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ME 306 Fluid Mechanics II Part 1 Potential Flow

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ME 306 Fluid Mechanics II Part 1 Potential Flow 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 Potential Flow 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 viscosities. 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 Vorticity (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 gradients. 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 pressure 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. Airfoil 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 field, 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 drag, 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 Velocity Potential (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, curl of the gradient of any scalar function 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 Continuity Equation : 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 : Laplace operator • 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 stream function 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 • Vortex 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.
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