NivaarExam PrepOfficial exam papers ↗

04-BS-7 · May 2016

Question 6 of 13: Sizing a Pipeline Diameter via the Colebrook/Moody Friction Relation

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

Notes on this paper

04-BS-7 Mechanics of Fluids — May 2016 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs "do seven"; Section B (Analytical/Graphical) offers 4 questions and instructs "do three." Every question is answered below (13 of 13), so students can use the full paper as a study resource. Constants used throughout (from the paper's own Constants page): g = 9.81 m/s², patm = 100 kPa (an atmospheric head of 10 m of water is specified separately for Question 1), ρwater = 1000 kg/m³, SGglycerine = 1.26, SGmercury = 13.56, ρconcrete = 2400 kg/m³, ρair = 1.19 kg/m³ (20°C) / 1.21 kg/m³ (15°C), μwater = 1.0×10⁻³ N·s/m², Rair = 287 J/kg·K.

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics and manometry (Ch. 2), hydrostatic forces and the middle-third rule (Ch. 2), dimensional analysis and drag (Ch. 5, 7), pipe friction and the Moody/Colebrook relation (Ch. 6), control-volume momentum (Ch. 3); B. R. Munson et al., Fundamentals of Fluid Mechanics — jets, orifices, and streamline patterns (Ch. 5, 8); J. D. Anderson, Fundamentals of Aerodynamics — wave/compressibility drag divergence (Ch. 5) for the Boeing 747 wind-tunnel chart used in Question 9.

Question 6: Sizing a Pipeline Diameter via the Colebrook/Moody Friction Relation (5 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
Pipe length L5000 m
Available head hf120 m
Flow rate Q1 m³/s
Roughness ε0.045 mm
Water viscosity μ1.0×10⁻³ N·s/m²

Find. A commercial pipe diameter D that delivers Q = 1 m³/s for hf = 120 m.

Approach. For each trial D, compute V=4Q/(πD²), Re=VD/ν, and f from the Colebrook–White relation (what the Moody chart plots), then hf=f(L/D)(V²/2g); interpolate/solve for the D that gives hf=120 m.

  1. Darcy–Weisbach and Colebrook–White. $$h_f = f\frac{L}{D}\frac{V^2}{2g}, \qquad \frac{1}{\sqrt f} = -2\log_{10}\left(\frac{\varepsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\right)$$
  2. Trial diameters (as the hint suggests) — V, Re, f, and resulting hf.
    D (m)V (m/s)Refhf (m)
    0.505.092.55×10⁶0.0125164.7
    0.603.542.12×10⁶0.012365.3
    0.702.601.82×10⁶0.012230.0
  3. Plot hf against D and read off the target. The three trial points plotted against the target hf=120 m line (figure below) bracket the answer between D=0.50 and D=0.60 m; interpolating (equivalently, solving the same f–Re–D system directly) gives $$\boxed{D \approx 0.53\ \text{m}}$$ The nearest commercial size above this is D = 0.55 m (550 mm), chosen so the achieved head loss is safely at or below the 120 m available.
trial diameter D (m)h_f (m)0.50.60.70.8060120180(0.50, 165)(0.60, 65)(0.70, 30)target h_f = 120 mD = 0.53 m
Fig. Q6 — trial points (D, h₀) plotted against the target head loss of 120 m; the curve crosses the target near D = 0.53 m.
QuantityResult
Required diameter (exact)0.53 m
Selected commercial pipe size550 mm
Velocity at this D4.50 m/s, Re ≈ 2.4×10⁶ (fully turbulent, f≈0.0124)