Question 1 of 6: Non-Newtonian Viscosity of a CMC Solution from Capillary Data
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Open-book, 3 hours; six numbered problems of equal value (20 points each), of which five constitute a complete paper (only the first five in the answer book are marked). All six problems are solved below so the set is complete for study.
Reference texts: Odian, Principles of Polymerization (4th ed., Wiley) — chain-growth & living/anionic kinetics, molecular-weight distributions; Rudin & Choi, The Elements of Polymer Science and Engineering (3rd ed., Academic Press) — dilute-solution rheology, MWD averages, capillary viscometry; Tadmor & Gogos, Principles of Polymer Processing (2nd ed., Wiley) — calendering, injection filling, die flow; Sperling, Introduction to Physical Polymer Science (4th ed., Wiley) — viscoelasticity; Young & Lovell, Introduction to Polymers (3rd ed.) — polyolefin processing; Middleman, Fundamentals of Polymer Processing — power-law tube/runner flow.
Question 1: Non-Newtonian Viscosity of a CMC Solution from Capillary Data (20 points)
Find. The apparent (non-Newtonian) viscosity η as a function of the true wall shear rate γ̇w.
Approach. Convert Q/R3 to the nominal Newtonian wall shear rate, extract the power-law index from the log–log slope of τw, apply the Weissenberg–Rabinowitsch correction to get the true wall shear rate, then divide stress by shear rate point-by-point.
Check
The fluid is assumed shear-thinning and power-law over the measured range (the log–log τw–shear-rate data fall on a straight line, so a single index n applies); if it were not, n would be evaluated locally at each point as the running slope.
Nominal Newtonian wall shear rate. For laminar capillary flow the apparent (Newtonian) wall shear rate is fixed by the throughput alone:$$\dot\gamma_{app}=\frac{4Q}{\pi R^{3}}=\frac{4}{\pi}\left(\frac{Q}{R^{3}}\right)=1.273\left(\frac{Q}{R^{3}}\right)$$so the tabulated Q/R3 maps directly onto γ̇app (e.g. 195 → 248 s-1, 19 645 → 25 013 s-1).
Power-law index from the log–log slope. For a power-law fluid $\tau_w=m'\,\dot\gamma_{app}^{\,n}$, so the slope of $\ln\tau_w$ against $\ln\dot\gamma_{app}$ is the flow index. A least-squares fit of the fifteen points gives$$n=\frac{d\ln\tau_w}{d\ln\dot\gamma_{app}}=\boxed{0.371}$$confirming a strongly shear-thinning liquid (n < 1).
Weissenberg–Rabinowitsch correction. The true wall shear rate exceeds the nominal value because the velocity profile is blunted:$$\dot\gamma_w=\frac{3n+1}{4n}\,\dot\gamma_{app}=\frac{3(0.371)+1}{4(0.371)}\,\dot\gamma_{app}=1.424\,\dot\gamma_{app}$$
Non-Newtonian viscosity. The apparent viscosity at each point is stress over true shear rate, $\eta=\tau_w/\dot\gamma_w$. Fitting $\tau_w=m\,\dot\gamma_w^{\,n}$ gives a consistency $m\approx26.6$ Pa·sn, so equivalently $\eta=m\,\dot\gamma_w^{\,n-1}$. At the extremes of the data:$$\eta(353\ \text{s}^{-1})=\frac{220}{353.5}=\boxed{0.622\ \text{Pa}\cdot\text{s}},\qquad \eta(35\,616\ \text{s}^{-1})=\frac{1200}{35616}=\boxed{0.0337\ \text{Pa}\cdot\text{s}}$$The viscosity falls roughly 18-fold as shear rate rises two decades, exactly the $\eta\propto\dot\gamma^{\,n-1}=\dot\gamma^{-0.629}$ signature of the power law.
Fig. 1 — apparent viscosity vs true wall shear rate on log–log axes; the straight line of slope n−1 = −0.63 is the power-law fingerprint.
