Question 2 of 6: A2 — Flow through a symmetric arterial stenosis
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. EGBC 16-CHEM-B1 Transport Phenomena, May 2017, 3 hours, open-book (one textbook permitted, no loose notes). Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer), each worth 25 marks; candidates attempt one problem from each section plus a fourth from any section, and only the first four in the answer book are marked. All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species-continuity tables) in rectangular, cylindrical and spherical coordinates; these are quoted rather than re-derived.
Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — laminar and turbulent boundary layers, flat-plate drag, and film transfer coefficients; W. M. Deen, Analysis of Transport Phenomena (Oxford) — the physiological transport problems (stenosis flow, reaction–diffusion of O&sub2; in tissue); F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — convection correlations and the mixed-convection $Gr/Re^2$ criterion; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.
Question 2: A2 — Flow through a symmetric arterial stenosis (25 marks)
Given. A parabolic (Poiseuille-shape) axial velocity $V_z(r)=V_{max}\{1-r^2/R^2\}$ where the local channel radius $R(z)=R_0\{1-0.5[1-4(z/L)^2]^{1/2}\}$; steady incompressible flow; the same volumetric flow $Q$ passes every cross-section.
Find. (a) $V_{max}(z)$ in terms of $Q$, $R_0$ and $z/L$; (b) the wall-shear ratio $\tau_w(z{=}0)/\tau_{w,\infty}$ between the throat and the unobstructed vessel.
Fig. A2: A symmetric constriction. Continuity fixes the same $Q$ at every section, so as the radius contracts to $R(0)=R_0/2$ at the throat, the centreline velocity and the wall shear both climb sharply.
Approach. This is a derivation: integrate the assumed parabolic profile to relate $V_{max}$ to the (constant) flow $Q$ at any section, then differentiate the profile to get the wall shear and form the throat-to-vessel ratio.
(a) Continuity fixes $V_{max}$ at each section. The volumetric flow through a section of radius $R$ is $$Q=\int_0^{R}V_z\,2\pi r\,dr=2\pi V_{max}\!\int_0^{R}\!\Big(1-\frac{r^2}{R^2}\Big)r\,dr=2\pi V_{max}\Big(\frac{R^2}{2}-\frac{R^2}{4}\Big)=\frac{\pi V_{max}R^2}{2}.$$ Solving, $$\boxed{V_{max}(z)=\frac{2Q}{\pi R(z)^2}.}$$
(a) Insert the stenosis geometry. With $R(z)=R_0\{1-0.5[1-4(z/L)^2]^{1/2}\}$, $$\boxed{V_{max}(z)=\frac{2Q}{\pi R_0^{2}\big\{1-0.5[1-4(z/L)^2]^{1/2}\big\}^{2}}.}$$ At the ends of the stenosis ($|z|=L/2$) the square root vanishes and $R=R_0$, so outside it $V_{max}=2Q/\pi R_0^2$ is constant (the printed $R_i(z)$ and $R(z)$ denote the same channel radius); inside it grows as the channel narrows.
(b) Wall shear stress from the profile. For this axial flow $\tau_{rz}=\mu\,dV_z/dr$ with $dV_z/dr=-2V_{max}r/R^2$, so the magnitude at the wall $r=R$ is $$\tau_w=\left|\mu\frac{dV_z}{dr}\right|_{r=R}=\frac{2\mu V_{max}}{R}=\frac{4\mu Q}{\pi R^{3}},$$ using $V_{max}=2Q/\pi R^2$ from step 1.
(b) Form the throat-to-vessel ratio. Because $\mu$ and $Q$ are common to both sections, $$\frac{\tau_w(0)}{\tau_{w,\infty}}=\left(\frac{R_0}{R(0)}\right)^{3}.$$ At the throat $z=0$: $R(0)=R_0\{1-0.5(1)\}=R_0/2$, hence $$\boxed{\frac{\tau_w(0)}{\tau_{w,\infty}}=\left(\frac{R_0}{R_0/2}\right)^{3}=2^{3}=8.}$$ The wall shear at the throat is eight times that in the healthy vessel.
