Offshore Hydromechanics Module 1

Transcription

1 Offshore Hydromechanics Module 1 Dr. ir. Pepijn de Jong 4. Potential Flows part 2

2 Introduction Topics of Module 1 Problems of interest Chapter 1 Hydrostatics Chapter 2 Floating stability Chapter 2 Constant potential flows Chapter 3 Constant real flows Chapter 4 Waves Chapter 5 2

3 Learning Objectives Chapter 3 Understand the basic principles behind potential flow To schematically model flows applying basic potential flow elements and the superposition principle To perform basic flow computations applying potential flow theory 3

4 Fluid Mechanics Laws Summarizing: Conservation of mass (continuity): Incompressible flow: ρ t + ρv = 0 u x + v y + w z = 0 V = 0 Conservation of momentum (inviscid flow): ρ DV = ρg p Dt ω x = 1 2 Rotation of a fluid element: w y v z ω y = 1 2 ω z = 1 u z w 2 x v x u y Vorticity: ζ = 2ω = V 4

5 Fluid Mechanics Laws Velocity Potential Assumptions Homogeneous Continuous Incompressible Non-viscous (inviscid) Irrotational flow The velocity potential is a function of time and position: Φ x, y, z, t The spatial derivatives of the velocity potential equal the velocity components at a time and position: Φ x = u Φ y = v Φ z = w 5

6 Fluid Mechanics Laws Velocity Potential Continuity equation for potential flow: 2 Φ x Φ y Φ z 2 = 0 2 Φ = 0 Laplace equation Irrotationality v x = u y in the (x,y) plane ω x = 1 2 w y v z w y = v z in the (y,z) plane ω y = 1 2 u z w x u z = w x in the (x,z) plane ω z = 1 2 v x u y 6

7 Fluid Mechanics Laws Bernoulli Equation Potential flow: V = Φ Φ t Φ 2 gz + p ρ = constant Bernoulli equation More general: also valid on a streamline 7

8 Potential Flow Stream funtion (2D) Ψ is the (2D) stream function, with: dψ dy = u dψ dx = v Difference of Ψ streamlines Orthogonality: between neighboring stream lines: rate of flow between dψ dy = dφ dx = u dψ dx = dφ dy = v Impervious boundaries equals streamline: dφ dn = 0 Ψ = constant 8

9 Potential flow elements Introduction Using the previous we can define 'flow elements' We can add these elements up to construct realistic flow patterns Modeling of submerged bodies by matching streamlines to body outline Using the velocity potential, stream function and Bernoulli equation to find velocities, pressures and eventually fluid forces on bodies Discussed: Uniform flow element Source/sink element Doublet element Vortex element 9

10 Potential flow elements Uniform flow y y U U Φ = U x x Φ = U x x Ψ = U y Ψ = U y u = dφ dx = dψ dy = U u = dφ dx = dψ dy = U 10

11 Potential flow elements Source and sink flow y y so r θ x si r θ x Φ = + Q 2π lnr = + Q 2π ln x2 + y 2 Φ = Q 2π lnr = Q 2π ln x2 + y 2 Ψ = + Q 2π θ = + Q 2π arctan y x Ψ = Q 2π θ = Q 2π arctan y x 11

12 Potential flow elements Source and sink flow y Φ = + Q 2π lnr Ψ = + Q 2π θ v r = Φ r = 1 r Ψ θ = Q 2πr so r θ x v θ = 1 r Φ θ = Ψ r = 0 12

13 Potential flow elements Circulation or vortex elements y Φ = + Γ 2π θ Ψ = + Γ 2π lnr v r = Φ r = 1 r Ψ θ = 0 Γ r θ x v θ = 1 r Φ θ = Ψ r = Γ 2πr 13

14 Potential flow elements Circulation or vortex elements Φ = + Γ 2π θ v r = Φ r = 1 r Ψ θ = 0 v θ = 1 r Φ θ = Ψ r = Ψ = + Γ 2π lnr Γ 2πr Circulation strenght constant: y Γ r θ x Γ = v θ ds = 2πr v θ = constant Therefore: no rotation, origin singular point: velocity infinite 14

15 Methodology (source in positive uniform flow) The resulting velocity fields, potential fields or stream function fields may be simply superposed to find the combined flow patterns (Using stream function values) Uniform flow field Source flow field 15

16 Methodology (source in positive uniform flow) The resulting velocity fields, potential fields or stream function fields may be simply superposed to find the combined flow patterns (Using stream function values) 16

17 Methodology (source in positive uniform flow) The resulting velocity fields, potential fields or stream function fields may be simply superposed to find the combined flow patterns (Using stream function values) 17

18 Sink in negative uniform flow Besides graphically this works also with formulas: Ψ = Q 2π arctan y x U y Φ = Q 2π ln x2 + y 2 U x For instance: Find location stagnation point (Blackboard...) 18

19 Separated source and sink Ψ source = + Q 2π θ 1 = + Q 2π arctan y x 1 Ψ sink = Q 2π θ 2 = Q 2π arctan y x 2 Ψ = Q 2π arctan 2ys x 2 + y 2 s 2 x 1 x 2 2s 19

20 Separated source and sink in uniform flow s s Ψ = Q 2π arctan 2ys x 2 + y 2 s 2 + U y 20

21 Separated source and sink in uniform flow s s Streamline resembles fixed boundary (Rankine oval) The flow outside this streamline resembles flow around solid boundary with this shape Shape can be changed by using more source-sinks along x-axis with different strenghts 21

22 Separated source and sink in uniform flow s s This approach can be extended to form ship forms in 2D or 3D: Rankine ship forms Useful for simple flow computations 22

23 Potential flow elements Doublet or dipole When distance 2s becomes zero a new basic flow element is produced: Doublet or dipole producing flow in positive x-direction Q Ψ = lim s 0 2π arctan 2ys x 2 + y 2 s 2 Ψ = lim s 0 Q π s y x 2 + y 2 s 2 s s Note: in book errors wrt to doublet and its orientation! 23

24 Potential flow elements Doublet or dipole When distance 2s becomes zero a new basic flow element is produced: Doublet or dipole producing flow in positive x-direction Q Ψ = lim s 0 2π arctan 2ys x 2 + y 2 s 2 Ψ = lim s 0 Q π s y x 2 + y 2 s 2 Set constant: μ = Q π s 24

25 Potential flow elements Doublet or dipole When distance 2s becomes zero a new basic flow element is produced: Doublet or dipole producing flow in positive x-direction Q Ψ = lim s 0 2π arctan 2ys x 2 + y 2 s 2 Ψ = lim s 0 Q π s y x 2 + y 2 s 2 Disappears when: s 0 Set constant: μ = Q π s 25

26 Potential flow elements Doublet or dipole When distance 2s becomes zero a new basic flow element is produced: Doublet or dipole producing flow in positive x-direction Ψ = lim s 0 Q 2π arctan 2ys x 2 + y 2 s 2 Ψ = μ Φ = μ y x 2 + y 2 = μ sinθ r x cosθ x 2 + y2 = μ r 26

27 Doublet in a uniform flow Φ = μ x x 2 + y 2 U x Ψ = μ y x 2 + y 2 U y Φ = μ cosθ r U rcosθ Wrong in book! Ψ = μ sinθ r U rsinθ Doublet pointing in positive x-direction, uniform flow in negative x- direction: + 27

28 Doublet in a uniform flow Φ = μ Φ = μ cosθ r x x 2 + y 2 U x U rcosθ Ψ = μ Ψ = μ sinθ r y x 2 + y 2 U y U rsinθ Set Ψ = 0 then: True when: Ψ = y μ x 2 + y 2 U = 0 y = 0 μ x 2 + y 2 U = 0 x 2 + y 2 = μ U 28

29 Doublet in a uniform flow: flow around a circle The radius of the circle: R = μ U Doublet strength needed for radius R: Φ = μ cosθ r Ψ = μ sinθ r U rcosθ U rsinθ μ = U R 2 This yields the following: Φ = U R 2 cosθ r U rcosθ = RU R r + r R cosθ Ψ = U R 2 sinθ r U rsinθ = RU R r r R sinθ 29

30 Doublet in a uniform flow: flow around a circle Φ = RU Φ = U R 2 R r + r R cosθ x x 2 + y 2 U x u = dφ dx = U R 2 x2 y 2 x 2 + y 2 2 U 30

31 Doublet in a uniform flow: flow around a circle Φ = RU Φ = U R 2 R r + r R cosθ x x 2 + y 2 U x u = dφ dx = U R 2 x 2 y 2 x 2 + y 2 2 U x = ±R, y = 0 u = dφ dx = U R 2 R2 0 2 R U = U R 2 R2 R 4 U = 0 Stagnation points! 31

32 Doublet in a uniform flow: flow around a circle Φ = RU Φ = U R 2 R r + r R cosθ x x 2 + y 2 U x u = dφ dx = U R 2 x 2 y 2 x 2 + y 2 2 U x = 0, y = ±R u = dφ dx = U R R R 2 2 U = U R 2 R2 R 4 U = 2U 32

33 Evaluate velocities on cylinder wall Generally, velocity on cylinder wall: Ψ = μ sinθ r U rsinθ v θ r = R = Ψ r r=r = r U R 2 sinθ r U rsinθ =... = 2U sinθ 33

34 Evaluate pressures on cylinder wall Use the Bernoulli equation: 1 2 ρu = p ρv θ 2 v θ = 2U sinθ Pressure at stagnation points: Pressure at cylinder boundary: v = 0 v r = 0 Assuming constant elevation Result: p = 1 2 ρu 2 1 4sin 2 θ 34

35 Evaluate pressures on cylinder wall Velocity profile: Pressure profile: v θ = 2U sinθ p = 1 2 ρu 2 1 4sin 2 θ stagn.pt stagn.pt stagn.pt 35

36 Evaluate pressures on cylinder wall Velocity profile: Pressure profile: v θ = 2U sinθ p = 1 2 ρu 2 1 4sin 2 θ No net resulting force!!! D'Alembert's Paradox stagn.pt stagn.pt stagn.pt 36

37 Pipeline near seabed How to calculate the flow around a pipeline near the seabed? y U y 0 37

38 Pipeline near seabed Mirror flow in seabed! Superpose flows: undisturbed cylinder mirrored cylinder y U y 0 y 0 38

39 Pipeline near seabed Mirror flow in seabed! Superpose flows: undisturbed cylinder mirrored cylinder y U Result is zero normal flow on seabed y 0 y 0 39

40 Pipeline near seabed Mirror flow in seabed! Superpose flows: undisturbed cylinder mirrored cylinder y U Result is zero normal flow on seabed y 0 Is flow exactly right? y 0 40

41 Pipeline near seabed Mirror flow in seabed! Superpose flows: undisturbed cylinder mirrored cylinder y U Result is zero normal flow on seabed y 0 Is flow exactly right? NO: interaction cylinders not modeled! y 0 41

42 Circulation Add circulation (or vortex flow element) to doublet in uniform flow Resulting velocity field: U r + Γ θ + 42

43 Circulation Add circulation (or vortex flow element) Resulting velocity field: 43

44 Circulation Add circulation (or vortex flow element) Resulting velocity field: 44

45 Circulation Add circulation (or vortex flow element) Resulting velocity field:

46 Circulation Now integration of pressure yields a net force perpendicular to the undisturbed flow direction: the lift force However: still no net force in the flow direction: no drag 46

47 Sources images All images are from the book Offshore Hydromechanics by Journée and Massie. 47