Question 4 of 4: Parallel galvanized-iron pipe loop, dead pump as a fixed loss
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2019 — 04-Bio-A7, 3 hours, open book (any non-communicating calculator permitted). Four questions constitute a complete exam paper, each of equal value, and all require calculation. All four are solved here.
Check — Question 4 roughness. The paper prints the roughness in Question 4 as “ε=0.15 mm”, and this value is used throughout.
Reference texts. White, Fluid Mechanics, 8th ed. (dimensional analysis and non-dimensionalizing the governing equations, manometry, laminar Hagen–Poiseuille pipe flow, and turbulent Darcy–Weisbach/Colebrook pipe networks — ch. 2, 5, 6).
Question 4: Parallel galvanized-iron pipe loop, dead pump as a fixed loss (25 marks)
Note — given data. This question uses $L_1=60$ m, $D_1=5$ cm; $L_2=55$ m, $D_2=4$ cm; $Q_{tot}=0.036$ m³/s; $K=1.5$; water at 20°C. The paper prints the roughness as “ε=0.15 mm”, and this value is used.
Given.
Quantity
Pipe 1
Pipe 2 (dead pump branch)
Length L
60 m
55 m
Diameter D
5 cm
4 cm
Roughness ε=0.15 mm (galvanized iron); water at 20°C, ρ=998 kg/m³, μ=0.001 Pa·s; total flow $Q_{tot}=0.036$ m$^3$/s splits between the two pipes and recombines; the idle pump in pipe 2's branch acts as a fixed minor loss, $K=1.5$.
Find. (a) $Q_1$ and $Q_2$; (b) the overall pressure drop across the parallel combination.
Parallel pipe loop: the two branches share the same start and end nodes, so their head losses must be equal.
Approach. The two pipes share both end nodes, so their head losses must be equal: $h_1(Q_1)=h_2(Q_{tot}-Q_1)$, each evaluated with the Darcy–Weisbach equation and the Colebrook friction factor (plus the pump's $K=1.5$ minor loss in branch 2), with $Q_1+Q_2=Q_{tot}$. Solve iteratively for the split, then recover the pressure drop from either branch.
Set up the governing equations.
$$h_1 = f_1\frac{L_1}{D_1}\frac{V_1^2}{2g}, \qquad h_2 = \left(f_2\frac{L_2}{D_2}+K\right)\frac{V_2^2}{2g}, \qquad h_1=h_2, \qquad Q_1+Q_2=0.036$$
with $f$ from Colebrook, $\dfrac{1}{\sqrt f}=-2\log_{10}\!\left(\dfrac{\varepsilon/D}{3.7}+\dfrac{2.51}{Re\sqrt f}\right)$, iterated to convergence at each trial split.
Iterate the flow split. Guessing $Q_1$, computing $V_1=Q_1/A_1$, $Re_1$, $f_1$ (Colebrook), and $h_1$; doing the same for pipe 2 with $Q_2=0.036-Q_1$ and the added $K=1.5$; and adjusting $Q_1$ until $h_1=h_2$ converges to
$$\boxed{Q_1 = 0.02296\ \text{m}^3/\text{s} = 22.96\ \text{L/s} \ (63.8\%), \qquad Q_2 = 0.01304\ \text{m}^3/\text{s} = 13.04\ \text{L/s}\ (36.2\%)}$$
which indeed sums to the given 0.036 m$^3$/s.
Report the converged branch state. Pipe 1: $V_1=11.69$ m/s, $Re_1\approx5.84\times10^5$, $f_1\approx0.0264$. Pipe 2: $V_2=10.38$ m/s, $Re_2\approx4.14\times10^5$, $f_2\approx0.0282$. Both flows are fully turbulent (rough-pipe regime), consistent with using Colebrook rather than a laminar formula. Pipe 1 carries the larger share because it has both the larger diameter and no added loss coefficient.
Compute the overall pressure drop (part b). The common head loss is $h=h_1=h_2\approx220.8$ m, so
$$\boxed{\Delta p = \rho g h = (998)(9.81)(220.8) = 2.162\times10^{6}\ \text{Pa} \approx 2.16\ \text{MPa}}$$
Check — velocity magnitude. The converged velocities (≈10–12 m/s) and pressure drop (≈2.16 MPa) are unusually high for a typical service pipe, but they follow directly and consistently from the stated total flow rate (0.036 m³/s, i.e. 36 L/s) forced through small-diameter (4–5 cm) pipes — both branches converge to the identical head loss (220.8 m) as required for a true parallel connection, and $Q_1+Q_2$ reproduces the given total exactly, so the result is taken as given rather than adjusted.
