Continuum Mechanics Lecture 5 Ideal fluids

Transcription

1 Continuum Mechanics Lecture 5 Ideal fluids Prof.

2 Outline - Helmholtz decomposition - Divergence and curl theorem - Kelvin s circulation theorem - The vorticity equation - Vortex dynamics and vortex flow - Bernoulli theorem and applications

3 Helmholtz decomposition of the velocity field For a continuous and differentiable velocity field, we have the following unique decomposition: v = φ + A with Gauge condition A =0 The scalar and vector potential are solutions of φ = v and with appropriate boundary conditions. The source terms for these 2 Poisson equations are respectively v : the divergence of the velocity field v : the curl of the velocity field A = v Limiting cases: 1- v =0 for an incompressible flow. The velocity field is solenoidal or divergence free. 2- v =0 for a potential flow, because in this case the velocity field derives from a scalar potential. The velocity is said to be curl free.

4 Physical interpretation of the divergence We have seen in the previous lecture that the variation of a Lagrangian volume is given by dv t dt = dx 3 v V t The rate of change of the specific volume V =1/ 1 DV is V Dt = v Using the divergence theorem, we can express the total volume variation as the net flux of volume across the outer surface as: dv t dt = v nds S t Let s consider the case of a point source (or sink) of divergence at r=0. v = Q δ(x = 0) We have a spherically symmetric velocity field around the source. Using the divergence theorem, we have: Q =4πr 2 v r A source (or sink) velocity field is thus v = Q 4πr 2 e r

5 Physical interpretation of the curl We define the vorticity as the following vector field y v z z v y In components, we have ω = z v x x v z x v y y v x ω = v For a rigid body motion v = v 0 + Ω r, we have ω = Ω r =2 Ω The vorticity is thus twice the local rotation rate in the fluid. A vortex is a vorticity line along the axis Let us consider a vortex line at r=0. A vortex velocity field is thus v = Ω z 2πr e θ t = ω / ω We introduce Stoke s theorem or curl s theorem. We define the circulation Γ as the integral of the parallel velocity along a closed contour. We have the following identity Γ = v d l = Using Stoke s theorem, we have L S Γ = L ( v) nds v = Ωz e z δ(r = 0) Γ =2πrv θ = Ω z v d l = L v dl

6 Velocity field induced by a vortex distribution This is the Biot-Savart law for vortices. We consider an incompressible fluid for which We know from the Helmholtz decomposition that together with the Gauge condition A =0 The potential vector satisfies the Poisson equation A = ω v(x) = 1 x x 4π V x x 3 ω (x )dx 3 The solution is We need to add the scalar potential contribution, solving φ =0 with the appropriate boundary conditions (see lecture on potential flows). We consider a filament of vorticity coordinate s. v(x) = Ω 4π For a vertical filament, we have v(x) = Ω 4π + v =0 v = A + φ ω (x )=Ωδ(r = 0) using the curvilinear x x L x x 3 t(x )ds x x rdz (r 2 + z 2 ) e 3/2 θ = Ω 2πr e θ = re r +(z z )e z with and t = e z e r e z = e θ

7 Kelvin s circulation theorem L t = φ (t,0) (L 0 ) L 0 We consider a closed contour evolving with the flow. We use the inverse Lagrangian mapping to compute the time derivative of the circulation. d dt Γ(t) = d v dx = d v(x(y, t),t) ( x dt L t dt L 0 y dy) d v 2 v dv = =0 dt Γ(t) = Dv L 0 Dt ( x y dy)+ v ( v L 0 y dy) = Dv L t Dt dx L 0 2 Dv Dt = F 1 We now inject the Euler equation for an ideal fluid P d dt Γ(t) = 1 F dx + P nds L t S t 2 If the external force derives from a potential F = Φ 1 d and if the fluid is barotropic P = Π then dt Γ =0 Lagrange theorem: if initially the vorticity is zero, then it remains zero everywhere.

8 The vorticity equation Dv Dt = F 1 We start with the Euler equation for ideal fluids P ω t + (v v) = F ( 1 Taking the curl leads to P ) Using the identity v v = ( v2 )+ω v we have 2 ω t + (ω v) = F + 1 P 2 Using the identity (ω v) =( v)ω +(v )ω (ω )v we find the vorticity equation: Dω Dt =(ω )v ( v)ω + F + 1 P 2 D ω ω = Dt For a barotropic fluid under gravity, we have v Helmholtz theorem: vortex lines move with the fluid. Proof: a line element that moves with the fluid satisfies D Dt δ =(δ )v

9 Vortex dynamics For a barotropic fluid, the vorticity equation writes in component form: Dω i Dt = ω x x v i + ω y y v i + ω z z v i ( v)ω i Let s consider a vertical vortex line ω = ω z e z Dω x Dt = ω z z v x Dω y Dt = ω z z v y vortex tilting due to shear Dω z Dt = ω z ( x v x + y v y ) vortex stretching due to 2D divergence The 2D divergence is the rate of change of the section of the vortex tube 1 DS S Dt =( xv x + y v y ) For a 2D velocity field, the total vorticity in the vortex tube is conserved. ω z S = constant

10 D Dt = v D Dt = P v Dv Dt = Φ 1 P Multiplying by velocity and defining the specific enthalpy as h = + P D v 2 = v Φ v Dt 2 P D Dt (Φ) = Φ t + v Φ First Bernoulli Theorem We start with the Euler equations in Lagrangian form with equations for the thermodynamical variables and D Dt (h) = 1 P t + v P Collecting everything, we have the following relation: D v 2 Dt 2 + Φ + h = 1 P t + Φ t Theorem follows trivially: in a stationary flow, the total enthalpy is conserved along streamlines. Validity: no viscosity, no dissipation (reversible isentropic flow), we have H = v2 2 + Φ + h

11 We consider a curl free flow v = φ in a barotropic fluid Using the now well known vector relation v v = ( v2 2 φ the Euler equation becomes t + v2 2 + Φ + Π =0 The theorem follows: Second Bernoulli Theorem φ t + v2 2 + Φ + Π = C(t) 1 P = Π )+ω v For a potential flow, we have everywhere in the flow (not only along streamlines): The constant depends only on time. The flow doesn t have to be stationary. For a stationary flow, the quantity H = v2 2 + Φ + Π For a curl free incompressible fluid, we have H = v2 2 + Φ + P is uniform everywhere.

12 Application of the Bernoulli Theorem: Pitot tube Ram pressure We would like to measure the velocity of the fluid at infinity. We consider a probe with section AC equal to section ED. The flow is stationary and incompressible: Mass conservation implies v A S A = v D S D 2 + P = constant so that v A = v D = v Point B, however, is a stagnation point with v B =0. We conclude that P B = v2. Using the probe, we measure P = P B P 2 + P 2 P The velocity is just v = and P is called the ram pressure. These probes (also called Pitot tube) are used in planes to measure the velocity. v2

13 Application of the Bernoulli Theorem: Venturi tube We would like to measure the incoming velocity in a pipe. We modify slightly the section of the pipe around point B. Mass conservation implies v A S A = v B S B. Bernoulli theorem implies v2 B 2 + P B = v2 A 2 + P A Assuming that S B = S A (1 ), if we measure P, we have: P v A = This probe is called a Venturi tube.

14 Hugoniot theorem For an stationary incompressible fluid, mass conservation implies vs = constant. dv If the section decreases, the velocity increases. v = ds S For a compressible fluid, we now have vs = constant. dv v + d = ds S The stationary Euler equation gives us vdv = 1. dp Introducing the sound speed c 2 = dp, d dv combining the 2 equations results in 1 v2 v c 2 = ds S The dimensionless number M = v is called the Mach number of the flow. c If M < 1, the fluid behaves qualitatively like an incompressible fluid. If M > 1, it is reversed: the velocity will increase if the section increases.