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Derivation of the turbulent kinetic energy equation

Introducing the Reynolds decomposition into the original x₁-momentum equation and multiplying the result by the fluctuation u₁’ yields

(1) u₁’(∂(U₁ + u₁’))/(∂t) + u₁’(U_j + u_j’)(∂(U₁ + u₁’))/(∂x_j) = − (u₁’)/(ρ)(∂(P + p’))/(∂x₁) + νu₁’(∂²(U₁ + u₁’))/(∂x_j∂x_j)

Eq.(1↑) is averaged to get

(2) u₁’(∂u₁’)/(∂t) + U_j u₁’(∂u₁’)/(∂x_j) + u₁’u_j’(∂U₁)/(∂x_j) + u₁’u_j’(∂u₁’)/(∂x_j) = − (1)/(ρ)u₁’(∂p’)/(∂x₁) + νu₁’(∂²u₁’)/(∂x_j∂x_j)

which can be rearranged as follows

(3) (1)/(2)(∂u₁’²)/(∂t) + (1)/(2)U_j(∂u₁’²)/(∂x_j) + u₁’u_j’(∂U₁)/(∂x_j) + (1)/(2)(∂u₁’²u_j’)/(∂x_j) − (1)/(2)u₁’²(∂u_j’)/(∂x_j) = − (1)/(ρ)u₁’(∂p’)/(∂x₁) + νu₁’(∂²u₁’)/(∂x_j∂x_j)

Multiplying the continuity equation by u₁’ and averaging the result, the following relation is obtained.

(4) u₁’²(∂)/(∂x_j)(U_j + u_j’) = u₁’²(∂U_j)/(∂x_j) + u₁’²(∂u_j’)/(∂x_j) = 0 so that u₁’²(∂u_j’)/(∂x_j) = 0

from the averaged continuity equation. Hence, this term can be removed from Eq.(3↑). Similarly, the same process can be followed with the x₂-momentum equation

(5) (1)/(2)(∂u₂’²)/(∂t) + (1)/(2)U_j(∂u₂’²)/(∂x_j) + u₂’u_j’(∂U₂)/(∂x_j) + (1)/(2)(∂u₂’²u_j’)/(∂x_j) = − (1)/(ρ)u₂’(∂p’)/(∂x₂) + νu₂’(∂²u₂’)/(∂x_j∂x_j)

and with the x₃-momentum equation

(6) (1)/(2)(∂u₃’²)/(∂t) + (1)/(2)U_j(∂u₃’²)/(∂x_j) + u₃’u_j’(∂U₃)/(∂x_j) + (1)/(2)(∂u₃’²u_j’)/(∂x_j) = − (1)/(ρ)u₃’(∂p’)/(∂x₃) + νu₃’(∂²u₃’)/(∂x_j∂x_j)

Adding Eq.(3↑), Eq.(5↑) and Eq.(6↑), the following equation is obtained.

(7) (1)/(2)(∂u_i’u_i’)/(∂t) + (1)/(2)U_j(∂u_i’u_i’)/(∂x_j) + u_i’u_j’(∂U_i)/(∂x_j) + (1)/(2)(∂u_i’u_i’u_j’)/(∂x_j) = − (1)/(ρ)u_j’(∂p’)/(∂x_j) + νu_i’(∂²u_i’)/(∂x_j∂x_j)

The pressure term can be rewritten by considering the continuity equation.

(8) − (1)/(ρ)u_j’(∂p’)/(∂x_j) = − (1)/(ρ)(∂u_j’p’)/(∂x_j) + (1)/(ρ)p’(∂u_j’)/(∂x_j) = − (1)/(ρ)(∂u_j’p’)/(∂x_j)

The viscous term can also be rewritten as follows

(9) νu_i’(∂²u_i’)/(∂x_j∂x_j) = ν(∂)/(∂x_j)⎡⎣(1)/(2)(∂u_i’u_i’)/(∂x_j)⎤⎦ − ν(∂u_i’)/(∂x_k)(∂u_i’)/(∂x_k)

The turbulent kinetic energy is defined by k = (1)/(2)u_i’u_i’. Finally, introducing the changes presented in Eq.(8↑) and Eq.(9↑) into Eq.(7↑), the turbulent kinetic energy (TKE) equation is obtained.

(10) (∂k)/(∂t) + U_j(∂k)/(∂x_j) = (∂)/(∂x_j)⎛⎝ν(∂k)/(∂x_j) − (1)/(2)u_i’u_i’u_j’ − (1)/(ρ)u_jp’⎞⎠ − u_i’u_j’(∂U_i)/(∂x_j) − ν(∂u_i’)/(∂x_k)(∂u_i’)/(∂x_k)