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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 τ\tau and rate of shearing strain dudy\frac{du}{dy} 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.

τ\tau (N/m²)0.040.060.120.180.30.521.122.1
dudy\frac{du}{dy} (s⁻¹)2.254.511.2522.54590225450

A Newtonian fluid obeys τ=μ dudy\tau=\mu\,\dfrac{du}{dy}, so the ratio τ/(du/dy)\tau/(du/dy) is constant and τ\tau against du/dydu/dy is a straight line through the origin.

Apparent viscosity μa=τ/dudy\mu_a=\tau\big/\dfrac{du}{dy} for each reading:

du/dydu/dy (s⁻¹)2.254.511.2522.54590225450
τ\tau (N/m²)0.040.060.120.180.300.521.122.10
μa\mu_a (Pa·s)0.01780.01330.01070.00800.00670.00580.00500.0047

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 τ\tau against du/dydu/dy is a curve, not a straight line.

A log-log fit of the data gives the power-law (Ostwald-de Waele) form

τ=K(dudy)n,n≈0.74,K≈0.0196 Pa⋅sn\tau=K\left(\frac{du}{dy}\right)^{n},\qquad n\approx0.74,\quad K\approx0.0196\ \text{Pa·s}^{n}

Since n<1n<1 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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