In this post, I present a little algebraic calculation to estimate the units so that the Reynolds number $N_{Re}$ remains dimensionless. You should recall that incorrect units on $m$ would lead you to numerical errors.
I shall then start with the formulas already presented for non-Newtonian fluids in the post: Hydraulic equations for non-Newtonian fluids. Then, the Reynolds number is defined as,
$N_{Re}=\dfrac{(4n)^{n}\,D^n\,V^{2-n}\rho}{g_c\,m\,(3n+1)^n8^{n-1}}$ Eq. (01)
where the constant $g_c$ is defined as $32.174\,lb_m\,\cdot \,ft/lb_f\, \cdot s^2$. Also, the diameter $D$ is used in $ft$, the fluid velocity is used in $ft/s$, and the fluid density must be used in $lb_m/ft^3$. The flow index $n$ is dimensionless.
Thus, we may envisage the units of $m$ by considering, from the Reynolds number in Eq. (01), solely,
$\dfrac{D^n\,V^{2-n}\rho}{g_c\,m}$ Eq. (02)
Next, if we substitute the units of all variables listed above, we obtain,
$\dfrac{(ft)^n\cdot \left(\dfrac{ft}{s}\right)^{2-n}\cdot \dfrac{lb_m}{ft^3}}{\dfrac{lb_m \cdot ft}{lb_f \cdot s^2}\cdot m}$ Eq. (03)
A simplification process leads to,
$\dfrac{\dfrac{lb_m}{ft \cdot s^{2-n}}}{\dfrac{lb_m \cdot ft\cdot m}{lb_f \cdot s^2}}$ Eq. (04)
$\Rightarrow \dfrac{lb_m \cdot lb_f \cdot s^2}{ft \cdot s^{2-n} \cdot lb_m \cdot ft\cdot m}$ Eq. (05)
Further simplification leads to,
$\dfrac{lb_f }{ft^2 \cdot s^{-n}\cdot m}$ Eq. (06)
From Eq. (06), it is easily seen that to get all units cancelled, the parameter $m$ must have units:
$m=\left[\dfrac{lb_f}{ft^2 \cdot s^{-n}}\right]$ Eq. (07)
or using slugs, since $1\, lb_f\cdot s/ft^2= 1\, slug/ft \cdot s$,
$m=\left[\dfrac{slug}{ft \cdot s^{2-n}}\right]$ Eq. (08)
This is the end of the post. I hope you find it useful.
Ildebrando.
