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Unformatted text preview: Fluid Dynamics 3 - Solutions to Sheet 7 1. (i) Since the pressure inside the bubble vanishes, it is also zero along a streamline that starts on the axis of symmetry, and continues along the upper surface of the bubble. Thus from steady Bernoulli we have q 2 / 2 + g ( z R ) = 0 , where we take z = 0 at the centre of the bubble. The constant on the right is zero, since the tip of the bubble z = R is a stagnation point, q = 0. But R z = R (1 cos ) on the surface, from which the result follows. (ii) According to the lecture, on the surface of a sphere in flow q 2 = 9 U 2 4 parenleftbigg 1 z 2 r 2 parenrightbigg , and so q = 3 U sin / 2. (iii) We assume that the results for q from (i) and (ii) agree for small (near the tip), and so q 3 U/ 2 = radicalbig 2 gR 2 / 2 , from which the result follows. 2. (i) In polar coordinates we have u r = r = U parenleftbigg 1 R 2 r 2 parenrightbigg cos = 1 r and u = 1 r = U parenleftbigg 1 + R 2 r 2 parenrightbigg sin = r Integrating the second equation, = U parenleftbigg r R 2 r parenrightbigg sin + f ( ) , it follows that 1 r = U parenleftbigg 1 R 2 r 2 parenrightbigg cos + f ( ) r , and thus f = 0, comparing to...
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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