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23-Chem-B1 Transport Phenomena · December 2019

Question 1 of 6: A1 — Flow in a slightly tapered tube

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

Notes on this paper

Paper format. 16-Chem-B1 Transport Phenomena, December 2019 — open-book, 3 hours, six problems (25 marks each) in three sections: A Fluid Mechanics (A1, A2), B Heat Transfer (B1, B2), C Mass Transfer (C1, C2). The rubric asks the candidate to attempt one problem from each section plus a fourth; all six are worked here as a complete study resource. Pages 9–12 of the paper are an appendix of the equations of change (continuity, Navier–Stokes, energy, species) and are used as the reference tables 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 film/annular differential balances (A1, C1, 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 internal-/external-flow convection correlations (A2, B1, B2); F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — radiation enclosures, Dittus–Boelter, Churchill–Chu natural convection and air properties (B1, B2); C. J. Geankoplis, Transport Processes and Separation Process Principles (Prentice Hall) — the film model for diffusion with reaction (C1); R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties of air and water.

Question 1: A1 — Flow in a slightly tapered tube (25 marks: 2 + 3 + 15 + setup)

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.

QuantitySymbolValue / relation
Entrance / exit radius$R_0,\ R_L$$R_L<R_0$ (slight taper)
Tube length$L$entrance $z=0$ to exit $z=L$
Fluid$\rho,\ \mu$constant (incompressible, Newtonian)
Local mass-flow relation$w$$\dfrac{\pi\rho R^4}{8\mu}\left(-\dfrac{dP}{dz}\right)$

Find. (a) $R(z)$; (b) the same balance with $R$ as the independent variable; (c) the closed-form mass flow rate $w$ as a function of the overall pressure drop $P_0-P_L$ and the two radii.

$R_0$ R₀ R₃ flow, w z=0 z=L z P = P₀ P = P₃
Fig. A1: The radius contracts linearly from $R_0$ to $R_L$. Because the local resistance scales as $R^{-4}$, most of the pressure drop occurs near the narrow exit; the problem is treated as a chain of Hagen–Poiseuille slices integrated along the taper (a lubrication approximation).

Approach. Write the linear radius profile, substitute it into the differential Hagen–Poiseuille balance, change the integration variable from $z$ to $R$ (the two are linearly related, so this is exact), then integrate the separated pressure equation from entrance to exit and collect the radii.

  1. (a) Linear radius profile. A straight-line contraction from $R_0$ at $z=0$ to $R_L$ at $z=L$ is $$\boxed{R(z)=R_0+\frac{R_L-R_0}{L}\,z.}$$
  2. Rearrange the local balance for the pressure gradient. Solving the given relation for $-dP/dz$, $$-\frac{dP}{dz}=\frac{8\mu w}{\pi\rho\,[R(z)]^4}.$$ This is a first-order ODE for $P(z)$ with a strongly varying coefficient.
  3. (b) Change the independent variable to $R$. From (a), $R$ is linear in $z$ with the constant slope $m\equiv\dfrac{dR}{dz}=\dfrac{R_L-R_0}{L}$, so $dz=dR/m$ and $-\dfrac{dP}{dz}=-m\,\dfrac{dP}{dR}$. The local balance becomes an expression in $R$ alone: $$w=\frac{\pi\rho R^4}{8\mu}\left(-\frac{dP}{dz}\right)=\frac{\pi\rho R^4}{8\mu}\left(\frac{R_L-R_0}{L}\right)\left(-\frac{dP}{dR}\right).$$ Equivalently $-\dfrac{dP}{dR}=\dfrac{8\mu w\,L}{\pi\rho\,(R_L-R_0)\,R^4}$, which is now directly separable in $R$.
  4. (c) Integrate across the tube. Integrate the pressure from $P_0$ (at $R=R_0$) to $P_L$ (at $R=R_L$): $$P_0-P_L=\frac{8\mu w\,L}{\pi\rho\,(R_L-R_0)}\int_{R_0}^{R_L}\frac{dR}{R^4}=\frac{8\mu w\,L}{\pi\rho\,(R_L-R_0)}\cdot\frac{1}{3}\!\left(\frac{1}{R_0^{3}}-\frac{1}{R_L^{3}}\right).$$
  5. Simplify the radius factor. Using $\dfrac{1}{R_0^3}-\dfrac{1}{R_L^3}=\dfrac{R_L^3-R_0^3}{R_0^3R_L^3}$ and the factorisation $R_L^3-R_0^3=(R_L-R_0)(R_0^2+R_0R_L+R_L^2)$, the awkward $(R_L-R_0)$ cancels, leaving $$P_0-P_L=\frac{8\mu w\,L}{3\pi\rho}\cdot\frac{R_0^2+R_0R_L+R_L^2}{R_0^3R_L^3}.$$
  6. Solve for the mass flow rate. $$\boxed{\,w=\frac{3\pi\rho\,(P_0-P_L)\,R_0^{3}R_L^{3}}{8\mu L\,(R_0^{2}+R_0R_L+R_L^{2})}=\frac{3\pi\rho\,(P_0-P_L)\,R_0^{4}}{8\mu L}\cdot\frac{(R_L/R_0)^{3}}{1+(R_L/R_0)+(R_L/R_0)^{2}}\,}$$
  7. Sanity check — uniform-tube limit. Setting $R_L=R_0$ gives $\dfrac{(1)^3}{1+1+1}=\dfrac13$, so $w=\dfrac{3\pi\rho(P_0-P_L)R_0^4}{8\mu L}\cdot\dfrac13=\dfrac{\pi\rho(P_0-P_L)R_0^4}{8\mu L}$ — exactly the Hagen–Poiseuille result for a straight tube, confirming the algebra.
  8. Match the printed target. With $\lambda=R_L/R_0$, the printed bracket simplifies as $$1-\frac{1+\lambda+\lambda^2-3\lambda^3}{1+\lambda+\lambda^2}=\frac{(1+\lambda+\lambda^2)-(1+\lambda+\lambda^2)+3\lambda^3}{1+\lambda+\lambda^2}=\frac{3\lambda^3}{1+\lambda+\lambda^2},$$ so the printed expression is $\dfrac{3\pi\rho(\mathcal P_0-\mathcal P_L)R_0^4}{8\mu}\cdot\dfrac{\lambda^3}{1+\lambda+\lambda^2}$ — exactly the boxed result above, apart from the tube length $L$ in the denominator (see the callout). The modified pressure $\mathcal P=P+\rho g z$ reduces to $P$ for a horizontal tube.
PartResult
(a) Radius profile$R(z)=R_0+\dfrac{R_L-R_0}{L}z$
(b) Balance in $R$$w=\dfrac{\pi\rho R^4}{8\mu}\dfrac{R_L-R_0}{L}\left(-\dfrac{dP}{dR}\right)$
(c) Mass flow rate$w=\dfrac{3\pi\rho(P_0-P_L)R_0^{3}R_L^{3}}{8\mu L\,(R_0^{2}+R_0R_L+R_L^{2})}$
Uniform-tube check$R_L\to R_0\Rightarrow w=\dfrac{\pi\rho(P_0-P_L)R_0^4}{8\mu L}$ (Hagen–Poiseuille)
Check — printed target on p.2 (the L is missing)

The printed part-(c) target, $w=\dfrac{\pi(\mathcal P_0-\mathcal P_L)R_0^{4}\rho}{8\mu}\left[1-\dfrac{1+\lambda+\lambda^2-3\lambda^3}{1+\lambda+\lambda^2}\right]$, is the same result as the derivation: the bracket reduces to $3\lambda^3/(1+\lambda+\lambda^2)$ (last step). Its prefactor, however, prints $8\mu$ where the integration gives $8\mu L$. The bracket is dimensionless, so without $L$ the right-hand side would have units of kg·m/s instead of kg/s, and it would not reduce to Hagen–Poiseuille, $w=\pi\rho(\mathcal P_0-\mathcal P_L)R_0^4/(8\mu L)$, when $R_L=R_0$. Treat the missing $L$ as a typo on the paper; the answer keeps it.

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