2.2 Determination of hydraulic shear stress
The fluid in the soil pipe generally follows the Bernoulli’s equation.
An idealized pipe model has been used to simulate the erosion process in
the HET. The cylindrical path with a length of L and a radius ofR is regarded as control volume, as indicated in Fig. 1 (b). The
fluid travels through soil body following the force-equilibrium equation
and Bernoulli’s Energy equation, which are given
by
\begin{equation}
P_{1}A-P_{2}A-\tau\bullet 2\pi RL=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [5]\nonumber \\
\end{equation}\begin{equation}
\left(Z_{1}+\frac{P_{1}}{\rho_{w}g}+\frac{V^{2}}{2g}\right)_{\text{in}}-h_{f}=\left(Z_{2}+\frac{P_{2}}{\rho_{w}g}+\frac{V^{2}}{2g}\right)_{\text{out}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [6]\nonumber \\
\end{equation}As designed in the HET, the eroded path in the experiential setup is
horizontal (i.e., \(Z=0\)). Therefore, the diameters of inlet and
outlet are expected to be the same with uniform erosion (i.e., the
velocities are constant). From equation [5], shear stress can be
obtained by
\begin{equation}
\tau=\frac{\left(P_{1}-P_{2}\right)\bullet R}{2L}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [7]\nonumber \\
\end{equation}Friction loss (\(h_{f}\)) is determined by
\begin{equation}
h_{f}=\frac{P_{1}-P_{2}}{\rho_{w}g}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [8]\nonumber \\
\end{equation}Combining equation [7] and equation [8],
\begin{equation}
h_{f}=\frac{2L\tau}{R\rho_{w}g}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [9]\nonumber \\
\end{equation}Friction loss in the eroded path is calculated by Darcy-Weishach
equation (Nieber et al. 2019),
\begin{equation}
h_{f}=\frac{32\nu\bullet LV}{D^{2}g}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [10]\nonumber \\
\end{equation}Where \(\nu\) is kinematic viscosity, and D is the diameter of
the eroded path.
Substitute equation [10] into equation [9], shear stress can be
further expressed as
\begin{equation}
\frac{32\nu\bullet LV}{D^{2}g}=\frac{2L\tau}{R\rho_{w}g}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [11]\nonumber \\
\end{equation}\begin{equation}
\tau=\frac{4\nu Q\rho_{w}}{\pi R^{3}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [12]\nonumber \\
\end{equation}