这是一份BATH巴斯大学ME20015作业代写的成功案例
$$
M e=M e_{(\lambda=1)} \cdot \lambda^{m}
$$
In Anglo-Saxon countries, the following nomenclature is often used for the description of the fill characteristics curves:
$$
\frac{K a V}{L}=f(L / G)=f(1 / \lambda)
$$
The convective heat transfer coefficient is calculated based on the Lewisanalogy taking into account a correction for one-directional diffusion at the phase interface:
$$
\frac{a}{\beta_{x}}=c_{p m t} \cdot L e^{(1-n)} \cdot \frac{\left(p_{\mathrm{S}, \mathrm{W}}^{\prime \prime}-p_{\mathrm{S}, \mathrm{A}}\right)}{\left(p-p_{\mathrm{S}, \mathrm{W}}^{\prime \prime}\right) \cdot \ln \left(p-p_{\mathrm{S}, \mathrm{A}} / p-p_{\mathrm{S}, \mathrm{W}}^{\prime \prime}\right)}
$$
Here, $c_{\mathrm{pm}}$ is the specific heat capacity of wet air related to the dry air mass:
$$
c_{p m}=c_{p, \mathrm{~A}}+c_{p, \mathrm{~S}} \cdot X_{\mathrm{A}}
$$
ME10305 COURSE NOTES ：
where the effective Nusselt number is
$$
N u_{\mathrm{a}}=\frac{N u_{\mathrm{D}}}{\sqrt{a_{1} R e_{D}}}
$$
the parameter $a_{1}$, the dimensionless strain rate, is $4.0$ for a cylinder in cross flow. The effective turbulence intensity, $\mathrm{Tu}{\lambda}$, is given by $$ T u{\lambda}=\frac{T u_{\mathrm{a}} \sqrt{L_{\mathrm{a}}}}{\left(1+0.004 L_{\mathrm{a}}^{2}\right)^{5 / 12}}
$$
the dimensionless length scale, $L_{\mathrm{a}}$, is given by
$$
L_{\mathrm{a}}=\frac{\Lambda_{x}}{D} \sqrt{a_{1} R e_{D}}
$$
where $\Lambda_{x}$ is the integral length scale based upon the streamwise velocity fluctuation component. The turbulence term is given as:
$$
T u_{\mathrm{a}}=\frac{T u \sqrt{R e_{D}}}{\sqrt{a_{1}}}
$$