p1148

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Unformatted text preview: 68 CHAPTER 11. POTENTIAL FLOW 11.48 Chapter 11, Problem 48 Problem: A uniform stream flows downward so that u = 0 and v = −V where V is constant. (a) What are the streamfunction and velocity potential for this flow in Cartesian coordinates? (b) What are the streamfunction and velocity potential for this flow in cylindrical coordinates? (c) Now add a source of strength Q located at the origin. What are the streamfunction and velocity potential in cylindrical coordinates? (d) For the streamfunction of Part (c), locate any stagnation points in the flow. Express your answer in Cartesian coordinates. (Note that most of the work is most conveniently done in cylindrical coordinates—just convert when you’re done.) (e) Now add a sink of strength Q located at x = a, y = −a to the flow of Part (c). What are the streamfunction and velocity potential in Cartesian coordinates? (f) Sketch what you think the flow of Part (e) looks like, i.e., draw streamlines and indicate flow direction on the streamlines. Don’t worry about precise location of stagnation point(s), etc., just indicate what the body shape might look like and give a rough idea of where the streamlines and stagnation point(s) might be. Solution: (a) First, consider the streamfunction. For this flow, we have u= ∂ψ ∂ψ = 0 and v = − = −V ∂y ∂x Integrating the second equation over x, we find ψ (x, y ) = V x + f (y ) where f (y ) is a function of integration. The, substituting into the first equation, ∂ψ = f (y ) = 0 ∂y =⇒ f (y ) = constant Therefore, dropping the constant for convenience, the streamfunction is ψ (x, y ) = V x 11.48. CHAPTER 11, PROBLEM 48 69 Similarly, for the velocity potential, u= ∂φ ∂φ = 0 and v = = −V ∂x ∂y Integrating the second equation over y , we find φ(x, y ) = −V y + g (x) where g (x) is a function of integration. The, substituting into the first equation, ∂φ = g (x) = 0 ∂x g (x) = constant =⇒ Again dropping the constant for convenience, the velocity potential is φ(x, y ) = −V y (b) To convert from Cartesian to cylindrical coordinates, note that x = r cos θ and y = r sin θ. Hence, ψ (r, θ) = V r cos θ and φ(r, θ) = −V r sin θ (c) For a source of strength Q, the streamfunction and velocity potential are ψ = Qθ/(2π ) and φ = (Q/2π ) nr, wherefore ψ (r, θ) = V r cos θ + Q θ 2π and φ(r, θ) = −V r sin θ + Q nr 2π (d) The velocity components for the superposition of Part (c) are ur = 1 ∂ψ Q = −V sin θ + r ∂θ 2π r and uθ = − ∂ψ = −V cos θ ∂r Now, at a stagnation point, ur = uθ = 0. Considering uθ first, we see that uθ = 0 =⇒ =⇒ cos θ = 0 x = r cos θ = 0 =⇒ y = r sin θ = From the radial velocity, we learn that ur = 0 =⇒ V sin θ = Q 2π r Therefore, there is a stagnation point at x = 0, y= Q 2π V Q 2π V 70 CHAPTER 11. POTENTIAL FLOW (e) First, observe that for a sink located at x = a, y = −a, the streamfunction and velocity potential are Q y+a ψsink (x, y ) = − tan−1 2π x−a Q φsink (x, y ) = − n (x − a)2 + (y + a)2 2π Therefore, the complete streamfunction and velocity potential for the superposition of a downward-flowing uniform stream with a source and a sink are ψ(x, y ) = V x + φ(x, y ) = −V y + Q y Q y+a tan−1 tan−1 − 2π x 2π x−a Q Q n x2 + y 2 − n (x − a)2 + (y + a)2 2π 2π (f) First of all, note that because the source is balanced by a sink of the same strength, we have a closed body. Also, note that the body should be expected to be nearly symmetric about an axis passing through the source and sink locations. In fact, we have constructed what could be described as flow past a Rankine oval at angle of attack. We expect to have leading and trailing stagnation points. The following figure is a sketch of the flowfield, including the stagnation points, external streamlines and internal streamlines. . . .. .. . . ... ... . .. . . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . . . . . . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . . . . .. . . . .. . . . . .. . . . . .. . . . .. . . . . .. . . . . .. . . . .. . . . . .. . . . . .. . .. .. . .. .. .. . . ... ... ... . ... ... .. .. .. .. .. . .. .. .. . .. .. .. .. .. . .. .. . .. .. .. .. . . . . . .. . . . . . . . . . . .. . . . . . .. . . . . . .. . . . . .. . . . . . . . .. . . . . . .. . . . . .. . . . . . . . . . .. . . . . . . . . .. .. . . . . .. . . . . . .. . . . .. . . . . .. . . . . .. . . . . . . . . . .. . . . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . ....... . . . . . . . . . ............ .. . ................ . . . . . . . . ......................... . . . . . ................................... . . . . ... . .. . .... ............................... . . . . .. . ........ ........ ............. . . . . . .................................... . . . . . . . . ..................................... .. . . . . . ............................................................... .. .. . . . . . . . . . .. . .. .. .......... ........ . . . .... . ............. .......... ....... . .... ...... ......... . . . .. . . . . ....... ....... .. . . . . .. .. . . .. . . . .. .. ... .......................... .... ...... . . . . ..... .... . ...... . .. ..... ......... . . . .. . .. . . .. .. .. .. . ..... . . ........ . .. . ....................................................................................................................................................................... .................................................................................................................................................................. .. . ............•. ................. . .. .. .. .. . . . .................................................. . .. . . .. . . .... . ... . . . .. .. . .. .. .. .. .. .. .............................................. .. .. ................................................... .. .. ... . . .. .. ......... .... .. . . . . . .. . .. .. ... ............................................... ................................................. . . ... ....... . ......... . .. .... .... . ..... . . . .. . . ... . . . . . . ............ ............. ..... ..... ... ............................................... . .. ..... . .. . .. . . .. ....... . ....•.. . .. . .. .. . . . . . . .............................................. . ...... ..... .. . . . .. .. . . .. . ...... .... ...... . .. . . . . . . . . . . . . . ... .. . . . .................................................... ............................................ . . .. .. .. .... . .... .. . .... . . ... . .. .. .. .. . ...... . .. ... ....... ... .. . . . .. .. .. . .. .............. ..................... ........ .. .. . .......................................... . .. . .. .. .. . .... .. ..... ... .... .. . . . . .. .. . ...................... ........ .......... .. .. . ....................................... .. .. .. ... .. ........... ..... ..... .. .... . . . . . . ..... ................................... .. . ...................................... . .. .. .. . .......... ............ ......... .. . .. .. .. ..... .. . .............................. .. . .. . . .. .. ....................... .. .. .. . ................... .. .. . ................. .. . ....... . . .. .. . .. .. . .. .. . .. . .. . . . .. . .. .. . .. . .. .. . .. . .. . . .. . . . .. .. . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . . .. . .. .. .. . .. .. .. ... .... . .... .... ... ... . ... . ... . ... .... . . .. .. . . .. .. . .. .. .. .. .. . . . . .. . . . . . . . .. . . . . . . . . . . .. . . . . .. . . . . .. . . . . . .. . . . . .. . . y • x • ...
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