这是一份exeter埃克塞特大学MTH1003作业代写的成功案

$$
y^{+}=\frac{1}{K} \frac{\mu+\mu_{t}}{\mu} .
$$
For the k- $\varepsilon$ model of turbulence this distance is:
$$
y^{+}=\frac{\rho \sqrt[4]{C_{\mu}} \sqrt{k_{p}} \delta_{n p}}{\mu} .
$$
In the previous equation, $k_{P}$ is the kinetic energy of turbulence at the centre of the boundary cell, while $\sigma_{n s}$ denotes the normal distance from the centre of the boundary cell to the wall.
The viscous sub-layer thickness is defined as the larger root of the equation:
$$
y_{r}^{+}=\frac{1}{K} \ln \left(\varepsilon y_{r}^{+}\right)
$$

MTH1003 COURSE NOTES ：

$\int_{V} \operatorname{grad} \psi \mathrm{d} V=\int_{S} \psi \mathrm{ds} \Rightarrow \quad(\operatorname{grad} \psi){\mathrm{P}{0}} \approx \frac{1}{V_{\mathrm{P}{0}}} \sum{j=1}^{n_{f}} \psi_{j} \mathbf{s}{j}$ Here, $\psi{j}$ is the value of variable $\psi$ at the cell face centre.
The first term in the prototype equation is different to the others because it contains an integral with respect to time. If the equation is rearranged into the following form:
$$
\frac{d \Psi}{d t}=F(\phi)
$$
where
$$
\Psi=\int_{\mathrm{V}} \rho B_{\phi} \mathrm{d} V \approx\left(\rho B_{\phi} V\right){\mathrm{P}{0}} \text { and } \phi=\phi(\mathbf{r}, t)
$$