 = ln 푟 , 휓 = 휃 Exercise : For the previous problem determine the equations of velocity potential 푟 2휋푟 휃 2휋 2휋 and streamfunction by superimposing elementary flows. Find the velocity at point A by differentiating both 휙 and 휓 equation. 1-25 1-26

Flow Past a Half Body (cont’d) Flow Past a Half Body (cont’d)

푦 Exercise : Consider the top part of a half body. Draw speed vs. 휃 and pressure vs. 휃 푞/2휋푟 using the following values: 휌 = 1000 kg/m3 , 푈 = 5 m/s , 푞 = 10 m2/s and 푈 푈 푝∞ = 100 kPa.

Stagnation 푠 Exercise: A 64 km/h wind blows toward a hill that can be approximated with the 푞 푥 streamline top part of a half body. The maximum height of the hill approaches 60 m. 휓 = 푞/2 a) What is the air speed at a point directly above the origin (at point 2)? b) What is the elevation of point 2? Stagnation point c) What is the pressure difference between point 2 and point 1 far from the hill? 푟푠 = 푞/2휋푈 Movie Reference: Munson’s book Flow Over Half Body 푦 • Flow outside the stagnation streamline resembles a flow over 64 km/h 푔 a body with a blunt nose. 2 60 m • Equation of the half body is given by the equation of the stagnation streamline. 푥 1 1-27 1-28 A Source and a Sink in Uniform Flow Flow Past a Rankine oval (cont’d)

(Flow Past a Rankine oval) 푦 • Superposition of 푦 푈 푈 • a source of strength 푞 at 푥 = −푐, • a sink of strength −푞 at 푥 = 푐 and 푐 푐 푥 푥 • uniform flow of magnitude 푈. 푞 −푞

푐 푐

푞 푞 • 휙 = 휙푢푛푖 + 휙푠표푢 + 휙푠푖푛푘 = 푈푥 + ln 푟1 − ln(푟2) 2휋 2휋 푦 Exercise : Determine the location of the stagnation points of the shown Rankine A 푞 푞 oval. Determine its length and thickness of the oval. Plot the variation of speed • 휓 = 휓푢푛푖 + 휓푠표푢 + 휓푠푖푛푘 = 푈푦 + 휃1 − 휃2 푟1 2휋 2휋 푟2 and pressure (with respect to 푝∞) on it as a function of 휃. 휃1 휃2 푐 푐 푥

1-29 1-30

Doublet A Doublet in Uniform Flow (Flow Past a Cylinder)

• Superposition of • Superposition of • a source of strength 푞 at the orgin (moved from – 푥 axis to the origin), • a doublet of strength 푑 at the origin. • a sink of strength −푞 at the origin (moved from +푥 axis to the origin) , • uniform flow of magnitude 푈 in +푥 direction. • Consider the limiting case of the source and sink of Slide 1-29 approaching to the origin. Skipping the details we can get 푦 푈 푑 푑 휙 = cos 휃 , 휓 = − sin 휃 푑표푢푏푙푒푡 2휋푟 푑표푢푏푙푒푡 2휋푟 푥 푑 where 푑 is the strength of the doublet. 푦 • Velocity field is given by Streamlines 푑 휕휙 푑 • 휙 = 휙푢푛푖 + 휙푑표푢푏푙푒푡 = 푈푥 + cos(휃) 푉 = = − cos 휃 Constant 휙 lines 2휋푟 푟 휕푟 2휋푟2 푑 푥 • 휓 = 휓 + 휓 = 푈푦 − sin(휃) 1 휕휙 푑 푢푛푖 푑표푢푏푙푒푡 2휋푟 푉 = = − sin 휃 휃 푟 휕휃 2휋푟2 1-31 1-32 Flow Past a Cylinder (cont’d) Flow Past a Cylinder (cont’d) 푈 Important results are as follows 푦 • Pressure distribution on the cylinder is 훽 (using BE with 푉 = 푈 and 푝 ) • Velocity components are ∞ ∞ 1 푅 휌푈2 푅2 푅2 푥 푝 = 푝 + 1 − 4 sin2(휃) 푑 푐푦푙 ∞ 푉푟 = 푈 1 − 2 cos 휃 , 푉휃 = −푈 1 + 2 sin 휃 2 푟 푟 0 Experimental • Pressure on the cylinder is symmetric

with 푅 = 푑/푈 with respect to both 푥 and 푦 axis. 푝푐푦푙 − 푝∞ 2 -1 • Pressure does not create any drag 휌푈 • Stagnation points are located at (−푅, 0) and (푅, 0). 2 force (in 푥 direction) or any force -2 • 푅 Stagnation streamline is a circle of radius . (in 푦 direction). Potential 2 -3 • Velocity distribution on the cylinder is 0 휋/2 휋 푦 푝 훽 2휋 푐푦푙 푉휃푐푦푙 푉휃푐푦푙 = −2푈 sin(휃) 1 퐹푑푟푎푔 = − 푝cyl cos 휃 푅푑휃 = 0 푈 0 푅푑휃 2휋 퐹푑푟푎푔 휃 푥 퐹 = − 푝 sin 휃 푅푑휃 = 0 0 푙푖푓푡 cyl 휋/2 휋 0 0 퐹푙푖푓푡 휃 1-33 1-34

Flow Past a Cylinder (cont’d) Flow Past a Cylinder (cont’d)

• As seen from the above exercise potential flow theory predicts ZERO DRAG FORCE 푅푒 = 1.5 on the cylinder. 퐷 푅푒퐷 = 26 • Actually this is the case for any closed body, irrespective of its shape. • This result is not physical and it is known as d’Alembert paradox (1752). • In a real viscous flow • shear stresses inside the will cause a frictional drag force. • viscous action will cause separation & the pressure at the front and back of the cylinder would not be symmetric.

Movie Movie Flow with Separation Potential vs. Viscous

푅푒퐷 = 2000 ‘‘An Album of Fluid Motion’’, by M. Van Dyke 푅푒퐷 = 10000 1-35 1-36 Flow Past a Cylinder (cont’d) Flow Past a Rotating Cylinder

Exercise : When a small circular cylinder is placed in a uniform stream, a • Superposition of stagnation point is created on the cylinder. If a small hole is located at this point, 푦 • a doublet of strength 푑 located at the origin, 푈 the stagnation pressure, can be measured and used to determine the approach velocity, 푈 (similar to a Pitot tube). • CCW rotating irrotational vortex of strength Γ 푑 Γ 푥 located at the origin, a) Show how 푝푠푡푎푔 and 푈 are related. Pressure far away is 푝∞. • uniform flow of magnitude 푈. b) If the cylinder is misaligned by an angle 훼, but the measured pressure is still interpreted as the stagnation pressure, use potential flow theory to determine an • This will result in expression for the ratio of the true velocity, 푈, to the predicted velocity, 푈′. Plot this ratio as a function of 훼 for the range 0° < 푎 < 20°. 푅2 Γ 휙 = 푈푟 1 + cos 휃 + 휃 푟2 2휋 Reference: Munson’s book. with 푅 = 푑/푈 푈 푈 푅2 Γ 휓 = 푈푟 1 − sin 휃 − ln 푟 푟2 2휋 훼

푅 • This is the potential flow that resembles the flow over a rotating cylinder of radius 푅. Stagnation point 1-37 1-38

Flow Past a Rotating Cylinder (cont’d) Flow Past a Rotating Cylinder (cont’d)

푦 푈 • Streamlines and stagnation points for different circulation values.

Γ 푑 푥 푅 푅 Γ Γ = 0 < 1 4휋푈푅

Exercise : For the flow shown above, obtain the following results Γ 푉 = −2푈 sin 휃 + White’s book 휃푐푦푙 2휋푅 휌푈2 2Γ Γ 2 푝 = 푝 + 1 − 4 sin2 휃 + sin 휃 − 푐푦푙 ∞ 2 휋푈푅 2휋푈푅 Γ Γ = 1 > 1 4휋푈푅 4휋푈푅 Integrate the above pressure distribution to get the following results

퐹푑푟푎푔 = 0 , 퐹푙푖푓푡 = −휌Γ푈 (per unit depth)

1-39 1-42 Magnus Effect Magnus Effect (cont’d)

• Magnus Effect: A rotating body in a uniform flow will have a net lift force on it (1853). Exercise : Magnus effect acts not only on cylinders but also on other rotating bodies such as spheres. It can be used to explain how a spinning ball moves in a curved • Direction of the lift force depends on the direction of 푈 and Γ. trajectory. A football player wants to make a penalty kick as sketched below. Will a CW or a CCW spin do the trick? Top Bottom Γ Γ 푉 = −2푈 + 푉 = 2푈 + 휃푡표푝 2휋푅 푈 휃푏표푡푡표푚 2휋푅

Opposite signs Γ 휌Γ푈 Same signs Velocity is low. Velocity is high. Pressure is high. Pressure is low. ?

Exercise : Watch the following movies 푈 https://www.youtube.com/watch?v=2OSrvzNW9FE (Suprising applications of Magnus effect) Exercise: Determine the direction of ? Γ http://www.youtube.com/watch?v=23f1jvGUWJs (Magnus force on Veritasium channel) the lift force. http://www.youtube.com/watch?v=2pQga7jxAyc (Enercon's rotor ship. Audio in German) http://www.youtube.com/watch?v=wb5tc_nnMUw (Roberto Carlos knows the Magnus force)

1-41 1-42

Magnus Effect (cont’d) (Lift on an Airfoil)

• Warning: Although potential flow theory can predict the direction of lift force • Magnus effect applies not only to cylinders but any closed shape. due to Magnus effect correctly, it may give quite inaccurate values for its • Consider the flow over a slender body with a sharp trailing edge, such as an airfoil. magnitude. We’ll come back to this in the next chapter. • An airfoil is designed to generate small drag and high lift force. Exercise: In 1920s Anton Flettner built a series of rotor ships that are propelled by rotating cylinders driven by electric motors. Read about Flettner’s ship at rexresearch.com/flettner/flettner.htm and understand how it works.

푠1 푠2

• There are two stagnation points, 푠1 and 푠2. • Experiments show that the streamlines leave the trailing edge smoothly as shown above, known as the Kutta condition.

One of Flettner’s original rotating cylinder ships https://en.wikipedia.org/wiki/E-Ship_1 1-43 1-44 Lift on an Airfoil (cont’d) Simulating Flows Near Walls using Mirror Images • Consider a line source located at a distance 푏 to a solid wall. • Potential flow theory will predict an unphysical location for point 푠2. • Fluid cannot pass across the wall and therefore it is a streamline. 푠2

• It is impossible for streamlines to make 푠1 • The effect of the wall can be simulated by using another source, which is the such a sharp turn at the trailing edge. mirror image of the original one with respect to the wall.

• If we add the correct amount of CW vortex to this flow field we can bring point 푠2 down to the trailing edge and obtain the correct streamline pattern.

+ = Γ Wrong Correct Çengel’s book • The magnitude, Γ, of the necessary vortex can be used to calculate the lift force generated on the airfoil.

퐹푙푖푓푡 = 휌 Γ 푈 Kutta-Joukowski Law (1902) 1-45 1-46

Simulating Flows Near Walls using Mirror Images (cont’d) Superposition Exercises Exercise : Consider a source of strength 푞 located close to two walls forming a 90o Exercise : We want to study the potential flow over the following bodies. If corner. possible, which elementary flows need to be superimposed to get the desired shape? a) How many and where the mirror images need to be placed to simulate existence of the walls? 푦 b) Locate the stagnation point(s). 푈 푈 푈 c) Draw the streamlines. Source (푞) 푎

푥 푎 푈 푈

Exercise : Repeat the previous exercise by replacing the source with a clockwise vortex of strength Γ.

1-47 1-48 Numerical Solution of Potential Flow Numerical Solution of Potential Flow (cont’d)

• Obtaining complicated flow fields by superposing elementary ones is limited. • Boundary conditions are (study in the given order) 휓 • To study potential flows on arbitrary geometries one can perform a numerical 2 Top wall is a streamline. should be constant solution. there. In order to have 10 m2/s flow rate per depth between the top and bottom walls 3 At the inlet 푢 = 10. • Consider a flow inside an expanding duct (coordinates are in meters). 휕휓 Therefore = 10. 휕푦 휓top = 10 (0,2) (3,2) 휓 varies linearly from 0 to 10. 10 m/s (0,1) 휓left = 10푦 − 10 (1,1) 5 m/s 푦 푥 (3,0) (2,0) At the exit 푢 = 5. 4 휕휓 Therefore = 5. 휕푦 휓 varies linearly from 0 to 10. Bottom wall is a streamline. 휓 • Potential flow inside the duct can be obtained by solving Laplace’s equation 1 should be constant there. Let’s set 휓right = 5푦 휕2휓 휕2휓 it to zero. 2 + 2 = 0 , with proper boundary conditions 휕푥 휕푦 휓bottom = 0

1-49 1-50

Numerical Solution of Potential Flow (cont’d) Numerical Solution of Potential Flow (cont’d)

• Finite Difference Method can be used to get the numerical solution. • Discretized form of the Laplace’s equation at node (푖, 푗) is • First we discretize the problem domain into a set of nodes. 휓푖+1,푗 − 2휓푖,푗 + 휓푖−1,푗 휓푖,푗+1 − 2휓푖,푗 + 휓푖,푗−1 + = 0 • Following mesh has 55 nodes with Δ푥 = Δ푦 = 1/3. Δ푥 2 Δ푦 2 • 27 of the nodes are at the boundaries and 휓 is known at these nodes. 휕2휓 휕2휓

• 28 of the nodes are inside the domain and 휓 needs to be calculated at them. 휕푥2 휕푦2 푖,푗 푖,푗

5 point computational stencil • For Δ푥 = Δ푦, discretized equation for node (푖, 푗) becomes 푖, 푗 + 1 휓푖+1,푗 + 휓푖−1,푗 + 휓푖,푗+1 + 휓푖,푗−1 − 4휓푖,푗 = 0 Δ푦 푖, 푗 푖 − 1, 푗 푖 + 1, 푗 • This equation needs to be written for all non-boundary nodes. • For nodes with boundary neighbors, some 휓 values are known and they need to be 푖, 푗 − 1 transferred to the right-hand-side of the equation. Δ푥 • At the end we’ll get a system of 28 equations for 28 unknowns and solve it. 1-51 1-52 Numerical Solution of Potential Flow (cont’d) Numerical Solution of Potential Flow (cont’d)

• Following solution is obtained using a mesh with Δ푥 = Δ푦 = 0.2. • After obtaining 푢 and 푣, 휓 = 10 can be calculated using the Bernoulli 푉푖푛 흍 = 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 1 0.00 10.00 10.00 10.00 8 푝푖푛 8.00 8.02 8.04 8.07 8.12 8.20 8.30 8.41 8.52 8.62 8.71 8.79 8.85 8.91 8.95 9.00 equation. 6 6.00 6.03 6.06 6.12 6.22 6.37 6.58 6.82 7.05 7.26 7.44 7.59 7.71 7.82 7.91 8.00 White’s book 4 4.00 4.03 4.07 4.13 4.26 4.48 4.84 5.24 5.61 5.93 6.19 6.41 6.59 6.74 6.88 7.00 2.00 2.02 2.05 2.09 2.20 2.44 3.08 3.69 4.22 4.65 5.00 5.28 5.50 5.69 5.85 6.00 2 흍 = 0.00 0.00 0.00 0.00 0.00 0.00 1.33 2.22 2.92 3.45 3.87 4.19 4.45 4.66 4.84 5.00 • For irregular geometries 0.00 1.00 1.77 2.37 2.83 3.18 3.45 3.66 3.84 4.00 modifications need to be done. Red ones are 0.00 0.80 1.42 1.90 2.24 2.50 2.70 2.86 3.00 푝 − 푝푖푛 0.00 0.63 1.09 1.40 1.61 1.77 1.89 2.00 퐶 = boundary values 푝 2 Black ones are 0.00 0.44 0.66 0.79 0.87 0.94 1.00 휌푉푖푛/2 calculated 0.00 0.00 0.00 0.00 0.00 0.00 White’s book 1.0 0.8 0.75 • After obtaining the 휓 values at the nodes, velocity components can be obtained 0.6 Upper 0.4 wall 휕휓 휓푖,푗+1 − 휓푖,푗−1 0.2 푢 = → 푢푖,푗 = • Red node is NOT at ∆푥 distance from 0.0 휕푦 2∆푦 the central node. What can be done? -0.2 -0.4 휕휓 휓푖+1,푗 − 휓푖−1,푗 Lower wall 푣 = − → 푣푖,푗 = − -0.6 휕푥 2∆푥 -0.8 • Different formulas need to be used at the boundary nodes. 1-53 1-54