Types of Potential Flows Simple Approximations of Fluid Flows – Lesson 4 Solving Potential Flow Problems • We now have all the tools for solving complete flow problems using the potential flow model. The essential steps are as follows: 1. Solve Laplace’s equation for either the velocity potential or stream function (in 2D), subject to appropriate boundary conditions. 2. Compute the velocity components from the derivatives of velocity potential or stream function. 3. Compute the pressure field using Bernoulli’s equation. Note that we must define the constant for Bernoulli’s equation from a known boundary condition (e.g., freestream conditions). • Question: How to solve Laplace’s Equation for flows of interest? ‐ Analytical Solutions: Employ exact, analytical solutions to create a flow field of interest. ‐ Numerical Solutions: Use well-known numerical algorithms to compute solutions to Laplace’s equations on a computer. 2 Exact Potential Flow Solutions • We will now examine some selected exact solutions to Laplace’s equation which we will later combine (using superposition) to create various basic flows. • The exact solutions we will work with are: ‐ Uniform Flow ‐ Source/Sink Flow ‐ Vortex Flow ‐ Doublet Flow • In all cases, the validity of the solutions can be proven by substituting them into Laplace’s equation. • The flow configuration we will assume in all cases is an unbounded domain, with one or more features of interest within the domain (e.g., a solid object, vortex, source, etc.). • We will also confine our attention to 2D solutions. 3 2D Laplace’s Equation for 휙 and 휓 휕2휙 휕2휙 휕휙 휕휓 + = 0 푢 = = − y 휕푥2 휕푦2 휕푥 휕푦 Cartesian 휕2휓 휕2휓 휕휙 휕휓 + = 0 푣 = = 휕푥2 휕푦2 휕푦 휕푥 풓 휽 2 휕 휕휙 1 휕 휙 휕휙 1 휕휓 x 푟 + 2 = 0 푣 = − = − origin 휕푟 휕푟 푟 휕휃 푟 휕푟 푟 휕휃 Cylindrical-Polar 휕 휕휓 1 휕2휓 1 휕휙 휕휓 푟 + = 0 푣휃 = = 휕푟 휕푟 푟 휕휃2 푟 휕휃 휕푥 4 Uniform Flow Equipotential Lines • A solution which describes a uniform flow field of magnitude 푉∞ is given by the velocity potential Uniform Flow 휙 푥, 푦 = 푢∞푥 + 푣∞푦 푽∞ 휓 푥, 푦 = −푢∞푦 + 푣∞푥 2 2 Streamlines 푉∞ = 푢∞ + 푣∞ • In cylindrical-polar coordinates 휙 푟, 휃 = 푉∞푟 cos 휃 휓 푟, 휃 = 푉∞푟 sin 휃 5 Potential Flow image courtesy – www.potentialflow.com Point Source/Sink Flow • Consider a 2D point source/sink flow of prescribed strength 푆ሶ (units m3/s per unit depth). We can think of this flow as an idealized fluid volumetric source/sink emerging from a point and issuing uniformly into the domain. Source Flow Sink Flow 퐒ሶ > ퟎ 퐒ሶ < ퟎ • The exact potential flow solution centered at point 푥0, 푦0 is given by: ሶ 푆ሶ 푦 − 푦 푆 2 2 −1 0 휙 푥, 푦 = ln 푥 − 푥0 + 푦 − 푦0 휓 푥, 푦 = tan 4휋 2휋 푥 − 푥0 6 Point Source/Sink Flow (cont.) • In cylindrical-polar coordinates, for a source/sink located at the origin: 푆ሶ 푆ሶ 휙 푟, 휃 = ln 푟 휓 푟, 휃 = 휃 2휋 2휋 Equipotential Lines Streamlines 7 Potential Flow image courtesy – www.potentialflow.com Superposition Example: The Rankine Oval • We can combine solutions to Laplace’s equation (by adding and subtracting them) to create a new solution. This concept is called superposition. • Using the exact solutions for the uniform and source/sink flows, we can create a flow-field around an oval shaped object by superposition. This object is called a Rankine oval. • The recipe for a Rankine oval solution consists of summing: ‐ Uniform flow = 푢∞ 푥 ‐ Source flow of strength +푆ሶ at 푥0 = −푎, 푦0 = 0 ‐ Sink flow of strength −푆ሶ at 푥0 = +푎, 푦0 = 0 • The resulting solutions for the velocity potential and stream function are: 푆ሶ 푥 + 푎 2 + 푦2 ሶ 푆 −1 푦 −1 푦 휙 푥, 푦 = 푢∞푥 + ln 휓 푥, 푦 = −푢 푦 + tan − tan 4휋 푥 − 푎 2 + 푦2 ∞ 2휋 푥 − 푎 푥 + 푎 8 Rankine Oval Solution (푆ሶ = 20, 푎 = ±2) Wall streamline Streamlines Equipotential Flows • Streamline forms a closed curve, which can be thought of as a “body” of the ellipse; streamlines outside of this curve represent potential flows around this body. 9 Potential Flow image courtesy – www.potentialflow.com Potential Doublet (Dipole) • The doublet is a special flow pattern that emerges if we take a source and sink of equal strength and constrain the product of source/sink strength and distance to be a constant (휅), as the distance approaches zero. The resulting solutions for velocity potential and stream function are: Equipotential Flows Streamlines 푑휃 휃2 휃1 푙 푆ሶ 푙 −푆ሶ 푆ሶ −푆ሶ 휅 = 푆ሶ푙 = 푐표푛푠푡푎푛푡 휅 휙 푟, 휃 = cos 휃 2휋푟 휅 휓 푟, 휃 = − sin 휃 2휋푟 10 Potential Flow image courtesy – www.potentialflow.com Potential Vortex Flow • A potential vortex is a rotational flow around a point. • The velocity is prescribed as a pure tangential velocity which is Potential Vortex inversely proportional to the radial distance from the center 퐶 푣 = 휃 푟 where 퐶 is a constant, known as the vortex strength. • It is customary to use the definition of circulation to define the constant as follows: 2휋 2휋 ퟐ Γ = ර 푉휃 푟푑휃 = ර 퐶푑휃 = 2휋퐶 휞 has units of 풎 /풔 0 0 hence Γ 푣 = 휃 2휋푟 11 Vortex Flow • The exact potential flow solution centered at the origin is given by: Streamlines Equipotential Flows Γ 휙 푟, 휃 = − 휃 2휋 Γ 휓 푟, 휃 = ln 푟 2휋 • In Cartesian form: Γ 푦 휙 푟, 휃 = − tan−1 2휋 푥 • Note the potential vortex formulation does not contradict the Γ 2 2 irrotational nature of the flow. The irrotational flow requires that fluid 휓 푟, 휃 = ln 푥 + 푦 elements do not rotate about their axis, which is satisfied in this case, 2휋 as fluid elements move along circular lines around the vortex center but do not rotate about their axes. 12 Potential Flow image courtesy – www.potentialflow.com Potential Flow Past a Circular Cylinder • Using superposition of a doublet and a uniform flow in the 푥 direction, we can construct a potential flow solution for flow around a circular cylinder of radius R: 휅 휙 푟, 휃 = 푉 푟 + cos 휃 ∞ 2휋푟 푦 휅 휓 푟, 휃 = 푉∞ 푟 − sin 휃 2휋푟 푅 • The velocity components are thus 푉∞ 푥 푅2 푣 푟, 휃 = 푉 1 − cos 휃 푟 ∞ 푟2 휅 Where 푅2 = 푅2 2휋푉 푣 푟, 휃 = −푉 1 + sin 휃 ∞ 휃 ∞ 푟2 13 Potential Flow Past a Circular Cylinder (cont.) • We can now apply Bernoulli’s equation to evaluate the pressure along the cylinder’s surface (푟 = 푅). Doing this yields: 1 푝 = 푝 + 휌푉2 1 − 4 sin2휃 푠 ∞ 2 ∞ • Note that the velocity reaches a maximum at the top/bottom of the cylinder and thus pressure is a minimum at those locations. • If we integrate the pressure force in the 푥-direction (drag) around the cylinder, we find that the total force is zero! ‐ This result is known as D’Alembert’s Paradox. A cylinder in potential flow experiences no drag. ‐ The paradox is resolved when we consider that we are not including viscous effects in our model. 14 Comparison with Real Flow • This figure shows a comparison of the potential flow surface pressure distribution for a circular cylinder with the actual flow observed in nature ‐ The agreement is reasonable only near the leading edge of the cylinder Potential flow Potential flow solutions do NOT predict separation. Flow separation is what causes drag on real objects. 15 Potential Flow Past a Rotating Circular Cylinder • Using superposition of a vortex of strength Γ , a doublet and a uniform flow in the 푥 direction, we can construct potential flow solutions for flow around a rotating circular cylinder of radius R: 푅2 Γ 휙 푟, 휃 = 푉∞ 1 + cos 휃 − 휃 푟2 2휋 푦 푅2 Γ 푟 휓 푟, 휃 = 푉 1 − sin 휃 + ln 푅 ∞ 푟2 2휋 푅 푉∞ 푥 • The velocity components are thus 푅2 푣 푟, 휃 = 푉 1 − cos 휃 푟 ∞ 푟2 푅2 Γ 푣 푟, 휃 = −푉 1 + sin 휃 − 휃 ∞ 푟2 2휋푟 16 Potential Flow Past a Rotating Circular Cylinder (cont.) • The velocity on the surface of the cylinder can be obtained by substituting 푟 = 푅 Γ 푣 푎, 휃 = −2푉 sin 휃 − 휃 ∞ 2휋푅 • Using this expression, we can now analyze the pressure coefficient for the rotating cylinder 2 2 푣 2 2Γ sin 휃 Γ 퐶푝 = 1 − = 1 − 4 sin 휃 + + 푉∞ 휋푅푉∞ 2휋푅푉∞ • Just like the previous case, if we integrate the pressure force in the 푥-direction (drag) around the cylinder, we find that the total force is zero! • However, this time due to the asymmetry induced along the 푦-axis, we get a net force along the positive 푦-direction, 퐿’ (Lift per unit span) ′ Kutta-Joukowski Theorem 퐿 = 휌∞푉∞Γ 17 Potential Flow and Lifting Bodies • Recall the definition of circulation given earlier Γ = − ර 푉 ∙ 푑푟Ԧ 퐶 • Kutta-Joukowski Theorem – The lift force exerted on an arbitrary 2D body in a potential flow is proportional to the circulation. C 퐹 퐿 = 휌푉 Γ 푏 ∞ ‐ Body shape is arbitrary ‐ Applications: airfoils, lifting surfaces, struts. ‐ Thin airfoil theory is based on using potential flow methods to determine the lift of an arbitrary thin airfoil shape. 18 Summary • Potential flow solutions are reasonable approximations to real flows if viscous effects can be ignored and flow upstream is undisturbed (irrotational). • Potential flow equations are linear and relatively easy to solve. • The solution techniques discussed in this lesson can be extended to compressible flows, unsteady flows and flows with body forces • Lift forces can be computed quite accurately, but drag forces will be totally missed, since drag is inherently a viscous phenomenon.