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Unformatted text preview: Fluid Dynamics 3  Solutions to Sheet 1 1. (i) For t = 0, C is a circle; the points a (1 , , 0) and a ( − 1 , , 0) remain stationary. For 0 < s < π xvalues are pushed to the right, for π < s < 2 π xvalues are pushed to the left. Thus one obtains the following picture: Figure 1: Deformation of C(t). (ii) Since s is a particle marker, v = ˙ x = a ( α sin s, , 0) is the Lagrangian velocity field. (ii) But a sin s is the coordinate y , and thus u = ( αy, , 0) is the Eulerian field. 2. (i) u = ( a cos πt, a sin πt, 0). Streamlines at t = t are d x a cos πt = d y a sin πt so y = x tan πt + C , C constant. Sketch: Particle paths found from d x d t = a cos πt , d y d t = a sin πt . I.e. x = a π sin πt + C 1 , y = − a π cos πt + C 2 . Constants C i determined by putting ( x, y ) = (1 , 1) when (a) t = 0 and when (b) t = 1. In case (a), we get x = 1 + ( a/π ) sin πt , y = 1 + ( a/π ) − ( a/π ) cos πt . In case (b), we get t = 0 t = 1 2 t = 1 Figure 2: Streamlines at different times x = 1 + ( a/π ) sin πt , y = 1 − ( a/π ) − ( a/π ) cos πt . Now sketch particle paths: 1 1 y x (a): t = 0 (b): t = 1 Figure 3: Particle paths for particles released at t = 0 and t = 1. Size of circle depends on a . (ii) u = ( x − V t, y, 0). Streamlines at t = t : d x x − V t = d y y so y = C ( x − V t ), C constant. Sketch: Particle paths: d x d t = x − V t , d y d t = y . So y x 1 2 V V Figure 4: Streamlines radial from x = V t at t = 0 , 1 2 , 1 1 get, for ( x, y ) = (1 , 1) at t = t : x = V ( t + 1) + e...
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This note was uploaded on 01/20/2012 for the course MATH 33200 taught by Professor Eggers during the Fall '11 term at University of Bristol.
 Fall '11
 Eggers

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