Derivation of the Dimensionless form of MPM from the Vanoni Form

Converting the HEC-RAS version of MPM (from Vanoni, 1975) to the dimensionless Form

$\begin{array}{l}\displaystyle \left( \frac {k_r}{k'_r} \right) \gamma RS = 0.047(\gamma_s - \gamma)d_m + 0.25 \left( \frac {\gamma}{g} \right) ^{\frac13} \left( \frac {\gamma_s - \gamma}{\gamma_s} \right) ^\frac23 g_s^{\frac23}\end{array}$

Get transport (gs) on the left side of the equation and the shear terms on the right.

$\begin{array}{l}\displaystyle 0.25 \left( \frac{\gamma}{g} \right)^{\frac13} \left(\frac{\gamma_s - \gamma}{\gamma_s} \right) ^{\frac23} g_s^{\frac23} = \left(\frac{k_r}{k'_r} \ right) \gamma RS - 0.047(\gamma_s - \gamma)d_m\end{array}$

Divide by 0.25 and by γs-γdm to convert shear to the dimensionless shields number and isolate the critical shear (0.047):

$\begin{array}{l}\displaystyle \frac { \left( \frac{\gamma}{g}\right)^{\frac13} \left( \frac{\gamma_s - \gamma}{\gamma_s} \right) ^{\frac23} g_s^{\frac23}} {(\gamma_s - \gamma)d_m} = 4 \left( \left( \frac {k_r}{k'_r}\right)^{\frac32} \tau^* - 0.047 \right)\end{array}$

Because the Shields number is:

$\begin{array}{l}\displaystyle τ^*= \frac {\gamma RS} {(\gamma_s- \gamma)d_m}\end{array}$

The right side of this equation is already looking familiar.
Next raise everything to the 3/2 power to prepare to isolate transport (gs):

$\begin{array}{l}\displaystyle \frac {\left( \frac {\gamma}{g}\right)^{\frac12} \left( \frac{\gamma_s-\gamma}{\gamma_s}g_s} {((\gamma_s-\gamma)d_m)^{\frac32}} = 8 \left( \left( \frac{k_r}{k'_r} \right) ^{\frac32}τ^*-0.04732\right)^\frac32\end{array}$

The Right side of the equation is in the final form, so simplify the left side of the equation into the dimensionless transport parameter (q*):

$\begin{array}{l}\displaystyle q^* = \frac {g_s/\gamma_s} {\sqrt{Rg}d_m^{\frac32} }\end{array}$

Where R is the difference between the specific gravity of the solids and liquid usually (S-1), (2.65-1) or 1.65 and g is the acceleration of gravity.
Simplify the expression until the dimensionless transport parameter emerges on the left side of the equation.

$\begin{array}{l}\displaystyle \frac{\gamma}{g}^{\frac12) \gamma (R) g_s} {\gamma_s ((\gamma_s-\gamma)d_m}^{\frac32} = 8\left( \left( \frac {k_r}{k'_r} \right) ^{\frac32} \tau^*-0.047 \right) ^{\frac32} \vspace{1} \frac{(\gamma)^{\frac12)} \gamma (R) g_s} {(g)^{\frac12} \gamma_s \gamma^{\frac32} R^{\frac32} d_m}^{\frac32} = 8\left( \left( \frac {k_r}{k'_r} \right) ^{\frac32} \tau^*-0.047 \right) ^{\frac32} \vspace{1} \frac {(\gamma)^{\frac12}\gamma}{\gamma_s \gamma^{\frac32}} \frac{g_s}{g^{\frac12} R^{\frac12} d_m ^{\frac32}}} = 8\left( \left( \frac {k_r}{k'_r} \right) ^{\frac32} \tau^*-0.047 \right) ^{\frac32}}\end{array}$

At this point, most of the left side of the equation collapses to the dimensionless transport parameter

$\begin{array}{l}\displaystyle \frac{g_s}{(R)^{\frac12}(g)^{\frac12}(d_m)^{\frac32}} = q^*\end{array}$

Leaving only the unit weight of the soil:

$\begin{array}{l}\displaystyle \frac {1} {\gamma_s}q^*= 8 \left( \left( \frac{k_r}{k'_r}\right)^{\frac32} \tau^* - 0.04732\right)^{\frac32}\end{array}$

But the original MPM equation computes the unit transport by weight while the dimensionless transport parameter calculates unit volume transport, so the unit weight of the solids converts the volume flux to a mass flux. But to leave it in the volume flux format, we can drop this conversion, yielding:

$\begin{array}{l}\displaystyle q^* = 8\left( \left( \frac{k_r}{k'_r}^{\frac32}\right)τ^*-0.047\right) ^{\frac32}\end{array}$