Question 9.8: Non-Newtonian Pipe Flow 1 A non-Newtonian fluid flows under......

To determine the rate of flow, it is necessary to first determine the variation of the velocity in the cross section of the pipe. A force balance on an element of the fluid along the centreline of a pipe (Figure 9.11) is

\Delta p\pi r^{2}=\tau2\pi r L                 (9.19)

The shear stress is therefore

\tau={\frac{r}{2}}{\frac{\Delta p}{L}}            (9.20)

since the shear stress related to shear rate as

\tau=k\left({\frac{d U_{x}}{d r}}\right)^{2}               (9.21)

From Equations 9.20 and 9.21, the change in velocity of the fluid is therefore

d U=\left({\frac{\Delta p}{2k L}}\right)^{1/2}r^{1/2}d r         (9.22)

The velocity of the fluid across the cross section can be found by integration:

\textstyle\int_{0}^{U_{x}}d U_{x}={\frac{2}{3}}{\bigg(}{\frac{\Delta p}{2k L}}{\bigg)}^{1/2}\int_{r}^{R}r^{1/2}d r           (9.23)

to give

U_{x}={\frac{2}{3}}\left({\frac{\Delta p}{2k L}}\right)^{1/2}\left(R^{3/2}-r^{3/2}\right)           (9.24)

The volumetric flow rate is found by integrating across the entire cross section:

\int^{{Q}}_0 d{\dot{Q}}=\int_{0}^{R}2\pi r U_{x}d r              (9.25)

to give

\dot{Q}=\frac{2\pi}{7}\left(\frac{\Delta p}{2k L}\right)^{1/2}R^{7/2}            (9.26)

From the data provided, the flow rate is therefore

{\dot{Q}}={\frac{2\pi}{7}}\times{\left({\frac{10\times10^{5}}{2\times1000\times5}}\right)}^{1/2}\times0.1^{7/2}=0.002837\,{\mathrm{m}}^{3}\mathrm{s}^{-1}             (9.27)

The average velocity is taken as the total flow divided by the cross sectional area of the pipe:

\bar{U}=\frac{\dot{Q}}{\pi r^{2}}=\frac{0.002837}{\pi\times0.1^{2}}=0.09\ \mathrm{ms}^{-1}              (9.28)