Turbulence kinetic energy

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Turbulence kinetic energy
Common symbols	TKE, k
In SI base units	J/kg = m2⋅s−2
Derivations from other quantities	${\displaystyle k={\frac {1}{2}}\left(\,{\overline {(u')^{2}}}+{\overline {(v')^{2}}}+{\overline {(w')^{2}}}\,\right)}$

In fluid dynamics, turbulence kinetic energy (TKE) is the mean kinetic energy per unit mass associated with eddies in turbulent flow. Physically, the turbulence kinetic energy is characterized by measured root-mean-square (RMS) velocity fluctuations. In the Reynolds-averaged Navier Stokes equations, the turbulence kinetic energy can be calculated based on the closure method, i.e. a turbulence model.

The TKE can be defined to be half the sum of the variances σ² (square of standard deviations σ) of the fluctuating velocity components: $${\displaystyle k={\frac {1}{2}}(\sigma _{u}^{2}+\sigma _{v}^{2}+\sigma _{w}^{2})={\frac {1}{2}}\left(\,{\overline {(u')^{2}}}+{\overline {(v')^{2}}}+{\overline {(w')^{2}}}\,\right),}$$ where each turbulent velocity component is the difference between the instantaneous and the average velocity: ${\displaystyle u'=u-{\overline {u}}}$ (Reynolds decomposition). The mean and variance are $${\displaystyle {\begin{aligned}{\overline {u'}}&={\frac {1}{T}}\int _{0}^{T}(u(t)-{\overline {u}})\,dt=0,\\[4pt]{\overline {(u')^{2}}}&={\frac {1}{T}}\int _{0}^{T}(u(t)-{\overline {u}})^{2}\,dt\geq 0=\sigma _{u}^{2},\end{aligned}}}$$ respectively.

TKE can be produced by fluid shear, friction or buoyancy, or through external forcing at low-frequency eddy scales (integral scale). Turbulence kinetic energy is then transferred down the turbulence energy cascade, and is dissipated by viscous forces at the Kolmogorov scale. This process of production, transport and dissipation can be expressed as: $${\displaystyle {\frac {Dk}{Dt}}+\nabla \cdot T'=P-\varepsilon ,}$$ where:^[1]

• ⁠${\displaystyle {\tfrac {Dk}{Dt}}}$⁠ is the mean-flow material derivative of TKE;
• ∇ · T′ is the turbulence transport of TKE;
• P is the production of TKE, and
• ε is the TKE dissipation.

Assuming that molecular viscosity is constant, and making the Boussinesq approximation, the TKE equation is: $${\displaystyle \underbrace {\frac {\partial k}{\partial t}} _{{\text{Local}} \atop {\text{derivative}}}\!\!\!+\ \underbrace {{\overline {u}}_{j}{\frac {\partial k}{\partial x_{j}}}} _{{\text{Advection}} \atop {}}=-\underbrace {{\frac {1}{\rho _{o}}}{\frac {\partial {\overline {u'_{i}p'}}}{\partial x_{i}}}} _{{\text{Pressure}} \atop {\text{diffusion}}}-\underbrace {{\frac {1}{2}}{\frac {\partial {\overline {u_{j}'u_{j}'u_{i}'}}}{\partial x_{i}}}} _{{{\text{Turbulent}} \atop {\text{transport}}} \atop {\mathcal {T}}}+\underbrace {\nu {\frac {\partial ^{2}k}{\partial x_{j}^{2}}}} _{{{\text{Molecular}} \atop {\text{viscous}}} \atop {\text{transport}}}-\underbrace {{\overline {u'_{i}u'_{j}}}{\frac {\partial {\overline {u_{i}}}}{\partial x_{j}}}} _{{\text{Production}} \atop {\mathcal {P}}}-\underbrace {\nu {\overline {{\frac {\partial u'_{i}}{\partial x_{j}}}{\frac {\partial u'_{i}}{\partial x_{j}}}}}} _{{\text{Dissipation}} \atop \varepsilon _{k}}-\underbrace {{\frac {g}{\rho _{o}}}{\overline {\rho 'u'_{i}}}\delta _{i3}} _{{\text{Buoyancy flux}} \atop b}}$$

By examining these phenomena, the turbulence kinetic energy budget for a particular flow can be found.^[2]