Webster s horn model on Bernoulli flow
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1 Webster s horn model on Bernoulli flow Aalto University, Dept. Mathematics and Systems Analysis January 5th, 2018
2 Incompressible, steady Bernoulli principle Consider a straight tube Ω R 3 havin circular intersections. By A(x) denote the areas of the intersection along the length variable x R. Let p 0 and v 0 denote the pressure and flow velocity as a function of x. Bernoulli principle: p 0 (x) + ρ 0 2 v 0(x) 2 = p stag, x R where p stag denotes the stagnation pressure. Steadyness everything is independent of time t. Incompressibility density ρ 0 is independent of x, too. Then the volume flow U 0 > 0 is constant and v 0 (x) = U 0 A(x), x R.
3 Lossless Webster s horn model For small pressure and velocity variations (p, v ) in a tubular domain Ω containing stationary fluid, the acoustic approximation for the longitudinal dynamics can be used: 2 φ t 2 c2 A(x) ( A(x) φ ) = 0. The sound pressure and the acoustic perturbation velocity are obtained from the velocity potential φ = φ(t, x) by the partial derivatives v = φ and p φ = ρ 0 t. This model can be derived for the Wave Equation in a tubular domain by averaging the 3D velocity potential ψ = ψ(t, x, y, z) over the transversal intersections having the area A(x).
4 Purpose of this work If the fluid column is moving in the tube Ω with a constant volume velocity U 0 in the longitudinal direction (i.e., x-axis), the fluid movement couples with the propagation of longitudinal acoustic waves. At narrow parts of the tube, the fluid velocity is higher, and the acoustic field gets transported faster. Can we write a PDE that would encompass both the incompressible Bernoulli flow and the Webster horn model at once? Doing so would help us studying the coupling between the transport and acoustic subsystems if the PDE can be kept simple enough for analysis. Let us see about that.
5 Equation of continuity (1) We assume that ρ = ρ 0 + ρ, v = v 0 i + v, p = p 0 + p where ρ 0 > 0 is constant, and v 0 = v 0 (x), p 0 = p 0 (x) are as above. The unit vector i is the direction of the steady flow. The perturbations ρ, v, p depend on both t and (x, y, z) Ω R 3. They define the acoustic field. Since the mass (i.e., the fluid volume) is conserved, we get by omitting the second order perturbation terms the Equation of Continuity 1 ρ ρ 0 t + v + v 0 = 0.
6 Euler s equation (2) Omitting the second order advection term and perturbations in the density, the Conservation of the Momentum takes the form v (v t + v ) 0 + v v 0 x i + 1 p = 0 ρ 0 where v x = v i.
7 Thermodynamics and the speed of sound (1) Assuming the locally isentropic thermodynamics, the perturbation pressure and density are connected by p = c 2 ρ for the speed of sound 0 < c <. Of course, having c < requires compressibility. Neglecting the density and temperature variation (i.e., compressibility) but taking into consideration the pressure variation in the underlying flow, we get c 2 = γ p 0 ρ 0 = c 2 0 γu2 0 2A(x) 2 0. Here c0 2 = γ pstag ρ 0 is the speed of sound at stagnation, and γ is the adiabatic index of the gas. For diatomic ideal gas, γ = 7/5.
8 Thermodynamics and the speed of sound (2) We can now eliminate ρ from the Equation of Continuity, and hence 1 p ρ 0 c 2 t + v + v 0 = 0. It does not complicate the final model much to have c = c(x). It is of some interest to compute the lower limit for constriction A(x) 0 where the underlying flow remains subsonic... but not today. You cannot see Mach = 1 condition from this argument since the density, the absolute temperature, and the speed of sound decrease significantly when A(x) 0 then you must consider the underlying steady flow as compressible to make sense.
9 Wave equation for pressure and velocity Proceeding as usual, the combination of the Continuity Equation, Euler s Equation and Isentropic Equation of State produce a variant of the Wave Equation for the sound pressure p and acoustic perturbation velocity component v x in the direction of the flow: 2 p t 2 c2 2 p + v 0 2 p t = ρ 0c 2 ( 2 v x ) v 0 + (v x v 0 ) 2 v 0 2. Bad news: It is not possible to directly eliminate v x = v i from this equation since the Continuity Equation only gives radially symmetric information about the divergence of v. Good news: We can do tricks by defining a velocity potential candidate and writing the wave equation in terms of it.
10 Velocity potential (1) Taking the i-component from Euler s Equation gives v x t + ) (v 0 v x + p = 0. (1) Note that both v x and v 0 v x + p ρ 0 are functions of t, x, y, z even though the differentiations concern only the variables x, t. For any twice differentiable ψ = ψ(t, x, y, z) satisfying v x = ψ we get Eq. (1) since vx t + ρ 0 and v 0 v x + p = ψ ρ 0 t ( ) v 0 v x + p ρ 0 = 2 ψ t + 2 ψ t = 0. Thus, any solution pair (v x, p) of Eq. (1) can be represented in terms of such ψ, but it is not a true velocity potential in R 3 since we have instead of there.
11 Averaged velocity potential For any f : Ω R, the planar averages are defined by the integral f (x 0 ) = 1 f (x 0, y, z) dx dy. A(x 0 ) Γ(x 0 ) By averaging the i-component of Euler s Equation, v x t + ) (v 0 v x + p = 0 (2) where v x = v x (t, x) and p = p (t, x) come from (v x, p ). Now Eq. (2) allows the 1D velocity potential φ = φ(t, x) v x = φ ( ) ( ) φ φ and p = ρ 0 t v 0 v x = ρ 0 t + v φ 0. ρ 0
12 Averaging the Wave Equation Similarly, averaging the Wave Equation for (v x, p ) introduced above, we get ) (A(x) p 2 p + v 0 t 2 p t 2 c2 A(x) ( = ρ 0 c 2 2 v x v 0 + ( v x v 0 ) 2 v 0 2 Observe that Webster s operator on the LHS W A := 1 ( A(x) ) A(x) ). is what you get by averaging the Laplacian on a tube Ω R 3 with the area function A(x) in the longitudinal direction.
13 Bernoulli Webster model (1) If (v x, p ) satisfy the Wave Equation, what is the PDE satisfied by the 1D velocity potential φ = φ(t, x)? There are several variants. First variant: ( 2 t 2 2 ) ( ) φ c2 W A + v 0 t t + v φ 0 c 2 v 0 B(x) ( W 1/A φ + W 1/B φ ) = c 2 v0 2 ( B (x) B(x) 2) where B(x) := A (x) A(x) is the logarithmic derivative of the area function.
14 Bernoulli Webster model (2) Second variant: ( ) ( t + v 2 φ 0 t 2 2 ) φ c2 W A φ + v 0 t c 2 v 0 B(x) ( 3W B φ + W 1/A φ ) = c 2 v0 2 ( B (x) B(x) 2). And then there are others, but none of them looks any nicer. Note that the equations become homogenous if B (x) = B(x) 2 which takes place if and only if A(x) = a b x for a, b R, a > 0. This defines a hyperbolic horn having its flare at x = b.
15 The combined flow and acoustic field If φ = φ(t, x) is a solution of the Bernoulli Webster model, then the acoustically perturbed steady Bernoulli flow is described by ( ρ(t, x) = ρ ( φ c 2 t (t, x) + U )) 0 φ (t, x), A(x) v(t, x) = U 0 A(x) φ (t, x), and p(t, x) = p stag + ρ 0 ( φ t (t, x) + U 0 A(x) ( φ (t, x) U )) 0. 2A(x)
16 Further developments: 1 The Lagrangian image: i.e., the acoustics in the coordinate system that moves along the flow? (Leads to complications at least if the model is not simplified further.) 2 The underlying steady flow modelled by the compressible Bernoulli principle? (Taking the variations in density and temperature of the underlying flow into account seems technically complicated however, necessary if the constrictions are severe; i.e., A(x) 0 for some x.) 3 Magnetohydrodynamical variants? Acoustical compression waves in moving plasma bounded by a static magnetic field? So little time, so much to do.
17 That s all, folks. Questions?
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