DeformingArea_Solution - MIT Department of Mechanical...

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MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Change of Volume/Area, ATP Consider a two–dimensional steady flow in a domain Ω described by an Eulerian velocity field V = ( u ( x, y ) , v ( x, y )). Find an expression for the instantaneous relative change in area — or volume per unit depth — of a material square element δx × δy (the lower left corner of which is) located at a position ( x, y ) as δx and δy tend to zero. 2.25 Advanced Fluid Mechanics 1 Copyright c 2010, MIT
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Continuum and Kinematics Change of Volume/Area, ATP Solution: Solution 1 Consider the originally rectangular differential particle with sides δ x and δ y, formed by the points a,b,c and d, located at (x,y), (x+ δ x,y), (x, y+ δ y) and (x+ δ x, y + δ y) respectively, shown in figure 1. Figure 1: Differential Area Element As the differential mass element moves accordingly to the velocity field, it’s area and geometrical shape changes. The corners of the original differential element, a,b,c,d, become a’, b’, c’ and d’. The distances travelled by the corners of the rectangle, a’-a, b’-b, c’-c, and d’-d, can be calculated from the velocity field (after all, the velocity field is the instantaneous velocity of the particles at the given point) and adding this amount to the original position, the final position of the points can be calculated. First, for the distances: A = ( u ( x, y ) , v ( x, y )) δt + O ( δt 2 ) (1) B = ( u ( x + δx, y ) , v ( x + δx, y )) δt + O ( δt 2 ) (2) C = ( u ( x, y + δy ) , v ( x, y + δy )) δt + O ( δt 2 ) (3) 2.25 Advanced Fluid Mechanics 2 Copyright c 2010, MIT
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Continuum and Kinematics Change of Volume/Area, ATP D = ( u ( x + δx, y + δy ) , v ( x + δx, y + δy )) δt + O ( δt 2 ) (4) Figure 2: Deformation of the Area Element Then, the final position of the Points are: a 0 = ( u ( x, y ) , v ( x, y )) δt + O ( δt 2 ) + ( x, y ) (5) b 0 = ( u ( x + δx, y ) , v ( x + δx, y )) δt + O ( δt 2 ) + ( x + δx, y ) (6) c 0 = ( u ( x, y + δy ) , v ( x, y + δy )) δt + O ( δt 2 ) + ( x, y + δy ) (7) d 0 = ( u ( x + δx, y + δy ) , v ( x + δx, y + δy )) δt + O ( δt 2 ) + ( x + δx, y + δy ) (8)
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