Question 9.5: Rheometer Data Analysis Test data were obtained from a rheo......
Rheometer Data Analysis
Test data were obtained from a rheometer for a non-Newtonian liquid. The data consist of the shear stress versus shear rate at a controlled temperature and pressure:
Determine the consistency coefficient k and power law index n for the liquid.
\begin{matrix} \text{Shear stress (Nm}^{-2}) & 4.5 & 7.92 & 11.6 & 15.5 \\ \text{Shear rate (s}^{-1}) & 2.5 & 4.0 & 5.5 & 7.0 \end{matrix}
The data of shear stress versus shear rate present a time-independent power law liquid for which the apparent viscosity decreases with an increasing shear rate. A power law relationship between shear stress and strain rate is given by Equation 9.11. This can be linearized by using natural logarithms to give
\tau = k \dot{\gamma}^2 (9.11)
\ln τ = \ln k + n \ln γ (9.12)
A plot of lnτ versus lnγ gives the straight line of gradient n and intercept lnk (Figure 9.10). The values are therefore calculated to be
From the straight-line relationship, the value of n is 1.2 and k is 1.5 (Nsm^{–2}) ^{1.2} .
\begin{matrix} \ln \tau \ (\ln \ Nm^{-2}) & 1.5 & 2.07 & 2.45 & 2.74 \\ \ln ϒ (\ln s^{-1}) & 0.92 & 1.39 & 1.70 & 1.95 \end{matrix}
