Exercise 8: Vorticity, Bernoulli and Stream ... - KTH Mechanics

yFigure 1: Streamlines above a hill with h = 100m and U ∞ = 5m/s. A paraglider pilot with a sink of 1m/swill find lift in the area within the dotted line, while soaring along the hill.xThe stream function satisfies continuity:The flow is irrotational:u r = 1 r∂ψ∂θ ,u θ = − ∂ψ∂rω = ∇ × ū = 0 ⇒ ∂ψ∂r + ψr∂2 ∂r 2 + 1 ∂ 2 ψr ∂θ 2 = 0 (1)Introduce the ansatz ψ = f(r)sin θ into equation (1):f ′ sin θ + rf ′′ sinθ − 1 r f sin θ = 0⇒Make the ansatz f = r n :So we havef ′ + rf ′′ − 1 r f = 0nr n−1 + rn(n − 1)r n−2 − 1 r rn = 0n + n 2 − n − 1 = 0 ⇒ n = ±1ψ =(Ar + B )sin θr⇒We need two boundary conditions.1. Free stream:Ar sin θ = U ∞ r sin θ ⇒ A = U ∞2. Streamline on the hill surface:U ∞ h + B h = 0 ⇒ B = −U ∞h 2So we have:( )ψ = U ∞ r − h2sinθrNow we can calculate the velocity field above the hill:u r = 1 )∂ψr ∂θ = U ∞(1 − h2r 2 cos θu θ = − ∂ψ)∂r = −U ∞(1 + h2r 2 sin θ2

) Derive an equation for the curve with constant vertical wind velocity V .Constant vertical wind velocity is described by:( )( )V = u r sin θ+u θ cos θ = U ∞ 1 − h2r 2 cos θ sinθ−U ∞ 1 + h2h 2r 2 sin θ cos θ = −2U ∞ sin θ cos θr2 ⇒√r = h −2 U ∞sin θ cos θVc) Assume that the density ρ and the gravitational acceleration g is constant. Calculate the atmosphericpressure at the top of the hill.Use the Bernoulli equation (valid everywhere) with free stream pressure p 0 at the ground:p o + 1 2 ρU2 ∞ = p + 1 2 ρ(2U ∞) 2 + ρgh⇒ p = p o − 3 2 ρU2 ∞ − ρghStokes stream functionConsider a 2D incompressible flowDefine the stream function Ψ such that:∇ · ū = 0oru = ∂Ψ∂y∂u∂x + ∂v∂y = 0.v = − ∂Ψ∂xThis means∂u∂x + ∂v∂y = ∂2 Ψ∂x∂y − ∂2 Ψ∂y∂x = 0,so continuity is always fulfilled. Now we can writeē x ē y ē z( )ū = ∇ × Ψē z =∂ ∂ ∂∂Ψ∂x ∂y ∂z∣ 0 0 Ψ ∣ = ∂y , −∂Ψ ∂x ,0AndFor irrotational flowē x ē y ē z(¯ω = ∇ × ū =∂ ∂ ∂∂x ∂y ∂z∣ ∂Ψ ∣ = 0,0, − ∂Ψ )∂x 2 − ∂2 Ψ∂y 2 = −∇ 2 Ψē z∂y− ∂Ψ∂x0∆Ψ = ∇ 2 Ψ = 0In spherical coordinates for axisymmetrical flow, define ΨIncompressibility is still validVelocityVorticityIrrotationalu r =1 ∂Ψr 2 sinθ ∂θū = ∇ ×∇ · ū = 0u θ = − 1 ∂Ψr sin θ ∂rΨr sinθēθ¯ω = − 1 [ ∂ 2 Ψr sin θ ∂r 2 + sin θr 2∂ 2 Ψ∂r 2 + sinθr 2(∂ 1∂θ sin θ(∂ 1∂θ sin θ)∂Ψ= 0∂θ)]∂Ψ∂θ3