Within the planetary boundary layer, particularly within the Prandtl layer, the vertical gradients of meteorological variables are relatively large. They can only be sampled sufficiently well by a model at vertical grid point distances of one meter. Therefore, the approaches valid in the free atmosphere for the vertical turbulent flows at the bottom of the lowest model layer are no longer applicable. Therefore, in this area, the knowledge about the planetary boundary layer, which was developed in Sect. 17.2.1, is used to determine the vertical turbulent flows.
According to Eq. (17.80) is the Monin-Obukhov length
\[ \begin{align} L \coloneqq -\frac{T_Au_\star^3}{kg\newoverline{\left(w'\theta'\right)}}. \end{align} \]
What is problematic here is the determination of $\newoverline{\left(w'\theta'\right)}_\text{sfc}$. If you put Eq. one, you get
\[ \begin{align} L = -\frac{T_Au_\star^3}{kg\frac{Q_H}{\rho c_v}} = -\frac{T_Au_\star^3}{kg\frac{T_G - T_A'}{r_H}} = -\frac{T_Au_\star^3}{kg\left(T_G - T_A'\right)}r_H. \end{align} \]
The friction speed $u_\star$ and the flow resistance of the sensible heat $r_H$ depend in turn on $L$:
\[ \begin{align} L = -\frac{T_Au_\star^3\left(L\right)}{kg\left(T_G - T_A'\right)}r_H\left(L\right).\tag{37.3}\label{eq:mo_length_model_ansatz} \end{align} \]
If you put Eq. one, you get
\[ \begin{align} L = -\frac{T_Au_\star^3}{g\left(T_G - T_A'\right)}\frac{1}{u_\star k^2}\left[\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)\right] = \frac{T_Au_\star^2}{k^2g\left(T_A' - T_G\right)}\left[\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)\right]. \end{align} \]
If you continue using Eq. (17.49), follows
\[ \begin{align} L = \frac{T_AU^2}{g\left(T_A' - T_G\right)}\frac{\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2}. \end{align} \]
This is about the determination $L$, but all other quantities are known. The size
\[ \begin{align} R_\text{i,b} \coloneqq \frac{zg\left(T_A' - T_G\right)}{T_AU^2}. \end{align} \]
is called Bulk Richardson number. With the help function
\[ \begin{align} s\left(L\right) &\coloneqq \frac{\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2} \end{align} \]
and
\[ \begin{align} a \coloneqq \frac{R_\text{i,b}}{z} \end{align} \]
you can do this in the form
\[ \begin{align} L = \frac{z}{R_\text{i,b}}s\left(L\right) = \frac{s\left(L\right)}{a} \Leftrightarrow aL - s\left(L\right) = 0\tag{37.9}\label{eq:mo_l_num_deriv_1} \end{align} \]
note down.
The sign of $L$ can already be determined in advance from Eq. (37.3) can be determined. In the case $T_G < T_A'$, which corresponds to stable conditions, $L > 0.$ In this case apply
\[ \begin{align} \psi_m\left(\frac{z}{L}\right) = \psi_h\left(\frac{z}{L}\right) = -4,7\cdot\frac{z}{L}. \end{align} \]
Now define
\[ \begin{align} h\left(L\right) \coloneqq \ln\left(\frac{z}{z_0}\right) + 4,7\cdot\frac{z}{L}, \end{align} \]
then it follows from Eq. (37.9)
\[ \begin{align} aL - s\left(L\right) &= aL - \frac{h\left(L\right) + \ln\left(7\right)}{h^2\left(L\right)} = 0\nonumber\\ \Leftrightarrow f_1\left(L\right) &\coloneqq aLh^2\left(L\right) - h\left(L\right) - \ln\left(7\right) = 0.\tag{37.12}\label{eq:mo_l_num_deriv_2} \end{align} \]
Eq. (37.12) cannot be solved analytically and must therefore be solved numerically. This corresponds to determining the zero of $f_1$. For this you use a Newton iteration with
\[ \begin{align} f_1\left(L_i\right) + f_1'\left(L_i\right)\cdot\left(L_{i+1} - L_i\right) = 0, \end{align} \]
here $i$ is the iteration step. It applies
\[ \begin{align} f_1'\left(L\right) = ah^2\left(L\right) + 2aLh\left(L\right)h'\left(L\right) - h'\left(L\right). \end{align} \]
Assuming $f_1' \not= 0$ this can be done
\[ \begin{align} \frac{f_1\left(L_i\right)}{f_1'\left(L_i\right)} + L_{i+1} - L_i &= 0\nonumber\\ \Leftrightarrow L_{i+1} &= L_i - \frac{f_1\left(L_i\right)}{f_1'\left(L_i\right)} \end{align} \]
rephrase.
In the case $T_G > T_A$, which corresponds to unstable conditions, $L < 0.$ In this case applies
\[ \begin{align} f_2\left(L\right) &\coloneqq aL - s\left(L\right) = aL - \frac{\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2} = 0. \end{align} \]
From this it follows
\[ \begin{align} f_2'\left(L\right) &= a - g'\left(L\right) = a - \frac{\frac{z}{L^2}\psi'_h\left(\frac{z}{L}\right)\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2 - \left[\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)\right]2\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]\frac{z}{L^2}\psi'_m\left(\frac{z}{L}\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^4}\nonumber\\ &= a - \frac{\frac{z}{L^2}\psi'_h\left(\frac{z}{L}\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^2} + \frac{2\left[\ln\left(\frac{z}{z_0}\right) - \psi_h\left(\frac{z}{L}\right) + \ln\left(7\right)\right]\frac{z}{L^2}\psi'_m\left(\frac{z}{L}\right)}{\left[\ln\left(\frac{z}{z_0}\right) - \psi_m\left(\frac{z}{L}\right)\right]^3}. \end{align} \]