Solution to Problem 6.4

MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Kundu & Cohen 6.4 This problem is from “Fluid Mechanics” by P. K. Kundu and I. M. Cohen (a) Take a plane source of strength m at point (−a, 0), a plane sink of equal strength at (a, 0), and superpose a uniform stream U directed along the x-axis. (b) Show that there are two stagnation points located on the x-axis at points ±a m +1 πaU 1/2 . (c) Show that the streamline passing through the stagnation points is given by ψ = 0. Verify that the line ψ = 0 represents a closed oval-shaped body, whose maximum width h is given by the solution of the equation πU h h = a cot . m The body generated by the superposition of a uniform stream and a source-sink pair is called a Rankine body. It becomes a circular cylinder as the source–sink pair approach each other. 2.25 Advanced Fluid Mechanics 1 c 2010, MIT Copyright @ Potential Flow Kundu & Cohen 6.4 Solution: (a) W (z) = Wuniform flow + Wsource + Wsink where W = φ + iψ, φ is the potential function, and ψ the stream function. Recap from Lecture: W satisfies the Laplace equation which is linear. Therefore, one can superimpose its solutions as above. y Wuniform flow = U∞ (x + iy)   m mθ m ln reiθ = ln r + i Wsource = 2π 2π 2π   m mθ m Wsink = − ln reiθ = − ln r + i 2π 2π 2π r θ y0 Substitute expressions for r and θ in terms of x and y (see figure):   m mθ m ln r + i ln reiθ = Wsource = 2π 2π 2π   m mθ m Wsink = − ln reiθ = − ln r + i 2π 2π 2π ⇒ Wtotal = U∞ x + arbitrary point (x, y) location of source/sink x0 x _ (x − x0 )2 + (y − y0 )2   y − y0 θ = arctan x − x0 r= m m (x + a)2 + y 2 ln +i U∞ y + arctan 2 2 2π 4π (x − a) + y  y x+a φ  − m arctan 2π  y x−a  ψ (b) Obtain the velocity field (vx , vy ) by invoking v = ∇φ vx = 2 �� + y2 ∂φ m � (x� −� a)� = U∞ + · ∂x 4π (x + a)2 + y 2 2(x + a) (x + a)2 + y 2 − · 2(x − a) � � 2 2 + y2 (x� −� a)� � [(x − a)2 + y 2 ] � m x+a x−a − (x − a)2 + y 2 2π (x + a)2 + y 2 2 �� ∂φ m � + y2 2y (x + a)2 + y 2 (x� −� a)� vy = = · − · 2y � � 2 2 2 2 � 2 ∂y 4π (x + a) + y (x� −� a) + y � [(x − a)2 + y 2 ] � ⇒ vx = U∞ + ⇒ vy = my 1 1 − (x − a)2 + y 2 2π (x + a)2 + y 2 Alternatively, one can find v by using: vx = ∂ψ , ∂y vy = − ∂ψ . ∂x Find the stagnation point(s) by finding (x, y) such that vx = vy = 0. vy = 0 at y = 0, ∀x 2.25 Advanced Fluid Mechanics 2 c 2010, MIT Copyright @ Potential Flow Kundu & Cohen 6.4 Plug in y = 0 into vx and find x that lets vx = 0:   m x+a x−a − =0 vx (x, y = 0) = U∞ + (x − a)2 2π (x + a)2   m 1 1 − =0 U∞ + 2π x + a x − a ma =0 OR (after some algebra. . . ) : x2 − a2 − πU∞ Using the quadratic formula1 , x = ±a 1 + m aπU∞ (c) Going back to ψ: θ1 (x, y) θ2     m y m y arctan − arctan 2π x+a 2π x−a   m 2ay arctan = U∞ y − 2π x2 + y 2 − a2 ψ = U∞ y + θ2 θ1 a a −(θ1 −θ2 ) A “Rankine oval” is defined by a curve ψ = 0, or U∞ y − m arctan 2π  2ay x2 + y 2 − a2  =0 (1) Maximum half-width, h, is obtained when x = 0:  2ah h2 − a2  2  h − a2 2πU∞ h = arccot m 2ah U∞ h = m arctan 2π  ⇒ h 1− h −a  a 2  h  = 2a cot 2πU∞ h m a  (2) D Problem Solution by Sungyon Lee, Fall 2005 1 ax2 + bx + c = 0, x = √ −b± b2 −4ac 2a 2.25 Advanced Fluid Mechanics 3 c 2010, MIT Copyright @ Potential Flow Kundu & Cohen 6.4 Figure 1: MATLAB® plot of streamlines for a Rankine oval. 2.25 Advanced Fluid Mechanics 4 c 2010, MIT Copyright @