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Unformatted text preview: Problem 2.17 [Difficulty: 4] Given: Pathlines of particles Find: Conditions that make them satisfy Problem 2.10 flow field; Also Problem 2.11 flow field; Plot pathlines Solution: The given pathlines are x p a − sin ω t ⋅ ( ) ⋅ = y p a cos ω t ⋅ ( ) ⋅ = The velocity field of Problem 2.12 is u K y ⋅ 2 π ⋅ x 2 y 2 + ( ) ⋅ − = v K x ⋅ 2 π ⋅ x 2 y 2 + ( ) ⋅ = If the pathlines are correct we should be able to substitute x p and y p into the velocity field to find the velocity as a function of time: u K y ⋅ 2 π ⋅ x 2 y 2 + ( ) ⋅ − = K a ⋅ cos ω t ⋅ ( ) ⋅ 2 π ⋅ a 2 sin ω t ⋅ ( ) 2 ⋅ a 2 cos ω t ⋅ ( ) 2 ⋅ + ( ) ⋅ − = K cos ω t ⋅ ( ) ⋅ 2 π ⋅ a ⋅ − = (1) v K x ⋅ 2 π ⋅ x 2 y 2 + ( ) ⋅ = K a − sin ω t ⋅ ( ) ⋅ ( ) ⋅ 2 π ⋅ a 2 sin ω t ⋅ ( ) 2 ⋅ a 2 cos ω t ⋅ ( ) 2 ⋅ + ( ) ⋅ − = K sin ω t ⋅ ( ) ⋅ 2 π ⋅ a ⋅ − = (2) We should also be able to find the velocity field as a function of time from the pathline equations (Eq. 2.9):We should also be able to find the velocity field as a function of time from the pathline equations (Eq....
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics

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