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Irrotational Flow Principles and Elementary Plane Flows
- 1. Irrotational Flow(Potential Flow) • Bernoulli Equation • Velocity Potential • Two-Dimensional, Irrotational, Incompressible Flow • Elementary Plane Flows • Superposition of Elementary Plane Flows
- 2. Irrotational Flow Fluid elements moving in the flow field do not undergo any rotation 0,0 =×∇=ω V 0= ∂ ∂ − ∂ υ∂ = ∂ ∂ − ∂ ∂ = ∂ υ∂ − ∂ ∂ y u xx w z u zy w 0 111 = θ∂ ∂ − ∂ ∂ = ∂ ∂ − ∂ ∂ = ∂ ∂ − θ∂ ∂ θθ rzrr V rr rV rr V z V z VV r Rectangular coordinate Cylindrical coordinate
- 3. Bernoulli Equation If the flow is irrotational, ( )VVkgp ˆ1 ∇⋅=−∇ ρ − ( ) ( ) ( )VVVVVV ×∇×−⋅∇=∇⋅ 2 1 = 0 ( ) ( )2 2 1 2 1ˆ1 VVVkgp ∇=⋅∇=−∇ ρ − [ ] rd ⋅ dzkdyjdxird ˆˆˆ ++= z k y j x i ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ ˆˆˆ ( )2 2 1 Vdgdz dp =− ρ − 0 2 2 = ++ ρ gz Vp d constant 2 2 =++ ρ gz Vp valid between any two points in an irrotational flow 0=×∇ V
- 4. Example: Flow field with tangential motion - Forced Vortex ( )rfVVr == θand0 Forced Vortex (rigid body rotation) ( ) rrfV ω==θ ( ) θ∂ ∂ − ∂ ∂ =×∇=ω θ r zz V rr rV r V 11 2 1 2 1 ( ) ω=ω= ∂ ω∂ = r rr r r 2 2 11 2 1 2 a b ra rb ∫ ρω=∫ ∂ ∂ =− ar br ar brba rdrdr r p pp 2 ( ) 0 2 22 2 >− ρω = ba rr
- 5. Example: Flow field with tangential motion - Free Vortex ( )rfVVr == θand0 Free Vortex (irrotational vortex) ( ) r r r C rfV a ω ===θ 2 ( ) θ∂ ∂ − ∂ ∂ =×∇=ω θ r zz V rr rV r V 11 2 1 2 1 0 1 2 1 = ∂ ∂ = r C r a b ra rb ( )22 2 abba VVpp − ρ =− − ρ = 22 2 11 2 ab rr C thenIf 2ω= arC ( ) 0 2 222 2 >− ρω =− babba rrrpp
- 6. Velocity Potential If the flow is irrotational, a potential function, φ, can be formulated to represent the velocity field. From the vector identity, 0=φ∇×∇ The velocity field of an irrotational flow can be defined by a potential function so that φ−∇≡V z w yx u ∂ φ∂ −= ∂ φ∂ −=υ ∂ φ∂ −= z V r V r V zr ∂ φ∂ −= θ∂ φ∂ −= ∂ φ∂ −= θ 1
- 7. Stream Function and Velocity Potential For a two-dimensional, incompressible, irrotational flow, the velocity field can be expressed in terms of both ψ and φ. xy u ∂ ψ∂ −=υ ∂ ψ∂ = yx u ∂ φ∂ −=υ ∂ φ∂ −= According to the irrotationality condition, 0= ∂ ∂ − ∂ υ∂ y u x 02 2 2 2 = ∂ ψ∂ + ∂ ψ∂ yx According to the continuity equation, 0= ∂ υ∂ + ∂ ∂ yx u 02 2 2 2 = ∂ φ∂ + ∂ φ∂ yx } Laplace’s equation Solution of Laplace’s equation represents a possible 2- D, incompressible, irrotational flow field.
- 8. Slope of Velocity Potential Line Along a line of constant φ, dφ=0 and 0= ∂ φ∂ + ∂ φ∂ =φ dy y dx x d The slope of a line of constant φ is given by υ −= ∂φ∂ ∂φ∂ −= φ u y x dx dy uuy x dx dy υ = υ− −= ∂ψ∂ ∂ψ∂ −= ψ 1−= υ × υ −= × ψφ u u dx dy dx dy The slope of a line of constant ψ is given by Lines of constant ψ and constant φ are orthogonal.
- 9. Example: Determine the Velocity Potential Line of a Flow The flow streamline function is .2axy=ψ ∂ ψ∂ ∂ ∂ − ∂ ψ∂ − ∂ ∂ = ∂ ∂ − ∂ υ∂ =ω yyxxy u x z2 ( ) ( ) 022 = ∂ ∂ −− ∂ ∂ = ax y ay x ay x ax y u 22 −= ∂ ψ∂ −=υ= ∂ ψ∂ = The flow is irrotational. In term of velocity potential, the velocity components are ay y ax x u 22 −= ∂ φ∂ −=υ= ∂ φ∂ −= ( ) ( ) 2 22 22 caxxgax dx dg uxgayay y +−=⇒=−=⇒+=φ⇒−= ∂ φ∂ −=υ ( ) ( ) 1 22 22 cayyfay dy df yfaxax x u +=⇒−=−=υ⇒+−=φ⇒= ∂ φ∂ −= caxay +−=φ 22
- 10. Elementary Plane Flows Uniform Flow 3c=ψ 2c=ψ 1c=ψ 0=ψ 1c−=ψ 2c−=ψ 3c−=ψ =φ 3k 2k 1k 0 1k− 2k− 3k− U x y x y Uu = 0=υ Uy=ψ U y u = ∂ ψ∂ = 0= ∂ ψ∂ −=υ x Ux−=φ U x u = ∂ φ∂ −= 0= ∂ φ∂ −=υ y Γ=0 around any closed curve
- 11. Inclined uniform flow =φ 3k 2k 1k 0 1k− 2k− 3k− x yα= cosVu α=υ sinV ( ) ( )xVyV α−α=ψ sincos α= ∂ ψ∂ = cosV y u α= ∂ ψ∂ −=υ sinV x ( ) ( )xVyV α−α−=φ cossin α= ∂ φ∂ −= cosV x u α= ∂ φ∂ −=υ sinV y V x y α 0 3c 2c 1c 1c− 2c− 3c− =ψ Γ=0 around any closed curve
- 12. Source Flow r q Vr π = 2 0=θV θ π =ψ 2 q r q ln 2π −=φ x y 3c=ψ 2c=ψ 1c=ψ 0=ψ4c=ψ 5c=ψ 6c=ψ 7c=ψ 1k−=φ 2k−=φ x y Origin is singular point q: volume flow rate per unit depth Γ=0 around any closed curve
- 13. Sink Flow r q Vr π −= 2 0=θV θ π −=ψ 2 q r q ln 2π =φ x y 3c−=ψ 2c−=ψ 1c−=ψ 0=ψ4c−=ψ 5c−=ψ 6c−=ψ 7c−=ψ 1k=φ 2k=φ x y Origin is singular point q: volume flow rate per unit depth Γ=0 around any closed curve
- 14. Irrotational Vortex x y 3c−=ψ 2c−=ψ 1c−=ψ 4c−=ψ 3k−=φ 2k−=φ 1k−=φ 0=φ4k−=φ 5k−=φ 6k−=φ 7k−=φ x y r K V π =θ 2 0=rV θ π −=φ 2 K r K ln 2π −=ψ Origin is singular point K: strength of the vortex Γ=K around any closed curve enclosing origin Γ=0 around any closed curve not enclosing origin
- 15. Doublet 3c−=ψ 2c−=ψ 1c−=ψ 0=ψ 1c=ψ 2c=ψ 3c=ψ 2k=φ 1k=φ 1k−=φ 2k−=φ x y x y θ Λ −=θ sin2 r Vθ Λ −= cos2 r Vr r θΛ −=φ cos r θΛ −=ψ sin Origin is singular point Λ: strength of the doublet Γ=0 around any closed curve
- 16. Superposition of Elementary Plane Flows ∀ φ and ψ satisfy Laplace’s equation. • Laplace’s equation is linear and homogeneous. • Solutions of Laplace’s equation may be added together (superposed) to develop more complex flow patterns. • If ψ1 and ψ2 satisfy Laplace’s equation, so does ψ3 = ψ1 + ψ2 . • Any streamline contour can be imagined to represent a solid surface (there is no flow across a streamline).
- 17. Source + Uniform Flow θ+θ π =+θ π =ψ+ψ=ψ sin 22 flowuniformsource Ur q Uy q θ− π −=− π −=φ+φ=φ cosln 2 ln 2 flowuniformsource Urr q Uxr q x y U Stagnation point Flow past a half body Solid surface formed by two streamlines
- 18. Source + Sink Flow ( )2121sinksource 222 θ−θ π =θ π −θ π =ψ+ψ=ψ qqq 1 2 21sinksource ln 2 ln 2 ln 2 r rq r q r q π = π + π −=φ+φ=φ Source and sink with equal strength, origin separated 2a apart
- 19. Source + Sink+Uniform Flow ( ) θ+θ−θ π =+θ π −θ π =ψ+ψ=ψ sin 222 2121flowuniformsource Ur q Uy qq θ− π =− π + π −=φ+φ=φ cosln 2 ln 2 ln 2 1 2 21flowuniformsource Ur r rq Uxr q r q x y U Stagnation point (Flow past a Rankine body) Solid surface formed by two streamlines Stagnation point
- 20. Doublet+Uniform Flow Λ −θ=θ+ θΛ −=+ θΛ −=ψ+ψ=ψ 2flowuniformdoublet 1sinsin sinsin r U UrUr r Uy r Λ +θ−=θ− θΛ −=− θΛ −=φ+φ=φ 2flowuniformdoublet 1coscos coscos r U UrUr r Ux r U Stagnation point Flow past a cylinder Solid surface formed by two streamlines Stagnation point a U a Λ =
- 21. Doublet+Vortex+Uniform Flow r K r a UrUrr K r ln 2 1sinsinln 2 sin 2 2 flowuniformvortexdoublet π + −θ=θ+ π + θΛ −=ψ+ψ+ψ=ψ θ π + +θ−=θ−θ π + θΛ −=φ+φ+φ=φ 2 1coscos 2 cos 2 2 flowuniformvortexdoublet K r a UrUr K r a U a Λ = Flow past a cylinder with circulation UaUK Λπ=π< 44
- 23. Vortex Pair 1 2 21vortex2vortex1 ln 2 ln 2 ln 2 r rK r K r K π = π + π −=ψ+ψ=ψ ( )1221vortex2vortex1 222 θ−θ π =θ π +θ π −=φ+φ=φ KKK
- 24. Example: Flow past a cylinder Λ −θ=ψ+ψ=ψ 2flowuniformdoublet 1sin r U Ur Λ +θ−=φ+φ=φ 2flowuniformdoublet 1cos r U Ur U Stagnation point (a,π) Stagnation point (a,0) a U a Λ = −θ= +θ ∂ ∂ = ∂ φ∂ −= 2 2 2 2 1cos1cos r a U r a Ur rr Vr +θ−= +θ θ∂ ∂ −= θ∂ φ∂ −=θ 2 2 2 2 1sin1cos 11 r a U r a Ur rr V
- 25. Example: Flow past a cylinder- pressure distribution on the surface U a U a Λ = −θ= 2 2 1cos r a UVr +θ−=θ 2 2 1sin r a UV ∞p gz V pgz U p ++=++∞ 22 22 ] ] θ=+= =θ= 22222 sin4UVVV arrar = 0 ( ) ( ) ( )θ−ρ=θ−ρ=−ρ=− ∞ 2222222 sin41 2 1 sin4 2 1 2 1 UUUVUpp θ−= ρ − ∞ 2 2 sin41 2 1 U pp
- 26. Pressure distribution on the surface of the cylinder 2 2 1 U pp ρ − ∞ θ θ−= ρ − ∞ 2 2 sin41 2 1 U pp
- 27. Pressure drag force acting on the cylinder U ∞p ApdFd = a − ρ += ∞ 2 22 1 2 U VU pp ] ( )∫ π ∞ θθ θ−ρ+−= 2 0 22 pressure cossin41 2 1 dabUpF Drag ] ππ π ∞ θρ+ θρ−θ−= 2 0 32 2 0 22 0 sin 3 4 2 1 sin 2 1 sin abUabUabp 0= θ ] ( ) ∫∫ π θθ−=θ−= 2 0 pressure coscos dpabpdAF ADrag
- 28. Lift force acting on the cylinder U ∞p ApdFd = a ] ( ) ∫∫ π θθ−=θ−= 2 0 pressure sinsin dpabpdAF ALift ] ( )∫ π ∞ θθ θ−ρ+−= 2 0 22 pressure sinsin41 2 1 dabUpF Drag ] ππ π ∞ θ− θ ρ+ θρ+θ= 2 0 3 2 2 0 22 0 cos4 3 cos4 2 1 cos 2 1 cos abUabUabp 0= θ
- 29. Example: Flow past a cylinder with circulation r K r a Ur ln 2 1sin 2 2 flowuniformvortexdoublet π + −θ=ψ+ψ+ψ=ψ θ π + +θ−=φ+φ+φ=φ 2 1cos 2 2 flowuniformvortexdoublet K r a Ur a U a Λ = UaUK Λπ=π< 44 −θ−= ∂ φ∂ −= 2 2 1cos r a U r Vr r K r a U r V π − +θ−= θ∂ φ∂ −=θ 2 1sin 1 2 2 Stagnation points ] a K UV ar π −θ−==θ 2 sin2 4 sinandat0 1- π =θ==θ Ua K arV
- 30. Flow past a cylinder with circulation - Surface pressure distribution a U a Λ = UaUK Λπ=π< 44 ] a K UV ar π −θ−==θ 2 sin2 ] ( )] 2 222 2 sin2 π −θ−=+= =θ= a K UVVV arrar ] 01cos 2 2 = −θ−== a a UV arr − ρ += ∞ 2 22 1 2 U VU pp 2 2 2 2 2 sin 2 sin4 π +θ π +θ= Ua K Ua K U V π −θ π −θ− ρ += ∞ 2 2 2 2 sin 2 sin41 2 Ua K Ua KU pp
- 31. Pressure distribution on the surface of the cylinder 2 2 1 U pp ρ − ∞ θ aUK π= aUK π= 2 aUK π= 3 π −θ π −θ− ρ += ∞ 2 2 2 2 sin 2 sin41 2 Ua K Ua KU pp
- 32. Pressure drag force acting on the cylinder ∞p − ρ += ∞ 2 22 1 2 U VU pp ] ∫ π ∞ θθ π −θ π −θ−ρ+−= 2 0 2 22 pressure cos 2 sin 2 sin41 2 1 dab aU K aU K UpF Drag ApdFd = θ ] ( ) ∫∫ π θθ−=θ−= 2 0 pressure coscos dpabpdAF ADrag a ( )∫ π ∞ θθ θ−ρ+−= 2 0 22 cossin41 2 1 dabUp ∫ π θθ π +θ π ρ+ 2 0 2 2 cos 2 sin 2 2 1 dab Ua K Ua K U 0= 0sin 22 sin2 2 0 22 = θ π + θ π = π ab Ua K Ua Kab
- 33. Lift force acting on the cylinder ∞p − ρ += ∞ 2 22 1 2 U VU pp ] ∫ π ∞ θθ π −θ π −θ−ρ+−= 2 0 2 22 pressure sin 2 sin 2 sin41 2 1 dab aU K aU K UpF Lift ApdFd = θ ] ( ) ∫∫ π θθ−=θ−= 2 0 pressure sinsin dpabpdAF ALift a ( )∫ π ∞ θθ θ−ρ+−= 2 0 22 sinsin41 2 1 dabUp ∫ π θθ π +θ π ρ+ 2 0 2 2 sin 2 sin 2 2 1 dab Ua K Ua K U 0= ρ = π π ρ = θ π + θ − θ π = π b U KU b U KU ab Ua K Ua Kab 2 22 2 2 cos 24 sin 2 2 22 2 0 22
- 34. Circulation around on the cylinder ∞p ( ) ∫∫∫ π θθθ π θ π θθθ ⋅θ+⋅θ=θ⋅+=Γ 2 0 2 0 2 0 ˆˆˆˆˆˆˆ eerdVeerdVerdeVeV rrrrr ∫ ⋅≡Γ sdV a ∫ π θ π −θ−= 2 0 2 sin2 rd Ur K U 0= r K UV π −θ−=θ 2 sin2 −θ−= 2 2 1cos r a UVr ] K K Ur −= θ π −θ= π π 2 0 2 0 2 cos2 ρ= U K U b FLift 2 2 1 2 Γρ−=ρ= UUK
- 35. Home work: 6.51, 6.66, 6.76, 6.93, 6.96, 6.98