Question 3.10: The rheological properties of a particular suspension may be......
The rheological properties of a particular suspension may be approximated reasonably well by either a powerlaw or a Bingham-plastic model over the shear rate range of 10 to 50 s^{-1}. If the consistency coefficient k is 10 N s^{n}/m^{-2} and the flow behaviour index n is 0.2 in the power law model, what will be the approximate values of the yield stress and of the plastic viscosity in the Bingham-plastic model?
What will be the pressure drop, when the suspension is flowing under laminar conditions in a pipe 200 m long and 40 mm diameter, when the centre line velocity is 1 m/s, according to the power-law model? Calculate the centre-line velocity for this pressure drop for the Bingham-plastic model.
Using the power-law model (equation 3.121):
|R_{y}|=k\left(\left|{\frac{\mathrm{d}u_{x}}{\mathrm{d}y}}\right|\right)^{n}=10\left(\left|{\frac{\mathrm{d}u_{x}}{\mathrm{d}y}}\right|\right)^{0.2}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}~}
\left|\frac{\mathrm{d}u_{x}}{\mathrm{d}y}\right|=50\mathrm{~s}^{-1}:\vert R_{y}\vert=10\times50^{0.2}=21.87\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)
|R_{y}|=R_{Y}+\mu_{p}\left|{\frac{\mathrm{d}u_{x}}{\mathrm{d}y}}\right|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:
|R_{y}|=14.35+0.150\left|{\frac{\mathrm{d}u_{x}}{\mathrm{d}y}}\right|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:
u_{C L}=1\ \mathrm{m/s},\ \ l=200\ \mathrm{m},\ \ r=0.02\ \mathrm{m},\ \ k=10\ \mathrm{Ns}^{n}\mathrm{m}^{-2},\ \ n=0.2and: -ΔP = \underline{\underline{626,000\ N/m^{2}}}
For a Bingham-plastic fluid:
The centre line velocity is given by equation 3.145:
u_{p}=\frac{-\Delta P r^{2}}{8\mu_{p}l}(2-4X+2X^{2})where: X={\frac{l}{r}}{\frac{2R_{Y}}{(-\Delta P)}}
=\frac{200}{0.02}\times\frac{2\times14.35}{626,000}=0.458\therefore 2 – 4X + 2X² = 0.589
u_{p}=\frac{626,000\times(0.02)^{2}}{8\times0.150\times200}\times0.589= \underline{\underline{0.61\ m/s}}