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Vortex topology and global-Bernoulli counterexamples — worked interpretation

Physical scope

Constructed numerical substitutions in exact inviscid model fields; not observations or predictions for a viscous material.

Example

In the steady incompressible Euler model, choose vortex circulation parameter $κ = 1 m^{2} s^{- 1}$ . Its circulation on a radius- $2 m$ circle is $2 π m^{2} s^{- 1}$ , although curl vanishes off the axis. In a separate solid-rotation field choose $Ω = 0.2 s^{- 1}$ and $ρ_{0} = 1000 k g m^{- 3}$ . For the steady solid-rotation field $u (x, y, z) = Ω (- y, x, 0)$ , a streamline is a $C^{1}$ curve $γ : J \to R^{3}$ , with $J \subseteq R$ an interval, satisfying $γ^{'} (s) = u (γ (s))$ . The Bernoulli value $B = p / ρ_{0} + ∣ u ∣^{2} / 2$ differs between radii $1$ m and $2$ m by $0.12 m^{2} s^{- 2}$ , while it remains constant along each circular streamline. These are conditional model calculations.

Example references

Stewart, MIT 8.09, Fluid Mechanics, §6.4, printed pp. 120–123

Facts & Assumptions

Given: The two exact steady incompressible Euler fields and selected values above.

[F1]

Vortex topology and global-Bernoulli counterexamples — conditional physical prediction: vortex circulation is $2 π κ$ , and solid-rotation Bernoulli is $p_{0} / ρ_{0} + Ω^{2} r^{2}$ , constant on but not across circles.

[F2]

Formal parameterized fluid systems and observables: Cartesian curl uses $curl u = (\partial_{y} u_{z} - \partial_{z} u_{y}, \partial_{z} u_{x} - \partial_{x} u_{z}, \partial_{x} u_{y} - \partial_{y} u_{x})$ . For a closed piecewise- $C^{1}$ oriented curve $γ : [a, b] \to Ω$ , circulation is $\oint_{γ} u \cdot d r = \int_{a}^{b} u (γ (s)) \cdot γ^{'} (s) d s$ .

Proof

1.1givenF1F2algebra

Using the closed-curve circulation convention in [F2], with $κ = 1$ and $R = 2$ , $Γ = 2 π κ = 2 π m^{2} s^{- 1}$ . Nonzero circulation contradicts the closed-loop integral of a global gradient, so the curl-free field has no global potential on the punctured domain.

2.1givenF1step 1.1algebra∎

The Bernoulli difference is $Ω^{2} (2^{2} - 1^{2}) = 0.04 \cdot 3 = 0.12 m^{2} s^{- 2}$ . For $γ (s) = (x (s), y (s), z (s))$ , the streamline ODE gives $\frac{d}{d s} (x (s)^{2} + y (s)^{2}) = 2 x (- Ω y) + 2 y (Ω x) = 0$ . Therefore each circular streamline has fixed radius and one Bernoulli value, while the selected radii have the displayed different values.

Depends on

Dependency tree · 2 levels

Sources

Stewart, MIT 8.09, Fluid Mechanics