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Irrotational Flow (Chapter 10) Natalie Carroll, Ph.D. YDAE & ABE 1 Irrotational Flow of an Ideal Fluid Studied to examine: 1. Flow around corners 2. Flow over weirs 3. Flow through constructions 4. Flow around airfoils 5. Flow of water through the earth • Flow under a dam • Regional aquifers 2 Irrotational Flow Assumptions: 1. Ideal Flow • No friction between fluid and surface • No flow perpendicular to the boundary 2. Irrotational Flow • No rotation or distortion of fluid particles occurs during movement 3 Flow of an Ideal Fluid 2-D Flow Formulations: 1. Stream Function y 2y 2y 2 0 x 2 y 2. Velocity Potential Function f 2f 2f 2 0 x 2 y 4 • G.D.E.s are identical Dx Dy 1 G Q0 • B.C.s are different • Lines of constant y are ^ to lines of constant f 5 Streamline Formulation • Lines of constant y are Streamlines • Streamlines are tangent to the velocity: No flow ^ to a streamline • Volume Flow Rate between any pair of streamlines Qij y i y j • Velocity Components y y Vx Vy y x 6 Streamline Formulation Assumptions: – Motion of fluid does not penetrate into surrounding body or separate from the surface of the body leaving empty spaces. – No flow ^ to fixed boundary \ no velocity ^ to boundary – Fixed boundaries and lines of symmetry || to flow are streamlines 7 Example: Flow Around a Cylinder 8 Streamline Formulation • Flow around a cylinder example: – 2 lines of symmetry • Volume flow rate assuming unit thickness: Qtop ,bottom 5cm / sec 12cm 1cm 60cm / sec 3 • B.C.s: – Horizontal axis and cylinder boundary (symmetry) – Top boundary 9 Streamline Formulation Boundary Conditions: 1 Q 2 Line of Symmetry 10 Streamline Formulation Boundary Conditions: Line of Symmetry ^ to flow No flow ^ streamlines Vy = 0 11 Streamline Formulation Left and Right Boundaries: – Left boundary: Vx is uniform 5 cm/sec (applied as y) \ y Vy 0 x – Right boundary: Line of symmetry streamlines must be symmetrical about edge \ y Vy 0 x 12 Derivative B.C.s Review – Chr. 9 13 Derivative B.C.s Review – Chr. 9 • Theoretical boundary condition: f Dx Mf S x f Dx S x • Actual boundary condition f S=0 0 x M=0 14 Example Mesh: Flow Around a Cylinder Length – long enough for streamlines to become parallel 15 Corrections to Table 10.1 • Node 8 y = 10.8 • Node 30 x = 9.64 • Node 32 y = 10.9 16 Stream Function (Streamlines) (cm2/s) 17 Streamlines (showing symmetry) 18 Velocity Vx (cm/s) 19 Velocity -Vy (cm/s) 20 Stream Function Review 2y 2y 2 0 x 2 y [KD] [KG] [KM] {fQ} {fS} Dx=Dy=1 G=0 M=0 Q=0 S=0 21 Velocity Potential Formulation • Velocity for an irrotational flow can be written as the gradient of a scalar potential function. • Velocity Components f f Vx Vy x y • Adaptable to 3-D problems 22 Velocity Potential Formulation • Model – 2 Lines of Symmetry • B.C.s: – No velocity ^ flow • Top Boundary and Horizontal Line of Symmetry: f Vy 0 y • Cylinder: f 0 (velocity normal to surface) n 23 Potential Formulation 24 Velocity Potential Formulation • Left Boundary: f M=0 Vx 5cm / sec x S = -5 • Right Boundary: – Streamlines are ^ to line of symmetry – Potential lines are ^ to streamlines \ must be potential line 25 Potential Formulation Arbitrary Value 26 Potential Function (cm2/s) 27 Potential Function (symmetry) 28 Velocity Vx (cm/s) 29 Velocity Vy (cm/s) 30 Plot of Stream and Potential Functions 31 Velocity Potential Review 2f 2f 2 0 x 2 y [KD] [KG] [KM] {fQ} {fS} Dx=Dy=1 G=0 M=0 Q=0 S=-Vx 32 B.C.s Irregular Shape Must include downstream region until uniform velocity is attained 33 Example: Flow Around a Corner 8 cm 14 cm Vx = 5 cm/s 6 cm 18 cm 34 Example: Flow Around a Corner Y formulation f formulation 35 Groundwater Flow • G.D.E. Seepage of groundwater under a dam 2f 2f Dx 2 Dy 2 0 x y • G.D.E. Drawdown at a well removing water from an aquifer 2f 2f Dx 2 Dy 2 Q 0 x y 36 Seepage Under a Dam G.D.E.: 2f 2f G=0 Dx 2 Dy 2 0 x y Q=0 f is piezometric head (m) measured from bottom of confined aquifer Dx and Dy are the coefficients of permeability (m/day) 37 Seepage Under a Dam • Boundary Conditions: – Known values (f) beneath the water – Zero seepage conditions on other boundaries • Fluid Velocity Components (Darcy’s Law) f f Vx Dx Vy Dy x y 38 Seepage Under a Dam Dx, Dy Impermeable Layer 39 Seepage Under a Dam f=a f=b 40 Seepage Under a Dam f=a f=b f 0 n f 0 n 41 Seepage Under a Dam f=a f=b f f 0 f 0 n 0 n n f 0 n 42 Example: Problem 10.7 1m 30 m 50 m 100 m Impermeable 43 Piezometric Head (m) 44 Hydraulic Gradient in x-direction (m/m) 45 Hydraulic Gradient in y-direction (m/m) 46 Hydraulic Gradient (m/m) 47 Hydraulic Flux (m/day) 48 Seepage Under a Dam Review 2f 2f Dx 2 Dy 2 0 x y [KD] [KG] [KM] {fQ} {fS} Dx, Dy G=0 M=0 Q=0 S=0 f Vx Dx x f Vy Dy y 49 Drawdown at a Well G.D.E.: 2f 2f Dx 2 Dy 2 Q 0 x y f is piezometric head (m) measured from bottom of confined aquifer Dx and Dy are coefficients of permeability (m/day) Q represents the well (a point sink) 50 Drawdown at a Well • Boundary Conditions: – Known values on all or a part of the boundary – Seepage of water into aquifer along boundary f f Dx cos Dy sin S x y • Fluid Velocity Components (Darcy’s Law) f f Vx Dx Vy Dy x y 51 Example: Regional Aquifer 52 Example: Regional Aquifer 53 Example: Regional Aquifer Mesh Pump Location 54 Example: Regional Aquifer Mesh 55 Example: Piezometric Head Contours 56 Piezometric Head (m) 57 Velocity Vx (m/day) 58 Velocity Vy (m/day) 59 Velocity Vmax (m/day) 60 Drawdown at a Well Review 2f 2f Dx 2 Dy 2 Q 0 x y [KD] [KG] [KM] {fQ} {fS} Dx, Dy G=0 M=0 Q=Q S = 0,S f Vx Dx x f Vy Dy y 61

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posted: | 11/24/2011 |

language: | English |

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