# Why is the turbulent kinetic energy $k=\overline{v_1'^2}+\overline{v_2'^2}+\overline{v_3'^2}$ even in 2D cases?

I'm writing an essay which partly includes writing about the Reynolds stress model where the dissipation term $$\varepsilon_{ij}$$ is assumed to be isotropic (since dissipation mostly occurs at small scale turbulence, which is isotropic). More specific, the dissipation term is modelled as

$$\varepsilon_{ij} = \frac{2}{3}\varepsilon\delta_{ij},$$

where $$\delta_{ij}$$ is the Kronecker delta and $$\varepsilon$$ is the dissipation term from the $$k$$ equation (ie turbulent kinetic energy equation)

\begin{align} \underbrace{\frac{\partial k}{\partial t}}_{\text{time term}} + \underbrace{\overline{v}_j\frac{\partial k}{\partial x_j}}_{C^k} = &\underbrace{-\overline{v_i'v_j'} \frac{\partial \overline{v}_i}{\partial x_j}}_{P^k} \underbrace{- \frac{\partial }{\partial x_j}\left( \overline{v'_j\left( \frac{p'}{\rho} + \frac{1}{2}v_i'v_i\right) }\right)}_{D_t^k} + \underbrace{\nu \frac{\partial^2 k}{\partial x_j \partial x_j}}_{D_\nu^k} \underbrace{- g_i\beta\overline{v_i'\theta'}}_{G^k} \underbrace{- \nu\overline{\frac{\partial v_i'}{\partial x_j} \frac{\partial v_i'}{\partial x_j}}}_{\varepsilon} . \end{align}

The two comes from that $$2\varepsilon$$ is the total dissipation. The division by $$3$$ comes from that in 3D we will have three normal dissipation terms $$\varepsilon_{11}$$, $$\varepsilon_{22}$$, and $$\varepsilon_{33}$$ and because of their isotropic behavior they should all be equal.

I thought that since for a 2D problem we'll only have two normal dissipation terms $$\varepsilon_{11}$$ and $$\varepsilon_{22}$$ our model should be

$$\varepsilon_{ij} = \frac{2}{2}\varepsilon\delta_{ij} = \varepsilon_{ij}\delta_{ij},$$

but this is apparently wrong. The answer I got was that this is wrong because "the turbulent kinetic energy $$k$$ will still be calculated as in 3D when in 2D." That is

$$k = \overline{v_1'^2}+\overline{v_2'^2}+\overline{v_3'^2}.$$

One person explained it as even though we're in 2D there will still exist a $$v_3'$$ stress. This seems very weird to me. Why would you calculate $$k$$ as in 3D even when you're in 2D?

OBS! Note that tensor notation is used.