23-Chem-A2 Unit Operations and Separation Processes · December 2014
Question 1 of 6: Parallel-Pipe Flow Split and Equivalent Single Pipe
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, December 2014. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt two from each section. All six are worked below for completeness.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, filtration, crystallisation); Perry's Chemical Engineers' Handbook (9th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.).
Question A1: Parallel-Pipe Flow Split and Equivalent Single Pipe (25 marks)
Given. Two pipes join the same two reservoirs, so they share the same length $L$ and the same pressure drop $\Delta P$. Turbulent friction with $f\propto \mathrm{Re}^{-1/4}$ (Blasius form).
Quantity
Value
Total flow $Q$
$1.4\ \mathrm{m^3/s}$
Pipe 1 diameter $D_1$
0.30 m
Pipe 2 diameter $D_2$
0.60 m
Friction law
$f = C\,\mathrm{Re}^{-1/4}$
Elevation / entrance / exit
negligible
Find. (a) the velocity in each parallel pipe; (b) the single-pipe diameter that carries the full $1.4\ \mathrm{m^3/s}$ at the same $\Delta P$.
Figure A1 — Two pipes in parallel between common reservoirs: identical $\Delta P$ and $L$, flows add to $1.4\ \mathrm{m^3/s}$.
Approach. Express the pressure drop in each pipe through the Blasius friction law, set the two drops equal (parallel branches), obtain a velocity–diameter scaling, then close with continuity; for (b) impose the same $\Delta P$-group on a single pipe carrying the full flow.
Write $\Delta P$ in terms of $v$ and $D$. For turbulent pipe flow $\Delta P = 4f\dfrac{L}{D}\dfrac{\rho v^2}{2}$ with $f=C\,\mathrm{Re}^{-1/4}=C\left(\dfrac{\rho v D}{\mu}\right)^{-1/4}$. Collecting the diameter and velocity powers, $$\Delta P \;\propto\; \frac{L}{D}\,v^{2}\,(vD)^{-1/4}\;=\;L\,v^{7/4}\,D^{-5/4}.$$
Equal drop across the parallel branches. Both pipes share $L$ and $\Delta P$, so $v^{7/4}D^{-5/4}$ is the same for each: $$v_1^{7/4}D_1^{-5/4}=v_2^{7/4}D_2^{-5/4}\;\Rightarrow\; \frac{v_2}{v_1}=\left(\frac{D_2}{D_1}\right)^{5/7}=2^{5/7}=1.641.$$
Close with continuity. The branch flows sum to the total, $\tfrac{\pi}{4}\!\left(D_1^2 v_1 + D_2^2 v_2\right)=1.4$. Substituting $v_2=1.641\,v_1$: $$\tfrac{\pi}{4}\big[0.09 + 0.36(1.641)\big]v_1 = 0.5346\,v_1 = 1.4 \;\Rightarrow\; v_1 = 2.62\ \mathrm{m/s}.$$ Hence $v_2 = 1.641(2.62)=4.30\ \mathrm{m/s}$. $v_1 = 2.62\ \mathrm{m/s}$ (0.30 m pipe), $v_2 = 4.30\ \mathrm{m/s}$ (0.60 m pipe) A check: $Q_1=0.185$, $Q_2=1.215\ \mathrm{m^3/s}$, summing to $1.40\ \mathrm{m^3/s}$.
(b) Single equivalent pipe at the same $\Delta P$. The new pipe must reproduce the branch $\Delta P$-group while carrying $Q=1.4$, so $v_s=\dfrac{4Q}{\pi D_s^2}$ and $v_s^{7/4}D_s^{-5/4}=v_1^{7/4}D_1^{-5/4}$. Substituting $v_s$ gives one equation in $D_s$: $$\left(\frac{4Q}{\pi D_s^2}\right)^{7/4}D_s^{-5/4}=v_1^{7/4}D_1^{-5/4}\;\Rightarrow\; D_s = 0.632\ \mathrm{m}.$$ The velocity is then $v_s = \dfrac{4(1.4)}{\pi(0.632)^2}=4.46\ \mathrm{m/s}$. $D_s = 0.63\ \mathrm{m}$ at $v_s = 4.46\ \mathrm{m/s}$
