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23-Chem-B1 Transport Phenomena · Undated paper

Question 2 of 6: A2 — Power-law melt flow between parallel plates

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper. National Examination (Engineers Canada) — 16-Chem-B1 Transport Phenomena, May 2019. Open book, 3 hours. Six 25-point problems in three sections — A Fluid Mechanics (A1, A2), B Heat Transfer (B1, B2), C Mass Transfer (C1, C2); one problem from each section must be attempted, plus a fourth from any section. All six problems are solved and fully worked below.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change, the power-law slit-flow momentum balance and the film/annular diffusion balances (A2, B2, C2); J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, the Moody chart, and the external-flow convection/mass-transfer correlations (A1, B1, C1); F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — Churchill–Chu free convection, horizontal-plate correlations, slug-flow internal convection and air properties (B1, B2); C. J. Geankoplis, Transport Processes and Separation Process Principles (Prentice Hall) — boundary-layer mass transfer and the film model (C1); R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.

Question 2: A2 — Power-law melt flow between parallel plates (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Constitutive law $\tau_{yx}=-\eta_o[dV_x/dY]^n$, read with the sign carried by the gradient, $\tau_{yx}=-\eta_o\lvert dV_x/dY\rvert^{n-1}(dV_x/dY)$ (so the stress always opposes the local shear); $\eta_o$ and $n$ are symbols only — no numbers are given. Assumed geometry: stationary plates a distance $2B$ apart at $Y=\pm B$, width $W$, flow driven by a constant pressure gradient $-\,dP/dx=\Delta P/L$.

Find. The fully-developed velocity profile $V_x(Y)$ and the volumetric flow rate $Q$.

Y = +B Y = −B Y = 0 Vₓ(Y) Vₓ,max flow → x (blunt, shear-thinning) −dP/dx = ΔP/L > 0
Fig. A2: Pressure-driven slit flow of a shear-thinning melt ($n<1$). Zero shear stress on the mid-plane and no-slip at the walls give a symmetric profile that is blunter than a parabola — the low-shear core near $Y=0$ is highly viscous and moves almost as a plug.

Approach. Integrate the $x$-momentum balance to get the (linear) shear-stress distribution, invert the power law to get the velocity gradient, integrate with no-slip to obtain $V_x(Y)$, and integrate $V_x$ over the gap for $Q$.

  1. Shear-stress distribution. For steady, fully-developed flow the $x$-momentum balance is $d\tau_{yx}/dY=-dP/dx=\text{const}$. Integrating and using symmetry ($\tau_{yx}=0$ at $Y=0$): $$\tau_{yx}=\frac{\Delta P}{L}\,Y.$$ The magnitude rises linearly from zero at the centre to a maximum at each wall.
  2. Invert the power law. In the upper half ($Y>0$) the velocity decreases outward, so $dV_x/dY<0$ and the printed law gives $\tau_{yx}=\eta_o\lvert dV_x/dY\rvert^{n}>0$. Setting this equal to $(\Delta P/L)\,Y$: $$\frac{dV_x}{dY}=-\left(\frac{\Delta P}{L\,\eta_o}\right)^{1/n}Y^{1/n}.$$
  3. Integrate to the velocity profile. With no-slip $V_x(B)=0$: $$\boxed{\;V_x(Y)=\frac{n}{n+1}\left(\frac{\Delta P}{L\,\eta_o}\right)^{1/n}\left[B^{\frac{n+1}{n}}-\lvert Y\rvert^{\frac{n+1}{n}}\right].\;}$$ The centreline maximum is $V_{x,\max}=\dfrac{n}{n+1}\left(\dfrac{\Delta P}{L\,\eta_o}\right)^{1/n}B^{\frac{n+1}{n}}.$
  4. Volumetric flow rate. Integrate over the gap (width $W$), using symmetry: $$Q=2W\!\int_{0}^{B}\!V_x\,dY=2W\frac{n}{n+1}\left(\frac{\Delta P}{L\eta_o}\right)^{1/n}\left[B^{\frac{2n+1}{n}}-\frac{n}{2n+1}B^{\frac{2n+1}{n}}\right],$$ $$\boxed{\;Q=\frac{2n}{2n+1}\,W\left(\frac{\Delta P}{L\,\eta_o}\right)^{1/n}B^{\frac{2n+1}{n}}.\;}$$
  5. Check the Newtonian limit and read the bluntness. For $n=1,\ \eta_o=\mu$ the flow rate becomes $Q=\tfrac{2}{3}W B^3\Delta P/(\mu L)$, the exact result for a Newtonian slit of half-gap $B$. The profile bluntness is $V_{x,\max}/\bar V=(2n+1)/(n+1)$: $1.5$ for a Newtonian fluid, falling toward plug flow ($\to1$) as $n\to0$ — the signature of shear-thinning ($n<1$).
QuantityResult
Velocity profile$V_x=\dfrac{n}{n+1}\left(\dfrac{\Delta P}{L\eta_o}\right)^{1/n}\!\left[B^{\frac{n+1}{n}}-\lvert Y\rvert^{\frac{n+1}{n}}\right]$
Maximum (centreline) velocity$V_{x,\max}=\dfrac{n}{n+1}\left(\dfrac{\Delta P}{L\eta_o}\right)^{1/n}B^{\frac{n+1}{n}}$
Volumetric flow rate$Q=\dfrac{2n}{2n+1}\,W\left(\dfrac{\Delta P}{L\eta_o}\right)^{1/n}B^{\frac{2n+1}{n}}$
Bluntness $V_{x,\max}/\bar V$$(2n+1)/(n+1)$
Check — reading of the printed law and assumptions

As printed, $-\eta_o[dV_x/dY]^n$ is undefined for a negative gradient and non-integer $n$, so it is used in its standard signed form $-\eta_o\lvert dV_x/dY\rvert^{n-1}(dV_x/dY)$ (the Ostwald–de Waele model with consistency $\eta_o$). The result assumes steady, laminar, fully-developed, isothermal flow driven by a constant axial pressure gradient with $\partial P/\partial Y=0$, stationary plates with no-slip, and a half-gap $B$ (full gap $2B$). If the full gap is called $B$, replace $B\to B/2$ throughout.