Rankine Ovals

1. y +Q -Q V x -xo xo Rankine Ovals • The previous flow example has limited application since the body isn’t closed but extends downstream to x. • A simple way of closing the body is to add a sink downstream of the source and of equal strength. • For symmetry purposes, we will also move the origin to be halfway between the source and sink. Thus: • Since the center of the source and sink are no longer on the origin, the definitions of r and  in terms of x and y also change. AE 401 Advanced Aerodynamics

2. (x,y) r2 1 r1 2 (-xo,0) (xo,0) Rankine Ovals [2] • For any point (x,y) in space, the following relations apply: • And the stream function solution is: AE 401 Advanced Aerodynamics

3. Rankine Ovals [3] • The stream (and potential) function solution to this case looks lie: • This body type is known as a Rankine Oval. AE 401 Advanced Aerodynamics

4. Rankine Ovals [4] • First lets find the length of this body. • Note that for this flow, there are two stagnations points – one at both the leading edge and trailing edges. • To determine the location of these points on the y axis, once again find where u=0. • At the stagnation point (xs,0) AE 401 Advanced Aerodynamics

5. Rankine Ovals [5] • With reduction, this expression becomes: • Or • The streamline which forms the body shape is found from the value of psi at the right stagnation point: • Since psi is zero on the surface, the following relationship exists: AE 401 Advanced Aerodynamics

6. Rankine Ovals [6] • Or, the difference in the two angles is: • Where a nondimensional source strength, q, is defined by: • Taking the tangent of the first equation, substituting for the tangent of the angles, and then solving for x, the following equation is found for the surface: • Which is really x(Y) rather than Y(x) – but is the best we can do. AE 401 Advanced Aerodynamics

7. Rankine Ovals [6] • Just for completeness, the u and v components of this flow velocity are: • And, once again, the pressure coefficients can be found from: • But I really doubt it this can be simplified to a short simple equation. AE 401 Advanced Aerodynamics

8. (x,y) r2 r1 2 1 (-xo,0) (xo,0) Doublets and Cylinders • While Rankine ovals are not a common aerodynamic shape, they do lead to one that is – cylinders. • First, consider a source/sink pair and consider the limit as they closer and closer (xo 0) while their strengths increase such that Q 2xo = = constant, we get: • This limit can be evaluated by looking at the details of the triangle formed… AE 401 Advanced Aerodynamics

9. d=1-2 r1 a 2 2xo Doublets and Cylinders [2] • By right triangle rules: • In the limit, d is very small, r1= r2 =r, and 1= 2 = . • Thus: • Or similarly for the potential function: AE 401 Advanced Aerodynamics

10. Doublets and Cylinders [3] • The streamlines defined by this Doublet look like: • One nice feature of a doublet is that it has a zero net mass flux – what flows out flows back in! • The other nice feature is when a doubled is combined with a uniform flow. AE 401 Advanced Aerodynamics

11. Doublets and Cylinders [4] • When a uniform flow is superimposed on a doublet, the flow looks like that around a circular cylinder. AE 401 Advanced Aerodynamics

12. Doublets and Cylinders [5] • The stream and potential functions for the flow just shown are usually written as: • As part of your homework, you will show that the velocity on the surface of the cylinder is given by: • And thus the surface pressure coefficient is: AE 401 Advanced Aerodynamics