Question 3.10: The rheological properties of a particular suspension may be......

Using the power-law model (equation 3.121):

When: $\left|\frac{\mathrm{d}u_{x}}{\mathrm{d}y}\right|=10\mathrm{~s}^{-1}:\vert R_{y}\vert=10\times10^{0.2}=15.85\mathrm{~N/m^{2}~}$

Using the Bingham-plastic model (equation 3.125):

$|R_{y}|-R_{Y}=\mu_{p}\left|\frac{\mathrm{d}u_{x}}{\mathrm{d}y}\right|\quad(|R_{y}|\geqslant R_{Y})$           (3.125)

When: $\left|\frac{du_{x}}{dy} \right|=10\ s^{-1}:\ \ 15.85=R_{Y}+10\mu _{p}$

$\left|\frac{du_{x}}{dy} \right|=50\ s^{-1}:\ \ \underline{21.87=R_{Y}+50\mu _{p}}$.

Subtracting:       6.02 = 40$\mu _{p}$.

Thus:                  $\mu _{p}$ = 0.150 N s/m²

and:                        $R_{Y}$ = 14.35 N/m²

Thus, the Bingham-plastic equation is:

For the power-law fluid:

Equation 3.131 gives: $u_{C L}=\left(\frac{-\Delta P}{2k l}\right)^{1/n}\frac{n}{n+1}r^{(n+1)/n}$                     (3.131)

Rearranging: $-\Delta P=2k l u_{C L}^{n}\left({\frac{n+1}{n}}\right)^{n}r^{-(n+1)}$.

The numerical values in SI units are:

and:  -ΔP = $\underline{\underline{626,000\ N/m^{2}}}$

For a Bingham-plastic fluid:

The centre line velocity is given by equation 3.145:

where:           $X={\frac{l}{r}}{\frac{2R_{Y}}{(-\Delta P)}}$

$\therefore$          2 – 4X + 2X² = 0.589

= $\underline{\underline{0.61\ m/s}}$