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Unformatted text preview: 2.10 Kelvins circulation theorem Defn : Circulation of a velocity field is defined to be = contintegraldisplay C ( t ) u d l where C ( t ) is a closed loop which moves with the fluid and d l is an infinitesimal line segment along C ( t ). Note: By Stokes theorem = contintegraldisplay C ( t ) u d l = integraldisplay S ( t ) ( u ) n dS = integraldisplay S ( t ) n dS where S ( t ) is a surface whose edges connect with C ( t ). Kelvins theorem states that D Dt = 0. Proof : The total derivative of is D Dt = contintegraldisplay C ( t ) D u Dt d l + contintegraldisplay C ( t ) u Dd l Dt . (25) We will show below that D Dt d l = ( d l ) u , (26) which comes from the fact that each line element is convected by the flow. Using this result, the second integrand in (25) becomes u Dd l Dt = u i D ( dl i ) Dt = u i dl j j u i = 1 2 u 2 d l . On the other hand, using Eulers equation, the first integrand becomes D u Dt d l = ( p/ + ) d l . But this means that D Dt = contintegraldisplay C ( t ) ( p/ + 1 2 u 2 ) d l = 0 , because contintegraldisplay v d l = integraldisplay S ( t ) ( v ) n dS = 0. Hence we obtain the result. Now we have to prove (26). Consider two points on C ( t ), x 1 and x 2 s.t. d l = x 2 x 1 . In a short time t , x 1 x 1 + u ( x 1 ) t x 2 x 2 + u ( x 2 ) t bracerightBigg d l ( t + t ) = d l ( t ) + ( u ( x 2 ) u ( x 1 )) t. Now, by Taylor expanding, u ( x 2 ) = u ( x 1 + d l ) = u ( x 1 ) + ( d l ) u , so that Dd l Dt = d l ( t + t ) d l ( t ) t = ( d l ) u , which is what we wanted to show. Example 1 Let a closed loop of particles C ( t ) be defined by x = a (cos s + t sin s, sin s, 0) , s < 2 , where each value of s corresponds to a different fluid particle, and a, > 0. In an exercise, we worked out how C ( t ) changed in time. We also showed that the Eulerian velocity field is u = ( y, , 0) , which clearly is a solution of Eulers equation. (Why?) Since s parameterizes the curve, we can calculate the circulation at any time as = integraldisplay C ( t ) u d l = integraldisplay 2 u x s ds. Now x s = a ( sin s + t cos s, cos s, 0) and u ( s ) = x t = a (sin s, , 0) , so that = a 2 integraldisplay 2 sin s ( sin s + t cos s ) ds = a 2 integraldisplay 2 sin 2 sds = a 2 . This is constant, consistent with Kelvins theorem. 3 Irrotational flows: potential theory Kelvins circulation theorem states that the circulation around any loop convected by the flow cannot change, if the fluid is inviscid . In particular, if the circulation is zero, it will remain zero. Now if there is no vorticity present in the flow at some intitial time, the circulation around any loop in the flow is zero. Hence Kelvins circulation theorem guarantees that there will never be any circulation in the flow, and thus...
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 Fall '11
 Eggers

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