Decay exponent of freely decaying homogeneous isotropic turbulence · revision history
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- Title
- Decay exponent of freely decaying homogeneous isotropic turbulence
- Plain
- When stirring stops, turbulence in a box dies away with its energy falling as a power of time. Measured powers differ by up to a factor of two between experiments, and theory gives competing values that depend on how the largest swirls were set up.
- Precise
- Freely decaying homogeneous isotropic Navier-Stokes turbulence at high Re has kinetic energy $K(t) \sim t^{-n}$. Kolmogorov's argument with a conserved Loitsyansky integral gives $n = 10/7$ for an initial spectrum $E(k) \sim k^4$ at small $k$; Saffman's conserved integral gives $n = 6/5$ for $E(k) \sim k^2$. Determine whether n is fixed by the low-wavenumber exponent of the initial spectrum alone as $\mathrm{Re} \to \infty \text{ and } t \to \infty$ in a domain much larger than the integral scale, whether the Loitsyansky integral is conserved, and the value of n for each class.
- Settled by
- DNS in domains much larger than the integral scale over long decay times at high Re, measuring n and the time dependence of the Loitsyansky and Saffman integrals for controlled initial spectra.
- Kind
- well-posed
- Answer type
- value
- References
- (empty)
- Literature status
- contested
- Status note
- 2022 DNS observe both exponents approximately and over limited times for suitably arranged initial conditions, but the low-wavenumber spectrum drifts, so the invariants are not strictly conserved (Panickacheril John, Donzis, Sreenivasan, Phil. Trans. R. Soc. A 2022).
- Posed since
- (empty)