Question 2 of 6: Two-Layer Tube Flow — Newtonian Core and Power-Law Annulus
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 marks 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- and step-growth kinetics, emulsion polymerization, molecular-weight distributions; Rudin & Choi, The Elements of Polymer Science and Engineering (3rd ed., Academic Press) — dilute-solution properties, osmometry, viscoelasticity; Sperling, Introduction to Physical Polymer Science (4th ed., Wiley) — linear viscoelasticity, terminal-zone moduli; Bird, Armstrong & Hassager, Dynamics of Polymeric Liquids, Vol. 1 — non-Newtonian tube flow; supporting polyolefin process detail from Young & Lovell, Introduction to Polymers (3rd ed.).
Find. (a) umax (centreline); (b) u at r = R1; (c) Q = QA + QB.
Approach. A momentum balance on a thin cylindrical fluid element fixes the shear stress $\tau_{rz}=\tfrac12 G r$ everywhere in the tube regardless of the constitutive law, so stress continuity across the interface is automatic; integrating each fluid's rate-of-strain from the wall inward, using velocity continuity at R1, gives the two velocity profiles and their integrals.
Check
The datum "zero-shear viscosity of B = 10 Pa·s" is read as the power-law consistency m = 10 Pa·sn (a true zero-shear viscosity is undefined for a shear-thinning power law, which diverges as γ̇→0). "i0 Pa.s" for A is read as 10 Pa·s.
Shear-stress distribution. For steady, fully-developed axial flow a momentum balance on a thin cylindrical fluid element gives, in both regions,
$$\tau_{rz}(r)=\frac{G\,r}{2},\qquad G=-\frac{dP}{dz}=5\times10^{4}\ \text{Pa}\cdot\text{m}^{-1}$$
so the stress is continuous at R1 by construction.
Annulus B (power-law), integrate from the wall. With $\tau=m|du/dr|^{n}$ and $du/dr<0$, $-du/dr=(Gr/2m)^{1/n}$. Integrating from R2 (u = 0) inward:
$$u_B(r)=\left(\frac{G}{2m}\right)^{1/n}\frac{n}{n+1}\left[R_2^{\frac{n+1}{n}}-r^{\frac{n+1}{n}}\right]$$
With n = 0.5 the exponents are 1/n = 2 and (n+1)/n = 3, and $(G/2m)^{2}=(2500)^2=6.25\times10^{6}$.
(b) Interface velocity at r = R1:
$$u_B(R_1)=\frac{6.25\times10^{6}}{3}\left[(0.005)^3-(0.004)^3\right]=2.083\times10^{6}\times6.1\times10^{-8}=\boxed{0.127\ \text{m}\cdot\text{s}^{-1}}$$
Core A (Newtonian), integrate from the interface. With $\tau=\mu_A(-du/dr)$, $-du/dr=Gr/2\mu_A$; integrating inward from R1 with velocity continuity $u_A(R_1)=u_B(R_1)$:
$$u_A(r)=u_B(R_1)+\frac{G}{4\mu_A}\left(R_1^2-r^2\right)$$
(a) Maximum velocity at the centreline r = 0, where $G/4\mu_A=1250$:
$$u_{max}=0.127+1250\,(0.004)^2=0.127+0.020=\boxed{0.147\ \text{m}\cdot\text{s}^{-1}}$$
(c) Volumetric flow rates, $Q=\int u\,2\pi r\,dr$ over each region:
$$Q_A=2\pi\!\left[\frac{u_B(R_1)R_1^2}{2}+\frac{G R_1^4}{16\mu_A}\right]=6.89\times10^{-6}\ \text{m}^3\text{s}^{-1}$$
$$Q_B=2\pi\!\left(\frac{G}{2m}\right)^{1/n}\!\frac{n}{n+1}\!\left[\frac{R_2^{3}(R_2^2-R_1^2)}{2}-\frac{R_2^{5}-R_1^{5}}{5}\right]=1.86\times10^{-6}\ \text{m}^3\text{s}^{-1}$$
$$Q=Q_A+Q_B=\boxed{8.75\times10^{-6}\ \text{m}^3\text{s}^{-1}=8.75\ \text{cm}^3\text{s}^{-1}}$$
Fig. 2 — axial velocity profile: parabolic Newtonian core (0–4 mm) joined with velocity continuity to the shear-thinning power-law annulus (4–5 mm), whose profile is flatter than a Newtonian annulus would be at the same wall stress; centreline umax = 0.147 m/s, interface u = 0.127 m/s.
