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23-Chem-A2 Unit Operations and Separation Processes · Undated paper

Question 2 of 6: Diameter of a Smooth Pipe from a Measured Pressure Drop

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

Notes on this paper

Paper format. National Exams — 16-CHEM-A2 Unit Operations and Separation Processes, May 2019 (this is the “undated” PDF; the cover sheet is a faint template reading MAY, while every interior page reads May 2019). 3 hours, open book. Six problems: Section A (Momentum-transfer / mechanical operations) A1–A3, and Section B (Separation processes) B1–B3, each 25 marks. Candidates attempt at least two from each section and only the first two per section are marked; all six are worked below for completeness.

Reference texts. Geankoplis, Transport Processes and Separation Process Principles (4th ed.) — filtration, packed beds, absorption, adsorption, packed-tower distillation; McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 2; Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (9th ed.).

Note on the data. The B1 feed is described as acetone–water but the equilibrium table supplied is explicitly acetone–ethanol — B1 is solved with the data actually provided (acetone–ethanol) and the discrepancy is flagged.

Question A2: Diameter of a Smooth Pipe from a Measured Pressure Drop (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.

QuantityValue
Density $\rho$ (SG 2.6)2600 kg/m³
Viscosity $\mu$$2\times10^{-3}$ Pa·s
Pressure drop $\Delta P$0.183 psi = 1262 Pa
Length $L$2.784 km = 2784 m
Mass flow $\dot m$3175 kg/h = 0.8819 kg/s
Pipe wallhydraulically smooth ($\varepsilon\to0$)

Find. The internal pipe diameter $D$ (in inches).

Unknown-diameter smooth pipe — $\Delta P$ over length $L$ $\dot m$=3175 kg/h $L$ = 2784 m,   $\Delta P$ = 1262 Pa,   $D=?$ $P_1$ $P_2$
Fully-developed flow in a horizontal smooth pipe: the friction pressure drop fixes $f/D^5$, and the smooth-pipe friction law closes the system.

Approach. Write $\Delta P$ in the Fanning form, express $Re$ and velocity in terms of $D$ and the fixed mass flow, then iterate the smooth-pipe friction law (Colebrook / Fanning chart) until the computed $\Delta P$ matches the measurement.

  1. Volumetric flow. $$Q=\frac{\dot m}{\rho}=\frac{0.8819}{2600}=3.39\times10^{-4}\ \mathrm{m^3/s}.$$
  2. Express $Re$ and $v$ through $D$. With $v=Q/(\tfrac{\pi}{4}D^2)$, the Reynolds number depends only on $D$: $$Re=\frac{\rho vD}{\mu}=\frac{4\dot m}{\pi D\mu}=\frac{4(0.8819)}{\pi D(2\times10^{-3})}=\frac{561}{D}\quad(D\text{ in m}).$$
  3. Pressure-drop constraint (Fanning). $$\Delta P=4f\,\frac{L}{D}\,\frac{\rho v^2}{2}\ \Rightarrow\ f=\frac{\Delta P}{4(L/D)(\rho v^2/2)},$$ which, with $v\propto D^{-2}$, reduces to a single relation between $f$ and $D$ (equivalently $f\propto D^5$).
  4. Close with the smooth-pipe law. Iterating the Colebrook smooth-wall equation (identical to reading the “smooth pipe” curve on the supplied Fanning chart) together with the pressure constraint converges to $$D=0.115\ \mathrm{m},\qquad Re=\frac{561}{0.115}=4.9\times10^{3},\qquad f\approx0.0095.$$ The Reynolds number ($\approx4900$) is just into the turbulent regime, so the smooth-turbulent curve — not the laminar $16/Re$ line — is the correct branch.
  5. Convert to inches. $$D=\frac{0.115}{0.0254}=4.53\ \mathrm{in}.$$ Required pipe diameter $\approx 4.5$ in (0.115 m).
QuantityValue
Volumetric flow $Q$$3.39\times10^{-4}$ m³/s
Reynolds number $Re$$\approx 4.9\times10^{3}$ (turbulent, smooth)
Fanning friction factor $f$$\approx 0.0095$
Pipe diameter $D$0.115 m ≈ 4.5 in
Check. $Re\approx4900$ sits near the laminar–turbulent transition. The laminar branch ($f=16/Re$) gives an inconsistent $Re>2100$ and is therefore rejected; the smooth-turbulent solution above is the physically admissible root. If the working fluid were near the transition on the exam-day chart, read the smooth curve at $Re\approx5\times10^3$.