Abstract

We explore theoretically non-Kolmogorov scaling and dissipation laws by employing continuous symmetry group transformations on statistical model equations for planar turbulent plume. The dissipation laws for mean turbulent K. E. (k¯) and mean thermal intensity (kθ¯) at infinite Reynolds number limit are obtained as ϵ∼(k¯)3/2/(δuReδum), ϵθ∼(k¯)1/2(kθ¯)/(δuReδum), where the exponent m=3a2/(a1+a2) preferably belongs to [−3/2, 1], a1, a2 the dilation symmetry group parameters. Here Reδu is the local Reynolds number based on plume velocity width δu. The Kolmogorov dissipation law will hold only when k¯∼u¯2∼u′v′¯, which implies the same streamwise variations of Reynolds stresses, while non-Kolmogorov dissipation can hold only when k¯∼u¯2≁u′v′¯ indicating different streamwise variations of Reynolds stresses. When m = −3/2, the plume scales as a non-equilibrium exponential function of streamwise distance x, while for m ∈ (−3/2, 0) ∪ (0, 1], there exist multiple non-equilibrium power law scalings and classical equilibrium scaling corresponds to m = 0. The turbulent kinetic energy dissipation law for m ≈ 1 is reported both experimentally and theoretically in axisymmetric turbulent wake and planar jet. When m = 1, the plume velocity and temperature widths (δu, δT) grow as ∼(x + a)2/5, centerline mean velocity grows as u¯c∼(x+a)1/5, and mean temperature decays as θc ∼ (x + a)−3/5, with a being a virtual origin. The non-equilibrium dissipation scaling instigates a variation of plume entrainment coefficient with streamwise distance. In particular, the region where dissipation scaling with exponent m = 1 holds, the entrainment coefficient in planar plume varies as ∼(x + a)−3/5, whereas the entrainment coefficient reported recently in planar jet varies as ∼(x + a)−1/3.

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