- 2078 Kartik · 4 marks
Newtonian or non-Newtonian fluid from data
Short answer
The apparent viscosity μ_a = τ/(du/dy) is not constant, decreasing from 0.0178 to 0.0047 Pa·s as shear rate increases. This indicates non-Newtonian, shear-thinning (pseudoplastic) behavior with power-law index n ≈ 0.74.
The viscosity of one of the liquids in a laboratory is determined by measurements of shear stress and rate of shearing strain tested in a suitable viscometer. Based on the following observations, determine if the given liquid is Newtonian or Non-Newtonian fluid. Explain how you arrive at your answer.
| (N/m²) | 0.04 | 0.06 | 0.12 | 0.18 | 0.3 | 0.52 | 1.12 | 2.1 |
|---|---|---|---|---|---|---|---|---|
| (s⁻¹) | 2.25 | 4.5 | 11.25 | 22.5 | 45 | 90 | 225 | 450 |
A Newtonian fluid obeys , so the ratio is constant and against is a straight line through the origin.
Apparent viscosity for each reading:
| (s⁻¹) | 2.25 | 4.5 | 11.25 | 22.5 | 45 | 90 | 225 | 450 |
|---|---|---|---|---|---|---|---|---|
| (N/m²) | 0.04 | 0.06 | 0.12 | 0.18 | 0.30 | 0.52 | 1.12 | 2.10 |
| (Pa·s) | 0.0178 | 0.0133 | 0.0107 | 0.0080 | 0.0067 | 0.0058 | 0.0050 | 0.0047 |
The ratio is not constant: it falls steadily from 0.0178 to 0.0047 Pa·s as the shear rate rises, so the plot of against is a curve, not a straight line.
A log-log fit of the data gives the power-law (Ostwald-de Waele) form
Since and the apparent viscosity decreases with shear rate, the liquid is non-Newtonian, of the pseudoplastic (shear-thinning) type.
✓ Answer checked by Claude Opus 5.5 (high effort)
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