4. Potential Flows Flashcards
Irrotational Flows
-flows with no vorticity, i.e. such that:
|ω = ∇x|u = 0
Velocity Potential
-if a velocity field |u is irrotational, i.e. if ∇x|u = 0, then there exists a velocity potential Φ(|x,t) such that:
|u = ∇Φ
Flows that are irrotational and incompressible
-if a flow is irrotational then it can be written in terms of a velocity potential:
|u = ∇Φ
-if in addition it is incompressible then the velocity potential Φ satisfies Laplace’s equation:
∇²Φ = 0
-hence, incompressible irrotational flows can be computed by solving Laplace’s equation and imposing appropriate boundary conditions on the solution
2D Incompressible Irrotational Flows
-both velocity potential Φ, and stream function ψ are solutions to Laplace’s equation;
∇²ψ = -ω = 0
AND
∇²Φ = 0
-however the boundary condition on ψ and Φ are different
Kinematic Boundary Conditions
-consider a flow past a solid body moving at velocity |U
-if ^n is the unit normal to the surface of the solid then, locally, the surface advances (moves in direction ^n) at the velocity:
(|U . ^n) ^n
-since the fluid cannot penetrate into the solid body, its velocity normal to the surface (|u.^n)^n must locally equal that of the solid;
|u . ^n = |U . ^n
-so since |u=∇Φ (for incompressible irrotational flow)
^n.∇Φ = ∂Φ/∂n = |U.^n
-the velocity satisfies Neumann boundary conditions at the solid body surface
Line Source / Sink
-consider an axisymmetric potential Φ=Φ(r)
-from Laplace’s equation in plane polar coordinates:
∇²Φ = 1/r d/dr(rdΦ/dr) = 0
<=>
dΦ/dr = m/r
-where m is a constant:
Φ(r) = mln(r) + C
-this potential produces the planar radial velocity:
|u = ∇Φ = m/r * ^er
-corresponding to a source for m>0 and a sink for m<0
-the constant C is arbitrary and doesn’t effect |u
-by convention m=Q/2π where Q is the flow rate
-this flow could be produced approximately using a perforated hose
Point Source / Sink
-consider a spherically symmetric potential Φ=Φ(r)
-from Laplace’s equation in spherical polar coordinates:
∇²Φ = 1/r² d/dr(r²*dΦ/dr)=0
<=>
dΦ/dr = m/r²
=>
Φ(r) = - m/r + C
-where m and C are integration constants
-this potential produces the 3D radial velocity:
|u = ∇Φ = m/r² * ^er
-corresponding to a source for m>0 and a sink for m<0
-by convention m=Q/4π where Q is the flow rate or volume flux
Line Vortex
-consider the potential Φ(θ) = kθ
-the solution to the Laplace equation in plane polar coordinates is:
ur = ∂Φ/∂r = 0
and
uθ = 1/r * ∂Φ/∂θ = k/r
-where the strength of flow k is a constant
-by convention k=Γ/2π where Γ is the circulation of the flow
-this represents a rotating fluid around a line vortex at r=0, it has zero vorticity but is singular at the origin
Uniform Stream
-for a uniform flow along the z axis, |u = (0,0,U) the velocity potential:
Φ(z) = Uz (+C)
Dipole (Doublet Flow)
Description
- since Laplace’s equation is linear, we can add two solutions together to form a new one
- a dipole is the superposition of a sink and a source of equal but opposite strength next to each other
Dipole (Doublet Flow)
3D Flow
-consider a point sink of strength -m at the origin and a point source of strength m at the positioni (0,0,𝛿)
-the velocity potential of the flow is formed by adding the potentials of the source and sink
-taking the limit as 𝛿->0 leads to the potential of a dipole:
Φ = - |μ .|r / r³
= |μ . ∇ (1/r)
-where |μ = μ ^ez is the dipole strength and |r=|x is the vector position
-fluid velocity for a dipole is then found using:
|u = ∇Φ
Dipole (Doublet Flow)
Planar Flow
-combining a line sink at the origin with a line source of equal but opposite strength at (𝛿,0) gives:
Φ = -m/2 * [ ln(x²+y²) - ln((x-𝛿)²+y²)]
-as in the 3D case, take the limit as 𝛿->0
-the expression for a 2D dipole of strength μ :
Φ = - |μ .|r / r²
= - |μ . ∇ (ln(r))
-where |μ= μ*^ex and |r=|x is the vector position
Uniqueness of Solutions to Laplace’s Equation
-given the value of the normal component of the fluid velocity, |u.|n on the surface S (i.e. the boundary condition), there exists a unique flow satisfying both ∇.|u=0 and ∇x|u=0 (i.e. incompressible and irrotational)
Properties of Laplace’s Equation
Identity From Vector Calculus
-let f(|x) be a function defined in a simply connected domain V with boundary S: ∇.(f∇f) = f∇²f + |∇f|² => ∫ ∇.(f∇f) dV = ∫ f∇²f dV + ∫ |∇f|²dV -where integrals are over volume V -using divergence theorem: ∫ (f∇f) dS = ∫ f∇²f dV + ∫ |∇f|²dV
Kelvin’s Minimum Energy Theorem
-of all possible fluid motions satisfying the boundary conditions for |u.|n on the surface S and ∇.|u in domain V, the potential flow is the flow with the smallest kinetic energy;
K = 1/2 ∫ρ ||u|² dV
