Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 73 / 3

Here $k$ is turbulent kinetic energy per unit mass, $k=\frac{1}{2}\overline{u^{\prime 2}+v^{\prime 2}+w^{\prime 2}}$, rather than the permeability of a porous medium used in the porous-flow questions. The K-epsilon turbulence model represents the high-Reynolds-number transport of this energy and its turbulent kinetic energy dissipation rate $ϵ$.

In the $k$ equation, $d[({\nu }_T/{\sigma }_k)k_y]/dy$ is the divergence of modeled turbulent diffusion, $\mathcal{P}$ is mean-shear production, and $-ϵ$ is molecular conversion of fluctuation energy into heat. Statistical stationarity and fully developed flow have removed storage and streamwise advection from this reduction.

In the $ϵ$ equation, $d[({\nu }_T/{\sigma }_{ϵ}){ϵ}_y]/dy$ transports dissipation intensity; $C_{ϵ1}\mathcal{P}ϵ/k$ models its generation associated with energy production; and $-C_{ϵ2}{ϵ}^2/k$ models its destruction. The time scale $k/ϵ$ supplies the source-rate dimension. The $ϵ$ equation is a closure, not another energy-conservation equation. Finally, ${\nu }_T=C_{\mu }{k}^2/ϵ$ has dynamic viscosity dimensions and parametrizes Reynolds stress transport; ${\sigma }_k,{\sigma }_{ϵ}$ set the relative diffusion strengths, and the remaining constants calibrate the closure.

Let the mean streamwise velocity be $U(y)$ and set $S=dU/dy>0$. The relevant covariance is $\overline{u^{\prime }{v}^{\prime }}<0$ on this half of the channel. The work of the Reynolds stress against the mean shear rate givesThe eddy viscosity relation is $-\overline{u^{\prime }{v}^{\prime }}={\nu }_{T}S$, soThese are the kinematic-stress conventions, with the covariance measured per unit mass; inserting a stress measured in force per area would require division by mass density.

The law of the wall in the logarithmic overlap region is $U/u_{\ast }={\kappa }^{-1}\log (y/y_0)+B$, with the wall-scale reference absorbed in $y_0$ and $B$. Differentiating gives the mean shear rateHere $\kappa$ is the Von Kármán constant. This form applies outside the viscous sublayer but well below the outer channel scale; it is not valid at the wall itself or at the channel centre.

Define the friction velocity by $u_{\ast }=(|{\tau }_w|/\rho {)}^{1/2}$, where ${\tau }_w$ is the total tangential wall stress. It is a stress-derived velocity scale, not the local mean velocity. In the overlap region, $y$ is small relative to the channel half-width, so the total stress is approximately its wall value. Molecular stress is small there, giving $-\overline{u^{\prime }{v}^{\prime }}\simeq u_{\ast }^2$.

Combining this Reynolds stress with the logarithmic shear gives $\mathcal{P}\simeq u_{\ast }^3/(\kappa y)$. For locally equilibrated wall turbulence, energy transfer to small scales occurs over the local eddy turnover time and is balanced by molecular dissipation; energy storage and the net energy-transport divergence are small in the leading overlap-layer balance. ThusThis is the local turbulent kinetic energy balance. It is a modeling approximation rather than a general identity for all wall positions. The constant-energy solution derived next is consistent with it: its modeled $k$ diffusion term is exactly zero, while the $ϵ$ diffusion term remains essential.

The constant Reynolds stress and the logarithmic shear imply the eddy viscosity ${\nu }_T=u_{\ast }\kappa y$. On the other hand the K-epsilon model gives ${\nu }_T=C_{\mu }{k}^2/ϵ$. Insert $ϵ=u_{\ast }^3/(\kappa y)$:Canceling the positive factors yields $C_{\mu }{k}^2=u_{\ast }^4$. Since energy is nonnegative, the log-layer solution of the k-epsilon model hasindependent of $y$ to this leading overlap-layer accuracy. In particular $k_y=0$, and the $k$ diffusion term vanishes, confirming the local production–dissipation balance in the energy equation.

Equating the two expressions for eddy viscosity gives directlyFrom the preceding part, $u_{\ast }^2=\sqrt{C_{\mu }}k$. Replace one factor of $k$ using that identity, and then replace $u_{\ast }$ itself by $C_{\mu }^{1/4}{k}^{1/2}$. This proves all three equivalent dissipation expressions:Each is the same turbulent kinetic energy dissipation rate, with dimensions of velocity cubed divided by length. Its equivalent friction-velocity form is $u_{\ast }^3/(\kappa y)$.

Use ${\nu }_T=u_{\ast }\kappa y$, $ϵ=u_{\ast }^3/(\kappa y)$, $k=u_{\ast }^2/\sqrt{C_{\mu }}$, and $\mathcal{P}=ϵ$ in the dissipation equation. Its diffusion term isThe net modeled source is $(C_{ϵ1}-C_{ϵ2}){ϵ}^2/k=-(C_{ϵ2}-C_{ϵ1})\sqrt{C_{\mu }}{u}_{\ast }^4/({\kappa }^2{y}^2)$. Dividing the equation by $u_{\ast }^4/y^2$ and solving givesThe physical Von Kármán constant is the positive root, requiring $C_{ϵ2}>C_{ϵ1}$ with positive $C_{\mu },{\sigma }_{ϵ}$. Neglecting dissipation diffusion would incorrectly require the two dissipation-equation constants to be equal. The parameter ${\sigma }_k$ does not enter because the constant $k$ profile has no energy-diffusion contribution.