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

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.