Question

The values of the consistency index ‘\(K\)’ and the flow behavior index ‘\(n\)’ of a dilatant fluid are 0.415 (in CGS units) and 1.23, respectively. The value of the apparent viscosity (in \( g \cdot cm^{-1} \cdot s^{-1} \)) of this fluid at a shear rate of 60 \( s^{-1} \) (rounded off to the nearest integer) is __________________.

Hint:

- Power-law fluids use \(\eta_{\text{app}}=K\,\dot{\gamma}^{\,n-1}\); check the exponent \(n-1\) carefully.
- \(n>1\) indicates a dilatant (shear-thickening) fluid; \(\eta_{\text{app}}\) increases with \(\dot{\gamma}\).

Answer 1

Correct Answer: 1

Step 1: For a power-law (Ostwald–de Waele) fluid, \(\tau = K\,(\dot{\gamma})^n\). The apparent viscosity is \[ \eta_{\text{app}}=\frac{\tau}{\dot{\gamma}}=K\,(\dot{\gamma})^{\,n-1}. \] Step 2: Substitute the given values \(K=0.415\), \(n=1.23\), and \(\dot{\gamma}=60~s^{-1}\): \[ \eta_{\text{app}}=0.415\,(60)^{1.23-1}=0.415\,(60)^{0.23}. \] Step 3: Evaluate the power: \[ (60)^{0.23} \approx e^{0.23\ln 60} \approx e^{0.23\times 4.0943} \approx e^{0.9417} \approx 2.564. \] Step 4: Compute the apparent viscosity: \[ \eta_{\text{app}} \approx 0.415 \times 2.564 \approx 1.064 \; g\cdot cm^{-1}\cdot s^{-1}. \] Step 5: Round to the nearest integer: \[ \boxed{1 \; g\cdot cm^{-1}\cdot s^{-1}}. \]