Question 9.7: Mixing of Non-Newtonian Fluids A mixing impeller with a dia......
Mixing of Non-Newtonian Fluids
A mixing impeller with a diameter of 300 mm with an average shear rate (s^{–1} )- to-speed proportionality constant 28 was used to test a Newtonian liquid with a density of 1120 kgm^{–3} and viscosity of 6.12 Nsm^{–2} followed by testing a pseudoplastic fluid. The following results of power uptake against speed were obtained for both fluids:
If the pseudoplastic liquid can be described by the power law model, determine the parameters for k and n.
Newtonian Liquid
\begin{matrix} \text{Power (W)} & 5.6 & 16.7 & 36.8 & 88.6 & 247 & 986 \\ N \ (rps) & 0.14 & 0.24 & 0.35 & 0.55 & 0.91 & 1.73 \end{matrix}Pseudoplastic Fluid
\begin{matrix} \text{Power (W)} & 18.3 & 38.4 & 80.0 & 127 & 293 \\ N \ (rps) & 0.29 & 0.44 & 0.68 & 0.89 & 1.46 \end{matrix}The power required for mixing is related to the reciprocal of the mixing Reynolds number by
\frac{P_{o}}{\rho N^{3}D^{5}}=\frac{K\mu}{\rho N D^{2}} (9.16)
Therefore,
P_{o}=K\mu N^{2}D^{3} (9.17)
A plot of power, P_o , with N² gives a linear relationship for which K is found to be 1978. Assuming that the same relationship holds with the pseudoplastic, then from Equation 9.16, the apparent viscosity is
{\mu}_{a p p}=\frac{P_{o}}{1978N^{2}D^{3}} (9.18)
The shear rate can be determined from the proportionality constant with rotational speed. Shear stress can therefore be determined using both the apparent viscosity and shear rate (Equation 9.11), and the following data can be obtained:
\tau=k\dot\gamma^{n} (9.11)
For the power law relationship for the pseudoplastic liquid, a plot of the log of the shear rate, \dot {γ} , with the shear stress, τ , gives a straight line as illustrated in Problem 9.5, for which the k and n are found to be 12.6 and 0.46, respectively