Subtopic: Power law apparent viscosity calculation
Question Statement:
A fruit puree follows the Ostwald-de Waele power law model: $\tau = K \dot{\gamma}^n$. The consistency index is $K = 2.5\text{ Pa}\,\text{s}^{0.4}$ and the flow behavior index is $n = 0.40$. At a shear rate of $\dot{\gamma} = 100\text{ s}^{-1}$, the apparent viscosity $\eta_{app}$ is ________ $\text{Pa}\,\text{s}$ (round off to three decimal places).
Answer Key:
0.158
Step-by-Step Solution & Explanation
The shear stress is:
$\tau = K \dot{\gamma}^n = 2.5 \times (100)^{0.40} = 2.5 \times 6.3096 = 15.77\text{ Pa}$
The apparent viscosity is defined as:
$\eta_{app} = \frac{\tau}{\dot{\gamma}} = \frac{15.77}{100} = 0.1577\text{ Pa}\,\text{s}$
Rounding to three decimal places: $0.158\text{ Pa}\,\text{s}$.